The Cleanest Village — NCERT Solutions

Given: Cost = ₹185, Paid = ₹200

Formula: Balance = Paid − Cost

$$200 - 185 = 15$$

Balance = ₹15

Given: Paid = ₹200, Balance = ₹0

Formula: Cost = Paid − Balance

$$200 - 0 = 200$$

Cost = ₹200

Given: Cost = ₹435, Paid = ₹500

Formula: Balance = Paid − Cost

$$500 - 435 = 65$$

Balance = ₹65

Given: Cost = ₹149, Paid = ₹500

Formula: Balance = Paid − Cost

$$500 - 149 = 351$$

Balance = ₹351

Given: Cost = ₹46

This problem has multiple valid answers depending on the note paid. One common possibility:

If Paid = ₹50, then Balance = ₹50 − ₹46 = ₹4

If Paid = ₹100, then Balance = ₹100 − ₹46 = ₹54

One valid answer: Paid = ₹50, Balance = ₹4

Given: Paid = ₹200, Balance = ₹75

Formula: Cost = Paid − Balance

$$200 - 75 = 125$$

Cost = ₹125

Given: Cost = ₹580, Balance = ₹120

Formula: Paid = Cost + Balance

$$580 + 120 = 700$$

Paid = ₹700

This is an open-ended question. Students may fill in any valid combination where:

$$\text{Balance} = \text{Paid} - \text{Cost}$$

For example: Cost = ₹250, Paid = ₹500, Balance = ₹250

Any correct combination satisfying Balance = Paid − Cost is acceptable.

Given: Total amount spent = ₹163

We need to find combinations of items that add up to ₹163. (Prices are based on the price list shown in the chapter.)

Possibility 1: 2 kg of beans (assume ₹80) + 1 kg of onion and 1 kg of potato (assume ₹83) = ₹163

Possibility 2: Any other combination of items from the price list that totals ₹163.

Note: Students should refer to the price list in their textbook and find at least two different combinations of items whose total cost equals ₹163.

Step 1: Find the cost of 2 oranges.

Cost of 1 orange = ₹21

Cost of 2 oranges = $$21 \times 2 = ₹42$$

Step 2: Find each child's balance.

Krishna:
Paid = ₹50
$$\text{Balance} = 50 - 42 = ₹8$$

Sudama:
Paid = ₹100
$$\text{Balance} = 100 - 42 = ₹58$$

Mala:
Paid = ₹200
$$\text{Balance} = 200 - 42 = ₹158$$

Neela:
Paid = ₹500
$$\text{Balance} = 500 - 42 = ₹458$$

Summary:

Concept: In a triangle with 3 corners and 3 sides, each number at a corner is counted twice (once for each side it belongs to).

Sum of all numbers 1 to 6 = $$1+2+3+4+5+6 = 21$$

If each side sums to 9 and there are 3 sides: Total counted = $$9 \times 3 = 27$$

Since corner numbers are counted twice: Sum of corner numbers = $$27 - 21 = 6$$

So the three corner numbers must add up to 6. One possibility: corners = 1, 2, 3 (since $1+2+3=6$).

Place 1, 2, 3 at the corners. Then fill the sides:

• Side with corners 1 and 2: middle number = $9 - 1 - 2 = 6$
• Side with corners 2 and 3: middle number = $9 - 2 - 3 = 4$
• Side with corners 1 and 3: middle number = $9 - 1 - 3 = 5$

Answer: Corners: 1, 2, 3; Middle of sides: 6, 4, 5 (each side sums to 9).

Concept: Each side sums to 10, 3 sides: Total counted = $$10 \times 3 = 30$$

Sum of corner numbers = $$30 - 21 = 9$$

Three numbers from 1–6 that add to 9: e.g., 2, 3, 4 (since $2+3+4=9$).

Place 2, 3, 4 at the corners. Fill the sides:

• Side with corners 2 and 3: middle = $10 - 2 - 3 = 5$
• Side with corners 3 and 4: middle = $10 - 3 - 4 = 3$... (3 is already used)

Try corners 1, 2, 6 ($1+2+6=9$):

• Side with 1 and 2: middle = $10-1-2=7$... 7 is not in 1–6.

Try corners 1, 3, 5 ($1+3+5=9$):

• Side with 1 and 3: middle = $10-1-3=6$ ✓
• Side with 3 and 5: middle = $10-3-5=2$ ✓
• Side with 1 and 5: middle = $10-1-5=4$ ✓

All numbers used: corners 1, 3, 5 and middles 6, 2, 4 — all six numbers 1–6 used once.

