Question

12. Find the least number which is exactly divisible by 12, 20, 36 leaving remainder 8.
13. The product of two numbers is 2160 and the L.C.M. is 180. Find H.C.F.​

Answer 1

Answer:

13)let the hcf of two nos be X

we know

$lcm \times hcf = \: product \: two \: numbers$

$180 \times \: x = 2160 \\ x = \frac{2160}{180} \\ x = 2$

12)

$36 - 8 = 28 \\ 20 - 8 = 12 \\ 12 - 8 = 4$

so we have to find the lcm of 28 12 and 4

$28 = 7 \times 4 \\ 12 = 3 \times 4 \\ 4 = 1 \times 4$

thus 4 being the hcf will be taken out and multiplied with rest of the nos

$lcm = 4 \times 7 \times 3 \\ lcm = 84$

hope so it helps