The Transport Museum

Given: A multiplication matrix where some products are given and we must find 1-digit multipliers/multiplicands to complete it.

Strategy: Work backwards from the given products to find 1-digit factors.

Step 1 – Identify factors from given products:
- $32 = 4 \times 8$ → row multiplier = 4, column multiplier = 8
- $42 = 6 \times 7$ → row multiplier = 6, column multiplier = 7
- $45 = 5 \times 9$ → row multiplier = 5, column multiplier = 9
- $21 = 3 \times 7$ → row multiplier = 3, column multiplier = 7

Step 2 – One possible completed matrix (header row: ×, 8, 9, 7, 6):

| × | 8 | 9 | 7 | 6 |
|---|---|---|---|---|
| 4 | 32 | 36 | 28 | 24 |
| 6 | 48 | 54 | 42 | 36 |
| 5 | 40 | 45 | 35 | 30 |
| 3 | 24 | 27 | 21 | 18 |

Note: Multiple valid answers are possible depending on the choice of 1-digit numbers, as long as the given products appear in the correct positions.

Given: Row products (orange) and column products (blue) are provided. We need to find the individual numbers.

Step 1 – Analyse the structure. The table appears to be a 3-row grid where the last column shows row products and the last row shows column products.

Step 2 – Find numbers for Row 1 (product = 42):
$42 = 6 \times 7$ or $2 \times 3 \times 7$, etc.
With the column product clue 56 in column 3: $56 = 7 \times 8$, so one entry is 7 or 8.
A possible row: $1, 6, 7$ → product = $1 \times 6 \times 7 = 42$ ✓

Step 3 – Find numbers for Row 2 (product = 50):
$50 = 2 \times 5 \times 5$ or $5 \times 10$
With column product clue 54 in column 3: $54 = 6 \times 9$
A possible row: $2, 5, 5$ → product = 50 ✓

Step 4 – Check column products against Row 3 (products: 63, 48, 60, 35):
- Column with product 63: $63 = 7 \times 9$ or $63 = 1 \times 63$
- Column with product 48: $48 = 6 \times 8$
- Column with product 60: $60 = 5 \times 12$ or $4 \times 15$
- Column with product 35: $35 = 5 \times 7$

Note: This is an open-ended puzzle with multiple valid solutions. The key method is to factorise each given product into 1-digit numbers and check consistency across rows and columns.

Concept: Multiplying any number by 10 gives that many tens.

$$2 \times 10 = 2 \text{ Tens} = \mathbf{20}$$

$$5 \times 10 = \mathbf{5} \text{ Tens} = \mathbf{50}$$

$$8 \times 10 = \mathbf{8} \text{ Tens} = \mathbf{80}$$

Concept: $10 \times 10$ means 10 groups of 10.

$$10 \times 10 = \mathbf{10} \text{ Tens} = \mathbf{100}$$

10 Tens make 1 Hundred = 100.

Given: $5 \times 15$

Method – Split 15 into 10 and 5:
$$5 \times 15 = 5 \times 10 + 5 \times 5$$
$$= 50 + 25 = \mathbf{75}$$

Method: Split 15 = 10 + 5, so $n \times 15 = n \times 10 + n \times 5$.

$$1 \times 15 = 10 + 5 = \mathbf{15}$$
$$2 \times 15 = 20 + 10 = \mathbf{30}$$
$$3 \times 15 = 30 + 15 = \mathbf{45}$$
$$4 \times 15 = 40 + 20 = \mathbf{60}$$
$$5 \times 15 = 50 + 25 = \mathbf{75}$$
$$6 \times 15 = 60 + 30 = \mathbf{90}$$
$$7 \times 15 = 70 + 35 = \mathbf{105}$$
$$8 \times 15 = 80 + 40 = \mathbf{120}$$
$$9 \times 15 = 90 + 45 = \mathbf{135}$$
$$10 \times 15 = 100 + 50 = \mathbf{150}$$

Pattern 1 – Patterns in the times-15 table:
- The products increase by 15 each time.
- All products end in either 0 or 5 (just like the times-5 table).
- The units digits follow the pattern: 5, 0, 5, 0, 5, 0, … (alternating).

Pattern 2 – Comparison with times-5 table:

| Times-5 | Times-15 | Difference |
|---|---|---|
| $1 \times 5 = 5$ | $1 \times 15 = 15$ | $15 - 5 = 10$ |
| $2 \times 5 = 10$ | $2 \times 15 = 30$ | $30 - 10 = 20$ |
| $3 \times 5 = 15$ | $3 \times 15 = 45$ | $45 - 15 = 30$ |

Similarity: Both tables end in 0 or 5.

