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NCERT Solutions Ganita Prakash (Part 1) Chapter 2 .4 Powers of 10 — In-text Questions

Book page 312 Updated on2026-09-05

Q1.
Write these numbers in the same way: (i) 172, (ii) 5642, (iii) 6374.
Answer

Each digit is multiplied by the power of 10 belonging to its place. The units place is 100 = 1.

(i) 172 = (1 × 100) + (7 × 10) + 2
= (1 × 102) + (7 × 101) + (2 × 100)
(ii) 5642 = (5 × 1000) + (6 × 100) + (4 × 10) + 2
= (5 × 103) + (6 × 102) + (4 × 101) + (2 × 100)
(iii) 6374 = (6 × 1000) + (3 × 100) + (7 × 10) + 4
= (6 × 103) + (3 × 102) + (7 × 101) + (4 × 100)
Why it happens: our numerals are a place-value system built on 10, so the expanded form is really a sum of powers of 10. Reading right to left the exponents run 0, 1, 2, 3, … — which is exactly why the units place needs 100 = 1. Without the convention n0 = 1 the pattern would break at the last digit.
Q2.
How can we write 561.903?
Answer

Continue the same pattern to the right of the decimal point, where the exponents go on decreasing: 2, 1, 0, then –1, –2, –3.

561.903 = (5 × 100) + (6 × 10) + 1 + (9 × 1/10) + (0 × 1/100) + (3 × 1/1000)
= (5 × 102) + (6 × 101) + (1 × 100) + (9 × 10–1) + (0 × 10–2) + (3 × 10–3)
Digit561.903
Place value10210110010–110–210–3
Why it happens: the decimal point is not a break in the system — it only marks where 100 sits. Each step to the right divides the place value by 10, which is precisely the step 100 → 10–1 → 10–2 on the power line. Negative exponents are what make the whole of the decimal system one single idea instead of two.
Q3.
Write the large-number facts we read just before in this form. [(i) The Sun is 30,00,00,00,00,00,00,00,00,000 m from the centre of our Milky Way galaxy. (ii) The number of stars in our galaxy is 1,00,00,00,00,000. (iii) The mass of the Earth is 59,76,00,00,00,00,00,00,00,00,00,000 kg.]
Answer

Count the digits after the leading figures; that count is the exponent.

(i) 30,00,00,00,00,00,00,00,00,000 m
= 3 followed by 20 zeros
= 3 × 1020 m
(ii) 1,00,00,00,00,000 stars
= 1 followed by 11 zeros
= 1 × 1011 (one hundred arab / one hundred billion stars)
(iii) 59,76,00,00,00,00,00,00,00,00,00,000 kg
= 5976 followed by 21 zeros
= 5976 × 1021 = 5.976 × 1024 kg
Why it happens: (iii) shows why standard form insists that the coefficient lie between 1 and 10. Written as 5976 × 1021, 59.76 × 1023 or 5.976 × 1024 the number is the same, but only the last form lets you compare it with another number at a glance, because then the exponent alone tells you the size.
Tip: to convert, move the decimal point until exactly one non-zero digit is left in front of it, and count the moves. Moving left raises the exponent; moving right lowers it.
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