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Translate these questions from English to Math.

Use an appropriate letter for your variable.

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If you can express one variable in terms of another, please do.

Solve the equations.

Interpret your solutions.

(1.) **ACT** Temperatures measured in degrees Fahrenheit ($F$) are related to temperatures
measured in degrees Celsius ($C$) by the formula $F = \dfrac{9}{5}C + 32$

There is $1$ value of $x$ for which $x$ degrees Fahrenheit equals $x$ degrees Celsius.

What is that value?

$ F.\:\: -72 \\[3ex] G.\:\: -40 \\[3ex] H.\:\: -32 \\[3ex] J.\:\: 0 \\[3ex] K.\:\: 32 $

We are going to use two methods to solve this question.

You may use any of the methods to solve it.

__First Method:__ Linear Equations

$ x^\circ F = x^\circ C \\[3ex] F = C \\[3ex] F = \dfrac{9}{5}C + 32 \\[5ex] \rightarrow C = \dfrac{9}{5}C + 32 \\[5ex] \dfrac{9}{5}C + 32 = C \\[5ex] LCD = 5 \\[3ex] 5\left(\dfrac{9}{5}C\right) + 5(32) = 5C \\[5ex] 9C + 160 = 5C \\[3ex] 9C - 5C = -160 \\[3ex] 4C = -160 \\[3ex] C = -\dfrac{160}{4} \\[5ex] C = -40 \\[3ex] $__Second Method:__ Testing the options

$ x^\circ F = x^\circ C \\[3ex] F = C \\[3ex] F = \dfrac{9}{5}C + 32 \\[5ex] C = -72 \\[3ex] F = \dfrac{9}{5} * -72 + 32 \\[5ex] F = -129.6 + 32 = -97.6 \\[3ex] F \ne C ...NO \\[5ex] C = -40 \\[3ex] F = \dfrac{9}{5} * -40 + 32 \\[5ex] F = 9(-8) + 32 = -72 + 32 = -40 \\[3ex] F = C = -40 ...YES $

There is $1$ value of $x$ for which $x$ degrees Fahrenheit equals $x$ degrees Celsius.

What is that value?

$ F.\:\: -72 \\[3ex] G.\:\: -40 \\[3ex] H.\:\: -32 \\[3ex] J.\:\: 0 \\[3ex] K.\:\: 32 $

We are going to use two methods to solve this question.

You may use any of the methods to solve it.

$ x^\circ F = x^\circ C \\[3ex] F = C \\[3ex] F = \dfrac{9}{5}C + 32 \\[5ex] \rightarrow C = \dfrac{9}{5}C + 32 \\[5ex] \dfrac{9}{5}C + 32 = C \\[5ex] LCD = 5 \\[3ex] 5\left(\dfrac{9}{5}C\right) + 5(32) = 5C \\[5ex] 9C + 160 = 5C \\[3ex] 9C - 5C = -160 \\[3ex] 4C = -160 \\[3ex] C = -\dfrac{160}{4} \\[5ex] C = -40 \\[3ex] $

$ x^\circ F = x^\circ C \\[3ex] F = C \\[3ex] F = \dfrac{9}{5}C + 32 \\[5ex] C = -72 \\[3ex] F = \dfrac{9}{5} * -72 + 32 \\[5ex] F = -129.6 + 32 = -97.6 \\[3ex] F \ne C ...NO \\[5ex] C = -40 \\[3ex] F = \dfrac{9}{5} * -40 + 32 \\[5ex] F = 9(-8) + 32 = -72 + 32 = -40 \\[3ex] F = C = -40 ...YES $

(2.) Five times a number is decreased by seven.

The result is eight.

What is the number?

$ Let\:\:the\:\:number = p \\[3ex] 5p - 7 = 8 \\[3ex] 5p - 7 = 8 \\[3ex] 5p = 8 + 7 \\[3ex] 5p = 15 \\[3ex] p = \dfrac{15}{5} \\[5ex] p = 3 \\[3ex] $ The number is $3$

The result is eight.

What is the number?

$ Let\:\:the\:\:number = p \\[3ex] 5p - 7 = 8 \\[3ex] 5p - 7 = 8 \\[3ex] 5p = 8 + 7 \\[3ex] 5p = 15 \\[3ex] p = \dfrac{15}{5} \\[5ex] p = 3 \\[3ex] $ The number is $3$

(3.) African elephants can weigh up to $16500$ pounds.

They weigh about $1.5$ times more than Indian elephants.

What is the weight of an Indian elephant?

Let the weight of an Indian elephant = $w$

So, the weight of an African elephant = $1.5w$

$ 1.5w = 16500 \\[3ex] 1.5w = 16500 \\[3ex] w = \dfrac{16500}{1.5} \\[5ex] w = 11000 \\[3ex] $ The weight of an Indian elephant is $11000$ pounds.

They weigh about $1.5$ times more than Indian elephants.

What is the weight of an Indian elephant?

Let the weight of an Indian elephant = $w$

So, the weight of an African elephant = $1.5w$

$ 1.5w = 16500 \\[3ex] 1.5w = 16500 \\[3ex] w = \dfrac{16500}{1.5} \\[5ex] w = 11000 \\[3ex] $ The weight of an Indian elephant is $11000$ pounds.

(4.) One number exceeds another number by eight.

The sum of the numbers is forty six.

What are the numbers?

Let the first number = $p$

The other number = $p + 8$

$ p + (p + 8) = 46 \\[3ex] p + p + 8 = 46 \\[3ex] 2p + 8 = 46 \\[3ex] 2p + 8 = 46 \\[3ex] 2p = 46 - 8 \\[3ex] 2p = 38 \\[3ex] p = \dfrac{38}{2} \\[5ex] p = 19 \\[3ex] p + 8 = 19 + 8 = 27 \\[3ex] $ The numbers are $19$ and $27$

The sum of the numbers is forty six.

What are the numbers?

Let the first number = $p$

The other number = $p + 8$

$ p + (p + 8) = 46 \\[3ex] p + p + 8 = 46 \\[3ex] 2p + 8 = 46 \\[3ex] 2p + 8 = 46 \\[3ex] 2p = 46 - 8 \\[3ex] 2p = 38 \\[3ex] p = \dfrac{38}{2} \\[5ex] p = 19 \\[3ex] p + 8 = 19 + 8 = 27 \\[3ex] $ The numbers are $19$ and $27$

(5.) __Observe the following scenario.__

Jeremiah came back home and was crying.

His mother asked him why he was crying.

The following conversation started.

Jeremiah: Mom, a dog ate five of my muffins.

Mom: What happened? Why? That was a quarter of all the muffins I made for you today.

How many muffins did Jeremiah's Mom make for him?

Let the number of muffins = $m$

A quarter of $m$ = $\dfrac{1}{4} * m = \dfrac{1m}{4}$

$ \dfrac{m}{4} = 5 \\[5ex] m = 4 * 5 \\[3ex] m = 20 \\[3ex] $ Jeremiah's Mom prepared $20$ muffins for him

Jeremiah came back home and was crying.

His mother asked him why he was crying.

The following conversation started.

Jeremiah: Mom, a dog ate five of my muffins.

Mom: What happened? Why? That was a quarter of all the muffins I made for you today.

How many muffins did Jeremiah's Mom make for him?

Let the number of muffins = $m$

A quarter of $m$ = $\dfrac{1}{4} * m = \dfrac{1m}{4}$

$ \dfrac{m}{4} = 5 \\[5ex] m = 4 * 5 \\[3ex] m = 20 \\[3ex] $ Jeremiah's Mom prepared $20$ muffins for him

(6.) A gallon of water is four times as much as one quart.

A gallon of water weighs about eight pounds.

About how much does a quart weigh?

Let the gallon of water = $g$

Let the quart = $k$

$ g = 4k \\[3ex] g = 8 \\[3ex] \therefore 4k = 8 \\[3ex] k = 2 \\[3ex] $ A quart weighs about two pounds.

A gallon of water weighs about eight pounds.

About how much does a quart weigh?

Let the gallon of water = $g$

Let the quart = $k$

$ g = 4k \\[3ex] g = 8 \\[3ex] \therefore 4k = 8 \\[3ex] k = 2 \\[3ex] $ A quart weighs about two pounds.

(7.) In Triangle $ABC$, the measure of angle $B$ is $4$ times the size of angle $A$.

The measure of angle $C$ is $12^o$ less than $5$ times the measure of angle $A$.

Calculate the measure of each angle.

Express angles $B$ and $C$ in terms of angle $A$

*Ask students to express angles $A$ and $C$ in terms of angle $B$*

*Ask students to express angles $A$ and $B$ in terms of angle $C$*

Let the measure of angle $A$ = $\angle A$ = A

Let the measure of angle $B$ = $\angle B$ = B

Let the measure of angle $C$ = $\angle C$ = C

The sum of the $3$ angles of a triangle = $180^o$...Triangle Theorem

$A + B + C = 180$ ...from Geometry (sum of angles of a triangle) ...eqn.(1)

$B = 4 * A$ ...the first sentence ...eqn.(2)

$C = 5 * A - 12$ ...the second sentence ...eqn.(3)

Let us express all of them in terms of $A$

Substitute eqns.(2) and (3) into eqn.(1)

$ A + B + C = 180...eqn.(1) \\[3ex] A + (4 * A) + (5 * A - 12) = 180 \\[3ex] A + 4A + 5A - 12 = 180 \\[3ex] 10A = 180 + 12 \\[3ex] 10A = 192 \\[3ex] A = \dfrac{192}{10} \\[5ex] A = 19.2\:\:units \\[3ex] $ Let us express all of them in terms of $B$

$ From\:\:eqn.(2) \\[3ex] B = 4 * A \\[3ex] 4 * A = B \\[3ex] A = \dfrac{B}{4} ...eqn.(4) \\[5ex] From\:\:eqn.(3) \\[3ex] C = 5 * \dfrac{B}{4} - 12 \\[5ex] C = \dfrac{5B}{4} - 12 ...eqn(5) \\[5ex] Substitute\:\:eqn.(4)\:\:and\:\:eqn.(5)\:\:into\:\:eqn.(1) \\[3ex] A + B + C = 180...eqn.(1) \\[3ex] \left(\dfrac{B}{4}\right) + B + \left(\dfrac{5B}{4} - 12\right) = 180 \\[5ex] LCD = 4 \\[3ex] Multiply\:\:each term\:\:by\:\:the\:\:LCD \\[3ex] 4 * \dfrac{B}{4} + 4 * B + 4 * \left(\dfrac{5B}{4} - 12\right) = 4 * 180 \\[5ex] B + 4B + 4 * \dfrac{5B}{4} - 4 * 12 = 720 \\[5ex] B + 4B + 5B - 48 = 720 \\[3ex] 10B = 720 + 48 \\[3ex] 10B = 768 \\[3ex] B = \dfrac{768}{10} \\[5ex] B = 76.8\:\:units \\[3ex] $ Let us express all of them in terms of $C$

$ From\:\:eqn.(3) \\[3ex] C = 5 * A - 12 \\[3ex] 5 * A - 12 = C \\[3ex] 5A = C + 12 \\[3ex] A = \dfrac{C + 12}{5} ...eqn.(6) \\[5ex] From\:\:eqn.(2) \\[3ex] B = 4 * A \\[3ex] Substitute\:\:eqn.(6)\:\:into\:\:eqn.(2) \\[3ex] B = 4 * A ...eqn.(2) \\[3ex] B = 4 * \left(\dfrac{C + 12}{5}\right) ...eqn.(7) \\[5ex] Substitute\:\:eqn.(6)\:\:and\:\:eqn.(7)\:\:into\:\:eqn.(1) \\[3ex] A + B + C = 180...eqn.(1) \\[3ex] \dfrac{C + 12}{5} + 4 * \left(\dfrac{C + 12}{5}\right) + C = 180 \\[5ex] LCD = 5 \\[3ex] Multiply\:\:each term\:\:by\:\:the\:\:LCD \\[3ex] 5 * \left(\dfrac{C + 12}{5}\right) + 5 * 4 * \left(\dfrac{C + 12}{5}\right) + 5 * C = 5 * 180 \\[5ex] (C + 12) + 4(C + 12) + 5C = 900 \\[3ex] C + 12 + 4C + 48 + 5C = 900 \\[3ex] 10C + 60 = 900 \\[3ex] 10C = 900 - 60 \\[3ex] 10C = 840 \\[3ex] C = \dfrac{840}{10} \\[5ex] C = 84\:\:units \\[3ex] $ The measure of angle $A$ is $19.2\:\:units$

The measure of angle $B$ is $76.8\:\:units$

The measure of angle $C$ is $84.0\:\:units$

The measure of angle $C$ is $12^o$ less than $5$ times the measure of angle $A$.

Calculate the measure of each angle.

Express angles $B$ and $C$ in terms of angle $A$

Let the measure of angle $A$ = $\angle A$ = A

Let the measure of angle $B$ = $\angle B$ = B

Let the measure of angle $C$ = $\angle C$ = C

The sum of the $3$ angles of a triangle = $180^o$...Triangle Theorem

$A + B + C = 180$ ...from Geometry (sum of angles of a triangle) ...eqn.(1)

$B = 4 * A$ ...the first sentence ...eqn.(2)

$C = 5 * A - 12$ ...the second sentence ...eqn.(3)

Let us express all of them in terms of $A$

Substitute eqns.(2) and (3) into eqn.(1)

$ A + B + C = 180...eqn.(1) \\[3ex] A + (4 * A) + (5 * A - 12) = 180 \\[3ex] A + 4A + 5A - 12 = 180 \\[3ex] 10A = 180 + 12 \\[3ex] 10A = 192 \\[3ex] A = \dfrac{192}{10} \\[5ex] A = 19.2\:\:units \\[3ex] $ Let us express all of them in terms of $B$

$ From\:\:eqn.(2) \\[3ex] B = 4 * A \\[3ex] 4 * A = B \\[3ex] A = \dfrac{B}{4} ...eqn.(4) \\[5ex] From\:\:eqn.(3) \\[3ex] C = 5 * \dfrac{B}{4} - 12 \\[5ex] C = \dfrac{5B}{4} - 12 ...eqn(5) \\[5ex] Substitute\:\:eqn.(4)\:\:and\:\:eqn.(5)\:\:into\:\:eqn.(1) \\[3ex] A + B + C = 180...eqn.(1) \\[3ex] \left(\dfrac{B}{4}\right) + B + \left(\dfrac{5B}{4} - 12\right) = 180 \\[5ex] LCD = 4 \\[3ex] Multiply\:\:each term\:\:by\:\:the\:\:LCD \\[3ex] 4 * \dfrac{B}{4} + 4 * B + 4 * \left(\dfrac{5B}{4} - 12\right) = 4 * 180 \\[5ex] B + 4B + 4 * \dfrac{5B}{4} - 4 * 12 = 720 \\[5ex] B + 4B + 5B - 48 = 720 \\[3ex] 10B = 720 + 48 \\[3ex] 10B = 768 \\[3ex] B = \dfrac{768}{10} \\[5ex] B = 76.8\:\:units \\[3ex] $ Let us express all of them in terms of $C$

$ From\:\:eqn.(3) \\[3ex] C = 5 * A - 12 \\[3ex] 5 * A - 12 = C \\[3ex] 5A = C + 12 \\[3ex] A = \dfrac{C + 12}{5} ...eqn.(6) \\[5ex] From\:\:eqn.(2) \\[3ex] B = 4 * A \\[3ex] Substitute\:\:eqn.(6)\:\:into\:\:eqn.(2) \\[3ex] B = 4 * A ...eqn.(2) \\[3ex] B = 4 * \left(\dfrac{C + 12}{5}\right) ...eqn.(7) \\[5ex] Substitute\:\:eqn.(6)\:\:and\:\:eqn.(7)\:\:into\:\:eqn.(1) \\[3ex] A + B + C = 180...eqn.(1) \\[3ex] \dfrac{C + 12}{5} + 4 * \left(\dfrac{C + 12}{5}\right) + C = 180 \\[5ex] LCD = 5 \\[3ex] Multiply\:\:each term\:\:by\:\:the\:\:LCD \\[3ex] 5 * \left(\dfrac{C + 12}{5}\right) + 5 * 4 * \left(\dfrac{C + 12}{5}\right) + 5 * C = 5 * 180 \\[5ex] (C + 12) + 4(C + 12) + 5C = 900 \\[3ex] C + 12 + 4C + 48 + 5C = 900 \\[3ex] 10C + 60 = 900 \\[3ex] 10C = 900 - 60 \\[3ex] 10C = 840 \\[3ex] C = \dfrac{840}{10} \\[5ex] C = 84\:\:units \\[3ex] $ The measure of angle $A$ is $19.2\:\:units$

The measure of angle $B$ is $76.8\:\:units$

The measure of angle $C$ is $84.0\:\:units$

(8.) Two pages that are back-to-back in Mr. Chukwuemeka's book have $205$ as the sum of their page numbers.

What are the page numbers?

"Back-to-back" pages means "consecutive" pages

Let the first of the two pages = $c$

The next page = $c + 1$

$ c + (c + 1) = 205 \\[3ex] c + c + 1 = 205 \\[3ex] 2c + 1 = 205 \\[3ex] 2c = 205 - 1 \\[3ex] 2c = 204 \\[3ex] c = 102 \\[3ex] c + 1 = 102 + 1 = 103 \\[3ex] $ The back-to-back page numbers are Pages $102$ and $103$

What are the page numbers?

"Back-to-back" pages means "consecutive" pages

Let the first of the two pages = $c$

The next page = $c + 1$

$ c + (c + 1) = 205 \\[3ex] c + c + 1 = 205 \\[3ex] 2c + 1 = 205 \\[3ex] 2c = 205 - 1 \\[3ex] 2c = 204 \\[3ex] c = 102 \\[3ex] c + 1 = 102 + 1 = 103 \\[3ex] $ The back-to-back page numbers are Pages $102$ and $103$

(9.) **WASSCE** The sum of three numbers is $81$

The second number is twice the first and the third number is six more than the second.

Find the numbers.

Let the first number = $c$

Let the second number = $d$

Let the third number = $e$

$ c + d + e = 81...eqn.(1) \\[3ex] d = 2c...eqn.(2) \\[3ex] e = 6 + d...eqn.(3) \\[3ex] From\:\: eqn.(2); e = 6 + 2c...eqn.(4) \\[3ex] Substitute\:\: eqns.(2) \:\:and\:\: (3) \:\:into\:\: eqn.(1) \\[3ex] c + (2c) + (6 + 2c) = 81 \\[3ex] c + 2c + 6 + 2c = 81 \\[3ex] 5c + 6 = 81 \\[3ex] 5c = 81 - 6 \\[3ex] 5c = 75\\[3ex] c = \dfrac{75}{5} \\[3ex] c = 15...eqn.(5) \\[3ex] Substitute\:\: eqn.(5) \:\:into\:\: eqn.(2) \\[3ex] d = 2(15) \\[3ex] d = 30...eqn.(6) \\[3ex] Substitute\:\: eqn.(6) \:\:into\:\: eqn.(3) \\[3ex] e = 6 + 30 \\[3ex] e = 36 \\[3ex] $ The first number is $15$

The second number is $30$

The third number is $36$

$15 + 30 + 36 = 81$

The second number is twice the first and the third number is six more than the second.

Find the numbers.

Let the first number = $c$

Let the second number = $d$

Let the third number = $e$

$ c + d + e = 81...eqn.(1) \\[3ex] d = 2c...eqn.(2) \\[3ex] e = 6 + d...eqn.(3) \\[3ex] From\:\: eqn.(2); e = 6 + 2c...eqn.(4) \\[3ex] Substitute\:\: eqns.(2) \:\:and\:\: (3) \:\:into\:\: eqn.(1) \\[3ex] c + (2c) + (6 + 2c) = 81 \\[3ex] c + 2c + 6 + 2c = 81 \\[3ex] 5c + 6 = 81 \\[3ex] 5c = 81 - 6 \\[3ex] 5c = 75\\[3ex] c = \dfrac{75}{5} \\[3ex] c = 15...eqn.(5) \\[3ex] Substitute\:\: eqn.(5) \:\:into\:\: eqn.(2) \\[3ex] d = 2(15) \\[3ex] d = 30...eqn.(6) \\[3ex] Substitute\:\: eqn.(6) \:\:into\:\: eqn.(3) \\[3ex] e = 6 + 30 \\[3ex] e = 36 \\[3ex] $ The first number is $15$

The second number is $30$

The third number is $36$

$15 + 30 + 36 = 81$

(10.) **ACT** In Cherokee County, the fine for speeding is $\$17$ for each mile per hour the
driver is traveling over the posted speed limit.

In Cherokee County, Kirk was fined $\$221$ for speeding on a road with a posted speed limit of $30$ mph.

Kirk was fined for traveling at what speed, in miles per hour?

mph means mile per hour

Let the speed(mph) traveled by Kirk be $p$

The speeding ticket is for speeds over $30$ mph Speed in excess of $30\:\: mph = p - 30$

At this speed, $p - 30$, the fine is $\$17$ per each mph

Total fine for Kirk = $\$221$

$ 17(p - 30) = 221 \\[3ex] 17p - 510 = 221 \\[3ex] 17p = 221 + 510 \\[3ex] 17p = 731 \\[3ex] p = \dfrac{731}{17} = 43\:\: mph \\[5ex] $ Kirk was driving $43\:\: mph$ on a $30\:\: mph$ zone.

That was an excess of $43 - 30 = 13\:\: mph$

He was fined $13 * 17 = \$221$

In Cherokee County, Kirk was fined $\$221$ for speeding on a road with a posted speed limit of $30$ mph.

Kirk was fined for traveling at what speed, in miles per hour?

mph means mile per hour

Let the speed(mph) traveled by Kirk be $p$

The speeding ticket is for speeds over $30$ mph Speed in excess of $30\:\: mph = p - 30$

At this speed, $p - 30$, the fine is $\$17$ per each mph

Total fine for Kirk = $\$221$

$ 17(p - 30) = 221 \\[3ex] 17p - 510 = 221 \\[3ex] 17p = 221 + 510 \\[3ex] 17p = 731 \\[3ex] p = \dfrac{731}{17} = 43\:\: mph \\[5ex] $ Kirk was driving $43\:\: mph$ on a $30\:\: mph$ zone.

That was an excess of $43 - 30 = 13\:\: mph$

He was fined $13 * 17 = \$221$

(11.) If six is subtracted from the third of three consecutive odd integers and the
result is multiplied by two, the answer is twenty three less than the sum of the first
and twice the second of the integers.

Find the integers.

Let the first odd integer = $c$

Second consecutive odd integer = $c + 2$

Third consecutive odd integer = $c + 4$

Let us begin from the LHS (Left Hand Side)

Six subtracted from the third consecutive odd integer = $(c + 4) - 6$

Result multiplied by $2$ = $2 * [(c + 4) - 6]$

$ LHS = 2[(c + 4) - 6] \\[3ex] LHS = 2[c + 4 - 6] \\[3ex] LHS = 2(c - 2) \\[3ex] LHS = 2c - 4 \\[3ex] $ "is" means =

Now, onto the RHS (Right Hand Side)

Depending on how the sentence was worded, it is sometomes better to work backwards.

Twice the second consecutive odd integer = $2 * (c + 2)$

Sum of the first and twice the second of the integers = $c + 2(c + 2)$

$23$ less than the sum of the first and twice the second = $[c + 2(c + 2)] - 23$

$ RHS = [c + 2(c + 2)] - 23 \\[3ex] RHS = c + 2c + 4 - 23 \\[3ex] RHS = 3c + 19 \\[3ex] \therefore 2c - 4 = 3c + 19 \\[3ex] 2c - 3c = 19 + 4 \\[3ex] -c = 23 \\[3ex] c = -\dfrac{23}{1} \\[5ex] c = -23 \\[3ex] $ The first odd integer = $-23$

The second consecutive odd integer = $c + 2 = -23 + 2 = -21$

The third consecutive odd integer = $c + 4 = -23 + 4 = -19$

The consecutive odd integers are: $-23, -21, -19$

Find the integers.

Let the first odd integer = $c$

Second consecutive odd integer = $c + 2$

Third consecutive odd integer = $c + 4$

Let us begin from the LHS (Left Hand Side)

Six subtracted from the third consecutive odd integer = $(c + 4) - 6$

Result multiplied by $2$ = $2 * [(c + 4) - 6]$

$ LHS = 2[(c + 4) - 6] \\[3ex] LHS = 2[c + 4 - 6] \\[3ex] LHS = 2(c - 2) \\[3ex] LHS = 2c - 4 \\[3ex] $ "is" means =

Now, onto the RHS (Right Hand Side)

Depending on how the sentence was worded, it is sometomes better to work backwards.

