NCERT Solutions Class 5 Maths Chapter 9 Coconut Farm

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 5 · M AT H S

NCERT Solutions

Chapter 9: Coconut Farm

NCERT Textbook — Math Mela

BOOK PAGES SECTIONS QUESTIONS MEDIUM

119 – 135 20 95 English

Solutions, notes, sample papers & more at 88 pages

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

CLASS 5 · MATHS · MATH MELA

NCERT Solutions — Chapter 9: Coconut Farm
Chapter 9 of Math Mela is about division. Susie and Sunitha's coconut farm in Kerala gives the story: 1,117
coconuts to share among customers, husk to weigh, oil to bottle. Along the way you learn that every
multiplication fact hides two division facts, that a big division can be done in easy chunks (partial quotients),
that place value lets you divide digit by digit, and that when the sharing does not come out even, what is left
over is called the remainder.

TEXTBOOK BOOK PAGES

Math Mela (Class 5) 119 – 135

SECTIONS QUESTIONS

20 95

MEDIUM

English

Division Facts — Page 119
Arrays of coconuts, and the rule N = D × Q

THINK AND ANSWER

Q1 Observe the following array of coconuts. Write two division facts using the given
multiplication fact. (The array shows 5 rows of 7 coconuts, and 5 × 7 = 35.)

35 ÷ 7 = 5 and 35 ÷ 5 = 7.
Look at the picture in your book. The coconuts are set out in 5 rows, with 7 coconuts in each
row. Things set out in equal rows like this are called an array.

Page 1 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

7 in each row

5 rows
of coconuts

5 × 7 = 35 coconuts in all

The same 35 coconuts can be seen as 5 rows of 7, or as 7 columns of 5.

Now share those 35 coconuts, in two different ways.

1. Make 7 groups — one group for each column. Each group has 5 coconuts. So 35 ÷ 7 = 5.
2. Make 5 groups — one group for each row. Each group has 7 coconuts. So 35 ÷ 5 = 7.

Multiplication fact: 5 × 7 = 35

First division fact: 35 ÷ 7 = 5

35 split into 7 groups has 5 in each group

Second division fact: 35 ÷ 5 = 7

35 split into 5 groups has 7 in each group

Why it happens: Multiplying puts equal groups together. Dividing takes them apart
again. The same three numbers 5, 7 and 35 are used both times — only the job
changes.

The book's names for the three numbers: the number being shared is the
dividend (N), the number of groups is the divisor (D), and the answer is the
quotient (Q). The rule is N = D × Q. Here 35 = 5 × 7. ✓

Page 2 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Q2 Think and answer: 35 ÷ 1 = ______

35 ÷ 1 = 35.
Picture it first. You have 35 coconuts and you must share them into 1 group. Nothing is taken
away — all 35 coconuts go into that one group.

35 ÷ 1 = 35

Check by multiplying back:

N=D×Q
1 × 35 = 35 ✓

Why it happens: Dividing by 1 means making groups of size 1, or making just 1
group. Either way nothing is broken up, so the number stays exactly as it was.

Try this too: 35 ÷ 35 = 1. If you make 35 groups out of 35 coconuts, each group gets
just 1 coconut.

Page 3 of 88

as e
Class 5 Maths Chapter 9 Coconut Farm
a g l AglaSem · NCERT Solutions

co m
m.
Write the appropriate multiplication fact for the array shown below. Write two
e
Q3

m l as
.co
division facts that follow from the multiplication fact.

a g
se m
g l a
a

co m
e m . ag
g l as
a

co m
em.
m l as
m .co a g
l a se
a g
m a s
m .co agl
l a se
a g
m
The array on page 119. Count the coconuts in one row and the number of rows, write

. co
m
those two numbers in the empty boxes, then fill the three sentences below.

as e
om × _____ = _____
.c_____ a g l
se m _____ ÷ _____ = _____
g l a
a _____ ÷ _____ = _____

se m
com g l a
m . a
e

Multiplication fact: 4 × 8 = 32.la s
a g Division facts: 32 ÷ 8 = 4 and 32 ÷ 4 = 8.
First count the array in the book. There are 4 rows and 8 coconuts in every row.

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 4 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

8 in each row

4 rows

4 × 8 = 32

Four rows of eight coconuts. Counting in fours or in eights gives the same total, 32.

1. Count the rows and the number in each row. 4 rows, 8 in each row.
2. Multiply. 4 × 8 = 32 coconuts altogether.
3. Share into 8 groups (the columns). Each group has 4. So 32 ÷ 8 = 4.
4. Share into 4 groups (the rows). Each group has 8. So 32 ÷ 4 = 8.

4 × 8 = 32

32 ÷ 8 = 4

32 ÷ 4 = 8

Check it yourself: N = D × Q. Here 32 = 8 × 4 ✓ and 32 = 4 × 8 ✓. Both division facts
pass the test.

Let Us Play — Page 120
Fill the circles so that the number in the square is the product or the quotient

LET US PLAY

Q1 Identify the numbers that can fill the circles such that the numbers in the squares
are the products of the numbers in the circles. The squares in the top row show 72
(with 36 × 2 already filled in), 60, 48, 36, 24 and 40.

In the top row the two circles are joined by ×, and the square above them holds the product. So
you need two numbers whose product is the number in the square.

Page 5 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

The first one is done for you: 36 × 2 = 72. Here is one good answer for each of the others.

SQUARE (PRODUCT) ONE ANSWER CHECK OTHER CORRECT ANSWERS

72 36 × 2 36 × 2 = 72 8 × 9, 12 × 6, 24 × 3, 18 × 4

60 30 × 2 30 × 2 = 60 6 × 10, 12 × 5, 15 × 4, 20 × 3

48 24 × 2 24 × 2 = 48 6 × 8, 12 × 4, 16 × 3

36 18 × 2 18 × 2 = 36 6 × 6, 9 × 4, 12 × 3

24 12 × 2 12 × 2 = 24 6 × 4, 8 × 3, 24 × 1

40 20 × 2 20 × 2 = 40 8 × 5, 10 × 4, 40 × 1

How to find a pair yourself, in three steps:

1. Say the number in the square out loud — for example 48.
2. Run through your tables and ask: which table has 48 in it? The 6 table (6 × 8), the 4 table (4 ×
12), the 3 table (3 × 16).
3. Write any one of those pairs in the two circles. Then multiply to check.

Tip: There is more than one right answer for every square. As long as the two circle
numbers multiply to give the square number, your answer is correct.

Q2 Identify the numbers that can fill the circles such that the numbers in the squares
are the quotients of the numbers in the circles. In the bottom row the squares hold
54, 42 and 56 (both circles empty), and then three empty squares whose first circle
already shows 54, 42 and 56.

In the bottom row the two circles are joined by ÷, and the square holds the quotient — the
answer of the division.
First three puzzles — the square is the answer, so you must find a division that gives it.

SQUARE (QUOTIENT) ONE ANSWER CHECK OTHER CORRECT ANSWERS

54 108 ÷ 2 2 × 54 = 108 ✓ 54 ÷ 1, 162 ÷ 3, 216 ÷ 4

42 84 ÷ 2 2 × 42 = 84 ✓ 42 ÷ 1, 126 ÷ 3, 420 ÷ 10

56 112 ÷ 2 2 × 56 = 112 ✓ 56 ÷ 1, 168 ÷ 3, 560 ÷ 10

Page 6 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Last three puzzles — here the first circle is already filled and the square is empty. Choose any
number that divides it exactly.

PUZZLE ONE ANSWER SQUARE BECOMES OTHER CORRECT ANSWERS

54 ÷ ○ 54 ÷ 6 9 54 ÷ 9 = 6, 54 ÷ 2 = 27, 54 ÷ 3 = 18

42 ÷ ○ 42 ÷ 6 7 42 ÷ 7 = 6, 42 ÷ 2 = 21, 42 ÷ 3 = 14

56 ÷ ○ 56 ÷ 7 8 56 ÷ 8 = 7, 56 ÷ 4 = 14, 56 ÷ 2 = 28

Why it works: Every division you write must pass the test N = D × Q. For 54 ÷ 6 = 9,
check 6 × 9 = 54 ✓. If the multiplication comes back to the dividend, your answer is
right.

Did you notice? The numbers 54, 42 and 56 appear twice in this row — first as an
answer, then as a number being shared. The same number can be a quotient in one
puzzle and a dividend in the next.

Let Us Do — Page 120
Solve the multiplication problem, then write two division statements

LET US DO

Q1 1. Solve the following multiplication problems. Write two division statements in
each case. 30 × 30 = ________

30 × 30 = 900. Division statements: 900 ÷ 30 = 30 (and again 900 ÷ 30 = 30).

1. Use the fact you already know. 3 × 3 = 9.
2. Count the zeros. There is one zero in each 30 — two zeros in all.
3. Put them back. 9 with two zeros is 900. So 30 × 30 = 900.
4. Turn it around. N = D × Q means 900 = 30 × 30, so 900 ÷ 30 = 30.

30 × 30 = 900

900 ÷ 30 = 30

900 ÷ 30 = 30

Page 7 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Something special here: Both circle numbers are the same, so the two division
statements come out the same as well. That only happens when a number is
multiplied by itself.

Q2 400 × 8 = ________ , and the two division statements.

400 × 8 = 3200. Division statements: 3200 ÷ 8 = 400 and 3200 ÷ 400 = 8.

1. Use the small fact. 4 × 8 = 32.
2. Count the zeros. 400 has two zeros.
3. Put them back. 32 with two zeros is 3200.
4. Write the two divisions by taking each factor away in turn.

400 × 8 = 3,200

3,200 ÷ 8 = 400

3,200 ÷ 400 = 8

Check it yourself: 8 × 400 = 3,200 ✓ and 400 × 8 = 3,200 ✓. Both division statements
pass the test N = D × Q.

Q3 15 × 60 = _______ , and the two division statements.

15 × 60 = 900. Division statements: 900 ÷ 15 = 60 and 900 ÷ 60 = 15.

