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NCERT Solutions

Coconut Farm

CBSE · Class 5 · Mathematics

NCERT Solutions for Coconut Farm — CBSE Class 5 Mathematics.

45 questions64 flashcards5 concepts

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51 Questions Solved · 14 Sections

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Let Us Play

1Identify the numbers that can fill the circles such that the numbers in the squares are the products or the quotients of the numbers in the circles.Show solution
Note: The figure is not visible in the OCR. However, the strategy is as follows:

Given: Numbers in squares are either the product or the quotient of the numbers in the adjacent circles.

Method:
- If the square shows a product, find two numbers whose multiplication gives that product and place them in the circles.
- If the square shows a quotient, find two numbers such that dividing one by the other gives that quotient.

Example: If a square shows 24, possible circle pairs are: (4,6)(4, 6), (3,8)(3, 8), (2,12)(2, 12), etc. since 4×6=244 \times 6 = 24.

Students should use multiplication tables to identify the correct pairs for each square shown in the figure.

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Let Us Do (Section 1 — Multiplication and Division Relationship)

1Solve the following multiplication problems. Write two division statements in each case. Observe the relationship between the divisor, dividend, and quotient.Show solution
Note: The specific multiplication problems are in images not visible in the OCR. The method is illustrated below with examples.

Concept: Every multiplication statement gives two related division statements.

Example 1: 6×7=426 \times 7 = 42
- Division statement 1: 42÷6=742 \div 6 = 7
- Division statement 2: 42÷7=642 \div 7 = 6

Example 2: 8×9=728 \times 9 = 72
- Division statement 1: 72÷8=972 \div 8 = 9
- Division statement 2: 72÷9=872 \div 9 = 8

Example 3: 5×12=605 \times 12 = 60
- Division statement 1: 60÷5=1260 \div 5 = 12
- Division statement 2: 60÷12=560 \div 12 = 5

Observation: If A×B=CA \times B = C, then C÷A=BC \div A = B and C÷B=AC \div B = A. The dividend is the product, and the divisor and quotient are the two factors.

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2Solve the following division problems. Notice the patterns and discuss in class. 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 = | | | |

Also fill:
| Problem | H | T | O |
|---|---|---|---|
| 110 ÷ 11 = | | | |
| 860 ÷ 86 = | | | |
| 7500 ÷ 750 = | | | |
| 8800 ÷ 88 = | | | |
| 2400 ÷ 24 = | | | |
| 440 ÷ 22 = | | | |Show solution
Part 1 — Place Value Chart (First Table):

40÷10=4⇒H: —, T: —, O: 440 \div 10 = 4 \quad \Rightarrow \text{H: —, T: —, O: 4}
400÷10=40⇒H: —, T: 4, O: 0400 \div 10 = 40 \quad \Rightarrow \text{H: —, T: 4, O: 0}
4000÷10=400⇒H: 4, T: 0, O: 04000 \div 10 = 400 \quad \Rightarrow \text{H: 4, T: 0, O: 0}
700÷70=10⇒H: —, T: 1, O: 0700 \div 70 = 10 \quad \Rightarrow \text{H: —, T: 1, O: 0}
1400÷100=14⇒H: —, T: 1, O: 41400 \div 100 = 14 \quad \Rightarrow \text{H: —, T: 1, O: 4}
220÷20=11⇒H: —, T: 1, O: 1220 \div 20 = 11 \quad \Rightarrow \text{H: —, T: 1, O: 1}
2200÷20=110⇒H: 1, T: 1, O: 02200 \div 20 = 110 \quad \Rightarrow \text{H: 1, T: 1, O: 0}

Pattern observed: When dividing by 10, the digits shift one place to the right (i.e., the number becomes 10 times smaller). When dividing by 100, digits shift two places to the right.

