# Quadratic Equation Questions For IBPS RRB PO

Download Top-20 IBPS RRB PO Quadratic Equation Questions PDF. Quadratic Equation questions based on asked questions in previous year exam papers very important for the IBPS RRB PO (Officer Scale-I, II & III) exam.

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**Instructions**

In the following questions 2 quantities X and Y are given. Select the option which best captures the relations between X and Y.

**Question 1:Â **3 dices are thrown.

X is the probability that sum of the number on the faces of the dice is 7.

Y is the probability that sum of the number on the faces of the dice is 14.

a)Â X>Y

b)Â X<Y

c)Â X$\geq$Y

d)Â X$\leq$Y

e)Â X=Y or Cannot be determined

**Question 2:Â **Solution A has p and q in the ratio 1:2. Solution B has p and q in the ratio 5:4. Solution A and Solution B are mixed in the ratio 3:4 to form the solution C. Solution A and Solution B are mixed in the ratio 5:3 to form the solution D.

X = the ratio of p and q in Solution C

Y = the ratio of p and q in Solution D.

a)Â X>Y

b)Â X

c)Â X$\geq$Y

d)Â X$\leq$Y

e)Â X=Y or Cannot be determined

**Question 3:Â **Cost price of 6 articles is equal to the Selling price of 4 articles. The shopkeeper Marked up the price by 100%. X is the % discount offered.

Y is the % profit obtained.

a)Â X>Y

b)Â X<Y

c)Â X$\geq$Y

d)Â X$\leq$Y

e)Â X=Y or Cannot be determined

**Question 4:Â **2 dices are thrown.

X is the probability that the sum of the number on the faces of the dice is 7.

Y is the probability that the sum of the number on the faces of the dice is 5.

a)Â X>Y

b)Â X

c)Â X$\geq$Y

d)Â X$\leq$Y

e)Â X=Y or Cannot be determined

**Question 5:Â **a and b are the roots of the quadratic equation $x^2-12x+25 = 0$

c and d are the roots of the equation $x^2-11x+15x = 0$

X = $a^3+b^3+4$

Y = $c^3+d^3-4$

a)Â X>Y

b)Â X

c)Â X$\geq$Y

d)Â X$\leq$Y

e)Â X=Y or Cannot be determined

**Instructions**

Calculate the quantity I and the quantity II on the basis of the given information then compare them and answer the following questions accordingly.

**Question 6:Â **Quantity 1: Selling price of an article on which a seller made a profit of 20% and he bought it for Rs. 800.

Quantity 2: Cost price of an article on which a seller made a loss of 40% by selling it for Rs. 576.

a)Â Quantity 1 > Quantity 2

b)Â Quantity 1 $\geq$ Quantity 2

c)Â Quantity 1 < Quantity 2

d)Â Quantity 1 $\leq$ Quantity 2

e)Â Quantity 1 = Quantity 2

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**Question 7:Â **Quantity 1: Simple interest earned by Ramesh by lending Rs. 6000 for 3 years at 20% per annum.

Quantity 2: Compound interest earned by Suresh by lending Rs. 5000 for 2 years at 25% per annum.

a)Â Quantity 1 > Quantity 2

b)Â Quantity 1 $\geq$ Quantity 2

c)Â Quantity 1 < Quantity 2

d)Â Quantity 1 $\leq$ Quantity 2

e)Â Quantity 1 = Quantity 2

**Question 8:Â **Quantity 1: Volume of a cylinder whose height is 10 cm and perimeter of the base is 88 cm.

Quantity 2: Volume of a sphere of 10.5 cm.

a)Â Quantity 1 > Quantity 2

b)Â Quantity 1 $\geq$ Quantity 2

c)Â Quantity 1 < Quantity 2

d)Â Quantity 1 $\leq$ Quantity 2

e)Â Quantity 1 = Quantity 2

**Question 9:Â **Quantity 1: The surface area of a cuboid whose side lengths are 20 cm, 30 cm and 50 cm.

Quantity 2: The curved surface area of a hemisphere of 35 cm radius.

a)Â Quantity 1 > Quantity 2

b)Â Quantity 1 $\geq$ Quantity 2

c)Â Quantity 1 < Quantity 2

d)Â Quantity 1 $\leq$ Quantity 2

e)Â Quantity 1 = Quantity 2

**Question 10:Â **Quantity 1: Cost price of an article on which a seller made a profit of 15% by selling it for Rs. 345.

Quantity 2: Selling price of an article on which a seller made a loss of 20% and he bought it for Rs. 400.

a)Â Quantity 1 > Quantity 2

b)Â Quantity 1 $\geq$ Quantity 2

c)Â Quantity 1 < Quantity 2

d)Â Quantity 1 $\leq$ Quantity 2

e)Â Quantity 1 = Quantity 2

**Instructions**

**Calculate the quantity I and the quantity II on the basis of the given information then compare them and answer the following questions accordingly.**