1. Drop the zero for a moment. 15 × 6 = 90.
2. Put the zero back. 60 has one zero, so 90 becomes 900.
3. Write the two divisions.

Page 8 of 88

as e
Class 5 Maths Chapter 9 Coconut Farm
a g l AglaSem · NCERT Solutions

co m
e m.
15 × 6 = 90
m l as
15 × 60 = 900
m .co a g
l a se
g
a900 ÷ 15 = 60

co m
. ag
900 ÷ 60 = 15
e m
g l as
a
Why it happens: 60 is 6 tens. Multiplying by 6 tens is the same as multiplying by 6
and then by 10, and multiplying by 10 just adds a zero.
co m
em.
m l as
.co a g
a s em
l
200 × 16 = _________ , and the two division statements.
g
Q4
a
a s
co÷m16 = 200 and 3200 ÷ 200 = 16. agl

.
em
200 × 16 = 3200. Division statements: 3200
a s
a gl
1. Use the small fact. 2 × 16 = 32.
2. Count the zeros. 200 has two zeros.
3. Put them back. 32 with two zeros is 3,200.
co m
m .
m as e
.co a g l
m
200 × 16 = 3,200

a s e3,200
agl ÷ 16 = 200

3,200 ÷ 200 = 16
se m
com g l a
m . a
e
asboth give 3,200. In Question 2 one number was
Notice: 400 × 8 and 200 × l16
a g
doubled and in this one the other was doubled — halving one factor and doubling

m
the other keeps the product the same.
. co
em
m l as
.co a g
a s em Us Do (Question 2) — Page 121
gl
Let
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 9 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Solve these division problems, notice the patterns and discuss in class

LET US DO

Q1 2. Solve the following division problems: 150 ÷ 3, 80 ÷ 4, 500 ÷ 5, 100 ÷ 10, 300 ÷ 100,
500 ÷ 50, 200 ÷ 20, 440 ÷ 44, 630 ÷ 63. (Hints in the book: ___ × 3 = 150, 5 × ___ = 500, 44 ×
__ = 440.)

Use the hint each time: think of the multiplication that goes with the division.

DIVISION THINK OF THIS MULTIPLICATION ANSWER

150 ÷ 3 50 × 3 = 150 50

80 ÷ 4 20 × 4 = 80 20

500 ÷ 5 5 × 100 = 500 100

100 ÷ 10 10 × 10 = 100 10

300 ÷ 100 100 × 3 = 300 3

500 ÷ 50 50 × 10 = 500 10

200 ÷ 20 20 × 10 = 200 10

440 ÷ 44 44 × 10 = 440 10

630 ÷ 63 63 × 10 = 630 10

The three hints filled in:

50 × 3 = 150

5 × 100 = 500

44 × 10 = 440

Check it yourself: Multiply the answer by the divisor and see if you get back the
dividend. 50 × 3 = 150 ✓, 100 × 5 = 500 ✓, 10 × 63 = 630 ✓.

Page 10 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Q2 What patterns do you notice here? What is happening to the quotients in each
case? Discuss.

Sample answer for the class discussion — three patterns.
Pattern 1 — a number divided by ten times itself always gives 10.

440 ÷ 44 = 10

630 ÷ 63 = 10

500 ÷ 50 = 10

200 ÷ 20 = 10

Look carefully: 440 is 44 with a zero added, 630 is 63 with a zero added. Adding a zero multiplies
a number by 10, so the answer must be 10 every time.
Pattern 2 — a bigger divisor gives a smaller quotient.

500 ÷ 5 = 100

500 ÷ 50 = 10

500 ÷ 500 = 1

The same 500 is being shared. When the divisor becomes 10 times bigger, the quotient becomes
10 times smaller.
Pattern 3 — you only need the small table fact.

15 ÷ 3 = 5, so 150 ÷ 3 = 50

8 ÷ 4 = 2, so 80 ÷ 4 = 20
3 ÷ 3 = 1, so 300 ÷ 100 = 3

Why it happens: A zero at the end of a number means ten times as much. If the
dividend gets ten times bigger and the divisor stays the same, the answer must also
be ten times bigger — because each group is now sharing ten times as many things.

Patterns in Division and Place Value — Page 121

Page 11 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

The blue boxes and the two place value charts

PATTERNS IN DIVISION AND PLACE VALUE

Q1 Solve: 1000 ÷ 10, 1000 ÷ 100, 1600 ÷ 4, 2000 ÷ 2, 2000 ÷ 20, 3700 ÷ 37, 3300 ÷ 3, 3300 ÷
300, 4000 ÷ 40. (Hints in the book: 10 × ___ = 1000 and 37 × ___ = 3700.)

DIVISION THINK OF THIS MULTIPLICATION ANSWER

1000 ÷ 10 10 × 100 = 1000 100

1000 ÷ 100 100 × 10 = 1000 10

1600 ÷ 4 4 × 400 = 1600 400

2000 ÷ 2 2 × 1000 = 2000 1000

2000 ÷ 20 20 × 100 = 2000 100

3700 ÷ 37 37 × 100 = 3700 100

3300 ÷ 3 3 × 1100 = 3300 1,100

3300 ÷ 300 300 × 11 = 3300 11

4000 ÷ 40 40 × 100 = 4000 100

The two hints filled in:

10 × 100 = 1000

37 × 100 = 3700

Look at the pair 3300 ÷ 3 and 3300 ÷ 300. The dividend is the same. The second
divisor is 100 times bigger, so the answer is 100 times smaller: 1,100 becomes 11.

Page 12 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Q2 Now fill the place value chart.

PROBLEM H T O

40 ÷ 10 = 4

400 ÷ 10 = 4 0

4000 ÷ 10 = 4 0 0

700 ÷ 70 =

1400 ÷ 100 =

220 ÷ 20 =

2200 ÷ 20 =

The first three rows are done in the book. Here is the whole chart. H means Hundreds, T means
Tens, O means Ones.

PROBLEM H T O QUOTIENT

40 ÷ 10 = 4 4

400 ÷ 10 = 4 0 40

4000 ÷ 10 = 4 0 0 400

700 ÷ 70 = 1 0 10

1400 ÷ 100 = 1 4 14

220 ÷ 20 = 1 1 11

2200 ÷ 20 = 1 1 0 110

How to do the last four, step by step. Take 2200 ÷ 20:

1. Cover one zero at the end of both numbers. You are left with 220 ÷ 2.
2. 220 ÷ 2 = 110.
3. Check: 20 × 110 = 2,200 ✓

Page 13 of 88

as e
Class 5 Maths Chapter 9 Coconut Farm
a g l AglaSem · NCERT Solutions

co m
m.
Why the digits shift: Dividing by 10 moves every digit one place to the right. The 4

as e
com is 400. l
in 4000 sits in the Thousands place; after dividing by 10 it sits in the Hundreds place,
. a g
em
and the number
a s
agl

co m
ag
Fill the second place value chart: 110 ÷ 11, 860 ÷ 86, 7500 ÷ 750, 8800 ÷ 88, 2400 ÷ 24,
.
Q3

e m
as
440 ÷ 22.

a g l

co m
m.
PROBLEM H T O QUOTIENT

m as e
.co a g l
m
110 ÷ 11 = 1 0 10

l a se
a g 860 ÷ 86 = 1 0 10

a s
com agl
7500 ÷ 750 = 1 0 10

m .0
8800 ÷ 88 = 1
as e 0 100

2400 ÷ 24 = a1gl 0 0 100

co m
.
440 ÷ 22 = 2 0 20

em
m l as
.co a g
The trick for the first five: the dividend is the divisor with one or two zeros stuck on the end.

se m
g l a
a 11 → 110 is one extra zero, so the answer is 10

se m
86 → 860 is one extra zero, so the answer is 10
com g l a
m . a
ase
750 → 7500 is one extra zero, so the answer is 10

agl
88 → 8800 is two extra zeros, so the answer is 100

m
24 → 2400 is two extra zeros, so the answer is 100
. co
e m
m l as
.co g
The last one is different. 440 ÷ 22:

em1. Cover one zero of 440 and think 44 ÷ 22 = 2. a
a s
agl
c
2. Put the zero back: the answer is 20.

m .
3. Check: 22 × 20 = 440 ✓
a s e
. com agl
a s
Watch out: 440 ÷ 44 = 10 butem440 ÷ 22 = 20. The divisor 22 is half of 44, so the
aglA smaller divisor always gives a bigger quotient.
answer is twice as big.

co m
m .
m ase
.co


a g l Page 14 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Q4 What is happening to the quotients in each case? Discuss.

Sample answer for the class discussion.
1. When the dividend grows ten times and the divisor stays the same, the quotient grows
ten times.

40 ÷ 10 = 4

400 ÷ 10 = 40

4000 ÷ 10 = 400

Ten times as many things shared among the same number of groups means ten times as much
for each group.
2. When both the dividend and the divisor grow ten times, the quotient does not change.

44 ÷ 22 = 2

440 ÷ 220 = 2
4400 ÷ 2200 = 2

Ten times as many sweets, but also ten times as many children — each child still gets the same.
3. When the divisor grows and the dividend stays the same, the quotient shrinks.

2200 ÷ 20 = 110

2200 ÷ 200 = 11

Why it happens: Division is sharing. More things to share means more each. More
people to share with means less each. That is the whole idea behind every one of
these patterns.

A quick way to use this: If both numbers end in a zero, cross one zero off each.
4000 ÷ 40 becomes 400 ÷ 4 = 100. Much easier, and the answer is exactly the same.

Let Us Do — Page 122

Page 15 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Word problems, the 1-to-8 puzzle, and fill in the blanks

LET US DO

Q1 1. Sabina cycles 160 km in 20 days and the same distance each day. How many
kilometres does she cycle each day?

Sabina cycles 8 km each day.
We need to find the distance for one day. The whole distance, 160 km, is shared equally over 20
days. Sharing equally means dividing.

