HCF and LCM – Railway Maths Part IV

Total Questions: 50

21. What is the smallest number which when increased by 3 is divisible by 27, 35, 25, and 21 ? [RRB NTPC 30/12/2020 (Morning)]

Correct Answer: (d) 4722
Solution:

The smallest number divisible by 27, 35, 25, and 21 is the LCM of these numbers.
LCM = 4725
So, the smallest number which when increased by 3, should give 4725.
i.e. the number will be (4725 – 3) = 4722

22. The sum of two positive numbers is 384 and their HCF is 24. How many pairs of such numbers can be formed? [RRB NTPC 30/12/2020 (Evening)]

Correct Answer: (a) 4
Solution:

We know, if two numbers are ax and bx, then their HCF = x,
Here, sum of the two numbers = 384 and HCF = x = 24
⇒ 24(a + b) = 384 ⇒ a + b = 16
So, the possible values of (a, b) are
(1, 15), (2, 14), (3, 13), (4, 12), (5, 11), (6, 10), (7, 9), (8, 8)
But from the above pairs only for the pairs (1, 15), (3, 13), (5, 11) and (7, 9) the HCF will be 24.
So, four pairs of such numbers can be formed.

23. The HCF of two numbers is 6 and their LCM is 84. If one of these numbers is 42, then the second number is. [RRB NTPC 04/01/2021 (Evening)]

Correct Answer: (c) 12
Solution:

HCF of two numbers = 6
LCM of two numbers = 84
1st number = 42
We have,
LCM × HCF = 1st number × 2nd number
84 × 6 = 42 × 2nd number
∴ 2nd number = (84 × 6) / 42 = 12

24. The sum of two numbers is 288 and their HCF is 16. How many pairs of such numbers can be formed? [RRB NTPC 04/01/2021 (Evening)]

Correct Answer: (a) 3
Solution:

ATQ,
Sum = 288 and HCF = 16
So the numbers can be 16x and 16y,
16x + 16y = 288
⇒ x + y = 18 ……… (where x and y are co-prime)
Numbers can be (1, 17), (5, 13) and (7, 11)
3 pairs are possible, so 3 pairs of such numbers can be formed.

25. Find the sum of the numbers between 400 and 500 such that when 8, 12, and 16 divide them, it leaves 5 as a remainder in each one. [RRB NTPC 04/01/2021 (Evening)]

Correct Answer: (b) 922
Solution:

LCM of (8, 12, 16) = 48
1st number that is multiple of 48 between 400 and 500 is = 432
And the 2nd number = 480
The remainder in each case should be 5
so the required numbers are 437 and 485.
Sum = 437 + 485 = 922

26. The HCF of two numbers is 15 and their LCM is 225. If one of the numbers is 45, then what is the other number? [RRB NTPC 05/01/2021 (Evening)]

Correct Answer: (d) (15 × 225) / 45
Solution:

HCF × LCM = Product of two numbers
15 × 225 = 45 × 2nd Number
∴ 2nd Number = (15 × 225) / 45 = 75

27. Five bells commence tolling together and toll at intervals of 2, 4, 6, 8, 20 seconds respectively. In 30 minutes, how many times do they toll together? [RRB NTPC 05/01/2021 (Evening)]

Correct Answer: (d) 16
Solution:

LCM of (2, 4, 6, 8, 20) = 120 seconds = 2 minutes
No. of times they toll together = (30 / 2) + 1 = 16

28. Since the HCF of (48, 144) is 48 therefore the LCM of (48, 144) = ? [RRB NTPC 07/01/2021 (Evening)]

Correct Answer: (c) 144
Solution:

HCF × LCM = Product of two numbers
⇒ 48 × LCM = 48 × 144
∴ LCM = (48 × 144) / 48 = 144

29. Find the sum of the numbers between 400 and 600 such that when they are divided by 6, 12 and 16, there will be no remainder. [RRB NTPC 08/01/2021 (Evening)]

Correct Answer: (d) 2016
Solution:

LCM of (6, 12, 16) = 48 The number which is divisible by 6, 12, and 16, must be divisible by 48.
The numbers between 400 and 600 which are divisible by 48 are = 432, 480, 528, 576
Sum = 432 + 480 + 528 + 576 = 2016

30. The least number which should be added to 4707 so that the sum is exactly divisible by 4, 5, 6 and 8 is [RRB NTPC 08/01/2021 (Evening)]

Correct Answer: (d) 93
Solution:

LCM of 4, 5, 6, 8 = 120
When we divide 4707 by 120 we get remainder = 27
Required number that should be added = 120 - 27 = 93