NCERT Solutions for Class 5th Maths Chapter 9 Solve the following word problems — Let Us Solve

Book page 126 Updated on2026-09-19

Q1.

1. Rani 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.

Answer

Packs of 6 are the better single choice — she needs 42 packs and only 2 samosas are left over.

She needs at least 250 samosas, one for each guest. Let us try each pack size.

If she buys packs of 6:

Step 1: 250 ÷ 6 → 6 × 41 = 246, and 250 − 246 = 4
Step 2: So 41 packs give only 246 samosas — 4 guests would get nothing
Step 3: She must buy 42 packs
Step 4: 42 × 6 = 252 samosas, so 2 samosas are extra

If she buys packs of 8:

Step 1: 250 ÷ 8 → 8 × 31 = 248, and 250 − 248 = 2
Step 2: So she must buy 32 packs
Step 3: 32 × 8 = 256 samosas, so 6 samosas are extra

Pack size	Packs needed	Samosas bought	Left over
6	42	252	2
8	32	256	6

Explanation: Both choices feed all 250 guests. Packs of 6 waste only 2 samosas, packs of 8 waste 6. So packs of 6 are the better buy. (If Rani cares more about carrying fewer packs, packs of 8 mean only 32 packs to carry instead of 42.)

Did you know? If Rani may buy both sizes, she can get exactly 250 with nothing wasted: 29 packs of 8 and 3 packs of 6.
29 × 8 = 232, 3 × 6 = 18, 232 + 18 = 250 ✓

Q2.

2. 342 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?

Answer

The school needs 9 buses.

We need to find how many buses of 41 seats hold 342 students. So we divide.

Step 1: 342 ÷ 41
Step 2: 41 × 8 = 328 (this fits)
Step 3: 41 × 9 = 369 (too many seats, but 8 buses are not enough)
Step 4: 342 − 328 = 14 students left over
Step 5: Those 14 students also need a bus

Buses = 8 full + 1 more = 9 buses

Why not 8? Eight buses carry only 328 students. Fourteen children would be left standing at the school gate. A part-full bus still counts as a bus.

Check by a second route: N = D × Q + R → 41 × 8 + 14 = 328 + 14 = 342 ✓
Answer: 9 buses.

Q3.

3. Sofia 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.

Answer

There are six ways. The neatest is 8 notes of ₹50 and 6 notes of ₹20.

How to find them, step by step.

Notice that ₹520 ends in a 0, and both ₹50 and ₹20 end in a 0. So work in tens: she needs 52 tens.
A ₹50 note is 5 tens. A ₹20 note is 2 tens.
Try 0 fifty-rupee notes, then 1, then 2 … and see what is left for the ₹20 notes.
Whatever is left must divide exactly by 20, so the number of ₹50 notes must be even.

₹50 notes	Value	Still to pay	₹20 notes	Check
0	₹0	₹520	26	26 × 20 = 520 ✓
2	₹100	₹420	21	100 + 420 = 520 ✓
4	₹200	₹320	16	200 + 320 = 520 ✓
6	₹300	₹220	11	300 + 220 = 520 ✓
8	₹400	₹120	6	400 + 120 = 520 ✓
10	₹500	₹20	1	500 + 20 = 520 ✓

Why the number of ₹50 notes must be even: One ₹50 note leaves ₹470, and 470 ÷ 20 = 23 with 10 left over — you cannot pay ₹10 with ₹20 notes. Two ₹50 notes make ₹100, which is a whole number of twenties. So the fifties must come in pairs.

Fewest notes: The row with 10 fifties and 1 twenty uses just 11 notes — the lightest purse. The row with 26 twenties uses the most.

Q4.

4. Three 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?

Answer

Each person should pay ₹139.

We need to find one person's share. First find the whole cost, then share it into 3 equal parts.

Step 1 — add up everything spent:
157 + 124 = 281
281 + 136 = ₹417 in all

Step 2 — share it among 3 friends:
417 ÷ 3
Split 417 into 300 + 117
300 ÷ 3 = 100
117 ÷ 3 = 39
100 + 39 = ₹139

Check by a second route: 3 × ₹139 = ₹417 ✓, and ₹417 is exactly what they spent.
Answer: ₹139 each.

Q5.

5. Identify the remainder, if any. Check if N = D × Q + R. (a) 887 ÷ 3 (b) 283 ÷ 8 (c) 745 ÷ 5 (d) 767 ÷ 26 (e) 530 ÷ 41 (f) 888 ÷ 67

Answer

Part	Division	Quotient (Q)	Remainder (R)	Check: N = D × Q + R
(a)	887 ÷ 3	295	2	3 × 295 + 2 = 885 + 2 = 887 ✓
(b)	283 ÷ 8	35	3	8 × 35 + 3 = 280 + 3 = 283 ✓
(c)	745 ÷ 5	149	0	5 × 149 + 0 = 745 ✓
(d)	767 ÷ 26	29	13	26 × 29 + 13 = 754 + 13 = 767 ✓
(e)	530 ÷ 41	12	38	41 × 12 + 38 = 492 + 38 = 530 ✓
(f)	888 ÷ 67	13	17	67 × 13 + 17 = 871 + 17 = 888 ✓

Two worked out in full.

(a) 887 ÷ 3

3) 887 ( 200 + 90 + 5
  −600 → 287 (3 × 200)
  −270 → 17 (3 × 90)
   −15 →   2 (3 × 5)

Q = 200 + 90 + 5 = 295, R = 2

(d) 767 ÷ 26

26) 767 ( 20 + 9
−520 → 247 (26 × 20)
−234 → 13 (26 × 9)

Q = 20 + 9 = 29, R = 13

Always true: the remainder is smaller than the divisor. Look down the table — 2 < 3, 3 < 8, 13 < 26, 38 < 41, 17 < 67. If your remainder is ever as big as the divisor, take away one more group.

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