Step 1: Distance for one day = 160 ÷ 20

Step 2: Cover one zero in each number → 16 ÷ 2

Step 3: 16 ÷ 2 = 8

Step 4: So 160 ÷ 20 = 8 km

Check by a second route: If she cycles 8 km a day for 20 days, she covers 8 × 20 =
160 km. That is exactly the distance given. ✓
Answer: 8 km each day.

Q2 2. How many notes of ₹100 does Seema need to carry if she wants to buy coconuts
worth ₹4200?

Seema needs 42 notes of ₹100.
We need to find how many ₹100 notes make ₹4200. So we ask: how many hundreds are there in
4200?

Step 1: Number of notes = 4200 ÷ 100

Step 2: Dividing by 100 removes two zeros

Step 3: 4200 ÷ 100 = 42 notes

Page 16 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Check by a second route: 42 notes × ₹100 = ₹4,200 ✓
Answer: 42 notes of ₹100.

Q3 3. The owner of an electric store has decided to distribute ₹5500 equally amongst 5
of his employees as a Diwali gift. What amount will each employee get? What will
happen if he distributes the same amount of money among 10 employees? Will each
employee get more or less? How much money would he have to distribute if
everyone must get the same amount as earlier?

With 5 employees each gets ₹1100. With 10 employees each gets only ₹550 — less. To give
₹1100 to all 10, he would need ₹11,000.
Part 1 — sharing ₹5500 among 5 people.

Step 1: Amount for one person = 5500 ÷ 5

Step 2: Split 5500 into 5000 + 500

Step 3: 5000 ÷ 5 = 1000 and 500 ÷ 5 = 100

Step 4: 1000 + 100 = ₹1,100 each

Part 2 — the same ₹5500 among 10 people.

Step 1: Amount for one person = 5500 ÷ 10

Step 2: Dividing by 10 removes one zero

Step 3: 5500 ÷ 10 = ₹550 each

So each employee gets less — in fact exactly half of ₹1,100.
Part 3 — how much money for 10 people at ₹1,100 each?

Step 1: Money needed = 1100 × 10

Step 2: Multiplying by 10 adds one zero

Step 3: Money needed = ₹11,000

Page 17 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Why each gets less: The money did not change, but the number of people doubled.
Twice as many people sharing the same money means each person gets half as
much.

Check it yourself: 10 × ₹550 = ₹5,500 ✓ and 10 × ₹1,100 = ₹11,000 ✓

Q4 4. Place the numbers 1 to 8 in the following boxes so that all the four operations,
division, multiplication, addition and subtraction are correct. No number must be
repeated. How did you think about solving this? Is there more than one answer?

One correct filling: 6 ÷ 3 = 2, then 6 − 5 = 1, 2 × 4 = 8, and 1 + 7 = 8.

6 ÷ 3 = 2
− ×

5 4

1 + 7 = 8

Read across the top row, down the two sides, and across the bottom row. All four sentences are true,
and 1 to 8 are each used once.

The four sentences to check:

Page 18 of 88

as e
Class 5 Maths Chapter 9 Coconut Farm
a g l AglaSem · NCERT Solutions

co m
e m.
Top row: 6 ÷ 3 = 2 ✓
m l as
.co
Left column: 6 − 5 = 1 ✓
m a g
l a se
g
Right column: 2 × 4 = 8 ✓
aBottom row: 1 + 7 = 8 ✓

co m
e m . ag
g l as
Numbers used: 6, 3, 2, 5, 4, 1, 7, 8 — that is 1 to 8, each once ✓
a
m
How to think about it (this answers "How did you think about solving this?"):

. c o
m a
Using only 1 to 8, the choices are few: 6 ÷ 3 = 2, 8 ÷ 4 =l2,s em
1. Start with the division. It is the fussiest sentence, because one number must divide the

.co
6 ÷ 3 = 2. Now 6, 3 and 2 are used up. Left over: 1, 4, 5, 7, 8. a
g
other exactly. 6 ÷ 2 = 3, 8 ÷ 2 = 4.
2. Trym
l a se the subtraction next. The top-left number is 6, so 6 − ? must be a leftover number. 6 − 5
a g 3. Do
= 1, and both 5 and 1 are free. Take it.

om a s
agl
4. Do the multiplication. The top-right number is 2, so 2 × ? must be a leftover number. Free
numbers left are 4, 7, 8. Take 2 × 4 = 8. .c
a s emrow is now 1 + ? = 8. The only number still free is 7, and
agl
5. Check the addition last. The bottom
1 + 7 = 8. It fits.

Is there more than one answer? Yes. Here is a second one:
co m
m .
m as e
.co a g l
e8m− 7 = 1
8÷4=2

a s
agl
2×3=6
se m
com g l a
1+5=6
m . a
e
Numbers used: 8, 4, 2, 7, 1, 3,a5,s 6 — all of 1 to 8 ✓
agl

co m
.
Tip: In a puzzle like this, always start with the hardest condition. Addition and
subtraction are easy to fix at the end; division hardly ever is. em
c o m g l as
m . a
l a se
ag
.c
m
5. Fill in the blanks: (a) _____ ÷ 18 = 100. (b) _____ ÷ 10 = 610. (c) _____ ÷ 100 = 72. (d) _____ ÷
e
Q5

m a s
co agl
100 = 10. (e) 870 ÷ _____ = 87. (f) _____ ÷ 100 = 70. (g) 200 ÷ _____ = 2. (h) 130 ÷ _____ = 13.

m .
ase
a g l
Use the rule N = D × Q every time.

co m
m .
m ase
.co


a g l Page 19 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

PART QUESTION WHAT TO DO ANSWER

(a) ___ ÷ 18 = 100 18 × 100 1,800

(b) ___ ÷ 10 = 610 10 × 610 6,100

(c) ___ ÷ 100 = 72 100 × 72 7,200

(d) ___ ÷ 100 = 10 100 × 10 1,000

(e) 870 ÷ ___ = 87 870 ÷ 87 10

(f) ___ ÷ 100 = 70 100 × 70 7,000

(g) 200 ÷ ___ = 2 200 ÷ 2 100

(h) 130 ÷ ___ = 13 130 ÷ 13 10

Two different jobs here — be careful which one you have.

1. If the dividend is missing (parts a, b, c, d, f), multiply the divisor by the quotient. For (a): 18
× 100 = 1,800.
2. If the divisor is missing (parts e, g, h), divide the dividend by the quotient. For (e): 870 ÷ 87
= 10.

Check them all: 1800 ÷ 18 = 100 ✓ 6100 ÷ 10 = 610 ✓ 7200 ÷ 100 = 72 ✓ 1000 ÷
100 = 10 ✓ 870 ÷ 10 = 87 ✓ 7000 ÷ 100 = 70 ✓ 200 ÷ 100 = 2 ✓ 130 ÷ 10 = 13 ✓

Mental Strategies for Division — Page 123
Splitting the number, and repeated halving

MENTAL STRATEGIES

Q1 Can you give 5 such examples where you can split the number conveniently?

Here are five. The idea is to break the dividend into parts that your tables already know.

Page 20 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

DIVISION SPLIT IT LIKE THIS DIVIDE EACH PART ANSWER

1236 ÷ 6 1200 + 36 200 + 6 206

728 ÷ 7 700 + 28 100 + 4 104

2408 ÷ 8 2400 + 8 300 + 1 301

4515 ÷ 5 4500 + 15 900 + 3 903

996 ÷ 4 1000 − 4 250 − 1 249

One worked out fully — 1236 ÷ 6:

1. Look for a nearby number that the divisor divides easily. 6 goes into 1200 nicely.
2. Split: 1236 = 1200 + 36.
3. Divide the first part: 1200 ÷ 6 = 200.
4. Divide the second part: 36 ÷ 6 = 6.
5. Add the two answers: 200 + 6 = 206.

Check by a second route: 6 × 206 = 1,236 ✓ 7 × 104 = 728 ✓ 8 × 301 = 2,408 ✓ 5 ×
903 = 4,515 ✓ 4 × 249 = 996 ✓

Why splitting is allowed: If you share 1236 sweets among 6 children, you can hand
out 1200 first and then the last 36. Each child ends up with the same total whichever
way you hand them out.

Q2 For which other divisors and dividends might this strategy of repeated halving
work?

Repeated halving works whenever the divisor is 2, 4, 8, 16, 32 … — the numbers you get by
doubling 2.

Page 21 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

DIVISOR HOW MANY TIMES TO HALVE EXAMPLE ANSWER

2 once 86 → 43 86 ÷ 2 = 43

4 twice 128 → 64 → 32 128 ÷ 4 = 32

8 three times 96 → 48 → 24 → 12 96 ÷ 8 = 12

16 four times 320 → 160 → 80 → 40 → 20 320 ÷ 16 = 20

Why it happens: 4 is 2 × 2, so dividing by 4 is dividing by 2 and then by 2 again. 8 is
2 × 2 × 2, so you halve three times. Each halving undoes one of the twos.

The dividend must cooperate too. Halving 100 gives 50, halving 50 gives 25 — after that you
cannot halve again and stay in whole numbers. So use it when the dividend keeps coming out
even.

Did you know? You can mix the tricks. For ÷ 6, halve once and then divide by 3. Take
132 ÷ 6: half of 132 is 66, and 66 ÷ 3 = 22. Check: 6 × 22 = 132 ✓. In the same way ÷
12 is halve, halve, then ÷ 3.

Try It! — Page 123
Fill the boxes: split the dividend, or halve again and again

TRY IT!

Q1 1. 64 ÷ 4 (split the number into two parts joined by +, divide each part by 4, then
add).

64 ÷ 4 = 16.

1. Split 64 into two parts that 4 divides easily: 64 = 40 + 24.
2. Divide the first part: 40 ÷ 4 = 10.
3. Divide the second part: 24 ÷ 4 = 6.
4. Add: 10 + 6 = 16.

Page 22 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

64 ÷ 4

40 + 24

÷4 ÷4

10 + 6 = 16

Split, divide each part, then add the two answers.

