HCF And LCM(Quantitative Aptitude ) Questions and answers for exam Preparation

Quantitative Aptitude

HCF AND LCM MCQs

Problems On Hcf And Lcm, H.C.F And L.C.M. Of Numbers, Lcm & Hcf, Hcf And Lcm

Total Questions : 1401 | Page 11 of 141 pages

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
None of these

Discuss Question

Answer: Option C. -> 85
- 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 = 19; Third number = 1073 = 37. 29 29
Required sum = (19 + 29 + 37) = 85.

Question 102.

Find the highest common factor of 36 and 84?

4
6
12
18
None of these

Discuss Question

Answer: Option C. -> 12
- 36 = 22 x 32
84 = 22 x 3 x 7
H.C.F. = 22 x 3 = 12.

Question 103.

Which of the following fraction is the largest?

7/8
13/16
31/40
63/80
None of these

Discuss Question

Answer: Option A. -> 7/8
- L.C.M. of 8, 16, 40 and 80 = 80.
7 = 70 ; 13 = 65 ; 31 = 62 8 80 16 80 40 80
Since, 70 > 65 > 63 > 62 , so 7 > 13 > 63 > 31 80 80 80 80 8 16 80 40
So, 7 is the largest. 8

Question 104.

The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:

504
536
544
548
None of these

Discuss Question

Answer: Option D. -> 548
-    Required number = (L.C.M. of 12, 15, 20, 54) + 8
   = 540 + 8
   = 548.

Question 105.

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:

101
107
111
185
None of these

Discuss Question

Answer: Option C. -> 111
-    Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
  ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
  Greater number = 111.

Question 106.

Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

40
80
120
200
None of these

Discuss Question

Answer: Option A. -> 40
- Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.

Question 107.

The G.C.D. of 1.08, 0.36 and 0.9 is:

0.03
0.9
0.18
0.108
None of these

Discuss Question

Answer: Option C. -> 0.18
- Given numbers are 1.08, 0.36 and 0.90.
H.C.F. of 108, 36 and 90 is 18,
H.C.F. of given numbers = 0.18.

Question 108.

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

1
2
3
4
None of these

Discuss Question

Answer: Option B. -> 2
- Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common
positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.

Question 109.

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

74
94
184
364
None of these

Discuss Question

Answer: Option D. -> 364
- 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 x 4) + 4 = 364.

Question 110.

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?

4
10
15
16
None of these

Discuss Question

Answer: Option D. -> 16
- L.C.M. of 2, 4, 6, 8, 10, 12 is 120. So, the bells will toll together after every 120 seconds, i.e, 2 minutes. In 30 minutes, they will toll together 30/2 + 1 = 16