NCERT MCQ Solutions for Class 5 Maths Mela Chapter 9 Coconut Farm

Every multiplication fact creates two related division facts. When 5 × 7 = 35, we can split 35 into 5 groups of 7 each, or split 35 into 7 groups of 5 each. This demonstrates the inverse relationship between multiplication and division operations.

This is the fundamental division formula. The dividend equals the divisor multiplied by the quotient plus the remainder. This relationship helps verify division calculations and shows how division and multiplication are inverse operations with an additional remainder component.

Using the splitting strategy, 1248 can be broken into 1200 + 48. Since 1200 ÷ 6 = 200 and 48 ÷ 6 = 8, the answer is 200 + 8 = 208. This mental strategy makes division easier by working with convenient number parts.

To divide by 4, we can halve twice. Half of 128 is 64, and half of 64 is 32. This strategy works because dividing by 4 is the same as dividing by 2 twice, making mental calculation much simpler.

To verify a division with remainder, we use the formula N = D × Q + R. Here, 16 × 56 + 6 should equal 902. This checks that our quotient and remainder are correct by reconstructing the original dividend.

When dividing multiples of 10 by 10, we essentially remove one zero from the dividend. For example, 400 ÷ 10 = 40, and 4000 ÷ 10 = 400. This pattern helps with quick mental calculations involving powers of ten.

To find the daily distance, we divide total distance by number of days: 160 ÷ 20 = 8 km per day. This is a practical application of division where we distribute a total quantity equally across a given number of units.

In place value division, we work with each place value separately. 324 has 3 hundreds and 24 ones (after regrouping 2 tens). Dividing gives us 1 hundred and 8 ones, making 108.

Partial quotient division breaks the division into smaller, easier steps. Instead of finding the complete quotient at once, we subtract manageable multiples of the divisor repeatedly. This reduces errors and makes mental calculation possible for larger numbers.

Since 50 × 3 = 150, we know that 150 ÷ 3 = 50. This demonstrates how division and multiplication are inverse operations, and knowing multiplication facts helps solve division problems quickly and accurately.

535 ÷ 25 = 21 remainder 10. This means 21 full bags plus one additional bag for the remaining 10 coconuts, totaling 22 bags. In real-world problems, we often need to round up to accommodate all items.

According to the division formula, the divisor times quotient plus remainder must equal the original dividend. So 3 × 295 + 2 = 885 + 2 = 887, which verifies our division calculation is correct.

We can rewrite 1992 as 2000 – 8. Then 2000 ÷ 4 = 500 and 8 ÷ 4 = 2, so 1992 ÷ 4 = 500 – 2 = 498. This strategy uses easier numbers to calculate mentally.

When a place value in the dividend is smaller than the divisor, we write 0 in that position of the quotient and regroup to the next place value. This maintains proper place value positioning in our answer.

475 ÷ 45 = 10 remainder 25. Since, we cannot have partial shows and must accommodate all 25 remaining people, we need 11 complete shows. This demonstrates how real-world division often requires rounding up.

Using partial quotients or standard division, 4376 ÷ 8 = 547 with no remainder. This can be verified by multiplying: 547 × 8 = 4376. The division represents extracting oil from coconuts at 8 coconuts per liter.

When the dividend is multiplied by 10 and the divisor remains the same, the quotient is also multiplied by 10. This pattern helps understand place value relationships and makes mental calculations with powers of ten easier.

Breaking 62 into 60 + 2, we find that 60 ÷ 5 = 12 with no remainder, but the additional 2 cannot be divided by 5, so it remains as the remainder. The complete answer is 12 remainder 2.

Since, 8 = 2³, we can divide by 2 three times: 168 ÷ 2 = 84, 84 ÷ 2 = 42, 42 ÷ 2 = 21. This repeated halving strategy is faster and less error-prone than other methods for powers of 2.

342 ÷ 41 = 8 remainder 14, meaning 8 full buses plus 14 additional students. Since, we cannot have partial buses and must transport all students, we need 9 buses total. Real-world division problems often require rounding up to ensure everyone is accommodated.