Twice the second consecutive odd integer = $2 * (c + 2)$

Sum of the first and twice the second of the integers = $c + 2(c + 2)$

$23$ less than the sum of the first and twice the second = $[c + 2(c + 2)] - 23$

$ RHS = [c + 2(c + 2)] - 23 \\[3ex] RHS = c + 2c + 4 - 23 \\[3ex] RHS = 3c + 19 \\[3ex] \therefore 2c - 4 = 3c + 19 \\[3ex] 2c - 3c = 19 + 4 \\[3ex] -c = 23 \\[3ex] c = -\dfrac{23}{1} \\[5ex] c = -23 \\[3ex] $ The first odd integer = $-23$

The second consecutive odd integer = $c + 2 = -23 + 2 = -21$

The third consecutive odd integer = $c + 4 = -23 + 4 = -19$

The consecutive odd integers are: $-23, -21, -19$

(12.) John sells cars on a salary-plus-commission basis.

He receives a weekly salary of $$400$ and a commission of $$200$ for each car he sells.

How many cars must he sell in a week to have a total weekly income of $$300$?

On a weekly basis;

Let the number of cars he must sell = $c$

Weekly salary = $$400$

Commission of $$200$ for each car sold = $200 * c = 200c$

$ \therefore 400 + 200c = 3000 \\[3ex] 200c = 3000 - 400 \\[3ex] 200c = 2600 \\[3ex] c = \dfrac{2600}{200} \\[5ex] c = 13 \\[3ex] $ John must sell $13$ cars in a week in order to have a total weekly income of $$300$

He receives a weekly salary of $$400$ and a commission of $$200$ for each car he sells.

How many cars must he sell in a week to have a total weekly income of $$300$?

On a weekly basis;

Let the number of cars he must sell = $c$

Weekly salary = $$400$

Commission of $$200$ for each car sold = $200 * c = 200c$

$ \therefore 400 + 200c = 3000 \\[3ex] 200c = 3000 - 400 \\[3ex] 200c = 2600 \\[3ex] c = \dfrac{2600}{200} \\[5ex] c = 13 \\[3ex] $ John must sell $13$ cars in a week in order to have a total weekly income of $$300$

(13.) The sum of the present ages of Naomi and her mother is fifty four years.

In five years, Naomi will be three-fifths as old as her mother at that time.

How old is Naomi?

How old is her mother?

Let Naomi's age = $n$

Let her mother's age = $m$

$n + m = 54 ...eqn.(1)$

In five years;

Naomi's age = $n + 5$

Her mother's age = $m + 5$

$ n + 5 = \dfrac{3}{5} * (m + 5) ...eqn.(2) \\[5ex] Use\:\:the\:\:Distributive\:\:Property \\[3ex] n + 5 = \dfrac{3m}{5} + 3 \\[5ex] LCD = 5 \\[3ex] Multiply\:\:both\:\:sides\:\:by\:\:the\:\:$LCD$ \\[3ex] 5 * (n + 5) = 5 * \dfrac{3m}{5} + 5 * 3 \\[5ex] 5(n + 5) = 3m + 15 \\[3ex] 5n + 25 = 3m + 15 ...eqn.(4) \\[3ex] From\:\:eqn.(1) \\[3ex] n + m = 54 \\[3ex] m = 54 - n ...eqn.(3) \\[3ex] Substitute\:\:eqn.(3)\:\:into\:\:eqn.(4) \\[3ex] 5n + 25 = 3(54 - n) + 15 \\[3ex] 5n + 25 = 162 - 3n + 15 \\[3ex] 5n + 3n = 162 + 15 - 25 \\[3ex] 8n = 152 \\[3ex] n = \dfrac{152}{8} \\[5ex] n = 19 \\[3ex] From\:\:eqn.(3) \\[3ex] m = 54 - n m = 54 - 19 \\[3ex] m = 35 \\[3ex] $ Naomi is $19$ years old

Her mother is $35$ years old

In five years, Naomi will be three-fifths as old as her mother at that time.

How old is Naomi?

How old is her mother?

Let Naomi's age = $n$

Let her mother's age = $m$

$n + m = 54 ...eqn.(1)$

In five years;

Naomi's age = $n + 5$

Her mother's age = $m + 5$

$ n + 5 = \dfrac{3}{5} * (m + 5) ...eqn.(2) \\[5ex] Use\:\:the\:\:Distributive\:\:Property \\[3ex] n + 5 = \dfrac{3m}{5} + 3 \\[5ex] LCD = 5 \\[3ex] Multiply\:\:both\:\:sides\:\:by\:\:the\:\:$LCD$ \\[3ex] 5 * (n + 5) = 5 * \dfrac{3m}{5} + 5 * 3 \\[5ex] 5(n + 5) = 3m + 15 \\[3ex] 5n + 25 = 3m + 15 ...eqn.(4) \\[3ex] From\:\:eqn.(1) \\[3ex] n + m = 54 \\[3ex] m = 54 - n ...eqn.(3) \\[3ex] Substitute\:\:eqn.(3)\:\:into\:\:eqn.(4) \\[3ex] 5n + 25 = 3(54 - n) + 15 \\[3ex] 5n + 25 = 162 - 3n + 15 \\[3ex] 5n + 3n = 162 + 15 - 25 \\[3ex] 8n = 152 \\[3ex] n = \dfrac{152}{8} \\[5ex] n = 19 \\[3ex] From\:\:eqn.(3) \\[3ex] m = 54 - n m = 54 - 19 \\[3ex] m = 35 \\[3ex] $ Naomi is $19$ years old

Her mother is $35$ years old

(14.) **ACT** The ABC Book Club charges a $\$40$ monthly fee, plus $\$2$ per book
read in that month.

The Easy Book Club charges a $\$35$ monthly fee, plus $\$3$ per book read in that month.

For each club, how many books must be read in $1$ month for the total charges from each club to be equal?

Let the number of books that must be read in $1$ month = $p$

ABC Book Club

Total charge for $1$ month = $40 + 2p$

Easy Book Club

Total charge for $1$ month = $35 + 3p$

The total charges should be equal

$ 40 + 2p = 35 + 3p \\[3ex] 40 - 35 = 3p - 2p \\[3ex] 5 = p \\[3ex] p = 5 \\[3ex] $ $5$ books need to read in $1$ month for the total charges of both clubs to be equal

The Easy Book Club charges a $\$35$ monthly fee, plus $\$3$ per book read in that month.

For each club, how many books must be read in $1$ month for the total charges from each club to be equal?

Let the number of books that must be read in $1$ month = $p$

ABC Book Club

Total charge for $1$ month = $40 + 2p$

Easy Book Club

Total charge for $1$ month = $35 + 3p$

The total charges should be equal

$ 40 + 2p = 35 + 3p \\[3ex] 40 - 35 = 3p - 2p \\[3ex] 5 = p \\[3ex] p = 5 \\[3ex] $ $5$ books need to read in $1$ month for the total charges of both clubs to be equal

(15.) **ACT** At a grocery store, Jo Ellen received $\$1.60$ when she returned her cans, glass bottles,
and plastic bottles.

Jo Ellen received $\$0.05$ for each can, $\$0.10$ for each glass bottle, and $\$0.05$ for each plastic bottle.

She knew she had returned $6$ cans and $8$ glass bottles.

How many plastic bottles did Jo Ellen return to the store?

$ A.\:\: 8 \\[3ex] B.\:\: 9 \\[3ex] C.\:\: 10 \\[3ex] D.\:\: 12 \\[3ex] E.\:\: 18 \\[3ex] $

$ Let\:\:the\:\:number\:\:of\:\:plastic\:\:bottles = g \\[3ex] Total\:\:amount\:\:she\:\:received = 1.6 \\[3ex] 6\:\:cans@0.05/can = 6(0.05) = 0.3 \\[3ex] 8\:\:glass\:\:bottles@0.1/glass\:\:bottle = 8(0.1) = 0.8 \\[3ex] g\:\:glass\:\:bottles@0.05/plastic\:\:bottle = g(0.05) = 0.05g \\[3ex] 0.3 + 0.8 + 0.05g = 1.6 \\[3ex] 1.1 + 0.05g = 1.6 \\[3ex] 0.05g = 1.6 - 1.1 \\[3ex] 0.05g = 0.5 \\[3ex] g = \dfrac{0.5}{0.05} \\[5ex] g = 10 \\[3ex] $ Jo Ellen returned $10$ plastic bottles.

Jo Ellen received $\$0.05$ for each can, $\$0.10$ for each glass bottle, and $\$0.05$ for each plastic bottle.

She knew she had returned $6$ cans and $8$ glass bottles.

How many plastic bottles did Jo Ellen return to the store?

$ A.\:\: 8 \\[3ex] B.\:\: 9 \\[3ex] C.\:\: 10 \\[3ex] D.\:\: 12 \\[3ex] E.\:\: 18 \\[3ex] $

$ Let\:\:the\:\:number\:\:of\:\:plastic\:\:bottles = g \\[3ex] Total\:\:amount\:\:she\:\:received = 1.6 \\[3ex] 6\:\:cans@0.05/can = 6(0.05) = 0.3 \\[3ex] 8\:\:glass\:\:bottles@0.1/glass\:\:bottle = 8(0.1) = 0.8 \\[3ex] g\:\:glass\:\:bottles@0.05/plastic\:\:bottle = g(0.05) = 0.05g \\[3ex] 0.3 + 0.8 + 0.05g = 1.6 \\[3ex] 1.1 + 0.05g = 1.6 \\[3ex] 0.05g = 1.6 - 1.1 \\[3ex] 0.05g = 0.5 \\[3ex] g = \dfrac{0.5}{0.05} \\[5ex] g = 10 \\[3ex] $ Jo Ellen returned $10$ plastic bottles.

(16.) Titus is paid double time for each hour he works over $40$ hours in a week.

Last week, he worked $55$ hours and earned $$1750$.

What is his normal hourly rate?

Let the normal hourly rate = $r$

How worked $55$ hours

So, overtime hours = $55 - 40 = 15$

Double means that you have to "multiply by $2$"

Double time = $2 * r = 2r$

$40$ hours @ $r / hour = $40 * r = 40r$

$15$ hours @ $2r / hour = $15 * 2r = 30r$

$ \therefore 40r + 30r = 1750 \\[3ex] 70r = 1750 \\[3ex] r = \dfrac{1750}{70} \\[5ex] r = 25 \\[3ex] $ Titus' normal hourly rate is $\$25.00$ per hour

Last week, he worked $55$ hours and earned $$1750$.

What is his normal hourly rate?

Let the normal hourly rate = $r$

How worked $55$ hours

So, overtime hours = $55 - 40 = 15$

Double means that you have to "multiply by $2$"

Double time = $2 * r = 2r$

$40$ hours @ $r / hour = $40 * r = 40r$

$15$ hours @ $2r / hour = $15 * 2r = 30r$

$ \therefore 40r + 30r = 1750 \\[3ex] 70r = 1750 \\[3ex] r = \dfrac{1750}{70} \\[5ex] r = 25 \\[3ex] $ Titus' normal hourly rate is $\$25.00$ per hour

(17.) **ACT** A box contains a combination of solid-colored tickets:

$\dfrac{1}{10}$ of the tickets are green

$\dfrac{1}{2}$ are red

$\dfrac{1}{4}$ are blue, and the remaining $30$ tickets are white.

How many blue tickets are in the box?

$ F.\:\: 10 \\[3ex] G.\:\: 20 \\[3ex] H.\:\: 50 \\[3ex] J.\:\: 100 \\[3ex] K.\:\: 200 \\[3ex] $

$ Let\:\:the\:\:number\:\:of\:\:tickets = p \\[3ex] Green = \dfrac{1}{10}p \\[5ex] Red = \dfrac{1}{2}p \\[5ex] Blue = \dfrac{1}{4}p \\[5ex] White = 30 \\[3ex] \dfrac{1}{10}p + \dfrac{1}{2}p + \dfrac{1}{4}p + 30 = p \\[5ex] LCD = 20 \\[3ex] Multiply\:\:both\:\:sides\:\:by\:\:the\:\:LCD \\[3ex] 20\left(\dfrac{1}{10}p + \dfrac{1}{2}p + \dfrac{1}{4}p + 30\right) = 20(p) \\[5ex] 20\left(\dfrac{1}{10}p\right) + 20\left(\dfrac{1}{2}p\right) + 20\left(\dfrac{1}{4}p\right) + 20(30) = 20p \\[5ex] 2p + 10p + 5p + 20(30) = 20p \\[3ex] 17p + 600 = 20p \\[3ex] 600 = 20p - 17p \\[3ex] 600 = 3p \\[3ex] 3p = 600 \\[3ex] p = \dfrac{600}{3} \\[5ex] p = 200 \\[3ex] Blue = \dfrac{1}{4} * 200 \\[5ex] Blue = 50 \\[3ex] $ There are $200$ tickets in the box.

$50$ of those tickets are blue.

$\dfrac{1}{10}$ of the tickets are green

$\dfrac{1}{2}$ are red

$\dfrac{1}{4}$ are blue, and the remaining $30$ tickets are white.

How many blue tickets are in the box?

$ F.\:\: 10 \\[3ex] G.\:\: 20 \\[3ex] H.\:\: 50 \\[3ex] J.\:\: 100 \\[3ex] K.\:\: 200 \\[3ex] $

$ Let\:\:the\:\:number\:\:of\:\:tickets = p \\[3ex] Green = \dfrac{1}{10}p \\[5ex] Red = \dfrac{1}{2}p \\[5ex] Blue = \dfrac{1}{4}p \\[5ex] White = 30 \\[3ex] \dfrac{1}{10}p + \dfrac{1}{2}p + \dfrac{1}{4}p + 30 = p \\[5ex] LCD = 20 \\[3ex] Multiply\:\:both\:\:sides\:\:by\:\:the\:\:LCD \\[3ex] 20\left(\dfrac{1}{10}p + \dfrac{1}{2}p + \dfrac{1}{4}p + 30\right) = 20(p) \\[5ex] 20\left(\dfrac{1}{10}p\right) + 20\left(\dfrac{1}{2}p\right) + 20\left(\dfrac{1}{4}p\right) + 20(30) = 20p \\[5ex] 2p + 10p + 5p + 20(30) = 20p \\[3ex] 17p + 600 = 20p \\[3ex] 600 = 20p - 17p \\[3ex] 600 = 3p \\[3ex] 3p = 600 \\[3ex] p = \dfrac{600}{3} \\[5ex] p = 200 \\[3ex] Blue = \dfrac{1}{4} * 200 \\[5ex] Blue = 50 \\[3ex] $ There are $200$ tickets in the box.

$50$ of those tickets are blue.

(18.) Can you calculate my age?

Three hundred reduced by three times my age is one hundred and ninety two.

What is my age?

Let my age = $a$

Three times my age = $3 * a = 3a$

Three hundred reduced by $3a$ = $300 - 3a$

"is" means =

$ \therefore 300 - 3a = 192 \\[3ex] 300 - 192 = 3a \\[3ex] 108 = 3a \\[3ex] 3a = 108 \\[3ex] a = \dfrac{108}{3} \\[5ex] a = 36 \\[3ex] $ Wow, that was my age last year ☺☺☺

Three hundred reduced by three times my age is one hundred and ninety two.

What is my age?

Let my age = $a$

Three times my age = $3 * a = 3a$

Three hundred reduced by $3a$ = $300 - 3a$

"is" means =

$ \therefore 300 - 3a = 192 \\[3ex] 300 - 192 = 3a \\[3ex] 108 = 3a \\[3ex] 3a = 108 \\[3ex] a = \dfrac{108}{3} \\[5ex] a = 36 \\[3ex] $ Wow, that was my age last year ☺☺☺

(19.) Mary worked 44 hours last week and earned a 1075.

She earns two times and a half of her regular pay for any overtime hours.

Overtime hours are hours in excess of 40 hours a week.

Calculate her regular hourly pay.

Let her regular hourly pay = $$p$

Regular weekly hours = $40$ hours a week Regular weekly pay = $40$ hours @ $$p$ per hour = $40 * p = 40p$

For a $44-hour$ week,

Overtime hours = $44 - 40 = 4$

Overtime weekly pay = $2.5 *$ regular hourly pay = $2.5 * p$ = $2.5p$

Overtime weekly pay = $4$ hours @ $$p$ per hour = $4 * 2.5p = 10p$

Total weekly pay for last week = Regular weekly pay + Overtime weekly pay

Total weekly pay = $40p + 10p = 50p$

Total weekly pay for last week = $1075$

$ 50p = 1075 \\[3ex] p = \dfrac{1075}{50} \\[5ex] p = 21.5 \\[3ex] $ Mary earns $\$21.50$ per hour

She earns two times and a half of her regular pay for any overtime hours.

Overtime hours are hours in excess of 40 hours a week.

Calculate her regular hourly pay.

Let her regular hourly pay = $$p$

Regular weekly hours = $40$ hours a week Regular weekly pay = $40$ hours @ $$p$ per hour = $40 * p = 40p$

For a $44-hour$ week,

Overtime hours = $44 - 40 = 4$

Overtime weekly pay = $2.5 *$ regular hourly pay = $2.5 * p$ = $2.5p$

Overtime weekly pay = $4$ hours @ $$p$ per hour = $4 * 2.5p = 10p$

Total weekly pay for last week = Regular weekly pay + Overtime weekly pay

Total weekly pay = $40p + 10p = 50p$

Total weekly pay for last week = $1075$

$ 50p = 1075 \\[3ex] p = \dfrac{1075}{50} \\[5ex] p = 21.5 \\[3ex] $ Mary earns $\$21.50$ per hour

(20.) In August 2009, the United States Senate had a total of 98 Democrats and Republicans.

There were 18 more Democrats than Republicans.

How many Democrats were there? (Source: www.thegreenpapers.com)

How many Independents were there?

Let the number of Democrats = $d$

The number of Republicans = $r$

$d + r = 98$ ... first sentence ...eqn.(1)

$d = 18 + r$ ... second sentence ...eqn.(2)

Substitute eqn.(2) into eqn.(1)

$ d + r = 98 \\[3ex] (18 + r) + r = 98 \\[3ex] 18 + r + r = 98 \\[3ex] 2r + 18 = 98 \\[3ex] 2r = 98 - 18 \\[3ex] 2r = 80 \\[3ex] r = \dfrac{80}{2} \\[5ex] r = 40 \\[3ex] 18 + r = 18 + 40 = 58 \\[3ex] $ In August $2009$, there were $40$ Republicans and $58$ Democrats in the United States Senate.

Because the United States Senate is made up of $100$ senators, there were $2$ Independents.

There were 18 more Democrats than Republicans.

How many Democrats were there? (Source: www.thegreenpapers.com)

How many Independents were there?

Let the number of Democrats = $d$

The number of Republicans = $r$

$d + r = 98$ ... first sentence ...eqn.(1)

$d = 18 + r$ ... second sentence ...eqn.(2)

Substitute eqn.(2) into eqn.(1)

$ d + r = 98 \\[3ex] (18 + r) + r = 98 \\[3ex] 18 + r + r = 98 \\[3ex] 2r + 18 = 98 \\[3ex] 2r = 98 - 18 \\[3ex] 2r = 80 \\[3ex] r = \dfrac{80}{2} \\[5ex] r = 40 \\[3ex] 18 + r = 18 + 40 = 58 \\[3ex] $ In August $2009$, there were $40$ Republicans and $58$ Democrats in the United States Senate.

Because the United States Senate is made up of $100$ senators, there were $2$ Independents.

(21.) An inheritance of $2,000,000$ is to be divided among Peter, James, and John.

James is to receive $\dfrac{9}{10}$ of what Peter receives.

John is to receive $\dfrac{1}{10}$ of what Peter receives.

How much does each receive?

*Ask students to express what each received in terms of the other two.*

*There should be three different answers.*

Let the amount received by Peter = $p$

The amount received by James = $x$

The amount received by John = $y$

$ p + x + y = 2000000 ...eqn.(1) \\[3ex] x = \dfrac{9}{10} * p \\[3ex] x = \dfrac{9p}{10} ...eqn.(2) \\[5ex] y = \dfrac{1}{10} * p \\[5ex] y = \dfrac{1p}{10} ...eqn.(3) \\[5ex] $ Let us express it in terms of what Peter received.

Substitute eqn.(2) and eqn.(3) into eqn.(1)

$ p + x + y = 2000000 ...eqn.(1) \\[3ex] p + \dfrac{9p}{10} + \dfrac{1p}{10} = 2000000 \\[5ex] LCD = 10 \\[3ex] Multiply\:\:each\:\:term\:\:by\:\:the\:\:LCD \\[3ex] 10 * p + 10 * \dfrac{9p}{10} + 10 * \dfrac{p}{10} = 10 * 2000000 \\[5ex] 10p + 9p + 1p = 20000000 \\[3ex] 20p = 20000000 \\[3ex] p = \dfrac{20000000}{20} \\[5ex] p = 1000000 \\[3ex] x = \dfrac{9}{10} * p \\[3ex] x = \dfrac{9}{10} * 1000000 \\[5ex] x = 900000 \\[3ex] y = \dfrac{1}{10} * p \\[3ex] y = \dfrac{1}{10} * 1000000 \\[3ex] y = 100000 \\[3ex] $ Peter received $\$1,000,000.00$

James received $\$900,000.00$

John received $\$100,000.00$

James is to receive $\dfrac{9}{10}$ of what Peter receives.

John is to receive $\dfrac{1}{10}$ of what Peter receives.

How much does each receive?

Let the amount received by Peter = $p$

The amount received by James = $x$

The amount received by John = $y$

$ p + x + y = 2000000 ...eqn.(1) \\[3ex] x = \dfrac{9}{10} * p \\[3ex] x = \dfrac{9p}{10} ...eqn.(2) \\[5ex] y = \dfrac{1}{10} * p \\[5ex] y = \dfrac{1p}{10} ...eqn.(3) \\[5ex] $ Let us express it in terms of what Peter received.

Substitute eqn.(2) and eqn.(3) into eqn.(1)

$ p + x + y = 2000000 ...eqn.(1) \\[3ex] p + \dfrac{9p}{10} + \dfrac{1p}{10} = 2000000 \\[5ex] LCD = 10 \\[3ex] Multiply\:\:each\:\:term\:\:by\:\:the\:\:LCD \\[3ex] 10 * p + 10 * \dfrac{9p}{10} + 10 * \dfrac{p}{10} = 10 * 2000000 \\[5ex] 10p + 9p + 1p = 20000000 \\[3ex] 20p = 20000000 \\[3ex] p = \dfrac{20000000}{20} \\[5ex] p = 1000000 \\[3ex] x = \dfrac{9}{10} * p \\[3ex] x = \dfrac{9}{10} * 1000000 \\[5ex] x = 900000 \\[3ex] y = \dfrac{1}{10} * p \\[3ex] y = \dfrac{1}{10} * 1000000 \\[3ex] y = 100000 \\[3ex] $ Peter received $\$1,000,000.00$

James received $\$900,000.00$

John received $\$100,000.00$

(22.) If twenty one is subtracted from twelve times a number, the result is
twenty seven.

Find the number.

Let the number = $n$

Twelve times the number = $12 * n = 12n$

Twenty one subtracted from $12n$ = $12n - 21$

"is" means =

$ \therefore 12n - 21 = 27 \\[3ex] 12n = 27 + 21 \\[3ex] 12n = 48 \\[3ex] n = \dfrac{48}{12} \\[5ex] n = 4 \\[3ex] $ The number is $4$

Find the number.

Let the number = $n$

Twelve times the number = $12 * n = 12n$

Twenty one subtracted from $12n$ = $12n - 21$

"is" means =

$ \therefore 12n - 21 = 27 \\[3ex] 12n = 27 + 21 \\[3ex] 12n = 48 \\[3ex] n = \dfrac{48}{12} \\[5ex] n = 4 \\[3ex] $ The number is $4$

(23.) At the $2008$ Beijing Summer Olympics, China earned one hundred medals.

The number of gold medals was twenty three more than the number of bronze medals.

The number of bronze medals was seven more than the number of silver medals.

How many gold medals did China earn? (Source: World Almanac)

Let the number of gold medals = $g$

Let the number of bronze medals = $b$

Let the number of silver medals = $s$

$ g = 23 + b ...eqn.(1) \\[3ex] b = 7 + s ...eqn.(2) \\[3ex] g + b + s = 100 ...eqn.(3) \\[3ex] Substitute\:\:eqn.(1)\:\:into\:\:eqn.(3) \\[3ex] 23 + b + b + s = 100 \\[3ex] 23 + 2b + s = 100... eqn.(4) \\[3ex] Substitute\:\:eqn.(2)\:\:into\:\:eqn.(4) \\[3ex] 23 + 2(7 + s) + s = 100 \\[3ex] 23 + 14 + 2s + s = 100 \\[3ex] 37 + 3s = 100 \\[3ex] 3s = 100 - 37 \\[3ex] 3s = 63 \\[3ex] s = \dfrac{63}{3} \\[5ex] s = 21 \\[3ex] b = 7 + s \\[3ex] b = 7 + 21 \\[3ex] b = 28 \\[3ex] g = 23 + b \\[3ex] g = 23 + 28 \\[3ex] g = 51 \\[3ex] 21 + 28 + 51 = 100 \\[3ex] $ At the $2008$ Beijing Summer Olympics, China earned $51$ gold medals, $21$ silver medals and $28$ bronze medals.

