Coconut Farm — NCERT Solutions

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: $(4, 6)$, $(3, 8)$, $(2, 12)$, etc. since $4 \times 6 = 24$.

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

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: $6 \times 7 = 42$

• Division statement 1: $42 \div 6 = 7$
• Division statement 2: $42 \div 7 = 6$

Example 2: $8 \times 9 = 72$

• Division statement 1: $72 \div 8 = 9$
• Division statement 2: $72 \div 9 = 8$

Example 3: $5 \times 12 = 60$

• Division statement 1: $60 \div 5 = 12$
• Division statement 2: $60 \div 12 = 5$

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

Part 1 — Place Value Chart (First Table):

$$40 \div 10 = 4 \quad \Rightarrow \text{H: —, T: —, O: 4}$$
$$400 \div 10 = 40 \quad \Rightarrow \text{H: —, T: 4, O: 0}$$
$$4000 \div 10 = 400 \quad \Rightarrow \text{H: 4, T: 0, O: 0}$$
$$700 \div 70 = 10 \quad \Rightarrow \text{H: —, T: 1, O: 0}$$
$$1400 \div 100 = 14 \quad \Rightarrow \text{H: —, T: 1, O: 4}$$
$$220 \div 20 = 11 \quad \Rightarrow \text{H: —, T: 1, O: 1}$$
$$2200 \div 20 = 110 \quad \Rightarrow \text{H: 1, T: 1, O: 0}$$

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):

$$110 \div 11 = 10 \quad \Rightarrow \text{H: —, T: 1, O: 0}$$
$$860 \div 86 = 10 \quad \Rightarrow \text{H: —, T: 1, O: 0}$$
$$7500 \div 750 = 10 \quad \Rightarrow \text{H: —, T: 1, O: 0}$$
$$8800 \div 88 = 100 \quad \Rightarrow \text{H: 1, T: 0, O: 0}$$
$$2400 \div 24 = 100 \quad \Rightarrow \text{H: 1, T: 0, O: 0}$$
$$440 \div 22 = 20 \quad \Rightarrow \text{H: —, T: 2, O: 0}$$

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

Given: Total distance = 160 km, Number of days = 20

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

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

$$= 8 \text{ km}$$

Answer: Sabina cycles 8 km each day.

Given: Total amount = ₹4200, Value of each note = ₹100

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

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

Answer: Seema needs 42 notes of ₹100.

Part (i): Amount each employee gets (5 employees)

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

Part (ii): If distributed among 10 employees:

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

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):

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

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.

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):
$$8 \div 4 = 2 \quad (\text{division})$$
$$2 \times 3 = 6 \quad (\text{multiplication})$$
$$6 + 1 = 7 \quad (\text{addition})$$
$$7 - 2 = 5 \quad (\text{subtraction})$$
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.

Concept: If $\_ \div D = Q$, then $\_ = D \times Q$.

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

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

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

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

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

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

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

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

Note: The specific problems are in images. The three strategies are illustrated below:

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

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

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

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

(a) $256 \div 4$ — Repeated halving:
$$\text{Half of } 256 = 128, \quad \text{Half of } 128 = 64$$
$$\therefore 256 \div 4 = \mathbf{64}$$

(b) $545 \div 5$ — Split:
$$545 = 500 + 45$$
$$500 \div 5 = 100, \quad 45 \div 5 = 9$$
$$\therefore 545 \div 5 = 100 + 9 = \mathbf{109}$$

(c) $147 \div 7$ — Split:
$$147 = 140 + 7$$
$$140 \div 7 = 20, \quad 7 \div 7 = 1$$
$$\therefore 147 \div 7 = 20 + 1 = \mathbf{21}$$

(d) $1212 \div 6$ — Split:
$$1212 = 1200 + 12$$
$$1200 \div 6 = 200, \quad 12 \div 6 = 2$$
$$\therefore 1212 \div 6 = 200 + 2 = \mathbf{202}$$

(e) $648 \div 12$ — Split:
$$648 = 600 + 48$$
$$600 \div 12 = 50, \quad 48 \div 12 = 4$$
$$\therefore 648 \div 12 = 50 + 4 = \mathbf{54}$$

(f) $9648 \div 48$ — Split:
$$9648 = 9600 + 48$$
$$9600 \div 48 = 200, \quad 48 \div 48 = 1$$
$$\therefore 9648 \div 48 = 200 + 1 = \mathbf{201}$$

(g) $775 \div 25$ — Split:
$$775 = 750 + 25$$
$$750 \div 25 = 30, \quad 25 \div 25 = 1$$
$$\therefore 775 \div 25 = 30 + 1 = \mathbf{31}$$

(h) $796 \div 4$ — Round up and subtract:
$$796 = 800 - 4$$
$$800 \div 4 = 200, \quad 4 \div 4 = 1$$
$$\therefore 796 \div 4 = 200 - 1 = \mathbf{199}$$

Part (i): Coconuts per customer

$$582 \div 6 = 97$$

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

Each customer gets 97 coconuts.