**Question 11:Â **Quantity 1: Simple interest charged by bank on a sum of Rs. 5000 at the rate of 23% annum for 3 years.

Quantity 2: Compound interest charged by another bank on a sum of Rs. 5000 at the rate of 30% annum for 2 years compounded annually.

a)Â Quantity 1 > Quantity 2

b)Â Quantity 1 $\geq$ Quantity 2

c)Â Quantity 1 < Quantity 2

d)Â Quantity 1 $\leq$ Quantity 2

e)Â Quantity 1 = Quantity 2

**Question 12:Â **Quantity 1: Volume of a cube whose length of a side is 50 cm.

Quantity 2: Volume of a right-circular cone whose height is 50 cm and radius of the circular base is 50 cm.

a)Â Quantity 1 > Quantity 2

b)Â Quantity 1 $\geq$ Quantity 2

c)Â Quantity 1 < Quantity 2

d)Â Quantity 1 $\leq$ Quantity 2

e)Â Quantity 1 = Quantity 2

**Question 13:Â **Quantity 1: Area of an equilateral triangle with side equals to 56 cm.

Quantity 2: Area of a square with side equals to 35 cm.

a)Â Quantity 1 > Quantity 2

b)Â Quantity 1 $\geq$ Quantity 2

c)Â Quantity 1 < Quantity 2

d)Â Quantity 1 $\leq$ Quantity 2

e)Â Quantity 1 = Quantity 2

**Instructions**

**In the following questions 2 quantities X and Y are given. Select the option which best captures the relations between X and Y.**

**Question 14:Â **Ramesh has 400L of milk with him. He sells 20% of the mixture and replaces it water. He then sells 25% of the mixture and replaces it water. Finally, he sells 40% of the mixture and replaces it water.

Suresh has 500L of milk with him. He sells 30% of the mixture and replaces it water. He then sells 40% of the mixture and replaces it water. Finally, he sells 30% of the mixture and replaces it water.

X = The amount of milk Ramesh is left with

Y = The amount of milk Suresh is left with

a)Â X>Y

b)Â X<Y

c)Â X$\geq$Y

d)Â X$\leq$Y

e)Â X=Y or Cannot be determined

**Question 15:Â **Solution A has p and q in the ratio 3:5 whereas solution B has p and q in the ratio 4:3.

Solution A and Solution B are mixed in the ratio m:n to form solution C. Solution C has p and q in ratio 53:59 . Solution A is mixed with solution B in ratio k:l to form solution D. Solution D has p and q and ratio 191:201

X is the ratio of m and n

Y is the ratio of k and l

a)Â X>Y

b)Â X<Y

c)Â X$\geq$Y

d)Â X$\leq$Y

e)Â X=Y or Cannot be determined

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**Question 16:Â **A shopkeeper marks up the price of an article by 40%. X is the Cost price of 3 articles.

He gives 1 item free for every 2 items bought. Y is the revenue earned by selling 3 articles.

a)Â X>Y

b)Â X<Y

c)Â X$\geq$Y

d)Â X$\leq$Y

e)Â X=Y or Cannot be determined

**Question 17:Â **Solution A has p and q in the ratio 3:2 whereas solution B has p and q in the ratio 1:4.

3 parts of Solution A is mixed with 5 parts of solution B to form solution C and 2 parts of Solution A is mixed with 3 parts of solution B to form solution D.

X is the ratio of p and q in solution C.

Y is the ratio of p and q in solution D.

a)Â X>Y

b)Â X<Y

c)Â X$\geq$Y

d)Â X$\leq$Y

e)Â X=Y or Cannot be determined

**Question 18:Â **5 men can complete a job in 6 days. 12 women and 6 children take $\frac{20}{7}$ days to complete the same job instead if 6 women and 12 children work it will take them $\frac{20}{91}$ more days to complete the work.

X is the number of it will take 8 women and 12 children to complete the work.

