Question

a rich merchant has 440kg first quality rice, 512 kg second quality and 416kg third quality of rice. He packs them and sells them in bags.each bag contains exactly one quality and has the same quantity of rice. 8kg of rice of each quality is left.What is the least total number of rice bags he packed

Answer 1

Lets subtract 8 from each number.

440-8= 432

512-8 = 504

416-8 = 408

ATQ,

All the three rice of 432 kgs, 504 kgs and 408 kgs are subdivided (packed) into smaller bags such that each group consists of equal amount of rice and no more rice is left.

So we have to find the HCF of these numbers

2 | 432, 504, 408

2 | 216, 252, 204

2 | 108, 126, 102

3 | 54, 63, 51

| 18, 21, 17

18, 21 and 17 don't share a common factor.

Hence HCF = 2 × 2 × 2 × 3 = 24

Least total of no.of rice bags packed

= 432/24 + 504/24 + 408/24

= 18 + 21 + 17

= 56

Hope it helps!!!