Rationale: D = 25 * 60 + 1300 = 2800 m
T = 2800/60 * 18/5 = 168 se
The answer: (C)
Question: A train of 24 carriages, each of 60 meters length, when an engine also of 60 meters length is running at a speed of 60 kmph. In what time will the train cross a bridge 1.3 km long?
Options:
(A) 2 mins 14 sec
(B) 2 mins 34 sec
(C) 168 sec
(D) 344 sec
(E) 2 mins 44 sec
Reasoning: The length of wall painted by one man in one day = 60 / 4 3 = 5 m
No. of days required to paint 50 m wall by 5 men = 50/ 5 5 = 2 day.
Ans: (B)
Question: If 4 men can paint 60 m long wall in 3 days, then 5 men can paint 50 m long wall in
Options:
(A) 3days
(B) 2days
(C) 5days
(D) 6days
(E) 1day
Rationale: 21-12 < no of hrs < 21-2
9< no of hrs < 19.
The answer: (C)
Question: Resident of town X participated in a survey to determine the number of hours per week each resident spent watching television. The distribution of the results of the survey had a mean of 21 hours and std deviation of 2 hrs. The number of hours that pat, a resident of town X, watched television last week was between 1 and 2 standard deviation below the mean. Which of the following could be the number of hours that Pat watched television last week?
Options:
(A) 30
(B) 20
(C) 18
(D) 22
(E) 6
Reasoning: 550 = 2*222 + (22*4) + 9*2
K weight of 600 = 2 +4+9 = 15
Ans: (E)
Question: K-numbers are positive integers with only 2's as their digits. For example, 2, 22, and 222 are K-numbers. The K-weight of a number n is the minimum number of K-numbers that must be added together to equal n. For example, the K-weight of 50 is 5, because 50 = 22 + 22 + 2 + 2 + 2. What is the K-weight of 550?
Options:
(A) 10
(B) 11
(C) 12
(D) 13
(E) 15
Rationale: Average score for first 4 rounds = 68
Average score after 5 rounds = 66
Score Scott needs on his 5th round to drop average score by 2 points = 68 - (2*5) = 58
The answer: (A)
Question: Scott's average (arithmetic mean) golf score on his first four rounds was 68.What score does he need on his fifth round to drop his average score by 2 points?
Options:
(A) 58
(B) 72
(C) 78
(D) 88
(E) 312
Rationale: We go on counting the odd days from 1991 onward still the sum is divisible by 7. The number of such days are 14 upto the year 2001. So, the calender for 1991 will be repeated in the year 2002.
The answer: (C)
Question: The year next to 1991 having the same calendar as that of 1990 is
Options:
(A) 1998
(B) 2001
(C) 2002
(D) 2003
(E) None of these