# Solved Examples: Word Problems on Quadratic 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 SAT Students
Any question labeled SAT-C is a question that allows a calculator.
Any question labeled SAT-NC is a question that does not allow a calculator.

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

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE and Malta Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve and reasonably check all these word problems.

(1.) A ladder leans against the side of a building.
The top of the ladder is 12 feet from the ground.
The bottom of the ladder is 10 feet from the side of the building.
Determine the length of the ladder.
Round you answer to the nearest tenth as needed.

Let us represent the information using a diagram

This is seen as a right triangle where:
hypothenus = hyp = the length of the ladder = x
leg = 12 feet
leg = 10 feet

$hyp^2 = leg^2 + leg^2 ...Pythagorean\;\;Theorem \\[3ex] x^2 = 12^2 + 10^2 \\[3ex] x^2 = 144 + 100 \\[3ex] x = \sqrt{244} \\[3ex] x = 15.62049935 \\[3ex] x \approx 15.6\;feet \\[3ex]$ The length of the ladder is approximately 15.6 feet
(2.) NZQA Teri is five years old.
Mari is four years older than Teri.
How many years will it take until Teri's and Mari's ages (in years) when multiplied together, make 77?

Let the number of years in which it will take the product of Teri's age and Mari's age to be $77$ be $p$

$Teri's\;\;age = 5\;years \\[3ex] Mari's\;\;age = 4 + 5 = 9\;years \\[3ex] In\;\; p\;\;years \\[3ex] Teri's\;\;age = 5 + p \\[3ex] Mari's\;\;age = 9 + p \\[3ex] Product\;\;should\;\;be\;\; 77 \\[3ex] (5 + p)(9 + p) = 77 \\[3ex] 45 + 5p + 9p + p^2 = 77 \\[3ex] p^2 + 14p + 45 - 77 = 0 \\[3ex] p^2 + 14p - 32 = 0 \\[3ex] (p + 16)(p - 2) = 0 \\[3ex] p = -16 \;\;OR\;\; p = 2 \\[3ex] years\;\;cannot\;\;be\;\;negative \\[3ex] \implies p \ne -16 \\[3ex] \implies p = 2 \\[3ex] \therefore Teri's\;\;age = 5 + 2 = 7 \;years \\[3ex] Mari's\;\;age = 9 + 2 = 11 \;years \\[3ex] \underline{Check} \\[3ex] 7 * 11 = 77\;years \\[3ex]$ In two years, the product of Teri's age and Mari's age would be $77$
(3.) ACT When asked his age, the algebra teacher said, "If you square my age, then subtract 23 times my age, the result is 50."
How old is he?

$F.\;\; 23 \\[3ex] G.\;\; 25 \\[3ex] H.\;\; 27 \\[3ex] J.\;\; 46 \\[3ex] K.\;\; 50 \\[3ex]$

$Let\;\;the\;\;age = p \\[3ex] square\;\;the\;\;age = p^2 \\[3ex] 23\;\;times\;\;the\;\;age = 23 * p = 23p \\[3ex] result = 50 \\[3ex] \implies \\[3ex] p^2 - 23p = 50 \\[3ex] p^2 - 23p - 50 = 0 \\[3ex] (p + 2)(p - 25) = 0 \\[3ex] p + 2 = 0 \;\;\;OR\;\;\;p - 25 = 0 \\[3ex] p = -2 \;\;\;OR\;\;\; p = 25 \\[3ex] years\;\;cannot\;\;be\;\;negative \\[3ex] \implies p \ne -2 \\[3ex] \implies p = 25 \\[3ex]$ The age of the algebra teacher is 25 years.
(4.) The hypotenuse of a right triangle is 13cm long.
The shorter leg is 7 cm shorter than the longer leg.
Find the side lengths of the triangle.

$hypotenuse = 13\;cm \\[3ex] Let\;\;the: \\[3ex] longer\;\;leg = x \\[3ex] shorter\;\;leg = x - 7 \\[3ex] hyp^2 = longer\;\;leg^2 + shorter\;\;leg^2 ...Pythagorean\;\;Theorem \\[3ex] 13^2 = x^2 + (x - 7)^2 \\[3ex] x^2 + (x - 7)^2 = 13^2 \\[3ex] x^2 + (x - 7)(x - 7) = 169 \\[3ex] x^2 + x^2 - 7x - 7x + 49 - 169 = 0 \\[3ex] 2x^2 - 14x - 120 = 0 \\[3ex] 2(x^2 - 7x - 60) = 0 \\[3ex] x^2 - 7x - 60 = 0 \\[3ex] (x - 12)(x + 5) = 0 \\[3ex] x - 12 = 0 \;\;\;OR\;\;\; x + 5 = 0 \\[3ex] x = 12 \;\;\;OR\;\;\; x = -5 \\[3ex] Length\;\;cannot\;\;be\;\;negative \\[3ex] x = 12 \\[3ex] x - 7 = 12 - 7 = 5 \\[3ex]$ The length of the hypotenuse is 13cm
The length of the longer leg is 12cm
The length of the shorter leg is 5cm
5 — 12 — 13 form a Pythagorean triple because $5^2 + 12^2 = 13^2$
(5.)

(6.)

(7.)

(8.)

(9.)

(10.)

(11.)

(12.)