Difference: Each product in the times-15 table is exactly 3 times the corresponding product in the times-5 table (since $15 = 3 \times 5$). The differences themselves form the times-10 table.

Question 3 – Times-tables for 11 to 20 (using splitting method):

For any number $n$ between 11 and 20, write $n = 10 + r$ where $r$ is the units digit.
Then: $k \times n = k \times 10 + k \times r$

Example – Times-12 table ($12 = 10 + 2$):
$$1 \times 12 = 10 + 2 = 12$$
$$2 \times 12 = 20 + 4 = 24$$
$$3 \times 12 = 30 + 6 = 36$$
$$\vdots$$
$$10 \times 12 = 100 + 20 = 120$$

Question 4 – Observations:
- Each product in the times-11 table = corresponding product in times-1 table + corresponding product in times-10 table.
- Each product in the times-12 table = corresponding product in times-2 table + corresponding product in times-10 table.
- In general: $k \times (n+10) = k \times n + k \times 10$, so every product in the higher table is exactly $k \times 10$ more than in the lower table.

Given: $3 \times 14$

Method – Split 14 into two equal groups of 7:
$$3 \times 14 = 3 \times 7 + 3 \times 7$$
$$= 21 + 21 = \text{double of } 21$$
$$= \mathbf{42}$$

Given: $6 \times 14$

Method – Split 14 into $7 + 7$:
$$6 \times 14 = 6 \times 7 + 6 \times 7$$
$$= 42 + 42 = \text{double of } 42$$
$$= \mathbf{84}$$

Times-14 table (using $14 = 7 + 7$, i.e., double of times-7):

$$1 \times 14 = 2 \times (1 \times 7) = 2 \times 7 = \mathbf{14}$$
$$2 \times 14 = 2 \times (2 \times 7) = 2 \times 14 = \mathbf{28}$$
$$3 \times 14 = 2 \times (3 \times 7) = 2 \times 21 = \mathbf{42}$$
$$4 \times 14 = 2 \times (4 \times 7) = 2 \times 28 = \mathbf{56}$$
$$5 \times 14 = 2 \times (5 \times 7) = 2 \times 35 = \mathbf{70}$$
$$6 \times 14 = 2 \times (6 \times 7) = 2 \times 42 = \mathbf{84}$$
$$7 \times 14 = 2 \times (7 \times 7) = 2 \times 49 = \mathbf{98}$$
$$8 \times 14 = 2 \times (8 \times 7) = 2 \times 56 = \mathbf{112}$$
$$9 \times 14 = 2 \times (9 \times 7) = 2 \times 63 = \mathbf{126}$$
$$10 \times 14 = 2 \times (10 \times 7) = 2 \times 70 = \mathbf{140}$$

Other tables that can be built by splitting and doubling:
- Times-6 = double of times-3
- Times-8 = double of times-4
- Times-10 = double of times-5
- Times-12 = double of times-6
- Times-16 = double of times-8
- Times-18 = double of times-9

Any even-numbered times-table can be constructed this way.

Concept: $n \times 10 = n$ Tens.

a) $15 \times 10 = \mathbf{15}$ Tens $= \mathbf{150}$

b) $16 \times 10 = \mathbf{16}$ Tens $= \mathbf{160}$

c) $19 \times 10 = \mathbf{19}$ Tens $= \mathbf{190}$

d) $20 \times 10 = \mathbf{20}$ Tens $= \mathbf{200}$

Concept: Multiplying a multiple of 10 by 10 — append a zero.

$$30 \times 10 = \mathbf{300}$$
$$40 \times 10 = \mathbf{400}$$
$$70 \times 10 = \mathbf{700}$$
$$50 \times 10 = \mathbf{500}$$
$$60 \times 10 = \mathbf{600}$$
$$80 \times 10 = \mathbf{800}$$

Given: 26 tempo travellers, each carrying 10 people.

$$26 \times 10 = 26 \text{ Tens} = 20 \text{ Tens} + 6 \text{ Tens} = 200 + 60 = \mathbf{260}$$

Answer: 260 people can travel.

Concept: $n \times 10$ = place a zero at the end of $n$.

a) $21 \times 10 = \mathbf{210}$

b) $42 \times 10 = \mathbf{420}$

c) $65 \times 10 = \mathbf{650}$

d) $38 \times 10 = \mathbf{380}$

e) $53 \times 10 = \mathbf{530}$

f) $87 \times 10 = \mathbf{870}$

Given: 12 buses, each seating 20 people.

