NCERT Solutions for Class 5 Maths Chapter 6 the Dairy Farm (PDF) – 2027

One set of numbers that works: green pentagon = 1, orange diamond = 2, pink circle = 3, blue rectangle = 4, yellow parallelogram = 6, purple triangle = 12, white shape = 16, blue half-circle = 18.

Put 53 in the two-digit box and 2 in the one-digit box: 53 × 2 = 106. That is the closest you can get to 100.

Reason: With the small digit 2 as the multiplier the product stays near 100, and 53 is the two-digit number closest to 50 (50 × 2 = 100).

All four arrangements hold 60 pouches. Only the way they are grouped changes.

You can make twelve different groupings of 60 pouches. Every pair of numbers that multiplies to 60 gives one.

The number is 18.

No. Clues (a), (d) and (e) never help, and clue (h) adds nothing once (c), (f), (g) and (b) have left only 18.

• (a) Every two-digit number is already greater than 8.
• (d) It cannot be used: it says the number is odd, but clue (g) says the ones digit is even.
• (e) The only two-digit number that is a multiple of both 9 and 11 is 99, and clue (f) already removes it.
• (h) After (c), (f), (g) and (b) only 18 is left, so it removes nothing more.

You can make every number from 1 to 20 except 19. The number 0 cannot be made either. So 19 numbers out of 21 are possible.

The numbers you could not make with two grid numbers are 0 and 19. With three numbers and two operations, both become easy.

Most of them. 0 and 19 cannot be made at all, and 11, 16, 17 and 18 can be made in only one way each. Every other number from 1 to 20 can be made in two or more ways.

Both arrangements in every pair hold the same number of packets. Here are all the answers.

The pattern: the number of groups and the group size are swapped, but the total never changes.

9 × 0 = 0 and 0 × 9 = 0. Yes, this is true for any two numbers: changing the order does not change the product, and any number multiplied by 0, in either order, gives 0.

Here are all the answers.

All six answers, with the easy way to get each one.

Write 4 in the hundreds column, 5 in the tens column and 0 in the ones column. The thousands column stays empty.

The completed table.

The pattern: count the zeros in the two numbers you are multiplying, do the small multiplication first, then write that many zeros after it.

Both tables, completed.

Every digit shifts three places to the left, and three zeros appear at the end.

The completed table.

The completed table.

18 × 5 = 90.

Yes, all five are correct. Every one of them ends at 90, and each one is built on a rule we can trust.

Reason: Each child only broke 18 × 5 into easier parts (halves, doubles, 10 + 8, or 20 − 2) and then put the parts back together. Nothing extra was added and nothing was lost, so all of them get 90.

They are the same because no packet is added and none is taken away — the block is only cut and re-stacked. 3 × 18 = 6 × 9 = 54.

11 × 10 = 110.

Halve the first number, double the second, then multiply.

Because doubling 5 gives 10, and doubling 25 twice gives 100 — and multiplying by 10 or 100 is the easiest job in arithmetic.

All six, each shown as a single easy step.

Pick problems where one number is even and the other is 5, 25 or 50 — those are the ones that climb to 10 or 100.

Because the picture shows 4 rows of 20 with one column rubbed out, and that column holds 4 packets.

Because 21 columns is 20 columns plus one extra, and that extra column holds 14 packets.

choose problems where one number is just below or just above a ten, a hundred or a nice multiple.

All six, with the friendly number named each time.

1,470 people can sit in the auditorium.

Priya will jog 124 kilometres in 31 days.

The school received 1,728 books.

She would need 32 metres of cloth.

The cows produce 145 litres of milk each day.

5,346 kg of fodder is needed every day.

The family produces 420 kg of kitchen waste in a year.

Both are the same idea; they only differ in how many lines they write. Both split 35 into 30 + 5 and 12 into 10 + 2, and both reach 420.

The family saves ₹3,600 in a year.

32 × 8 = 256. Here is every blank filled in.

69 × 45 = 3,105. Every blank filled in.

Nida’s way is to draw a grid, split each number into its place values, multiply every box, then add.

a) 78 × 4 = 312

b) 83 × 9 = 747

c) 67 × 28 = 1,876

d) 53 × 37 = 1,961

Kanti’s way is to write every small piece on its own line, smallest first, then add them all.

John’s way is one line for each digit of the second number, with the pieces of that line already added together.

48 people are watching. After 3 more rows fill up, there will be 84 people.

Runs in 4s = 96. Runs in 6s = 108. Total runs in the two innings = 461.

Anjali should give ₹783 to the shopkeeper.

He earned ₹9,800.

The library has 13,932 books.

The dairy cooperative gets milk from at least 1,072 cows.

In Mili’s father’s method you multiply one column at a time, from the right, and any extra tens or hundreds are carried to the next column — where they are added to the product, not to the original digit.

The cooperative receives 5,889 litres of milk from Gir cows every day.

The cooperative earns ₹71,750 in a month.

Digits of the same colour stand in the same place value, so they are added together to make one digit of that row’s answer. The 2 of 20, the 3 of 350 and the 1 of 1,400 are the carries of the column method.

Nida’s grid: split both numbers into place values, multiply every box, add everything.

a) 548 × 6 = 3,288

b) 682 × 3 = 2,046

c) 324 × 18 = 5,832

d) 507 × 23 = 11,661

e) 190 × 65 = 12,350

John’s way: one line for each digit of the multiplier, with that line’s pieces already added.

806 × 9 = 7,254; 455 × 26 = 11,830; 143 × 208 = 29,744. Here is every box filled.