The number of gold medals was twenty three more than the number of bronze medals.

The number of bronze medals was seven more than the number of silver medals.

How many gold medals did China earn? (Source: World Almanac)

Let the number of gold medals = $g$

Let the number of bronze medals = $b$

Let the number of silver medals = $s$

$ g = 23 + b ...eqn.(1) \\[3ex] b = 7 + s ...eqn.(2) \\[3ex] g + b + s = 100 ...eqn.(3) \\[3ex] Substitute\:\:eqn.(1)\:\:into\:\:eqn.(3) \\[3ex] 23 + b + b + s = 100 \\[3ex] 23 + 2b + s = 100... eqn.(4) \\[3ex] Substitute\:\:eqn.(2)\:\:into\:\:eqn.(4) \\[3ex] 23 + 2(7 + s) + s = 100 \\[3ex] 23 + 14 + 2s + s = 100 \\[3ex] 37 + 3s = 100 \\[3ex] 3s = 100 - 37 \\[3ex] 3s = 63 \\[3ex] s = \dfrac{63}{3} \\[5ex] s = 21 \\[3ex] b = 7 + s \\[3ex] b = 7 + 21 \\[3ex] b = 28 \\[3ex] g = 23 + b \\[3ex] g = 23 + 28 \\[3ex] g = 51 \\[3ex] 21 + 28 + 51 = 100 \\[3ex] $ At the $2008$ Beijing Summer Olympics, China earned $51$ gold medals, $21$ silver medals and $28$ bronze medals.

(24.) **ACT** A triangle has a perimeter of $26$ cm and sides of length $x$ cm,
$(x + 3)$ cm, and $(x + 5)$ cm.

What is the value of $x$?

Let the perimeter = $P$

The side lengths are $x$, $x + 3$, and $x + 5$

$ P = x + (x + 3) + (x + 5) \\[3ex] P = x + x + 3 + x + 5 \\[3ex] P = 3x + 8 \\[3ex] P = 26 \\[3ex] 3x + 8 = 26 \\[3ex] 3x = 26 - 8 \\[3ex] 3x = 18 \\[3ex] x = \dfrac{18}{3} \\[3ex] x = 6\: cm $

What is the value of $x$?

Let the perimeter = $P$

The side lengths are $x$, $x + 3$, and $x + 5$

$ P = x + (x + 3) + (x + 5) \\[3ex] P = x + x + 3 + x + 5 \\[3ex] P = 3x + 8 \\[3ex] P = 26 \\[3ex] 3x + 8 = 26 \\[3ex] 3x = 26 - 8 \\[3ex] 3x = 18 \\[3ex] x = \dfrac{18}{3} \\[3ex] x = 6\: cm $

(25.) Ifa's gym membership costs a one-time membership fee of $$120$ and
a monthly fee of $$30$.

Godwin has $$300$ and wants to become a member.

How many months would his membership be valid?

Let the number of months = $m$

One-time membership fee = $$120$

A monthly fee of $$30$ means $30 * m = 30m$

$m$ number of months @ $$30$ per month = $30 * m = 30m$

$ \therefore 120 + 30m = 300 \\[3ex] 30m = 300 - 120 \\[3ex] 30m = 180 \\[3ex] m = \dfrac{180}{30} \\[5ex] m = 6 \\[3ex] $ With the sum of $\$300$, Ifa's gym membership would be valid for $6$ months.

Godwin has $$300$ and wants to become a member.

How many months would his membership be valid?

Let the number of months = $m$

One-time membership fee = $$120$

A monthly fee of $$30$ means $30 * m = 30m$

$m$ number of months @ $$30$ per month = $30 * m = 30m$

$ \therefore 120 + 30m = 300 \\[3ex] 30m = 300 - 120 \\[3ex] 30m = 180 \\[3ex] m = \dfrac{180}{30} \\[5ex] m = 6 \\[3ex] $ With the sum of $\$300$, Ifa's gym membership would be valid for $6$ months.

(26.) The sum of three consecutive integers is $72$.

Find the integers.

Let the first integer = $p$

The second integer = $p + 1$

The third integer = $p + 2$

$ p + (p + 1) + (p + 2) = 72 \\[3ex] p + p + 1 + p + 2 = 72 \\[3ex] 3p + 3 = 72 \\[3ex] 3p = 72 - 3 \\[3ex] 3p = 69 \\[3ex] p = \dfrac{69}{3} \\[5ex] p = 23 \\[3ex] p + 1 = 23 + 1 = 24 \\[3ex] p + 2 = 23 + 2 = 25 \\[3ex] $ The integers are: $23, 24, 25$

Find the integers.

Let the first integer = $p$

The second integer = $p + 1$

The third integer = $p + 2$

$ p + (p + 1) + (p + 2) = 72 \\[3ex] p + p + 1 + p + 2 = 72 \\[3ex] 3p + 3 = 72 \\[3ex] 3p = 72 - 3 \\[3ex] 3p = 69 \\[3ex] p = \dfrac{69}{3} \\[5ex] p = 23 \\[3ex] p + 1 = 23 + 1 = 24 \\[3ex] p + 2 = 23 + 2 = 25 \\[3ex] $ The integers are: $23, 24, 25$

(27.) The temperature in Samdom City rose $30^oF$ from $7:00\: am$ to $3:00\: pm$.

The temperature at $3:00\: pm$ was $70^oF$.

What was the temperature at $7:00\: am$?

Let the temperature at $7:00\: am$ = $t$

It rose by $30^oF$ to $70^oF$ at $3:00\: pm$

$ t + 30 = 70 \\[3ex] t = 70 - 30 \\[3ex] t = 40 \\[3ex] $ The temperature at $7:00\: am$ is $40^oF$

The temperature at $3:00\: pm$ was $70^oF$.

What was the temperature at $7:00\: am$?

Let the temperature at $7:00\: am$ = $t$

It rose by $30^oF$ to $70^oF$ at $3:00\: pm$

$ t + 30 = 70 \\[3ex] t = 70 - 30 \\[3ex] t = 40 \\[3ex] $ The temperature at $7:00\: am$ is $40^oF$

(28.) **ACT** Two sides of a triangle are equal in length.

The third side is $3$ centimeters longer than either of the other $2$ sides.

Given that the perimeter of the triangle is $93$ centimeters, what is the length, in centimeters, of the longest side?

Let the perimeter = $P$

Let the length of each of the two equal sides = $x$

Two equal side lengths = $x + x = 2x$

The length of the third side = $3 + x$

$ P = 2x + (3 + x) \\[3ex] P = 2x + 3 + x \\[3ex] P = 3x + 3 \\[3ex] P = 93 \\[3ex] 3x + 3 = 93 \\[3ex] 3x = 93 - 3 \\[3ex] 3x = 90 \\[3ex] x = \dfrac{90}{3} \\[3ex] x = 30 \\[3ex] 3 + x = 3 + 30 = 33 \\[3ex] $ The longest side of the triangle is $33$ centimeters

The third side is $3$ centimeters longer than either of the other $2$ sides.

Given that the perimeter of the triangle is $93$ centimeters, what is the length, in centimeters, of the longest side?

Let the perimeter = $P$

Let the length of each of the two equal sides = $x$

Two equal side lengths = $x + x = 2x$

The length of the third side = $3 + x$

$ P = 2x + (3 + x) \\[3ex] P = 2x + 3 + x \\[3ex] P = 3x + 3 \\[3ex] P = 93 \\[3ex] 3x + 3 = 93 \\[3ex] 3x = 93 - 3 \\[3ex] 3x = 90 \\[3ex] x = \dfrac{90}{3} \\[3ex] x = 30 \\[3ex] 3 + x = 3 + 30 = 33 \\[3ex] $ The longest side of the triangle is $33$ centimeters

(29.) Rebecca has three hundred and ninety coins consisting of pennies, nickels, and dimes.

The number of nickels is ten less than twice the number of pennies.

The number of dimes is twenty less than three times the number of pennies.

How many coins of each kind does she have?

Let the number of pennies = $p$

Let the number of nickels = $n$

Let the number of dimes = $d$

$p + n + d = 390 ...eqn.(1)$

Twice the number of pennies = $2 * p = 2p$

Ten less than $2p$ = $2p - 10$

$n = 2p - 10 ...eqn.(2)$

Three times the number of pennies = $3 * p = 3p$

Twenty less than $3p$ = $3p - 20$

$d = 3p - 20 ...eqn.(3)$

Substitute eqn.(2) and eqn.(3) into eqn.(1)

$ p + (2p - 10) + (3p - 20) = 390 \\[3ex] p + 2p - 10 + 3p - 20 = 390 \\[3ex] 6p - 30 = 390 \\[3ex] 6p = 390 + 30 \\[3ex] 6p = 420 \\[3ex] p = \dfrac{420}{6} \\[5ex] p = 70 \\[3ex] d = 3p - 20 \\[3ex] d = 3(70) - 20 \\[3ex] d = 210 - 20 \\[3ex] d = 190 \\[3ex] n = 2p - 10 \\[3ex] n = 2(70) - 10 \\[3ex] n = 140 - 10 \\[3ex] n = 130 \\[3ex] 70 + 190 + 130 = 390 \\[3ex] $ Rebecca has $70$ pennies, $190$ dimes, and $130$ nickels

The number of nickels is ten less than twice the number of pennies.

The number of dimes is twenty less than three times the number of pennies.

How many coins of each kind does she have?

Let the number of pennies = $p$

Let the number of nickels = $n$

Let the number of dimes = $d$

$p + n + d = 390 ...eqn.(1)$

Twice the number of pennies = $2 * p = 2p$

Ten less than $2p$ = $2p - 10$

$n = 2p - 10 ...eqn.(2)$

Three times the number of pennies = $3 * p = 3p$

Twenty less than $3p$ = $3p - 20$

$d = 3p - 20 ...eqn.(3)$

Substitute eqn.(2) and eqn.(3) into eqn.(1)

$ p + (2p - 10) + (3p - 20) = 390 \\[3ex] p + 2p - 10 + 3p - 20 = 390 \\[3ex] 6p - 30 = 390 \\[3ex] 6p = 390 + 30 \\[3ex] 6p = 420 \\[3ex] p = \dfrac{420}{6} \\[5ex] p = 70 \\[3ex] d = 3p - 20 \\[3ex] d = 3(70) - 20 \\[3ex] d = 210 - 20 \\[3ex] d = 190 \\[3ex] n = 2p - 10 \\[3ex] n = 2(70) - 10 \\[3ex] n = 140 - 10 \\[3ex] n = 130 \\[3ex] 70 + 190 + 130 = 390 \\[3ex] $ Rebecca has $70$ pennies, $190$ dimes, and $130$ nickels

(30.) The perimeter of a field is $100$ meters.

The length is $16$ meters more than the width.

Calculate the length and the width of the field.

Let the length of the field = $L$

Let the width of the field = $W$

A field is rectangular in shape.

Perimeter = $2 * L + 2 * W$

$100 = 2L + 2W$ ... first sentence ...eqn.(1)

$ 2L + 2W = 100 \\[3ex] 2(L + W) = 100 \\[3ex] L + W = \dfrac{100}{2} \\[5ex] L + W = 50 ...eqn.(2) \\[3ex] $ $L = 16 + W$ ... second sentence ...eqn.(3)

Substitute eqn.(3) into eqn.(2)

$ L + W = 50 \\[3ex] 16 + W + W = 50 \\[3ex] 16 + 2W = 50 \\[3ex] 2W = 50 - 16 \\[3ex] 2W = 34 \\[3ex] W = \dfrac{34}{2} \\[5ex] W = 17 \\[3ex] L = 16 + W \\[3ex] L = 16 + 17 \\[3ex] L = 33 \\[3ex] $ The length of the field is $33$ meters

The width of the field is $17$ meters

The length is $16$ meters more than the width.

Calculate the length and the width of the field.

Let the length of the field = $L$

Let the width of the field = $W$

A field is rectangular in shape.

Perimeter = $2 * L + 2 * W$

$100 = 2L + 2W$ ... first sentence ...eqn.(1)

$ 2L + 2W = 100 \\[3ex] 2(L + W) = 100 \\[3ex] L + W = \dfrac{100}{2} \\[5ex] L + W = 50 ...eqn.(2) \\[3ex] $ $L = 16 + W$ ... second sentence ...eqn.(3)

Substitute eqn.(3) into eqn.(2)

$ L + W = 50 \\[3ex] 16 + W + W = 50 \\[3ex] 16 + 2W = 50 \\[3ex] 2W = 50 - 16 \\[3ex] 2W = 34 \\[3ex] W = \dfrac{34}{2} \\[5ex] W = 17 \\[3ex] L = 16 + W \\[3ex] L = 16 + 17 \\[3ex] L = 33 \\[3ex] $ The length of the field is $33$ meters

The width of the field is $17$ meters

(31.) Find three consecutive even integers such that five times the first minus the third, is eight more than twice the second.

Let the first even integer = $e$

The second consecutive even integer = $e + 2$

The third consecutive even integer = $e + 4$

Five times the first even integer = $5 * e = 5e$

Minus the third = $5e - (e + 4)$

$= 5e - e - 4$

$= 4e - 4$

$"is" = equal$

Twice the second = $2 * (e + 2)$

$= 2e + 4$

Eight more than twice the second $= 8 + (2e + 4)$

$= 8 + 2e + 4$

$= 2e + 12$

$ \therefore 4e - 4 = 2e + 12 \\[3ex] 4e - 2e = 12 + 4 \\[3ex] 2e = 16 \\[3ex] e = \dfrac{16}{2} \\[5ex] e = 8 \\[3ex] e + 2 = 8 + 2 = 10 \\[3ex] e + 4 = 8 + 4 = 12 \\[3ex] $ The consecutive even integers are: $8, 10, 12$

Let the first even integer = $e$

The second consecutive even integer = $e + 2$

The third consecutive even integer = $e + 4$

Five times the first even integer = $5 * e = 5e$

Minus the third = $5e - (e + 4)$

$= 5e - e - 4$

$= 4e - 4$

$"is" = equal$

Twice the second = $2 * (e + 2)$

$= 2e + 4$

Eight more than twice the second $= 8 + (2e + 4)$

$= 8 + 2e + 4$

$= 2e + 12$

$ \therefore 4e - 4 = 2e + 12 \\[3ex] 4e - 2e = 12 + 4 \\[3ex] 2e = 16 \\[3ex] e = \dfrac{16}{2} \\[5ex] e = 8 \\[3ex] e + 2 = 8 + 2 = 10 \\[3ex] e + 4 = 8 + 4 = 12 \\[3ex] $ The consecutive even integers are: $8, 10, 12$

(32.) **ACT** The perimeter of a certain scalene triangle is $100$ inches.

The side lengths of the triangle are represented by $5x$, $3x + 30$, and $2x + 10$, respectively.

What is the length, in inches, of the longest side of the triangle?

Let the perimeter = $P$

The side lengths are $5x$, $3x + 30$, and $2x + 10$

$ P = 5x + (3x + 30) + (2x + 10) \\[3ex] P = 5x + 3x + 30 + 2x + 10 \\[3ex] P = 10x + 40 \\[3ex] P = 100 \\[3ex] 10x + 40 = 100 \\[3ex] 10x = 100 - 40 \\[3ex] 10x = 60 \\[3ex] x = \dfrac{60}{10} \\[3ex] x = 6 \\[3ex] 5x = 5(6) = 30 \\[3ex] 3x + 30 = 3(6) + 30 = 18 + 30 = 48 \\[3ex] 2x + 10 = 2(6) + 10 = 12 + 10 = 22 \\[3ex] $ The longest side of the triangle is $48$ inches

The side lengths of the triangle are represented by $5x$, $3x + 30$, and $2x + 10$, respectively.

What is the length, in inches, of the longest side of the triangle?

Let the perimeter = $P$

The side lengths are $5x$, $3x + 30$, and $2x + 10$

$ P = 5x + (3x + 30) + (2x + 10) \\[3ex] P = 5x + 3x + 30 + 2x + 10 \\[3ex] P = 10x + 40 \\[3ex] P = 100 \\[3ex] 10x + 40 = 100 \\[3ex] 10x = 100 - 40 \\[3ex] 10x = 60 \\[3ex] x = \dfrac{60}{10} \\[3ex] x = 6 \\[3ex] 5x = 5(6) = 30 \\[3ex] 3x + 30 = 3(6) + 30 = 18 + 30 = 48 \\[3ex] 2x + 10 = 2(6) + 10 = 12 + 10 = 22 \\[3ex] $ The longest side of the triangle is $48$ inches

(33.) A triangle is such that it's medium side is twice as long as it's shortest side.

It's longest side is seven meters less than three times it's shortest side.

The perimeter of the triangle is sixty five meters.

Calculate the lengths of the three sides of the triangle.

Let the shortest side = $S$

Let the medium side = $M$

Let the longest side = $L$

Medium side is twice as long as the shortest side.

$M = 2 * S$

$M = 2S ...eqn.(1)$

Three times the shortest side = $3 * S = 3S$

Seven less than $3S$ = $3S - 7$

$ L = 3S - 7 ...eqn.(2) \\[3ex] Perimeter = S + M + L \\[3ex] S + M + G = 65 ...eqn.(3) \\[3ex] Substitute\:\:eqn.(1)\:\:and\:\:eqn.(2)\:\:into\:\:eqn.(3) \\[3ex] S + (2S) + (3S - 7) = 65 \\[3ex] S + 2S + 3S - 7 = 65 \\[3ex] 6S - 7 = 65 \\[3ex] 6S = 65 + 7 \\[3ex] 6S = 72 \\[3ex] S = \dfrac{72}{6} \\[5ex] S = 12 \\[3ex] M = 2 * S \\[3ex] M = 2 * 12 \\[3ex] M = 24 \\[3ex] L = 3S - 7 \\[3ex] L = 3(12) - 7 \\[3ex] L = 36 - 7 \\[3ex] L = 29 \\[3ex] $ The side lengths of the triangle are: $12\:meters,\:\:24\:meters,\:\:and\:\:29\:meters$

It's longest side is seven meters less than three times it's shortest side.

The perimeter of the triangle is sixty five meters.

Calculate the lengths of the three sides of the triangle.

Let the shortest side = $S$

Let the medium side = $M$

Let the longest side = $L$

Medium side is twice as long as the shortest side.

$M = 2 * S$

$M = 2S ...eqn.(1)$

Three times the shortest side = $3 * S = 3S$

Seven less than $3S$ = $3S - 7$

$ L = 3S - 7 ...eqn.(2) \\[3ex] Perimeter = S + M + L \\[3ex] S + M + G = 65 ...eqn.(3) \\[3ex] Substitute\:\:eqn.(1)\:\:and\:\:eqn.(2)\:\:into\:\:eqn.(3) \\[3ex] S + (2S) + (3S - 7) = 65 \\[3ex] S + 2S + 3S - 7 = 65 \\[3ex] 6S - 7 = 65 \\[3ex] 6S = 65 + 7 \\[3ex] 6S = 72 \\[3ex] S = \dfrac{72}{6} \\[5ex] S = 12 \\[3ex] M = 2 * S \\[3ex] M = 2 * 12 \\[3ex] M = 24 \\[3ex] L = 3S - 7 \\[3ex] L = 3(12) - 7 \\[3ex] L = 36 - 7 \\[3ex] L = 29 \\[3ex] $ The side lengths of the triangle are: $12\:meters,\:\:24\:meters,\:\:and\:\:29\:meters$

(34.) A trimaran boat with a top speed of $25$ knots won the tournament prize in $2012$.

The top speed of the boat is $7$ knots more than $1.5$ times the top speed of the winner monohull in the $2010$ tournament.

Calculate the top speed of the monohull in the $2010$ tournament.

Let the speed of the winner monohull = $p$

Let the speed of the trimaran boat = $t$

$ t = 25 \\[3ex] 25 = 7 + 1.5p \\[3ex] 7 + 1.5p = 25 \\[3ex] 1.5p + 7 = 25 \\[3ex] 1.5p = 25 - 7 \\[3ex] 1.5p = 18 \\[3ex] p = \dfrac{18}{1.5} \\[5ex] p = 12 \\[3ex] $ The top speed of the monohull in the $2010$ tournament is $12$ knots.

The top speed of the boat is $7$ knots more than $1.5$ times the top speed of the winner monohull in the $2010$ tournament.

Calculate the top speed of the monohull in the $2010$ tournament.

Let the speed of the winner monohull = $p$

Let the speed of the trimaran boat = $t$

$ t = 25 \\[3ex] 25 = 7 + 1.5p \\[3ex] 7 + 1.5p = 25 \\[3ex] 1.5p + 7 = 25 \\[3ex] 1.5p = 25 - 7 \\[3ex] 1.5p = 18 \\[3ex] p = \dfrac{18}{1.5} \\[5ex] p = 12 \\[3ex] $ The top speed of the monohull in the $2010$ tournament is $12$ knots.

(35.) The sum of three times a number and seven more than the number is the same
as the difference of negative seventeen and twice the number.

Find the number.

Let the number = $n$

"is" means =

Three times the number = $3 * n = 3n$

Seven more than the number = $7 + n$

Sum of $3n$ and $7 + n$ = $3n + 7 + n$

Negative seventeen = $-17$

Twice the number = $2 * n = 2n$

Difference of $-17$ and $2n$ = $-17 - 2n$

$ \therefore 3n + 7 + n = -17 - 2n \\[3ex] 4n + 7 = -17 - 2n \\[3ex] 4n + 2n = -17 - 7 \\[3ex] 6n = -24 \\[3ex] n = -\dfrac{24}{6} \\[5ex] n = -4 \\[3ex] $ The number is $-4$

Find the number.

Let the number = $n$

"is" means =

Three times the number = $3 * n = 3n$

Seven more than the number = $7 + n$

Sum of $3n$ and $7 + n$ = $3n + 7 + n$

Negative seventeen = $-17$

Twice the number = $2 * n = 2n$

Difference of $-17$ and $2n$ = $-17 - 2n$

$ \therefore 3n + 7 + n = -17 - 2n \\[3ex] 4n + 7 = -17 - 2n \\[3ex] 4n + 2n = -17 - 7 \\[3ex] 6n = -24 \\[3ex] n = -\dfrac{24}{6} \\[5ex] n = -4 \\[3ex] $ The number is $-4$

(36.) **ACT** What is the sum of $3$ consecutive odd integers whose mean is $27$?

$ A.\:\: 39 \\[3ex] B.\:\: 75 \\[3ex] C.\:\: 81 \\[3ex] D.\:\: 87 \\[3ex] E.\:\: 93 \\[3ex] $

To find only the sum and to save time - this is ACT test

$ Mean = \bar{x} \\[3ex] \bar{x} = \dfrac{Sum}{3} \\[5ex] 27 = \dfrac{Sum}{3} \\[5ex] \dfrac{Sum}{3} = 27 \\[5ex] Sum = 3(27) \\[3ex] Sum = 81 \\[3ex] $ To find the three integers;

Let the first odd integer = $x$

The second consecutive odd integer = $x + 2$

The third consecutive odd integer = $x + 4$

Mean = $\bar{x}$

$ \bar{x} = \dfrac{x + (x + 2) + (x + 4)}{3} \\[5ex] \bar{x} = \dfrac{x + x + 2 + x + 4}{3} \\[5ex] \bar{x} = \dfrac{3x + 6}{3} \\[5ex] \bar{x} = 27 \\[3ex] \rightarrow \dfrac{3x + 6}{3} = 27 \\[5ex] Cross\:\:Multiply \\[3ex] 3x + 6 = 3(27) \\[3ex] 3x + 6 = 81 \\[3ex] 3x = 81 - 6 \\[3ex] 3x = 75 \\[3ex] x = \dfrac{75}{3} \\[5ex] x = 25 \\[3ex] x + 2 = 25 + 2 = 27 \\[3ex] x + 4 = 25 + 4 = 29 \\[3ex] Sum = 25 + 27 + 29 \\[3ex] Sum = 81 \\[3ex] \underline{Check} \\[3ex] \bar{x} = \dfrac{81}{3} = 27 $

$ A.\:\: 39 \\[3ex] B.\:\: 75 \\[3ex] C.\:\: 81 \\[3ex] D.\:\: 87 \\[3ex] E.\:\: 93 \\[3ex] $

To find only the sum and to save time - this is ACT test

$ Mean = \bar{x} \\[3ex] \bar{x} = \dfrac{Sum}{3} \\[5ex] 27 = \dfrac{Sum}{3} \\[5ex] \dfrac{Sum}{3} = 27 \\[5ex] Sum = 3(27) \\[3ex] Sum = 81 \\[3ex] $ To find the three integers;

Let the first odd integer = $x$

The second consecutive odd integer = $x + 2$

The third consecutive odd integer = $x + 4$

Mean = $\bar{x}$

$ \bar{x} = \dfrac{x + (x + 2) + (x + 4)}{3} \\[5ex] \bar{x} = \dfrac{x + x + 2 + x + 4}{3} \\[5ex] \bar{x} = \dfrac{3x + 6}{3} \\[5ex] \bar{x} = 27 \\[3ex] \rightarrow \dfrac{3x + 6}{3} = 27 \\[5ex] Cross\:\:Multiply \\[3ex] 3x + 6 = 3(27) \\[3ex] 3x + 6 = 81 \\[3ex] 3x = 81 - 6 \\[3ex] 3x = 75 \\[3ex] x = \dfrac{75}{3} \\[5ex] x = 25 \\[3ex] x + 2 = 25 + 2 = 27 \\[3ex] x + 4 = 25 + 4 = 29 \\[3ex] Sum = 25 + 27 + 29 \\[3ex] Sum = 81 \\[3ex] \underline{Check} \\[3ex] \bar{x} = \dfrac{81}{3} = 27 $

(37.) **CSEC** The table below shows the number of tickets sold for a bus tour.

