HCF and LCM – Railway Maths Part III

Total Questions: 52

31. Let n be the least number which when divided by 15, 18, 20 and 27, the remainder in each case is 3 and n is divisible by 53. What is the sum of the digits of n? [Group D 07/10/2022 (Afternoon)]

Correct Answer: (b) 12
Solution:

LCM of 15, 18, 20 and 27 = 540
So, number = (540 × x) + 3 / 53
Put the value of x = 5
number = (540 × 5) + 3 / 53 = 2703 / 53 = 51
So sum of the digits of number (2703) = 2 + 7 + 0 + 3 = 12

32. Two numbers, whose LCM is 416 and HCF is 2, are in the ratio 13 : 16. Find the greater of these two numbers. [Group D 07/10/2022 (Afternoon)]

Correct Answer: (a) 32
Solution:

LCM = 416 and HCF = 2
Ratio = 13 : 16
According to question,
13 × 16 × k × k = 416 ⇒ k = 2
Greater number = 16k = 16 × 2 = 32

33. The HCF of two numbers is 47. If the product of the two numbers is 6627, then the larger number between these two numbers is: [Group D 07/10/2022 (Evening)]

Correct Answer: (c) 141
Solution:

Let the simplest ratio of numbers be a : b
HCF = 47
Numbers = 47a, 47b
Product = 6627
47a × 47b = 662 ⇒ ab = 3
Possible pair = (1 : 3)
So, larger number = 47 × 3 = 141

34. If the HCF of 408 and 1032 is expressed as 516m − 252 × 4, then m is: [Group D 11/10/2022 (Morning)]

Correct Answer: (d) 2
Solution:

If the HCF of 408 and 1032 is expressed as 516m − 252 × 4, then m is:
HCF of 408 and 1032 = 24
516m − 252 × 4 = 24
516m − 1008 = 24
516m = 1032 ⇒ m = 2

35. If a and b are two co-prime numbers, then their LCM is: [Group D 11/10/2022 (Morning)]

Correct Answer: (b) a × b 
Solution:

If a and b are co-prime then LCM = a × b

36. A tailor has 22 metres of cloth and he has to cut each metre of cloth into 5 pieces. How many pieces can he get from the 22 metres of cloth he has? [Group D 11/10/2022 (Afternoon)]

Correct Answer: (c) 110
Solution:

Given, tailor has 22 m cloth
As per question 1 meter cloth has 5 pieces
So 22 meter cloth has = 5 × 22 = 110 pieces.

37. Find the smallest number, which when tripled, will leave the remainder as zero on being divided by 5, 15 and 35. [Group D 11/10/2022 (Afternoon)]

Correct Answer: (c) 35 
Solution:

LCM of 5, 15 and 35 = 105
So, number will be 105k
Possible value of k = 1, 2, 3,...
Possible numbers = 105, 210, 315, 420, …
Tripled the 35 = 35 × 3 = 105
When 105 is divided by 35 then remainder = 0
So, smallest number = 35

38. While packing birthday caps for a party in packs of 8 or 10, one cap was always left out. How many caps were there if there were more than 250 but less than 300 caps in the lot? [Group D 11/10/2022 (Evening)]

Correct Answer: (c) 281 
Solution:

L.C.M. of 8 and 10 = 40
Min. number of cap for required arrangements = 40 + 1 = 41
But the number of caps lies between 250 and 300.
40 × 7 + 1 = 281

39. The HCF of 1638, 2244 and 5049 is x. Then sum of digits of x is: [Level 4 (10/05/2022) Shift 1]

Correct Answer: (d) 12
Solution:

1683, 2244, 5049
By difference method:
we take difference of these numbers
2244 − 1683 = 561
5049 − 2244 = 2805 = 561 × 5
So 561 is highest common factor of 561 and 2805

40. k is the greatest number which, when divides 2996, 4752 and 7825, the remainder in each case is the same. The product of the digit of k is [Level 4 (10/05/2022) Shift 1]

Correct Answer: (d) 108
Solution:

By difference method
4756 − 2996 = 1756 = 439 × 4
7825 − 4752 = 3073 = 439 × 7
here highest common factor is 439
4 × 3 × 9 = 108