Y is the number of days it will take 2 men and 15 children to complete the work.

a)Â X>Y

b)Â X<Y

c)Â X$\geq$Y

d)Â X$\leq$Y

e)Â X=Y or Cannot be determined

**Instructions**

In the following questions, two equations numbered I and II are given. You have to establish a relation between the two variables and select the option accordingly:

**Question 19:Â **I: $x^2 – 11x + 30 = 0$

II: $y^2 + 6y – 16 = 0$

a)Â $x > y$

b)Â $x \geq y$

c)Â $y > x$

d)Â $y \geq x$

e)Â $x = y$ or No relationship can be established

**Question 20:Â **I: $2x^2 – 33x + 133 = 0$

II: $y^2 – 4y – 21 = 0$

a)Â $x > y$

b)Â $x \geq y$

c)Â $y > x$

d)Â $y \geq x$

e)Â $x = y$ or No relationship can be established

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**Answers & Solutions:**

**1)Â AnswerÂ (E)**

7 = 1 + 1 + 5 = 2 + 2 + 3 = 3 + 3 + 1 = 1 + 2 + 4

Thus, the probability that sum of the number on the faces of the dice is 7 = (3C1+3C1+3C1+3!)/216 = 15/216 = X

10 = 6 + 3 + 1 = 6 + 2 + 2 = 5 + 3 + 2 = 4 + 4 + 2 = 4 + 3 + 3 = 1 + 4 + 5

Thus, the probability that sum of the number on the faces of the dice is 14 = (3C1+3C1+3C1+3!)/216 = 15/216 = Y

Hence, X=Y

Hence, option E is the correct answer.

**2)Â AnswerÂ (A)**

Let the quantity of A and B be 9L.

Thus, the amount of p and q in A is 3L and 6L respectively.

Also, the amount of p and q in B is 5L and 4L respectively.

A and B are mixed in the ratio 3:4 to form solution C.

Thus, the amount of p and q in solution C = (3*3+5*4):(6*3+4*4) = 29:34=X

Solution A and Solution B are mixed in the ratio 5:3 to form the solution D.

Thus, the amount of p and q in solution D = (5*3+5*3):(6*5+4*3) = 30:42=Y

Hence, X>Y

Hence, option A is the correct answer.

**3)Â AnswerÂ (B)**

Let the CP be 10 Rs.

Thus, 6*10 = 4*SP. Thus, SP = 15 Rs

MP = 2*10 = 20 Rs

% discount = $\frac{20-15}{20}$ = 25% = X

% profit = $\frac{15-10}{10}$ = 50% = Y

Hence, option B is the correct answer.

**4)Â AnswerÂ (A)**

Total possible combinations = 36.

Sum of the number on the faces of the dice is 7 will be for (1,6),(6,1),(2,5),(5,2),(3,4) and (4,3)

Thus, X = 6/36

Sum of the number on the faces of the dice is 5 will be for (1,4),(4,1),(2,3) and (3,2).

Thus, Y = 4/36

Hence, X>Y

Hence, option A is the correct answer.

**5)Â AnswerÂ (E)**

a and b are the roots of the quadratic equation $x^2-12x+25 = 0$

Thus, a+b = 12 and ab = 25

Thus, $a^3+b^3+4$ = $(a+b)^3-3ab(a+b)+4 = 13^3-3*25*13 = 832$

c and d are the roots of the equation $x^2-11x+15x = 0$

Thus, c+d = 11 and cd=15

Thus, $c^3+d^3-4$ = $(c+d)^3-3cd(c+d)-4$ = $11^3-3*11*15-4 = 832$

Thus, X=Y.

Hence, option E is the correct answer.

**6)Â AnswerÂ (E)**

Quantity 1: Selling price of the article = $\dfrac{100+20}{100}\times 800$ = Rs. 960

Quantity 2: Cost price of the article = $\dfrac{100}{100-40}\times 576$ = Rs. 960

We can see that Quantity 1 = Quantity 2. Hence, option E is the correct answer.

**7)Â AnswerÂ (A)**

Quantity 1: Simple interest earned by Ramesh = 6000*0.20*3 = Rs. 3600.

Quantity 2: Compound interest earned by Suresh = $5000(1 +\dfrac{25}{100})^2 – 5000$ = Rs. 2812.50.

Hence, we can say that Quantity 1 > Quantity 2. Option A is the correct answer.

**8)Â AnswerÂ (A)**

Quantity 1: Radius of the cylinder = $\dfrac{88}{2\pi}$ = 14 cm

Therefore, the volume of the cylinder = $\pi*14^2*10$ = 6160 $\text{cm}^3$

Quantity 2: Volume of the sphere = $\dfrac{4\pi}{3}10.5^3$ = 4849 $\text{cm}^3$

Hence, we can say that Quantity 1 > Quantity 2. Option A is the correct answer.

**9)Â AnswerÂ (C)**

Quantity 1: The surface area of a cuboid whose side lengths are 20 cm, 30 cm and 50 cm = 2(20*30 + 30*50 + 50*20) = 6200 $\text{cm}^2$.

Quantity 2: The curved surface area of a hemisphere of 35 cm radius = $2\pi*(35)^2$ = 7700 $\text{cm}^2$.

Hence we can say that Quantity 1 < Quantity 2. Option C is the correct answer.