Some items in the table are missing.

(i) Calculate the value of $P$

(ii) Calculate the value of $Q$

(iii) An adult ticket is TWICE the cost of a youth ticket.

Calculate the value of $R$

(iv) The bus company pays taxes of $15\%$ on each ticket sold.

Calculate the taxes paid by the bus company.

Total Cost in $\$$ is equal to the__product__ of the Number of Tickets Sold __and__ the Cost per Ticket in $\$$

$ (i) \\[3ex] 5 * P = 130.50 \\[3ex] P = \dfrac{130.50}{5} \\[5ex] P = \$26.10 \\[3ex] (ii) \\[3ex] 14 * 44.35 = Q \\[3ex] 620.9 = Q \\[3ex] Q = \$620.90 \\[3ex] (iii) \\[3ex] Cost\:\:of\:\:Youth\:\:ticket = \$44.35 \\[3ex] Cost\:\:of\:\:Adult\:\:ticket = 2(44.35) = \$88.70 \\[3ex] \rightarrow R * 88.70 = 2483.60 \\[3ex] R = \dfrac{2483.60}{88.70} \\[5ex] R = 28\:\:tickets \\[3ex] (iv) \\[3ex] Total\:\:Cost = 130.50 + Q + 2483.60 \\[3ex] Total\:\:Cost = 130.50 + 620.90 + 2483.60 \\[3ex] Total\:\:Cost = \$3235.00 \\[3ex] 15\%\:\:tax = \dfrac{15}{100} * 3235 = 0.15 * 3235 = 485.25 \\[3ex] Taxes\:\:paid = \$485.25 $

Some items in the table are missing.

Category | Number of Tickets Sold | Cost per Ticket in $ | Total Cost in $ |
---|---|---|---|

Juvenile | $5$ | $P$ | $130.50$ |

Youth | $14$ | $44.35$ | $Q$ |

Adult | $R$ | $2483.60$ |

(i) Calculate the value of $P$

(ii) Calculate the value of $Q$

(iii) An adult ticket is TWICE the cost of a youth ticket.

Calculate the value of $R$

(iv) The bus company pays taxes of $15\%$ on each ticket sold.

Calculate the taxes paid by the bus company.

Total Cost in $\$$ is equal to the

$ (i) \\[3ex] 5 * P = 130.50 \\[3ex] P = \dfrac{130.50}{5} \\[5ex] P = \$26.10 \\[3ex] (ii) \\[3ex] 14 * 44.35 = Q \\[3ex] 620.9 = Q \\[3ex] Q = \$620.90 \\[3ex] (iii) \\[3ex] Cost\:\:of\:\:Youth\:\:ticket = \$44.35 \\[3ex] Cost\:\:of\:\:Adult\:\:ticket = 2(44.35) = \$88.70 \\[3ex] \rightarrow R * 88.70 = 2483.60 \\[3ex] R = \dfrac{2483.60}{88.70} \\[5ex] R = 28\:\:tickets \\[3ex] (iv) \\[3ex] Total\:\:Cost = 130.50 + Q + 2483.60 \\[3ex] Total\:\:Cost = 130.50 + 620.90 + 2483.60 \\[3ex] Total\:\:Cost = \$3235.00 \\[3ex] 15\%\:\:tax = \dfrac{15}{100} * 3235 = 0.15 * 3235 = 485.25 \\[3ex] Taxes\:\:paid = \$485.25 $

(38.) **WASSCE** When a number is subtracted from $2$, the result equals $4$ less than one-fifth
of the number.

Find the number.

$ A.\:\: 11 \\[3ex] B.\:\: \dfrac{15}{2} \\[5ex] C.\:\: 5 \\[3ex] D.\:\: \dfrac{5}{2} \\[5ex] $

Let the number = $p$

number is subtracted from $2$ = $2 - p$

one-fifth of the number = $\dfrac{1}{5}p$

$4$ less than one-fifth of the number = $\dfrac{1}{5}p - 4$

$ \rightarrow 2 - p = \dfrac{1}{5}p - 4 \\[5ex] 2 + 4 = \dfrac{1}{5}p + p \\[5ex] 6 = \dfrac{1p}{5} + \dfrac{5p}{5} \\[5ex] 6 = \dfrac{p + 5p}{5} \\[5ex] \dfrac{6}{1} = \dfrac{6p}{5} \\[5ex] Cross\:\:Multiply \\[3ex] 1(6p) = 6(5) \\[3ex] 6p = 30 \\[3ex] p = \dfrac{30}{6} \\[5ex] p = 5 $

Find the number.

$ A.\:\: 11 \\[3ex] B.\:\: \dfrac{15}{2} \\[5ex] C.\:\: 5 \\[3ex] D.\:\: \dfrac{5}{2} \\[5ex] $

Let the number = $p$

number is subtracted from $2$ = $2 - p$

one-fifth of the number = $\dfrac{1}{5}p$

$4$ less than one-fifth of the number = $\dfrac{1}{5}p - 4$

$ \rightarrow 2 - p = \dfrac{1}{5}p - 4 \\[5ex] 2 + 4 = \dfrac{1}{5}p + p \\[5ex] 6 = \dfrac{1p}{5} + \dfrac{5p}{5} \\[5ex] 6 = \dfrac{p + 5p}{5} \\[5ex] \dfrac{6}{1} = \dfrac{6p}{5} \\[5ex] Cross\:\:Multiply \\[3ex] 1(6p) = 6(5) \\[3ex] 6p = 30 \\[3ex] p = \dfrac{30}{6} \\[5ex] p = 5 $

(39.) **ACT** Linb has $\$3.67$ in quarters $(\$0.25)$, dimes $(\$0.10)$, nickels $(\$0.05)$,
and pennies $(\$0.01)$.

She arranges these coins in rows and notices that there are $5$ more dimes than quarters, $1$ more nickel than quarters, and $25$ more pennies than quarters.

How many pennies does Linb have?

$ F.\:\: 7 \\[3ex] G.\:\: 12 \\[3ex] H.\:\: 25 \\[3ex] J.\:\: 31 \\[3ex] K.\:\: 32 \\[3ex] $

$ Let\:\:the\:\:number\:\:of\:\:pennies = p \\[3ex] Let\:\:the\:\:number\:\:of\:\:nickels = n \\[3ex] Let\:\:the\:\:number\:\:of\:\:dimes = d \\[3ex] Let\:\:the\:\:number\:\:of\:\:quarters = q \\[3ex] 0.01p + 0.05n + 0.1d + 0.25q = 3.67 ...eqn.(1) \\[3ex] d = q + 5...eqn.(2) \\[3ex] n = q + 1...eqn.(3) \\[3ex] p = q + 25...eqn.(4) \\[3ex] Substitute\:\:eqns(1.), (2.), (3.), (4.)\:\:into\:\:eqn.(1) \\[3ex] 0.01(q + 25) + 0.05(q + 1) + 0.1(q + 5) + 0.25q = 3.67 \\[3ex] 0.01q + 0.25 + 0.05q + 0.05 + 0.1q + 0.5 + 0.25q = 3.67 \\[3ex] 0.41q + 0.8 = 3.67 \\[3ex] 0.41q = 3.67 - 0.8 \\[3ex] 0.41q = 2.87 \\[3ex] q = \dfrac{2.87}{0.41} \\[5ex] q = 7 \\[3ex] Substitute\:\:q = 7\:\:into\:\:eqn.(4) \\[3ex] p = 7 + 25 \\[3ex] p = 32 \\[3ex] $ Linb has $32$ pennies

She arranges these coins in rows and notices that there are $5$ more dimes than quarters, $1$ more nickel than quarters, and $25$ more pennies than quarters.

How many pennies does Linb have?

$ F.\:\: 7 \\[3ex] G.\:\: 12 \\[3ex] H.\:\: 25 \\[3ex] J.\:\: 31 \\[3ex] K.\:\: 32 \\[3ex] $

$ Let\:\:the\:\:number\:\:of\:\:pennies = p \\[3ex] Let\:\:the\:\:number\:\:of\:\:nickels = n \\[3ex] Let\:\:the\:\:number\:\:of\:\:dimes = d \\[3ex] Let\:\:the\:\:number\:\:of\:\:quarters = q \\[3ex] 0.01p + 0.05n + 0.1d + 0.25q = 3.67 ...eqn.(1) \\[3ex] d = q + 5...eqn.(2) \\[3ex] n = q + 1...eqn.(3) \\[3ex] p = q + 25...eqn.(4) \\[3ex] Substitute\:\:eqns(1.), (2.), (3.), (4.)\:\:into\:\:eqn.(1) \\[3ex] 0.01(q + 25) + 0.05(q + 1) + 0.1(q + 5) + 0.25q = 3.67 \\[3ex] 0.01q + 0.25 + 0.05q + 0.05 + 0.1q + 0.5 + 0.25q = 3.67 \\[3ex] 0.41q + 0.8 = 3.67 \\[3ex] 0.41q = 3.67 - 0.8 \\[3ex] 0.41q = 2.87 \\[3ex] q = \dfrac{2.87}{0.41} \\[5ex] q = 7 \\[3ex] Substitute\:\:q = 7\:\:into\:\:eqn.(4) \\[3ex] p = 7 + 25 \\[3ex] p = 32 \\[3ex] $ Linb has $32$ pennies

(40.) **ACT** The equation $t = -0.0066a + 15$ models the noon temperature, $t$ degrees Celsius,
$a$ meters above sea level, on a certain day on Laurel Mountain.

According to the equation, what would be the noon temperature for that certain day on Laurel Mountain at sea level?

$ F.\:\: 0^oC \\[3ex] G.\:\: 0.0066^oC \\[3ex] H.\:\: 14.9934^oC \\[3ex] J.\:\: 15^oC \\[3ex] K.\:\: 15.0066^oC \\[3ex] $

$ t = -0.0066a + 15 \\[3ex] At\:\:sea\:\:level,\:\: a = 0 \\[3ex] t = -0.0066 * 0 + 15 \\[3ex] t = 0 + 15 \\[3ex] t = 15^oC $

According to the equation, what would be the noon temperature for that certain day on Laurel Mountain at sea level?

$ F.\:\: 0^oC \\[3ex] G.\:\: 0.0066^oC \\[3ex] H.\:\: 14.9934^oC \\[3ex] J.\:\: 15^oC \\[3ex] K.\:\: 15.0066^oC \\[3ex] $

$ t = -0.0066a + 15 \\[3ex] At\:\:sea\:\:level,\:\: a = 0 \\[3ex] t = -0.0066 * 0 + 15 \\[3ex] t = 0 + 15 \\[3ex] t = 15^oC $

(41.) Cosmas and Damian were driving in opposite directions on a straight highway.

Cosmas was driving $6$ mph faster than Damian.

After $2.5$ hours, they were $270$ miles apart.

Calculate their different driving speeds.

$mph$ means miles per hour

Remember:**s t d**

distance = speed * time

Let the driving speed of Damian = $x$

Therefore, the driving speed of Cosmas = $x + 6$

driving time of Damian = $2.5\: hours$

driving time of Cosmas = $2.5\: hours$

distance covered by Damian = $2.5x$

distance covered by Cosmas = $2.5(x + 6)$

Let us represent this information in a table.

Their total distance after $2.5\: hours$ = $2.5x + 2.5(x + 6)$

$ Total\:\:Distance = 2.5x + 2.5x + 15 \\[3ex] Total\:\:Distance = 5x + 15 \\[3ex] $ This should be equal to $270\: miles$

$ 5x + 15 = 270 \\[3ex] 5x = 270 - 15 \\[3ex] 5x = 255 \\[3ex] x = \dfrac{255}{5} \\[3ex] x = 51 \\[3ex] x + 6 = 51 + 6 = 57 \\[3ex] $ The driving speed of Damian is $51$ mph

The driving speed of Cosmas is $57$ mph

Cosmas was driving $6$ mph faster than Damian.

After $2.5$ hours, they were $270$ miles apart.

Calculate their different driving speeds.

$mph$ means miles per hour

Remember:

distance = speed * time

Let the driving speed of Damian = $x$

Therefore, the driving speed of Cosmas = $x + 6$

driving time of Damian = $2.5\: hours$

driving time of Cosmas = $2.5\: hours$

distance covered by Damian = $2.5x$

distance covered by Cosmas = $2.5(x + 6)$

Let us represent this information in a table.

$Scenario$ | $s(mph)$ | $t(hours)$ | $d(miles)$ |
---|---|---|---|

$Damian$ | $x$ | $2.5$ | $2.5x$ |

$Cosmas$ | $x + 6$ | $2.5$ | $2.5(x + 6)$ |

Their total distance after $2.5\: hours$ = $2.5x + 2.5(x + 6)$

$ Total\:\:Distance = 2.5x + 2.5x + 15 \\[3ex] Total\:\:Distance = 5x + 15 \\[3ex] $ This should be equal to $270\: miles$

$ 5x + 15 = 270 \\[3ex] 5x = 270 - 15 \\[3ex] 5x = 255 \\[3ex] x = \dfrac{255}{5} \\[3ex] x = 51 \\[3ex] x + 6 = 51 + 6 = 57 \\[3ex] $ The driving speed of Damian is $51$ mph

The driving speed of Cosmas is $57$ mph

(42.) Suppose that a chemist is mixing two acid solutions, one of $45\%$ concentration and the other
of $55\%$ concentration, which of the following concentrations could not be obtained?

$ A.\:\: 47\% \\[3ex] B.\:\: 49\% \\[3ex] C.\:\: 53\% \\[3ex] D.\:\: 57\% \\[3ex] $

When one mixes a $45\%$ concentration with a $55\%$ concentration, the concentration of the resulting mixture should be more than $45\%$ but less than $55\%$

So, $57\%$ cannot be obtained.

$ A.\:\: 47\% \\[3ex] B.\:\: 49\% \\[3ex] C.\:\: 53\% \\[3ex] D.\:\: 57\% \\[3ex] $

When one mixes a $45\%$ concentration with a $55\%$ concentration, the concentration of the resulting mixture should be more than $45\%$ but less than $55\%$

So, $57\%$ cannot be obtained.

(43.) A truck, driving at a speed of $60\: mph$ enters a highway.

$3$ hours later; a car, driving at a speed of $70 \: mph$ enters the same highway and drives in the same direction as the truck.

How long will it take the car to pass the truck?

$mph$ means miles per hour

Remember:**s t d**

distance = speed * time

Let the time driven by the car on the highway = $t$

$3$ hours earlier, the truck has been driving on the same highway

The truck drove on the highway first; then after $3$ hours, the car entered.

So, the truck had gained $3$ hours on the highway before the car entered.

Therefore, the time driven by the truck = $t + 3$

Speed of the truck = $60\: mph$

Speed of the car = $70\: mph$

Let us represent this information in a table.

In how many hours will it take for the car to catch up with the truck?

For that to happen, this means that their distances should be the same.

$ 70t = 60(t + 3) \\[3ex] 70t = 60t + 180 \\[3ex] 70t - 60t = 180 \\[3ex] 10t = 180 \\[3ex] t = \dfrac{180}{10} \\[5ex] t = 18 \\[3ex] $ In $18$ hours, the car will catch up with the truck.__After__ $18$ hours, the car will pass the truck .

$3$ hours later; a car, driving at a speed of $70 \: mph$ enters the same highway and drives in the same direction as the truck.

How long will it take the car to pass the truck?

$mph$ means miles per hour

Remember:

distance = speed * time

Let the time driven by the car on the highway = $t$

$3$ hours earlier, the truck has been driving on the same highway

The truck drove on the highway first; then after $3$ hours, the car entered.

So, the truck had gained $3$ hours on the highway before the car entered.

Therefore, the time driven by the truck = $t + 3$

Speed of the truck = $60\: mph$

Speed of the car = $70\: mph$

Let us represent this information in a table.

$Scenario$ | $s(mph)$ | $t(hours)$ | $d(miles)$ |
---|---|---|---|

$Car$ | $70$ | $t$ | $70t$ |

$Truck$ | $60$ | $t + 3$ | $60(t + 3)$ |

In how many hours will it take for the car to catch up with the truck?

For that to happen, this means that their distances should be the same.

$ 70t = 60(t + 3) \\[3ex] 70t = 60t + 180 \\[3ex] 70t - 60t = 180 \\[3ex] 10t = 180 \\[3ex] t = \dfrac{180}{10} \\[5ex] t = 18 \\[3ex] $ In $18$ hours, the car will catch up with the truck.

(44.) Suppose that a pure alcohol is added to a $34\%$ alcohol mixture.

Which concentration cannot be obtained?

$ A.\:\: 32\% \\[3ex] B.\:\: 36\% \\[3ex] C.\:\: 38\% \\[3ex] D.\:\: 40\% \\[3ex] $

Pure alchol means $100\%$ alcohol

When one mixes a $100\%$ concentration with a $34\%$ concentration, the concentration of the resulting mixture should be less than $100\%$ but more than $34\%$

So, $32\%$ cannot be obtained.

Which concentration cannot be obtained?

$ A.\:\: 32\% \\[3ex] B.\:\: 36\% \\[3ex] C.\:\: 38\% \\[3ex] D.\:\: 40\% \\[3ex] $

Pure alchol means $100\%$ alcohol

When one mixes a $100\%$ concentration with a $34\%$ concentration, the concentration of the resulting mixture should be less than $100\%$ but more than $34\%$

So, $32\%$ cannot be obtained.

(45.) How many gallons of a $50\%$ antifreeze solution must be mixed with $70$ gallons of a $10\%$
antifreeze solution to get a mixture that is $40\%$ antifreeze?

What are we looking for?

Let the number of gallons of the $50\%$ antifreeze solution be $x$

$ C-V-n \\[3ex] n = C * V \\[3ex] Let\:\:the\:\:10\%\:\:antifreeze\:\:solution = Antifreeze\:\:A \\[3ex] Let\:\:the\:\:50\%\:\:antifreeze\:\:solution = Antifreeze\:\:B \\[3ex] 10\% = \dfrac{10}{100} = 0.1 \\[5ex] 50\% = \dfrac{50}{100} = 0.5 \\[5ex] 40\% = \dfrac{40}{100} = 0.4 \\[5ex] $

$ \rightarrow 0.4(70 + x) = 7 + 0.5x \\[3ex] 28 + 0.4x = 7 + 0.5x \\[3ex] 28 - 7 = 0.5x - 0.4x \\[3ex] 21 = 0.1x \\[3ex] 0.1x = 21 \\[3ex] x = \dfrac{21}{0.1} \\[5ex] x = 210\:\:gallons \\[3ex] $ $210$ gallons of a $50\%$ antifreeze solution must be mixed with $70$ gallons of a $10\%$ antifreeze solution to get a mixture of $40\%$ antifreeze.

What are we looking for?

Let the number of gallons of the $50\%$ antifreeze solution be $x$

$ C-V-n \\[3ex] n = C * V \\[3ex] Let\:\:the\:\:10\%\:\:antifreeze\:\:solution = Antifreeze\:\:A \\[3ex] Let\:\:the\:\:50\%\:\:antifreeze\:\:solution = Antifreeze\:\:B \\[3ex] 10\% = \dfrac{10}{100} = 0.1 \\[5ex] 50\% = \dfrac{50}{100} = 0.5 \\[5ex] 40\% = \dfrac{40}{100} = 0.4 \\[5ex] $

Antifreeze | $C$ | $V$ | $n = C * V$ |
---|---|---|---|

$A$ | $0.1$ | $70$ | $7$ |

$B$ | $0.5$ | $x$ | $0.5x$ |

$Mixture$ | $0.4$ | $70 + x$ | $0.4(70 + x)$ |

$ \rightarrow 0.4(70 + x) = 7 + 0.5x \\[3ex] 28 + 0.4x = 7 + 0.5x \\[3ex] 28 - 7 = 0.5x - 0.4x \\[3ex] 21 = 0.1x \\[3ex] 0.1x = 21 \\[3ex] x = \dfrac{21}{0.1} \\[5ex] x = 210\:\:gallons \\[3ex] $ $210$ gallons of a $50\%$ antifreeze solution must be mixed with $70$ gallons of a $10\%$ antifreeze solution to get a mixture of $40\%$ antifreeze.

(46.) **ACT** The dimensions of equilateral triangle $\triangle ABC$ are given in centimeters in
the figure below.

What is the value of*y*?

$ F.\:\: 2 \\[3ex] G.\:\: 5 \\[3ex] H.\:\: 8 \\[3ex] J.\:\: 13 \\[3ex] K.\:\: 16 \\[3ex] $

An Equilateral Triangle has $3$ equal sides

This implies that all the sides are equal to one another

$ 5x - 9 = 3x + 1 \\[3ex] 5x - 3x = 1 + 9 \\[3ex] 2x = 10 \\[3ex] x = \dfrac{10}{2} \\[5ex] x = 5 \\[3ex] y + 3 = 3x + 1 \\[3ex] y + 3 = 3(5) + 1 \\[3ex] y + 3 = 15 + 1 \\[3ex] y + 3 = 16 \\[3ex] y = 16 - 3 \\[3ex] y = 13 $

What is the value of

$ F.\:\: 2 \\[3ex] G.\:\: 5 \\[3ex] H.\:\: 8 \\[3ex] J.\:\: 13 \\[3ex] K.\:\: 16 \\[3ex] $

An Equilateral Triangle has $3$ equal sides

This implies that all the sides are equal to one another

$ 5x - 9 = 3x + 1 \\[3ex] 5x - 3x = 1 + 9 \\[3ex] 2x = 10 \\[3ex] x = \dfrac{10}{2} \\[5ex] x = 5 \\[3ex] y + 3 = 3x + 1 \\[3ex] y + 3 = 3(5) + 1 \\[3ex] y + 3 = 15 + 1 \\[3ex] y + 3 = 16 \\[3ex] y = 16 - 3 \\[3ex] y = 13 $

(47.) Philemon invested a total of $\$5000.00$, part at $4\%$ simple interest and part at $5\%$
simple interest.

At the end of $1$ year, the investments had earned $\$226.00$ interest.

How much was invested at each rate?

What are we looking for?

Two things

$(1.)$ Let the investment (Principal) at the $4\%$ rate be $x$

$(2.)$ Let the investment (Principal) at the $5\%$ rate be $y$

$ x + y = 5000 \\[3ex] \rightarrow y = 5000 - x \\[3ex] $ $(2.)$ So, the investment on the $5\%$ rate is $5000 - x$

$ P-r-t-I \\[3ex] I = P * r * t \\[3ex] Let\:\:the\:\:4\%\:\:investment\:\:rate = Bank\:\:A \\[3ex] Let\:\:the\:\:5\%\:\:investment\:\:rate = Bank\:\:B \\[3ex] 4\% = \dfrac{4}{100} = 0.04 \\[5ex] 5\% = \dfrac{5}{100} = 0.05 \\[5ex] t = 1\:\:year \\[3ex] $

$ \rightarrow 0.04x + 0.05(5000 - x) = 226 \\[3ex] 0.04x + 250 - 0.05x = 226 \\[3ex] -0.01x = 226 - 250 \\[3ex] -0.01x = -24 \\[3ex] x = \dfrac{-24}{-0.01} \\[5ex] x = \$2400.00 \\[3ex] y = 5000 - x \\[3ex] y = 5000 - 2400 \\[3ex] y = \$2600.00 \\[3ex] $ Philemon invested $\$2,400.00$ at $4\%$ interest rate and $\$2,600.00$ at $5\%$ interest rate in order to earn $\$226.00$ interest in one year.

