Quadratic Equation Quiz Set -3 IBPS Clerk

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Quadratic Equation Question for IBPS Clerk 2017. Welcome to the letsstudytogether.co online Reasoning section. If you are preparing for IBPS PO/Clerk 2017 exam, you will come across a section on Reasoning Ability Section.Quadratic Equation Quiz Set -3 IBPS Clerk.

Here we are providing you with Reasoning quiz “Quadratic Equation for IBPS Clerk 2017″based on the latest pattern for your daily practice.This “Quadratic Equation for IBPS Clerk 2017” is also important for other banking exams such as IBPS PO, IBPS Clerk, SBI Clerk, IBPS RRB Officer, IBPS RRB Office Assistant, IBPS SO, SBI SO and other competitive exams.

Quadratic Equation Question for IBPS Clerk 2017


1.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 + 17x + 52 = 0
II. y2 + 16y + 39 = 0

A. x ≥ y

B. x > y

C. x = y or no specific relation can be established

D. x ≤ y

E. x < y

Show Correct Answers

C. x = y or no specific relation can be established

I. x2 + 17x + 52 = 0
⇒ x2 + 13x + 4x + 52 = 0
Splitting the middle term
⇒ x (x + 13) + 4 (x + 13) = 0
⇒ (x + 4) (x + 13) = 0
So, (x + 4) = 0
⇒ x = – 4
Or, (x + 13) = 0
⇒ x = – 13
II. y2 + 16y + 39 = 0
⇒ y2 + 13y + 3y + 39 = 0
Splitting the middle term
⇒ y (y + 13) + 3 (y + 13) = 0
⇒ (y + 3) (y + 13) = 0
So, (y + 3) = 0
⇒ y = – 3
Or, (y + 13) = 0
⇒ y = – 13
So, when x = – 4, x < y for y = – 3 and x > y for y = – 13
And when x = – 13, x < y for y = – 3 and x = y for y = – 13
∴ so, we can observe that no clear relationship cannot be determined between x and y.

2.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. 5x2 + 1 = 6x
II. 16y2 + 1 = 8y

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. x < y

Show Correct Answers

C. x = y or no specific relation can be established

I. 5x2 + 1 = 6x
⇒ 5x2 – 6x + 1 = 0
⇒ 5x2 – 5x – x + 1 = 0
Now splitting middle term,
⇒ 5x (x – 1) – 1 (x – 1) = 0
⇒ (5x – 1) (x – 1) = 0
So, (5x – 1) = 0
⇒ x = 1/5
Or, (x – 1) = 0
⇒ x = 1
II. 16y2 + 1 = 8y
⇒ 16y2 – 8y + 1 = 0
Now splitting middle term
⇒ 16y2 – 4y – 4y + 1 = 0
⇒ 4y (4y – 1) – 1 (4y – 1) = 0
⇒ (4y – 1) (4y – 1) = 0
Then, y = ¼
So, when x = 1/5, x < y for y = ¼
And when x = 1, x > y for y = ¼
∴ So, we can observe that no clear relationship cannot be determined between x and y

3.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x3 – 120 = 392
II. y3 – 217 = 512

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. y > x

Show Correct Answers

E. y > x

I. x3 – 120 = 392
⇒ x3 = 392 + 120
⇒ x3 = 512
⇒ x = 8
II. y3 – 217 = 512
⇒ y3 = 512 + 217
⇒ y3 = 729
⇒ y = 9
∴ We can observe that y > x.

4.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. x < y

Show Correct Answers

C. x = y or no specific relation can be established

5.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 – 26x = – 5x – 68
II. y2 + 18y = 2y – 48

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. x < y

Show Correct Answers

D. x > y

I. x2 – 26x = – 5x – 68
⇒ x2 – 26x + 5x + 68 = 0
⇒ x2 – 21x + 68 = 0
Splitting middle term
⇒ x2 – 17x – 4x + 68 = 0
⇒ x (x – 17) – 4 (x – 17) = 0
⇒ (x – 4) (x – 17) = 0
So, x = + 4 or x = + 17
II. y2  + 18y = 2y – 48
⇒ y2 + 18y – 2y + 48 = 0
⇒ y2 + 16y + 48 = 0
⇒ y2 + 12y + 4y + 48 = 0
⇒ y (y + 12) + 4 (y + 12) = 0
⇒ (y + 4) (y + 12) = 0
So, y = – 4 or y = – 12
∴ As, we can observe that both of the values of x are positive and both of the values of y are negative then we can clearly say that x > y

