NCERT Solutions for Class 4th Maths Chapter 13 Chinnu’s Coins — Chinnu’s Coins
Book page 201 Updated on2026-09-19
Q1.
1. Five friends plan to visit an amusement park nearby. … The cost of the ticket is ₹ 750. Bujji has brought all notes of ₹ 200. Munna has brought all notes of ₹ 50. Balu has brought all notes of ₹ 20. Chinnu has all coins of ₹ 5. Sansu has all coins of ₹ 2. a) Find out how many notes/coins each child has to bring to buy the ticket.
Answer
The price board at the amusement park ticket counter.
Child
Note / coin
Working
How many
Bujji
₹200 notes
3 notes = ₹600, 4 notes = ₹800
4 notes
Munna
₹50 notes
750 ÷ 50 = 15
15 notes
Balu
₹20 notes
37 notes = ₹740, 38 notes = ₹760
38 notes
Chinnu
₹5 coins
750 ÷ 5 = 150
150 coins
Sansu
₹2 coins
750 ÷ 2 = 375
375 coins
Bujji: 750 ÷ 200 = 3, and ₹150 is still left. So 3 notes are not enough. He must bring 4 notes (₹800). Balu: 750 ÷ 20 = 37, and ₹10 is still left. So he must bring 38 notes (₹760).
Why it happens: ₹200 and ₹20 do not fit into ₹750 exactly, so the child must bring one extra note and take change.
Q2.
b) Which of these children will not receive any change from the cashier?
Answer
The price board at the amusement park ticket counter.
Munna, Chinnu and Sansu will not get any change.
Child
Money given
Ticket
Change
Bujji
₹800
₹750
₹50
Munna
₹750
₹750
nil
Balu
₹760
₹750
₹10
Chinnu
₹750
₹750
nil
Sansu
₹750
₹750
nil
Bujji’s change:
7
10
₹
8
0
0
−
₹
7
5
0
₹
5
0
Balu’s change:
₹
7
6
0
−
₹
7
5
0
₹
1
0
Why it happens: ₹50, ₹5 and ₹2 all divide ₹750 exactly, so those three can pay the exact amount.
Tip: Look at the last digits. 750 ends in 50, and 50, 5 and 2 all go into it with nothing left over.
Q3.
c) How long would the cashier take to count Chinnu’s coins?
Answer
Chinnu brings 150 coins of ₹5.
Sample answer. Say the cashier counts 1 coin every second.
If the cashier counts 2 coins a second, it takes 75 seconds — about 1 minute 15 seconds.
Try This: Count 20 coins (or 20 pebbles) yourself with a clock. Then work out your own answer for 150.
Why it happens: Counting time depends on how fast we count, so we first decide a speed and then multiply.
Q4.
2. Observe the following multiplications. The answers have been provided. (12 × 13 = 156, 11 × 14 = 154, 13 × 13 = 169, 11 × 12 = 132) In each case, do you see any pattern in the two numbers and their product? (Hint: Look at the coloured digits!)The four multiplications printed in the book, with the coloured digits.
Answer
The four multiplications printed in the book, with the coloured digits.
Both numbers start with 1. Look only at the ones digits.
Problem
Ones digits
Their sum
Their product
Answer
12 × 13
2 and 3
5
6
156
11 × 14
1 and 4
5
4
154
13 × 13
3 and 3
6
9
169
11 × 12
1 and 2
3
2
132
The pattern Hundreds digit of the answer = 1 Tens digit = sum of the two ones digits Ones digit = product of the two ones digits
Example: 12 × 13 2 + 3 = 5 and 2 × 3 = 6 So the answer is 1 5 6 = 156
Why it happens: 12 × 13 = 12 × 10 + 12 × 3 = 120 + 36. The 100 gives the 1, the 20 and the 30 make the tens, and 2 × 3 makes the ones.
Q5.
For what other multiplication problems will this pattern hold? Find 5 such examples.The four multiplications printed in the book, with the coloured digits.
Answer
The pattern works when both numbers are in the teens (11 to 19) and
the two ones digits add to less than 10, and
the two ones digits multiply to less than 10.
Five examples:
Problem
Sum of ones digits
Product of ones digits
Answer
11 × 13
1 + 3 = 4
1 × 3 = 3
143
12 × 12
2 + 2 = 4
2 × 2 = 4
144
11 × 15
1 + 5 = 6
1 × 5 = 5
165
12 × 14
2 + 4 = 6
2 × 4 = 8
168
11 × 17
1 + 7 = 8
1 × 7 = 7
187
Where it breaks: 14 × 14 4 + 4 = 8 but 4 × 4 = 16, which is not one digit. The true answer is 196, not 1–8–16.
Why it happens: The ones digit has room for only one digit. If the product of the ones digits is 10 or more, it carries into the tens and the neat pattern is spoilt.