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Quantitative Aptitude DI

Here we are providing new series of Quantitative Aptitude Questions for upcoming exams, so the aspirants can practice it on a daily basis.

Study the following data carefully and answer the questions:

An exam in conducted that consists of three sections: Quant, Reasoning and English. Sectional marked can be calculated as mentioned below:

Sectional marks = 5 * Right answers – 2 * Wrong answers – 0.5 * Un-attempted questions

Normalized marks in any section (If Right answers are > 80%) = 1.2 * Sectional marks

Normalized marks in any section (If Right answers are > 60% and < 80%) = 1 * Sectional marks

Normalized marks in any section (If Right answers are < 60%) = 0.8 * Sectional marks

Total marks in the exam = Normalized marks in Quant + Normalized marks in Reasoning + Normalized marks in English

Total questions in the test are 100 out of which 35 are from Quant section, 35 from Reasoning section and remaining 30 are from English section.

Table given below shows the ratio of number of Right answers to the number of Wrong answers marked by five different students A, B, C, D and E in all the three sections:

1) If student A attempts 80 questions and ratio of attempted questions in Quant, Reasoning and English is 7: 8: 5 respectively, then what is the total marks obtained by him in the test?

a) 252.8

b) 272.9

c) 264.7

d) 246.4

e) None of these

2) If student B left 19 questions in Quant, 23 questions in Reasoning and 20 questions in English un-attempted, then what will be the ratio of sections marks of him in Quant, Reasoning and English respectively?

a) 127: 69: 38

b) 87: 39: 28

c) 27: 9: 13

d) 135: 89: 58

e) None of these

3) If person C attempted, 60% of total quant questions in quant, 71(3/7)% of total questions in Reasoning and 70% of total questions in English, then total right answers are what per cent more than total wrong answers marked by him?

a) 56%

b) 73%

c) 64%

d) 68%

e) None of these

4) Students D marked 15 questions in Quant, 20 questions in reasoning and 17 questions in English correctly, then what will be the ratio of normalized marks obtained by him in Quant, Reasoning and English respectively?

a) 3: 2: 4

b) 1: 3: 2

c) 3: 1: 2

d) 1: 2: 3

e) None of these

5) If the ratio of number of attempted questions by student E in Quant, Reasoning and English is 6: 4: 5 respectively and total marks obtained by him in the test is 147, then how many did he left un-attempted in the test?

a) 50

b) 40

c) 45

d) 75

e) None of these

Directions (1-5):

1) Answer: c)

Total attempted questions in Quant = 80 * (7/20) = 28

Total attempted questions in Reasoning = 80 * (8/20) = 32

Total attempted questions in English = 80 * (5/20) = 20

Normalized marks in Quant = 1.2 * [{5 * 28 * (6/7)} – {2 * 28 * (1/7))} – {0.5 * (35 – 28)}] = 130.2

Normalized marks in Reasoning = [{5 * 32 * (5/8)} – {2 * 32 * (3/8))} – {0.5 * (35 – 32)}] = 74.5

Normalized marks in English = [{5 * 20 * (3/4)} – {2 * 20 * (1/4))} – {0.5 * (30 – 20)}] = 60

Total marks in the exam = 130.2 + 74.5 + 60 = 264.7

2) Answer: a)

Attempted questions in Quant = 35 – 19 = 16

Attempted questions in Reasoning = 35 – 23 = 12

Attempted questions in English = 30 – 20 = 10

Sectional marks in Quant = 5 * 16 * (15/16) – 2 * 16 * (1/16) – 0.5 * 19 = 63.5

Sectional marks in Reasoning = 5 * 12 * (5/6) – 2 * 12 * (1/6) – 0.5 * 23 = 34.5

Sectional marks in English = 5 * 10 * (7/10) – 2 * 10 * (3/10) – 0.5 * 20 = 19

Required ratio = 63.5: 34.5: 19 = 127: 69: 38

3) Answer: d)

Total attempted questions in Quant = 60% of 35 = 21

Total attempted questions in Reasoning = 71(3/7)% of 35 = 25

Total attempted questions in English = 70% of 30 = 21

Total right answers marked by him = 21 * (5/7) + 25 * (3/5) + 21 * (4/7) = 15 + 15 + 12 = 42

Total wrong answers marked by him = (21 + 25 + 21) – 42 = 25

Required per cent = [(42 – 25)/25] * 100 = 68%

4) Answer: b)

Normalized marks of D in Quant

= 0.8 * [{5 * 15} – {2 * 15} – {0.5 * (35 – 30)}] = 34

Normalized marks of D in Reasoning

= 1.2 * [{5 * 20} – {2 * 20 * (1/4)} – {0.5 * (35 – 25)}] = 102

Normalized marks of D in English

= [{5 * 17} – {2 * 7} – {0.5 * (30 – 24)}] = 68

Required ratio = 34: 102: 68 = 1: 3: 2

5) Answer: b)

Let number of questions attempted by E in Quant, Reasoning and English is ‘6x’, ‘4x’ and ‘5x’ respectively.

Total un-attempted questions by E = 100 – (6x + 4x + 5x) = (100 – 15x)

Normalized marks in Quant = [5 * 4x – 2 * 2x – 0.5 * (35 – 6x)] = (19x – 17.5)

Normalized marks in Reasoning = [5 * 2.5x – 2 * 1.5x – 0.5 * (35 – 4x)]

= (11.5x – 17.5)

Normalized marks in English = [5 * 3.75x – 2 * 1.25x – 0.5 * (30 – 5x)]

= (18.75x – 15)

Total marks in the test = (19x – 17.5) + (11.5x – 17.5) + (18.75x – 15) = 147

49.25x = 147 + 50

x = 4

Total un-attempted questions by E = (100 – 15x) = 40