1.

The L.C.M of two numbers is 50 and their H.C.F is 5, if the sum of the numbers is 50, then their Difference is:

**Answer: Option 'B'**

**1500**

2.

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.**

3.

A, B and C start at the same time in the same direction to run around a circular stadium. 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 'C'**

**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**

4.

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.**

5.

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.
**

6.

The H.C.F of the fractions 7/3, 8/10, 9/5 is

**Answer: Option 'D'**

**Answer: Option 'D'**

7.

252 can be expressed as a product of primes as:

**Answer: Option 'D'**

**Clearly, 252 = 2 × 2 × 3 × 3 × 7.**

8.

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

**Answer: Option 'A'**

**Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43) = H.C.F. of 48, 92 and 140 = 4.**

9.

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.**

10.

The L.C.M of two prime numbers x and y (x>y) is 203, the value of 2y-x

**Answer: Option 'B'**

**-15**

11.

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.**

12.

Find the highest common factor of 36 and 84.

**Answer: Option 'C'**

**36 = 2 ^{2} x 3^{2}**

84 = 2^{2} x 3 x 7 H.C.F. = 2^{2} x 3 = 12.

13.

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 'A'**

**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.
Required sum = (19 + 29 + 37) = 85.**

14.

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.**

15.

The ratio of the two numbers is 9:11 and their L.C.M is 999999. The numbers are

**Answer: Option 'C'**

**90909,111111**

16.

Find the L.C.M of 582, 432 and 156:

**Answer: Option 'C'**

**544752**

17.

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.**

18.

Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

**Answer: Option 'D'**

**N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305) = H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4**

19.

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 'C'**

**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.**

20.

Third of three fifth of one fourth of a number is 30. What is 80% of that number?

**Answer: Option 'C'**

**240**

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