# 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 - H.C.F of 4 x 27 x 3125, 8 x 9 x 25 x 7 & 16 x 81 x 5 x 11 49 is?

### Answer : C

### Explanation

4 x 27 x 3125 = 2^{2}x 3^{3}x 5^{5}; 8 x 9 x 25 x 7 = 2^{3}x 3^{2}x 5^{2}x 7; 16 x 81 x 5 x 11 x 49 = 2^{4}x 3^{4}x 5 x 7^{2}x 11. Therefore H.C.F = 180.

Q 2 - H.C.F of 3240, 3600 and a third number is 36 and their L.C.M is 2^{4} x 3 ^{5} x 5^{2} x 7^{2}?

### Answer : B

### Explanation

3240 = 2^{3}x 3^{2}x 5; 3600 = 2^{4}x 3^{2}5^{2}H.C.F = 36 = 2^{2}x 3^{2}. Since H.C.F is the product of the lowest powers of common factors so the the third number must have 2^{2}x 3^{2}as its factor Since L.C.M is the product of highest powers of common prime factors so the third number must have 3^{5}x 7^{2}as its factor Therefore third number = 2^{2}x 3^{5}x 7^{2}

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 H.C.F and L.C.M of the two numbers are 84 and 21 respectively. If the ratio of the two numbers is 1:4, then the larger number of the twi is?

### Answer : A

### Explanation

Let the number be z and 4z. Then z x 4z = 84 x 21 = z^{2}=^{84 x 21}⁄_{4}= z = 21 Hence larger number is 4z = 4 x 21 = 84

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.

### Answer : B

### Explanation

H.C.F. of given fractions = (H.C.F.of denominators)/(L.C.M.of numerators)=2/105 ( H.C.F.of numerators = 2 L.C.M.of denominators =105)

Q 7 - Three number are in the ratio of 5: 6: 7 and their L.C.M. is 2100. Their H.C.F. is:

### Answer : D

### Explanation

Let the numbers be 5X, 6X and 7X Then, their L.C.M. = 210X. So, 210X = 2100 or X = 10. The numbers are (5 x 10), (6 x 10) and (7 x 10).Hence, required H.C.F. = 10.

Q 8 - If two numbers are greater than 13 and the H.C.F of two numbers be 13, L.C.M 273, then the sum of the numbers is:

### Answer : B

### Explanation

Let the number be 13 a and 13 b, where a and b are co-primes. Then, 13a * 13b= (13* 273) ⇒ab= 21 Two co-primes with product 21are 3 and 7. ∴ numbers are (13*3, 13*7) i.e , 39 and 91. Their sum = (39+91) = 130

Q 9 - The H.C.F of three numbers is 12 and they are in the ratio of 1:2:3, find the numbers.

### Answer : C

### Explanation

Let the numbers be a, 2a , 3a . Then , their H.C.F is a. ∴ a =12 and hence , the numbers are 12, 24, 36.

Q 10 - If the product of two co-primes is 117 then their L.C.F should be

### Answer : B

### Explanation

H.C.F of co-primes=1 H.C.F * L.C.M = Their product = 117 ∴ 1* L.C.M = 117 ⇒L.C.M = 117