Part (ii): Bags needed for 97 coconuts

$$97 \div 25 = 3 \text{ remainder } 22$$
$$\because 25 \times 3 = 75, \quad 97 - 75 = 22$$

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

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

Step 1: Find remaining coconuts.
$$1117 - 582 = 535 \text{ coconuts}$$

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

Using partial quotients:
$$25 \times 20 = 500$$
$$535 - 500 = 35$$
$$25 \times 1 = 25$$
$$35 - 25 = 10$$

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

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

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

Division:
$$726 \div 4$$

$$4 \times 100 = 400, \quad 726 - 400 = 326$$
$$4 \times 80 = 320, \quad 326 - 320 = 6$$
$$4 \times 1 = 4, \quad 6 - 4 = 2$$

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

Check: Is $726 = 4 \times 181$?
$$4 \times 181 = 724 \neq 726 \quad \Rightarrow \text{No}$$

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

Verification using $N = D \times Q + R$:
$$4 \times 181 + 2 = 724 + 2 = 726 \checkmark$$

Division:
$$902 \div 16$$

$$16 \times 50 = 800, \quad 902 - 800 = 102$$
$$16 \times 6 = 96, \quad 102 - 96 = 6$$

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

Check: Is $902 = 16 \times 56$?
$$16 \times 56 = 896 \neq 902 \quad \Rightarrow \text{No}$$

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

Verification using $N = D \times Q + R$:
$$16 \times 56 + 6 = 896 + 6 = 902 \checkmark$$

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

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

For packs of 8:
$$250 \div 8 = 31 \text{ remainder } 2$$
She needs 32 packs of 8 → $32 \times 8 = 256$ 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.

Given: Total students = 342, Capacity per bus = 41

$$342 \div 41 = 8 \text{ remainder } 14$$
$$\because 41 \times 8 = 328, \quad 342 - 328 = 14$$

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

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

Given: Total = ₹520, using ₹50 and ₹20 notes.

Let number of ₹50 notes = $a$, number of ₹20 notes = $b$.
$$50a + 20b = 520$$
$$5a + 2b = 52$$

Finding combinations:

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

Step 1: Find total cost.
$$157 + 124 + 136 = ₹417$$

Step 2: Divide equally among 3 friends.
$$417 \div 3 = 139$$
$$\because 3 \times 139 = 417$$

Answer: Each person should pay ₹139.

(a) $887 \div 3$:
$$3 \times 295 = 885, \quad 887 - 885 = 2$$
$$Q = 295, \quad R = 2$$
$$\text{Check: } 3 \times 295 + 2 = 885 + 2 = 887 \checkmark$$

(b) $283 \div 8$:
$$8 \times 35 = 280, \quad 283 - 280 = 3$$
$$Q = 35, \quad R = 3$$
$$\text{Check: } 8 \times 35 + 3 = 280 + 3 = 283 \checkmark$$

(c) $745 \div 5$:
$$5 \times 149 = 745, \quad R = 0$$
$$Q = 149, \quad R = 0$$
$$\text{Check: } 5 \times 149 + 0 = 745 \checkmark$$

(d) $767 \div 26$:
$$26 \times 29 = 754, \quad 767 - 754 = 13$$
$$Q = 29, \quad R = 13$$
$$\text{Check: } 26 \times 29 + 13 = 754 + 13 = 767 \checkmark$$

(e) $530 \div 41$:
$$41 \times 12 = 492, \quad 530 - 492 = 38$$
$$Q = 12, \quad R = 38$$
$$\text{Check: } 41 \times 12 + 38 = 492 + 38 = 530 \checkmark$$

(f) $888 \div 67$:
$$67 \times 13 = 871, \quad 888 - 871 = 17$$
$$Q = 13, \quad R = 17$$
$$\text{Check: } 67 \times 13 + 17 = 871 + 17 = 888 \checkmark$$

Given: Total coconuts = 4376, Coconuts per litre = 8

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

Using partial quotients:
$$8 \times 500 = 4000, \quad 4376 - 4000 = 376$$
$$8 \times 40 = 320, \quad 376 - 320 = 56$$
$$8 \times 7 = 56, \quad 56 - 56 = 0$$

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

Answer: They extracted 547 litres of coconut oil.

Using partial quotients:
$$6 \times 1000 = 6000, \quad 7032 - 6000 = 1032$$
$$6 \times 100 = 600, \quad 1032 - 600 = 432$$
$$6 \times 70 = 420, \quad 432 - 420 = 12$$
$$6 \times 2 = 12, \quad 12 - 12 = 0$$

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

Standard algorithm check: $6 \times 1172 = 7032$ ✓

Using partial quotients:
$$5 \times 600 = 3000, \quad 3005 - 3000 = 5$$
$$5 \times 1 = 5, \quad 5 - 5 = 0$$

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

Check: $5 \times 601 = 3005$ ✓

Using partial quotients:
$$14 \times 200 = 2800, \quad 2874 - 2800 = 74$$
$$14 \times 5 = 70, \quad 74 - 70 = 4$$

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

Check using $N = D \times Q + R$:
$$14 \times 205 + 4 = 2870 + 4 = 2874 \checkmark$$

Using partial quotients:
$$32 \times 300 = 9600, \quad 9805 - 9600 = 205$$
$$32 \times 6 = 192, \quad 205 - 192 = 13$$

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

Check using $N = D \times Q + R$:
$$32 \times 306 + 13 = 9792 + 13 = 9805 \checkmark$$

Note: The specific problems are in images not visible in the OCR. The general method is:

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

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: $\_ \div 6 = 12 \Rightarrow \_ = 6 \times 12 = 72$

Example: $84 \div \_ = 7 \Rightarrow \_ = 84 \div 7 = 12$

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

Clue 1: $\_ \div 5 = 42$
$$\_ = 5 \times 42 = 210$$

Verification with Clue 2: $210 \times 2 = 420$ ✓

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