# 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.

### 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 - The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is?

### Answer : B

### Explanation

Required number = H.C.F of (1657 - 6) = (2037 - 5) = H.C.F of 1651 and 2032 = 127.

Q 3 - 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?

### Answer : C

### Explanation

L.C.M of 5,6,7,8 = 840 Therefore Required number is of the form (840k + 3) Least value of k which (840k + 3) is divisible by 9 is k = 2. Therefore Required number = (840 x 2 + 3) = 1683

Q 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?

### Answer : B

### 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, the 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.

Q 5 - L.C.M of two prime numbers a and b(a>b) is 161. The value of 3b - a is?

### Answer : B

### Explanation

H.C.F of two prime numbers is 1. Product of numbers = (1 x 161) = 161 Let the numbers be a and b. Then a x b = 161 now co-primes with product 161 are (1, 161) (7,23). Since a and b are prime numbers and a > b, we have a = 23 and b = 7. Therefore 3b - a = (3 x 7) - 23 = -2.

Q 6 - Find the greatest number which can divide 103 and 199 leaving the same remainder 7 in each case.

### Answer : D

### Explanation

Required number = H.C.F. of [(103-7) and (199-7)] = H.C.F. of (96 and 192) = H.C.F. of (25*3 and 26*3) = H.C.F. of (25*3 and 25*3*2) = 25*3=96

Q 7 - The ratio between the LCM and HCF of two numbers is 12:1. If the numbers be 48 and 64, find their LCM and HCF.

### Answer : A

### Explanation

Let the LCM & HCF be 12X & X respectively LCM x HCF =12X x X=48x64 ⇒X2= (48x64)/12⇒X=16 ⇒HCF=16 & LCM=192

Q 8 - The L.C.M of two numbers is 12 times their H.C.F and sum of H.C.F and L.C.M is 403. Having one number as 93, find the other one.

### Answer : B

### Explanation

Let L.C.M =x and H.C.F = y . Then, X= 12y and x+y = 403 ⇒12y+y = 403 ⇒13y= 403 ⇒y = 403/13 = 31 ∴ x= (12*31 ) = 372 Let the other number be z. then, 372*31 = 93*z ⇒ z = (372*31)/93 =124 Hence , the other number is 124 .

Q 9 - If ratio of two numbers is 2:3 and the product of their H.C.F and L.C.M is 33750, find the sum of the numbers:

### Answer : C

### Explanation

Let the numbers be 2x and 3x. Then, H.C.F = x and L.C.M = 6x ∴ 6x*x = 33750 ⇒ x^{2}= 5625 ⇒ &sqrt;5625= 75 So, the numbers are 150 and 225 . Their sum is 375.

Q 10 - What is the greatest length possible of a scale that can be used to measure exactly 3m , 5m 10 cm and 12m 90 cm length?

### Answer : D

### Explanation

Required length = H.C.F of 300 cm, 510 cm, 1290 cm =30 cm