1.  The traffic lights at three different road crossing change after every 48 sec., 72 sec. and 108 sec. respectively. If they all change simultaneously at 8:20:00 hours, then at what time will they again change simultaneously?


9:30:15
8:28:16
8:27:12
9:32:14


Answer

 Option

Interval of change = (L.C.M. of 48, 72, 108) sec. = 432 sec. So, the lights will again change simultaneously after every 432 seconds i.e., 7 min.12 sec. Hence, next simultaneous change will take place at 8:27:12 hrs.

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2.  Which of the following has most number of divisors?


99
101
176
182


Answer

 Option

99 = 1*3*3*11; 101= 1*101 176 = 1*2*2*2*2*11; 182 = 1*2*7*13 So, divisors of 99 are 1,3,9,11,33 and 99; divisor of 101 are 1 and 101; divisor of 176 are 1,2,4,8,16,22,44,88 and 176; divisors of 182 are 1,2,7,13,14,26,91 and 182. Hence, 176 has the most number of divisors.

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3.  Find the largest number of four digits exactly divisible by 12,15,18 and 27?


9720
9850
9385
9430


Answer

 Option

The largest number of four digits is 9999. Required number must be divisible by L.C.M of 12,15,18,27 i.e.,540 On dividing 9999 by 540, we get 279 as remainder. Therefore required number = (9999 - 279) = 9720.

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4.  Find the H.C.F of 2^3*3^2*5*7^4, 2^2*3^5*5^2*7^6, 2^3*5^3*7^2


750
870
980
890


Answer

 Option

The prime numbers common to given numbers are 2,5 and 7 H.C.F = 2^2 * 5 * 7^2 = 980

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5.  Three numbers which are co - prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is?


75
81
85
89


Answer

 Option

Since the numbers are co - prime, they contain only 1 as the common factor. Also, the given two products have the middle number in common. So, middle number = H.C.F. of 551 and 1073 = 29; First number = (551/29) = 19; Third number = (1073/29) = 37. Therefore required sum = (19 + 29 + 37) = 85.

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