**10)Â AnswerÂ (C)**

Quantity 1: Cost price of the article = $\dfrac{100}{100+15}\times 345$ = Rs. 300

Quantity 2: Selling price of the article = $\dfrac{100-20}{100}\times 400$ = Rs. 320

We can see that Quantity 1 < Quantity 2. Hence, option C is the correct answer.

**11)Â AnswerÂ (E)**

Simple interest paid to bank = 5000*0.23*3 = Rs. 3450

Compound interest paid to bank = $5000(1 + \dfrac{30}{100})^2 – 5000$ = Rs. 3450

Hence, we can say that Quantity 1 = Quantity 2. Option E is the correct answer.

**12)Â AnswerÂ (C)**

Volume of a cube whose length of a side is 50 cm = $50^3$ = $125000$

Volume of right-circular cone whose height is 100 cm and radius of circular base is 50 cm = $\dfrac{\pi}{3}\times 50^2*50$ = $\dfrac{125000\pi}{3}$

Hence, we can say that Quantity 1 < Quantity 2. Option C is the correct answer.

**13)Â AnswerÂ (A)**

Quantity 1: Area of the equilateral triangle with side equals to 56 cm = $\dfrac{\sqrt{3}}{4}\times 56^2$ = 1357.93 sq.cm

Quantity 2: Area of square with side equals to 35 cm = $35^2$ = 1225 sq. cm

Hence, we can say that Quantity 1 > Quantity 2. Option A is the correct answer.

**14)Â AnswerÂ (B)**

The amount of milk with Ramesh at the end = 400*0.8*0.75*0.6 = 144L= X

The amount of milk Suresh is left with at the end = 500*0.7*0.6*0.7 = 147L = Y

Hence, Y>X

Thus, option B is the correct answer.

**15)Â AnswerÂ (A)**

Let the amount of solution A and solution B be 56L.

In solution A p is 21L and q is 35L.

In solution B p is 32L and q is 24L.

In solution C p:q = 53:59 = (21*m+32*n):(35*m+24*n)

Solving we get, m:n = 1:1 = X

In solution D p:q = 191:201 = (21*k+32*l):(35*k+24*l)

Solving we get k:l = 3:4 = Y

Thus, X > Y

Hence, option A is the correct answer.

**16)Â AnswerÂ (A)**

Let the CP of the article be x.

Thus CP of 3 articles = 3x.

MP = 1.4x

He gives 1 item free for every 2 items bought. Y is the revenue earned by selling 3 articles.

Y = 2.8x

Thus, X>Y.

hence, option A is the correct answer.

**17)Â AnswerÂ (B)**

Let the amount of A and B be 5L each.

Thus, the amount of p and q in A is 3L and 2L and the amount of p and q is 1L and 4L.

15L of A and 25L of B is taken to form C.

Thus, in solution C the ratio of p:q = (3*3+1*5):(2*3+4*5) = 14:26 = 7:13 = X

10L of A and 15L of B is taken to form D

Thus, in solution D the ratio of p:q = (3*2+1*3):(2*2+4*3) = 9:16= Y

Thus, Y>X.

Hence, option B is the correct answer.

**18)Â AnswerÂ (B)**

5 men can complete a job in 6 days. Thus, 1 men will take 30 days to complete the work. In 1 day 1 man will complete $\frac{1}{30}$ of the work.

Let W be the number of days 1 women will take to complete the work and C be the number of days 1 children will take to complete the work

As per the given Condition,

$\frac{12}{W}+\frac{6}{C} = \frac{7}{20}$

$\frac{20}{7}+\frac{20}{91} = \frac{13}{40}$ = number of days 6 women and 12 children will take.

and $\frac{6}{W}+\frac{12}{C} = \frac{13}{40}$

Solving these 2 equation we get W = 48 and C = 60

Thus, the number of it will take 8 women and 12 children to complete the work = $\dfrac{1}{\dfrac{8}{48}+\dfrac{12}{60}} = \dfrac{60}{22}$ = X

the number of days it will take 2 men and 15 children to complete the work =

$\dfrac{1}{\dfrac{2}{30}+\dfrac{15}{60}} = \dfrac{60}{19}$ = Y

Thus, Y>X

Hence, option B is the correct answer.

**19)Â AnswerÂ (A)**

On solving $x^2 – 11x + 30 = 0$,

$x = 5$ or $x = 6$

On solving $y^2 + 6y – 16 = 0$,

$y = 2$ or $y = -8$

We can see that $x > y$ for all values of x and y.

Hence, option A is the correct answer.

**20)Â AnswerÂ (B)**

On solving $2x^2 – 33x + 133 = 0$,

$x = 7$ or $x = 8.5$

On solving $y^2 – 4y – 21 = 0$,

$y = 7$ or $y = -3$

We can see that $x \geq y$ for all values of x and y.

Hence, option B is the correct answer.