Part 2 — Place Value Chart (Second Table):

110÷11=10⇒H: —, T: 1, O: 0110 \div 11 = 10 \quad \Rightarrow \text{H: —, T: 1, O: 0}
860÷86=10⇒H: —, T: 1, O: 0860 \div 86 = 10 \quad \Rightarrow \text{H: —, T: 1, O: 0}
7500÷750=10⇒H: —, T: 1, O: 07500 \div 750 = 10 \quad \Rightarrow \text{H: —, T: 1, O: 0}
8800÷88=100⇒H: 1, T: 0, O: 08800 \div 88 = 100 \quad \Rightarrow \text{H: 1, T: 0, O: 0}
2400÷24=100⇒H: 1, T: 0, O: 02400 \div 24 = 100 \quad \Rightarrow \text{H: 1, T: 0, O: 0}
440÷22=20⇒H: —, T: 2, O: 0440 \div 22 = 20 \quad \Rightarrow \text{H: —, T: 2, O: 0}

Pattern observed: When the divisor is exactly 110\frac{1}{10} of the dividend, the quotient is always 10. When the divisor is exactly 1100\frac{1}{100} of the dividend, the quotient is 100. The quotients are multiples of 10 or 100 in these cases.

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Let Us Do (Word Problems and Puzzles)

1Sabina cycles 160 km in 20 days and the same distance each day. How many kilometres does she cycle each day?Show solution
Given: Total distance = 160 km, Number of days = 20

Formula: Distance per day =Total distance÷Number of days= \text{Total distance} \div \text{Number of days}

Distance per day=160÷20\text{Distance per day} = 160 \div 20

=8 km= 8 \text{ km}

Answer: Sabina cycles 8 km each day.

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2How many notes of ₹100 does Seema need to carry if she wants to buy coconuts worth ₹4200?Show solution
Given: Total amount = ₹4200, Value of each note = ₹100

Formula: Number of notes =Total amount÷Value of one note= \text{Total amount} \div \text{Value of one note}

Number of notes=4200÷100=42\text{Number of notes} = 4200 \div 100 = 42

Answer: Seema needs 42 notes of ₹100.

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3The 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?Show solution
Part (i): Amount each employee gets (5 employees)

Amount per employee=5500÷5=₹1100\text{Amount per employee} = 5500 \div 5 = ₹1100

Part (ii): If distributed among 10 employees:

Amount per employee=5500÷10=₹550\text{Amount per employee} = 5500 \div 10 = ₹550

Each employee gets less (₹550 instead of ₹1100) because the same amount is shared among more people.

Part (iii): If each of the 10 employees must get ₹1100 (same as before):

Total amount needed=1100×10=₹11000\text{Total amount needed} = 1100 \times 10 = ₹11000

Answer: Each of the 5 employees gets ₹1100. With 10 employees, each gets only ₹550. To give ₹1100 to each of 10 employees, the owner would need to distribute ₹11,000.

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4Place the numbers 1 to 8 in the boxes so that all the four operations — division, multiplication, addition and subtraction — are correct. No number must be repeated.Show solution
Note: The exact box arrangement is in images not visible in the OCR. The general approach is:

Strategy:
- List numbers 1 to 8: {1, 2, 3, 4, 5, 6, 7, 8}
- Try combinations that satisfy all four operations simultaneously.
- Use trial and error, starting with division (most restrictive).

One possible solution (commonly accepted):
8÷4=2(division)8 \div 4 = 2 \quad (\text{division})
2×3=6(multiplication)2 \times 3 = 6 \quad (\text{multiplication})
6+1=7(addition)6 + 1 = 7 \quad (\text{addition})
7−2=5(subtraction)7 - 2 = 5 \quad (\text{subtraction})
Numbers used: 8, 4, 2, 3, 6, 1, 7, 5 — all different, all from 1–8. ✓

Note: There may be more than one valid answer. Students should verify that each number from 1 to 8 is used exactly once.

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5Fill in the blanks:
(a) _÷18=100\_ \div 18 = 100
(b) _÷10=610\_ \div 10 = 610
(c) _÷100=72\_ \div 100 = 72
(d) _÷100=10\_ \div 100 = 10
(e) 870÷_=87870 \div \_ = 87
(f) _÷100=70\_ \div 100 = 70
(g) 200÷_=2200 \div \_ = 2
(h) 130÷_=13130 \div \_ = 13Show solution
Concept: If _÷D=Q\_ \div D = Q, then _=D×Q\_ = D \times Q.