Method: $12 \times 20 = 12 \times 2 \text{ Tens} = 24 \text{ Tens}$

$$12 \times 20 = 24 \times 10 = \mathbf{240}$$

Answer: 240 people can be seated in 12 buses.

Method: $a \times (b \times 10) = (a \times b) \times 10$

$$24 \times 40 = 24 \times 4 \times 10 = 96 \times 10 = \mathbf{960}$$

$$13 \times 30 = 13 \times 3 \times 10 = 39 \times 10 = \mathbf{390}$$

$$50 \times 60 = 5 \times 6 \times 100 = 30 \times 100 = \mathbf{3000}$$

$$43 \times 60 = 43 \times 6 \times 10 = 258 \times 10 = \mathbf{2580}$$

$$70 \times 80 = 7 \times 8 \times 100 = 56 \times 100 = \mathbf{5600}$$

Given: 15 coaches, each seating 14 children.

Method – Split using a grid (15 = 10 + 5, 14 = 10 + 4):

| × | 10 | 4 |
|---|---|---|
| 10 | $10 \times 10 = 100$ | $10 \times 4 = 40$ |
| 5 | $5 \times 10 = 50$ | $5 \times 4 = 20$ |

$$15 \times 14 = 100 + 40 + 50 + 20 = \mathbf{210}$$

Answer: 210 children can be seated in the toy train.

Given: 324 children, each coach seats 14.

Method – Repeated subtraction / long division:

$$14 \times 10 = 140$$
$$14 \times 20 = 280$$
$$14 \times 23 = 322$$

| No. of children | No. of coaches | Remaining |
|---|---|---|
| — | — | 324 |
| 140 | 10 | 184 |
| 140 | 10 | 44 |
| 14 | 1 | 30 |
| 28 | 2 | 2 |

$$324 \div 14 = 23 \text{ remainder } 2$$

Total coaches needed = 23 (for 322 children) + 1 extra coach for the remaining 2 children.

Total coaches = 24

Remainder = 2 (2 children still need a seat, so one more coach is required even though it won't be full).

a) $25 \times 34$

Split: $25 \times 34 = 25 \times 30 + 25 \times 4$
$$= 750 + 100 = \mathbf{850}$$

b) $16 \times 43$

Split: $16 \times 43 = 16 \times 40 + 16 \times 3$
$$= 640 + 48 = \mathbf{688}$$

c) $68 \times 12$

Split: $68 \times 12 = 68 \times 10 + 68 \times 2$
$$= 680 + 136 = \mathbf{816}$$

d) $39 \times 13$

Split: $39 \times 13 = 39 \times 10 + 39 \times 3$
$$= 390 + 117 = \mathbf{507}$$

e) $125 \div 15$

$15 \times 8 = 120$; $125 - 120 = 5$
$$125 \div 15 = \mathbf{8} \text{ remainder } \mathbf{5}$$

f) $94 \div 11$

$11 \times 8 = 88$; $94 - 88 = 6$
$$94 \div 11 = \mathbf{8} \text{ remainder } \mathbf{6}$$

g) $440 \div 22$

$22 \times 20 = 440$; $440 - 440 = 0$
$$440 \div 22 = \mathbf{20} \text{ remainder } \mathbf{0}$$

h) $508 \div 18$

$18 \times 28 = 504$; $508 - 504 = 4$
$$508 \div 18 = \mathbf{28} \text{ remainder } \mathbf{4}$$

Concept: $n \times 100 = n$ Hundreds.

$$3 \times 100 = \mathbf{3} \text{ Hundreds} = \mathbf{300}$$

$$5 \times 100 = \mathbf{5 \text{ Hundreds}} = \mathbf{500}$$

$$8 \times 100 = \mathbf{8 \text{ Hundreds}} = \mathbf{800}$$

Given: $10 \times 100 = 10$ Hundreds

$$10 \times 100 = \mathbf{1000}$$

When we put 10 Hundreds together, we get 1 Thousand = 1000.