Check by a second route: 4 is 2 × 2, so halve twice. Half of 64 is 32, half of 32 is 16.
Same answer, 16. ✓

Q2 2. 265 ÷ 5 (split, divide each part by 5, then add).

265 ÷ 5 = 53.

1. Split 265 into 250 + 15. Both are easy for the 5 table.
2. Divide: 250 ÷ 5 = 50.
3. Divide: 15 ÷ 5 = 3.
4. Add: 50 + 3 = 53.

265 = 250 + 15

250 ÷ 5 = 50

15 ÷ 5 = 3

50 + 3 = 53

Check by a second route: 5 × 53 = 265 ✓ (because 5 × 50 = 250 and 5 × 3 = 15, and
250 + 15 = 265).

Page 23 of 88

as e
Class 5 Maths Chapter 9 Coconut Farm
a g l AglaSem · NCERT Solutions

co m
m.
3. 1560 ÷ 8 (the two boxes are joined by −, so divide each part by 8 and subtract).
e
Q3

m l as
.co a g
a
s em
agl÷ 8 = 195.
1560
This time the boxes are joined by a minus sign. So look for a nearby number bigger than 1560

com
. ag
that 8 divides neatly.

e m
l as
1. Choose 1600 — the 8 table reaches it easily.
g
a
2. Write the split: 1560 = 1600 − 40.
3. Divide: 1600 ÷ 8 = 200.

co m
m.
4. Divide: 40 ÷ 8 = 5.

m
5. Subtract: 200 − 5 = 195.
as e
.co a g l
a s em= 1600 − 40
gl
1560
a 1600 ÷ 8 = 200
m a s
.co agl
40 ÷ 8 = 5
se m
200 − 5 = 195
g l a
a

co m
.
Check by a second route: 8 × 195 = 8 × 200 − 8 × 5 = 1600 − 40 = 1,560 ✓
e m
m l as
.co a g
a s em
agl Q4 4. 4824 ÷ 24 (split, divide each part by 24, then add).
se m
com g l a
. a

m
ase
agl
4824 ÷ 24 = 201.

1. Split 4824 into 4800 + 24.

m
2. Divide the first part: 4800 ÷ 24. Think 48 ÷ 24 = 2, so 4800 ÷ 24 = 200.

. co
m
3. Divide the second part: 24 ÷ 24 = 1.

m as e
.co l
4. Add: 200 + 1 = 201.
a g
se m
g l a
a
4824 = 4800 + 24
.c
4800 ÷ 24 = 200
s e m
m a
24 ÷ 24 = 1
e m . co agl
g l as
a
200 + 1 = 201

co m
m .
m ase
.co


a g l Page 24 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Check by a second route: 24 × 201 = 24 × 200 + 24 × 1 = 4800 + 24 = 4,824 ✓

Q5 5. 168 ÷ 8 — Halve 168 → ____ , Halve ____ → ____ , Halve ____ → ____

168 ÷ 8 = 21.
8 is 2 × 2 × 2, so halve three times.

halve halve halve
168 84 42 21
Three halvings, because 8 = 2 × 2 × 2
Each arrow cuts the number in half. After three cuts you have divided by 8.

Halve 168 → 84

Halve 84 → 42

Halve 42 → 21

So 168 ÷ 8 = 21

Check by a second route: 8 × 21 = 168 ✓

Q6 6. 144 ÷ 4 — Halve 144 → ____ , Halve ____ → ____

144 ÷ 4 = 36.
4 is 2 × 2, so halve twice.

Page 25 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Halve 144 → 72

Halve 72 → 36

So 144 ÷ 4 = 36

Check by a second route: Split instead. 144 = 120 + 24, so 120 ÷ 4 = 30 and 24 ÷ 4 =
6, and 30 + 6 = 36. Same answer ✓ Also 4 × 36 = 144 ✓

Let Us Solve — Page 124
Solve using the strategies used in the previous question

LET US SOLVE

Q1 (a) 256 ÷ 4

256 ÷ 4 = 64.

1. Split 256 into 240 + 16.
2. 240 ÷ 4 = 60.
3. 16 ÷ 4 = 4.
4. 60 + 4 = 64.

Check by a second route: Halve twice — 256 → 128 → 64. Same answer ✓ And 4 ×
64 = 256 ✓

Q2 (b) 545 ÷ 5

545 ÷ 5 = 109.

1. Split 545 into 500 + 45.
2. 500 ÷ 5 = 100.
3. 45 ÷ 5 = 9.
4. 100 + 9 = 109.

Page 26 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Check: 5 × 109 = 545 ✓

Q3 (c) 147 ÷ 7

147 ÷ 7 = 21.

1. Split 147 into 140 + 7.
2. 140 ÷ 7 = 20.
3. 7 ÷ 7 = 1.
4. 20 + 1 = 21.

Check: 7 × 21 = 147 ✓

Q4 (d) 1212 ÷ 6

1212 ÷ 6 = 202.

1. Split 1212 into 1200 + 12.
2. 1200 ÷ 6 = 200.
3. 12 ÷ 6 = 2.
4. 200 + 2 = 202.

Check: 6 × 202 = 1,212 ✓ Notice there are 0 tens in the answer — 200 and 2, nothing
in between.

Q5 (e) 648 ÷ 12

648 ÷ 12 = 54.

1. Split 648 into 600 + 48.
2. 600 ÷ 12 = 50 (because 12 × 5 = 60, so 12 × 50 = 600).
3. 48 ÷ 12 = 4.
4. 50 + 4 = 54.

Page 27 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Check by a second route: 12 is 4 × 3. Halve 648 twice: 648 → 324 → 162, then 162 ÷
3 = 54. Same answer ✓ And 12 × 54 = 648 ✓

Q6 (f) 9648 ÷ 48

9648 ÷ 48 = 201.

1. Split 9648 into 9600 + 48.
2. 9600 ÷ 48 = 200 (because 48 × 2 = 96, so 48 × 200 = 9600).
3. 48 ÷ 48 = 1.
4. 200 + 1 = 201.

Check: 48 × 201 = 9600 + 48 = 9,648 ✓

Q7 (g) 775 ÷ 25

775 ÷ 25 = 31.

1. Split 775 into 750 + 25.
2. 750 ÷ 25 = 30 (because 25 × 3 = 75, so 25 × 30 = 750).
3. 25 ÷ 25 = 1.
4. 30 + 1 = 31.

Check: 25 × 31 = 775 ✓ Tip: Four 25s make 100, so ₹775 is 31 coins of ₹25.

Q8 (h) 796 ÷ 4

796 ÷ 4 = 199.
796 is very close to 800, so use the take away split.

1. Write 796 = 800 − 4.
2. 800 ÷ 4 = 200.
3. 4 ÷ 4 = 1.

Page 28 of 88

as e
Class 5 Maths Chapter 9 Coconut Farm
a g l AglaSem · NCERT Solutions

4. 200 − 1 = 199.
co m
e m.
m l as
.co a g
Check by a second route: Halve twice — 796 → 398 → 199. Same answer ✓ And 4 ×

se m
199 = 796 ✓
g l a
a
m
.co ag
em coconuts into bags
Susie's Farm in Kerala — Pages 124–125
a s
agl
Susie's and Sunitha's ways of dividing, and packing

LET US SOLVE

. c om
m a s em 1,117
1. Susie and Sunitha have a large coconut farm and they have harvested

gl customers. How many
co in April. They sold 582 coconuts equally to 6 regular
Q1

.
coconuts
a
a s em
coconuts did each customer get? Estimate the answer first. Do you realise that each

agl customer will likely get less than 100 coconuts?
m a s
.co agl

Each customer got 97 coconuts.
a s em
agl600 is more than 582. So each customer must get a little less
Estimate first. 6 × 100 = 600, and
than 100 coconuts. Now do the division and see if the answer matches the estimate.

co m
.
Susie's way — take away small chunks of 6. She takes away 120 coconuts (that is 6 × 20) again

e m
as
and again.
m l
.co a g
a s e6)m582 ( 20 + 20 + 20 + 20 + 10 + 7
agl
m
−120 → 462
a se
−120 → 342
. com a g l
m
ase
agl
−120 → 222

−120 → 102

co m
.
−60 → 42
em
m l as
.co
−42 → 00
a g
se m
g l a
a c
.
20 + 20 + 20 + 20 + 10 + 7 = 97

s e m
m a
m . co
Sunitha's way — take away one big chunk. She sees straight away that 6 × 90 = 540.
e agl
g l as
a

co m
m .
m ase
.co


a g l Page 29 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

6) 582 ( 90 + 7

−540 → 42

−42 → 00

90 + 7 = 97

Why both work: Every chunk you take away is a whole number of sixes. Adding up
how many sixes you removed gives the quotient. Susie removed the sixes in six little
bites; Sunitha removed them in two big bites. The total is the same.

Check by a second route: 6 × 97 = 582 ✓ And 97 is just under 100, exactly as the
estimate said.
Answer: 97 coconuts for each customer.

Q2 Do you think Sunitha's method is better? Discuss which one you would prefer and
why.

Sunitha's method is better for a quick answer, because it needs only 2 steps instead of 6.

SUSIE'S WAY SUNITHA'S WAY

Chunks taken away 120, 120, 120, 120, 60, 42 540, 42

Number of steps 6 2

Table facts needed only 6 × 20 6 × 90

Chances of a mistake more, as there are more subtractions fewer

Sample answer for the class discussion: Both methods give 97, so both are correct. I would
prefer Sunitha's method because it is shorter and there are fewer subtractions to get wrong. But
Susie's method is safer when I am not sure how big a chunk to take — I can always take a small
chunk and go again. The more of my tables I know, the bigger the chunks I can take, and the
closer I get to Sunitha's way.

Page 30 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

The habit to build: Before you start, ask "what is the biggest chunk I can take away
in one go?" Estimating first (6 × 90 = 540, close to 582) is exactly what lets you take a
big chunk.