$ \underline{Check} \\[3ex] SI = Prt \\[3ex] SI = 2400 * 0.04 * 1 = \$96 ...Bank\:\:A \\[3ex] SI = 2600 * 0.05 * 1 = \$130 ...Bank\:\:B \\[3ex] \$96 + \$130 = \$226 $

At the end of $1$ year, the investments had earned $\$226.00$ interest.

How much was invested at each rate?

What are we looking for?

Two things

$(1.)$ Let the investment (Principal) at the $4\%$ rate be $x$

$(2.)$ Let the investment (Principal) at the $5\%$ rate be $y$

$ x + y = 5000 \\[3ex] \rightarrow y = 5000 - x \\[3ex] $ $(2.)$ So, the investment on the $5\%$ rate is $5000 - x$

$ P-r-t-I \\[3ex] I = P * r * t \\[3ex] Let\:\:the\:\:4\%\:\:investment\:\:rate = Bank\:\:A \\[3ex] Let\:\:the\:\:5\%\:\:investment\:\:rate = Bank\:\:B \\[3ex] 4\% = \dfrac{4}{100} = 0.04 \\[5ex] 5\% = \dfrac{5}{100} = 0.05 \\[5ex] t = 1\:\:year \\[3ex] $

Bank | $P$ | $r$ | $t$ | $I = P * r * t$ |
---|---|---|---|---|

$A$ | $x$ | $0.04$ | $1$ | $0.04x$ |

$B$ | $5000 - x$ | $0.05$ | $1$ | $0.05(5000 - x)$ |

$Total$ | $226$ |

$ \rightarrow 0.04x + 0.05(5000 - x) = 226 \\[3ex] 0.04x + 250 - 0.05x = 226 \\[3ex] -0.01x = 226 - 250 \\[3ex] -0.01x = -24 \\[3ex] x = \dfrac{-24}{-0.01} \\[5ex] x = \$2400.00 \\[3ex] y = 5000 - x \\[3ex] y = 5000 - 2400 \\[3ex] y = \$2600.00 \\[3ex] $ Philemon invested $\$2,400.00$ at $4\%$ interest rate and $\$2,600.00$ at $5\%$ interest rate in order to earn $\$226.00$ interest in one year.

$ \underline{Check} \\[3ex] SI = Prt \\[3ex] SI = 2400 * 0.04 * 1 = \$96 ...Bank\:\:A \\[3ex] SI = 2600 * 0.05 * 1 = \$130 ...Bank\:\:B \\[3ex] \$96 + \$130 = \$226 $

(48.) Deborah's uncle visited her family.

The following conversation began.

**Deborah:** Good afternoon, Uncle Joseph.

**Uncle Joseph:** Good afternoon, my dear.

How are you?

**Deborah:** Hmmm...I think I am doing great.

How are you too?

**Uncle Joseph:** You think?

**Deborah:** Yes, I'm always thinking.

**Uncle Joseph:** Interesting. It is much better than watching TV.

How old are you?

**Deborah:** Hmmmm...I'm still very young.

**Uncle Joseph:** I know. What is your age?

**Deborah:** In ten years, I will be six years more than twice my age.

How old is Deborah?

$ Let\:\:Deborah's\:\:age = p \\[3ex] Twice\:\:her\:\:age = 2 * p = 2p \\[3ex] Six\:\:more\:\:than\:\:twice\:\:her\:\:age = 6 + 2p \\[3ex] In\:\:10\:\:years, \\[3ex] Deborah's\:\:age = p + 10 \\[3ex] p + 10 = 6 + 2p \\[3ex] 6 + 2p = p + 10 \\[3ex] 2p - p = 10 - 6 \\[3ex] p = 4 \\[3ex] $ Deborah is $4$ years old.

The following conversation began.

How are you?

How are you too?

How old are you?

How old is Deborah?

$ Let\:\:Deborah's\:\:age = p \\[3ex] Twice\:\:her\:\:age = 2 * p = 2p \\[3ex] Six\:\:more\:\:than\:\:twice\:\:her\:\:age = 6 + 2p \\[3ex] In\:\:10\:\:years, \\[3ex] Deborah's\:\:age = p + 10 \\[3ex] p + 10 = 6 + 2p \\[3ex] 6 + 2p = p + 10 \\[3ex] 2p - p = 10 - 6 \\[3ex] p = 4 \\[3ex] $ Deborah is $4$ years old.

(49.) How much pure alcohol should be added to $2mL$ of a $10\%$ alcohol solution to obtain a $30\%$
alcohol solution? Round your answer to the nearest tenth.

What are we looking for?

Let the volume of the pure alcohol be $x$

Pure alcohol is $100\%$ alcohol

$ C-V-n \\[3ex] n = C * V \\[3ex] Let\:\:the\:\:10\%\:\:alcohol = Solution\:\:A \\[3ex] Let\:\:the\:\:100\%\:\:alcohol (pure\:\:alcohol) = Solution\:\:B \\[3ex] 10\% = \dfrac{10}{100} = 0.1 \\[5ex] 100\% = \dfrac{100}{100} = 1 \\[5ex] 30\% = \dfrac{30}{100} = 0.3 \\[5ex] $

$ \rightarrow 0.3(2 + x) = 0.2 + x \\[3ex] 0.6 + 0.3x = 0.2 + x \\[3ex] 0.2 + x = 0.6 + 0.3x \\[3ex] x - 0.3x = 0.6 - 0.2 \\[3ex] 0.7x = 0.4 x = \dfrac{0.7}{0.4} \\[5ex] x = 0.571428571 \\[3ex] x \approx 0.6mL \\[3ex] $ $0.6mL$ of pure alcohol should be added to $2mL$ of a $10\%$ alcohol solution to obtain a $30\%$ alcohol solution.

What are we looking for?

Let the volume of the pure alcohol be $x$

Pure alcohol is $100\%$ alcohol

$ C-V-n \\[3ex] n = C * V \\[3ex] Let\:\:the\:\:10\%\:\:alcohol = Solution\:\:A \\[3ex] Let\:\:the\:\:100\%\:\:alcohol (pure\:\:alcohol) = Solution\:\:B \\[3ex] 10\% = \dfrac{10}{100} = 0.1 \\[5ex] 100\% = \dfrac{100}{100} = 1 \\[5ex] 30\% = \dfrac{30}{100} = 0.3 \\[5ex] $

Alcohol | $C$ | $V$ | $n = C * V$ |
---|---|---|---|

$A$ | $0.1$ | $2$ | $0.2$ |

$B$ | $1$ | $x$ | $x$ |

$Mixture$ | $0.3$ | $2 + x$ | $0.3(2 + x)$ |

$ \rightarrow 0.3(2 + x) = 0.2 + x \\[3ex] 0.6 + 0.3x = 0.2 + x \\[3ex] 0.2 + x = 0.6 + 0.3x \\[3ex] x - 0.3x = 0.6 - 0.2 \\[3ex] 0.7x = 0.4 x = \dfrac{0.7}{0.4} \\[5ex] x = 0.571428571 \\[3ex] x \approx 0.6mL \\[3ex] $ $0.6mL$ of pure alcohol should be added to $2mL$ of a $10\%$ alcohol solution to obtain a $30\%$ alcohol solution.

(50.) A pharmacist wishes to mix a solution that is $6\%$ Minoxidil.

She has on hand, a $60mL$ of a $4\%$ solution and wishes to add some $8\%$ solution to obtain the desired $6\%$ solution.

How much $8\%$ solution should she add?

This is too much English.

Can we paraphrase?

How much $8\%$ should be added to $60mL$ of a $4\%$ solution to get $6\%$ solution?

What are we looking for?

Let the volume of the $8\%$ solution be $x$

$ C-V-n \\[3ex] n = C * V \\[3ex] Let\:\:the\:\:4\%\:\:solution = Solution\:\:A \\[3ex] Let\:\:the\:\:8\%\:\:solution = Solution\:\:B \\[3ex] 4\% = \dfrac{4}{100} = 0.04 \\[5ex] 8\% = \dfrac{8}{100} = 0.08 \\[5ex] 6\% = \dfrac{6}{100} = 0.06 \\[5ex] $

$ \rightarrow 0.06(60 + x) = 2.4 + 0.08x \\[3ex] 3.6 + 0.06x = 2.4 + 0.08x \\[3ex] 3.6 - 2.4 = 0.08x - 0.06x \\[3ex] 1.2 = 0.02x \\[3ex] 0.02x = 1.2 \\[3ex] x = \dfrac{1.2}{0.02} \\[5ex] x = 60mL \\[3ex] $ She needs to mix $60mL$ of an $8\%$ solution with $60mL$ of a $4\%$ solution in order to obtain $120mL$ of the desired $6\%$ solution.

She has on hand, a $60mL$ of a $4\%$ solution and wishes to add some $8\%$ solution to obtain the desired $6\%$ solution.

How much $8\%$ solution should she add?

This is too much English.

Can we paraphrase?

How much $8\%$ should be added to $60mL$ of a $4\%$ solution to get $6\%$ solution?

What are we looking for?

Let the volume of the $8\%$ solution be $x$

$ C-V-n \\[3ex] n = C * V \\[3ex] Let\:\:the\:\:4\%\:\:solution = Solution\:\:A \\[3ex] Let\:\:the\:\:8\%\:\:solution = Solution\:\:B \\[3ex] 4\% = \dfrac{4}{100} = 0.04 \\[5ex] 8\% = \dfrac{8}{100} = 0.08 \\[5ex] 6\% = \dfrac{6}{100} = 0.06 \\[5ex] $

Solution | $C$ | $V$ | $n = C * V$ |
---|---|---|---|

$A$ | $0.04$ | $60$ | $2.4$ |

$B$ | $0.08$ | $x$ | $0.08x$ |

$Mixture$ | $0.06$ | $60 + x$ | $0.06(60 + x)$ |

$ \rightarrow 0.06(60 + x) = 2.4 + 0.08x \\[3ex] 3.6 + 0.06x = 2.4 + 0.08x \\[3ex] 3.6 - 2.4 = 0.08x - 0.06x \\[3ex] 1.2 = 0.02x \\[3ex] 0.02x = 1.2 \\[3ex] x = \dfrac{1.2}{0.02} \\[5ex] x = 60mL \\[3ex] $ She needs to mix $60mL$ of an $8\%$ solution with $60mL$ of a $4\%$ solution in order to obtain $120mL$ of the desired $6\%$ solution.

(51.) Water must be added to $20mL$ of a $4\%$ Minoxidil solution to dilute it to a $2\%$ solution.

How many milliliters of water should be used?

What are we looking for?

Let the volume of water be $x$

Water is $0\%$ Minoxidil solution

$ C-V-n \\[3ex] n = C * V \\[3ex] Let\:\:the\:\:4\%\:\:Minoxidil\:\:solution = Solution\:\:A \\[3ex] Let\:\:the\:\:Water(0\%\:\:Minoxidil\:\:solution) = Solution\:\:B \\[3ex] 4\% = \dfrac{4}{100} = 0.04 \\[5ex] 0\% = \dfrac{0}{100} = 0 \\[5ex] 2\% = \dfrac{2}{100} = 0.02 \\[5ex] $

$ \rightarrow 0.02(20 + x) = 0.8 + 0 \\[3ex] 0.4 + 0.02x = 0.8 \\[3ex] 0.02x = 0.8 - 0.4 \\[3ex] 0.02x = 0.4 \\[3ex] x = \dfrac{0.02}{0.4} \\[5ex] x = 20mL \\[3ex] $ $20mL$ of water added to $20mL$ of a $4\%$ Minoxidil solution dilutes it to a $2\%$ solution.

How many milliliters of water should be used?

What are we looking for?

Let the volume of water be $x$

Water is $0\%$ Minoxidil solution

$ C-V-n \\[3ex] n = C * V \\[3ex] Let\:\:the\:\:4\%\:\:Minoxidil\:\:solution = Solution\:\:A \\[3ex] Let\:\:the\:\:Water(0\%\:\:Minoxidil\:\:solution) = Solution\:\:B \\[3ex] 4\% = \dfrac{4}{100} = 0.04 \\[5ex] 0\% = \dfrac{0}{100} = 0 \\[5ex] 2\% = \dfrac{2}{100} = 0.02 \\[5ex] $

Solution | $C$ | $V$ | $n = C * V$ |
---|---|---|---|

$A$ | $0.04$ | $20$ | $0.8$ |

$B$ | $0$ | $x$ | $0$ |

$Mixture$ | $0.02$ | $20 + x$ | $0.02(20 + x)$ |

$ \rightarrow 0.02(20 + x) = 0.8 + 0 \\[3ex] 0.4 + 0.02x = 0.8 \\[3ex] 0.02x = 0.8 - 0.4 \\[3ex] 0.02x = 0.4 \\[3ex] x = \dfrac{0.02}{0.4} \\[5ex] x = 20mL \\[3ex] $ $20mL$ of water added to $20mL$ of a $4\%$ Minoxidil solution dilutes it to a $2\%$ solution.

(52.) How much water should be added to $30$ ounces of a $7\%$ salt solution to make it a $2.1\%$
solution?

What are we looking for?

Let the volume of water be $x$

Water is $0\%$ salt solution

$ C-V-n \\[3ex] n = C * V \\[3ex] Let\:\:the\:\:7\%\:\:salt\:\:solution = Solution\:\:A \\[3ex] Let\:\:the\:\:Water(0\%\:\:salt\:\:solution) = Solution\:\:B \\[3ex] 7\% = \dfrac{7}{100} = 0.07 \\[5ex] 0\% = \dfrac{0}{100} = 0 \\[5ex] 2.1\% = \dfrac{2.1}{100} = 0.021 \\[5ex] $

$ \rightarrow 0.021(30 + x) = 2.1 + 0 \\[3ex] 0.63 + 0.021x = 2.1 \\[3ex] 0.021x = 2.1 - 0.63 \\[3ex] 0.021x = 1.47 \\[3ex] x = \dfrac{1.47}{0.021} \\[5ex] x = 70\:ounces \\[3ex] $ $70\:ounces$ of water should be added to $30\:ounces$ of a $7\%$ salt solution to make it a $2.1\%$ solution.

What are we looking for?

Let the volume of water be $x$

Water is $0\%$ salt solution

$ C-V-n \\[3ex] n = C * V \\[3ex] Let\:\:the\:\:7\%\:\:salt\:\:solution = Solution\:\:A \\[3ex] Let\:\:the\:\:Water(0\%\:\:salt\:\:solution) = Solution\:\:B \\[3ex] 7\% = \dfrac{7}{100} = 0.07 \\[5ex] 0\% = \dfrac{0}{100} = 0 \\[5ex] 2.1\% = \dfrac{2.1}{100} = 0.021 \\[5ex] $

Solution | $C$ | $V$ | $n = C * V$ |
---|---|---|---|

$A$ | $0.07$ | $30$ | $2.1$ |

$B$ | $0$ | $x$ | $0$ |

$Mixture$ | $0.021$ | $30 + x$ | $0.021(30 + x)$ |

$ \rightarrow 0.021(30 + x) = 2.1 + 0 \\[3ex] 0.63 + 0.021x = 2.1 \\[3ex] 0.021x = 2.1 - 0.63 \\[3ex] 0.021x = 1.47 \\[3ex] x = \dfrac{1.47}{0.021} \\[5ex] x = 70\:ounces \\[3ex] $ $70\:ounces$ of water should be added to $30\:ounces$ of a $7\%$ salt solution to make it a $2.1\%$ solution.

(53.) $Some$ gallons of a $50\%$ antifreeze solution must be mixed with $some$ gallons of a $10\%$
antifreeze solution to give $280$ gallons of a $40\%$ antifreeze mixture?

Determine the number of gallons of each antifreeze solution.

*
Did you notice that this question is a re-modeled form of ***Question $45$?**

So, we would have to confirm our answers with Question $45$.

What are we looking for?

Two things

$(1.)$ Let the number of gallons of the $50\%$ antifreeze solution be $x$

$(2.)$ Let the number of gallons of the $10\%$ antifreeze solution be $y$

$ x + y = 280 \\[3ex] \rightarrow y = 280 - x \\[3ex] $ $(2.)$ So, the number of gallons of the $10\%$ antifreeze solution be $280 - x$

$ C-V-n \\[3ex] n = C * V \\[3ex] Let\:\:the\:\:50\%\:\:antifreeze\:\:solution = Antifreeze\:\:A \\[3ex] Let\:\:the\:\:10\%\:\:antifreeze\:\:solution = Antifreeze\:\:B \\[3ex] 50\% = \dfrac{50}{100} = 0.5 \\[5ex] 10\% = \dfrac{10}{100} = 0.1 \\[5ex] 40\% = \dfrac{40}{100} = 0.4 \\[5ex] $

$ \rightarrow 112 = 0.5x + 0.1(280 - x) \\[3ex] 112 = 0.5x + 28 - 0.1x \\[3ex] 112 - 28 = 0.4x \\[3ex] 84 = 0.4x \\[3ex] 0.4x = 84 \\[3ex] x = \dfrac{84}{0.4} \\[5ex] x = 210\:\:gallons \\[3ex] y = 280 - x \\[3ex] y = 280 - 210 \\[3ex] y = 70\:\:gallons \\[3ex] $ $210$ gallons of a $50\%$ antifreeze solution must be mixed with $70$ gallons of a $10\%$ antifreeze solution to give $280$ gallons of a $40\%$ antifreeze mixture.

Determine the number of gallons of each antifreeze solution.

So, we would have to confirm our answers with Question $45$.

What are we looking for?

Two things

$(1.)$ Let the number of gallons of the $50\%$ antifreeze solution be $x$

$(2.)$ Let the number of gallons of the $10\%$ antifreeze solution be $y$

$ x + y = 280 \\[3ex] \rightarrow y = 280 - x \\[3ex] $ $(2.)$ So, the number of gallons of the $10\%$ antifreeze solution be $280 - x$

$ C-V-n \\[3ex] n = C * V \\[3ex] Let\:\:the\:\:50\%\:\:antifreeze\:\:solution = Antifreeze\:\:A \\[3ex] Let\:\:the\:\:10\%\:\:antifreeze\:\:solution = Antifreeze\:\:B \\[3ex] 50\% = \dfrac{50}{100} = 0.5 \\[5ex] 10\% = \dfrac{10}{100} = 0.1 \\[5ex] 40\% = \dfrac{40}{100} = 0.4 \\[5ex] $

Antifreeze | $C$ | $V$ | $n = C * V$ |
---|---|---|---|

$A$ | $0.5$ | $x$ | $0.5x$ |

$B$ | $0.1$ | $280 - x$ | $0.1(280 - x)$ |

$Mixture$ | $0.4$ | $280$ | $112$ |

$ \rightarrow 112 = 0.5x + 0.1(280 - x) \\[3ex] 112 = 0.5x + 28 - 0.1x \\[3ex] 112 - 28 = 0.4x \\[3ex] 84 = 0.4x \\[3ex] 0.4x = 84 \\[3ex] x = \dfrac{84}{0.4} \\[5ex] x = 210\:\:gallons \\[3ex] y = 280 - x \\[3ex] y = 280 - 210 \\[3ex] y = 70\:\:gallons \\[3ex] $ $210$ gallons of a $50\%$ antifreeze solution must be mixed with $70$ gallons of a $10\%$ antifreeze solution to give $280$ gallons of a $40\%$ antifreeze mixture.

(54.) **CSEC** A drinking straw of length $21\:cm$ is cut into $3$ pieces.

The length of the first piece is $x\:cm$

The second piece is $3\:cm$ shorter than the first piece.

The third piece is twice as long as the first piece.

(i) State, in terms of $x$, the length of EACH of the piece.

(ii) Write an**expression**, in terms of $x$, to represent the sum of the lengths of the three
pieces of drinking straw.

(iii)**Hence**, calculate the value of $x$

$ (i) \\[3ex] First\:\:piece = x \\[3ex] Second\:\;piece = x - 3 \\[3ex] Third\:\:piece = 2x \\[3ex] (ii) \\[3ex] Sum = x + (x - 3) + 2x \\[3ex] Sum = x + x - 3 + 2x \\[3ex] Sum = 4x - 3 \\[3ex] (iii) \\[3ex] Sum = 21 \\[3ex] \rightarrow 4x - 3 = 21 \\[3ex] 4x = 21 + 3 \\[3ex] 4x = 24 \\[3ex] x = \dfrac{24}{4} \\[5ex] x = 6\:cm $

The length of the first piece is $x\:cm$

The second piece is $3\:cm$ shorter than the first piece.

The third piece is twice as long as the first piece.

(i) State, in terms of $x$, the length of EACH of the piece.

(ii) Write an

(iii)

$ (i) \\[3ex] First\:\:piece = x \\[3ex] Second\:\;piece = x - 3 \\[3ex] Third\:\:piece = 2x \\[3ex] (ii) \\[3ex] Sum = x + (x - 3) + 2x \\[3ex] Sum = x + x - 3 + 2x \\[3ex] Sum = 4x - 3 \\[3ex] (iii) \\[3ex] Sum = 21 \\[3ex] \rightarrow 4x - 3 = 21 \\[3ex] 4x = 21 + 3 \\[3ex] 4x = 24 \\[3ex] x = \dfrac{24}{4} \\[5ex] x = 6\:cm $

(55.) Ruth invested some money at $9\%$, and $\$1600$ less than that amount at $4\%$.

The investments produced a total of $\$274$ interest in $1$ year.

How much did she invest at each rate?

What are we looking for?

Two things

$(1.)$ Let the investment (Principal) at the $9\%$ rate be $x$

$(2.)$ The investment (Principal) at the $4\%$ rate be $x - 1600$

$ P-r-t-I \\[3ex] I = P * r * t \\[3ex] Let\:\:the\:\:9\%\:\:investment\:\:rate = Bank\:\:A \\[3ex] Let\:\:the\:\:4\%\:\:investment\:\:rate = Bank\:\:B \\[3ex] 9\% = \dfrac{9}{100} = 0.09 \\[5ex] 4\% = \dfrac{4}{100} = 0.04 \\[5ex] t = 1\:\:year \\[3ex] $

$ \rightarrow 0.09x + 0.04(x - 1600) = 274 \\[3ex] 0.09x + 0.04x - 64 = 274 \\[3ex] 0.13x = 274 + 64 \\[3ex] 0.13x = 338 \\[3ex] x = \dfrac{338}{0.13} \\[5ex] x = \$2600.00 \\[3ex] x - 1600 = 2600 - 1600 = \$1000.00 \\[3ex] $ Ruth invested $\$2,600.00$ at $9\%$ interest rate and $\$1,000.00$ at $4\%$ interest rate in order to earn $\$274.00$ interest in one year

$ \underline{Check} \\[3ex] SI = Prt \\[3ex] SI = 2600 * 0.09 * 1 = \$234 ...Bank\:\:A \\[3ex] SI = 1000 * 0.04 * 1 = \$40 ...Bank\:\:B \\[3ex] \$234 + \$40 = \$274 $

The investments produced a total of $\$274$ interest in $1$ year.

How much did she invest at each rate?

What are we looking for?

Two things

$(1.)$ Let the investment (Principal) at the $9\%$ rate be $x$

$(2.)$ The investment (Principal) at the $4\%$ rate be $x - 1600$

$ P-r-t-I \\[3ex] I = P * r * t \\[3ex] Let\:\:the\:\:9\%\:\:investment\:\:rate = Bank\:\:A \\[3ex] Let\:\:the\:\:4\%\:\:investment\:\:rate = Bank\:\:B \\[3ex] 9\% = \dfrac{9}{100} = 0.09 \\[5ex] 4\% = \dfrac{4}{100} = 0.04 \\[5ex] t = 1\:\:year \\[3ex] $

Bank | $P$ | $r$ | $t$ | $I = P * r * t$ |
---|---|---|---|---|

$A$ | $x$ | $0.09$ | $1$ | $0.09x$ |

$B$ | $x - 1600$ | $0.04$ | $1$ | $0.04(x - 1600)$ |

$Total$ | $274$ |

$ \rightarrow 0.09x + 0.04(x - 1600) = 274 \\[3ex] 0.09x + 0.04x - 64 = 274 \\[3ex] 0.13x = 274 + 64 \\[3ex] 0.13x = 338 \\[3ex] x = \dfrac{338}{0.13} \\[5ex] x = \$2600.00 \\[3ex] x - 1600 = 2600 - 1600 = \$1000.00 \\[3ex] $ Ruth invested $\$2,600.00$ at $9\%$ interest rate and $\$1,000.00$ at $4\%$ interest rate in order to earn $\$274.00$ interest in one year

$ \underline{Check} \\[3ex] SI = Prt \\[3ex] SI = 2600 * 0.09 * 1 = \$234 ...Bank\:\:A \\[3ex] SI = 1000 * 0.04 * 1 = \$40 ...Bank\:\:B \\[3ex] \$234 + \$40 = \$274 $

(56.) **JAMB** John gives one-third of his money to Janet who has $₦105.00$

He finds that his money is reduced to one-fourth of what Janet now has.

