HCF and LCM

Explanation / Important formulas:

Highest Common Factor (H.C.F) or Greatest Common Divisor (G.C.D.):

The H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly. There are two methods of finding the H.C.F. of a given set of numbers:

  • Factorization Method: Express the each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F.
  • Division Method: Suppose we have to find the H.C.F. of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.

Finding the H.C.F. of more than two numbers: Suppose we have to find the H.C.F. of three numbers then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given number. Similarly, the H.C.F. of more than three numbers may be obtained.

Least Common Multiple (L.C.M.):

The least number which is exactly divisible by each one of the given numbers is called their L.C.M. There are two methods of finding the L.C.M. of a given set of numbers:

  • Factorization Method: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors.
  • Division Method (short-cut): Arrange the given numbers in a rwo in any order. Divide by a number which divided exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers.

Product of two numbers = Product of their H.C.F. and L.C.M.

Co-primes: Two numbers are said to be co-primes if their H.C.F. is 1.

H.C.F. and L.C.M. of Fractions:

  • H.C.F = H.C.F of Numerators / L.C.M of Denominators
  • L.C.M = L.C.M of Numerators / H.C.F of Denominators

Instructions to take Aptitude Test

  • Click on start to start taking the test.
  • Click on the option (A, B, C or D) to figure out the right answer.
  • You can answer multiple times till you get the right answer.
  • Once you get the right answer, explanations (if any) for the same will be showcased down.
  • On click of list, you get to see total no of questions, no of questions you answered and no of questions pending to answer.
  • On click of question number you will go to that particular question.
  • On click of END your test will end.
  • On click of ‘Get Results’ you will get to see correct answer for each questions.

Test



Hcf and Lcm - Question and Answers

Please wait while the activity loads.
If this activity does not load, try refreshing your browser. Also, this page requires javascript. Please visit using a browser with javascript enabled.

If loading fails, click here to try again

Congratulations - you have completed Hcf and Lcm - Question and Answers.

You scored %%SCORE%% out of %%TOTAL%%.

Your performance has been rated as %%RATING%%

Your answers are highlighted below.

Question 1
A teacher can divide her class into groups of 5,13, and 17. What is the smallest possible strength of the class?
A
835
B
940
C
1105
D
1220
Question 1 Explanation: 
Smallest possible strength of the class will be L.C.M of 5,13,17 ans:1105
Question 2
The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:
A
544
B
548
C
504

D
536
Question 2 Explanation: 
Required number = (L.C.M. of 12, 15, 20, 54) + 8

= 540 + 8

= 548.

Question 3
Find the highest common factor of 36 and 84.
A
4
B
6
C
12
D
18
Question 3 Explanation: 
36 = 22 x 32

84 = 22 x 3 x 7

H.C.F. = 22 x 3 = 12.

Question 4
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
85
B
75
C
65
D
60
Question 4 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 5
The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is:
A
15 cm
B
35 cm
C
25 cm
D
52 cm
Question 5 Explanation: 
Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.

Question 6
The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively is:
A
127
B
305
C
235
D
123
Question 6 Explanation: 
Required number = H.C.F. of (1657 - 6) and (2037 - 5)

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

Question 7
Which of the following has the most number of divisors?
A
176
B
182
C
99
D
101
Question 7 Explanation: 
Option A, 176 = 1 x 2 x 2 x 2 x 2 x 11.

Option B, 182 = 1 x 2 x 7 x 13.

Option C, 99 = 1 x 3 x 3 x 11.

Option D, 101 = 1 x 101.

Divisors of 99 are 1, 3, 9, 11, 33, 99.

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 have the most number of divisors

Question 8
The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
A
30
B
40
C
38
D
36
Question 8 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.

Question 9
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
267
B
318
C
308
D
279
Question 9 Explanation: 
Other number = (11 x 7700/275) = 308.
Question 10
What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30?
A
1260
B
630
C
2524
D
None of these
Question 10 Explanation: 
L.C.M. of 12, 18, 21 30 2 | 12 - 18 - 21 - 30
----------------------------
= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 - 9 - 21 - 15
----------------------------
Required number = (1260 ÷ 2) | 2 - 3 - 7 - 5

= 630.

Question 11
The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:
A
12
B
68
C
48
D
78
Question 11 Explanation: 
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.

Question 12
The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:
A
1008
B
1032
C
1015
D
1022
Question 12 Explanation: 
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7

= 1008 + 7

= 1015

Question 13
252 can be expressed as a product of primes as:
A
2 x 3 x 3 x 3 x 7
B
3 x 3 x 3 x 3 x 7
C
2 x 2 x 2 x 3 x 7
D
2 x 2 x 3 x 3 x 7
Question 13 Explanation: 
Clearly, 252 = 2 x 2 x 3 x 3 x 7.
Question 14
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
45 minutes
C
46 minutes and 12 seconds
D
42 minutes and 36 second
Question 14 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.

Question 15
The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leave no remainder, is:
A
1677
B
2523
C
1683
D
3363
Question 15 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 least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
A
13
B
33
C
23
D
43
Question 16 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 17
The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
A
114
B
294
C
364
D
364
Question 17 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 18
The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
A
123
B
127
C
235
D
305
Question 18 Explanation: 
Required number = H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.
Question 19
Which of the following has the most number of divisors?
A
99
B
101
C
176
D
182
Question 19 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 20
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
Question 20 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.
Question 21
The H.C.F.