Q3 Each bag can hold 25 coconuts. How many bags would be needed to pack 97
coconuts?

4 bags.
We need to find how many bags of 25 hold 97 coconuts. So we divide 97 by 25.

1. Count in 25s: 25, 50, 75, 100. Three bags hold 25 + 25 + 25 = 75 coconuts.
2. What is left? 97 − 75 = 22 coconuts.
3. Those 22 still need a bag. A fourth bag is needed, even though it will not be full.
4. So the answer is 3 + 1 = 4 bags.

97 ÷ 25 → quotient 3, remainder 22

N=D×Q+R

97 = 25 × 3 + 22 ✓

Bags needed = 3 full + 1 part-full = 4 bags

Why we round up: You cannot leave 22 coconuts lying loose. Every leftover coconut
still has to go into a bag, so a part-full bag counts as a whole bag.

Careful: The answer to the division is 3, but the answer to the question is 4. Always
read the question again once you have the quotient.

Q4 2. They pack the remaining coconuts for drying and extracting oil. They can pack 25
coconuts in each bag. How many bags will they need to pack the remaining
coconuts? (Guess the number of bags needed first. Use the strategies learnt.)

22 bags.
Step 1 — how many coconuts are left? They harvested 1,117 and sold 582.

Page 31 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

1117 − 582 = 535 coconuts

Step 2 — guess first. 25 × 20 = 500, which is close to 535. So the answer will be a little more
than 20 bags.
Step 3 — divide 535 by 25 using big chunks.

25) 535 ( 20 + 1

−500 → 35 (25 × 20 = 500)

−25 → 10 (25 × 1 = 25)

Quotient = 20 + 1 = 21, Remainder = 10

The book asks: can we write 30 here? No. 25 × 30 = 750, which is bigger than 535. You can never
take away more than you have.
Step 4 — read the question again. 21 bags are full and 10 coconuts are still loose. Those 10
need one more bag.

Bags = 21 full + 1 for the leftover 10 = 22 bags

Check by a second route: 21 bags hold 21 × 25 = 525 coconuts. 525 + 10 = 535 ✓
And in the rule N = D × Q + R: 535 = 25 × 21 + 10 ✓
Answer: 22 bags.

Let Us Learn to Divide — Page 125
Filling the partial-quotient ladders, and the rule N = D × Q + R

LET US LEARN TO DIVIDE

Q1 726 ÷ 4 — complete: 4) 726 (100 + ____ + ____ . Could we have written 200 here?

726 ÷ 4 = 181 with remainder 2. The blanks are 80 and 1.

Page 32 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

4) 726 ( 100 + 80 + 1

−400 → 326 (4 × 100 = 400)

−320 → 6 (4 × 80 = 320)

−4 → 2 (4 × 1 = 4)

Quotient = 100 + 80 + 1 = 181

Remainder = 2

How each blank was found:

1. After taking away 400, we have 326 left. Ask: how many fours in 326? 4 × 80 = 320, which fits.
4 × 90 = 360, too big. So the chunk is 80.
2. After taking away 320, we have 6 left. 4 × 1 = 4 fits, 4 × 2 = 8 is too big. So the chunk is 1.
3. Now 2 is left, and 2 is smaller than 4. We cannot take away another whole 4. So 2 is the
remainder.

Could we have written 200 here? No.

4 × 200 = 800

800 is bigger than 726

So 200 is too big a chunk to take away

Check by a second route: N = D × Q + R → 4 × 181 + 2 = 724 + 2 = 726 ✓

Q2 902 ÷ 16 — complete: 16) 902 (____ + 6. What should we write here so that we get a
number close to 902 but less than it? Could we have multiplied 16 by a larger tens?

902 ÷ 16 = 56 with remainder 6. The blank is 50.

Page 33 of 88

as e
Class 5 Maths Chapter 9 Coconut Farm
a g l AglaSem · NCERT Solutions

co m
em.
16) 902 ( 50 + 6
m l as
.co
−800 → 102 (16 × 50 = 800)
m a g
l a se
g
−96 → 6 (16 × 6 = 96)
a

co m
. ag
Quotient = 50 + 6 = 56
e m
Remainder = 6
g l as
a
m
Finding the first chunk — count in tens of 16.
co
e m.
m l as
.co
TRY 16 × THAT IS IT LESS THAN 902?

a g
se m
l a
40 640 Yes, but far below 902

g
a 50 800 Yes — close to 902 ✓ best choice

m a s
em
.co agl
60 960 No, bigger than 902

a s
Could we have multiplied 16 bygal larger tens? No. The next tens up is 60, and 16 × 60 = 960,
a
which is more than 902. So 50 is the largest tens that fits.

c o m
Check by a second route: N = D × Q + R → 16 × 56 + 6 = 896 + 6 =m .
a s e 902 ✓ And the

. com 6 is smaller than the divisor 16, as it always must
remainder
a glbe.
m
ase
agl
se m
com between N, D, Q and R? Is 726 = 4 × 181? a
Sometimes, the divisor (D) does not completely divide the dividend (N) and leaves a
l
Q3

. a g
em Is 902 = 16 × 56? Yes/No. So, 902 = 16 × 56 + ______.
remainder (R). What is the relationship
Yes/No. So, 726 = 4 × 181 +s______.
a
agl

co m
The relationship is N = D × Q + R.
m .
m ase
.co
First one: Is 726 = 4 × 181? No.
a g l
se m
g l a
a 4 × 181 = 724
c
m .
724 is 2 less than 726
m a s e
So 726 = 4 × 181 + 2
e m . co agl
g l as
Second one: Is 902 = 16 × 56? No.a

co m
m .
m as e
.co


a g l Page 34 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

16 × 56 = 896

896 is 6 less than 902

So 902 = 16 × 56 + 6

the part that shares out evenly

D × Q = 4 × 181 = 724 2

remainder
Whole strip = N = 726

The dividend is made of the part that shares out evenly plus the little bit left over.

The four names, in one sentence: the dividend (N) is what you start with, the divisor (D) is
how you share it, the quotient (Q) is what each group gets, and the remainder (R) is what
would not fit.

Two rules to remember:
1. N = D × Q + R — use it to check every division you do.
2. The remainder is always smaller than the divisor. If your remainder is bigger, the
quotient should have been larger.

Let Us Solve — Page 126
Solve the following word problems

LET US SOLVE

Q1 1. Rani is planning to host a party. She estimates that 250 guests will attend. She
plans to serve one samosa to each guest. Samosas are available in packs of 6 or 8.
Which pack should Rani buy? Explain your answer.

Packs of 6 are the better single choice — she needs 42 packs and only 2 samosas are left
over.
She needs at least 250 samosas, one for each guest. Let us try each pack size.

Page 35 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

If she buys packs of 6:

Step 1: 250 ÷ 6 → 6 × 41 = 246, and 250 − 246 = 4

Step 2: So 41 packs give only 246 samosas — 4 guests would get nothing

Step 3: She must buy 42 packs

Step 4: 42 × 6 = 252 samosas, so 2 samosas are extra

If she buys packs of 8:

Step 1: 250 ÷ 8 → 8 × 31 = 248, and 250 − 248 = 2

Step 2: So she must buy 32 packs

Step 3: 32 × 8 = 256 samosas, so 6 samosas are extra

PACK SIZE PACKS NEEDED SAMOSAS BOUGHT LEFT OVER

6 42 252 2

8 32 256 6

Explanation: Both choices feed all 250 guests. Packs of 6 waste only 2 samosas, packs of 8
waste 6. So packs of 6 are the better buy. (If Rani cares more about carrying fewer packs, packs
of 8 mean only 32 packs to carry instead of 42.)

Did you know? If Rani may buy both sizes, she can get exactly 250 with nothing
wasted: 29 packs of 8 and 3 packs of 6.
29 × 8 = 232, 3 × 6 = 18, 232 + 18 = 250 ✓

Q2 2. 342 students from a school are going on a trip to the Science Park. Each bus can
carry a maximum of 41 students. How many buses does the school need to arrange?

The school needs 9 buses.
We need to find how many buses of 41 seats hold 342 students. So we divide.

Page 36 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Step 1: 342 ÷ 41

Step 2: 41 × 8 = 328 (this fits)

Step 3: 41 × 9 = 369 (too many seats, but 8 buses are not enough)

Step 4: 342 − 328 = 14 students left over

Step 5: Those 14 students also need a bus

Buses = 8 full + 1 more = 9 buses

Why not 8? Eight buses carry only 328 students. Fourteen children would be left
standing at the school gate. A part-full bus still counts as a bus.

Check by a second route: N = D × Q + R → 41 × 8 + 14 = 328 + 14 = 342 ✓
Answer: 9 buses.

Q3 3. Sofia has only ₹50 and ₹20 notes. She needs to pay ₹520 using these notes. How
many ₹50 and ₹20 notes does she need to make ₹520? Find out the different
possible combinations.

There are six ways. The neatest is 8 notes of ₹50 and 6 notes of ₹20.
How to find them, step by step.

1. Notice that ₹520 ends in a 0, and both ₹50 and ₹20 end in a 0. So work in tens: she needs 52
tens.
2. A ₹50 note is 5 tens. A ₹20 note is 2 tens.
3. Try 0 fifty-rupee notes, then 1, then 2 … and see what is left for the ₹20 notes.
4. Whatever is left must divide exactly by 20, so the number of ₹50 notes must be even.

Page 37 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

₹50 NOTES VALUE STILL TO PAY ₹20 NOTES CHECK

0 ₹0 ₹520 26 26 × 20 = 520 ✓

2 ₹100 ₹420 21 100 + 420 = 520 ✓

4 ₹200 ₹320 16 200 + 320 = 520 ✓

6 ₹300 ₹220 11 300 + 220 = 520 ✓

8 ₹400 ₹120 6 400 + 120 = 520 ✓

10 ₹500 ₹20 1 500 + 20 = 520 ✓

Why the number of ₹50 notes must be even: One ₹50 note leaves ₹470, and 470 ÷
20 = 23 with 10 left over — you cannot pay ₹10 with ₹20 notes. Two ₹50 notes make
₹100, which is a whole number of twenties. So the fifties must come in pairs.

