model 2 find lcm of numbers Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (b)

Using Rule 4,

LCM of 4, 5, 6, 7 and 8

=

= 2 × 2 × 2 × 3 × 5 × 7 = 840.

let required number be 840 K + 2 which is multiple of 13.

Least value of K for which (840 K + 2) is divisible by 13 is K = 3

∴ Required number

= 840 × 3 + 2

= 2520 + 2 = 2522

Answer: (c)

Using Rule 8,

Greatest n digit number which when divided by three numbers A, B, C leaves no remainder will be

Required Number = (n – digit greatest number) – R

R is the remainder obtained on dividing greatest n digit number by L.C.M of A, B, C.

The largest number of 4-digits is 9999. L.C.M. of divisors

LCM = 2 × 2 × 3 × 3 × 3 × 5 = 540

Divide 9999 by 540, now we get 279 as remainder.

9999 – 279 = 9720

Hence, 9720 is the largest 4-digit number exactly divisible by each of 12, 15, 18 and 27.

Answer: (a)

Required time = LCM of 252, 308 and 198 seconds

∴ LCM = 2 × 2 × 7 × 9 × 11 = 2772 seconds

= 46 minutes 12 seconds

Answer: (c)

Required answer = LCM of 36, 40 and 48 seconds

= 720 seconds

= $(720/60)$minutes = 12 minutes

Illustration :

∴ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720

Answer: (d)

Using Rule 5,

When a number is divided by P, Q or R leaving remainders X, Y or Z respectively such that the difference between divisor and remainder in each case is the same i

.e., (P – X) = (Q – Y) = (R – Z) = T

(say) then that (least) number must be in the form of (K – T),

Where K is LCM of P, Q and R

Here,12 – 2 = 10; 16 – 6 = 10; 24 – 14 = 10

Now, LCM of 12, 16 and 24 = 48

∴ The greatest 4–digit number exactly divisible by 48 = 9984

∴ Required number = 9984 – 10 = 9974