Partial quotients

In your book, Maths Mela (Chapter 9), Susie and Sunitha have a large coconut farm in Kerala. In April they harvest 1,117 coconuts, and they sell 582 coconuts equally to 6 regular customers. How many does each customer get? We need 582 ÷ 6.

Susie takes away groups of 6 in easy steps. 20 groups of 6 is 120, so she takes away 120 again and again: 582 − 120 = 462, then 342, then 222, then 102. That is 20 + 20 + 20 + 20 = 80 groups so far. Then 10 groups (60) leaves 42, and 7 groups (42) leaves 0. She adds the parts: 20 + 20 + 20 + 20 + 10 + 7 = 97. Each customer gets 97 coconuts. The numbers she wrote, 20, 10, 7, are called partial quotients: parts of the quotient.

Sunitha says she has a better way: estimate first! 100 groups of 6 would be 600, which is more than 582. So each customer gets less than 100. She takes away 90 groups of 6 at once: 90 × 6 = 540, and 582 − 540 = 42. Then 7 groups of 6 = 42, leaving 0. So 582 ÷ 6 = 90 + 7 = 97. Same answer, fewer steps!

Bigger numbers work the same way, using hundreds and tens. The farm uses 4376 coconuts to make oil, and 8 coconuts give 1 l. 4376 ÷ 8: take away 500 × 8 = 4000, leaving 376. Take away 40 × 8 = 320, leaving 56. Take away 7 × 8 = 56, leaving 0. So 4376 ÷ 8 = 500 + 40 + 7 = 547, and they get 547 l of oil.

They sell coconut husk (used to make coir) at ₹23 a kilogram and earn ₹9913. 9913 ÷ 23: take away 400 × 23 = 9200, leaving 713; take away 30 × 23 = 690, leaving 23; take away 1 × 23 = 23, leaving 0. So they sold 400 + 30 + 1 = 431 kg of husk.