Given: $80 \times 100 = 8000$

Finding $80 \times 50$:
$50 = \frac{100}{2}$, so $80 \times 50 = \frac{80 \times 100}{2} = \frac{8000}{2} = \mathbf{4000}$

Finding $40 \times 50$:
$40 \times 50 = 40 \times 5 \times 10 = 200 \times 10 = \mathbf{2000}$

Alternatively: $40 \times 50 = \frac{80 \times 50}{2} = \frac{4000}{2} = \mathbf{2000}$

$$11 \times 100 = \mathbf{1100}$$
$$22 \times 100 = \mathbf{2200}$$
$$11 \times 200 = \mathbf{2200}$$
$$22 \times 200 = \mathbf{4400}$$

Observation:
- $22 \times 100 = 11 \times 200 = 2200$ (doubling the multiplier is the same as doubling the multiplicand).
- $22 \times 200 = 2 \times (11 \times 200) = 2 \times 2200 = 4400$.
- When the number of groups doubles OR the group size doubles, the product doubles.

$$44 \times 10 = \mathbf{440}$$

$$16 \times 100 = \mathbf{1600}$$

$$22 \times 20 = 22 \times 2 \times 10 = 44 \times 10 = \mathbf{440}$$

$$4 \times 400 = 4 \times 4 \times 100 = 16 \times 100 = \mathbf{1600}$$

Observation: $44 \times 10 = 22 \times 20 = 440$ and $16 \times 100 = 4 \times 400 = 1600$ — halving one factor and doubling the other keeps the product the same.

Given: 64 flights, 152 people each.

Method – Split using grid (64 = 60 + 4, 152 = 100 + 50 + 2):

| × | 100 | 50 | 2 |
|---|---|---|---|
| 60 | 6000 | 3000 | 120 |
| 4 | 400 | 200 | 8 |
| Total | 6400 | 3200 | 128 |

$$64 \times 152 = 6400 + 3200 + 128 = \mathbf{9728}$$

Answer: 9728 people travelled in the first week of the Vande Bharat Mission.

Given: 960 participants, 64 per boat.

Method – Repeated subtraction:

| No. of boats | Participants used | Remaining |
|---|---|---|
| — | — | 960 |
| 10 | 640 | 320 |
| 5 | 320 | 0 |

$$960 \div 64 = \mathbf{15} \text{ remainder } \mathbf{0}$$

Answer: 15 boats will be needed.

a) $237 \times 28$

$237 \times 28 = 237 \times 20 + 237 \times 8$
$= 4740 + 1896 = \mathbf{6636}$

b) $140 \times 16$

$140 \times 16 = 140 \times 10 + 140 \times 6$
$= 1400 + 840 = \mathbf{2240}$

c) $389 \times 57$

$389 \times 57 = 389 \times 50 + 389 \times 7$
$= 19450 + 2723 = \mathbf{22173}$

d) $807 \div 24$

$24 \times 30 = 720$; $807 - 720 = 87$
$24 \times 3 = 72$; $87 - 72 = 15$
$$807 \div 24 = \mathbf{33} \text{ remainder } \mathbf{15}$$

e) $692 \div 33$

$33 \times 20 = 660$; $692 - 660 = 32$
$$692 \div 33 = \mathbf{20} \text{ remainder } \mathbf{32}$$

f) $996 \div 45$

$45 \times 22 = 990$; $996 - 990 = 6$
$$996 \div 45 = \mathbf{22} \text{ remainder } \mathbf{6}$$

1. Bunches of bananas:

Given: 870 bananas, 10 per bunch.

Division statement: $870 \div 10 = ?$

$$870 \div 10 = \mathbf{87} \text{ bunches}$$

2. Money for each grandchild:

Given: ₹1000 shared among 10 grandchildren.

Division statement: $1000 \div 10 = ?$

$$1000 \div 10 = \mathbf{₹100}$$

Each grandchild will get ₹100.

a) People per coach:

Given: 3 passenger coaches carried 96 passengers in total.

$$96 \div 3 = \mathbf{32} \text{ people per coach}$$

b) Total passengers now:

Given: 24 coaches, 72 passengers each.

$$24 \times 72 = 24 \times 70 + 24 \times 2$$
$$= 1680 + 48 = \mathbf{1728} \text{ passengers}$$

Given:
- Total students = Amala + 35 classmates = 36 children
- Teachers = 6
- Total people = 36 + 6 = 42
- Seats per deck: Lower = 15, Upper = 10 → Total seats = 25
- Each seat holds 2 people → Total capacity = $25 \times 2 = 50$ people

a) Seats needed:

Total people = 42
Each seat holds 2, so seats needed = $42 \div 2 = \mathbf{21}$ seats

Total available seats = 25

$25 \geq 21$ → Yes, there are enough seats for everyone.

b) Cost of tickets for all children:
*(Ticket prices depend on the image. Assuming Child ticket = ₹50)*
$$36 \times 50 = \mathbf{₹1800}$$

c) Cost of tickets for all teachers:
*(Assuming Teacher/Adult ticket = ₹100)*
$$6 \times 100 = \mathbf{₹600}$$

Note: Students should use the actual ticket prices shown in their textbook image to calculate the correct answers.