Answer: Corners: 1, 3, 5; Middle of sides: 6, 2, 4 (each side sums to 10).

Finding possible sums:

For a sum $S$ on each side:

• Sum of corner numbers = $3S - 21$
• Corner numbers must be 3 distinct numbers from 1–6 with sum = $3S - 21$
• Minimum sum of 3 numbers from 1–6 = $1+2+3=6$, so $3S-21 \geq 6 \Rightarrow S \geq 9$
• Maximum sum of 3 numbers from 1–6 = $4+5+6=15$, so $3S-21 \leq 15 \Rightarrow S \leq 12$

So possible sums are 9, 10, 11, and 12.

Sum = 12: Corner sum = $3(12)-21=15$, so corners = 4, 5, 6.

• Side with 4 and 5: middle = $12-4-5=3$ ✓
• Side with 5 and 6: middle = $12-5-6=1$ ✓
• Side with 4 and 6: middle = $12-4-6=2$ ✓

Yes, sum of 12 is possible.

Sum = 13: Corner sum = $3(13)-21=18$, but maximum sum of 3 numbers from 1–6 is 15. So sum of 13 is NOT possible.

Given: Teachers from School 1 = 24, Teachers from School 2 = 28

Total teachers = $$24 + 28 = 52$$

There are 52 teachers accompanying the children.

Given: Children from School 1 = 438, Children from School 2 = 476

Total children:
$$438 + 476$$

Column addition:
$$\begin{array}{r} 438 \\ +476 \\ \hline 914 \end{array}$$

Ones: $8+6=14$, write 4 carry 1
Tens: $3+7+1=11$, write 1 carry 1
Hundreds: $4+4+1=9$

914 children are going on the trip.

Rounding to nearest hundred:

$438 \approx 400$ and $476 \approx 500$

$$400 + 500 = 900$$

Estimated number of children ≈ 900

(Actual answer is 914, which is close to the estimate.)

Given: Cost of large piece = ₹38, Cost of small piece = ₹16

Total spent:
$$38 + 16$$

$$\begin{array}{r} 38 \\ +16 \\ \hline 54 \end{array}$$

Ones: $8+6=14$, write 4 carry 1
Tens: $3+1+1=5$

They spent ₹54 on pusaw.

Given: Money in piggy bank = ₹185, Money from mother = ₹125

Total money for trip:
$$185 + 125$$

$$\begin{array}{r} 185 \\ +125 \\ \hline 310 \end{array}$$

Ones: $5+5=10$, write 0 carry 1
Tens: $8+2+1=11$, write 1 carry 1
Hundreds: $1+1+1=3$

Daisy and Lou took ₹310 for the trip.

Estimate: $185 \approx 200$, $125 \approx 100$; $200+100=300$ ≈ ₹300 (close to actual ₹310).

Total boys = $78 + 532$
$$\begin{array}{r} 532 \\ +78 \\ \hline 610 \end{array}$$
Ones: $2+8=10$, write 0 carry 1; Tens: $3+7+1=11$, write 1 carry 1; Hundreds: $5+0+1=6$
Total boys = 610

Total girls = $95 + 346$
$$\begin{array}{r} 346 \\ +95 \\ \hline 441 \end{array}$$
Ones: $6+5=11$, write 1 carry 1; Tens: $4+9+1=14$, write 4 carry 1; Hundreds: $3+0+1=4$
Total girls = 441

Total tabla players = $78 + 95$
$$\begin{array}{r} 78 \\ +95 \\ \hline 173 \end{array}$$
Ones: $8+5=13$, write 3 carry 1; Tens: $7+9+1=17$
Total tabla players = 173

Total singers = $532 + 346$
$$\begin{array}{r} 532 \\ +346 \\ \hline 878 \end{array}$$
Total singers = 878

Girls playing tabla after 15 more join = $95 + 15 = 110$
110 girls play tabla now.

$$\begin{array}{r} 32 \\ +47 \\ \hline 79 \end{array}$$
Answer: 79

$$\begin{array}{r} 654 \\ +95 \\ \hline 749 \end{array}$$
Ones: $4+5=9$; Tens: $5+9=14$, write 4 carry 1; Hundreds: $6+0+1=7$
Answer: 749

$$\begin{array}{r} 286 \\ +123 \\ \hline 409 \end{array}$$
Ones: $6+3=9$; Tens: $8+2=10$, write 0 carry 1; Hundreds: $2+1+1=4$
Answer: 409

$$\begin{array}{r} 476 \\ +324 \\ \hline 800 \end{array}$$
Ones: $6+4=10$, write 0 carry 1; Tens: $7+2+1=10$, write 0 carry 1; Hundreds: $4+3+1=8$
Answer: 800

$$\begin{array}{r} 700 \\ +289 \\ \hline 989 \end{array}$$
Answer: 989

$$\begin{array}{r} 534 \\ +483 \\ \hline 1017 \end{array}$$
Ones: $4+3=7$; Tens: $3+8=11$, write 1 carry 1; Hundreds: $5+4+1=10$
Answer: 1017

Given: Preeti's school = 423 children; Cousin's school has 178 more children.