Asma is correct. 46 × 59 = 2,714.

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

1. He treated 203 as 20 + 3. His rows are 4 × 20 + 4 × 3 = 92 and 5 × 20 + 5 × 3 = 115, which is 4 × 23 and 5 × 23. The 2 in 203 is 2 hundreds = 200.
2. He used 5 instead of 50. The 5 in 54 is 5 tens, so the second row must be 50 × 203.
3. His addition is wrong too: 92 + 115 = 207, not 135.

Lado is correct. 38 × 150 = 5,700.

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

The error is in row 2. The 7 in 272 is 7 tens, so the row must be 70 × 193 = 13,510, not 7 × 193 = 1,351. Rows 1 and 3 are correct.

Asher is wrong. The correct answer is 626 × 323 = 2,02,198.

The error: the rows are not moved to the left. Row 2 is 2 tens × 626 = 12,520 and row 3 is 3 hundreds × 626 = 1,87,800. (His own addition is also wrong: 1,878 + 1,252 + 1,878 = 5,008, not 5,018.)

Same answer as the top of the box: 6 × 34; 26 × 10 and 26 × 1; 20 × 11 and 6 × 11; 9 × 8; 20 × 4 − 8; 50 × 9 and 5 × 9; 55 × 10 − 55; 100 × 42 and 42; 250 × 8 − 24; 1,000 × 5 and 5; 2,000 × 2 − 2.

All three become easy if you climb to 100, 500 or 1,000 by doubling.

the easiest problems are the ones where a number is close to 10, 100 or 1,000, or where doubling turns it into one of those.

(b) 408 (c) 374 (d) 368. You do not need to multiply again — just add or take away one whole group.

1. (b) 17 × 24. This is one more group of 17 than 17 × 23. So 391 + 17 = 408.
2. (c) 17 × 22. This is one group of 17 fewer. So 391 − 17 = 374.
3. (d) 16 × 23. Here the other number went down by 1, so we lose one group of 23. So 391 − 23 = 368.

(f) 162 (g) 252 (h) 972 (i) 414.

1. (f) 18 × 9. 18 is 8 + 10, so add ten more 9s. 72 + 90 = 162.
2. (g) 28 × 9. 28 is 18 + 10, so add ten more 9s again. 162 + 90 = 252. (Or from 8 × 9: 72 + 180 = 252.)
3. (h) 108 × 9. 108 is 8 + 100, so add a hundred 9s. 72 + 900 = 972.
4. (i) 18 × 23. Go back to 17 × 23 = 391. One more row of 23 gives 391 + 23 = 414.

Add 17. 17 × 24 has one more column than 17 × 23, and each column has 17 squares. So 17 × 24 = 391 + 17 = 408.

Add 23. 18 × 23 has one more row than 17 × 23, and each row has 23 squares. So 18 × 23 = 391 + 23 = 414.

(a) 11 × 3 = 33, 22 × 3 = 66, 33 × 3 = 99
(b) 111 × 3 = 333, 222 × 3 = 666, 333 × 3 = 999
(c) 1,111 × 3 = 3,333, 2,222 × 3 = 6,666, 3,333 × 3 = 9,999

Reason: all the answer boxes are orange, so every digit of the product is the same. In (a) the product must be 11, 22, …, 99 and also in the 3 times table: only 33, 66 and 99 are (3 × 11, 3 × 22, 3 × 33). (b) and (c) work the same way. The top boxes have no colour, so they need not match.

(d) any even two-digit number from 20 to 38. (e) 110 × 5 = 550, the only answer. (f) 2,222 × 5 = 11,110 or 3,332 × 5 = 16,660.

(g) 202 × 9 = 1,818, 303 × 9 = 2,727, … up to 909 × 9 = 8,181 (eight answers). (h) 857 × 7 = 5,999. (i) no answer — see About the book.

Round each number to a friendly one, then multiply.

Sample answer: I would choose Choice 3 — take 1 gold coin and multiply the number of coins by 5 every day for 7 days. It ends with by far the most gold.

Minister 3 received the most: 78,125 gold coins. Minister 1 got 640 and Minister 2 got 6,561.

3) 16 × 22 = 352 and 4) 32 × 44 = 1,408.

1) 96 2) 384 3) 1,536 4) 192

1111 × 1111 = 12,34,321.

11 × 54 = 594 and 11 × 82 = 902.

1. 11 × 54. Split 54 into 5 and 4. Write 5, then 5 + 4 = 9, then 4. That gives 594.
2. 11 × 82. Split 82 into 8 and 2. Write 8, then 8 + 2 = 10 — this is more than 9, so keep the 0 and carry the 1 to the 8, making 9. Then write 2. That gives 902.

45 × 45 = 2,025 and 55 × 55 = 3,025.

1,234 × 9 + 4 = 11,110.

1. Multiply first. 1,234 × 9. Use the nearest multiple: 1,234 × 10 = 12,340, then take away one 1,234.
2. 12,340 − 1,234 = 11,106.
3. Now add 4. 11,106 + 4 = 11,110.

Smallest product: 11 × 14 = 154. Largest product: 35 × 75 = 2,625. No, you do not need to work out every product — you can see it by looking.

Reason: the smallest product comes from the two smallest numbers that are neighbours (11 and 14); the largest from the two largest neighbours (35 and 75).

Mala had ₹920 at the beginning.

The club collected ₹11,352 in all.

The total expenses were ₹2,879.

(a) 30 rows (b) 41 students (c) 1,230 students.

All six products, each with a short method.

67 × 68 = 4,556.

99 × 99 = 9,801.