IBPS PO Quantitative Aptitude (Miscellaneous) Quiz For Prelims: 27th March 2019

Dear Aspirants,

Quantitative Aptitude Quiz For IBPS 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.

Q1. A trader bought two buffalo for Rs 19,500. He sold one at a loss of 20% and the other at a profit of 15%. If the selling price of each buffalo is the same, then their cost prices are respectively?

10,000 and 9,500

11,500 and 8,000

12,000 and 7,500

10,500 and 9,000

15,100 and 5,700

Solution:

Q2. A bath tub can be filled by the cold water pipe in 40 minutes and by the hot water pipe in 60 minutes. A person leaves the bathroom after turning on both pipes simultaneously and returns at the moment when the bath should be full. Finding, however, that the waste pipe has been open, he now closes it. In 12 minutes more the bath tub is full. In what time would the waste pipe empty it?

32 min

58 min

48 min

54 min

24 min

Solution:

Q3. 25 men can do a work in 20 days, when should 15 men leave the work, if the whole work is to be completed in

after they leave the work?

3 days

4 days

5 days

6 days

2 days

Solution:

Q4. Two equal sums of money are lent at the same times at 8% and 7% per annum simple interest. The former is recovered 6 months earlier than the latter and the amount in each case is Rs. 2560. The sums of money are lent out are :

Rs. 2000

Rs. 1500

Rs. 2500

Rs. 3000

Rs. 2800

Solution:

Q5. A person invested some amount at the rate of 12% simple interest and a certain amount at the rate of 10% simple interest. He received yearly interest of Rs. 130. But if he had interchanged the amounts invested, he would have received Rs. 4 more as interest. How much did he invest at 12% simple interest?

Rs. 700

Rs. 500

Rs. 800

Rs. 400

Rs. 200

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
x>y

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): Find the approximate value of the following questions.

Q11. 21 + 4.9 × 7.9 + 9.88 =?

Solution:

?≃21+5×8+10
≃ 71

Q12.