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NCERT Solutions for Class 5th Maths Chapter 6 Are these children's solutions correct? — Check, Check!

Book page 85 Updated on2026-09-19

Q1.
Check if Asma's solution is correct. Asma's solution for 46 × 59: 2,000 + 300 + 360 + 54 = 2,714. If correct, explain why. If incorrect, identify the error and correct the solution.
Answer

Asma is correct. 46 × 59 = 2,714.

She split both numbers into place values and wrote all four pieces. Let us name them.

Her lineWhere it comes fromCheck
2,00050 × 402,000 ✓
30050 × 6300 ✓
3609 × 40360 ✓
549 × 654 ✓
2,000 + 300 = 2,300
2,300 + 360 = 2,660
2,660 + 54 = 2,714 ✓
Why her solution works: 46 is 40 + 6 and 59 is 50 + 9. Two parts times two parts gives four pieces, and she found all four. None is missing and none is counted twice, so the total must be right.
Check it by a second route: 46 × 59 = 46 × 60 − 46 = 2,760 − 46 = 2,714 ✓
Q2.
Check if Pankaj's solution is correct. Pankaj's solution for 203 × 54: 80 + 12 = 92; 100 + 15 = 115; = 135. If incorrect, identify the error and correct the solution.
Answer

Pankaj is wrong. The correct answer is 203 × 54 = 10,962. He made three mistakes.

  1. He used 20 instead of 200. In 203 the 2 stands for 2 hundreds, not 2 tens. So the first row should have 4 × 200 = 800, not 4 × 20 = 80.
  2. He used 5 instead of 50. In 54 the 5 stands for 5 tens. So the second row should be 50 × 203, not 5 × 203. He wrote 100 (which is 50 × 2, not 50 × 200) and 15 (which is 5 × 3, not 50 × 3).
  3. His final addition is wrong too. Even his own numbers give 92 + 115 = 207, not 135.
The correct working
203 (200 + 0 + 3) × 54 (50 + 4)

Row 1 — the 4
4 × 200 = 800
4 × 0 = 0
4 × 3 = 12
800 + 0 + 12 = 812

Row 2 — the 5 tens
50 × 200 = 10,000
50 × 0 = 0
50 × 3 = 150
10,000 + 0 + 150 = 10,150

Total = 812 + 10,150 = 10,962
The lesson here: Before you multiply, say out loud what each digit is worth. "The 2 is two hundred. The 5 is fifty." If you say it, you will not shrink it.
Estimate to spot the error fast: 203 is about 200 and 54 is about 50, so the answer should be near 200 × 50 = 10,000. An answer of 135 is nowhere near — that alone tells you something has gone wrong.
Q3.
Check if Lado's solution is correct. Lado's solution for 38 × 150: 1,500 + 400 = 1,900; 3,000 + 800 = 3,800; = 5,700.
Answer

Lado is correct. 38 × 150 = 5,700.

She split 38 into 30 + 8 and 150 into 100 + 50, then made one row for each part of 150.

Her rowWhere it comes fromRow total
1,500 + 40050 × 30 and 50 × 81,900 (= 50 × 38)
3,000 + 800100 × 30 and 100 × 83,800 (= 100 × 38)
1,900 + 3,800 = 5,700 ✓
Why her solution works: 150 groups of 38 is 50 groups plus 100 groups. She found each part correctly and kept every place value: 50 × 30 really is 1,500, and 100 × 30 really is 3,000.
Check it with doubling and halving: 38 × 150 = 19 × 300 = 19 × 3 × 100 = 57 × 100 = 5,700 ✓
Q4.
Check if Kira's solution is correct. Kira's solution for 193 × 272: 200 + 180 + 6 = 386; 700 + 630 + 21 = 1,351; 20,000 + 18,000 + 600 = 38,600; = 40,337.
Answer

Kira is wrong. The correct answer is 193 × 272 = 52,496.

Her first and third rows are perfect. The middle row is where she slipped.

  1. Row 1 is right. 2 × 193 = 200 + 180 + 6 = 386 ✓
  2. Row 2 is wrong. The 7 in 272 stands for 7 tens = 70, not 7. She worked out 7 × 193 = 1,351 instead of 70 × 193 = 13,510. Every part of her row is ten times too small.
  3. Row 3 is right. 200 × 193 = 20,000 + 18,000 + 600 = 38,600 ✓
The correct working
Row 1 (the 2): 2 × 193 = 386
Row 2 (the 7 tens): 70 × 193 = 7,000 + 6,300 + 210 = 13,510
Row 3 (the 2 hundreds): 200 × 193 = 38,600

386 + 13,510 = 13,896
13,896 + 38,600 = 52,496
How to avoid Kira's mistake: Before writing a row, say which place the digit is in. "This 7 is 7 tens, so my row must end in a zero." A missing zero is the most common error in long multiplication.
Estimate to check: 193 is about 200 and 272 is about 270. 200 × 270 = 54,000. Our answer 52,496 sits close to that, so it is believable. Kira's 40,337 is far too small ✓
Q5.
Check if Asher's solution is correct. Asher's solution for 626 × 323 gives the rows 1878, 1252, 1878 and a total of 5,018.
Answer

Asher is wrong. The correct answer is 626 × 323 = 202,198.

Asher multiplied every digit correctly, but he never moved the rows across.

  1. Row 1 is right. 3 × 626 = 1,878 ✓
  2. Row 2 should be shifted one place. The 2 in 323 means 2 tens. So the row is 20 × 626 = 12,520, not 1,252.
  3. Row 3 should be shifted two places. The first 3 in 323 means 3 hundreds. So the row is 300 × 626 = 187,800, not 1,878.
  4. His addition is wrong as well. Even 1,878 + 1,252 + 1,878 comes to 5,008, not the 5,018 he wrote.
The correct working
Row 1 (the 3 ones): 3 × 626 = 1,878
Row 2 (the 2 tens): 20 × 626 = 12,520
Row 3 (the 3 hundreds): 300 × 626 = 187,800

1,878 + 12,520 = 14,398
14,398 + 187,800 = 202,198
Why each row moves left: A row is not "3 times 626" and "2 times 626". It is 3 ones times 626, then 2 tens times 626, then 3 hundreds times 626. Every step left multiplies the row by another 10, so a zero must appear at its end.
Estimate to check: 626 is about 600 and 323 is about 300, so expect something near 600 × 300 = 180,000. Our 202,198 is close. Asher's 5,018 is nowhere near ✓
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