Coconut Farm

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.

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:

| ₹50 notes ($a$) | ₹20 notes ($b$) | Check |
|---|---|---|
| 0 | 26 | $0 + 520 = 520$ ✓ |
| 2 | 21 | $100 + 420 = 520$ ✓ |
| 4 | 16 | $200 + 320 = 520$ ✓ |
| 6 | 11 | $300 + 220 = 520$ ✓ |
| 8 | 6 | $400 + 120 = 520$ ✓ |
| 10 | 1 | $500 + 20 = 520$ ✓ |

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.

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: $5 \times 601 = 3005$ ✓

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 $\mathbf{210}$.

(a) Number of shows needed:
$$475 \div 45 = 10 \text{ remainder } 25$$
$$\because 45 \times 10 = 450, \quad 475 - 450 = 25$$

10 shows seat 450 people. The remaining 25 people need 1 more show.
$$\therefore \text{Total shows needed} = 10 + 1 = \mathbf{11 \text{ shows}}$$

(b) Number of days needed (2 shows per day):
$$11 \div 2 = 5 \text{ remainder } 1$$

5 days have 10 shows. The 11th show needs 1 more day.
$$\therefore \text{Total days needed} = 5 + 1 = \mathbf{6 \text{ days}}$$

Step 1: Convert total ice cream to grams.
$$5 \text{ kg} = 5000 \text{ g}$$

Step 2: Find ice cream distributed.
$$5000 - 400 = 4600 \text{ g}$$

Step 3: Divide equally among 23 friends.
$$4600 \div 23 = 200$$
$$\because 23 \times 200 = 4600$$

Answer: Each friend ate 200 g of ice cream.

Step 1: Find total number of biscuits.
$$15 \times 8 = 120 \text{ biscuits}$$

Step 2: Find biscuits per person for the whole trip.
$$120 \div 6 = 20 \text{ biscuits per person}$$

Step 3: Find biscuits per person per day.
$$20 \div 4 = 5 \text{ biscuits per person per day}$$

Answer: Each person can have 5 biscuits per day.

Part A:

(a) $340 \div 34 = \mathbf{10}$ (given)

(b) $340 \div 17 = \_$: 17 is half of 34, so quotient doubles → $\mathbf{20}$
$$340 \div 17 = 20$$

(c) $680 \div 17 = \_$: 680 is double 340, divisor same → quotient doubles → $\mathbf{40}$
$$680 \div 17 = 40$$

(d) $680 \div 34 = \_$: 680 is double 340, divisor same as (a) → quotient doubles → $\mathbf{20}$
$$680 \div 34 = 20$$

(e) $170 \div 17 = \_$: 170 is half of 340, divisor same as (b) → quotient halves → $\mathbf{10}$
$$170 \div 17 = 10$$

(f) $680 \div 68 = \_$: 680 is double 340, 68 is double 34 → quotient same as (a) → $\mathbf{10}$
$$680 \div 68 = 10$$

Part B:

(a) $192 \div 4 = \mathbf{48}$ (given)

(b) $192 \div 8 = \_$: 8 is double 4 → quotient halves → $\mathbf{24}$

(c) $384 \div 8 = \_$: 384 is double 192, 8 is double 4 → quotient same as (a) → $\mathbf{48}$

(d) $384 \div 4 = \_$: 384 is double 192, divisor same → quotient doubles → $\mathbf{96}$

(e) $384 \div 8 = \mathbf{48}$ (same as (c))

(f) $86 \div 2 = \_$: $\mathbf{43}$

Part C:

(a) $352 \div 11 = \mathbf{32}$ (given)

(b) $704 \div 22 = \_$: 704 is double 352, 22 is double 11 → quotient same → $\mathbf{32}$

(c) $704 \div 11 = \_$: 704 is double 352, divisor same → quotient doubles → $\mathbf{64}$

(d) $352 \div 22 = \_$: dividend same, 22 is double 11 → quotient halves → $\mathbf{16}$

(e) $1408 \div 44 = \_$: 1408 is 4 times 352, 44 is 4 times 11 → quotient same → $\mathbf{32}$

(a) Distance per day (12 days):
$$576 \div 12 = 48 \text{ km}$$
$$\because 12 \times 48 = 576$$

Answer: They should cycle 48 km per day.