Given: 125 bricks per day, 1 month = 30 days.

a) Bricks in a month:
$$125 \times 30 = 125 \times 3 \times 10 = 375 \times 10 = \mathbf{3750} \text{ bricks}$$

b) Money earned in a month:
$$3750 \times 9 = 3750 \times 10 - 3750 \times 1 = 37500 - 3750 = \mathbf{₹33750}$$

Given:
- Number of people = 8
- Bus ticket per person = ₹60
- Boat ride for 8 people = ₹1200

Step 1 – Total bus cost for 8 people:
$$8 \times 60 = \mathbf{₹480}$$

Step 2 – Boat cost per person:
$$1200 \div 8 = \mathbf{₹150}$$

Step 3 – Total cost per person:
$$60 + 150 = \mathbf{₹210}$$

Answer: Each person needs to spend ₹210 (₹60 for bus + ₹150 for boat).

Given examples (already done):
- $250 \times 4 = 1000$ ✓
- $50 \times 20 = 1000$ ✓
- $525 \div 5 = 105$ ✓

Additional sentences that can be found in the grid:

- $5 \times 20 = 100$ (5, 20, 100 appear in grid)
- $3 \times 6 = 18$ — check grid
- $6 \times 22 = 132$ (6, 22, 132 appear in grid)
- $16 \times 200 = 3200$ (16, 200, 3200 appear in grid)
- $30 \times 20 = 600$ (30, 20, 600 appear in grid)
- $14 \times 8 = 112$ — check grid
- $6000 \div 200 = 30$ (6000, 200, 30 appear in grid)
- $1200 \div 8 = 150$ — check grid
- $104 \div 4 = 26$ (104, 4, 26 appear in grid)
- $9 \times 7 = 63$ — check grid

Note: Students should scan each row, column, and diagonal of the grid to find all valid multiplication and division sentences.

Multiplication:

a) $35 \times 76$
$= 35 \times 70 + 35 \times 6 = 2450 + 210 = \mathbf{2660}$

b) $267 \times 38$
$= 267 \times 30 + 267 \times 8 = 8010 + 2136 = \mathbf{10146}$

c) $498 \times 9$
$= 500 \times 9 - 2 \times 9 = 4500 - 18 = \mathbf{4482}$

d) $89 \times 42$
$= 89 \times 40 + 89 \times 2 = 3560 + 178 = \mathbf{3738}$

e) $55 \times 23$
$= 55 \times 20 + 55 \times 3 = 1100 + 165 = \mathbf{1265}$

f) $345 \times 17$
$= 345 \times 10 + 345 \times 7 = 3450 + 2415 = \mathbf{5865}$

g) $66 \times 22$
$= 66 \times 20 + 66 \times 2 = 1320 + 132 = \mathbf{1452}$

h) $704 \times 11$
$= 704 \times 10 + 704 \times 1 = 7040 + 704 = \mathbf{7744}$

i) $319 \times 26$
$= 319 \times 20 + 319 \times 6 = 6380 + 1914 = \mathbf{8294}$

---

Division:

j) $459 \div 3$
$3 \times 153 = 459$
$$459 \div 3 = \mathbf{153}, \text{ remainder } \mathbf{0}$$

k) $774 \div 18$
$18 \times 43 = 774$
$$774 \div 18 = \mathbf{43}, \text{ remainder } \mathbf{0}$$

l) $864 \div 26$
$26 \times 33 = 858$; $864 - 858 = 6$
$$864 \div 26 = \mathbf{33}, \text{ remainder } \mathbf{6}$$

m) $304 \div 12$
$12 \times 25 = 300$; $304 - 300 = 4$
$$304 \div 12 = \mathbf{25}, \text{ remainder } \mathbf{4}$$

n) $670 \div 9$
$9 \times 74 = 666$; $670 - 666 = 4$
$$670 \div 9 = \mathbf{74}, \text{ remainder } \mathbf{4}$$

o) $584 \div 25$
$25 \times 23 = 575$; $584 - 575 = 9$
$$584 \div 25 = \mathbf{23}, \text{ remainder } \mathbf{9}$$

p) $900 \div 15$
$15 \times 60 = 900$
$$900 \div 15 = \mathbf{60}, \text{ remainder } \mathbf{0}$$

q) $658 \div 32$
$32 \times 20 = 640$; $658 - 640 = 18$
$$658 \div 32 = \mathbf{20}, \text{ remainder } \mathbf{18}$$

r) $974 \div 9$
$9 \times 108 = 972$; $974 - 972 = 2$
$$974 \div 9 = \mathbf{108}, \text{ remainder } \mathbf{2}$$