(a) _÷18=100\_ \div 18 = 100
_=18×100=1800\_ = 18 \times 100 = \mathbf{1800}

(b) _÷10=610\_ \div 10 = 610
_=10×610=6100\_ = 10 \times 610 = \mathbf{6100}

(c) _÷100=72\_ \div 100 = 72
_=100×72=7200\_ = 100 \times 72 = \mathbf{7200}

(d) _÷100=10\_ \div 100 = 10
_=100×10=1000\_ = 100 \times 10 = \mathbf{1000}

(e) 870÷_=87870 \div \_ = 87
_=870÷87=10\_ = 870 \div 87 = \mathbf{10}

(f) _÷100=70\_ \div 100 = 70
_=100×70=7000\_ = 100 \times 70 = \mathbf{7000}

(g) 200÷_=2200 \div \_ = 2
_=200÷2=100\_ = 200 \div 2 = \mathbf{100}

(h) 130÷_=13130 \div \_ = 13
_=130÷13=10\_ = 130 \div 13 = \mathbf{10}

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Let Us Solve (Mental Strategies for Division)

1Try It! — Solve the division problems shown (images not fully visible). The mental strategies demonstrated are: (1) splitting the dividend into convenient parts, (2) rounding up and subtracting, (3) repeated halving.Show solution
Note: The specific problems are in images. The three strategies are illustrated below:

Strategy 1 — Split into convenient parts:
1248÷6:1248=1200+481248 \div 6: \quad 1248 = 1200 + 48
1200÷6=200,48÷6=81200 \div 6 = 200, \quad 48 \div 6 = 8
∴1248÷6=200+8=208\therefore 1248 \div 6 = 200 + 8 = \mathbf{208}

Strategy 2 — Round up and subtract:
1992÷4:1992=2000−81992 \div 4: \quad 1992 = 2000 - 8
2000÷4=500,8÷4=22000 \div 4 = 500, \quad 8 \div 4 = 2
∴1992÷4=500−2=498\therefore 1992 \div 4 = 500 - 2 = \mathbf{498}

Strategy 3 — Repeated halving (for divisor 4):
128÷4:Half of 128=64,Half of 64=32128 \div 4: \quad \text{Half of } 128 = 64, \quad \text{Half of } 64 = 32
∴128÷4=32\therefore 128 \div 4 = \mathbf{32}

Students should apply these strategies to the problems shown in the images.

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2Solve the following problems using mental strategies:
(a) 256÷4256 \div 4
(b) 545÷5545 \div 5
(c) 147÷7147 \div 7
(d) 1212÷61212 \div 6
(e) 648÷12648 \div 12
(f) 9648÷489648 \div 48
(g) 775÷25775 \div 25
(h) 796÷4796 \div 4Show solution
**(a) 256÷4256 \div 4 — Repeated halving:**
Half of 256=128,Half of 128=64\text{Half of } 256 = 128, \quad \text{Half of } 128 = 64
∴256÷4=64\therefore 256 \div 4 = \mathbf{64}

**(b) 545÷5545 \div 5 — Split:**
545=500+45545 = 500 + 45
500÷5=100,45÷5=9500 \div 5 = 100, \quad 45 \div 5 = 9
∴545÷5=100+9=109\therefore 545 \div 5 = 100 + 9 = \mathbf{109}

**(c) 147÷7147 \div 7 — Split:**
147=140+7147 = 140 + 7
140÷7=20,7÷7=1140 \div 7 = 20, \quad 7 \div 7 = 1
∴147÷7=20+1=21\therefore 147 \div 7 = 20 + 1 = \mathbf{21}

**(d) 1212÷61212 \div 6 — Split:**
1212=1200+121212 = 1200 + 12
1200÷6=200,12÷6=21200 \div 6 = 200, \quad 12 \div 6 = 2
∴1212÷6=200+2=202\therefore 1212 \div 6 = 200 + 2 = \mathbf{202}

