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# MATHS SOLVED IN STEPS: HCM & LCM - 05

36. What is the least number which when divided by 8, 12, 15 and 20 leaves in
each case a remainder of 5 ?
A. 125          B. 117        C. 132       D. 112
Explanation :
LCM of 8, 12, 15 and 20 = 120
Required Number = 120 + 5 = 125

37. Three numbers which are co-prime to each other are such that the product of the first two is 119 and that of the last two is 391. What is the sum of the three numbers?
A. 47                B. 43         C. 53            D. 51
Explanation :
Since the numbers are co-prime, their HCF = 1
Product of first two numbers = 119
Product of last two numbers = 391
The middle number is common in both of these products.
Hence if we take HCF of 119 and 391, we get the common middle number
HCF of 119 and 391 = 17
=> Middle Number = 17
First Number = 119?17 = 7
Last Number = 391?17 = 23
Sum of the three numbers = 7+17+23 = 47

38. What is the greatest number which divides 24, 28 and 34 and leaves the same remainder in each case?
A. 1                 B. 2                  C. 3                 D. 4
Explanation :
If the remainder is same in each case and remainder is not given, HCF of the
differences of the numbers is the required greatest number
34 - 24 = 10
34 - 28 = 6
28 - 24 = 4
Hence, the greatest number which divides 24, 28 and 34 and gives the same remainder
= HCF of 10, 6, 4
= 2

39. Six bells start ringing together and ring at intervals of 4, 8, 10, 12, 15 and 20 seconds respectively. how many times will they ring together in 60 minutes ?
A. 31                         B. 15                        C. 16                   D. 30
Explanation :
LCM of 4, 8, 10, 12, 15 and 20 = 120
120 seconds = 2 minutes
Hence all the six bells will ring together in every 2 minutes
Hence, number of times they will ring together in 60 minutes
=1+(60÷2)=31

40. The HCF of two numbers is 23 and the other two factors of their LCM are 13 and 14. What is the largest number?
A. 312                     B. 282                   C. 299                    D. 322
Explanation :
The HCF of a group of numbers will be always a factor of their LCM
HCF is the product of all common prime factors using the least power of each
common prime factor.
LCM is the product of highest powers of all prime factors
HCF of the two numbers = 23
=> Highest Common Factor in the numbers = 23
Since HCF will be always a factor of LCM,
23 is a factor of the LCM.
Other two factors in the LCM are 13 and 14.
Hence factors of the LCM are 23, 13, 14
So, numbers can be taken as (23 × 13) and (23 × 14) = 299 and 322
Hence, largest number = 322

41. What is the smallest number which when diminished by 12, is divisible 8, 12, 22 and 24?
A. 276                   B. 264             C. 272            D. 268
Explanation :
Required Number = (LCM of 8, 12, 22 and 24) + 12 = 264 + 12 = 276
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LDC/ LGS/ VEO Questions and Answers