Given: Ticket cost = ₹750

a) Notes/coins needed:

- Bujji (₹200 notes):
$750 \div 200 = 3$ notes = ₹600; needs 1 more ₹200 note → brings 4 notes (₹800), gets ₹50 change.

- Munna (₹50 notes):
$750 \div 50 = 15$ notes exactly → brings 15 notes (₹750).

- Balu (₹20 notes):
$750 \div 20 = 37$ notes = ₹740; needs 1 more → brings 38 notes (₹760), gets ₹10 change.

- Chinnu (₹5 coins):
$750 \div 5 = 150$ coins exactly → brings 150 coins (₹750).

- Sansu (₹2 coins):
$750 \div 2 = 375$ coins = ₹750; but 750 is odd, so $2 \times 375 = 750$ — actually $750 \div 2 = 375$ exactly → brings 375 coins (₹750).

b) Children who will NOT receive any change:

Children who pay exactly ₹750:
- Munna (15 × ₹50 = ₹750) ✓
- Chinnu (150 × ₹5 = ₹750) ✓
- Sansu (375 × ₹2 = ₹750) ✓

Bujji (pays ₹800, gets ₹50 change) and Balu (pays ₹760, gets ₹10 change) will receive change.

c) Time to count Chinnu's coins:

Chinnu has 150 coins of ₹5 each.
If the cashier counts 1 coin per second, it takes 150 seconds = 2 minutes 30 seconds.
If counting at 2 coins per second, it takes 75 seconds ≈ 1 minute 15 seconds.

*(The exact time depends on counting speed — this is an open discussion question.)*

Pattern Observation:

Look at the tens digit and units digit of the product:

- $12 \times 13 = 156$: $1 \times 1 = 1$ (hundreds), $2 + 3 = 5$ (tens), $2 \times 3 = 6$ (units) → 156 ✓
- $11 \times 14 = 154$: $1 \times 1 = 1$, $1 + 4 = 5$, $1 \times 4 = 4$ → 154 ✓
- $13 \times 13 = 169$: $1 \times 1 = 1$, $3 + 3 = 6$, $3 \times 3 = 9$ → 169 ✓
- $11 \times 12 = 132$: $1 \times 1 = 1$, $1 + 2 = 3$, $1 \times 2 = 2$ → 132 ✓

Pattern: For two-digit numbers of the form $1a$ and $1b$ (i.e., both in the teens):
$$1a \times 1b = 100 + (a+b) \times 10 + a \times b$$

5 more examples:

1. $11 \times 13 = 100 + (1+3) \times 10 + 1 \times 3 = 100 + 40 + 3 = \mathbf{143}$ ✓
2. $12 \times 14 = 100 + (2+4) \times 10 + 2 \times 4 = 100 + 60 + 8 = \mathbf{168}$ ✓
3. $11 \times 15 = 100 + (1+5) \times 10 + 1 \times 5 = 100 + 60 + 5 = \mathbf{165}$ ✓
4. $12 \times 12 = 100 + (2+2) \times 10 + 2 \times 2 = 100 + 40 + 4 = \mathbf{144}$ ✓
5. $13 \times 14 = 100 + (3+4) \times 10 + 3 \times 4 = 100 + 70 + 12 = \mathbf{182}$ ✓

Note: This pattern works perfectly when a \times b < 10 (single digit). When $a \times b \geq 10$, carry-over must be added.

Note: This question requires matching vehicles shown in the textbook image (which is not visible in the OCR).

Method to solve: For each vehicle, multiply the number of rows/seats by the number of people per seat (or use the given capacity).

General approach:
- Read the capacity of each vehicle from the image.
- Multiply: Number of seats × People per seat = Total capacity.
- Match the vehicle to its total passenger count.

Example (typical vehicles in such problems):
- Auto-rickshaw: 3 passengers
- Car: 4–5 passengers
- Mini-bus: 20–30 passengers
- Bus: 40–60 passengers
- Train coach: 72 passengers

Students should refer to the actual image in their textbook to complete the matching.