Practice MCQ Questions and Answer on Problems on H.C.F and L.C.M
181.
The greatest number which when divides 989 and 1327 leave remainder 5 and 7 respectively = ?
(A) 8
(B) 16
(C) 24
(D) 32
Solution:
989 - 5 = 984 1327 - 7 = 1320 Subtract the remainder from the number. HCF = (984, 1320) = 24 for greatest number take HCF of the numbers.
182.
The product of two numbers is 396 ÃÂÃÂÃÂÃÂÃÂÃÂ 576 and their LCM is 6336. Find their HCF = ?
(A) 36
(B) 34
(C) 63
(D) 43
Solution:
We, know that (a × b) = (HCF and LCM) 396 × 576 = HCF × 6336 HCF = 36
183.
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 again at the starting point ?
(A) 26 minutes and 18 seconds
(B) 42 minutes and 36 seconds
(C) 45 minutes
(D) 46 minutes and 12 seconds
Solution:
L.C.M. of 252, 308 and 198 = 2772 So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.
184.
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) 1683
(C) 2523
(D) 3363
Solution:
LCM of 5, 6, 7, 8 = 840 ∴ Required number is of the from 840k + 3 Least value of k for which (840k + 3) is divisible by 9 is k = 2 ∴ Required number = (840 × 2 + 3) = 1683
185.
The LCM of two numbers is 20 times their HCF. The sum of HCF ans LCM is 2520. If one of the number 480, the other number is = ?
The smallest number, which when divided by 5, 10, 12 and 15, leaves remainder 2 in each case , but when divided by 7 leaves no remainder , is = ?
(A) 189
(B) 182
(C) 175
(D) 91
Solution:
LCM of (5, 10, 12, 15) = 5 × 2 × 6 = 60 smallest number divided by (5, 10, 12, 15) leaves remainder 2 and when divided by 7 leaves no remainder is $$\eqalign{ & {\text{ = }}\frac{{60{\text{K}} + 2}}{7} = \frac{{4{\text{K}} + 2}}{7} \cr & {\text{At K = 3, }}\frac{{4{\text{K}} + 2}}{7} \cr & \Rightarrow {\text{remainder = 0}} \cr & \therefore {\text{number = 60K + 2 }} \cr & {\text{ = 60}} \times {\text{3 + 2 }} \cr & {\text{ = 182}} \cr} $$
187.
21 mango trees, 42 apple trees and 56 orange trees have to be planted in rows such that each row contains the same numbers of trees of one variety only. Minimum number of row in which the trees may be planted is = ?
(A) 3
(B) 15
(C) 17
(D) 20
Solution:
For the minimum number of rows, the numbers of trees in each row must be the maximum. Number of trees in each row = HCF of 21, 42, 56 = 7 Hence number of rows $$\eqalign{ & {\text{ = }}\left( {\frac{{21 + 42 + 56}}{7}} \right) \cr & = \frac{{119}}{7} \cr & = 17 \cr} $$