# Quantitative Aptitude Quiz for IBPS Clerk Prelims: 13th November 2018

**Dear Students,**

**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): Study the following table chart carefully and answer the question given below. The following table shows marks obtained by 6 students in 5 different subjects in a class.**

**Q1. Find the ratio between the marks obtained by Abhishek in Maths and Shivam in English together to the marks obtained by Prabhat in Physics and Aman in Hindi together ?**

9 : 16

7 : 8

14 : 15

7 : 11

8 : 7

Solution:

Marks obtained by Abhishek in Maths and Shivam in English = 72 + 82 = 154

Marks obtained by Prabhat in Physics and Aman in Hindi = 82 + 94 = 176

Required ratio = 154 : 176 = 7 : 8

Marks obtained by Prabhat in Physics and Aman in Hindi = 82 + 94 = 176

Required ratio = 154 : 176 = 7 : 8

**Q2. Among these students who scored highest in all the five subjects ?**

Prabhat

Aman

Shivam

Harshit

Akash

Solution:

Abhishek’s score = 76 + 66 + 72 + 80 + 60 = 354

Harshit’s score = 64 + 78 + 82 + 48 + 90 = 362

Akash’s score = 83 + 56 + 68 + 62 + 86 = 355

Shivam’s score = 89 + 82 + 78 + 54 + 66 = 369

Aman’s score = 94 + 90 + 96 + 70 + 52 = 402

Prabhat’s score = 92 + 88 + 75 + 82 + 86 = 423

Harshit’s score = 64 + 78 + 82 + 48 + 90 = 362

Akash’s score = 83 + 56 + 68 + 62 + 86 = 355

Shivam’s score = 89 + 82 + 78 + 54 + 66 = 369

Aman’s score = 94 + 90 + 96 + 70 + 52 = 402

Prabhat’s score = 92 + 88 + 75 + 82 + 86 = 423

**Q3. What is the difference between the total marks obtained by all the students in Physics and Hindi ?**

96

112

90

102

110

Solution:

Total marks in Hindi = 498

Total marks in Physics = 396

Required difference = 498 – 396 = 102

Total marks in Physics = 396

Required difference = 498 – 396 = 102

**Q4. Marks obtained by Harshit in all the five subjects together is what approximate percent more or less than the marks obtained by Shivam in all the five subjects together ?**

2% more

3% less

5% more

6% less

2% less

**Q5. Find the percentage of marks obtained by Akash in all the subjects together.**

70%

71%

75%

74%

69%

**Q6. 8 years ago, the ratio of the ages of Akshita and Prerana was 4 : 3. If the ratio of their present ages is 6 : 5 respectively, what is the ratio of the sum to the difference of their present ages?**

11 : 2

11 : 4

11 : 1

11 : 3

7: 11

**Q7. In a 63 litre mixture of honey and water, the ratio of honey to water is 5 : 4 . What amount of honey should be added to this mixture to make the ratio of water to honey 5 : 7?**

6.2 li

5.4 li

4.2 li

4.8 li

5.6 li

**Q8. The percentage of metals in a mine of lead ore is 60%. Now the percentage of silver is 3/4% of metals and the rest is lead. If the mass of ore extracted from the mine is 8000 kg, the mass (in kg) of lead is:**

4763

4764

4762

4761

4264

Solution:

Mass of lead ore = 8000 kg

⇒Mass of metal in lead ore = 60% of 8000 = 4800 kg

⇒ Mass of silver in metal = (3/4)% of 4800=36 kg

⇒ Mass of lead in ore = 4800 – 36 = 4764 kg

⇒Mass of metal in lead ore = 60% of 8000 = 4800 kg

⇒ Mass of silver in metal = (3/4)% of 4800=36 kg

⇒ Mass of lead in ore = 4800 – 36 = 4764 kg

**Q9. Sum of area of a square and a rectangle is 644 sq. cm and perimeter of rectangle is 88 cm. If ratio of length to breadth of rectangle is 7 : 4, then find perimeter of square ?**

72 cm

81 cm

48 cm

64 cm

56 cm

Solution:

Perimeter of rectangle = 2(â„“ + b) 2 (7X + 4X) = 88 cm

11X = 44

X = 4

â„“ = 28, b = 16

Area of rectangle = 28 × 16 = 448 sq.

Area of square = 644 – 448 = 196 sq. cm

a² = 196

a = 14

perimeter of square = 4 × 14 = 56cm

11X = 44

X = 4

â„“ = 28, b = 16

Area of rectangle = 28 × 16 = 448 sq.

