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H.C.F & L.C.M. - Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to H.C.F & L.C.M.. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Q 1 - Find L.C.M and H.C.F of 0.63, 1.05 and 2.1.

A - H.C.F 0.26 and L.C.M 6.35

B - H.C.F 0.25 and L.C.M 6.00

C - H.C.F 0.23 and L.C.M 7.30

D - H.C.F 0.21 and L.C.M 6.30

Answer : D

Explanation

Making the same number of decimal places, the given numbers are 0.63, 1.05 and 2.10.
without decimal places, these numbers are 63, 105, and 210
Now, H.C.F of 63, 105 and 210 is 21 so H.C.F of 0.63, 1.05 and 2.1 is 0.21 
 L.C.M of 63, 105 and 210 is 630 so L.C.M of 0.63, 1.05 and 2.1 is 6.30

Q 2 - What will be the least number which when doubled will be exactly divisible by 12, 18,21 and 30?

Answer : A

Explanation

L.C.M of 12, 18, 21, 30 = 2 x 3 x 2 x 3 x 7 x 5 = 1260.
Required number =(¹²⁶⁰⁄₂) = 630

Q 3 - The least multiple of 7, which leaves a remainder 4, when divided by 6,9,15 and 18 is?

Answer : A

Explanation

L.C.M  of 6,9,15 and 18 is 90.
Let required number be 90k + 4, which is a multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4. 
 Therefore Required number 90 x 4 + 4 = 364.

Q 4 - The product of two numbers is 4107. If the H.C.F of these numbers is 37, then the greater number is?

Answer : B

Explanation

Let the numbers be 37a and 37b. then 37a x 37b = 4107
a x b = 3.
Now, co-primes with product 3 are (1,3)
So, the required numbers are (37 x 1, 37 x 3) i.e. (1, 111)
Therefore greater number = 111.

Q 5 - The maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and pencils?

Answer : B

Explanation

Required number of students = H.C.F of 1001 and 910 = 91.

Q 6 - The number of prime factors in the expression 2⁵x2⁶x3⁶x5³

Answer : A

Explanation

2, 3, 5 are the prime numbers and the given expression is 2⁵x2⁶x3⁶x5³ (2⁵x6⁶x5³).
So the total would be the sum of powers in the expression that is 5+6+6+3=20

Q 7 - A milk vendor has three kinds of milk: 68 litres, 119 litres and 153 litres. Find the least number of casks of equal size required to store all the milk without mixing.

Answer : B

Explanation

Size of the cask (G.C.D. of 68,119 and 153) =17
Number of casks =68/17+119/17+153/17=20

Q 8 - The H.C.F and L.C.M of two numbers is 11 and 7700. If one of the numbers is 275, find the other number:

Answer : C

Explanation

Let the other number be x. than ,
11 * 7700 =  275 * X ⇒ x = (11* 7700)/275 = 308
Hence,  the number is 308.

Q 9 - H.C.F and L.C.F of two numbers x and y are 3 and 105 respectively. If x+y=36 then figure out 1/x + 1/y.

Answer : D

Explanation

We have X *y =3 *105 ⇒ xy = 315
∴  x+y/xy =36/315  = 4/35 ⇒ 1/y+1/x = 4/35 ⇒ 1/x+1/y = 4/35.

Q 10 - Find the largest divisor of 210, 315, 147 and 168.

Answer : C

Explanation

210  = 2*3*5*7,   315 =3*5*3*7 , 147 = 7*7*3  , 168 = 2*2*2*3*7
Required number = H.C.F of given  numbers=  (3*7) =21

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