SBI Clerk Quantitative Aptitude Quiz: 13th May 2019

Dear Aspirants,

Quantitative Aptitude Quiz For SBI PO/Clerk Prelims

Numerical Ability or Quantitative Aptitude Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions.

Directions (1-5): In each of these questions, two equations are given. You have to solve these equations and find out the values of x and y and give answer

Q1. I. 4x + 7y = 209
II. 12x - 14y = – 38

if x < y
if x > y
if x ≤ y
if x ≥ y
if x = y or no relation can be established
Solution:

I. 4x+7y=209
II. 12x-14y=-38
or 6x-7y=-19
Adding I and II we get
10x = 190
⇒ x = 19
∴ 7y = 114 + 19
⇒ y = 19
x = y

Q2. I. 17x² + 48x = 9
II. 13y² = 32y – 12

if x < y
if x > y
if x ≤ y
if x ≥ y
if x = y or no relation can be established
Solution:

Q3. I. 16x² + 20x + 6 = 0
II. 10y² + 38y + 24 = 0

if x < y
if x > y
if x ≤ y
if x ≥ y
if x = y or no relation can be established
Solution:

Q4. I. 8x² + 6x = 5
II. 12y² – 22y + 8 = 0

if x < y
if x > y
if x ≤ y
if x ≥ y
if x = y or no relation can be established
Solution:

Q5. I. 18x² + 18x + 4 = 0
II. 12y² + 29y + 14 = 0

if x < y
if x > y
if x ≤ y
if x ≥ y
if x = y or no relation can be established
Solution:

Directions (6-10): In each of the following questions two equations are given. You have to solve the equations and Give answer —

Q6. I. 9x² – 36x + 35 = 0
II. 2y² – 15y – 17 = 0

if x < y
if x ≤ y
relationship between x and y cannot be determined
if x ≥ y
if x > y
Solution:

I. 9x² - 36x + 35 = 0
⇒ 9x² - 21x – 15x + 35 = 0
⇒ 3x (3x – 7) -5 (3x – 7) = 0
⇒ (3x- 7) (3x – 5) = 0
⇒ x = 5/3, 7/3
II. 2y² – 15y – 17 = 0
⇒ 2y² - 17y + 2y – 17 = 0
⇒ (y + 1) (2y – 17) = 0
⇒ y= -1, 17/2
No relation

Q7. I. 2x² – 7x + 3 = 0
II. 2y² – 7y + 6 = 0

if x < y
if x ≤ y
relationship between x and y cannot be determined
if x ≥ y
if x > y
Solution:

I. 2x² - 7x + 3 = 0
⇒ 2x² - 6x – x +3 = 0
⇒ (x – 3) (2x – 1) = 0
⇒ x = 3, 1/2,
II. 2y² - 7y +6 = 0
⇒ 2y² - 4y – 3y + 6 =0
⇒ (y – 2) (2y – 3) = 0
⇒ y = 2, 3/2
No relation

Q8. I. 4x² + 16x + 15 = 0
II. 2y² + 3y + 1 = 0

if x < y
if x ≤ y
relationship between x and y cannot be determined
if x ≥ y
if x > y
Solution:

I. 4x² + 16x + 15 = 0
⇒ 4x² + 10x + 6x + 15 = 0
⇒ 2x (2x + 5) + 3 (2x + 5) = 0
⇒ (2x + 5) (2x + 3) = 0
⇒ x= -5/2, -3/2
II. 2y² + 3y + 1 = 0
⇒ 2y² + 2y + y + 1 = 0
⇒ (y + 1) (2y + 1) = 0
⇒ y = -1, -1/2
y > x

Q9. I. 9x² – 45x + 56 = 0
II. 4y² – 17y + 18 = 0

if x < y
if x ≤ y
relationship between x and y cannot be determined
if x ≥ y
if x > y
Solution:

