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NCERT Solutions Ganita Prakash (Part 1) Chapter 3 In-text Questions — A Hundredth Part

Book page 58 Updated on2026-09-05

Q1.
Solve this by converting to hundredths.
Answer

The difference asked for is 25 9⁄10 – 6 4⁄10 7⁄100. Write both numbers in hundredths.

25 9⁄10 = 2500⁄100 + 90⁄100 = 2590⁄100
6 4⁄10 7⁄100 = 600⁄100 + 40⁄100 + 7⁄100 = 647⁄100
2590⁄100 – 647⁄100 = 1943⁄100
= 1900⁄100 + 40⁄100 + 3⁄100
= 19 4⁄10 3⁄100  (19.43)
Why it happens: converting both numbers to the same unit turns the whole problem into the plain subtraction 2590 – 647 = 1943. The answer is the same as by the regrouping method, as it must be.
Q2.
What is the difference 15 3/10 4/100 – 2 6/10 8/100 ?
Answer

Regroup twice, then subtract place by place.

15 3⁄10 4⁄100
= 15 2⁄10 14⁄100  (1 tenth → 10 hundredths)
= 14 12⁄10 14⁄100  (1 unit → 10 tenths)
Hundredths: 14 – 8 = 6
Tenths: 12 – 6 = 6
Units: 14 – 2 = 12
= 12 6⁄10 6⁄100  (12.66)

In hundredths: 1534⁄100 – 268⁄100 = 1266⁄100 = 12 66⁄100.

Why it happens: 8 hundredths cannot come out of 4 hundredths, and after the first exchange 6 tenths cannot come out of 2 tenths — so a second exchange is needed. Both exchanges leave the value untouched.
Check it yourself: 12.66 + 2.68 = 15.34 ✓
Q3.
Observe the subtraction done below for 653 – 268. Do you see any similarities with the methods shown above?
Answer

Yes — it is the same double regrouping, one place value higher.

(600 + 50 + 3) – (200 + 60 + 8)
= (600 – 200) + (50 – 60) + (3 – 8)
= (600 – 200) + (40 – 60) + (13 – 8)  (1 ten → 10 ones)
= (500 – 200) + (140 – 60) + 5  (1 hundred → 10 tens)
= 300 + 80 + 5
= 385
653 – 26815 3⁄10 4⁄100 – 2 6⁄10 8⁄100
Ones short → borrow a tenHundredths short → borrow a tenth
Tens short → borrow a hundredTenths short → borrow a unit
Answer 385Answer 12.66
Why it happens: the borrowing rule only needs one fact — each place is worth 10 of the place on its right. That fact holds on both sides of the decimal point, so the procedure never changes.
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