(b) Distance per day after Ratnagiri:

Distance covered up to Ratnagiri (before rest day):
- They have been cycling for some days. The problem states they rest after reaching Ratnagiri and then need 4 more days to reach Goa.
- Total trip = 12 days cycling + 1 rest day.
- Assume Ratnagiri is at the halfway point or that the remaining distance after Ratnagiri needs to be covered in 4 days.

Distance covered in first part (12 - 4 = 8 days at 48 km/day):
$$8 \times 48 = 384 \text{ km}$$

Remaining distance:
$$576 - 384 = 192 \text{ km}$$

Distance per day for remaining 4 days:
$$192 \div 4 = 48 \text{ km per day}$$

Answer: They should still cycle 48 km per day to cover the remaining distance in 4 days.

*(Note: The daily distance remains the same because the total remaining distance and remaining days are proportional to the original plan.)*

(a) Missing information: Price per mango (or price per basket)

Assuming price per mango = ₹10 (student fills this in):
$$\text{Total mangoes} = 6 \times 12 = 72$$
$$\text{Total earnings} = 72 \times 10 = ₹720$$

(b) Missing information: Total number of desks in the school

Assuming total desks = 240 (student fills this in):
$$\text{Desks per classroom} = 240 \div 8 = 30$$

(c) No missing information — this problem is complete.
$$\text{Cost of one bat} = 3500 \div 5 = ₹700$$

Answer: One bat costs ₹700.

(d) Missing information: Price per plate of idlis

The problem gives total earnings (₹6250) and number of plates (125), and asks for idlis per plate — but to find idlis per plate, we need the price per idli.

Assuming price per idli = ₹10 (student fills this in):
$$\text{Price per plate} = 6250 \div 125 = ₹50$$
$$\text{Idlis per plate} = 50 \div 10 = 5$$

Answer: There are 5 idlis per plate (based on assumed price of ₹10 per idli).

Find how many bookshelves each material allows:

$$\text{Long panels: } 264 \div 4 = 66 \text{ bookshelves}$$
$$\text{Short panels: } 306 \div 8 = 38 \text{ bookshelves (remainder 2)}$$
$$\text{Small clips: } 2400 \div 16 = 150 \text{ bookshelves}$$
$$\text{Large clips: } 120 \div 4 = 30 \text{ bookshelves}$$
$$\text{Screws: } 2800 \div 32 = 87 \text{ bookshelves (remainder 16)}$$

The limiting factor is the material that allows the fewest bookshelves:
$$\min(66, 38, 150, 30, 87) = 30$$

Answer: The carpenter can make 30 bookshelves (limited by the large clips).

Row 1 — Radish:
$$\text{Total} = 26 \times 78$$
$$= 26 \times 80 - 26 \times 2 = 2080 - 52 = ₹\mathbf{2028}$$

Row 2 — Potato:
$$\text{Quantity} = 2240 \div 20 = \mathbf{112} \text{ kg}$$

Row 3 — Cabbage:
$$\text{Total} = 32 \times 56$$
$$= 32 \times 50 + 32 \times 6 = 1600 + 192 = ₹\mathbf{1792}$$

Row 4 — Green Peas:
$$\text{Cost per kg} = 3125 \div 125 = ₹\mathbf{25}$$

Total money earned:
$$2028 + 2240 + 1792 + 3125$$
$$= 2028 + 2240 = 4268$$
$$4268 + 1792 = 6060$$
$$6060 + 3125 = ₹\mathbf{9185}$$

Completed Table:
| S.No. | Vegetable | Cost/kg | Quantity (kg) | Total |
|---|---|---|---|---|
| 1 | Radish | ₹26 | 78 | ₹2028 |
| 2 | Potato | ₹20 | 112 | ₹2240 |
| 3 | Cabbage | ₹32 | 56 | ₹1792 |
| 4 | Green peas | ₹25 | 125 | ₹3125 |
| | Total | | | ₹9185 |

(a) LHS: $8 \times 9 = 72$; RHS: $70 + 2 = 72$
$$72 = 72 \quad \Rightarrow \mathbf{True (T)}$$

(b) LHS: $20 - 6 = 14$; RHS: $7 \times 3 = 21$
$$14 \neq 21 \quad \Rightarrow \mathbf{False (F)}$$

(c) LHS: $48 \div 3 = 16$; RHS: $4 \times 4 = 16$
$$16 = 16 \quad \Rightarrow \mathbf{True (T)}$$