Fewest notes: The row with 10 fifties and 1 twenty uses just 11 notes — the lightest
purse. The row with 26 twenties uses the most.

Q4 4. Three friends decide to split the money spent on their picnic equally. They buy
snacks and sweets for ₹157, juice and fruits for ₹124 and pulav and paratha for
₹136. How much should each person pay to share the cost equally?

Each person should pay ₹139.
We need to find one person's share. First find the whole cost, then share it into 3 equal parts.

Page 38 of 88

as e
Class 5 Maths Chapter 9 Coconut Farm
a g l AglaSem · NCERT Solutions

co m
e m.
Step 1 — add up everything spent:
m l as
.co
157 + 124 = 281
m a g
l a se
281 + 136 = ₹417 in all
a g

co m
. ag
Step 2 — share it among 3 friends:
e m
417 ÷ 3
g l as
Split 417 into 300 + 117 a

co m
m.
300 ÷ 3 = 100

m as e
.co l
117 ÷ 3 = 39
a g
se m
100 + 39 = ₹139
g l a
a
m a s
agl
Check by a second route: 3 × ₹139 = ₹417 ✓, and ₹417 is exactly what they spent.
Answer: ₹139 each.
m .co
l a se
a g

co m
5. Identify the remainder, if any. Check if N = D × Q + R. (a) 887 ÷ 3 (b) 283 ÷ 8 (c) 745 ÷
.
Q5

e m
as
5 (d) 767 ÷ 26 (e) 530 ÷ 41 (f) 888 ÷ 67
m l
m .co a g
l a se

a g
m
PART DIVISION QUOTIENT (Q) REMAINDER (R) CHECK: N = D × Q + R

a se
com 2 g l
m. a
(a) 887 ÷ 3 295 3 × 295 + 2 = 885 + 2 = 887 ✓

ase
agl
(b) 283 ÷ 8 35 3 8 × 35 + 3 = 280 + 3 = 283 ✓

m
(c) 745 ÷ 5 149 0 5 × 149 + 0 = 745 ✓

26 m . c o
(d)
s e
m la
767 ÷ 26 29 13 × 29 + 13 = 754 + 13 = 767 ✓

. co ag
m
ase
(e) 530 ÷ 41 12 38 41 × 12 + 38 = 492 + 38 = 530 ✓

agl c
.
(f) 888 ÷ 67 13 17 67 × 13 + 17 = 871 + 17 = 888 ✓

s e m
m a
. co agl
Two worked out in full.

e m
l as
(a) 887 ÷ 3

a g

co m
m .
m ase
.co


a g l Page 39 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

3) 887 ( 200 + 90 + 5

−600 → 287 (3 × 200)

−270 → 17 (3 × 90)
−15 → 2 (3 × 5)

Q = 200 + 90 + 5 = 295, R = 2

(d) 767 ÷ 26

26) 767 ( 20 + 9

−520 → 247 (26 × 20)

−234 → 13 (26 × 9)

Q = 20 + 9 = 29, R = 13

Always true: the remainder is smaller than the divisor. Look down the table — 2 < 3,
3 < 8, 13 < 26, 38 < 41, 17 < 67. If your remainder is ever as big as the divisor, take
away one more group.

Kalpavruksha Coconut Oil — Pages 126–127
Oil, coir and tender coconuts from Susie and Sunitha's farm

KALPAVRUKSHA COCONUT OIL

Q1 1. In a particular year, Susie and Sunitha used 4376 coconuts for extracting coconut
oil. They can extract 1 l of oil from 8 coconuts. What quantity of oil were they able to
extract?

They extracted 547 litres of oil.
Every 8 coconuts make 1 litre. So the number of litres is the number of groups of 8 inside 4376.
That is a division.
Susie's way — small chunks.

Page 40 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

8) 4376 ( 200 + 200 + 100 + 40 + 7

−1600 → 2776

−1600 → 1176

−800 → 376

−320 → 56

−56 → 00

200 + 200 + 100 + 40 + 7 = 547

Sunitha's way — the biggest chunk first. She thinks: 8 × 500 = 4000, and 4000 is close to 4376.

8) 4376 ( 500 + 40 + 7

−4000 → 376

−320 → 56

−56 → 00

500 + 40 + 7 = 547

Check by a second route: 8 × 547 = 8 × 500 + 8 × 40 + 8 × 7 = 4000 + 320 + 56 = 4,376
✓
Answer: 547 litres of oil.

Q2 How much will they earn if they sell the oil at ₹175 for 1 l? They will earn ₹ 547 × 175.
Find out.

They will earn ₹95,725.
547 litres, each sold at ₹175. So multiply.

Page 41 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Step 1 — break 175 into 100 + 70 + 5

Step 2 — multiply each part:
547 × 100 = 54,700

547 × 70 = 38,290

547 × 5 = 2,735

Step 3 — add them up:

54,700 + 38,290 = 92,990

92,990 + 2,735 = ₹95,725

The same sum written in columns:

547

×175
─────────

2 7 3 5 (547 × 5)

3 8 2 9 0 (547 × 70)

5 4 7 0 0 (547 × 100)

─────────

95725

Check with an estimate: 550 × 175 is about 96,250 — very close to ₹95,725, so the
answer is sensible.
Answer: ₹95,725.

Page 42 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Q3 2. Coconut husk is used for making coir. Susie and Sunitha's farm sells coconut husk
at ₹23 per kilogram. They earned ₹9913 from the sale of husk in May. What quantity
of husk did they sell in May? What would happen if 23 is multiplied by 300 or 500?

They sold 431 kg of coconut husk in May.
Each kilogram brings ₹23. So the number of kilograms is how many 23s there are in 9913.
Make a guess first. 23 × 400 = 9200, which is close to 9913. So the answer is a bit more than
400 kg.

23) 9913 ( 400 + 30 + 1

−9200 → 713 (23 × 400)

−690 → 23 (23 × 30)

−23 → 00 (23 × 1)

Quotient = 400 + 30 + 1 = 431 kg, Remainder = 0

What would happen if 23 is multiplied by 300 or 500?

CHUNK 23 × WHAT HAPPENS
TRIED CHUNK

300 6,900 It fits, but 3,013 would still be left — you would need extra steps. Too small
a chunk.

400 9,200 Fits and comes closest to 9,913. Best choice.

500 11,500 Bigger than 9,913 — you cannot take away more than you have.

Check by a second route: 23 × 431 = 23 × 400 + 23 × 30 + 23 × 1 = 9200 + 690 + 23 =
9,913 ✓
Answer: 431 kg.

Page 43 of 88

as e
Class 5 Maths Chapter 9 Coconut Farm
a g l AglaSem · NCERT Solutions

co m
m.
3. In the hot summer months, tender coconuts are sold for ₹35. Ibrahim earns ₹8890
e
Q4

comsold ______ tender coconuts. l as
in a week. How many tender coconuts did he sell? Complete: 35) 8890 (____ + 50 + ____
g
. a
em
. Ibrahim

l a s
ag

m
Ibrahim sold 254 tender coconuts. The blanks are 200 and 4.
. co ag
e m
Every coconut brings ₹35, so the number of coconuts is how many 35s make ₹8890.

g l as
35) 8890 ( 200 + 50 + 4
a

co m
m.
−7000 → 1890 (35 × 200 = 7000)

m as e
.co
−1750 → 140 (35 × 50 = 1750)
a g l
se m
a
−140 → 00 (35 × 4 = 140)

ag l

m a s
.co agl
Quotient = 200 + 50 + 4 = 254, Remainder = 0

se m
g l a
a
How each chunk was chosen:

1. 35 × 200 = 7,000 fits inside 8,890. 35 × 300 = 10,500 would be too big. So the first chunk is
200.
. c om
2. 1,890 is left. 35 × 50 = 1,750 fits. 35 × 60 = 2,100 is too big. So the next m
s e chunk is 50.

. com
3. 140 is
a glaremains.
left. 35 × 4 = 140 exactly. So the last chunk is 4, and nothing

a s em
ag l Check by a second route: 35 × 254 = 35 × 250 + 35 × 4 = 8,750 + 140 = ₹8,890 ✓
Answer: 254 tender coconuts.
se m
com g l a
m . a
ase
Q5 agl
Ibrahim had bought the tender coconuts for ₹20 each. How much extra money did
he earn by selling the coconuts at ₹35? The cost of _______ coconuts at ₹20 each =

co m
.
_______ × ₹20 = ₹_______. He earned ₹8890 from the sale. The extra amount he earned is
m
as e
com
₹8890 – ₹_______ = ₹_______.

. a g l
e m
as
agl

He earned ₹3,810 extra.
.c
s e m
m a
co agl
From the last question we know he sold 254 tender coconuts.

m .
ase
a g l

com
m .
m ase
.co


a g l Page 44 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

The cost of 254 coconuts at ₹20 each

= 254 × ₹20

= ₹5,080

He earned ₹8,890 from the sale.

Extra amount = ₹8,890 − ₹5,080 = ₹3,810

Working out 254 × 20, step by step:

1. 254 × 2 = 508.
2. Multiplying by 20 is multiplying by 2 and then by 10, so add a zero: 5,080.

Working out the subtraction:

8890 − 5080

= (8890 − 5000) − 80

= 3890 − 80

= ₹3,810

Why it works out this way: On every single coconut he made ₹35 − ₹20 = ₹15
profit. For 254 coconuts that is 254 × 15.

Check by that second route: 254 × 15 = 254 × 10 + 254 × 5 = 2,540 + 1,270 = ₹3,810
✓ — the same answer.
Answer: ₹3,810 extra.

Division Using Place Value — Pages 128–129

Page 45 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Sunitha's mother shares the candies; dividing Hundreds, Tens and Ones in turn

DIVISION USING PLACE VALUE

Q1 Sunitha's mother has 62 candies to be distributed equally among 5 children. How
many candies would each child get? 1. 62 ÷ 5 → Divide 62 into 5 equal parts.