**(e) 648÷12648 \div 12 — Split:**
648=600+48648 = 600 + 48
600÷12=50,48÷12=4600 \div 12 = 50, \quad 48 \div 12 = 4
∴648÷12=50+4=54\therefore 648 \div 12 = 50 + 4 = \mathbf{54}

**(f) 9648÷489648 \div 48 — Split:**
9648=9600+489648 = 9600 + 48
9600÷48=200,48÷48=19600 \div 48 = 200, \quad 48 \div 48 = 1
∴9648÷48=200+1=201\therefore 9648 \div 48 = 200 + 1 = \mathbf{201}

**(g) 775÷25775 \div 25 — Split:**
775=750+25775 = 750 + 25
750÷25=30,25÷25=1750 \div 25 = 30, \quad 25 \div 25 = 1
∴775÷25=30+1=31\therefore 775 \div 25 = 30 + 1 = \mathbf{31}

**(h) 796÷4796 \div 4 — Round up and subtract:**
796=800−4796 = 800 - 4
800÷4=200,4÷4=1800 \div 4 = 200, \quad 4 \div 4 = 1
∴796÷4=200−1=199\therefore 796 \div 4 = 200 - 1 = \mathbf{199}

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Susie's Farm in Kerala

1Susie and Sunitha harvested 1,117 coconuts in April. They sold 582 coconuts equally to 6 regular customers. How many coconuts did each customer get? Each bag can hold 25 coconuts. How many bags would be needed to pack 97 coconuts?Show solution
Part (i): Coconuts per customer

582÷6=97582 \div 6 = 97

Verification using partial quotients:
6×90=540,582−540=426 \times 90 = 540, \quad 582 - 540 = 42
6×7=426 \times 7 = 42
∴582÷6=90+7=97\therefore 582 \div 6 = 90 + 7 = \mathbf{97}

Each customer gets 97 coconuts.

Part (ii): Bags needed for 97 coconuts

97÷25=3 remainder 2297 \div 25 = 3 \text{ remainder } 22
∵25×3=75,97−75=22\because 25 \times 3 = 75, \quad 97 - 75 = 22

3 full bags hold 75 coconuts. The remaining 22 coconuts need 1 more bag.

∴Total bags needed=3+1=4 bags\therefore \text{Total bags needed} = 3 + 1 = \mathbf{4 \text{ bags}}

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2They 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? (Remaining = 1117−582=5351117 - 582 = 535 coconuts)Show solution
Step 1: Find remaining coconuts.
1117−582=535 coconuts1117 - 582 = 535 \text{ coconuts}

Step 2: Find number of bags needed.
535÷25=?535 \div 25 = ?

Using partial quotients:
25×20=50025 \times 20 = 500
535−500=35535 - 500 = 35
25×1=2525 \times 1 = 25
35−25=1035 - 25 = 10

∴535÷25=21 remainder 10\therefore 535 \div 25 = 21 \text{ remainder } 10

21 full bags hold 21×25=52521 \times 25 = 525 coconuts. The remaining 10 coconuts need 1 more bag.

∴Total bags needed=21+1=22 bags\therefore \text{Total bags needed} = 21 + 1 = \mathbf{22 \text{ bags}}

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Let Us Learn to Divide

1Divide 726÷4726 \div 4. Find the remainder. Check: Is 726=4×181726 = 4 \times 181? Complete: 726=4×181+_726 = 4 \times 181 + \_Show solution
Division:
726÷4726 \div 4

4×100=400,726−400=3264 \times 100 = 400, \quad 726 - 400 = 326
4×80=320,326−320=64 \times 80 = 320, \quad 326 - 320 = 6
4×1=4,6−4=24 \times 1 = 4, \quad 6 - 4 = 2

∴Quotient=100+80+1=181,Remainder=2\therefore \text{Quotient} = 100 + 80 + 1 = 181, \quad \text{Remainder} = 2

Check: Is 726=4×181726 = 4 \times 181?
4×181=724≠726⇒No4 \times 181 = 724 \neq 726 \quad \Rightarrow \text{No}

