Best LCM and HCF Quizzes - Free job deliver

Before attempting the LCM and HCF quizzes, take some time to read and learn the important rules and concepts on the subject.

Here are some important rules and concepts related to LCM and HCF that you should know:

Rules for Least Common Multiple (LCM)

LCM Definition: The LCM is the smallest shared multiple of two or more numbers.
- Example: LCM of 4 and 6 is 12 because 12 is the smallest number divisible by both 4 and 6.
Prime Factorization Method: Find prime factors of numbers and use the highest powers to get LCM.
- Example: For 8 and 10, prime factors are 2x2x2 and 2×5. Highest powers give LCM = 2x2x2x5 = 40.
Multiplication-Division Method: LCM(a, b) = (a * b) / HCF(a, b).
- Example: LCM of 5 and 7 is (5 * 7) / 1 = 35.
LCM of Fractions: LCM of denominators for fractions.
- Example: LCM of 1/3 and 1/5 is LCM of 3 and 5, which is 15. So, the common denominator is 15.

Rules for Highest Common Factor (HCF) / Greatest Common Divisor (GCD)

HCF Definition: The HCF is the largest shared factor of two or more numbers.
- Example: HCF of 12 and 18 is 6, as it’s the largest number dividing both 12 and 18.
Prime Factorization Method: Find prime factors and use the lowest powers for HCF.
- Example: For 24 and 36, prime factors are 2x2x2x3 and 2x2x3x3. The lowest powers give HCF = 2x2x3 = 12.
Division Method: Subtract smaller from larger until equal, the remaining number is HCF.
- Example: HCF of 48 and 60: 60 – 48 = 12. Then, 48 – 12 = 36. Finally, 36 – 12 = 24. So, HCF is 12.
HCF and LCM Relationship: HCF(a, b) * LCM(a, b) = a * b.
- Example: HCF of 15 and 20 is 5. LCM of 15 and 20 is 60. And, 5 * 60 = 15 * 20 = 300.
HCF of Fractions: Simplify fractions by canceling common factors in the numerator and denominator.
- Example: Simplifying 4/8, the common factor is 4. Divide both by 4 to get 1/2.

LCM and HCF Quizzes

LCM & HCF 1

1 / 10

The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:

(A) 16

(B) 24

(D) 20

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.

2 / 10

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

(A) 1009

(B) 1015

(D) 1002

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

= 1008 + 7

= 1015

3 / 10

Let x be the smallest number, which when added to 2000 makes the resulting number divisible by 12, 16, 18, and 21. The sum of the digits of x is

1) 7

2) 5

3) 6

4) 4

LCM of 12, 16, 18 and 21 is 1008.
Then multiple of 1008 = 2 × 1008 = 2016
Required number = 2016 – 2000 = 16
Sum of the digits = 1 + 6 = 7

4 / 10

The number between 3000 and 4000 which is exactly divisible by 30, 36 and 80 is

1) 3525

2) 3150

3) 3400

4) 3600

The LCM of 30, 36 and 80 is = 720.
Required number = Multiple of 720 = 720 × 5 = 3600 Since 3000 < 3600 < 4000

5 / 10

The LCM of the two numbers is 44 times of their HCF. The sum of the LCM and HCF is 1125. If one number is 25, then the other number is

1) 1100

2) 925

3) 905

4) 800

Let HCF be H then LCM = 44H
44H + H = 1125 ⇒ H = 25
Therefore, LCM = 44 × 25 = 1100
LCM x HCM = First number x Second number
1100 x 25 = 25 x Second number
Second number = 1100

6 / 10

Which least number should be subtracted from the number 36798 so that the new number is completely divisible by 78?

1) 58

2) 60

3) 68

4) 68

Divide 36798 by 78, remainder is 60.

7 / 10

The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is:

25 cm

45 cm

35 cm

42 cm

Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.

8 / 10

Find the highest common factor of 36 and 84.

(A) 4

(B) 8

(D) 12

36 = 2² x 3²

84 = 2² x 3 x 7

H.C.F. = 2² x 3 = 12.

9 / 10

Which of the following fractions is the largest?

(A) 7/80

(B) 13/16

(D) 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

10 / 10

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

(A) 504

(B) 538

(D) 548

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

= 540 + 8

= 548.