Each child gets 12 candies, and 2 candies are left over.
Think of the 62 candies as 6 bundles of ten and 2 loose candies.

6 Tens and 2 Ones = 62

5 children take 1 ten each.
1 ten is left, so break it
into 10 Ones.
10 + 2 = 12 Ones to share
12 ÷ 5 = 2 each, 2 left over

Share the tens first, then break what is left into ones.

1. Share the Tens. 6 Tens ÷ 5 = 1 Ten each, and 1 Ten is left.
2. Regroup. That leftover 1 Ten becomes 10 Ones. With the 2 Ones already there, we now have
12 Ones.
3. Share the Ones. 12 Ones ÷ 5 = 2 Ones each, and 2 Ones are left.
4. Read the answer. Each child gets 1 Ten and 2 Ones, that is 12 candies. The remainder is 2.

TO

5) 62 ( 1 2

−5

──

12 (Ones)

−10

──

2 (remainder)

Page 46 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Check by a second route: N = D × Q + R → 5 × 12 + 2 = 60 + 2 = 62 ✓
Answer: 12 candies each, 2 left over.

Q2 2. 75 ÷ 8 → Divide 75 into 8 equal parts. Can we divide this into 8 equal parts without
breaking them? What can we do?

75 ÷ 8 = 9 with remainder 3.
Can we share the 7 Tens among 8 without breaking them? No. There are only 7 tens and 8
children, so not even one ten each. Every child gets 0 Tens.

1. Regroup everything into Ones. 7 Tens = 70 Ones. Add the 5 Ones already there: 75 Ones.
2. Share the 75 Ones among 8. 8 × 9 = 72, which fits. 8 × 10 = 80 is too many.
3. So each part gets 9, and 75 − 72 = 3 are left.

TO

8) 75 ( 0 9

−72 (Ones)
──

3 (remainder)

Why the 0 in the Tens place: It records that nobody got any tens. We do not write it
in the final answer (we say 9, not 09) but writing it while working keeps every digit in
its correct column.

Check by a second route: 8 × 9 + 3 = 72 + 3 = 75 ✓

Q3 3. 324 ÷ 3 → Divide 324 into 3 equal parts. Why do we put a 0 here?

324 ÷ 3 = 108, with no remainder.
Break 324 into place values first: 3 Hundreds + 2 Tens + 4 Ones.

1. Share the Hundreds. 3 Hundreds ÷ 3 = 1 Hundred each. Nothing left over.

Page 47 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

2. Share the Tens. 2 Tens shared among 3 — not possible without breaking them. So everyone
gets 0 Tens.
3. Regroup. The 2 Tens become 20 Ones. With the 4 Ones already there, we have 20 + 4 = 24
Ones.
4. Share the Ones. 24 Ones ÷ 3 = 8 Ones each. Nothing left over.
5. Read the quotient down the columns: 1 Hundred, 0 Tens, 8 Ones = 108.

HTO
3) 324 ( 1 0 8

−3 (Hundreds)

──

2 (Tens)

−0

──

24 (Ones)

−24

──

00

Why do we put a 0 here?

Because there really are zero Tens in each part. The 0 is not decoration — it is a
number. If you leave it out, the 1 and the 8 slide together and you write 18. But 18 is
far too small: 3 × 18 = 54, not 324. The 0 holds the Tens place open so the 1 stays a
Hundred.

Check by a second route: 3 × 108 = 324 ✓ Or split: 324 = 300 + 24, and 300 ÷ 3 =
100, 24 ÷ 3 = 8, so 100 + 8 = 108. Same answer ✓

Q4 4. 136 ÷ 6 → Divide 136 into 6 equal parts.

136 ÷ 6 = 22 with remainder 4.
Break 136 into 1 Hundred + 3 Tens + 6 Ones.

Page 48 of 88

as e
Class 5 Maths Chapter 9 Coconut Farm
a g l AglaSem · NCERT Solutions

1. Try the Hundreds. 1 Hundred among 6 — not possible. Each part gets 0 Hundreds.
co m
2. Regroup. 1 Hundred becomes 10 Tens. With the 3 Tens already there, we have 13 Tens.
e m.
m l as
.co a g
3. Share the Tens. 13 Tens ÷ 6 = 2 Tens each (6 × 2 = 12), and 1 Ten is left.

se m
a
4. Regroup again. That 1 Ten becomes 10 Ones. With the 6 Ones already there, we have 16

a g l
Ones.
5. Share the Ones. 16 Ones ÷ 6 = 2 Ones each (6 × 2 = 12), and 4 Ones are left.

co m
ag
6. Read the quotient: 0 Hundreds, 2 Tens, 2 Ones = 22, remainder 4.

m .
as e
HTO
a g l
6) 136 ( 0 2 2
co m
−12 (Tens)
em.
m l as
──
m .co a g
a 16e (Ones)
s
a gl
−12
m a s
.co agl
──

se m
4 (remainder)
g l a
a

co m
.
Check by a second route: N = D × Q + R → 6 × 22 + 4 = 132 + 4 = 136 ✓
e m
m l as
.co a g
a s em Can you tell just by looking at the divisor and dividend, how many digits the
a gl Q5
m
quotient would have? Look at the problems above and find this out. Explain your

a se
com l
thoughts.
. a g
m
ase
agl

Yes. Compare the divisor with the front of the dividend.

om
The test, in two steps:
. c
s e
1. Look at the first digit of the dividend. If the divisor is smaller than mor equal to it, the
. om has the same number of digits as the dividend.agla
cquotient
a s em2. If the divisor is bigger than that first digit, the quotient has one digit fewer.
agl c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 49 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

PROBLEM DIVISOR VS FIRST DIGIT DIGITS IN THE QUOTIENT QUOTIENT

324 ÷ 3 3 is not bigger than 3 3 digits 108

136 ÷ 6 6 is bigger than 1 2 digits 22

62 ÷ 5 5 is not bigger than 6 2 digits 12

75 ÷ 8 8 is bigger than 7 1 digit 9

Explain your thoughts — sample answer: I look at the first digit of the dividend. In 136 ÷ 6,
the 1 stands for 1 Hundred, and 1 Hundred cannot be shared among 6 parts. So there is no digit
in the Hundreds place of the answer, and the quotient must be a 2-digit number. In 324 ÷ 3, the
3 Hundreds can be shared among 3, so the answer does have a Hundreds digit and is a 3-digit
number.

For a two-digit divisor, compare it with the first two digits. In 902 ÷ 16, we ask
whether 16 fits into 90 — it does, so the quotient has 2 digits (56). In 2874 ÷ 14, 14
fits into 28, so the quotient has 3 digits (205).

Let Us Divide — Pages 130–131
Each division done twice: with partial quotients and with place value

LET US DIVIDE

Q1 (a) 7,032 ÷ 6 — complete both solutions: 6) 7,032 (1,000 + ____ + 70 + ____ , and the
place value method.

7,032 ÷ 6 = 1,172, with no remainder. The blanks are 100 and 2.
First way — partial quotients (take away chunks).

Page 50 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

6) 7032 ( 1,000 + 100 + 70 + 2

−6000 → 1032 (6 × 1000)

−600 → 432 (6 × 100)

−420 → 12 (6 × 70)

−12 → 0 (6 × 2)

1000 + 100 + 70 + 2 = 1,172

Second way — place value (Thousands, Hundreds, Tens, Ones).

Th H T O

6) 7032 ( 1 1 7 2

−6 (Thousands) → 1 Th left

10 (Hundreds: 1 Th regrouped = 10 H, plus 0 H)

−6 → 4 H left

43 (Tens: 4 H = 40 T, plus 3 T)

−42 → 1 T left

12 (Ones: 1 T = 10 O, plus 2 O)

−12 → 0

1. 7 Thousands ÷ 6 = 1 Thousand, 1 Thousand left.
2. Regroup: 1 Th = 10 Hundreds. 10 Hundreds ÷ 6 = 1 Hundred, 4 Hundreds left.
3. Regroup: 4 H = 40 Tens, plus the 3 Tens = 43 Tens. 43 ÷ 6 = 7 Tens, 1 Ten left.
4. Regroup: 1 Ten = 10 Ones, plus the 2 Ones = 12 Ones. 12 ÷ 6 = 2 Ones, nothing left.

Check by a second route: 6 × 1,172 = 6 × 1,000 + 6 × 100 + 6 × 72 = 6,000 + 600 + 432
= 7,032 ✓
Both methods gave 1,172 — that is the point of doing it twice.

Page 51 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Q2 (b) 3,005 ÷ 5 — complete both solutions: 5) 3,005 (____ + ____ , and the place value
method. Discuss why we have to write this 0 here.

3,005 ÷ 5 = 601, with no remainder. The blanks are 600 and 1.
First way — partial quotients.

5) 3005 ( 600 + 1

−3000 → 5 (5 × 600)

−5 → 0 (5 × 1)

600 + 1 = 601

Second way — place value.

Th H T O

5) 3005 ( 0 6 0 1

3 Thousands ÷ 5 → not possible, so 0 Th

−30 (Hundreds: 3 Th = 30 H) → 0 H left

00 (Tens) − 00 → 0

5 (Ones) − 5 → 0

1. 3 Thousands cannot be shared among 5. Write 0 in the Thousands place of the quotient.
2. Regroup: 3 Th = 30 Hundreds. 30 ÷ 5 = 6 Hundreds, nothing left.
3. There are 0 Tens to share. 0 ÷ 5 = 0 Tens. Write the 0.
4. 5 Ones ÷ 5 = 1 One, nothing left.
5. Quotient = 0 Th, 6 H, 0 T, 1 O = 601.

Discuss why we have to write this 0 here — sample answer:

The 0 in the Tens place says "each part gets no tens". It keeps the 6 sitting in the
Hundreds column and the 1 sitting in the Ones column. If we skipped it, the digits
would slide together and we would write 61 — but 5 × 61 is only 305, nowhere near
3,005. The zero is what makes 601 mean six hundred and one instead of sixty-one.

