Haryana CET Group-D Exam, 21.10.2023 (Shift-II)

Total Questions: 100

41. (0.5 x 0.5 + 0.5 ÷ 5) is equal to:

Correct Answer: (1) 0.35
Solution:

=

=

=

= 35/100

42. The L.C.M. of three numbers 2 p, 5 p and 7 p is:

Correct Answer: (3) 70 p
Solution:LCM of 2p,5p & 7p

= 2p × 5 × 7

= 70p

43. For which sum, the simple interest at the rate of 3×3/4% per annum will be Rs. 210 in 2 ×1/3 years?

Correct Answer: (4) Rs. 2,400
Solution:P = I × 100/R × T

= 210 × 100/15/4 × 7/3

= 210 × 100 × 12/15 × 7

= Rs.2400

44. If a and b are two distinct integers, then which of the following is incorrect?

Correct Answer: (4) a-b = b-a
Solution:If a and b are two distinct integers then

a-b = b-a is correct

45. A certain number Is divided Into two parts such that 5 times the first part added to 11 times the second part makes 7 times the number. The ratio of the first part to the second part is:

Correct Answer: (3) 2:1
Solution:Let the number be x

First part = k

Second part = x-k

According to question,

5k + 11 (x-k) = 7x

=> 5k + 11x-11k = 7x

=> 11x - 7x = 6k

=> 4x = 6k

=> k = 4x/6 = 2x/3

46. Half of a certain number is equal to 65% of the second number. The ratio, of the first number to the second number is:

Correct Answer: (2) 13:10
Solution:Let first number = x

Second number = y

According to question

x/2 = 65% of y

=> x/2 =  y × 65/100

=> x/y = 2 × 65/100

=> x/y = 13/10

=> x : y = 13 : 10

47. The length of the longest rod that can be placed in a room of dimensions 10m x 10 m x 5 m is:

Correct Answer: (4) 15 m
Solution:Length of longest rod that can be placed in a room

= √l² + b² + h²

= √10² + 10² + 5²

= √100 + 100 +25

= √225 = 15 m

48. In how many years will a sum of Rs. 800 at 10% per annum compound interest, compounded semi-annually become Rs. 926.10?

Correct Answer: (4) 1×1/2 years
Solution:Interest is compounded half  yearly.

r = 10/2 = 5%

A = P (l + r/100)ⁿ

926.10 = 800 (1 + 5/100)ⁿ

=> 926.10/800 = (21/20)ⁿ

=> 9261/8000 = (21/20)ⁿ

=> (21/20)³ = (21/20)ⁿ

=> n = 3

Required time

= 3/2 = 1 × 1/2 years

49. 20 litres of a mixture contains milk and water in the ratio 3: 1. The amount of milk to be added to the mixture so as to make the ratio 4: 1, is:

Correct Answer: (1) 5 litres
Solution:Quantity of milk = 3/4 × 20 = 15 litres

Quantity of water = 1/4 × 20 = 5 litres

According to question,

15 + x/5 = 4/1

=> 15 + x = 20

=> x = 20 - 15 = 5

50. Simplification of 0.2 x 0.2+0.01 / 0.1×0.1+0.02 gives:

Correct Answer: (3) 5/3
Solution: