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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.
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 - Three numbers are in the ratio 1:2:3 and their H.C.F is 12. The numbers are?
Answer : B
Explanation
Let the required numbers be z, 2z, and 3z. Then, their H.C.F = z So, z = 12. Therefore The number are 12, 24 and 36.
Q 3 - The least number which should be added to 2497 so that the sum is exactly divisible by 5,6,4 and 3 is?
Answer : B
Explanation
L.C.M of 5,6,4 and 3 = 60 On dividing 2497 by 60, remainder is 37. Therefore Number to be added = (60 - 37) = 23.
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 least number which when divided separately by 13, 26, 39 and 52 leaves 5 as remainder in each case.
Answer : C
Explanation
Required number = L.C.M. of (13, 26,39 and 52) +5 = 156 + 5 = 161
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 - The sum of two numbers is 45 and difference is 1/9 of their sum. What is their L.C.M:
Answer : A
Explanation
Let the numbers be x and y . Then,
{x+y= 45 , x= (1/9*45)= 5} ⇒x = 25, y =20
L.C.M of 25 and 20 = (5*5*4)= 100
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 ⇒ x2 = 5625 ⇒ &sqrt;5625= 75 So, the numbers are 150 and 225 . Their sum is 375.
Q 10 - Three numbers are in the ratio of 3:4 :5 , their L.C.M is 2400. Their H.C.F is :
Answer : A
Explanation
Let the numbers be 3x, 4x and 5x. Then L.C.M = 60x 60x= 2400 ⇒ x= 40 ∴ The numbers are 120, 160 and 200. Clearly , their H.C.F is 40