(d) LHS: $89 - 9 = 80$; RHS: $90 + 0 = 90$
$$80 \neq 90 \quad \Rightarrow \mathbf{False (F)}$$

(e) LHS: $25 + 10 = 35$; RHS: $45 - 10 = 35$
$$35 = 35 \quad \Rightarrow \mathbf{True (T)}$$

(a) $7 \times 6 = 42$
$$42 = \_ + 17 \Rightarrow \_ = 42 - 17 = \mathbf{25}$$

(b) $87 + 6 = 93$
$$93 = \_ \times 31 \Rightarrow \_ = 93 \div 31 = \mathbf{3}$$

(c) $74 - 4 = 70$
$$63 + \_ = 70 \Rightarrow \_ = 70 - 63 = \mathbf{7}$$

(d) $16 \div 2 = 8$
$$\_ \div 9 = 8 \Rightarrow \_ = 8 \times 9 = \mathbf{72}$$

(a) 'Sum of two odd numbers is even' — Always True

Examples:
$$1 + 3 = 4 \text{ (even)} \checkmark$$
$$5 + 7 = 12 \text{ (even)} \checkmark$$
$$9 + 11 = 20 \text{ (even)} \checkmark$$
$$13 + 15 = 28 \text{ (even)} \checkmark$$
$$21 + 19 = 40 \text{ (even)} \checkmark$$

This is always true. We cannot find a counter-example because odd + odd = even always.

(b) 'Multiplying a number by 2 gives an odd number' — Never True

Any number $\times$ 2 is always even (it is a multiple of 2).
$$2 \times 1 = 2, \quad 2 \times 3 = 6, \quad 2 \times 5 = 10$$
All results are even. We cannot find any example where $n \times 2$ is odd.

(c) 'Halving a number always gives an even number' — Sometimes True

Examples where it IS true (halving gives even):
$$16 \div 2 = 8 \text{ (even)} \checkmark$$
$$20 \div 2 = 10 \text{ (even)} \checkmark$$
$$28 \div 2 = 14 \text{ (even)} \checkmark$$

Examples where it is NOT true (halving gives odd):
$$6 \div 2 = 3 \text{ (odd)}$$
$$10 \div 2 = 5 \text{ (odd)}$$
$$14 \div 2 = 7 \text{ (odd)}$$

(i) Adding 10 to a number gives a multiple of ten.

Example: $13 + 10 = 23$ (not a multiple of 10); $20 + 10 = 30$ (multiple of 10).
$$\Rightarrow \mathbf{Sometimes\ True}$$

(ii) Changing the order of the numbers in subtraction makes no difference.

$8 - 3 = 5$ but $3 - 8 = -5$ (different result).
$$\Rightarrow \mathbf{Never\ True}$$

(iii) In multiplication, doubling one number and halving the other keeps the product the same.

$4 \times 6 = 24$; double 4 and halve 6: $8 \times 3 = 24$ ✓. This works for all even numbers being halved.
However, if the number being halved is odd (e.g., $4 \times 5 = 20$; $8 \times 2.5 = 20$), it still holds mathematically but may not give whole numbers.
For whole number contexts: Sometimes True (works when the number being halved is even).
$$\Rightarrow \mathbf{Sometimes\ True}$$

(iv) Multiplication by an odd number gives an even number.

$3 \times 4 = 12$ (even); $3 \times 3 = 9$ (odd). Depends on the other number.
$$\Rightarrow \mathbf{Sometimes\ True}$$

(v) Multiplying a number by 5 leads to numbers which have '0' in the Ones place.

$5 \times 2 = 10$ (ends in 0); $5 \times 3 = 15$ (ends in 5); $5 \times 4 = 20$ (ends in 0).
Multiples of 5 end in either 0 or 5, not always 0.
$$\Rightarrow \mathbf{Sometimes\ True}$$

Summary Table:
| Statement | Always True | Sometimes True | Never True |
|---|---|---|---|
| (i) Adding 10 gives multiple of 10 | | ✓ | |
| (ii) Order in subtraction doesn't matter | | | ✓ |
| (iii) Double one, halve other — product same | | ✓ | |
| (iv) Multiplying by odd gives even | | ✓ | |
| (v) Multiplying by 5 gives '0' in ones place | | ✓ | |