HCF and LCM – Railway Maths Part Ⅴ

LCM of 8, 12 and 16 = 48
A/Q
48x + 3
let x = 1, 2, 3 …
48 × 1 + 3 = 51
48 × 2 + 3 = 99
48 × 3 + 3 = 147
147 divided by 7 leaves no remainder.

(x + 3) exactly divides 1391
LCM (7, 9, 11) → 693
Now, 1391 is divided by 693 then we get the remainder of 5.
A/Q,
(x + 3) = 5 ⇒ x = 2

LCM of 2, 6, 4 and 3 = 12
Now 3496 is divided by 12 then we get 4 remainder.
So, 12 − 4 = 8, when we added 8 in 3496 then 3496 + 8 = 3504 is completely divisible by 2, 6, 4, and 3.

The ratio of two numbers is 5 : 6 and their HCF is 6. Their LCM is:
Let the numbers be 5x and 6x
So the HCF of 5x and 6x is 6
So, x = 6,
Numbers 5x and 6x are 30, 36
LCM (30, 36) = 180

One number = 2 × 44 = 88
HCF × LCM = Product of two numbers
⇒ 44 × 264 = 88 × 2nd number
⇒ 2nd number = (44 × 264) ÷ 88 = 132

Factor of 21 = 3 × 7
Factor of 24 = 2 × 2 × 2 × 3
Factor of 30 = 2 × 5 × 3
Factor of 35 = 7 × 5
LCM of 21, 24, 30 and 35
= 2³ × 3 × 5 × 7 = 840
Then number = 840 + 3 = 843

HCF [(a - b), (b - c)] will be the greatest number that gives the same remainder when we divide a, b and c by the greatest number. So,
Greatest no. = HCF [(226 - 151), (151 - 76)]
HCF = [75, 75] = 75
Hence, 75 will be the greatest number such that when 76, 151 and 226 are divided by it, the remainder is alike.

LCM of 16, 18, 24 and 28 = 1008
A/Q

⇒ 1008 - 7 = 1001

LCM of 4, 6, 7, 8 and 12 seconds = 168 seconds
After 168 seconds, they will toll together again.

HCF × LCM = Product of two numbers
⇒ 11 × 330 = 55 × 2nd no.
⇒ 2nd number = (11 × 330) ÷ 55 = 66