problems on hcf and lcm

26.

The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
A.     123    B.     127
C.     235    D.     305
Answer & Explanation

Answer: Option B

Explanation:

Required number = H.C.F. of (1657 – 6) and (2037 – 5)

= H.C.F. of 1651 and 2032 = 127.

27.

Which of the following has the most number of divisors?
A.     99    B.     101
C.     176    D.     182
Answer & Explanation

Answer: Option C

Explanation:

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.

28.

The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
A.     28    B.     32
C.     40    D.     64
Answer & Explanation

Answer: Option C

Explanation:

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.

29.

The H.C.F. of     9     ,     12     ,     18     and     21     is:
10     25     35     40
A.
3
5
B.
252
5
C.
3
1400
D.
63
700
Answer & Explanation

Answer: Option C

Explanation:

Required H.C.F. =     H.C.F. of 9, 12, 18, 21     =     3
L.C.M. of 10, 25, 35, 40     1400

30.

If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:
A.
55
601
B.
601
55
C.
11
120
D.
120
11
Answer & Explanation

Answer: Option C

Explanation:

Let the numbers be a and b.

Then, a + b = 55 and ab = 5 x 120 = 600.

The required sum =     1     +     1     =     a + b     =     55     =     11
a     b     ab     600     120

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