Word Problems that can be solved using Quadratic Equation Techniques

Monday, August 17th, 2020

Word Problems that can be solved using Quadratic Equation Techniques

In the quadratic equations word problems, the equations wouldn’t be given directly, in fact, you have to deduct the equation from the given facts within the equations. There could be many different traits of question which can even include all the linear equations type questions into the quadratic form. It can also include profit maximization or loss minimization questions in which you have to find either minimum or maximum value of the equation.
Here are a few examples of the word problems in quadratic equations:

Example 1: Four friends have some coins. Pavan has 2 less than Samir, who has 2 less than Tarun. If Pranav has 2 less than Pavan, and the product of the number of coins that each of them has is 5760, how many coins in all do they have among themselves?

Sol: Let the number of coins of Pranav be x;

Then, the number of coins with Samir will be x+2;
Pranav will be x-2;
Tarun will be x+4;

Given the product of the number of each of them =5760
∴, (x-2)(x)(x+2)(x+4) =5760;
Now either the above equation can be done by multiplying all terms and then solve them

We can use hit and trial. Since the product is ending with 0 then
there must be a pair of 5 and 2. However, 5 is not possible as it will make all terms odd, therefore one of the term will be 10 or multiple of 10.

By observation, x can’t be equal to 10 as it will make one term =14
which is not a factor of 5760. On more similar trials we will deduct, 6,8, 10, and 12 are satisfying this equation.

Therefore, total number of coins = 6+8+10+12 =36 Ans.

Example 2: The hypotenuse of right-angled triangle is 2 more than twice of one of the other side while the third side is 13 more than half of the hypotenuse. Find the length of the median to the hypotenuse.

Sol: Let one of the sides of the right-angled triangle be x, hypotenuse is 2x + 2 and the other side is x + 14

∴, x2 + (x+14)2 = (2x+2)2
⇒ x2+ x2 +28x + 196 = 4x2 + 8x + 4

⇒ x2 -20x – 192 =0

⇒ x2-10x -96 = 0

⇒ (x-16)(x+10) =0 ⇒ x=16.

The sides of the right-angled triangle are 16, 30, 34.
The median to the hypotenuse is half the hypotenuse, i.e., 17.

Example 3: The difference between the squares of the two numbers is 72. Eight times the numerically smaller number is 1 more than 5 times the other number. Find the numerically greater number.

Sol: Let the numerically greater and smaller numbers be x, y

∴ x2 –y2 =72 ……..(1)
8y= 5x +1…(2)

Though we use (2) to eliminate one unknown, it is more convenient to work with (1) (x+y)(x-y) =2(36) =4(18) = 6(12)

If we assume that x, y are integers, x + y and x – y have to be both even numbers.

Getting x + y equal to the bigger factor and x – y equal to the smaller one, we get x =19, y =17 or x =11, y=7 or x =9, y=3.

We see that (2) is satisfied only for x =11, y =7, the numerically greater one is 11.

Example 4: A total of Rs. 3,300 is raised by collecting equal amounts from a certain number of people. If there were 22 people more, each person would have to contribute Rs.200 less to raise the same amount, how many people actually contributed?

Sol: Let the number of people, and the contribution of each person be n and a respectively.

∴, (n+22)(a-200) = na
⇒ 22a -200n =4400 ….(1)
and na= 3300… (2)

Substituting the value of a from (2) in (1)
22(3300/n) -200n =4400
⇒ 11(33)/n -n =22
⇒ n2 +22n -11(33) = 0
⇒ (n+33)(n-11)=0
⇒ n=11.


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Example 5: A & B, together buy 45 notebooks. Each of them bought a different number of books and both of them spend exactly the same amount. If A has bought his notebooks at the price at which B had bought and B had bought his notebooks at the price where A had bought, they would have spent Rs. 160 and Rs. 250 respectively. How many notebooks did B buy?

Sol:                                  A             B
No. of notebooks          m             n
Price          a               b

We have m+n = 45…(1)
ma =nb ….(2)
mb= 160… (3)
na =250… (4)

From (2) ⇒ m/n = b/a
(3) + (4) ⇒ (m/n)2 = (16/25) = (4/5)2
∴ m :n = 4: 5

And n = 5/9(45) = 25

Example 6: A fraction plus 4 times its reciprocal is 533/77. What is the value of 4 times the fraction plus the reciprocal of the fraction?

Sol: Let the fraction be x then,

x + 4/x = 533/77

⇒ 77x2 -533x +308 = 0

⇒ (11x – 7)(7x-44) =0
⇒ x =7/11 or 44/7

∴, 4x +1/x =28/11 +11/7 or 176/7 + 7/44

= 317/77 or 7793/308

Example 7: The perimeter and the sum of the products of the sides of a triangle taken two at a time are denoted by P and Q respectively. Then what would be the maximum of Q in terms of P?

Sol: Clearly, sides of the triangle would always be positive. Suppose a, b, and c are sides. Then, a, b, & c is positive.

a2 + b2 ≥ 2ab,
c2 + b2 ≥ 2cb,
a2 + c2 ≥ 2ac,

∴, 2(a2 + b2 + c2)≥ 2(ab +bc +ca)

⇒ a2 + b2 + c2 ≥ ab +bc +ca

∴,  a2 + b2 + c2 + 2ab + 2bc +2ca ≥ 3(ab +bc +ca)

⇒ ab +bc +ca ≤ (a + b + c)3/ 3

Hence, the maximum value of (ab + bc + ca) = P2/3

Example 8: If the speed of a vehicle decreases by 10kmph, it takes 2 hours more than what it usually takes to cover a distance of 1800km. Find the time it usually takes.

Sol: The data is tabulated below:

Speed         Time            Distance
u                     t                 1800 ⇒ ut = 1800
u – 10           t+2             1800

∴ ut= (u-10)(t+2) =ut + 2u -10t -20

2u -10(1800/u) =20,

u2 -10u -9000 = 0,

(u -100)(u+90) =0, that is u =100

And t =1800/100 = 18

Example 9: Joe is selling tickets on a boat tour around the bay of islands. If he charges Rs. 30, he will sell 50 tickets. For Re.1 decrease in price, 5 more tickets would get sell. What is the price that will maximize his revenue.

Sol: Let n denotes the decrease in price.
Revenue = price x no. of tickets sold

R = (50+5n)(30-n)
= 1500 – 50n + 150n – 5n2
⇒R = -5n2 + 100n +1500

Now we have to maximize R.

Above is a quadratic equation in n with coefficient of n2 , its maximum value will be (-b/2a) i.e

{-100/2(-5)}= 10

At n=10; R = -5(10)2 +100(10) +1500

= 2000 Ans.

Example 10: A fruit grower has 400 crates of grapes ready for market and will have 20 more crates each day shipment is delayed, the price per crate decreases by $2. Write a quadratic function to model the grower’s revenue and use it to determine how many days the grower should delay shipment in order to maximize revenue.

Sol: Let n be the no of days by which the crates get delayed.

Revenue = selling price x no. of crates

=       (400+20n)(60-2n)

= 24000 -800n +1200n -40n2

= -40n2 + 400n +24000

Similar to above question, the maximum of the revenue would be at (-b/2a) i.e.

{-400/2(-40)} = 5

At n=5; Revenue = 500 x 50 = 25000

The difficulty level of quadratic word problem can vary arbitrarily in CAT. The best would be to get familiar all type of problems and keep practicing.

How to Construct a Quadratic Equation Based on Given Conditions in the CAT?
Quadratic Equations based on x + 1/x type and other algebraic identities

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