H.C.F and L.C.M Problems Questions And Answers.
Exercise ::
H.C.F and L.C.M Problems
1. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
A. 4 B. 7 C. 9 D. 11
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Answer: Option A
Explanation:
Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)
= H.C.F. of 48, 92 and 140 = 4.
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2. The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is?
A. 276 B. 299 C. 322 D. 345
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Answer: Option C
Explanation:
Clearly, the numbers are (23 x 13) and (23 x 14).
Hence: Larger number = (23 x 14) = 322.
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3. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?
A. 4 B. 10 C. 15 D. 16
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Answer: Option D
Explanation:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together
30
+ 1 = 16 times.
2
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4. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is?
A. 4 B. 5 C. 6 D. 8
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Answer: Option A
Explanation:
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
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5. The greatest number of four digits which is divisible by 15, 25, 40 and 75 is?
A. 9000 B. 9400 C. 9600 D. 9800
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Answer: Option C
Explanation:
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Hence: Required number (9999 - 399) = 9600.
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Published by:Michael Daani