∴726=4×181+2\therefore 726 = 4 \times 181 + \mathbf{2}

**Verification using N=D×Q+RN = D \times Q + R:**
4×181+2=724+2=726✓4 \times 181 + 2 = 724 + 2 = 726 \checkmark

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2Divide 902÷16902 \div 16. Find the remainder. Check: Is 902=16×56902 = 16 \times 56? Complete: 902=16×56+_902 = 16 \times 56 + \_Show solution
Division:
902÷16902 \div 16

16×50=800,902−800=10216 \times 50 = 800, \quad 902 - 800 = 102
16×6=96,102−96=616 \times 6 = 96, \quad 102 - 96 = 6

∴Quotient=50+6=56,Remainder=6\therefore \text{Quotient} = 50 + 6 = 56, \quad \text{Remainder} = 6

Check: Is 902=16×56902 = 16 \times 56?
16×56=896≠902⇒No16 \times 56 = 896 \neq 902 \quad \Rightarrow \text{No}

∴902=16×56+6\therefore 902 = 16 \times 56 + \mathbf{6}

**Verification using N=D×Q+RN = D \times Q + R:**
16×56+6=896+6=902✓16 \times 56 + 6 = 896 + 6 = 902 \checkmark

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Let Us Solve (Word Problems — Party, Trip, Money)

1Rani 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.Show solution
Given: 250 guests, 1 samosa each, packs of 6 or 8.

For packs of 6:
250÷6=41 remainder 4250 \div 6 = 41 \text{ remainder } 4
She needs 42 packs of 6 → 42×6=25242 \times 6 = 252 samosas (2 extra)

For packs of 8:
250÷8=31 remainder 2250 \div 8 = 31 \text{ remainder } 2
She needs 32 packs of 8 → 32×8=25632 \times 8 = 256 samosas (6 extra)

Comparison:
- Packs of 6: 42 packs, 252 samosas, 2 extra
- Packs of 8: 32 packs, 256 samosas, 6 extra

Answer: Rani should buy packs of 6 because she will have fewer extra samosas (only 2 extra vs 6 extra), which means less wastage. She needs to buy 42 packs of 6.

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2342 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?Show solution
Given: Total students = 342, Capacity per bus = 41

342÷41=8 remainder 14342 \div 41 = 8 \text{ remainder } 14
∵41×8=328,342−328=14\because 41 \times 8 = 328, \quad 342 - 328 = 14

8 buses carry 328 students. The remaining 14 students need 1 more bus.

∴Total buses needed=8+1=9 buses\therefore \text{Total buses needed} = 8 + 1 = \mathbf{9 \text{ buses}}

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3Sofia 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.Show solution
Given: Total = ₹520, using ₹50 and ₹20 notes.

Let number of ₹50 notes = aa, number of ₹20 notes = bb.
50a+20b=52050a + 20b = 520
5a+2b=525a + 2b = 52

Finding combinations:

| ₹50 notes (aa) | ₹20 notes (bb) | Check |
|---|---|---|
| 0 | 26 | 0+520=5200 + 520 = 520 ✓ |
| 2 | 21 | 100+420=520100 + 420 = 520 ✓ |
| 4 | 16 | 200+320=520200 + 320 = 520 ✓ |
| 6 | 11 | 300+220=520300 + 220 = 520 ✓ |
| 8 | 6 | 400+120=520400 + 120 = 520 ✓ |
| 10 | 1 | 500+20=520500 + 20 = 520 ✓ |

Answer: There are 6 possible combinations. For example, 8 notes of ₹50 and 6 notes of ₹20.

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4Three 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?Show solution
Step 1: Find total cost.
157+124+136=₹417157 + 124 + 136 = ₹417

Step 2: Divide equally among 3 friends.
417÷3=139417 \div 3 = 139
∵3×139=417\because 3 \times 139 = 417

Answer: Each person should pay ₹139.