Your score is

The average score is 0%

LCM & HCF 2

1 / 10

The LCM of two numbers is 48. The numbers are in the ratio 2 : 3. The sum of the numbers is

1) 28

2) 32

3) 40

4) 64

Assume that the numbers are 2x and 3x.
LCM = 2 × 3 × x = 48
i.e., 6x = 48
i.e., x = 8
Therefore, the sum of the numbers = 2 ×8 + 3 × 8 = 16 + 24 = 40

2 / 10

The ratio of two numbers is 4 : 5 and their LCM is 120. The numbers are

1) 30 and 40

2) 40 and 32

3) 24 and 30

4) 36 and 20

Assume that, the numbers are 4x and 5x respectively.
LCM = 4 × 5 × x = 120
i.e., 20x = 120
i.e., x = 6
Number is 24 and 30

3 / 10

If the ratio of two numbers is 2 : 3 and their LCM is 54, then the sum of the two numbers is

1) 5

2) 15

3) 45

4) 270

Assume that, the numbers are 2x and 3x respectively.
LCM = 2 × 3 × x = 54
i.e., 6 x = 54

i.e., x = 9
Therefore, the numbers are 2 × 9 = 18 and3 × 9 = 27.
And sum of the numbers is 18 + 27 = 45.

4 / 10

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

(D) 33

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.

5 / 10

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

(D) 3363

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.

6 / 10

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 ?

26 minutes and 18 seconds

42 minutes and 36 seconds

45 minutes

46 minutes and 12 seconds

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.

7 / 10

The number between 4000 and 5000 that is divisible by each of 12, 18, 21 and 32 is

1) 4023

2) 4032

3) 4302

4) 4203

LCM of 12, 18, 21 and 32 is 2016.
Then, required number = 2016 × 2 = 4032

8 / 10

A number between 1000 and 2000 which when divided by 30, 36, and 80 gives a remainder of 11 in each case is

1) 1451

2) 1641

3) 1712

4) 1523

LCM of 30, 36, 80 is 720.
Required number = 2 × 720 + 11 = 1451

9 / 10

The greatest four-digit number which is exactly divisible by each one of the numbers 12, 18, 21, and 28 is

1) 9828

2) 9288

3) 9882

4) 9928

LCM of 12, 18, 21 and 28 is 252. If we divide the largest 4-digit number 9999 by 252, it gives a remainder 171.
Then, required number = 9999 – 171 = 9828

10 / 10

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) 308

(B) 312

(D) 310

Other number =(11 x 7700)/275= 308.275

Your score is

The average score is 0%

LCM & HCF 3

1 / 10

If x : y be the ratio of two whole numbers and z be their HCF, then the LCM of those two numbers is

1) yz

2) xz/y

3) xy/z

4) xyz

numbers are zx and zy.
Product of two numbers = LCM × HCF
i.e., zx × zy = z × LCM
i.e., LCM = xyz

2 / 10

Three numbers are in the ratio 1 : 2 : 3 and their HCF is 12. The numbers are

1) 12, 24, 36

2) 5, 10, 15

3) 4, 8, 12

4) 10, 20, 30

Assume that the numbers are x, 2x and 3x.
Since, HCF = 12 then x = 12
Therefore, the numbers are 12, 2 × 12 = 24 and 3 × 12 = 36

3 / 10

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

(D) 16

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	+ 1 = 16 times.
	2

4 / 10

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

(A) 9000

(B) 9400

(D) 9800

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.

5 / 10

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

(D) 185

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.

6 / 10

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

(D) 200

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.

7 / 10

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

(A) 1

(B) 2

(D) 4

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.

8 / 10

Find the lowest common multiple of 24, 36 and 40.

(A) 120

(B) 240

(D) 480

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

9 / 10

The ratio of two numbers is 3 : 4 and their HCF is 5. Their LCM is

1) 10

2) 60

3) 15

4) 12

Assume that the numbers are 3x and 4x.
Since, HCF = 5 then x = 5
Therefore, the numbers are 3 × 5 = 15 and 4 × 5 = 20
The LCM of 15 and 20 is 60.

