Samuel Dominic Chukwuemeka (SamDom For Peace)

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka



Word Problems on Linear Equations

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

Verify Answers with Calculators:
For Mathematics of Finance problems: Mathematics of Finance Word Problems Calculator

Translate these questions from English to Math.
Use an appropriate letter for your variable.
Use as very few variables as possible.
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 \\[3ex] 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 $3$
(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.
(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 $27$
(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
(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.
(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 $84.0\:\:units$
(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 $103$
(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 $36$
$15 + 30 + 36 = 81$
(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 $13 * 17 = \$221$
(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: $-23, -21, -19$
(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$
(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
(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
(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.
(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
(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.
(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 ☺☺☺
(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
(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.




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(21.) An inheritance of $2,000,000$ is to be divided among Peter, James, and John.
James is to receive ${9\over10}$ of what Peter receives.
John is to receive ${1\over10}$ 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 $\$100,000.00$
(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 $4$
(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.
(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 $
(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.
(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: $23, 24, 25$
(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 $40^oF$
(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
(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
(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
(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: $8, 10, 12$
(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
(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: $12\:meters,\:\:24\:meters,\:\:and\:\:29\:meters$
(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.
(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 $-4$
(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 $
(37.) CSEC The table below shows the number of tickets sold for a bus tour.
Some items in the table are missing.

Tickets Sold for Bus Tour
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 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 $
(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 $
(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
(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 $




Top




(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.
$Scenario$ $s(mph)$ $t(hours)$ $d(miles)$
$Damian$ $x$ $2.5$ $2.5x$
$Cosmas$ $x + 6$ $2.5$ $2.5(x + 6)$

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.
(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.
$Scenario$ $s(mph)$ $t(hours)$ $d(miles)$
$Car$ $70$ $t$ $70t$
$Truck$ $60$ $t + 3$ $60(t + 3)$

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. After $18$ hours, the car will pass 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.
(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] $
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.
WPLE-46
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 $
(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] $
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.
(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] $
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] $
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] $
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] $
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] $
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 $
(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] $
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
$ \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] $
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 drained so it can be replaced with the pure antifreeze solution.
(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 $
(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] $
$Scenario$ $s(mph)$ $t(hours)$ $d(miles)$
$First$ $5$ $x$ $5x$
$Second$ $6$ $1.2 - x$ $6(1.2 - x)$
$Total$ $1.2$ $7$

$ 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.




Top




(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] $
$Scenario$ $s(mph)$ $t(hours)$ $d(miles)$
$Peter\rightarrow East$ $x$ $4$ $4x$
$Paul\rightarrow West$ $3 + x$ $4$ $4(3 + x)$
$Total$ $108$

$ 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 $\$9960.00$
(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] $
$Scenario$ $s(mph)$ $t(hours)$ $d(miles)$
$Freeway$ $x$ $5$ $5x$
$Country\:\:Road$ $x - 20$ $3$ $3(x - 20)$
$Total$ $436$

$ 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.
(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] $
$Scenario$ $s(km/hr)$ $t(hr)$ $d(km)$
$First$ $x$ $4$ $4x$
$Second$ $2x$ $3$ $6x$
$Total$ $600$

$ 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 $
(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

$Scenario$ 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] $

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.
(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 $20$
(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 $
(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
(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] $
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 $
(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] $
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