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5Identify the remainder, if any. Check if N=D×Q+RN = D \times Q + R.
(a) 887÷3887 \div 3
(b) 283÷8283 \div 8
(c) 745÷5745 \div 5
(d) 767÷26767 \div 26
(e) 530÷41530 \div 41
(f) 888÷67888 \div 67Show solution
**(a) 887÷3887 \div 3:**
3×295=885,887−885=23 \times 295 = 885, \quad 887 - 885 = 2
Q=295,R=2Q = 295, \quad R = 2
Check: 3×295+2=885+2=887✓\text{Check: } 3 \times 295 + 2 = 885 + 2 = 887 \checkmark

**(b) 283÷8283 \div 8:**
8×35=280,283−280=38 \times 35 = 280, \quad 283 - 280 = 3
Q=35,R=3Q = 35, \quad R = 3
Check: 8×35+3=280+3=283✓\text{Check: } 8 \times 35 + 3 = 280 + 3 = 283 \checkmark

**(c) 745÷5745 \div 5:**
5×149=745,R=05 \times 149 = 745, \quad R = 0
Q=149,R=0Q = 149, \quad R = 0
Check: 5×149+0=745✓\text{Check: } 5 \times 149 + 0 = 745 \checkmark

**(d) 767÷26767 \div 26:**
26×29=754,767−754=1326 \times 29 = 754, \quad 767 - 754 = 13
Q=29,R=13Q = 29, \quad R = 13
Check: 26×29+13=754+13=767✓\text{Check: } 26 \times 29 + 13 = 754 + 13 = 767 \checkmark

**(e) 530÷41530 \div 41:**
41×12=492,530−492=3841 \times 12 = 492, \quad 530 - 492 = 38
Q=12,R=38Q = 12, \quad R = 38
Check: 41×12+38=492+38=530✓\text{Check: } 41 \times 12 + 38 = 492 + 38 = 530 \checkmark

**(f) 888÷67888 \div 67:**
67×13=871,888−871=1767 \times 13 = 871, \quad 888 - 871 = 17
Q=13,R=17Q = 13, \quad R = 17
Check: 67×13+17=871+17=888✓\text{Check: } 67 \times 13 + 17 = 871 + 17 = 888 \checkmark

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Kalpavruksha Coconut Oil

1Susie and Sunitha used 4376 coconuts for extracting coconut oil. They can extract 1 litre of oil from 8 coconuts. What quantity of oil were they able to extract?Show solution
Given: Total coconuts = 4376, Coconuts per litre = 8

Litres of oil=4376÷8\text{Litres of oil} = 4376 \div 8

Using partial quotients:
8×500=4000,4376−4000=3768 \times 500 = 4000, \quad 4376 - 4000 = 376
8×40=320,376−320=568 \times 40 = 320, \quad 376 - 320 = 56
8×7=56,56−56=08 \times 7 = 56, \quad 56 - 56 = 0

∴4376÷8=500+40+7=547\therefore 4376 \div 8 = 500 + 40 + 7 = \mathbf{547}

Answer: They extracted 547 litres of coconut oil.

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Let Us Divide

a7,032÷67{,}032 \div 6Show solution
Using partial quotients:
6×1000=6000,7032−6000=10326 \times 1000 = 6000, \quad 7032 - 6000 = 1032
6×100=600,1032−600=4326 \times 100 = 600, \quad 1032 - 600 = 432
6×70=420,432−420=126 \times 70 = 420, \quad 432 - 420 = 12
6×2=12,12−12=06 \times 2 = 12, \quad 12 - 12 = 0

∴7032÷6=1000+100+70+2=1172\therefore 7032 \div 6 = 1000 + 100 + 70 + 2 = \mathbf{1172}

Standard algorithm check: 6×1172=70326 \times 1172 = 7032 ✓

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b3,005÷53{,}005 \div 5Show solution
Using partial quotients:
5×600=3000,3005−3000=55 \times 600 = 3000, \quad 3005 - 3000 = 5
5×1=5,5−5=05 \times 1 = 5, \quad 5 - 5 = 0

