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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.
Answer : B
Explanation
99 = 1 x 3 x 3 x 11; 101 = 1 x 101; 176 = 1 x 2 x 2 x 2 x 2 x 11; 182 = 1 x 2 x 7 x 13. So, divisor of 99 are 1, 3, 9, 11, 33, 99 divisor of 101 are 1, 101 divisor of 176 are 1, 2, 4, 8, 16, 22, 44, 88, 176 divisor of 182 are 1, 7, 2, 13, 14, 26, 91, 182 Hence 176 has the most number of divisors.
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 - A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point. After what time will they meet again at the starting point?
Answer : D
Explanation
L.C.M of 252, 308 and 198 = 2772, So A,B,C will again meet at the starting point in 2772 sec. i.e 46 minutes 12 seconds
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 - Two numbers, both greater than 29, have H.C.F 29 and L.C.M 4147. the sum of the numbers is?
Answer : B
Explanation
Product of numbers = 29 x 4147. let the numbers be 29a and 29b. then 29a x 29b = (24 x 4147) = ab = 143 Now co-primes with product 143 are (1, 143) (11, 13). So the numbers are (29 x 1, 29 x 143) and (29 x 11, 29 x 13). Since both numbers are greater than 29 the suitable pair (29 x 11, 29 x 13) i.e (319, 377). Required sum = (319 + 377) = 696.
Q 6 - The L.C.M. of two numbers is 72. Their H.C.F. is 12. If one number is 12, the other is
Answer : D
Explanation
Let the other number be X HCF*LCM=Product of two numbers 12*72=24*X ⇒X=(12*72)/24=36
Q 7 - If the sum of two numbers is 175 and the H.C.F. and L.C.M. of these numbers are 35 and 140 respectively, then the sum of the reciprocals of the numbers is equal to:
Answer : A
Explanation
Let the numbers be X and Y. Then, X+Y = 175 and XY = 35x140 = 4900. The required sum = 1/X+1/Y=(X+Y)/XY=175/4900 = 1/28
Q 8 - Having H.C.F and L.C.M of two numbers as 21 and 4641 respectively. If one of the numbers is in between 200 and 300, then the two numbers are :
Answer : D
Explanation
Let the numbers be 21 a and 21 b, where a and b are co-primes. Then ,21 a* 21 b= (21* 4641)⇒ab= 221. Two co-primes with product 221 are 13 and 17. ∴ Numbers are (21*13, 21*17) , i.e (273,357)
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 - 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