6.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 + 96 = 321
II. y + 1296 = 1312

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. x < y

Show Correct Answers

E. x < y

I. x2 + 96 = 321
⇒ x2 = 321 – 96 = 225
⇒ x = √225
⇒ x = ± 15
II. y +1296 = 1312
⇒ y = 1312 – 1296
⇒ y = 16
So, when x = + 15, x < y for y = 16
And when x = – 15, x < y for y = 16
Hence, we can conclude x < y.

7.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 – 16x + 28 = 0
II. y2  – 14y + 13 = 0

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. x < y

Show Correct Answers

C. x = y or no specific relation can be established

I. x2 – 16x + 28 = 0
⇒ x2 – 14x – 2x + 28 = 0
⇒ x(x – 14) – 2(x – 14) = 0
⇒ (x – 14)(x – 2) = 0
Then, x = + 14 or x = + 2
II.y2 – 14y + 13 = 0
⇒ y2 – 13y – y + 13 = 0
⇒ y(y – 13) – 1(y – 13) = 0
⇒ (y – 13)(y – 1)
Then, y = + 13 or y = + 1
So, when x = + 14, x > y for y = + 13 and x > y for y = + 1
And when x = + 2, x < y for y = + 13 and x > y for y = + 1
∴ So, we can observe that no clear relationship cannot be determined between x and y.

8.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 + 6x + 8 = 0
II. y2 + 13y +42 = 0

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. x < y

Show Correct Answers

D. x > y

I. x2 + 6x + 8 = 0
⇒ x2 + 4x + 2x + 8 = 0
⇒ x(x + 4) + 2(x + 4) = 0
⇒ (x + 4)(x + 2) = 0
Then, x = – 4 or x = – 2
II.y2 + 13y + 42 = 0
⇒ y2 + 6y + 7y + 42 = 0
⇒ y(y + 6) + 7(y + 6) = 0
⇒ (y + 7)(y + 6) = 0
Then, y = – 7 or y = – 6
So, when x = – 4, x > y for y = – 7 and x > y for y = – 6
And when x = – 2, x > y for y = – 7 and x > y for y = – 6
∴ We can clearly observe that x > y.

9.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 – 17x + 72 = 0
II. y2 – 23y + 130 = 0

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. x < y

Show Correct Answers

E. x < y

I. x2 – 17x + 72 = 0
⇒ x2 – 9x – 8x + 72 = 0
⇒ x(x – 9) – 8(x – 9) = 0
⇒ (x – 8)(x – 9) = 0
Then, x = 8 or x = 9
II. y2 – 23y + 130 = 0
⇒ y2 – 13y – 10y + 130 = 0
⇒ y(y – 13) – 10(y – 13) = 0
⇒ (y – 10)(y – 13) = 0
Then, y = 10 or y = 13
As, we can observe that both the values of x are less than both values of y then we can conclude x < y.

10.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. 3x2 – 51x + 180 = 0
II. 2y2 – 24y + 64 = 0

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. x < y

Show Correct Answers

C. x = y or no specific relation can be established

I. 3x2 – 51x + 180 = 0
⇒ x2 – 17x + 60 = 0 [Dividing both sides by 3] ⇒ x2 – 12x – 5x + 60 = 0
⇒ x(x – 12) – 5(x – 12) = 0
⇒ (x – 12)(x – 5) = 0
Then, x = 12 or x = 5
II. 2y2 – 24y + 64 = 0
⇒ y2 – 12y + 32 = 0 [Dividing both sides by 2] ⇒ y2 – 8y – 4y + 32 = 0
⇒ y(y – 8) – 4(y – 8) = 0
⇒ (y – 8)(y – 4) = 0
Then, y = 8 or y = 4
So, when x = 12, x > y for y = 8 and x > y for y = 4
And when x = 5, x < y for y = 8 and x > y for y = 4
∴ So, we can observe that no clear relationship cannot be determined between x and y.


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