# SBI PO Questions On Quadratic Equations PDF

Quadratic Questions for SBI PO Prelims and Mains exam. Download SBI PO Quadratic questions on with solutions.

Download SBI PO Questions On Quadratic Equations PDF

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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: **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) X>Y

b) X<Y

c) X$\geq$Y

d) X$\leq$Y

e) X=Y or Cannot be determined

**Question 2: **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) X>Y

b) X

c) X$\geq$Y

d) X$\leq$Y

e) X=Y or Cannot be determined

**Question 3: **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) X>Y

b) X<Y

c) X$\geq$Y

d) X$\leq$Y

e) X=Y or Cannot be determined

**Question 4: **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) X>Y

b) X

c) X$\geq$Y

d) X$\leq$Y

e) X=Y or Cannot be determined

**Question 5: **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) X>Y

b) X

c) X$\geq$Y

d) X$\leq$Y

e) X=Y or Cannot be determined

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**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: **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) Quantity 1 > Quantity 2

b) Quantity 1 $\geq$ Quantity 2

c) Quantity 1 < Quantity 2

d) Quantity 1 $\leq$ Quantity 2

e) Quantity 1 = Quantity 2

**Question 7: **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) Quantity 1 > Quantity 2

b) Quantity 1 $\geq$ Quantity 2

c) Quantity 1 < Quantity 2

d) Quantity 1 $\leq$ Quantity 2

e) Quantity 1 = Quantity 2

**Question 8: **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) Quantity 1 > Quantity 2

b) Quantity 1 $\geq$ Quantity 2

c) Quantity 1 < Quantity 2

d) Quantity 1 $\leq$ Quantity 2

e) Quantity 1 = Quantity 2

**Question 9: **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) Quantity 1 > Quantity 2

b) Quantity 1 $\geq$ Quantity 2

c) Quantity 1 < Quantity 2

d) Quantity 1 $\leq$ Quantity 2

e) Quantity 1 = Quantity 2

**Question 10: **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) Quantity 1 > Quantity 2

b) Quantity 1 $\geq$ Quantity 2

c) Quantity 1 < Quantity 2

d) Quantity 1 $\leq$ Quantity 2

e) Quantity 1 = Quantity 2

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**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: **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) Quantity 1 > Quantity 2

b) Quantity 1 $\geq$ Quantity 2

c) Quantity 1 < Quantity 2

d) Quantity 1 $\leq$ Quantity 2

e) Quantity 1 = Quantity 2

**Question 12: **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) Quantity 1 > Quantity 2

b) Quantity 1 $\geq$ Quantity 2

c) Quantity 1 < Quantity 2

d) Quantity 1 $\leq$ Quantity 2

e) Quantity 1 = Quantity 2

**Question 13: **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) Quantity 1 > Quantity 2

b) Quantity 1 $\geq$ Quantity 2

c) Quantity 1 < Quantity 2

d) Quantity 1 $\leq$ Quantity 2

e) 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: **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) X>Y

b) X<Y

c) X$\geq$Y

d) X$\leq$Y

e) X=Y or Cannot be determined

**Question 15: **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) X>Y

b) X<Y

c) X$\geq$Y

d) X$\leq$Y

e) X=Y or Cannot be determined

**Question 16: **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) X>Y

b) X<Y

c) X$\geq$Y

d) X$\leq$Y

e) X=Y or Cannot be determined

**Question 17: **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) X>Y

b) X<Y

c) X$\geq$Y

d) X$\leq$Y

e) X=Y or Cannot be determined

**Question 18: **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) X>Y

b) X<Y

c) X$\geq$Y

d) X$\leq$Y

e) 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: **I: $x^2 – 11x + 30 = 0$

II: $y^2 + 6y – 16 = 0$

a) $x > y$

b) $x \geq y$

c) $y > x$

d) $y \geq x$

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

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

II: $y^2 – 4y – 21 = 0$

a) $x > y$

b) $x \geq y$

c) $y > x$

d) $y \geq x$

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

**Answers & Solutions:**

**1) Answer (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) Answer (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) Answer (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) Answer (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) Answer (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.

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**6) Answer (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) Answer (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) Answer (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) Answer (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) Answer (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.

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**11) Answer (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) Answer (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) Answer (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) Answer (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) Answer (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) Answer (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) Answer (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) Answer (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) Answer (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) Answer (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.