Quantitative Aptitude Quiz for IBPS Clerk Prelims: 12th November 2018 |

Dear Students,

Quantitative Aptitude Quiz for IBPS Clerk

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.

Watch the Video Solutions

Directions (1-5): What should come in place of question mark (?) in the following questions?

Q1. √225÷ ∛125× 40% of 550 =?

670

660

650

640

630

Solution:

Quantitative Aptitude Quiz for IBPS Clerk Prelims: 12th November 2018 |_3.1

Q2. 150% of 220+120% of 140+?=850

342

355

352

360

370

Solution:

Quantitative Aptitude Quiz for IBPS Clerk Prelims: 12th November 2018 |_4.1

Q3. 30% of 340 - 70% of 110 = 65 - ?

Solution:

Quantitative Aptitude Quiz for IBPS Clerk Prelims: 12th November 2018 |_5.1

Q4. [7 × (4.4 + 8.6)] - 8² = ?

Solution:

? = 91 – 64
= 27

Q5. (5555 ÷ 55) + (625 ÷ 25) + (120 ÷ 24) = ?

120

125

131

141

121

Solution:

? = 101 + 25 + 5
= 131

Q6. In a class, there are 15 girls and 10 boys. Three students are selected at random. The probability that only girls or only boys get selected is:

1/3

2/5

1/4

1/5

3/5

Solution:

Quantitative Aptitude Quiz for IBPS Clerk Prelims: 12th November 2018 |_6.1

Q7. Two letters are chosen out of the alphabets of the English language. Find the probability that both the letters are vowels.

2/65

3/65

1/65

3/5

0.2

Solution:

Quantitative Aptitude Quiz for IBPS Clerk Prelims: 12th November 2018 |_7.1

Q8. How many three digits number can be formed by using the digits 0, 2, 4, 6, 7 if repetition of digits is allowed.

100

Solution:

Total digits = 5
First place can be filled up by using only one of 4 digits (except 0, since 0 at the first place is meaning less).
Second place can be filled up by using any one of out of 5 digits (because repetition is allowed).
Similarly, third place can be filled up by using any one of out of all the five digits.
Thus,
Places: 0 0 0
Digit: 4 5 5
Total numbers = 4 × 5 × 5 = 100

Q9. Find the numbers between 100 and 1000 in which all digits are distinct.

548

648

748

448

684

Solution:

There are three digits numbers between 100 and 1000.
Total digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 which are 10.
First place can be filled up by using any one of 9 digits (except 0, since 0 at the first place is meaningless).
Second place can be filled up by using any one of 9 digits (as one digit has been used at first place)
Third place can be filled up by using only one of 8 digits.
Thus,
Places: 0 0 0
Digits: 9 9 8
Total number = 9 × 9 × 8 = 648

Q10. The letters of the word ‘PROMISE’ are to be arranged so that all the vowels should not come together. Find the number of arrangements.

3420

4320

5320

6420

4420

Solution:

Quantitative Aptitude Quiz for IBPS Clerk Prelims: 12th November 2018 |_8.1

Directions (11-15): Solve the following equations and mark the correct option given below.

Q11. I. x² – 27x + 180 = 0
II. y² – 7y = 60

if x>y

if x≥y

if y>x

if y≥x

if x=y or no relation can be established

Solution:

I. x² – 27x + 180 = 0
x² – 12x – 15x + 180 = 0
x (x – 12) –15(x – 12) = 0
(x – 15) (x – 12) = 0
x = 15, 12
II. y² – 7y – 60 = 0
y² – 12y + 5y – 60 = 0
y (y – 12) +5 (y – 12) = 0
(y + 5) (y – 12) = 0
y= -5, 12
⇒ x ≥ y

Q12. I. x² – 59x + 868 = 0
II. y² – 53y + 702 = 0

if x>y

if x≥y

if y>x

if y≥x

if x=y or no relation can be established

Solution:

I. x² – 59x + 868 = 0
x² – 28x – 31x + 868 = 0
x (x – 28) – 31 (x – 28) = 0
(x – 31) (x – 28) = 0
x = 28, 31
II. y² – 53y + 702 = 0
y² – 27y – 26y + 702 = 0
y (y – 27) – 26(y – 27) = 0
(y– 27) (y – 26) = 0
y = 26, 27
⇒ x > y

Q13. I. 100x² – 120x + 32 = 0
II. 10y² – 17y + 6 = 0

if x>y

if x≥y

if y>x

if y≥x

if x=y or no relation can be established

Solution:

I. 100x² – 120x + 32 = 0
100x² – 40x – 80x + 32 = 0
20x (5x – 2) – 16(5x – 2) = 0
(20x – 16) (5x – 2) = 0
x=4/5, 2/5
II. 10y² – 17y + 6 = 0
10y² – 12y – 5y + 6 = 0
2y (5y – 6) – 1 (5y – 6) = 0
(2y – 1) (5y – 6) = 0
y=1/2, 6/5
⇒ No relation

Q14. I. 15x² – 22x + 8 = 0
II. 12y² – 5y – 2 = 0

if x>y

if x≥y

if y>x

if y≥x

if x=y or no relation can be established

Solution:

I. 15x² – 22x + 8 = 0
15x² – 12x – 10x + 8 = 0
3x (5x – 4) –2 (5x – 4) = 0
(5x – 4) (3x– 2) = 0
x=4/5, 2/3
II. 12y² – 5y – 2 = 0
12y² – 8y + 3y – 2 = 0
4y (3y – 2) + 1 (3y – 2) = 0
(4y + 1) (3y – 2) = 0
y= -1/4, 2/3
⇒ x ≥ y

Q15. I. x² + 8x + 15 = 0
II. y² – 2y – 8 = 0

if x>y

if x≥y

if y>x

if y≥x

if x=y or no relation can be established

Solution:

I. x²+ 8x + 15 = 0
x² + 5x + 3x + 15 = 0
x (x + 5) + 3 (x+ 5) = 0
(x + 5) (x + 3) = 0
x = –5, –3
II. y² – 2y – 8 = 0
y² – 4y + 2y – 8 = 0
y (y – 4) + 2(y – 4) = 0
(y – 4) (y + 2) = 0
y= –2, 4
⇒ x < y