10 / 10

Two numbers are in the ratio 3 : 4. If their HCF is 4, then their LCM is

1) 48

2) 42

3) 36

4) 24

Assume that, the numbers are 3x and 4x respectively.
Since, HCF = 4 then x = 4
Therefore, the numbers are 3 × 4 = 12 and 4 × 4 = 16.
LCM of 12 and 16 is 48

Your score is

The average score is 0%

LCM & HCF 4

1 / 10

The LCM of two numbers is 48. The numbers are in the ratio 2 : 3. The sum of the numbers is

1) 28

2) 32

3) 40

4) 64

Assume that the numbers are 2x and 3x.
LCM = 2 × 3 × x = 48
i.e., 6x = 48
i.e., x = 8
Therefore, the sum of the numbers = 2 ×8 + 3 × 8 = 16 + 24 = 40

2 / 10

The ratio of two numbers is 4 : 5 and their LCM is 120. The numbers are

1) 30 and 40

2) 40 and 32

3) 24 and 30

4) 36 and 20

Assume that, the numbers are 4x and 5x respectively.
LCM = 4 × 5 × x = 120
i.e., 20x = 120
i.e., x = 6
Number is 24 and 30

3 / 10

If the ratio of two numbers is 2 : 3 and their LCM is 54, then the sum of the two numbers is

1) 5

2) 15

3) 45

4) 270

Assume that, the numbers are 2x and 3x respectively.
LCM = 2 × 3 × x = 54
i.e., 6 x = 54

i.e., x = 9
Therefore, the numbers are 2 × 9 = 18 and3 × 9 = 27.
And sum of the numbers is 18 + 27 = 45.

4 / 10

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

(D) 33

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.

5 / 10

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

(D) 3363

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.

6 / 10

26 minutes and 18 seconds

42 minutes and 36 seconds

45 minutes

46 minutes and 12 seconds

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.

7 / 10

The number between 4000 and 5000 that is divisible by each of 12, 18, 21 and 32 is

1) 4023

2) 4032

3) 4302

4) 4203

LCM of 12, 18, 21 and 32 is 2016.
Then, required number = 2016 × 2 = 4032

8 / 10

A number between 1000 and 2000 which when divided by 30, 36, and 80 gives a remainder of 11 in each case is

1) 1451

2) 1641

3) 1712

4) 1523

LCM of 30, 36, 80 is 720.
Required number = 2 × 720 + 11 = 1451

9 / 10

The greatest four-digit number which is exactly divisible by each one of the numbers 12, 18, 21, and 28 is

1) 9828

2) 9288

3) 9882

4) 9928

LCM of 12, 18, 21 and 28 is 252. If we divide the largest 4-digit number 9999 by 252, it gives a remainder 171.
Then, required number = 9999 – 171 = 9828

10 / 10

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) 308

(B) 312

(D) 310

Other number =(11 x 7700)/275= 308.275

Your score is

The average score is 0%

LCM & HCF 5

1 / 10

The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:

(A) 16

(B) 24

(D) 20

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.

2 / 10

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

(A) 1009

(B) 1015

(D) 1002

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

= 1008 + 7

= 1015

3 / 10

Let x be the smallest number, which when added to 2000 makes the resulting number divisible by 12, 16, 18, and 21. The sum of the digits of x is

1) 7

2) 5

3) 6

4) 4

LCM of 12, 16, 18 and 21 is 1008.
Then multiple of 1008 = 2 × 1008 = 2016
Required number = 2016 – 2000 = 16
Sum of the digits = 1 + 6 = 7

4 / 10

The number between 3000 and 4000 which is exactly divisible by 30, 36 and 80 is

1) 3525

2) 3150

3) 3400

4) 3600

The LCM of 30, 36 and 80 is = 720.
Required number = Multiple of 720 = 720 × 5 = 3600 Since 3000 < 3600 < 4000

5 / 10

The LCM of the two numbers is 44 times of their HCF. The sum of the LCM and HCF is 1125. If one number is 25, then the other number is

1) 1100

2) 925

3) 905

4) 800

Let HCF be H then LCM = 44H
44H + H = 1125 ⇒ H = 25
Therefore, LCM = 44 × 25 = 1100
LCM x HCM = First number x Second number
1100 x 25 = 25 x Second number
Second number = 1100

6 / 10

Which least number should be subtracted from the number 36798 so that the new number is completely divisible by 78?

1) 58

2) 60

3) 68

4) 68

Divide 36798 by 78, remainder is 60.

7 / 10

The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is:

25 cm

45 cm

35 cm

42 cm

Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.

8 / 10

Find the highest common factor of 36 and 84.

(A) 4

(B) 8

(D) 12

36 = 2² x 3²

84 = 2² x 3 x 7

H.C.F. = 2² x 3 = 12.

9 / 10

Which of the following fractions is the largest?

(A) 7/80

(B) 13/16

(D) 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

10 / 10

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

(A) 504

(B) 538

(D) 548

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

= 540 + 8

= 548.

Your score is

The average score is 0%

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