HCF and LCM – Railway Maths Part III

Total Questions: 52

11. The product of the HCF and the LCM of 4, 20, 28 is: [Group D 19/09/2022 (Evening)]

Correct Answer: (d) 560
Solution:

LCM of 4, 20 and 28 = 140
HCF of 4, 20 and 28 = 4
Product = 140 × 4 = 560

12. If the HCF of two numbers is 6 and the product of the numbers is 2520, find their LCM. [Group D 20/09/2022 (Evening)]

Correct Answer: (b) 420
Solution:

Product of the two numbers
= L.C.M × H.C.F
⇒ 6 × L.C.M = 2520 ⇒ L.C.M = 420

13. The product of two numbers is 2160 and their HCF is 12. One of the possible pairs of numbers is: [Group D 22/09/2022 (Morning)]

Correct Answer: (d) (36, 60)
Solution:

Let the no. is 12x and 12y
According to the question,
12x × 12y = 2160
xy = 15
Co-prime factor of 15 = 3 and 5
So, the number are 36 and 60.

Short tricks:-
from the given options , only option (d) has the products equal to 2160.
or , the H.C.F. of the given no. in option (d) equals to 12.

14. Find the smallest natural number which, when divided by 9, 24 and 12, leaves a remainder 4 in each case but when divided by 5 leaves no remainder. [Group D 22/09/2022 (Afternoon)]

Correct Answer: (a) 220
Solution:

L.C.M of 9, 24, 12 = 72
Let the number be 72k + 4.
Put k = 3, then the number becomes 220 which is divisible by 5.

15. Three bells A, B and C toll every 15 seconds, 24 seconds, and 42 seconds, respectively. If all of them toll together at 8 a.m., then between 8 a.m. and 10 a.m. on the same day, how many times will they all toll together, including the one at 8 a.m.? [Group D 26/09/2022 (Morning)]

Correct Answer: (c) 9
Solution:

L.C.M of 15, 24 and 42 is 840
840 second means 14 min, they will toll after each 14 minutes.
So, between 8 a. m and 10 a. m ,
they will rang = (120/14) + 1 = 9 times.

16. In a seminar, the number of participants in subjects A, B and C are 96, 160 and 224, respectively. Find the minimum number of rooms required, if in each room the same number of participants are seated and all participants in a room are from the same subject. [Group D 26/09/2022 (Afternoon)]

Correct Answer: (d) 15
Solution:

H.C.F of 96, 160 and 224 = 32
Hence, required minimum number of
rooms = 96/32 + 160/32 + 224/32
= 3 + 5 + 7 = 15

17. 44, 96 and 184, when divided by a number x, leaves the same remainder in each case. The largest value of x is: [Group D 26/09/2022 (Afternoon)]

Correct Answer: (b) 4
Solution:

H.C.F of 44 , 96 and 184 = 4
Hence, the required number = 4

18. The LCM and the HCF of two numbers are 5005 and 77, respectively. When one of the two numbers is divided by 55, the quotient is 18 and the remainder is 11. The other number is: [Group D 26/09/2022 (Evening)]

Correct Answer: (b) 385 
Solution:

Given, L.C. M = 5005 , H.C.F = 77
First number = 18 × 55 + 11 = 1001
Hence, the other number = (L.C.M × H.C.F) / first no.
= (5005 × 77) / 1001 = 385

19. Three numbers are in the ratio 3 : 5 : 7 and their LCM is 5250. What is the middle number? [Group D 27/09/2022 (Morning) ]

Correct Answer: (b) 250
Solution:

Given ratio → 3 : 5 : 7 and LCM = 5250
⇒ 3 × 5 × 7 × k = 5250 (where k is constant)
⇒ 105k = 5250 ⇒ k = 50
So, middle number = 50 × 5 = 250

20. If the sum of two numbers is 48, while the HCF of these two numbers is 6, then the total number of such pairs of numbers (x, y), where x > y, is: [Group D 27/09/2022 (Evening)]

Correct Answer: (d) 2
Solution:

Let the two numbers be 6x and 6y.
6x + 6y = 48 ⇒ x + y = 8
So, (5,3) and (7,1) are only two pairs.