I. 9x² - 45x + 56 = 0
⇒ 9x² - 24x – 21x + 56 = 0
⇒ 3x (3x – 8) – 7 (3x – 8) = 0
⇒ (3x – 8) (3x – 7) = 0
⇒ x = 8/3, 7/3
II. 4y² - 17y + 18 = 0
⇒ 4y² - 8y – 9y + 18 = 0
⇒ (y – 2) (4y – 9) = 0
⇒ y = 2, 9/4

Q10. I. 2x² + 11x + 14 = 0
II. 2y² + 15y + 28 = 0

if x < y
if x ≤ y
relationship between x and y cannot be determined
if x ≥ y
if x > y
Solution:

I. 2x² + 11x + 14 = 0
⇒ 2x² + 4x + 7x + 14= 0
⇒ (x + 2) (2x + 7) = 0
⇒ x = -2, -7/2
II. 2y² + 15y + 28= 0
⇒ 2y² + 8y + 7y + 28 = 0
⇒ (y + 4) (2y + 7) = 0
⇒ y = -4, -7/2
⇒ x≥ y

Directions (11 - 15): In the following Questions, two equations I and II are given. You have to solve both the equations and Give answer

Q11. I. x² – 7x + 6 = 0
II. 2y² – 8y + 6 = 0

If x > y
if x ≥ y
if x < y
if x ≤ y
if x = y or the relationship between x and y cannot be established.
Solution:

I. x² – 7x + 6 = 0
x² – 6x– x + 6 = 0
(x – 6) (x – 1) = 0
x= 1, 6
II. 2y² – 8y + 6 = 0
⇒ y² – 4y + 3 = 0
⇒ y² – 3y – y + 3 = 0
⇒ (y – 1) (y – 3) = 0
⇒ y = 1, 3
No relation

Q12. I. 3x² + 13x – 16 = 0
II. y² – 5y + 4 = 0

If x > y
if x ≥ y
if x < y
if x ≤ y
if x = y or the relationship between x and y cannot be established.
Solution:

I. 3x² + 13x – 16 = 0
⇒ 3x² + 16x – 3x – 16 = 0
⇒ (3x + 16) (x – 1) = 0
⇒ x=1, -16/3
II. y² – 5y + 4 = 0
⇒ y² – 4y – y + 4 = 0
⇒ (y – 4) (y – 1) = 0
⇒ y = 4, 1
y ≥ x

Q13. I. y² + 17y + 72= 0
II. x² + 11x + 30 = 0

If x > y
if x ≥ y
if x < y
if x ≤ y
if x = y or the relationship between x and y cannot be established.
Solution:

I. y² + 17y+ 72 = 0
⇒ y² + 8y + 9y + 72 = 0
⇒ (y + 8) (y + 9) = 0
⇒ y = –8, –9
II. x² + 11x + 30 = 0
⇒ x² + 5x + 6x + 30 = 0
⇒ (x + 5) (x + 6) = 0
⇒ x = –5, –6
x > y

Q14. I. x + 3y = 8
II. 2x +y = 6

If x > y
if x ≥ y
if x < y
if x ≤ y
if x = y or the relationship between x and y cannot be established.
Solution:

I. x + 3y = 8
II. 2x + y = 6
Multiplying equation (ii) by 2 and then substracting (ii) from (i) we get
x = 2, y = 2

Q15. I. 2x² – 9x + 10 = 0
II. 3y² – 14y + 16 = 0

If x > y
if x ≥ y
if x < y
if x ≤ y
if x = y or the relationship between x and y cannot be established.
Solution:

I. 2x² – 9x + 10 = 0
⇒ 2x² – 4x – 5x + 10 = 0
⇒ 2x (x – 2) – 5 (x – 2) = 0
⇒ (x – 2) (2x – 5) = 0
x=2, 5/2
II. 3y² – 14y + 16 = 0
⇒ 3y² – 6y – 8y + 10 = 0
⇒ 3y (y – 2) – 8 (y – 2) = 0
⇒ (y – 2) (3y – 8) = 0
⇒ y = 2, 8/3
No relation

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