Page 52 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Check by a second route: 5 × 601 = 3,005 ✓ Also 601 × 5 = 600 × 5 + 1 × 5 = 3,000 +
5 = 3,005 ✓

Q3 (c) 2,874 ÷ 14 — complete both solutions: 14) 2,874 (____ + ____ , and the place value
method.

2,874 ÷ 14 = 205 with remainder 4. The blanks are 200 and 5.
First way — partial quotients.

14) 2874 ( 200 + 5

−2800 → 74 (14 × 200 = 2800)

−70 → 4 (14 × 5 = 70)

Quotient = 200 + 5 = 205, Remainder = 4

Second way — place value.

Th H T O

14) 2874 ( 0 2 0 5

2 Thousands ÷ 14 → not possible, so 0 Th

−28 (Hundreds: 2 Th = 20 H, plus 8 H = 28 H) → 0 H left

7 (Tens) − 0 → 7 Tens left, so 0 Tens in the quotient

74 (Ones: 7 T = 70 O, plus 4 O)

−70 → 4 (remainder)

1. 2 Thousands cannot be shared among 14. Write 0 in the Thousands place.
2. 28 Hundreds ÷ 14 = 2 Hundreds, nothing left.
3. 7 Tens ÷ 14 — not possible, so 0 Tens. Write the 0.
4. Regroup: 7 Tens = 70 Ones, plus 4 Ones = 74 Ones. 74 ÷ 14 = 5, with 4 left over.

Page 53 of 88

as e
Class 5 Maths Chapter 9 Coconut Farm
a g l AglaSem · NCERT Solutions

co m
m.
Check by a second route: N = D × Q + R → 14 × 205 + 4 = 2,870 + 4 = 2,874 ✓ And 4

m as e
l
is smaller than 14, so the remainder is right.

m .co a g
l a se
a g
Q4 (d) 9,805 ÷ 32 — complete both solutions: 32) 9,805 (____ + ____ , and the place value

co m
ag
method. Compare both solutions. Also, remember to put 0 in the right places.

m .
as e
a g l
9,805 ÷ 32 = 306 with remainder 13. The blanks are 300 and 6.

co m
m.
First way — partial quotients.

as e
. co(m300 + 6 a g l
em
32) 9805
a s
a gl −9600 → 205 (32 × 300 = 9600)

s
−192 → 13 (32 × 6 = 192)
m a
m .co agl
l a se
a
Quotient = 300 + 6 = 306, Remainder = 13
g
Why 300 and not 400? 32 × 400 = 12,800, which is bigger than 9,805. Why 6 and not 7? 32 × 7 =
co m
m .
e
224, bigger than 205.
m l as
.co
Second way — place value.
m a g
l a se
ag Th H T O

se m
32) 9805 ( 0 3 0 6
com g l a
m . a
ase
9 Thousands ÷ 32 → not possible, so 0 Th

−96 (Hundreds: 98 H) → 2 H left agl
m
20 (Tens: 2 H = 20 T, plus 0 T) − 0 → 0 Tens in the quotient
. co
em
as
205 (Ones: 20 T = 200 O, plus 5 O)
om → 13 (remainder)
.c−192 a g l
se m
g l a
a c
Compare both solutions — sample answer:
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 54 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

PARTIAL QUOTIENTS PLACE VALUE

What you take away whole chunks like 32 × 300 one place value at a time

Number of steps 2 4 (Th, H, T, O)

Chance of a zero mistake low — you add the chunks up higher — the 0 in the Tens place is easy to forget

Answer 306 R 13 306 R 13 — the same

The zero warning: In the place value method, 20 Tens shared among 32 gives 0
Tens. That 0 must be written. Drop it and the answer becomes 36 instead of 306 —
ten times too small.
Check: 32 × 306 + 13 = 9,792 + 13 = 9,805 ✓

Let Us Do — Page 131

Page 55 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Find the missing numbers, and the 3-digit number riddle

LET US DO

Q1 1. Find the missing numbers such that there is no remainder. Remember, there
could be more than one solution. The ten puzzles are: 4)480(□□0 ; 3)906(□0□ ;
20)400(□0 ; 50)100□(□0 ; □)8□8(21□ ; 3)□36(3□□ ; □)88(□□ ; □)9□0(□□□ ; □)180(□□□ ;
□)6,480(□□□

PUZZLE FILLED IN HOW TO FIND IT OTHER ANSWERS

4) 480 ( □ □ 0 4) 480 ( 1 2 0 480 ÷ 4 = 120 only one

3) 906 ( □ 0 □ 3) 906 ( 3 0 2 906 ÷ 3 = 302 only one

20) 400 ( □ 0 20) 400 ( 2 0 400 ÷ 20 = 20 only one

50) 100 □ ( □ 0 50) 1000 ( 2 50 × 20 = 1000, so the missing digit only one
0 is 0

□) 8 □ 8 ( 2 1 □ 4) 848 ( 2 1 2 4 × 212 = 848 4) 868 ( 2 1 7

3) □ 3 6 ( 3 □ □ 3) 936 ( 3 1 2 3 × 312 = 936 only one

□) 8 8 ( □ □ 4) 88 ( 2 2 4 × 22 = 88 2) 88 ( 4 4 and 8) 88 ( 1 1

□) 9 □ 0 ( □ □ □ 3) 900 ( 3 0 0 3 × 300 = 900 2) 900 ( 4 5 0, 5) 900 ( 1 8 0, 4) 920 ( 2
30…

□) 1 8 0 ( □ □ □ 1) 180 ( 1 8 0 only a divisor of 1 keeps the quotient only one
3-digit

□) 6,4 8 0 ( □ □ 8) 6480 ( 8 1 8 × 810 = 6480 9) 6480 ( 7 2 0
□ 0

How to attack a puzzle like this — the method.

1. Count the boxes in the quotient. That tells you how many digits the answer must have.
2. Use any digit already printed. In 3) □36 ( 3□□, the quotient starts with 3 and the dividend
ends in 36. Since 3 × 3 hundreds = 9 hundreds, the dividend must start with 9.
3. Multiply back to test. 3 × 312 = 936 ✓ — the printed digits all match, so it is right.

One shown in full — □) 8 □ 8 ( 2 1 □:

Page 56 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

The quotient is between 210 and 219

The dividend starts with 8 and ends with 8

Try divisor 4: 4 × 212 = 848 ✓ (8 _ 8 with the middle digit 4)

Also 4 × 217 = 868 ✓ (8 _ 8 with the middle digit 6)

No other divisor works

About □) 180 ( □□□: the quotient must have 3 digits. 180 ÷ 2 = 90 has only 2 digits,
and every larger divisor gives an even smaller answer. So the divisor has to be 1,
giving 180 ÷ 1 = 180.

Q2 I am a 3-digit number. If you divide me by 5, you get 42. If you multiply me by 2, you
get 420. What number am I?

The number is 210.
Use each clue on its own, then see that both give the same answer.
Clue 1 — divide me by 5 and you get 42.

Number ÷ 5 = 42

So Number = 5 × 42

5 × 42 = 5 × 40 + 5 × 2

= 200 + 10

= 210

Clue 2 — multiply me by 2 and you get 420.

Number × 2 = 420

So Number = 420 ÷ 2

= 210

Both clues point to the same number, and 210 has three digits. So the answer is 210.

Page 57 of 88

Class 5 Maths Chapter 9 Coconut Farm AglaSem · NCERT Solutions

Why the two clues agree: Undo whatever was done. If the number was divided by
5, multiply by 5 to get it back. If it was multiplied by 2, divide by 2 to get it back.
Multiplication and division undo each other.

Check both clues on 210: 210 ÷ 5 = 42 ✓ and 210 × 2 = 420 ✓ And 210 is a 3-digit
number ✓

Let Us Solve — Pages 132–133
Shows, ice cream, biscuits, remainders, relations, the cycle rally and the carpenter

LET US SOLVE

Q1 1. A theatre company can accommodate 45 people during one show. (a) A total of
475 people bought tickets for a puppet show. How many shows are needed to seat
all the people who bought tickets? (b) There are 2 shows in a day. How many days
will be needed to accommodate all the people?

(a) 11 shows. (b) 6 days.
(a) How many shows? Each show seats 45 people, and 475 people have tickets.

Step 1: 475 ÷ 45

Step 2: 45 × 10 = 450, which fits

Step 3: 475 − 450 = 25 people still without a seat

Step 4: Those 25 need one more show

Shows = 10 full + 1 more = 11 shows

(b) How many days? There are 2 shows in a day, and 11 shows are needed.

Page 58 of 88

as e
Class 5 Maths Chapter 9 Coconut Farm
a g l AglaSem · NCERT Solutions

co m
e m.
Step 1: 11 ÷ 2
m l as
.co
Step 2: 2 × 5 = 10, and 11 − 10 = 1
m a g
l a se
g
Step 3: After 5 days, 10 shows are done and 1 show is still left
aStep 4: That last show needs one more day

co m
e m . ag
Days = 5 + 1 = 6 days
g l as
a

. c om
Why we round up twice: You cannot run 10 and a half shows, and you cannot use

s e
half a day. Whenever people or events are left over, you need one more m whole show,
.
one more comwhole day. a gla
a s em
a gl Check by a second route: 11 shows × 45 seats = 495 seats, enough for 475 people ✓

m a s
agl
And 6 days × 2 shows = 12 shows, enough for 11 ✓

m .co
l a se
a g
Q2 2. Naina bought 5 kg of ice cream as a birthday treat for her 23 friends. 400 g ice
cream was left after everyone had an equal share. How much ice cream did each of
co m
m .
e
her friends eat?
m l as
.co a g
a s em

agl Each friend ate 200 g of ice cream.
Be careful: the ice cream is in kilograms but the leftover is in grams. Change everything to
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grams first.
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