Find how much money John had at first.

$ A.\:\:₦45.00 \\[3ex] B.\:\:₦48.00 \\[3ex] C.\:\:₦52.00 \\[3ex] D.\:\:₦58.00 \\[3ex] E.\:\:₦60.00 \\[3ex] $

Let John's initial amount of money be $x$

$ one-third\:\:of\:\:x = \dfrac{1}{3}x \\[5ex] He\:\:gave\:\:\dfrac{1}{3}x\:\:to\:\:Janet \\[5ex] balance = x - \dfrac{1}{3}x \\[5ex] balance = \dfrac{3}{3}x - \dfrac{1}{3}x \\[5ex] balance = \dfrac{3 - 1}{3}x \\[5ex] balance = \dfrac{2}{3}x \\[5ex] He\:\:is\:\:left\:\:with\:\:\dfrac{2}{3}x \\[5ex] Janet\:\:had\:\:₦105 \\[3ex] Janet\:\:now\:\:has\:\:105 + \dfrac{1}{3}x \\[5ex] one-fourth\:\:of\:\:it = \dfrac{1}{4}\left(105 + \dfrac{1}{3}x\right) \\[5ex] This\:\:is\:\:what\:\:John\:\:has\:\:now \\[3ex] \rightarrow \dfrac{2}{3}x = \dfrac{1}{4}\left(105 + \dfrac{1}{3}x\right) \\[5ex] LCD = 12 \\[3ex] 12\left(\dfrac{2}{3}x\right) = 12 * \dfrac{1}{4}\left(105 + \dfrac{1}{3}x\right) \\[5ex] 4(2x) = 3\left(105 + \dfrac{1}{3}x\right) \\[5ex] 8x = 3(105) + 3\left(\dfrac{1}{3}x\right) \\[5ex] 8x = 315 + x \\[3ex] 8x - x = 315 \\[3ex] 7x = 315 \\[3ex] x = \dfrac{315}{7} \\[5ex] x = ₦45.00 \\[3ex] $__Check__

He finds that his money is reduced to one-fourth of what Janet now has.

Find how much money John had at first.

$ A.\:\:₦45.00 \\[3ex] B.\:\:₦48.00 \\[3ex] C.\:\:₦52.00 \\[3ex] D.\:\:₦58.00 \\[3ex] E.\:\:₦60.00 \\[3ex] $

Let John's initial amount of money be $x$

$ one-third\:\:of\:\:x = \dfrac{1}{3}x \\[5ex] He\:\:gave\:\:\dfrac{1}{3}x\:\:to\:\:Janet \\[5ex] balance = x - \dfrac{1}{3}x \\[5ex] balance = \dfrac{3}{3}x - \dfrac{1}{3}x \\[5ex] balance = \dfrac{3 - 1}{3}x \\[5ex] balance = \dfrac{2}{3}x \\[5ex] He\:\:is\:\:left\:\:with\:\:\dfrac{2}{3}x \\[5ex] Janet\:\:had\:\:₦105 \\[3ex] Janet\:\:now\:\:has\:\:105 + \dfrac{1}{3}x \\[5ex] one-fourth\:\:of\:\:it = \dfrac{1}{4}\left(105 + \dfrac{1}{3}x\right) \\[5ex] This\:\:is\:\:what\:\:John\:\:has\:\:now \\[3ex] \rightarrow \dfrac{2}{3}x = \dfrac{1}{4}\left(105 + \dfrac{1}{3}x\right) \\[5ex] LCD = 12 \\[3ex] 12\left(\dfrac{2}{3}x\right) = 12 * \dfrac{1}{4}\left(105 + \dfrac{1}{3}x\right) \\[5ex] 4(2x) = 3\left(105 + \dfrac{1}{3}x\right) \\[5ex] 8x = 3(105) + 3\left(\dfrac{1}{3}x\right) \\[5ex] 8x = 315 + x \\[3ex] 8x - x = 315 \\[3ex] 7x = 315 \\[3ex] x = \dfrac{315}{7} \\[5ex] x = ₦45.00 \\[3ex] $

$ \underline{John} \\[3ex] \dfrac{1}{3} * 45 = 15 \\[5ex] 45 - 15 = 30 $ | $ \underline{Janet} \\[3ex] 105 + 15 = 120 \\[3ex] \dfrac{1}{4} * 120 = 30 $ |

(57.) The capacity of the radiator of a Lamborghini is $90L$.

It contains a $20\%$ antifreeze solution.

But, it needs a $70\%$ antifreeze solution.

So, a pure antifreeze solution is needed.

How many liters of the $20\%$ solution needs to be__drained__ so it can be replaced with the pure
antifreeze solution?

The radiator of the Lamborghini holds $90L$

We shall drain out the $30\%$ antifreeze solution

Then, we shall add the pure antifreeze solution

Pure antifreeze is $100\%$ antifreeze solution

What are we looking for?

Let the volume of the $20\%$ antifreeze solution to be drained be $x$

That $x$ liters has to be filled with the pure antifreeze $(100\%)$ antifreeze solution

After $x$ liters has been drained, you have $90 - x$ liters remaining.

The remaining $90 - x$ liters still contains the $30\%$ antifreeze solution

$ C-V-n \\[3ex] n = C * V \\[3ex] Let\:\:the\:\:20\%\:\:antifreeze = Solution\:\:A \\[3ex] Let\:\:the\:\:100\%\:\:antifreeze (pure\:\:antifreeze) = Solution\:\:B \\[3ex] 20\% = \dfrac{20}{100} = 0.2 \\[5ex] 100\% = \dfrac{100}{100} = 1 \\[5ex] 70\% = \dfrac{60}{100} = 0.7 \\[5ex] $

$ \rightarrow 63 = 0.2(90 - x) + x \\[3ex] 0.2(90 - x) + x = 63 \\[3ex] 18 - 0.2x + x = 63 \\[3ex] 0.8x = 63 - 18 \\[3ex] 0.8x = 45 \\[3ex] x = \dfrac{45}{0.8} \\[5ex] x = 56.25L \\[3ex] $ $56.25L$ of the $20\%$ antifreeze solution needs to be__drained__ so it can be replaced with the pure
antifreeze solution.

It contains a $20\%$ antifreeze solution.

But, it needs a $70\%$ antifreeze solution.

So, a pure antifreeze solution is needed.

How many liters of the $20\%$ solution needs to be

The radiator of the Lamborghini holds $90L$

We shall drain out the $30\%$ antifreeze solution

Then, we shall add the pure antifreeze solution

Pure antifreeze is $100\%$ antifreeze solution

What are we looking for?

Let the volume of the $20\%$ antifreeze solution to be drained be $x$

That $x$ liters has to be filled with the pure antifreeze $(100\%)$ antifreeze solution

After $x$ liters has been drained, you have $90 - x$ liters remaining.

The remaining $90 - x$ liters still contains the $30\%$ antifreeze solution

$ C-V-n \\[3ex] n = C * V \\[3ex] Let\:\:the\:\:20\%\:\:antifreeze = Solution\:\:A \\[3ex] Let\:\:the\:\:100\%\:\:antifreeze (pure\:\:antifreeze) = Solution\:\:B \\[3ex] 20\% = \dfrac{20}{100} = 0.2 \\[5ex] 100\% = \dfrac{100}{100} = 1 \\[5ex] 70\% = \dfrac{60}{100} = 0.7 \\[5ex] $

Antifreeze | $C$ | $V$ | $n = C * V$ |
---|---|---|---|

$A$ | $0.2$ | $90 - x$ | $0.2(90 - x)$ |

$B$ | $1$ | $x$ | $x$ |

$Mixture$ | $0.7$ | $90$ | $63$ |

$ \rightarrow 63 = 0.2(90 - x) + x \\[3ex] 0.2(90 - x) + x = 63 \\[3ex] 18 - 0.2x + x = 63 \\[3ex] 0.8x = 63 - 18 \\[3ex] 0.8x = 45 \\[3ex] x = \dfrac{45}{0.8} \\[5ex] x = 56.25L \\[3ex] $ $56.25L$ of the $20\%$ antifreeze solution needs to be

(58.) **CSEC** $500$ tickets were sold for a concert.

Of these, $x$ tickets were sold at $\$6$ each, and the remainder at $\$10$ each.

(i) Write an expression, in terms of $x$, for

a) the number of tickets sold at $\$10$ each

b) the TOTAL amount of money collected for the sale of the $500$ tickets.

(ii) The sum of $\$4108$ was collected for the sale of the $500$ tickets.

Calculate the number of tickets sold at $\$6$ each.

$ (i) \\[3ex] 500\:\:tickets \\[3ex] x\:\:tickets\:\:sold\:\:at\:\:\$6\:\:each \\[3ex] remaining = 500 - x \\[3ex] (500 - x)\:\:tickets\:\:sold\:\:at\:\:\$10\:\:each \\[3ex] a) \\[3ex] Number\:\:of\:\:tickets\:\:sold\:\:at\:\:\$10\:\:each = 500 - x \\[3ex] b) \\[3ex] x\:\:tickets\:\:@\:\:\$6\:\:each = x(6) = 6x \\[3ex] (500 - x)\:\:tickets\:\:@\:\:\$10\:\:each = 10(500 - x) \\[3ex] Total\:\:amount = 6x + 10(500 - x) \\[3ex] Total\:\:amount = 6x + 5000 - 10x \\[3ex] Total\:\:amount = 5000 - 4x \\[3ex] (ii) \\[3ex] Total\:\:amount = \$4108 \\[3ex] \rightarrow 4108 = 5000 - 4x \\[3ex] 4x = 5000 - 4108 \\[3ex] 4x = 892 \\[3ex] x = \dfrac{892}{4} \\[5ex] x = 223 \\[3ex] 500 - x = 500 - 223 = 277 \\[3ex] $ $223$ tickets were sold at $\$6$ each

$277$ tickets were sold at $\$10$ each

$ \underline{Check} \\[3ex] 223(6) + 277(10) \\[3ex] = 1338 + 2770 \\[3ex] = \$4108 $

Of these, $x$ tickets were sold at $\$6$ each, and the remainder at $\$10$ each.

(i) Write an expression, in terms of $x$, for

a) the number of tickets sold at $\$10$ each

b) the TOTAL amount of money collected for the sale of the $500$ tickets.

(ii) The sum of $\$4108$ was collected for the sale of the $500$ tickets.

Calculate the number of tickets sold at $\$6$ each.

$ (i) \\[3ex] 500\:\:tickets \\[3ex] x\:\:tickets\:\:sold\:\:at\:\:\$6\:\:each \\[3ex] remaining = 500 - x \\[3ex] (500 - x)\:\:tickets\:\:sold\:\:at\:\:\$10\:\:each \\[3ex] a) \\[3ex] Number\:\:of\:\:tickets\:\:sold\:\:at\:\:\$10\:\:each = 500 - x \\[3ex] b) \\[3ex] x\:\:tickets\:\:@\:\:\$6\:\:each = x(6) = 6x \\[3ex] (500 - x)\:\:tickets\:\:@\:\:\$10\:\:each = 10(500 - x) \\[3ex] Total\:\:amount = 6x + 10(500 - x) \\[3ex] Total\:\:amount = 6x + 5000 - 10x \\[3ex] Total\:\:amount = 5000 - 4x \\[3ex] (ii) \\[3ex] Total\:\:amount = \$4108 \\[3ex] \rightarrow 4108 = 5000 - 4x \\[3ex] 4x = 5000 - 4108 \\[3ex] 4x = 892 \\[3ex] x = \dfrac{892}{4} \\[5ex] x = 223 \\[3ex] 500 - x = 500 - 223 = 277 \\[3ex] $ $223$ tickets were sold at $\$6$ each

$277$ tickets were sold at $\$10$ each

$ \underline{Check} \\[3ex] 223(6) + 277(10) \\[3ex] = 1338 + 2770 \\[3ex] = \$4108 $

(59.) Jeremiah jogged at $5$ miles per hour and then jogged at $6$ miles per hour, traveling $7$ miles
in $1.2$ hours.

How long did he jog at each speed?

Let the first scenario be the $5$ miles per hour jog

Let the second scenario be the $6$ miles per hour jog

Let the time to jog $5$ miles per hour = $x$

Let the time to jog $6$ miles per hour = $y$

$ x + y = 1.2 \\[3ex] y = 1.2 - x \\[3ex] s\:\:\:t\:\:\:d \\[3ex] $

$ 5x + 6(1.2 - x) = 7 \\[3ex] 5x + 7.2 - 6x = 7 \\[3ex] -x = 7 - 7.2 \\[3ex] -x = -0.2 \\[3ex] x = \dfrac{-0.2}{-1} \\[5ex] x = 0.2\:hours \\[3ex] 1.2 - x = 1.2 - 0.2 = 1\:hour \\[3ex] $ Jeremiah jogged $5$ miles per hour for $0.2$ hour

Jeremiah jogged $6$ miles per hour for $1$ hour.

$ \underline{Check} \\[3ex] 5\:mph\:\:for\:\:0.2\:hour = 5(0.2) = 1\:mile \\[3ex] 6\:mph\:\:for\:\:1\:hour = 6(1) = 6\:miles \\[3ex] Total\:\:distance = 1 + 6 = 7\:miles $

How long did he jog at each speed?

Let the first scenario be the $5$ miles per hour jog

Let the second scenario be the $6$ miles per hour jog

Let the time to jog $5$ miles per hour = $x$

Let the time to jog $6$ miles per hour = $y$

$ x + y = 1.2 \\[3ex] y = 1.2 - x \\[3ex] s\:\:\:t\:\:\:d \\[3ex] $

$Scenario$ | $s(mph)$ | $t(hours)$ | $d(miles)$ |
---|---|---|---|

$First$ | $5$ | $x$ | $5x$ |

$Second$ | $6$ | $1.2 - x$ | $6(1.2 - x)$ |

$Total$ | $1.2$ | $7$ |

$ 5x + 6(1.2 - x) = 7 \\[3ex] 5x + 7.2 - 6x = 7 \\[3ex] -x = 7 - 7.2 \\[3ex] -x = -0.2 \\[3ex] x = \dfrac{-0.2}{-1} \\[5ex] x = 0.2\:hours \\[3ex] 1.2 - x = 1.2 - 0.2 = 1\:hour \\[3ex] $ Jeremiah jogged $5$ miles per hour for $0.2$ hour

Jeremiah jogged $6$ miles per hour for $1$ hour.

$ \underline{Check} \\[3ex] 5\:mph\:\:for\:\:0.2\:hour = 5(0.2) = 1\:mile \\[3ex] 6\:mph\:\:for\:\:1\:hour = 6(1) = 6\:miles \\[3ex] Total\:\:distance = 1 + 6 = 7\:miles $

(60.) Phoebe has a choice between receiving a monthly salary of $\$1560$ from a company OR a base
salary of $\$1400$ and a $4\%$ commission on the amount of furniture she sells during the month.

For what amount of sales will the two choices be equal?

Let the amount of sales = $x$

$ \underline{First\:\:Choice} \\[3ex] Monthly\:\:salary = 1560 \\[3ex] \underline{Second\:\:Choice} \\[3ex] Base\:\:salary = 1400 \\[3ex] 4\%\:\:commission\:\:on\:\:x = \dfrac{4}{100} * x = 0.04x \\[5ex] \rightarrow 1400 + 0.04x = 1560 \\[3ex] 0.04x = 1560 - 1400 \\[3ex] 0.04x = 160 x = \dfrac{160}{0.04} \\[5ex] x = 4000 \\[3ex] \underline{Check} \\[3ex] 1400 + 0.04(4000) = 1400 + 160 = 1560 \\[3ex] $ Phoebe has to sell $\$4000.00$ worth of furniture for the two choices to be equal.

For what amount of sales will the two choices be equal?

Let the amount of sales = $x$

$ \underline{First\:\:Choice} \\[3ex] Monthly\:\:salary = 1560 \\[3ex] \underline{Second\:\:Choice} \\[3ex] Base\:\:salary = 1400 \\[3ex] 4\%\:\:commission\:\:on\:\:x = \dfrac{4}{100} * x = 0.04x \\[5ex] \rightarrow 1400 + 0.04x = 1560 \\[3ex] 0.04x = 1560 - 1400 \\[3ex] 0.04x = 160 x = \dfrac{160}{0.04} \\[5ex] x = 4000 \\[3ex] \underline{Check} \\[3ex] 1400 + 0.04(4000) = 1400 + 160 = 1560 \\[3ex] $ Phoebe has to sell $\$4000.00$ worth of furniture for the two choices to be equal.

(61.) Peter and Paul left a city at the same time.
Peter went East.

Paul went West.

Paul traveled $3$ miles per hour faster than Peter.

After $4$ hours, they were $108$ miles apart.

How fast were they traveling?

miles per hour = $mph$

Let the speed of Peter = $x$

So, the speed of Paul = $3 + x$

$ s\:\:\:t\:\:\:d \\[3ex] $

$ 4x + 4(3 + x) = 108 \\[3ex] 4x + 12 + 4x = 108 \\[3ex] 8x = 108 - 12 \\[3ex] 8x = 96 \\[3ex] x = \dfrac{96}{8} \\[5ex] x = 12mph \\[3ex] 3 + x = 3 + 12 = 15mph \\[3ex] $ Peter traveled at $12mph$

Paul traveled at $15mph$

Paul went West.

Paul traveled $3$ miles per hour faster than Peter.

After $4$ hours, they were $108$ miles apart.

How fast were they traveling?

miles per hour = $mph$

Let the speed of Peter = $x$

So, the speed of Paul = $3 + x$

$ s\:\:\:t\:\:\:d \\[3ex] $

$Scenario$ | $s(mph)$ | $t(hours)$ | $d(miles)$ |
---|---|---|---|

$Peter\rightarrow East$ | $x$ | $4$ | $4x$ |

$Paul\rightarrow West$ | $3 + x$ | $4$ | $4(3 + x)$ |

$Total$ | $108$ |

$ 4x + 4(3 + x) = 108 \\[3ex] 4x + 12 + 4x = 108 \\[3ex] 8x = 108 - 12 \\[3ex] 8x = 96 \\[3ex] x = \dfrac{96}{8} \\[5ex] x = 12mph \\[3ex] 3 + x = 3 + 12 = 15mph \\[3ex] $ Peter traveled at $12mph$

Paul traveled at $15mph$

(62.) Haggai earns a base salary of $\$900$ per month and a commission of $5\%$ on the amount of sales
he makes.

One month, he received a $\$1398$ paycheck.

Calculate the amount of his sales for the month.

Let his monthly sales for that month be $x$

$ Base\:\:salary = 900 \\[3ex] 5\%\:\:commission\:\:on\:\:x = \dfrac{5}{100} * x = 0.05x \\[5ex] Monthly\:\:salary(for\:\:that\:\:month) = 1398 \\[3ex] \rightarrow 900 + 0.05x = 1398 \\[3ex] 0.05x = 1398 - 900 \\[3ex] 0.05x = 498 \\[3ex] x = \dfrac{498}{0.05} \\[5ex] x = \$9960 \\[3ex] $ Haggai's monthly sales for that month is $\$9960.00$

One month, he received a $\$1398$ paycheck.

Calculate the amount of his sales for the month.

Let his monthly sales for that month be $x$

$ Base\:\:salary = 900 \\[3ex] 5\%\:\:commission\:\:on\:\:x = \dfrac{5}{100} * x = 0.05x \\[5ex] Monthly\:\:salary(for\:\:that\:\:month) = 1398 \\[3ex] \rightarrow 900 + 0.05x = 1398 \\[3ex] 0.05x = 1398 - 900 \\[3ex] 0.05x = 498 \\[3ex] x = \dfrac{498}{0.05} \\[5ex] x = \$9960 \\[3ex] $ Haggai's monthly sales for that month is $\$9960.00$

(63.) Malachi drove for $5$ hours on the freeway, then decreased his speed by $20$ miles per hour
and drove for $3$ more hours on a country road.

If his total trip was $436$ miles, then what was his speed on the freeway?

miles per hour = $mph$

Let the speed on the freeway = $x$

So, the speed on the country road = $x - 20$

$ s\:\:\:t\:\:\:d \\[3ex] $

$ 5x + 3(x - 20) = 436 \\[3ex] 5x + 3x - 60 = 436 \\[3ex] 8x = 436 + 60 \\[3ex] 8x = 496 \\[3ex] x = \dfrac{496}{8} \\[5ex] x = 62mph \\[3ex] x - 20 = 62 - 20 = 42mph \\[3ex] $ Malachi traveled at $62mph$ on the freeway

He traveled at $42mph$ on the country road

If his total trip was $436$ miles, then what was his speed on the freeway?

miles per hour = $mph$

Let the speed on the freeway = $x$

So, the speed on the country road = $x - 20$

$ s\:\:\:t\:\:\:d \\[3ex] $

$Scenario$ | $s(mph)$ | $t(hours)$ | $d(miles)$ |
---|---|---|---|

$Freeway$ | $x$ | $5$ | $5x$ |

$Country\:\:Road$ | $x - 20$ | $3$ | $3(x - 20)$ |

$Total$ | $436$ |

$ 5x + 3(x - 20) = 436 \\[3ex] 5x + 3x - 60 = 436 \\[3ex] 8x = 436 + 60 \\[3ex] 8x = 496 \\[3ex] x = \dfrac{496}{8} \\[5ex] x = 62mph \\[3ex] x - 20 = 62 - 20 = 42mph \\[3ex] $ Malachi traveled at $62mph$ on the freeway

He traveled at $42mph$ on the country road

(64.) Love Thy Neighbor Cabs charges $\$2.25$ pickup fee and $\$1.75$ per mile traveled.

Judith's fare for a cross-town cab ride is $\$19.75$.

How far did she travel in the cab?

Let the distance Judith traveled in the cab be $p$

$ Pickup\:\:Fee = 2.25 \\[3ex] 1.75\:\:per\:\:mile\:\:for\:\:p\:\:miles = 1.75 * p = 1.75p \\[3ex] Total\:\:charge = 19.75 \\[3ex] \rightarrow 2.25 + 1.75p = 19.75 \\[3ex] 1.75p = 19.75 - 2.25 \\[3ex] 1.75p = 17.5 \\[3ex] p = \dfrac{17.5}{1.75} \\[5ex] p = 10\:\:miles \\[3ex] $ Judith traveled for $10$ miles in the cab.

Judith's fare for a cross-town cab ride is $\$19.75$.

How far did she travel in the cab?

Let the distance Judith traveled in the cab be $p$

$ Pickup\:\:Fee = 2.25 \\[3ex] 1.75\:\:per\:\:mile\:\:for\:\:p\:\:miles = 1.75 * p = 1.75p \\[3ex] Total\:\:charge = 19.75 \\[3ex] \rightarrow 2.25 + 1.75p = 19.75 \\[3ex] 1.75p = 19.75 - 2.25 \\[3ex] 1.75p = 17.5 \\[3ex] p = \dfrac{17.5}{1.75} \\[5ex] p = 10\:\:miles \\[3ex] $ Judith traveled for $10$ miles in the cab.

(65.) **JAMB** A man drove for $4$ hours at a certain speed.

He then doubled his speed and drove for another $3$ hours.

Altogether, he covered $600\:km$.

At what speed did he drive for the last $3$ hours?

$ A.\:\:120\:km/hr \\[3ex] B.\:\:60\:km/hr \\[3ex] C.\:\:600\:km/hr \\[3ex] D.\:\:50\:km/hr \\[3ex] E.\:\:100\:km/hr \\[3ex] $

kilometers per hour = $km/hr$

Let the first scenario (for the first speed) = $x$

So, the second scenario (for the second speed) = $2x$

$ s\:\:\:t\:\:\:d \\[3ex] $

$ 4x + 6x = 600 \\[3ex] 10x = 600 \\[3ex] x = \dfrac{600}{10} \\[5ex] x = 60\:km/hr \\[3ex] 2x = 2(60) = 120\:km/hr \\[3ex] $ The man drove $60\:km/hr$ for $4$ hours

He drove $120\:km/hr$ for $3$ hours

He then doubled his speed and drove for another $3$ hours.

Altogether, he covered $600\:km$.

At what speed did he drive for the last $3$ hours?

