1.
The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:
Answer: Option 'C'
Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.
2.
A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point?
Answer: Option 'B'
L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.
3.
Which of the following has the most number of divisors ?
Answer: Option 'B'
99 = 1 x 3 x 3 x 11
101 = 1 x 101
176 = 1 x 2 x 2 x 2 x 2 x 11
182 = 1 x 2 x 7 x 13
So, divisors of 99 are 1, 3, 9, 11, 33, .
99 Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 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.
4.
Third of three fifth of one fourth of a number is 30. What is 80% of that number?
Answer: Option 'C'
240
5.
The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
Answer: Option 'B'
Let the numbers be 2x and 3x.
Then, their L.C.M. = 6x.
So, 6x = 48 or x = 8
The numbers are 16 and 24.
Hence, required sum = (16 + 24) = 40.
6.
The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
Answer: Option 'C'
L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 - 37) = 23.
7.
The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
Answer: Option 'C'
Let the numbers be 2x and 3x.
Then, their L.C.M. = 6x. So, 6x = 48 or x = 8.
The numbers are 16 and 24.
Hence, required sum = (16 + 24) = 40.
8.
The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:
Answer: Option 'B'
Required number = (L.C.M. of 12, 15, 20, 54) + 8 = 540 + 8 = 548.
9.
The ratio of the two numbers is 9:11 and their L.C.M is 999999. The numbers are
Answer: Option 'C'
90909,111111
10.
The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:
Answer: Option 'B'
L.C.M. of 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3 Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 x 2 + 3) = 1683.
11.
The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is:
Answer: Option 'B'
Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.
12.
The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:
Answer: Option 'C'
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015
13.
The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
Answer: Option 'C'
Answer: Option 'C'
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 × 4) + 4 = 364.
14.
The H.C.F of the fractions 7/3, 8/10, 9/5 is
Answer: Option 'D'
Answer: Option 'D'
15.
Four prime numbers are arranged in ascending order according to their magnitude. Product of first Three is 20,677 and the product of last three 33,263
Answer: Option 'D'
37
16.
Find the L.C.M of the fractions 7/3, 8/10, and 9/15:
Answer: Option 'C'
504
17.
What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?
Answer: Option 'B'
L.C.M. of 12, 18, 21 30
= 2 x 3 x 2 x 3 x 7 x 5 = 1260.
Required number = (1260 ÷ 2)
= 630.
18.
If 3/5 of 1/5 of a number is 12, then the sum of digits of the number is
Answer: Option 'D'
1
19.
Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?
Answer: Option 'D'
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes). In 30 minutes, they will toll together 30/2 + 1 = 16 times
20.
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:
Answer: Option 'C'
ince 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.