CAPF (AC) 2020 (Paper-1) (Question 101-125)

Total Questions: 25

11. In an exam, a candidate attempts 20 questions and scores 72 marks. If 5 marks are awarded for each correct answer and 2 marks are deducted for each wrong answer, then how many questions were answered correctly by him?

(a) 18
(b) 17
(c) 16
(d) 15

Correct Answer: (c) 16

Solution:the x number of questions were answered correct by the candidate.

∴ Number of wrong answered questions = 20 - x

According to the question,

x × 5 - (20-x) × 2 = 72

⇒ 5x - 40 + 2x = 72

⇒ 7x = 112

⇒ x = 16

So, 16 questions were answered correctly by him.

12. Which one of the following is the greatest number by which the product of three consecutive even numbers would be exactly divisible?

(a) 12
(b) 24
(c) 48
(d) 64

Correct Answer: (c) 48

Solution:Let three consecutive even numbers are 2n, (2n + 2) and (2n + 4)

∴ Product of the numbers = 8[n(n + 1)(n + 2)]

∴ Number would definetly divisible by 8 and atleast one more factor n, n + 1, n + 2 is even and divisible by 2 and in the three consecutive number atleast one is divisible by 3.

So, in this case the largest divisior of the product would be 8 × 3 × 2 = 48

13. If 15% of A is double of 30% of B, then what is the ratio of A to B?

(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1

Correct Answer: (d) 4:1

Solution:A × 15/100 = 2 × B × 30/100

A/B = 4/1

Required ratio = 4/1

14. The cost of gold varies directly as the cube of its weight. A gold piece weighing 20 decigram costs ₹1,000. If it is broken into two pieces whose weights are in the ratio 2 : 3, then what is the profit or loss incurred?

(a) ₹280 profit
(b) ₹280 loss
(c) ₹720 profit
(d) ₹720 loss

Correct Answer: (d) ₹720 loss

Solution:Let weight of pieces be 2x and 3x.

The total weight of statue before breaking = 2x + 3x = 5x

The cost of initial piece = (5x)³ = 125x³

Now, let us find the cost of pieces after breaking weight of pieces = P₁ = 2x

Cost of P₁ = (2x)³ = 8x³

Weight of piece = P₂ = 3x

Cost of P₂ = (3x)³ = 27x³

Total cost of all pieces = 8x³ + 27x³ = 35x³

When the gold price cost ₹1000

⇒ 125x³ = 1000

⇒ x³ = 8

The price of golden piece sold = 35x³

= 35 × 8 = ₹280

The loss = 1000 - 280 - ₹ 720

∴ The loss incurred is ₹ 720.

15. The average age of the boys in a class is 12 years. The average age of the girls in the class is 11 years. There are 50% more girls than boys in the class. Which one of the following is the average age of the class (in years)?

(a) 11.2 years
(b) 11.4 years
(c) 11.6 years
(d) 11.8 years

Correct Answer: (b) 11.4 years

Solution:Let the number of boys in the class is 2x.

Number of girls in the class,

= 2x + 2x × 50/100 = 2x + x = 3x

Average age of the boys = 12 years

∴ Total age of boys in the class

= 12 × 2x = 24x years

Average age of the girls = 11 years

∴ Total age of girls in the class

= 11 × 3x = 33x years

Total age of the class = 24x + 33x = 57x years.

Total number of students = 2x + 3x = 5x

Required average = 57x/5x = 11.4 years

16. A sum triples in ten years under compound interest at a certain rate of interest, the interest is being compounded annually. In how many years, it would become nine times?

(a) 20 years
(b) 30 years
(c) 40 years
(d) 50 years

Correct Answer: (a) 20 years

Solution:Let sum is A and rate is r%

According to the question,

3A = A × (1 + r/100)¹⁰

17. The number of ways by which 6 distinct balls can be put in 5 distinct boxes are

(a) 7776
(b) 15625
(c) 720
(d) 120

Correct Answer: (b) 15625

Solution:In every box, 6 different balls can be put.

∴ Required number of ways = 5⁶ = 15625

18. A wire of length 6 m is stretched such that its radius is reduced by 20%. Which one of the following is the value of increase in its length?

(a) 50%
(b) 56.25%
(c) 62.25%
(d) 75%

Correct Answer: (b) 56.25%

Solution:Length of the wire = 6 m.

Let radius of the wire is 5 × r' m.

Then, area of the wire = π (5r)² × 6

= 150 πr²m²

When radius is reduced by 20%

Then, radius = 5r x 80/100 = '4r' m.

Let, now length of the wire is h m.

∴ 150 πr²= π(4r)² × h

⇒ 150 r² = 16r² × h

⇒ h = 75/8 × m

% increase in the length of the wire

= (75/8 - 6)/6 × 100 = 27/(8 × 6) × 100 = 56.25%

19. The alphabets from A to J are numbered from 0 to 9 respectively. Which one of the following is the value of AGJ-CEG + EDB?

(a) CEF
(b) DGF
(c) GFD
(d) FCE

Correct Answer: (a) CEF

Solution:

20. A is the smallest positive integer which when divided by 9 and 12 leaves remainder 8. B is the smallest positive integer which when divided by 9 and 12 leaves remainder 5. Which one of the following is the value of A - B?

(a) 3
(b) 2
(c) 1
(d) 0

Correct Answer: (a) 3

Solution:LCM of 9 and 12 is 36.

Let A = 36 + 8 = 44 and B = 36 + 5 = 41

[∴ If a number is divisible by 36, then it will be also divisible by 9, 12]

∴ A - B = 44 - 41 = 3

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