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Problem on HCF and LCM – 20+ Imp Aptitude Questions and Answers

LCM and HCF Questions

Practice Problem on HCF and LCM by solving the important Aptitude Questions and Answers given here. LCM and HCF Questions provided on this page with their explanations will help you in understanding the basic concepts. These questions are useful in all competitive examinations such as placement or entrance test.

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For the easiness of contenders preparing for the examination, we have provided LCM and HCF Questions with Solutions, so that in case they are not able to solve the questions at their own, they can take help from there. So, let’s start solving the questions one by one:

Problem on HCF and LCM

LCM and HCF Questions and Answers are mentioned here.

Question 1) Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

(a) 40

(b) 80

(c) 120

(d) 200

Answer 1) (a) 40

Explanation:

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

(a) 74

(b) 94

(c) 184

(d) 364

Answer 2) (d) 364

Explanation:

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 3)   The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

(a) 276

(b) 299

(c) 322

(d) 345

Answer 3) (c) 322

Explanation:

Clearly, the numbers are (23 x 13) and (23 x 14).

 Larger number = (23 x 14) = 322.

Must Check Out: LCM and HCF Shortcut Tricks

Question 4) The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

(a) 9000

(b) 9400

(c) 9600

(d) 9800

Answer 4) (c) 9600

Explanation:

Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

 Required number (9999 – 399) = 9600.

Question 5)   The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

(a) 3

(b) 13

(c) 23

(d) 33

Answer 5) (c) 23

Explanation:

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.

Question 6) What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?

(a) 196

(b) 630

(c) 1260

(d) 2520

Answer 6) (b) 630

Explanation:

L.C.M. of 12, 18, 21

   = 2 x 3 x 2 x 3 x 7 x 5 = 1260.

   Required number = (1260 / 2)  = 630.

Question 7) 252 can be expressed as a product of primes as:

(a) 2 x 2 x 3 x 3 x 7

(b) 2 x 2 x 2 x 3 x 7

(c) 3 x 3 x 3 x 3 x 7

(d) 2 x 3 x 3 x 3 x 7

Answer 7) (a) 2 x 2 x 3 x 3 x 7

Explanation:

Clearly, 252 = 2 x 2 x 3 x 3 x 7

Know Here: How to Prepare Maths for Competitive Exams

Question 8) Find the highest common factor of 36 and 84.

(a) 4

(b) 6

(c) 12

(d) 18

Answer 8) (c) 12

Explanation:

36 = 22 x 32

84 = 22 x 3 x 7

 H.C.F. = 22 x 3 = 12.

Question 9) The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:

(a) 504

(b) 536

(c) 544

(d) 548

Answer 9) (d) 548

Explanation:

Required number = (L.C.M. of 12, 15, 20, 54) + 8

   = 540 + 8

   = 548.

Question 10) Which of the following has the most number of divisors?

(a) 99

(b) 101

(c) 176

(d) 182

Answer 10) (c) 176

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.

Question 11) 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 11) (c) 40

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.

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Question 12) 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:

(a) 4

(b) 5

(c) 6

(d) 8

Answer 12) (a) 4

Explanation:

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

Question 13) The G.C.D. of 1.08, 0.36 and 0.9 is:

(a) 0.03

(b) 0.9

(c) 0.18

(d) 0.108

Answer 13) (c) 0.18

Explanation:

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

(a) 101

(b) 107

(c) 111

(d) 185

Answer 14) (c) 111

Explanation:

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 15) 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:

(a) 1677

(b) 1683

(c) 2523

(d) 3363

Answer 15) (b) 1683

Explanation:

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.

Question 16) The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:

(a) 1008

(b) 1015

(c) 1022

(d) 1032

Answer 16) (b) 1015

Explanation:

Required number = (L.C.M. of 12,16, 18, 21, 28) + 7

   = 1008 + 7

   = 1015

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Question 17) 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:

(a) 279

(b) 283

(c) 308

(d) 318

Answer 17) (b) 1015

Explanation: (c) 308

Other number = (11 x 7700)/ 275   = 308.

Question 18) 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:

(a) 75

(b) 81

(c) 85

(d) 89

Answer 18) (c) 85

Explanation:

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.

Question 19) The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

(a) 1

(b) 2

(c) 3

(d) 4

Answer 19) (b) 2

Explanation:

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.

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Question 20) 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 ?

(a) 26 minutes and 18 seconds

(b) 42 minutes and 36 seconds

(c) 45 minutes

(d) 46 minutes and 12 seconds

Answer 20) (d) 46 minutes and 12 seconds

Explanation:

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.

Hope that LCM and HCF Questions provided on this page of recruitmentresult.com are useful for the students preparing for the competitive examination. The given Problem on HCF and LCM are solved using easy tricks and shortcuts.

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