∴3005÷5=600+1=601\therefore 3005 \div 5 = 600 + 1 = \mathbf{601}

Check: 5×601=30055 \times 601 = 3005 ✓

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c2,874÷142{,}874 \div 14Show solution
Using partial quotients:
14×200=2800,2874−2800=7414 \times 200 = 2800, \quad 2874 - 2800 = 74
14×5=70,74−70=414 \times 5 = 70, \quad 74 - 70 = 4

∴2874÷14=200+5=205,R=4\therefore 2874 \div 14 = 200 + 5 = 205, \quad R = 4

**Check using N=D×Q+RN = D \times Q + R:**
14×205+4=2870+4=2874✓14 \times 205 + 4 = 2870 + 4 = 2874 \checkmark

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d9,805÷329{,}805 \div 32Show solution
Using partial quotients:
32×300=9600,9805−9600=20532 \times 300 = 9600, \quad 9805 - 9600 = 205
32×6=192,205−192=1332 \times 6 = 192, \quad 205 - 192 = 13

∴9805÷32=300+6=306,R=13\therefore 9805 \div 32 = 300 + 6 = 306, \quad R = 13

**Check using N=D×Q+RN = D \times Q + R:**
32×306+13=9792+13=9805✓32 \times 306 + 13 = 9792 + 13 = 9805 \checkmark

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Let Us Do (Missing Numbers and Riddle)

1Find the missing numbers such that there is no remainder. (The division problems are shown in images — general method given.)Show solution
Note: The specific problems are in images not visible in the OCR. The general method is:

Concept: If N÷D=QN \div D = Q with no remainder, then N=D×QN = D \times Q.

Method:
- If the dividend is missing: Multiply divisor × quotient.
- If the divisor is missing: Divide dividend by quotient.
- If the quotient is missing: Divide dividend by divisor.

Example: _÷6=12⇒_=6×12=72\_ \div 6 = 12 \Rightarrow \_ = 6 \times 12 = 72

Example: 84÷_=7⇒_=84÷7=1284 \div \_ = 7 \Rightarrow \_ = 84 \div 7 = 12

Students should apply this method to each problem shown in the images.

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2I 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?Show solution
Clue 1: _÷5=42\_ \div 5 = 42
_=5×42=210\_ = 5 \times 42 = 210

Verification with Clue 2: 210×2=420210 \times 2 = 420 ✓

Answer: The number is 210\mathbf{210}.

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Let Us Solve (Theatre, Ice Cream, Biscuits, Division)

1A 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?
2Naina 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 her friends eat?
3Megha packs 15 packets of ragi-oats biscuits for a 4-day group trip. Each packet contains 8 biscuits. There are 6 people in the group. If distributed evenly, how many biscuits can one person have each day?
4Solve the following and identify the remainder, if any. Check whether N=D×Q+RN = D \times Q + R in each case.
(a) 9,045÷59{,}045 \div 5
(b) 1,034÷41{,}034 \div 4
(c) 2,504÷72{,}504 \div 7
(d) 8,900÷158{,}900 \div 15
(e) 9,876÷329{,}876 \div 32
(f) 7,506÷247{,}506 \div 24
5Find the solutions for part A. Observe the relations between the quotient, divisor and dividend and use it to answer parts B and C.

A: (a) 340÷34=10340 \div 34 = 10, (b) 340÷17=_340 \div 17 = \_, (c) 680÷17=_680 \div 17 = \_, (d) 680÷34=_680 \div 34 = \_, (e) 170÷17=_170 \div 17 = \_, (f) 680÷68=_680 \div 68 = \_

B: (a) 192÷4=48192 \div 4 = 48, (b) 192÷8=_192 \div 8 = \_, (c) 384÷8=_384 \div 8 = \_, (d) 384÷4=_384 \div 4 = \_, (e) 384÷8=_384 \div 8 = \_, (f) 86÷2=_86 \div 2 = \_