Children in cousin's school = $423 + 178$

$$\begin{array}{r} 423 \\ +178 \\ \hline 601 \end{array}$$

Ones: $3+8=11$, write 1 carry 1; Tens: $2+7+1=10$, write 0 carry 1; Hundreds: $4+1+1=6$

601 children study in Preeti's cousin's school.

Given: Dolls made = 639, Dolls sold = 531

Dolls left = $639 - 531$

$$\begin{array}{r} 639 \\ -531 \\ \hline 108 \end{array}$$

Ones: $9-1=8$; Tens: $3-3=0$; Hundreds: $6-5=1$

Estimate: $639 \approx 600$, $531 \approx 500$; $600-500=100$ ≈ 108 ✓

Number of dolls left = 108

$$\begin{array}{r} 83 \\ -29 \\ \hline 54 \end{array}$$
Ones: $3-9$ not possible, regroup: $13-9=4$; Tens: $7-2=5$
Answer: 54

$$\begin{array}{r} 345 \\ -123 \\ \hline 222 \end{array}$$
Ones: $5-3=2$; Tens: $4-2=2$; Hundreds: $3-1=2$
Answer: 222

$$\begin{array}{r} 763 \\ -437 \\ \hline 326 \end{array}$$
Ones: $3-7$ not possible, regroup: $13-7=6$; Tens: $5-3=2$; Hundreds: $7-4=3$
Answer: 326

$$\begin{array}{r} 803 \\ -350 \\ \hline 453 \end{array}$$
Ones: $3-0=3$; Tens: $0-5$ not possible, regroup from hundreds: $10-5=5$; Hundreds: $7-3=4$
Answer: 453

$$\begin{array}{r} 900 \\ -328 \\ \hline 572 \end{array}$$
Ones: $0-8$ not possible, regroup (borrow from tens, but tens is 0 too, so borrow from hundreds): $10-8=2$; Tens: $9-2-1=6$...

Step by step: Regroup 900 as 8 hundreds, 9 tens, 10 ones.
Ones: $10-8=2$; Tens: $9-2=7$...

Actually: 900 → regroup to 8H 10T 0O → regroup to 8H 9T 10O
Ones: $10-8=2$; Tens: $9-2=7$; Hundreds: $8-3=5$
Answer: 572

Rami's pages: 236

Kesu's pages: $96 + 30 + 90$
$$96 + 30 = 126$$
$$126 + 90 = 216$$

Kesu read 216 pages.

Comparison: Rami read 236 pages, Kesu read 216 pages.

$$236 - 216 = 20$$

Rami read more. Rami read 20 more pages than Kesu.

Given: Total pages = 128, Pages read = 23

Pages left = $128 - 23$

$$\begin{array}{r} 128 \\ -23 \\ \hline 105 \end{array}$$

Ones: $8-3=5$; Tens: $2-2=0$; Hundreds: $1-0=1$

105 more pages are left to complete the book.

Step 1: Total pages of all books.
$$179 + 236 + 30 + 96 + 90 + 128$$
$$= 179 + 236 = 415$$
$$415 + 30 = 445$$
$$445 + 96 = 541$$
$$541 + 90 = 631$$
$$631 + 128 = 759$$
Total pages = 759

Step 2: Pages already read.
Swami and Friends = 179
Akbar Birbal = 96
50 pages of Feluda = 50
$$179 + 96 + 50 = 325$$

Step 3: Pages left to read.
$$759 - 325 = 434$$

$$\begin{array}{r} 759 \\ -325 \\ \hline 434 \end{array}$$

Jaggu has to read 434 more pages.

Note: The exact numbers at each station are shown in the figure (img_28) which is not visible here. Students should refer to their textbook figure.

General method: Passengers when leaving Dadri = (Passengers when arriving at Dadri) + (Passengers who boarded at Dadri) − (Passengers who got off at Dadri)

Students should substitute the values from the figure and compute accordingly.