$ A.\:\:120\:km/hr \\[3ex] B.\:\:60\:km/hr \\[3ex] C.\:\:600\:km/hr \\[3ex] D.\:\:50\:km/hr \\[3ex] E.\:\:100\:km/hr \\[3ex] $

kilometers per hour = $km/hr$

Let the first scenario (for the first speed) = $x$

So, the second scenario (for the second speed) = $2x$

$ s\:\:\:t\:\:\:d \\[3ex] $

$Scenario$ | $s(km/hr)$ | $t(hr)$ | $d(km)$ |
---|---|---|---|

$First$ | $x$ | $4$ | $4x$ |

$Second$ | $2x$ | $3$ | $6x$ |

$Total$ | $600$ |

$ 4x + 6x = 600 \\[3ex] 10x = 600 \\[3ex] x = \dfrac{600}{10} \\[5ex] x = 60\:km/hr \\[3ex] 2x = 2(60) = 120\:km/hr \\[3ex] $ The man drove $60\:km/hr$ for $4$ hours

He drove $120\:km/hr$ for $3$ hours

(66.) **CSEC** (c) Adam, Imran, and Shakeel were playing a card game.

Adam scored $x$ points

Imran scored $3$ points fewer than Adam

Shakeel scored twice as many as many points as Imran

Together, they scored $39$ points.

(i) Write down, in terms of $x$, an expression for the number of points scored by Shakeel.

(ii) Write an equation which may be used to find the value of $x$

$ Adam's\:\:points = x \\[3ex] 3\:\:points\:\:fewer\:\:than\:\:x = x - 3 \\[3ex] \rightarrow Imran's\:\:points = x - 3 \\[3ex] twice\:\:as\:\:many\:\:points\:\:as\:\:(x - 3) = 2(x - 3) \\[3ex] \rightarrow Shakeel's\:\:points = 2(x - 3) \\[3ex] (i) \\[3ex] Shakeel's\:\:points = 2(x - 3) \\[3ex] (ii) \\[3ex] \therefore x + (x - 3) + 2(x - 3) = 39 \\[3ex] x + x - 3 + 2x - 6 = 39 \\[3ex] 4x - 9 = 39 \\[3ex] 4x = 39 + 9 \\[3ex] 4x = 48 \\[3ex] x = \dfrac{48}{4} \\[5ex] x = 12 $

Adam scored $x$ points

Imran scored $3$ points fewer than Adam

Shakeel scored twice as many as many points as Imran

Together, they scored $39$ points.

(i) Write down, in terms of $x$, an expression for the number of points scored by Shakeel.

(ii) Write an equation which may be used to find the value of $x$

$ Adam's\:\:points = x \\[3ex] 3\:\:points\:\:fewer\:\:than\:\:x = x - 3 \\[3ex] \rightarrow Imran's\:\:points = x - 3 \\[3ex] twice\:\:as\:\:many\:\:points\:\:as\:\:(x - 3) = 2(x - 3) \\[3ex] \rightarrow Shakeel's\:\:points = 2(x - 3) \\[3ex] (i) \\[3ex] Shakeel's\:\:points = 2(x - 3) \\[3ex] (ii) \\[3ex] \therefore x + (x - 3) + 2(x - 3) = 39 \\[3ex] x + x - 3 + 2x - 6 = 39 \\[3ex] 4x - 9 = 39 \\[3ex] 4x = 39 + 9 \\[3ex] 4x = 48 \\[3ex] x = \dfrac{48}{4} \\[5ex] x = 12 $

(67.) **ACT** A copy machine makes $60$ copies per minute.

A second copy machine makes $80$ copies per minute.

The second machine starts making copies $2$ minutes after the first machine starts.

Both machines stop making copies $8$ minutes after the first machine started.

Together, the $2$ machines made how many copies?

$ A.\:\:480 \\[3ex] B.\:\:600 \\[3ex] C.\:\:680 \\[3ex] D.\:\:720 \\[3ex] E.\:\:960 \\[3ex] $

We can solve this question just as we did for Motion, Mixtures, and the Mathematics of Finance problems

It is best solved by setting it up as a table

Let the number of minutes for which the first copier started making copies = $p$

This implies that $2$ minutes later, for which the second copier starts = $p - 2$

number of copies = copies per minute * number of minutes

Both machines stop making copies $8$ minutes after the first machine started.

This means that the number of minutes is $8$

$ p = 8 \\[3ex] First\:\:Machine:\:\:60p = 60(8) = 480 \\[3ex] Second\:\:Machine:\:\:80(p - 2) = 80(8 - 2) = 80(6) = 480 \\[3ex] Total\:\:number\:\:of\:\:copies = 480 + 480 = 960 \\[3ex] $ $8$ minutes after the first machine started, both copy machines made $960$ copies.

A second copy machine makes $80$ copies per minute.

The second machine starts making copies $2$ minutes after the first machine starts.

Both machines stop making copies $8$ minutes after the first machine started.

Together, the $2$ machines made how many copies?

$ A.\:\:480 \\[3ex] B.\:\:600 \\[3ex] C.\:\:680 \\[3ex] D.\:\:720 \\[3ex] E.\:\:960 \\[3ex] $

We can solve this question just as we did for Motion, Mixtures, and the Mathematics of Finance problems

It is best solved by setting it up as a table

Let the number of minutes for which the first copier started making copies = $p$

This implies that $2$ minutes later, for which the second copier starts = $p - 2$

number of copies = copies per minute * number of minutes

$Scenario$ | Copies per minutes | minutes | copies |
---|---|---|---|

First Copy Machine | $60$ | $p$ | $60p$ |

Second Copy Machine | $80$ | $p - 2$ | $80(p - 2)$ |

Both machines stop making copies $8$ minutes after the first machine started.

This means that the number of minutes is $8$

$ p = 8 \\[3ex] First\:\:Machine:\:\:60p = 60(8) = 480 \\[3ex] Second\:\:Machine:\:\:80(p - 2) = 80(8 - 2) = 80(6) = 480 \\[3ex] Total\:\:number\:\:of\:\:copies = 480 + 480 = 960 \\[3ex] $ $8$ minutes after the first machine started, both copy machines made $960$ copies.

(68.) **JAMB** Three boys shared some oranges.

The first received $\dfrac{1}{3}$ of the oranges.

The second received $\dfrac{2}{3}$ of the remainder.

If the third boy received the remaining $12$ oranges, how many oranges did they share?

$ A.\:\: 60 \\[3ex] B.\:\: 54 \\[3ex] C.\:\: 48 \\[3ex] D.\:\: 42 \\[3ex] $

We can solve this in at least two ways.

One method is to use the*Proportional Reasoning Method*...Question $25$ of Applications of Fractions, Ratios, and Proportions

Another method is to solve it Algebraically (as shown below)

Let the number of oranges that they shared be $p$

$ First\:\:boy\:\:received\:\: \dfrac{1}{3}p \\[5ex] Remainder = p - \dfrac{1}{3}p = \dfrac{3}{3}p - \dfrac{1}{3}p = \dfrac{3 - 1}{3}p = \dfrac{2}{3}p \\[5ex] Second\:\:boy\:\:received\:\: \dfrac{2}{3} * \dfrac{2}{3}p = \dfrac{2 * 2}{3 * 3}p = \dfrac{4}{9}p \\[5ex] Remaining = p - \left(\dfrac{1}{3}p + \dfrac{4}{9}p\right) \\[5ex] \dfrac{1}{3}p + \dfrac{4}{9}p \\[5ex] = \dfrac{p}{3} + \dfrac{4p}{9} = \dfrac{3p}{9} + \dfrac{4p}{9} \\[5ex] = \dfrac{3p + 4p}{9} \\[5ex] = \dfrac{7p}{9} \\[5ex] p - \left(\dfrac{1}{3}p + \dfrac{4}{9}p\right) \\[5ex] = p - \dfrac{7p}{9} \\[5ex] = \dfrac{9p}{9} - \dfrac{7p}{9} \\[5ex] = \dfrac{9p - 7p}{9} \\[5ex] = \dfrac{2p}{9} \\[5ex] Remaining = \dfrac{2p}{9} \\[5ex] Remaining\:\:is\:\:also\:\:12 \\[3ex] \rightarrow \dfrac{2p}{9} = 12 \\[5ex] 2p = 12(9) \\[3ex] p = \dfrac{12(9)}{2} \\[5ex] p = 6(9) \\[3ex] p = 54 \\[3ex] The\:\:boys\:\:shared\:\:54\:\:oranges \\[3ex] First\:\:boy\:\:received\:\: \dfrac{1}{3}p = \dfrac{1}{3} * 54 = 18 \\[5ex] Second\:\:boy\:\:received\:\: \dfrac{4}{9}p = \dfrac{4}{9} * 54 = 4(6) = 24 \\[5ex] Third\:\:boy\:\:received\:\: 12 \\[3ex] 18 + 24 + 12 = 54 \\[3ex] $ $54$ oranges were shared among the three boys

The first boy received $18$ oranges

The second boy received $24$ oranges

The third boy received $12$ oranges

The first received $\dfrac{1}{3}$ of the oranges.

The second received $\dfrac{2}{3}$ of the remainder.

If the third boy received the remaining $12$ oranges, how many oranges did they share?

$ A.\:\: 60 \\[3ex] B.\:\: 54 \\[3ex] C.\:\: 48 \\[3ex] D.\:\: 42 \\[3ex] $

We can solve this in at least two ways.

One method is to use the

Another method is to solve it Algebraically (as shown below)

Let the number of oranges that they shared be $p$

$ First\:\:boy\:\:received\:\: \dfrac{1}{3}p \\[5ex] Remainder = p - \dfrac{1}{3}p = \dfrac{3}{3}p - \dfrac{1}{3}p = \dfrac{3 - 1}{3}p = \dfrac{2}{3}p \\[5ex] Second\:\:boy\:\:received\:\: \dfrac{2}{3} * \dfrac{2}{3}p = \dfrac{2 * 2}{3 * 3}p = \dfrac{4}{9}p \\[5ex] Remaining = p - \left(\dfrac{1}{3}p + \dfrac{4}{9}p\right) \\[5ex] \dfrac{1}{3}p + \dfrac{4}{9}p \\[5ex] = \dfrac{p}{3} + \dfrac{4p}{9} = \dfrac{3p}{9} + \dfrac{4p}{9} \\[5ex] = \dfrac{3p + 4p}{9} \\[5ex] = \dfrac{7p}{9} \\[5ex] p - \left(\dfrac{1}{3}p + \dfrac{4}{9}p\right) \\[5ex] = p - \dfrac{7p}{9} \\[5ex] = \dfrac{9p}{9} - \dfrac{7p}{9} \\[5ex] = \dfrac{9p - 7p}{9} \\[5ex] = \dfrac{2p}{9} \\[5ex] Remaining = \dfrac{2p}{9} \\[5ex] Remaining\:\:is\:\:also\:\:12 \\[3ex] \rightarrow \dfrac{2p}{9} = 12 \\[5ex] 2p = 12(9) \\[3ex] p = \dfrac{12(9)}{2} \\[5ex] p = 6(9) \\[3ex] p = 54 \\[3ex] The\:\:boys\:\:shared\:\:54\:\:oranges \\[3ex] First\:\:boy\:\:received\:\: \dfrac{1}{3}p = \dfrac{1}{3} * 54 = 18 \\[5ex] Second\:\:boy\:\:received\:\: \dfrac{4}{9}p = \dfrac{4}{9} * 54 = 4(6) = 24 \\[5ex] Third\:\:boy\:\:received\:\: 12 \\[3ex] 18 + 24 + 12 = 54 \\[3ex] $ $54$ oranges were shared among the three boys

The first boy received $18$ oranges

The second boy received $24$ oranges

The third boy received $12$ oranges

(69.) A $150-lb$ person who runs at $6mph$ for $1$ hour burns about $720$ calories.

The same person, walking at $2mph$ for $90$ minutes, burns about $240$ calories.

Suppose a $150-lb$ person runs at $6mph$ for $75$ minutes.

How far would the person have to walk at $2mph$ in order to burn the same number of calories as burned when running?

Please do not let the $150-lb$ confuse you. You will not use $150$ in your calculation.

This question deviated from the normal $s\:\:\:t\:\:d$ that we have been doing

We have three variables: speed (mph), time (hours), and calories.

We shall use $s\:\:\:t\:\:d$ partially (to help us do some steps)

We shall not be using $s\:\:\:t\:\:d$ completely

*
Teacher: So, how should we solve it? *

What do you think?

Wait for responses from students

Use Direct Questioning technique to encourage critical thinking

__First:__ We need to determine the number of calories burned per mile by running

In other words, how many calories does a $150-lb$ person burn by running one mile?

__Second:__ We need to determine the number of calories burned per mile by walking

In other words, how many calories does a $150-lb$ person burn by walking one mile?

__Third:__ We need to determine the number of calories burned by a $150-lb$ person
for running $6mph$ for $75$ minutes.

__Fourth:__ We need to determine the number of miles that $150-lb$ person needs to walk at
$2mph$ to burn the same number of calories as he/she burned by running $6mph$ for $75$ minutes.

$ \underline{First} \\[3ex] d = s * t \\[3ex] s = 6mph \\[3ex] t = 1\:\:hour \\[3ex] d = 6(1) \\[3ex] d = 6\:\:miles \\[3ex] Running\:\:6\:\:miles\:\:burned\:\:720\:\:calories \\[3ex] Running\:\:1\:\:mile\:\:burns\:\:x\:\:calories \\[3ex] \underline{Method\:\:of\:\:Proportional\:\:Reasoning} \\[3ex] \dfrac{6}{1} = \dfrac{720}{x} \\[5ex] Cross\:\:Multiply \\[3ex] 6(x) = 1(720) \\[3ex] 6x = 720 \\[3ex] x = \dfrac{720}{6} \\[5ex] x = 120\:\:calories \therefore Running\:\:1\:\:mile\:\:burns\:\:120\:\:calories \\[3ex] \underline{Second} \\[3ex] d = s * t \\[3ex] s = 2mph \\[3ex] t = 90\:\:minutes = \dfrac{90}{60}\:\:hours \\[3ex] t = 1.5\:\:hours \\[3ex] d = 2(1.5) \\[3ex] d = 3\:\:miles \\[3ex] Walking\:\:3\:\:miles\:\:burned\:\:240\:\:calories \\[3ex] Walking\:\:1\:\:mile\:\:burns\:\:y\:\:calories \\[3ex] \underline{Method\:\:of\:\:Proportional\:\:Reasoning} \\[3ex] \dfrac{3}{1} = \dfrac{240}{y} \\[5ex] Cross\:\:Multiply \\[3ex] 3(y) = 1(240) \\[3ex] 3y = 240 \\[3ex] y = \dfrac{240}{3} \\[5ex] y = 80\:\:calories \therefore Walking\:\:1\:\:mile\:\:burns\:\:80\:\:calories \\[3ex] \underline{Third} \\[3ex] d = s * t \\[3ex] s = 6mph \\[3ex] t = 75\:\:minutes = \dfrac{75}{60}\:\:hours \\[3ex] t = 1.25\:\:hours \\[3ex] d = 6(1.25) \\[3ex] d = 7.5\:\:miles \\[3ex] Running\:\:1\:\:mile\:\:burns\:\:120\:\:calories \\[3ex] Running\:\:7.5\:\:mile\:\:burns\:\:p\:\:calories \\[3ex] \underline{Method\:\:of\:\:Proportional\:\:Reasoning} \\[3ex] \dfrac{1}{7.5} = \dfrac{120}{p} \\[5ex] Cross\:\:Multiply \\[3ex] 1(p) = 7.5(120) \\[3ex] p = 900\:\:calories \therefore Running\:\:6\:\:miles\:\:for\:\:75\:\:minutes\:\:burned\:\:900\:\:calories \\[3ex] \underline{Fourth} \\[3ex] Walking\:\:1\:\:mile\:\:burns\:\:80\:\:calories \\[3ex] Walking\:\:k\:\:miles\:\:burns\:\:900\:\:calories \\[3ex] \underline{Method\:\:of\:\:Proportional\:\:Reasoning} \\[3ex] \dfrac{1}{k} = \dfrac{80}{900} \\[5ex] Cross\:\:Multiply \\[3ex] k(80) = 1(900) \\[3ex] 80k = 900 \\[3ex] k = \dfrac{900}{80} \\[5ex] k = 11.25\:\:miles \\[3ex] $ The $150-lb$ person would have to walk $11.25\:\:miles$ at $2mph$ in order to burn the same number of calories $900\:\:calories$ as burned when running.

The same person, walking at $2mph$ for $90$ minutes, burns about $240$ calories.

Suppose a $150-lb$ person runs at $6mph$ for $75$ minutes.

How far would the person have to walk at $2mph$ in order to burn the same number of calories as burned when running?

Please do not let the $150-lb$ confuse you. You will not use $150$ in your calculation.

This question deviated from the normal $s\:\:\:t\:\:d$ that we have been doing

We have three variables: speed (mph), time (hours), and calories.

We shall use $s\:\:\:t\:\:d$ partially (to help us do some steps)

We shall not be using $s\:\:\:t\:\:d$ completely

What do you think?

Wait for responses from students

Use Direct Questioning technique to encourage critical thinking

In other words, how many calories does a $150-lb$ person burn by running one mile?

In other words, how many calories does a $150-lb$ person burn by walking one mile?

$ \underline{First} \\[3ex] d = s * t \\[3ex] s = 6mph \\[3ex] t = 1\:\:hour \\[3ex] d = 6(1) \\[3ex] d = 6\:\:miles \\[3ex] Running\:\:6\:\:miles\:\:burned\:\:720\:\:calories \\[3ex] Running\:\:1\:\:mile\:\:burns\:\:x\:\:calories \\[3ex] \underline{Method\:\:of\:\:Proportional\:\:Reasoning} \\[3ex] \dfrac{6}{1} = \dfrac{720}{x} \\[5ex] Cross\:\:Multiply \\[3ex] 6(x) = 1(720) \\[3ex] 6x = 720 \\[3ex] x = \dfrac{720}{6} \\[5ex] x = 120\:\:calories \therefore Running\:\:1\:\:mile\:\:burns\:\:120\:\:calories \\[3ex] \underline{Second} \\[3ex] d = s * t \\[3ex] s = 2mph \\[3ex] t = 90\:\:minutes = \dfrac{90}{60}\:\:hours \\[3ex] t = 1.5\:\:hours \\[3ex] d = 2(1.5) \\[3ex] d = 3\:\:miles \\[3ex] Walking\:\:3\:\:miles\:\:burned\:\:240\:\:calories \\[3ex] Walking\:\:1\:\:mile\:\:burns\:\:y\:\:calories \\[3ex] \underline{Method\:\:of\:\:Proportional\:\:Reasoning} \\[3ex] \dfrac{3}{1} = \dfrac{240}{y} \\[5ex] Cross\:\:Multiply \\[3ex] 3(y) = 1(240) \\[3ex] 3y = 240 \\[3ex] y = \dfrac{240}{3} \\[5ex] y = 80\:\:calories \therefore Walking\:\:1\:\:mile\:\:burns\:\:80\:\:calories \\[3ex] \underline{Third} \\[3ex] d = s * t \\[3ex] s = 6mph \\[3ex] t = 75\:\:minutes = \dfrac{75}{60}\:\:hours \\[3ex] t = 1.25\:\:hours \\[3ex] d = 6(1.25) \\[3ex] d = 7.5\:\:miles \\[3ex] Running\:\:1\:\:mile\:\:burns\:\:120\:\:calories \\[3ex] Running\:\:7.5\:\:mile\:\:burns\:\:p\:\:calories \\[3ex] \underline{Method\:\:of\:\:Proportional\:\:Reasoning} \\[3ex] \dfrac{1}{7.5} = \dfrac{120}{p} \\[5ex] Cross\:\:Multiply \\[3ex] 1(p) = 7.5(120) \\[3ex] p = 900\:\:calories \therefore Running\:\:6\:\:miles\:\:for\:\:75\:\:minutes\:\:burned\:\:900\:\:calories \\[3ex] \underline{Fourth} \\[3ex] Walking\:\:1\:\:mile\:\:burns\:\:80\:\:calories \\[3ex] Walking\:\:k\:\:miles\:\:burns\:\:900\:\:calories \\[3ex] \underline{Method\:\:of\:\:Proportional\:\:Reasoning} \\[3ex] \dfrac{1}{k} = \dfrac{80}{900} \\[5ex] Cross\:\:Multiply \\[3ex] k(80) = 1(900) \\[3ex] 80k = 900 \\[3ex] k = \dfrac{900}{80} \\[5ex] k = 11.25\:\:miles \\[3ex] $ The $150-lb$ person would have to walk $11.25\:\:miles$ at $2mph$ in order to burn the same number of calories $900\:\:calories$ as burned when running.

(70.) **ACT** For $2$ consecutive integers, the result of adding the smaller integer and triple the larger
integer is $79$.

What are the $2$ integers?

$ A.\:\: 18, 19 \\[3ex] B.\:\: 19, 20 \\[3ex] C.\:\: 20, 21 \\[3ex] D.\:\: 26, 27 \\[3ex] E.\:\: 39, 40 \\[3ex] $

$ Let\:\:the\:\:first\:\:integer(smaller\:\:integer) = p \\[3ex] Next\:\:consecutive\:\:integer(larger\:\:integer) = p + 1 \\[3ex] Triple\:\:the\:\:larger\:\:integer = 3(p + 1) \\[3ex] p + 3(p + 1) = 79 \\[3ex] p + 3p + 3 = 79 \\[3ex] 4p = 79 - 3 \\[3ex] 4p = 76 \\[3ex] p = \dfrac{76}{4} \\[5ex] p = 19 \\[3ex] p + 1 = 19 + 1 = 20 \\[3ex] $ The two integers are $19$ and $20$

What are the $2$ integers?

$ A.\:\: 18, 19 \\[3ex] B.\:\: 19, 20 \\[3ex] C.\:\: 20, 21 \\[3ex] D.\:\: 26, 27 \\[3ex] E.\:\: 39, 40 \\[3ex] $

$ Let\:\:the\:\:first\:\:integer(smaller\:\:integer) = p \\[3ex] Next\:\:consecutive\:\:integer(larger\:\:integer) = p + 1 \\[3ex] Triple\:\:the\:\:larger\:\:integer = 3(p + 1) \\[3ex] p + 3(p + 1) = 79 \\[3ex] p + 3p + 3 = 79 \\[3ex] 4p = 79 - 3 \\[3ex] 4p = 76 \\[3ex] p = \dfrac{76}{4} \\[5ex] p = 19 \\[3ex] p + 1 = 19 + 1 = 20 \\[3ex] $ The two integers are $19$ and $20$

(71.) **ACT** Yolanda collects trading cards, and she has started her younger brothers, Xavier
and Zach, collecting cards as well.

As of today, Zach owns $5$ more cards than Xavier, and Yolanda owns twice as many cards as Xavier and Zach combined.

Which of the following equations expresses the relationship between $y$, the number of cards Yolanda owns, and $x$, the number of cards Xavier owns?

$ F.\:\: y = x - 5 \\[3ex] G.\:\: y = x + 5 \\[3ex] H.\:\: y = 2x \\[3ex] J.\:\: y = 4x - 10 \\[3ex] K.\:\: y = 4x + 10 \\[3ex] $

$ Number\:\:of\:\:cards\:\:owned\:\:by\:\:Xavier = x \\[3ex] Zach\:\:owns\:\:5\:\:more\:\:cards\:\:than\:\:Xavier \\[3ex] Number\:\:of\:\:cards\:\:owned\:\:by\:\:Zach = x + 5 \\[3ex] Number\:\:of\:\:cards\:\:owned\:\:by\:\:Yolanda = y \\[3ex] Yolanda\:\:owns\:\:twice\:\:as\:\:many\:\:as\:\:Xavier\:\:and\:\:Zach \\[3ex] Number\:\:of\:\:cards\:\:owned\:\:by\:\:Xavier\:\:and\:\:Zach = x + x + 5 = 2x + 5 \\[3ex] y = 2(2x + 5) \\[3ex] y = 4x + 10 $

As of today, Zach owns $5$ more cards than Xavier, and Yolanda owns twice as many cards as Xavier and Zach combined.

Which of the following equations expresses the relationship between $y$, the number of cards Yolanda owns, and $x$, the number of cards Xavier owns?