Area of square = 644 – 448 = 196 sq. cm

a² = 196

a = 14

perimeter of square = 4 × 14 = 56cm

**Q10. What is the difference between compound interest on Rs. 5,000 for**

**years at 4% per annum if the interest is compounded yearly and half yearly?**

**(**It is given that)

Rs. 3.04

Rs. 5.04

Rs. 8.30

Rs. 4.80

Rs. 6.80

**Directions (11-15): In each of the following questions two equations are given. You have to solve the equations and Give answer –**

**Q11. I. 8x² + 6x = 5**

**II. 12y² – 22y + 8 = 0**

if x < y

if x ≤ y

relationship between x and y cannot be determined

if x ≥ y

if x > y

Solution:

I. 8x² + 6x – 5 = 0

⇒ 8x² + 10x – 4x – 5 = 0

⇒ (4x + 5) (2x – 1) = 0

⇒ x = ½, –5/4

II. 12y² – 22y + 8 = 0

⇒ 6y² – 11y + 4 = 0

⇒ 6y² – 3y – 8y + 4 = 0

⇒ (2y – 1) (3y – 4) = 0

⇒ y = 1/2, 4/3

y ≥ x

⇒ 8x² + 10x – 4x – 5 = 0

⇒ (4x + 5) (2x – 1) = 0

⇒ x = ½, –5/4

II. 12y² – 22y + 8 = 0

⇒ 6y² – 11y + 4 = 0

⇒ 6y² – 3y – 8y + 4 = 0

⇒ (2y – 1) (3y – 4) = 0

⇒ y = 1/2, 4/3

y ≥ x

**Q12. I. 17x² + 48x = 9**

**II. 13y² = 32y – 12**

if x < y

if x ≤ y

relationship between x and y cannot be determined

if x ≥ y

if x > y

Solution:

I. 17x² + 48x – 9 = 0

⇒ 17x² + 51x – 3x – 9 = 0

⇒ (x + 3) (17x – 3) = 0

⇒ x = 3/17, – 3

II. 13y² – 32y + 12 = 0

⇒ 13y² – 26y – 6y + 12 = 0

⇒ (y – 2) (13y – 6) = 0

⇒ y = 2, 6/13

y > x

⇒ 17x² + 51x – 3x – 9 = 0

⇒ (x + 3) (17x – 3) = 0

⇒ x = 3/17, – 3

II. 13y² – 32y + 12 = 0

⇒ 13y² – 26y – 6y + 12 = 0

⇒ (y – 2) (13y – 6) = 0

⇒ y = 2, 6/13

y > x

**Q13. I. 8x² + 26x + 15 = 0**

**II. 4y² + 24y + 35 = 0**

if x < y

if x ≤ y

relationship between x and y cannot be determined

if x ≥ y

if x > y

Solution:

I. 8x² + 26x + 15 = 0

⇒ 8x² + 20x + 6x + 15 = 0

⇒ 4x (2x + 5) + 3(2x + 5) = 0

⇒ (2x + 5) (4x + 3) = 0

⇒ x = – 5/2, –3/4

II. 4y² + 24y + 35 = 0

⇒ 4y² + 10y + 14y + 35 = 0

⇒ 2y (2y + 5) + 7 (2y + 5) = 0

⇒ (2y + 5) (2y + 7) = 0

⇒ y = –5/2, –7/2

x ≥ y

⇒ 8x² + 20x + 6x + 15 = 0

⇒ 4x (2x + 5) + 3(2x + 5) = 0

⇒ (2x + 5) (4x + 3) = 0

⇒ x = – 5/2, –3/4

II. 4y² + 24y + 35 = 0

⇒ 4y² + 10y + 14y + 35 = 0

⇒ 2y (2y + 5) + 7 (2y + 5) = 0

⇒ (2y + 5) (2y + 7) = 0

⇒ y = –5/2, –7/2

x ≥ y

**Q14. I. 6x² + 19x + 15 = 0**

**II. 24y² + 11y + 1 = 0**

if x < y

if x ≤ y

relationship between x and y cannot be determined

if x ≥ y

if x > y

Solution:

I. 6x² + 19x + 15 = 0

⇒ 6x² + 9x + 10x + 15 = 0

⇒ (2x + 3) (3x + 5) = 0

⇒ x = –3/2, –5/3

II. 24y² + 11y + 1 = 0

⇒ 24y² + 8y + 3y + 1= 0

⇒ (3y + 1) (8y + 1) = 0

⇒ y = –1/3, –1/8

y > x

⇒ 6x² + 9x + 10x + 15 = 0

⇒ (2x + 3) (3x + 5) = 0

⇒ x = –3/2, –5/3

II. 24y² + 11y + 1 = 0

⇒ 24y² + 8y + 3y + 1= 0

⇒ (3y + 1) (8y + 1) = 0

⇒ y = –1/3, –1/8

y > x

**Q15. I. 2x² + 11x + 15 = 0**

**II. 4y² + 22y + 24 = 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 + 15 = 0

⇒ 2x² + 6x + 5x + 15 = 0

⇒ (x + 3) (2x + 5) = 0

⇒ x = – 3, –5/2

II. 4y² + 22y + 24 = 0

⇒ 2y² + 11y + 12 = 0

⇒ 2y² + 8y + 3y + 12 = 0

⇒ (y + 4) (2y + 3) = 0

⇒ y = –4, –3/2

No relation

⇒ 2x² + 6x + 5x + 15 = 0

⇒ (x + 3) (2x + 5) = 0

⇒ x = – 3, –5/2

II. 4y² + 22y + 24 = 0

⇒ 2y² + 11y + 12 = 0

⇒ 2y² + 8y + 3y + 12 = 0

⇒ (y + 4) (2y + 3) = 0

⇒ y = –4, –3/2

No relation

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