9 , 12 , 18 and 21 is:
10 25 35 40

 

A
3/5
B
252/5
C
3/1400
D
63/700
Question 21 Explanation: 
Required H.C.F. = H.C.F. of 9, 12, 18, 21 = 3
L.C.M. of 10, 25, 35, 40 1400
Question 22
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
Question 22 Explanation: 
Let the numbers be a and b.
Then, a + b = 55 and ab = 5 x 120 = 600.
The required sum = 1/a+1/b = a+b/ab == 55/600 = 11/200
Question 23
The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is:
A
15 cm
B
25 cm
C
35 cm
D
42 cm
Question 23 Explanation: 
Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.
Question 24
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
Question 24 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 25
Find the highest common factor of 36 and 84.
A
4
B
6
C
12
D
18
Question 25 Explanation: 
36 = 22 x 32
84 = 22 x 3 x 7
H.C.F. = 22 x 3 = 12.
Question 26
Which of the following fraction is the largest ?
A
7/8
B
13/16
C
31/40
D
63/80
Question 26 Explanation: 
L.C.M. of 8, 16, 40 and 80 = 80.
7/8 = 70/80 ; 13/16 = 65/80 ; 31/40 = 62/80

Since,70/80>65/80>63/80>62/80>, so 7/8,>13/16>63/80>31/40
So, 7/8 is the largest.

Question 27
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
Question 27 Explanation: 
Required number = (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8
= 548.
Question 28
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
Question 28 Explanation: 
Other number = (11 x 7700/275) = 308.
Question 29
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
Question 29 Explanation: 
L.C.M. of 12, 18, 21 30 2 | 12 - 18 - 21 - 30

= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 - 9 - 21 - 15
----------------------------
Required number = (1260 ÷ 2) | 2 - 3 - 7 - 5

= 630.

Question 30
The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:
A
12
B
16
C
24
D
48
Question 30 Explanation: 
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.
Question 31
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
Question 31 Explanation: 
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015
Question 32
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
Question 32 Explanation: 
Clearly, 252 = 2 x 2 x 3 x 3 x 7.
Question 33
Find the lowest common multiple of 24, 36 and 40.
A
120
B
240
C
360
D
480
Question 33 Explanation: 
2 | 24 - 36 - 40
--------------------
2 | 12 - 18 - 20
--------------------
2 | 6 - 9 - 10
-------------------
3 | 3 - 9 - 5
-------------------
| 1 - 3 - 5

L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.

Question 34
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
Question 34 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 35
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
Question 35 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 36
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
Question 36 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.
Question 37
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
Question 37 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 38
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
Question 38 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 39
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
Question 39 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 40
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
Question 40 Explanation: 
Clearly, the numbers are (23 x 13) and (23 x 14).
Larger number = (23 x 14) = 322.
Question 41
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?
A
4
B
10
C
15
D
16
Question 41 Explanation: 
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together 30/2 + 1 = 16 times.
Question 42
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
Question 42 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 43
The least number which when divided by 16, 18 and 21, leave the remainder 3, 5 and 8 respectively is:
A
982
B
893
C
1024
D
995
Question 43 Explanation: 
16----------3
18----------5 CD = 13
21----------8

LCM od 16,18,21 = 1008

1008-13 = 995

Question 44
Four bells begin to toll together respectively at the intervals of 8, 10, 12 and 16 seconds. After how many seconds will they toll together again?
A
246 seconds
B
242 seconds
C
240 seconds
D
243 seconds
Question 44 Explanation: 
LCM = 240
Question 45
A heap of coconuts is divided into groups of 2, 3 and 5 and each time one coconut is left over. The least number of Coconuts in the heap is?
A
31
B
41
C
51
D
61
Question 45 Explanation: 
LCM = 30 => 30 + 1 = 31
Question 46
Five bells first begin to toll together and then at intervals of 5, 10, 15, 20 and 25 seconds respectively. After what interval of time will they toll again together? s?
A
5 min
B
5.5 min
C
5.2 min
D
61
Question 46 Explanation: 
LCM = 300/60 = 5 min
Question 47
HCF and LCM two numbers are 12 and 396 respectively. If one of the numbers is 36, then the other number is?
A
36
B
66
C
132
D
264
Question 47 Explanation: 
12 * 396 = 36 * x
x = 132
Question 48
The L.C.M of two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 10, then their difference is:
A
10
B
46
C
70
D
90
Question 48 Explanation: 
Let the numbers be x and (100 - x).
Then, x(100 - x) = 5 * 495
x2 - 100x + 2475 = 0
(x - 55)(x - 45) = 0
x = 55 or 45
The numbers are 45 and 55.
Required difference = 55 - 45 = 10.
Question 49
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 reciprocal of the numbers is equal to:
A
55/601
B
601/55
C
11/120
D
120/11
Question 49 Explanation: 
Let the numbers be a and b.
Then, a + b = 55 and ab = 5 * 120 = 600.
Required sum = 1/a + 1/b = (a + b)/ab = 55/600 = 11/120.

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
Get Results


There are 49 questions to complete.


More Aptitude Question

RELATED ARTICLESMORE FROM AUTHOR

LEAVE A REPLY

Please enter your comment!
Please enter your name here