HCF and LCM – Railway Maths

Total Questions: 50

11. The traffic lights at three different crossings turn red after every 30 sec, 45 sec and 60 sec respectively. If they all turn red together at 8.30 am, at what time will they turn red together again? [Group D 24/08/2022 (Afternoon)]

Correct Answer: (a) 8.33 a.m.
Solution:

L.C.M of 30, 45, 60 = 180 sec or 3 min
Three lights will red together again, after 3 minutes, 8:30 am + 3 min = 8:33 am

12. The least number which, when diminished by 7, is divisible by 12,16,18 and 21 is: [Group D 25/08/2022 (Evening)]

Correct Answer: (c) 1015
Solution:

L.C.M of 12, 16, 18, 21 = 1008
Required number = 1008 + 7 = 1015

13. If the lowest common multiple (LCM) of 27 and n is 54, and the highest common factor (HCF) is 9, then find the value of n. [Group D 26/08/2022 (Afternoon)]

Correct Answer: (b) 18
Solution:

We know that , First number × second number = LCM × HCF
Now,
27 × n = 54 × 9 so , n = 18

14. The LCM and HCF of two numbers are in the ratio 3: 1, and the product of those numbers is 432. Find the value of least common multiple (LCM) and highest common factor (HCF) respectively. [Group D 26/08/2022 (Afternoon)]

Correct Answer: (a) 36 and 12
Solution:

We know that
HCF × LCM = First number × second number
3x × x = 432 (where x is common constant)
⇒ x² = 144 then x = 12
Therefore LCM = 12 × 3 = 36
HCF = 12 × 1 = 12

15. If GCD(108, 36) = GCD(x, 72) then what is the minimum possible value of x? [Group D 29/08/2022 (Morning)]

Correct Answer: (b) 36
Solution:

G.C.D. (108, 36) = G.C.D. (x , 72)
36 = G.C.D. (x , 72) x must 36.

16. The sum of two natural numbers, X and y, is 320 and the HCF of x and y is 20. If x>y, how many such possible pairs of x and y are there? [Group D 29/08/2022 (Evening)]

Correct Answer: (b) 4
Solution:

x + y = 320
20 (a + b) = 320 ⇒ a + b = 16
Where ‘a’ and ‘b’ are co-prime numbers.
There are 4 pairs, x > y ..
(15, 1), (13, 3), (11, 5), (9, 7).

17. If X represents the smallest prime number multiplied by 4 and added to the least common multiple of 5 and 10; Y represents the smallest odd prime number multiplied by 5 and added to the 2nd smallest odd prime, then which of the following is true? [Group D 30/08/2022 (Evening)]

Correct Answer: (d) The difference of Y and X is 2.
Solution:

X = Smallest prime number × 4
+    (L.C.M of 5,10) = 2 × 4 + 10 = 18
Y = Smallest odd prime number × 5
+ 2nd smallest odd prime number
= 3 × 5 + 5 = 20
X = 18 and Y = 20
Hence, the difference of Y and X is 2.

18. What is the smallest natural number that should be added to 1225 such that a remainder of 3 is left when the resulting number is divided by each of the numbers 12, 18, 21 and 28? [Group D 05/09/2022 (Evening)]

Correct Answer: (a) 38
Solution:

LCM of (12, 18, 21 and 28) = 252
252 × 5 = 1260
1260 + 3 = 1263
Nearest number to 1225, which when divided by 12, 18, 21 and 28 leaves a remainder of 3 = 1263
Smallest number to be added = 1263 - 1225 = 38

19. Nandan had some marbles. When he distributed those marbles equally among 15 children, he found that he was left with 2 marbles. If he had distributed those marbles equally among 24 children and 32 children, he would have been left with 11 and 19 marbles respectively. But when he distributed those marbles equally among the 31 children, he was not left with a single marble. Initially the number of marbles that Nandan had can be between _________. [Group D 06/09/2022 (Morning)]

Correct Answer: (d) 2380 and 2390
Solution:

The common difference between number and remainder

         

The LCM of 15, 24 and 32 = 480
Number of marbles = 480 × 5 - 13 = 2387
So option (d) is right answer

20. Two positive numbers differ by 3422. When the greater number is divided by the smaller number, the quotient is 4 and the remainder is 290. What is the HCF of the greater of the two given numbers and 4292? [Group D 08/09/2022 (Morning)]

Correct Answer: (d) 58
Solution:

Let the greater number = x
And the smaller number = y
ATQ,
x - y = 3422 ⇒ X = 3422 + y
And x = 4y + 290
By putting value of x
⇒ 3422 + y = 4y + 290
⇒ 3y = 3422 - 290 ⇒ 3y = 3132
⇒ y = 1044
And x = 3422 + 1044 = 4466
Therefore HCF of 4466 and 4292 = 58