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Problems On H.C.F And L.C.M - Aptitude MCQ Questions and Solutions with Explanations

Problems On H.C.F And L.C.M MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Problems On H.C.F And L.C.M with easy and logical explanations. Problems On H.C.F And L.C.M MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-2 section-1

Each section contains 100 MCQs on Problems on H.C.F and L.C.M .
11
Find the lowest common multiple of 24, 36 and 40.
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & 2\,\,\,|\,\,\,24\,\,\,\,\, - \,\,\,\,\,36\,\,\,\,\, - \,\,\,\,\,40 \cr & - - - - - - - - - - - - - \cr & 2\,\,\,|\,\,\,12\,\,\,\,\, - \,\,\,\,\,18\,\,\,\,\, - \,\,\,\,\,20 \cr & - - - - - - - - - - - - - \cr & 2\,\,\,|\,\,\,\,\,\,6\,\,\,\,\, - \,\,\,\,\,\,\,\,9\,\,\,\,\, - \,\,\,\,\,10 \cr & - - - - - - - - - - - - - \cr & 3\,\,\,|\,\,\,\,\,\,\,3\,\,\,\,\, - \,\,\,\,\,\,\,\,9\,\,\,\,\, - \,\,\,\,\,\,\,\,5 \cr & - - - - - - - - - - - - - \cr & \,\,\,\,\,\,\,|\,\,\,\,\,1\,\,\,\, - \,\,\,\,\,\,\,\,\,\,3\,\,\,\,\, - \,\,\,\,\,\,\,\,5 \cr & L.C.M = 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 360 \cr} $$
12
The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
Discuss
Answer & Solution
Answer: Option C
Solution:
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
13
Reduce $$\frac{{128352}}{{238368}}$$  to its lowest terms.
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{128352}}\overline {)238368{\text{(}}} 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{128352}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline {110016{\text{)128352(}}} 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,110016 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline {183336{\text{)110016(}}} {\text{6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{110016}}{ \times } \cr & {\text{So, H}}{\text{.C}}{\text{.F of 128352 and 238368}} \cr & {\text{ = 18336}} \cr & \therefore \frac{{128352}}{{238368}} = \frac{{128352 \div 18336}}{{238368 \div 18336}} = \frac{7}{{13}} \cr} $$
14
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:
Discuss
Answer & Solution
Answer: Option B
Solution:
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
15
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 ?
Discuss
Answer & Solution
Answer: Option D
Solution:
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.
16
The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{Other number}} \cr & = {\frac{{11 \times 7700}}{{275}}} \cr & = 308 \cr} $$
17
What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?
Discuss
Answer & Solution
Answer: Option B
Solution:
12 = 2 × 2 × 3
18 = 2 × 3 × 3
21 = 3 × 7
30 = 2 × 3 × 5

L.C.M. of 12, 18, 21 30 = 2 × 3 × 2 × 3 × 7 × 5 = 1260
Required number = $$\frac{{1260}}{2}$$  = 630
18
The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:
Discuss
Answer & Solution
Answer: Option D
Solution:
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
19
The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:
Discuss
Answer & Solution
Answer: Option B
Solution:
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
   = 1008 + 7
   = 1015
20
252 can be expressed as a product of primes as:
Discuss
Answer & Solution
Answer: Option A
Solution:
Clearly, 252 = 2 × 2 × 3 × 3 × 7