NCERT Solutions for Class 7th Maths Chapter 4 In-text Questions — A Quick Recap of Decimals

Book page 67 – 68 Updated on2026-09-19

Q1.
Jonali and Pallabi play a game. Jonali says a fraction and Pallabi gives the equivalent decimal. Write Pallabi's answer in the blank spaces. (Fractions: 3/10, 4/100, 67/1000, 457/100, 71/100, 43/100, 9/100)
Answer

Read the denominator first. It tells you how many digits must sit after the point.

FractionWhat it meansDecimal
3/103 tenths0.3
4/1000 tenths and 4 hundredths0.04
67/10000 tenths, 6 hundredths, 7 thousandths0.067
457/1004 ones, 5 tenths, 7 hundredths4.57
71/1007 tenths and 1 hundredth0.71
43/1004 tenths and 3 hundredths0.43
9/1000 tenths and 9 hundredths0.09
Why it happens: a denominator of 10 means the digits stop at the tenths place, 100 means they stop at the hundredths place, 1000 at the thousandths place. So the number of zeroes in the denominator is exactly the number of digits after the point. In 4/100 there is no tenth at all, so a 0 has to hold the tenths place: 0.04, not 0.4.
Check it yourself: 457/100 is more than 4 because 400/100 = 4. The answer 4.57 sits between 4 and 5, as it should.
Q2.
Jonali goes to the market to buy spices. She purchases 50 g of Cinnamon, 100 g of Cumin seeds, 25 g of Cardamom and 250 g of Pepper. Express each of the quantities in kilograms by writing them in terms of fractions as well as decimals.
Answer

1 kg = 1000 g, so every weight in grams becomes that many thousandths of a kilogram.

SpiceWeightFraction of a kgIn lowest termsDecimal (kg)
Cinnamon50 g50/10005/1000.05 kg
Cumin seeds100 g100/10001/100.1 kg
Cardamom25 g25/100025/10000.025 kg
Pepper250 g250/100025/1000.25 kg
Total = 50 + 100 + 25 + 250 = 425 g
= 425/1000 kg = 0.425 kg
Why it happens: writing 50 g as 50/1000 kg already gives a decimal fraction, so it can be read straight off as 0.050 = 0.05. Cardamom cannot be simplified to a denominator of 10 or 100, so its decimal needs all three places: 0.025.
Q3.
Write the following fractions as a sum of fractions and also as decimals: 254/1000 [done: 200/1000 + 50/1000 + 4/1000 = 2/10 + 5/100 + 4/1000 = 0.2 + 0.05 + 0.004 = 0.254], 847/10000, 173/100, 23/1000.
Answer

Split the numerator place by place, then simplify each piece.

FractionExpanding the numeratorSum of tenths, hundredths, …Decimal
254/1000200/1000 + 50/1000 + 4/10002/10 + 5/100 + 4/10000.254
847/10000800/10000 + 40/10000 + 7/100008/100 + 4/1000 + 7/100000.0847
173/100100/100 + 70/100 + 3/1001 + 7/10 + 3/1001.73
23/100020/1000 + 3/10002/100 + 3/10000.023
847/10000 = 0.08 + 0.004 + 0.0007 = 0.0847
173/100 = 1 + 0.7 + 0.03 = 1.73
23/1000 = 0.02 + 0.003 = 0.023
Why it happens: 800/10000 cancels to 8/100 because both are divided by 100. Each digit of the numerator therefore lands in its own place — the leftmost digit of 847 is worth hundredths, not tenths, because 10000 has four zeroes and 847 has only three digits. That missing digit is why a 0 appears right after the point in 0.0847.
Tip: count the zeroes in the denominator, then count the digits of the numerator. If the numerator is short, pad it in front with zeroes: 847 → 0847 → 0.0847.
Q4.
Math Talk: Can you give a simple rule to divide any number by a number of the form 1 followed by zeroes — 10, 100, 1000, etc.? For example, 123/10, 24/100 or 678/1000? Look for a pattern in the previous problems.
Answer

Rule: write the number with a decimal point at its end, then move the point left by as many places as there are zeroes in the divisor. Put in extra zeroes in front if you run out of digits.

Step 1: 123 → 123.
Step 2: 10 has 1 zero
Step 3: move the point 1 place left → 12.3
DivisionZeroesMove the pointAnswer
123 ÷ 1011 place left12.3
24 ÷ 10022 places left0.24
678 ÷ 100033 places left0.678
12 ÷ 100033 places left (pad one 0)0.012
12345 ÷ 100033 places left12.345
Why it happens: dividing by 10 makes every digit worth one-tenth of what it was. A digit in the Tens place becomes a Ones digit, a Ones digit becomes a Tenths digit, and so on. Sliding the point one place left is exactly this drop in value for all the digits at once. Each extra zero in the divisor repeats the drop once more.
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