$ F.\:\: y = x - 5 \\[3ex] G.\:\: y = x + 5 \\[3ex] H.\:\: y = 2x \\[3ex] J.\:\: y = 4x - 10 \\[3ex] K.\:\: y = 4x + 10 \\[3ex] $

$ Number\:\:of\:\:cards\:\:owned\:\:by\:\:Xavier = x \\[3ex] Zach\:\:owns\:\:5\:\:more\:\:cards\:\:than\:\:Xavier \\[3ex] Number\:\:of\:\:cards\:\:owned\:\:by\:\:Zach = x + 5 \\[3ex] Number\:\:of\:\:cards\:\:owned\:\:by\:\:Yolanda = y \\[3ex] Yolanda\:\:owns\:\:twice\:\:as\:\:many\:\:as\:\:Xavier\:\:and\:\:Zach \\[3ex] Number\:\:of\:\:cards\:\:owned\:\:by\:\:Xavier\:\:and\:\:Zach = x + x + 5 = 2x + 5 \\[3ex] y = 2(2x + 5) \\[3ex] y = 4x + 10 $

(72.) **ACT** The text message component of each of Juan's monthly phone bills consists of
$\$10.00$ for the first $300$ text messages sent that month, plus $\$0.10$
for each additional text message sent that month. On Juan's most recent phone bill, he was charged a
total of $\$16.50$ for text messages. For how many text messages in total was Juan charged
on this bill?

We know the cost of the first $300$ messages. It is $\$10.00$

We do not know the number of "additional text messages". Let it be $x$

We do know that the cost of each additional text message = $\$0.10$

So, the cost of the additional messages = $0.1(x) = 0.1x$

We also know the total cost. It is $\$16.50$

Basic cost (for the first $300$ messages) + Cost of "additional" text messages = Total cost

$ 10 + 0.1x = 16.50 \\[3ex] 0.1x = 16.5 - 10 \\[3ex] 0.1x = 6.5 \\[3ex] x = \dfrac{6.5}{0.1} \\[5ex] x = 65 \\[3ex] $ Additional text messages = $65$

"Basic" text messages = $300$

Total text messages for that bill = $65 + 300 = 365$ messages

We know the cost of the first $300$ messages. It is $\$10.00$

We do not know the number of "additional text messages". Let it be $x$

We do know that the cost of each additional text message = $\$0.10$

So, the cost of the additional messages = $0.1(x) = 0.1x$

We also know the total cost. It is $\$16.50$

Basic cost (for the first $300$ messages) + Cost of "additional" text messages = Total cost

$ 10 + 0.1x = 16.50 \\[3ex] 0.1x = 16.5 - 10 \\[3ex] 0.1x = 6.5 \\[3ex] x = \dfrac{6.5}{0.1} \\[5ex] x = 65 \\[3ex] $ Additional text messages = $65$

"Basic" text messages = $300$

Total text messages for that bill = $65 + 300 = 365$ messages

(73.) Samson invested a total of $\$12,900$ in two accounts.

The first account earned an annual rate of return of $12\%$.

However, the second account suffered a $3\%$ loss in the same period.

At the end of one year, the total amount of money gained was $\$243$

How much was invested into each account?

What are we looking for?

Two things

$(1.)$ Let the investment (Principal) at the $12\%$ rate be $x$

$(2.)$ Let the investment (Principal) at the $3\%$ rate be $y$

$ x + y = 12900 \\[3ex] \rightarrow y = 12900 - x \\[3ex] $ $(2.)$ So, the investment on the $3\%$ rate is $12900 - x$

What other thing should we note?

The interest rate in the account that earned a gain (the 12\% rate account) is positive.

It is positive because of the gain.

The interest rate in the account that suffered a loss (the 3\% account) is negative.

It is negative because of the loss.

$ P-r-t-I \\[3ex] I = P * r * t \\[3ex] Let\:\:the\:\:12\%\:\:investment\:\:rate = Bank\:\:A \\[3ex] Let\:\:the\:\:3\%\:\:investment\:\:rate = Bank\:\:B \\[3ex] 12\% = \dfrac{12}{100} = 0.12 \\[5ex] 3\% = \dfrac{3}{100} = 0.03 \\[5ex] t = 1\:\:year \\[3ex] $

$ \rightarrow 0.12x + -0.03(12900 - x) = 243 \\[3ex] 0.12x - 0.03(12900 - x) = 243 \\[3ex] 0.12x - 387 + 0.03x = 243 \\[3ex] 0.12x + 0.03x = 243 + 387 \\[3ex] 0.15x = 630 \\[3ex] x = \dfrac{630}{0.15} \\[5ex] x = \$4200.00 \\[3ex] y = 12900 - x \\[3ex] y = 12900 - 4200 \\[3ex] y = \$8700.00 \\[3ex] $ Samson invested $\$4,200.00$ at $12\%$ interest rate and $\$8,700.00$ at negative $3\%$ interest rate in order to earn $\$243.00$ interest in one year.

$ \underline{Check} \\[3ex] SI = Prt \\[3ex] SI = 4200 * 0.12 * 1 = \$504 ...Bank\:\:A \\[3ex] SI = 8700 * -0.03 * 1 = \$-261 ...Bank\:\:B \\[3ex] \$504 + -\$261 = \$504 - \$261 = \$243 $

The first account earned an annual rate of return of $12\%$.

However, the second account suffered a $3\%$ loss in the same period.

At the end of one year, the total amount of money gained was $\$243$

How much was invested into each account?

What are we looking for?

Two things

$(1.)$ Let the investment (Principal) at the $12\%$ rate be $x$

$(2.)$ Let the investment (Principal) at the $3\%$ rate be $y$

$ x + y = 12900 \\[3ex] \rightarrow y = 12900 - x \\[3ex] $ $(2.)$ So, the investment on the $3\%$ rate is $12900 - x$

What other thing should we note?

The interest rate in the account that earned a gain (the 12\% rate account) is positive.

It is positive because of the gain.

The interest rate in the account that suffered a loss (the 3\% account) is negative.

It is negative because of the loss.

$ P-r-t-I \\[3ex] I = P * r * t \\[3ex] Let\:\:the\:\:12\%\:\:investment\:\:rate = Bank\:\:A \\[3ex] Let\:\:the\:\:3\%\:\:investment\:\:rate = Bank\:\:B \\[3ex] 12\% = \dfrac{12}{100} = 0.12 \\[5ex] 3\% = \dfrac{3}{100} = 0.03 \\[5ex] t = 1\:\:year \\[3ex] $

Bank | $P$ | $r$ | $t$ | $I = P * r * t$ |
---|---|---|---|---|

$A$ | $x$ | $0.12$ | $1$ | $0.12x$ |

$B$ | $12900 - x$ | $-0.03$ | $1$ | $-0.03(12900 - x)$ |

$Total$ | $243$ |

$ \rightarrow 0.12x + -0.03(12900 - x) = 243 \\[3ex] 0.12x - 0.03(12900 - x) = 243 \\[3ex] 0.12x - 387 + 0.03x = 243 \\[3ex] 0.12x + 0.03x = 243 + 387 \\[3ex] 0.15x = 630 \\[3ex] x = \dfrac{630}{0.15} \\[5ex] x = \$4200.00 \\[3ex] y = 12900 - x \\[3ex] y = 12900 - 4200 \\[3ex] y = \$8700.00 \\[3ex] $ Samson invested $\$4,200.00$ at $12\%$ interest rate and $\$8,700.00$ at negative $3\%$ interest rate in order to earn $\$243.00$ interest in one year.

$ \underline{Check} \\[3ex] SI = Prt \\[3ex] SI = 4200 * 0.12 * 1 = \$504 ...Bank\:\:A \\[3ex] SI = 8700 * -0.03 * 1 = \$-261 ...Bank\:\:B \\[3ex] \$504 + -\$261 = \$504 - \$261 = \$243 $

(74.) **ACT** Jeralyn purchases $1$ box of granola bars and $6$ boxes of chocolate bars for a
total price of $\$22.00$.

The price of each box of granola bars is $\$2.50$, and the price of each box of chocolate bars is $n$ dollars.

Which of the following equations models Jeralyn's purchase?

$ F.\:\: 2.50 + 22.00 = 6n \\[3ex] G.\:\: 2.50n + 22.00 = 6n \\[3ex] H.\:\: 2.50(6n) = 22.00 \\[3ex] J.\:\: 2.50 + 6n = 22.00 \\[3ex] K.\:\: 2.50n + 6n = 22.00 \\[3ex] $

$1$ box of granola bar @ $\$2.50$ per box = $1 * 2.50 = 2.50$

$6$ boxes of chocolate bars @ $n$ dollars per box = $6 * n = 6n$

Total price = $\$22.00$

$ \therefore 2.50 + 6n = 22.00 $

The price of each box of granola bars is $\$2.50$, and the price of each box of chocolate bars is $n$ dollars.

Which of the following equations models Jeralyn's purchase?

$ F.\:\: 2.50 + 22.00 = 6n \\[3ex] G.\:\: 2.50n + 22.00 = 6n \\[3ex] H.\:\: 2.50(6n) = 22.00 \\[3ex] J.\:\: 2.50 + 6n = 22.00 \\[3ex] K.\:\: 2.50n + 6n = 22.00 \\[3ex] $

$1$ box of granola bar @ $\$2.50$ per box = $1 * 2.50 = 2.50$

$6$ boxes of chocolate bars @ $n$ dollars per box = $6 * n = 6n$

Total price = $\$22.00$

$ \therefore 2.50 + 6n = 22.00 $

(75.) Phoebe invests her savings in two accounts, one paying $6\%$ and the other paying $10\%$ simple
interest per year.

She puts twice as much in the lower-yielding account because it is less risky.

Her annual interest is $\$4950$.

How much did she invest at each rate?

Let the investment (Principal) at the $10\%$ rate (the higher-yielding account) be $x$

This means that the investment at the $6\%$ rate (the lower-yielding account) = $2x$

$ P-r-t-I \\[3ex] I = P * r * t \\[3ex] Let\:\:the\:\:6\%\:\:investment\:\:rate = Bank\:\:A \\[3ex] Let\:\:the\:\:10\%\:\:investment\:\:rate = Bank\:\:B \\[3ex] 6\% = \dfrac{6}{100} = 0.06 \\[5ex] 10\% = \dfrac{10}{100} = 0.1 \\[5ex] t = 1\:\:year \\[3ex] $

$ \rightarrow 0.12x + 0.1x = 4950 \\[3ex] 0.22x = 4950 \\[3ex] x = \dfrac{4950}{0.22} \\[5ex] x = \$22500.00 \\[3ex] 2x = 2(22500) = \$45000 \\[3ex] $ Phoebe invested $\$45,000.00$ at $6\%$ interest rate and $\$22,500.00$ at $10\%$ interest rate in order to earn $\$4,950.00$ interest in one year.

$ \underline{Check} \\[3ex] SI = Prt \\[3ex] SI = 45000 * 0.06 * 1 = \$2700 ...Bank\:\:A \\[3ex] SI = 22500 * 0.1 * 1 = \$2250 ...Bank\:\:B \\[3ex] \$2700 + \$2250 = \$4950 $

She puts twice as much in the lower-yielding account because it is less risky.

Her annual interest is $\$4950$.

How much did she invest at each rate?

Let the investment (Principal) at the $10\%$ rate (the higher-yielding account) be $x$

This means that the investment at the $6\%$ rate (the lower-yielding account) = $2x$

$ P-r-t-I \\[3ex] I = P * r * t \\[3ex] Let\:\:the\:\:6\%\:\:investment\:\:rate = Bank\:\:A \\[3ex] Let\:\:the\:\:10\%\:\:investment\:\:rate = Bank\:\:B \\[3ex] 6\% = \dfrac{6}{100} = 0.06 \\[5ex] 10\% = \dfrac{10}{100} = 0.1 \\[5ex] t = 1\:\:year \\[3ex] $

Bank | $P$ | $r$ | $t$ | $I = P * r * t$ |
---|---|---|---|---|

$A$ | $2x$ | $0.06$ | $1$ | $0.12x$ |

$B$ | $x$ | $0.1$ | $1$ | $0.1x$ |

$Total$ | $4950$ |

$ \rightarrow 0.12x + 0.1x = 4950 \\[3ex] 0.22x = 4950 \\[3ex] x = \dfrac{4950}{0.22} \\[5ex] x = \$22500.00 \\[3ex] 2x = 2(22500) = \$45000 \\[3ex] $ Phoebe invested $\$45,000.00$ at $6\%$ interest rate and $\$22,500.00$ at $10\%$ interest rate in order to earn $\$4,950.00$ interest in one year.

$ \underline{Check} \\[3ex] SI = Prt \\[3ex] SI = 45000 * 0.06 * 1 = \$2700 ...Bank\:\:A \\[3ex] SI = 22500 * 0.1 * 1 = \$2250 ...Bank\:\:B \\[3ex] \$2700 + \$2250 = \$4950 $

(76.) **WASSCE** A petrol tanker is $\dfrac{2}{5}$ full.

When $35,000$ litres of petrol are added, the tanker will be $\dfrac{3}{4}$ full.

What is the capacity of the tanker in litres?

$ A.\:\: 70,000 \\[3ex] B.\:\: 75,000 \\[3ex] C.\:\: 90,000 \\[3ex] D.\:\: 100,000 \\[3ex] $

We can solve this question in two ways.

Use any method you prefer.

One method will be explained here.

The other method is explained here (Question $8$)

Litres or Liters...no worries

United State: Liters

Nigeria, Britain: Litres

WASSCE is West African question...so I am using litres

$ Let\:\:x = capacity\:\:of\:\:the tank \\[3ex] Initial\:\:level:\:\: \dfrac{2}{5}x \\[5ex] Added:\:\:35000\:litres \\[3ex] New\:\:level:\:\: \dfrac{3}{4}x \\[5ex] \rightarrow \dfrac{2}{5}x + 35000 = \dfrac{3}{4}x \\[5ex] 35000 = \dfrac{3}{4}x - \dfrac{2}{5}x \\[5ex] 35000 = \dfrac{15}{20}x - \dfrac{8}{20}x \\[5ex] 35000 = \dfrac{7}{20}x \\[5ex] \dfrac{7}{20}x = 35000 \\[5ex] 7x = 20(35000) \\[3ex] x = \dfrac{20(35000)}{7} \\[5ex] x = 20(5000) \\[3ex] x = 100000 \\[3ex] $ The capacity of the tank is $100,000$ litres

When $35,000$ litres of petrol are added, the tanker will be $\dfrac{3}{4}$ full.

What is the capacity of the tanker in litres?

$ A.\:\: 70,000 \\[3ex] B.\:\: 75,000 \\[3ex] C.\:\: 90,000 \\[3ex] D.\:\: 100,000 \\[3ex] $

We can solve this question in two ways.

Use any method you prefer.

One method will be explained here.

The other method is explained here (Question $8$)

Litres or Liters...no worries

United State: Liters

Nigeria, Britain: Litres

WASSCE is West African question...so I am using litres

$ Let\:\:x = capacity\:\:of\:\:the tank \\[3ex] Initial\:\:level:\:\: \dfrac{2}{5}x \\[5ex] Added:\:\:35000\:litres \\[3ex] New\:\:level:\:\: \dfrac{3}{4}x \\[5ex] \rightarrow \dfrac{2}{5}x + 35000 = \dfrac{3}{4}x \\[5ex] 35000 = \dfrac{3}{4}x - \dfrac{2}{5}x \\[5ex] 35000 = \dfrac{15}{20}x - \dfrac{8}{20}x \\[5ex] 35000 = \dfrac{7}{20}x \\[5ex] \dfrac{7}{20}x = 35000 \\[5ex] 7x = 20(35000) \\[3ex] x = \dfrac{20(35000)}{7} \\[5ex] x = 20(5000) \\[3ex] x = 100000 \\[3ex] $ The capacity of the tank is $100,000$ litres

(77.) Daniel left the White House and drove toward the recycling plant at an average speed of 40 km/h.

Ezekiel left some time later driving in the same direction at an average speed of 48 km/h.

After driving for five hours Ezekiel caught up with Daniel.

How long did Daniel drive before Ezekiel caught up?

distance (d) = speed (s) * time (t)

Let the driving time for Daniel before Ezekiel caught up with him =*t*

The distance is the same: distance covered by Daniel and the distance covered by Ezekiel.

$ 40t = 240 \\[3ex] t = \dfrac{240}{40} \\[5ex] t = 6\;hours \\[3ex] $ Daniel drove for six hours before Ezekiel caught up with him.

Ezekiel left some time later driving in the same direction at an average speed of 48 km/h.

After driving for five hours Ezekiel caught up with Daniel.

How long did Daniel drive before Ezekiel caught up?

distance (d) = speed (s) * time (t)

Let the driving time for Daniel before Ezekiel caught up with him =

Person | speed (km/hr) | time (hr) | distance (km) |
---|---|---|---|

Daniel | 40 | t |
40t |

Ezekiel | 48 | 5 | 240 |

The distance is the same: distance covered by Daniel and the distance covered by Ezekiel.

$ 40t = 240 \\[3ex] t = \dfrac{240}{40} \\[5ex] t = 6\;hours \\[3ex] $ Daniel drove for six hours before Ezekiel caught up with him.

(78.) A passenger plane made a trip to the City of Las Vegas, State of Nevada of the country of the United States of America and back.

On the trip there it flew 432 miles per hour (mph).

On the return trip it went 480 miles per hour (mph).

How long did the trip there take if the return trip took nine hours?

distance (d) = speed (s) * time (t)

distance for the forward trip is the same distance for the return trip.

$ \underline{Return\;\;Trip} \\[3ex] d = s * t \\[3ex] d = 480 * 9 \\[3ex] d = 4320\;miles \\[3ex] \underline{Forward\;\;Trip} \\[3ex] d = s * t \\[3ex] s * t = d \\[3ex] t = \dfrac{d}{s} \\[5ex] t = \dfrac{4320}{432} \\[5ex] t = 10\;hours \\[3ex] $ The forward trip took ten hours.

On the trip there it flew 432 miles per hour (mph).

On the return trip it went 480 miles per hour (mph).

How long did the trip there take if the return trip took nine hours?

distance (d) = speed (s) * time (t)

distance for the forward trip is the same distance for the return trip.

$ \underline{Return\;\;Trip} \\[3ex] d = s * t \\[3ex] d = 480 * 9 \\[3ex] d = 4320\;miles \\[3ex] \underline{Forward\;\;Trip} \\[3ex] d = s * t \\[3ex] s * t = d \\[3ex] t = \dfrac{d}{s} \\[5ex] t = \dfrac{4320}{432} \\[5ex] t = 10\;hours \\[3ex] $ The forward trip took ten hours.

(79.) Judith left the airport and traveled toward the mountains.

Esther left 2.1 hours later traveling 35 miles per hour (mph) faster in an effort to catch up to her.

After 1.2 hours, Esther finally caught up.

Determine Judith's average speed.

Let Judith's average speed =*x*

This implies that Esther's speed = 35 +*x*

The entire time for Judith to travel toward the mountains = 2.1 + 1.2 = 3.3 hours

Esther's time to catch up to Judith = 1.2 hours

distance (d) = speed (s) * time (t)

The distance is the same: distance for Judith's trip is the same distance for Esther's trip.

$ 3.3x = 1.2(35 + x) \\[3ex] 3.3x = 42 + 1.2x \\[3ex] 3.3x - 1.2x = 42 \\[3ex] 2.1x = 42 \\[3ex] x = \dfrac{42}{2.1} \\[5ex] x = 20\;mph \\[3ex] $ Judith travelled at an average speed of 20 miles per hour (mph).

Esther left 2.1 hours later traveling 35 miles per hour (mph) faster in an effort to catch up to her.

After 1.2 hours, Esther finally caught up.

Determine Judith's average speed.

Let Judith's average speed =

This implies that Esther's speed = 35 +

The entire time for Judith to travel toward the mountains = 2.1 + 1.2 = 3.3 hours

Esther's time to catch up to Judith = 1.2 hours

distance (d) = speed (s) * time (t)

Person | speed (mph) | time (hr) | distance (miles) |
---|---|---|---|

Judith | x |
3.3 | 3.3x |

Esther | 35 + x |
1.2 | 1.2(35 + x) |

The distance is the same: distance for Judith's trip is the same distance for Esther's trip.

$ 3.3x = 1.2(35 + x) \\[3ex] 3.3x = 42 + 1.2x \\[3ex] 3.3x - 1.2x = 42 \\[3ex] 2.1x = 42 \\[3ex] x = \dfrac{42}{2.1} \\[5ex] x = 20\;mph \\[3ex] $ Judith travelled at an average speed of 20 miles per hour (mph).

(80.) **ACT** The book *Fahrenheit 451* by Ray Bradbury is about a society in which all books are banned and burned.

The title of the book gives the approximate temperature at which paper starts to burn.

Since Fahrenheit,*F*, and Celsius, *C*, temperatures are related by the formula $C = \dfrac{5}{9}(F - 32)$,
which of the following would make an equivalent title for the book?

$ A.\;\; Celsius\;219 \\[3ex] B.\;\; Celsius\;233 \\[3ex] C.\;\; Celsius\;268 \\[3ex] D.\;\; Celsius\;754 \\[3ex] E.\;\; Celsius\;844 \\[3ex] $

$ F = 451^\circ F \\[3ex] C = \dfrac{5}{9}(F - 32) \\[5ex] C = \dfrac{5}{9}(451 - 32) \\[5ex] C = \dfrac{5}{9} * 419 \\[5ex] C = 232.7777778 \\[3ex] C \approx 233^\circ C \\[3ex] Celsius\;233 $

The title of the book gives the approximate temperature at which paper starts to burn.

Since Fahrenheit,

$ A.\;\; Celsius\;219 \\[3ex] B.\;\; Celsius\;233 \\[3ex] C.\;\; Celsius\;268 \\[3ex] D.\;\; Celsius\;754 \\[3ex] E.\;\; Celsius\;844 \\[3ex] $

$ F = 451^\circ F \\[3ex] C = \dfrac{5}{9}(F - 32) \\[5ex] C = \dfrac{5}{9}(451 - 32) \\[5ex] C = \dfrac{5}{9} * 419 \\[5ex] C = 232.7777778 \\[3ex] C \approx 233^\circ C \\[3ex] Celsius\;233 $

(81.)

(82.)

(83.) **ACT** Bo and Tia will balance perfectly on the seesaw shown below if $w_1 * d_1 = w_2 * d_2$, where
$w_1$ and $d_1$ are the weight of Bo and his distance from the fulcrum, respectively, and $w_2$ and $d_2$ are
similarly defined for Tia.

Bo weighs 60 pounds and is sitting $3\dfrac{1}{2}$ feet from the fulcrum.

Tia weighs $\dfrac{2}{3}$ Bo's weight.

What distance, in feet, from the fulcrum must Tia sit in order for Bo and Tia to balance perfectly on the seesaw?

$ F.\;\; 1\dfrac{1}{6} \\[5ex] G.\;\; 2\dfrac{1}{3} \\[5ex] H.\;\; 5\dfrac{1}{4} \\[5ex] J.\;\; 6\dfrac{5}{12} \\[5ex] K.\;\; 10\dfrac{1}{2} \\[5ex] $

$ Tia's\;\;weight = \dfrac{2}{3} * 60 = 40 \\[5ex] \implies \\[3ex] 60 * 3\dfrac{1}{2} = 40 * d_2 \\[5ex] 60 * \dfrac{7}{2} = 40 * d_2 \\[5ex] 210 = 40 * d_2 \\[3ex] d_2 = \dfrac{210}{40} \\[5ex] d_2 = \dfrac{21}{4} = 5\dfrac{1}{4} \\[5ex] $ Tia had to sit $5\dfrac{1}{4}$ feet from the fulcrum in order for both of them to balance perfectly on the seesaw.

Bo weighs 60 pounds and is sitting $3\dfrac{1}{2}$ feet from the fulcrum.

Tia weighs $\dfrac{2}{3}$ Bo's weight.

What distance, in feet, from the fulcrum must Tia sit in order for Bo and Tia to balance perfectly on the seesaw?

$ F.\;\; 1\dfrac{1}{6} \\[5ex] G.\;\; 2\dfrac{1}{3} \\[5ex] H.\;\; 5\dfrac{1}{4} \\[5ex] J.\;\; 6\dfrac{5}{12} \\[5ex] K.\;\; 10\dfrac{1}{2} \\[5ex] $

$ Tia's\;\;weight = \dfrac{2}{3} * 60 = 40 \\[5ex] \implies \\[3ex] 60 * 3\dfrac{1}{2} = 40 * d_2 \\[5ex] 60 * \dfrac{7}{2} = 40 * d_2 \\[5ex] 210 = 40 * d_2 \\[3ex] d_2 = \dfrac{210}{40} \\[5ex] d_2 = \dfrac{21}{4} = 5\dfrac{1}{4} \\[5ex] $ Tia had to sit $5\dfrac{1}{4}$ feet from the fulcrum in order for both of them to balance perfectly on the seesaw.

(84.)

(85.)

(86.)

(87.)

(88.)

(89.)

(90.)