model 2 find lcm of numbers Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (d)

LCM of x and y = 161

∴ xy = 23 × 7

∴ x = 23; y = 7

∴ 3y – x = 3 × 7 – 23

= 21 – 23 = – 2

Answer: (d)

LCM of 4, 6, 8, 14

= 168 seconds

= 2 minutes 48 seconds

They ring again at 12 + 2 min. 48 sec.

= 12 hrs. 2 min. 48 sec.

Answer: (a)

Using Rule 4,

i.e. When a number is divided by x, y or z leaving same remainder ‘R’ in each case then that number must be K + R where k is LCM of x, y and z.

Required number = (LCM of 15, 20, 36 and 48) + 3

∴ LCM = 2 × 2 × 3 × 5 × 3 × 4 = 720

∴ Required number = 720 + 3 = 723

Answer: (c)

∴ LCM = 2 × 2 × 5 × 7 × 8 = 1120

∴ Required number

= 5834 – 1120 = 4714

Answer: (c)

Using Rule 9,

The n digit largest number which when divided by a, b, c, leaves remainder ‘x’ will be

Required number = [n – digit largest number – R] + x

where, R is the remainder obtained when n – digit largest number is divided by the L.C.M of a, b, c.

We will find the LCM of 16, 24, 30 and 36

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

The largest number of five digits = 99999

On dividing 99999 by 720, the remainder = 639

∴ The largest five-digit number divisible by 720

= 99999 – 639 = 99360

∴ Required number = 99360 + 10

= 99370

Answer: (a)

We find LCM of 5, 6 and 8

5=5

6=3×2

8=$2^3$

= $2^3$ ×3 × 5 = 8 × 15 = 120

Required number = 120K + 3

∴ when K = 2, 120 × 2 + 3 = 243 required no.

It is completely divisible by 9