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


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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.

Questions and Answers

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?

A - 123

B - 127

C - 152

D - 125

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?

A - 1677

B - 2535

C - 1683

D - 3363

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?

A - 45

B - 85

C - 60

D - 65

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 = 55129 = 19; Third number = 107329 = 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?

A - -1

B - -2

C - 1

D - 2

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.

A - 211

B - 89

C - 206

D - 96

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.

A - 192, 16

B - 144, 12

C - 120, 10

D - 48, 4

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.

A - 128

B - 124

C - 134

D - None of these

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:

A - 250

B - 325

C - 375

D - 425

Answer : C

Explanation

Let the numbers be 2x and 3x.  Then, H.C.F = x and L.C.M = 6x
∴ 6x*x = 33750  ⇒ x2 = 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?

A - 10 cm

B - 20 cm

C - 25 cm

D - 30 cm

Answer : D

Explanation

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


aptitude_hcf_lcm.htm

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