C: (a) 352÷11=32352 \div 11 = 32, (b) 704÷22=_704 \div 22 = \_, (c) 704÷11=_704 \div 11 = \_, (d) 352÷22=_352 \div 22 = \_, (e) 1408÷44=_1408 \div 44 = \_
6A company in Mumbai organises cycle rallies from Mumbai to Panjim, Goa every year. They aim to cover 576 km in 12 days.
(a) How much distance should they cycle every day, to cover the distance evenly?
(b) After reaching Ratnagiri, they rest for 1 day. How much distance should they cycle each day to reach Goa in 4 days? Assume that they cover the distance evenly.
7Given below are a few problems. Identify the missing information. Write the missing information and find the answer.
(a) A fruit vendor sells 6 baskets of mangoes. Each basket contains 12 mangoes. How much did the vendor earn in total?
(b) A school has 8 classrooms, and each classroom has an equal number of desks. How many desks are there in each classroom?
(c) Rahul buys 5 cricket bats for his team. The total bill is ₹3500. How much does one bat cost?
(d) A restaurant serves 125 plates of idlis in a day. The total earnings from selling all the idli plates is ₹6250. How many idlis are there in each plate?
8To make one bookshelf, a carpenter needs: 4 long wooden panels, 8 short wooden panels, 16 small clips, 4 large clips, 32 screws. The carpenter has: 264 long panels, 306 short panels, 2400 small clips, 120 large clips, 2800 screws. How many bookshelves can the carpenter make?

Vegetable Market — Complete the Table

1Help Munshi Lal complete the table:
1. Radish: ₹26 per kg, 78 kg supplied, Total = ₹____
2. Potato: ₹20 per kg, ____ kg supplied, Total = ₹2,240
3. Cabbage: ₹32 per kg, 56 kg supplied, Total = ₹____
4. Green peas: ₹____ per kg, 125 kg supplied, Total = ₹3,125
Total money earned = ₹____

Let Us Solve (Final Division Practice)

1506÷5506 \div 5
2918÷8918 \div 8
38,126÷78{,}126 \div 7
49,324÷49{,}324 \div 4
5876÷6876 \div 6
67,008÷37{,}008 \div 3
7934÷12934 \div 12
8829÷23829 \div 23
9705÷18705 \div 18
108,704÷328{,}704 \div 32
116,790÷456{,}790 \div 45
125,074÷215{,}074 \div 21

Mathematical Statements

1Find out whether the following statements are True (T) or False (F).
(a) 8×9=70+28 \times 9 = 70 + 2
(b) 20−6=7×320 - 6 = 7 \times 3
(c) 48÷3=4×448 \div 3 = 4 \times 4
(d) 89−9=90+089 - 9 = 90 + 0
(e) 25+10=45−1025 + 10 = 45 - 10
2Complete the following statements such that they are true.
(a) 7×6=_+177 \times 6 = \_ + 17
(b) 87+6=_×3187 + 6 = \_ \times 31
(c) 63+_=74−463 + \_ = 74 - 4
(d) _÷9=16÷2\_ \div 9 = 16 \div 2
3Think about the following statements:
(a) 'When two odd numbers are added, the sum is even.' Find 5 examples. Can you find an example to show it is false? (Always true)
(b) 'Multiplying a number by 2 can give an odd number.' Give examples. Can you find any? (Never true)
(c) 'Halving a number always leads to an even number.' Give 3 examples for and 3 against. (Sometimes true)
4Tick in the appropriate cell for the following statements:
(i) Adding 10 to a number gives a multiple of ten.
(ii) Changing the order of the numbers in subtraction makes no difference.
(iii) In multiplication, doubling one number and halving the other keeps the product the same.
(iv) Multiplication by an odd number gives an even number.
(v) Multiplying a number by 5 leads to numbers which have '0' in the Ones place.

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Frequently Asked Questions

What are the important topics in Coconut Farm for CBSE Class 5 Mathematics?
Coconut Farm covers several key topics that are frequently asked in CBSE Class 5 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Coconut Farm — CBSE Class 5 Mathematics?
Understand the core concepts first, then work through the 45 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
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This page has free step-by-step NCERT Solutions for every exercise question in Coconut Farm (CBSE Class 5 Mathematics) — written the way examiners award marks: given, formula, working, answer.

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