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NCERT Solutions for Class 5th Maths Chapter 6 Coloured box puzzles and estimating products — Let Us Think

Book page 87–88 Updated on2026-09-19

Q1.
Find the possible values of the coloured boxes in each of the following problems. The same colour indicates the same number in a problem. Some problems can have more than one answer. a) (2-digit) × 3 = (2-digit) b) (3-digit) × 3 = (3-digit) c) (4-digit) × 3 = (4-digit)
Answer

Each puzzle asks: which numbers can be tripled without growing an extra digit?

  1. a) A two-digit number × 3 must stay two-digit. The largest two-digit number is 99, and 99 ÷ 3 = 33. So the number can be anything from 10 to 33. Example: 33 × 3 = 99. Example: 21 × 3 = 63.
  2. b) A three-digit number × 3 must stay three-digit. The largest three-digit number is 999, and 999 ÷ 3 = 333. So the number can be anything from 100 to 333. Example: 333 × 3 = 999. Example: 214 × 3 = 642.
  3. c) A four-digit number × 3 must stay four-digit. The largest four-digit number is 9999, and 9999 ÷ 3 = 3333. So the number can be anything from 1000 to 3333. Example: 3333 × 3 = 9999. Example: 2104 × 3 = 6312.
Test the biggest one each time
33 × 3 = 99 ✓ still two digits
34 × 3 = 102 ✗ three digits

333 × 3 = 999 ✓
334 × 3 = 1,002 ✗

3333 × 3 = 9,999 ✓
3334 × 3 = 10,002 ✗
Why the cut-off is always a 3: Multiplying by 3 makes a number three times bigger. It grows an extra digit as soon as it passes the biggest number with that many digits. So the largest starting number is that biggest number divided by 3 — 99 ÷ 3, 999 ÷ 3, 9999 ÷ 3.
Q2.
d) (2 boxes) × 5 = 1 (box) 0 e) (3 boxes, first two same colour) × 5 = (2 same-colour boxes) 0 f) (4 boxes, first three same colour) × 5 = 1 (3 same-colour boxes) 0
Answer

(d) any even two-digit number from 20 to 38. (e) 110 × 5 = 550, the only answer. (f) 2222 × 5 = 11,110 or 3332 × 5 = 16,660.

d) two-digit × 5 = a three-digit number of the form 1 _ 0

  1. The answer ends in 0, so the two-digit number must be even (an odd number times 5 ends in 5).
  2. The answer starts with 1, so it is between 100 and 190. Divide by 5: the number is between 20 and 38.
  3. So the answer can be 20, 22, 24, 26, 28, 30, 32, 34, 36 or 38.
  4. Example: 26 × 5 = 130. Here the middle green box is 3.

e) a three-digit number whose first two digits are the same, × 5, gives a three-digit number whose first two digits are the same and which ends in 0

  1. The answer looks like 110, 220, 330 … 990.
  2. Divide each by 5: 22, 44, 66, 88, 110, 132, 154, 176, 198.
  3. Only 110 has three digits with its first two the same.
  4. Check: 110 × 5 = 550. First two digits of 550 are both 5 ✓ Only one answer.

f) a four-digit number whose first three digits are the same, × 5, gives 1 _ _ _ 0 with the three middle digits the same

  1. The answer looks like 10,000; 11,110; 12,220; … 19,990.
  2. Divide each by 5: 2000, 2222, 2444, 2666, 2888, 3110, 3332, 3554, 3776, 3998.
  3. Only 2222 and 3332 have their first three digits the same.
  4. Check: 2222 × 5 = 11,110 ✓ and 3332 × 5 = 16,660 ✓ Two answers.
The trick that cracks all three: Work backwards. Instead of hunting for the starting number, list every answer that could fit the pattern, then divide each by 5 and see which one has the right shape.
Q3.
g) (3 boxes, first and third same colour) × 9 = (4 boxes, alternating two colours) h) (3 boxes) × (box, same colour as the last one) = 5,999 i) (3 boxes) × (box, same colour as the middle one) = 2,000
Answer

(g) 202 × 9 = 1818, and seven more like it. (h) 857 × 7 = 5,999. (i) see the note — the colours do not fit.

g) a three-digit number of the shape ABA × 9 gives a four-digit answer of the shape CDCD

  1. Try 202 × 9. That is 1,818 — which reads C D C D with C = 1 and D = 8 ✓
  2. Try 303 × 9 = 2,727 ✓, and the pattern keeps going.
  3. All eight answers: 202 × 9 = 1818, 303 × 9 = 2727, 404 × 9 = 3636, 505 × 9 = 4545, 606 × 9 = 5454, 707 × 9 = 6363, 808 × 9 = 7272, 909 × 9 = 8181.

h) a three-digit number × a one-digit number = 5,999, where the multiplier equals the ones digit of the big number

  1. Find the factors of 5,999. Try dividing by each one-digit number. 5,999 ÷ 7 = 857 exactly.
  2. Check the colour rule. The ones digit of 857 is 7, and the multiplier is 7. Same number, same colour ✓
  3. So the answer is 857 × 7 = 5,999. Red box = 8, green box = 5, purple boxes = 7.

i) a three-digit number × a one-digit number = 2,000, where the multiplier equals the tens digit

  1. List every way to write 2,000 as a three-digit number times a one-digit number. 500 × 4, 400 × 5, 250 × 8. Those are the only three.
  2. Check the colour rule for each. 500 has 0 tens but the multiplier is 4. 400 has 0 tens but the multiplier is 5. 250 has 5 tens but the multiplier is 8. None of them matches.
  3. So write down the three products anyway: 500 × 4 = 2,000, 400 × 5 = 2,000, 250 × 8 = 2,000.
About puzzle (i): If the two green boxes really must hold the same digit, this puzzle has no answer — we checked all three possibilities. It looks like the colouring in the book is a printing slip. Give the three factor pairs and explain your check; that is the honest, complete answer.
How to attack puzzles like (h) and (i): Do not guess the boxes. Take the finished answer and find its factors, because a product can only be built from its own factors. Then test each factor pair against the colour rule.
Q4.
Estimate the products on the left and match them to the numbers given on the right. (25 × 31, 132 × 19, 101 × 11, 248 × 49, 12 × 25) with (2,600; 12,500; 300; 750; 1,000). Discuss how you estimated.
Answer

Round each number to a friendly one, then multiply.

ProblemRound it toEstimateMatchesExact answer
25 × 3125 × 30750750775
132 × 19130 × 202,6002,6002,508
101 × 11100 × 101,0001,0001,111
248 × 49250 × 5012,50012,50012,152
12 × 2512 × 25300300300
How to estimate 248 × 49
Step 1. 248 is very close to 250.
Step 2. 49 is very close to 50.
Step 3. 250 × 50 = 25 × 5 × 100 = 125 × 100 = 12,500
Why an estimate is useful: It gives you a size to expect. If your careful answer comes out as 1,215 instead of 12,152, the estimate warns you that a digit has gone missing.
Sample answer for the class discussion: We rounded each number to the nearest ten or hundred, whichever was closer. For 132 we used 130 because 132 is nearer to 130 than to 100. For 19 we used 20. Then we multiplied the easy numbers. The estimates were never exactly right, but each one was clearly closest to just one number on the list.
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