Q. 131. In a rectangle, length is 8 cm, breadth is 6 cm, then the its diagonal is:
(A) 9 cm
(B) 14 cm
(C) 10 cm
(D) 12 cm
Answer:
(C) 10 cm
Explanation:
By using the Pythagorean theorem for the right-angled triangle formed by the sides and diagonal: Diagonal $= \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$ cm.
Q. 132. A’s age is 3 times that of B. In 12 years, A will be double the age of B. A’s present age is:
(A) 27 years
(B) 32 years
(C) 36 years
(D) 40 years
Answer:
(C) 36 years
Explanation:
Let B’s current age be $x$, making A’s age $3x$. In 12 years: $3x + 12 = 2(x + 12)$. Solving this yields $3x + 12 = 2x + 24$, meaning $x = 12$. A’s present age is $3 \times 12 = 36$ years.
Q. 133. A man sells his bike to his friend at 10% loss. If the friend sells it for Rs. 54,000 and gains 20%, the original CP of the bike was:
(A) Rs. 25,000
(B) Rs. 37,500
(C) Rs. 50,000
(D) Rs. 60,000
Answer:
(C) Rs. 50,000
Explanation:
Let the friend’s Cost Price (which is the man’s Selling Price) be $C$. Since the friend sells it at a 20% profit for 54,000: $C \times 1.2 = 54,000$, so $C = 45,000$. The man sold it to his friend for 45,000 at a 10% loss. Thus, Original CP $\times 0.9 = 45,000$, resulting in an Original CP of Rs. 50,000.
Q. 134. If the probability of winning a game is 0.3, the probability of losing it is:
(A) 0.1
(B) 0.7
(C) 1
(D) 1.3
Answer:
(B) 0.7
Explanation:
The sum of all mutually exclusive probabilities for an event is exactly 1. If the probability of winning is 0.3, the probability of not winning (losing) is $1 – 0.3 = 0.7$.
Q. 135. Mean of 100 numbers was calculated as 49. Later it was found that three numbers should have been 60, 70, 80 were wrongly read as 40, 20, 50 respectively. The correct mean is:
(A) 48
(B) 49
(C) 50
(D) 60
Answer:
(C) 50
Explanation:
The original total sum was $100 \times 49 = 4900$. The difference between the correct numbers and the wrong ones is $(60-40) + (70-20) + (80-50) = 20 + 50 + 30 = 100$. The new correct sum is $4900 + 100 = 5000$. Correct mean = $5000 / 100 = 50$.
