Class 7 Maths Ganita Prakash Chapter 12 Solutions (NCERT 2026–27) – Another Peek Beyond the Point

These Class 7 Maths Ganita Prakash Chapter 12 solutions cover Another Peek Beyond the Point — Chapter 4 of Ganita Prakash Part II (continuous chapter 12 on ClearStudy). Every Figure it Out question, Math Talk and Try This task is solved step by step, with verified decimal multiplication, decimal division, long division, recurring-decimal patterns and the leap-year calculation, so you can master the chapter and revise it quickly.

Class: 7 Subject: Mathematics Book: Ganita Prakash (Part II) Chapter: 12 (Part II, Chapter 4) Exercises: Figure it Out (p. 73), (p. 83), (p. 86), (p. 93–95) Session: 2026–27
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Chapter 12 Overview

Chapter 12 of Ganita Prakash, Another Peek Beyond the Point (Part II, Chapter 4), continues the story of decimals begun in A Peek Beyond the Point. It shows that multiplying and dividing decimals are natural extensions of the same operations on counting numbers. You learn to multiply decimals (multiply as whole numbers, then place the point so that the number of decimal places in the product equals the sum of decimal places in the factors), to divide decimals by powers of ten and by other decimals using the division-by-place-value (long division) method, to spot when a quotient never ends (recurring decimals like 10 ÷ 3 and the “magic number” 142857 from 1 ÷ 7), and finally to apply decimals to the real-life design of the leap-year calendar. The Class 7 Maths Ganita Prakash Chapter 12 solutions below work through every Figure it Out, Math Talk and Try This task step by step.

Key Concepts & Definitions

Decimal: an extension of the place-value system to show tenths, hundredths, thousandths, … e.g. 27.53 = 2 Tens + 7 Ones + 5 Tenths + 3 Hundredths.

Multiplying decimals: remove the points, multiply as counting numbers, then put the point back so the product has as many decimal places as the two factors put together.

Dividing by 10, 100, 1000: shift the decimal point to the left by as many places as there are zeros in the divisor (e.g. 18.7 ÷ 100 = 0.187).

Multiplying by 10, 100, 1000: shift the decimal point to the right by as many places as there are zeros (e.g. 5.7 × 100 = 570).

Division by place value (long division): share Thousands, Hundreds, Tens and Ones; when Ones are regrouped into Tenths, put a decimal point in the quotient and carry on into Tenths, Hundredths, …

Decimal divisor: multiply both the dividend and the divisor by 10, 100, 1000… until the divisor becomes a whole number, then divide normally — the quotient is unchanged.

Recurring (non-terminating) decimal: a quotient whose digits never end and repeat in a cycle, e.g. 10 ÷ 3 = 3.333… and 1 ÷ 7 = 0.142857142857…

Important Formulas & Patterns (Chapter 12)

Decimal-place rule for products: (places in product) = (places in multiplier) + (places in multiplicand). Example: 5.96 (2 places) × 24.8 (1 place) → 147.808 (3 places).

Same-digits shortcut: if a × b = N, then shifting any point only moves the point in the answer, e.g. 18 × 12 = 216 ⇒ 1.8 × 1.2 = 2.16.

Division by powers of 10: N ÷ 10k moves the point k places left; multiplication by 10k moves it k places right.

Decimal-divisor trick: a ÷ b = (a × 10k) ÷ (b × 10k), e.g. 4.68 ÷ 0.13 = 468 ÷ 13.

Relationship product vs factors: two factors > 1 → product > both; two factors between 0 and 1 → product < both; one of each → product lies in between.

Leap-year rule: a year is a leap year (366 days) if it is divisible by 4, except century years, which are leap only if divisible by 400.

Figure it Out — Decimal Multiplication (Page 73)

Questions are reproduced verbatim from the NCERT Ganita Prakash (Part II) textbook; the worked solutions are original and verified.

1. Recall that a tenth is 0.1, a hundredth is 0.01, and so on. Find the following products in tenths, hundredths and so on: (a) 6 × 4 tenths = 24 tenths (b) 7 × 0.3 (c) 9 × 5 hundredths

SOLUTION (a) 6 × 4 tenths = 24 tenths = 2.4. (b) 7 × 0.3 = 7 × 3 tenths = 21 tenths = 2.1. (c) 9 × 5 hundredths = 45 hundredths = 0.45.

2. Find the products: (a) 27.34 × 6 (b) 4.23 × 3.7 (c) 0.432 × 0.23

SOLUTION (a) 2734 × 6 = 16404; 27.34 has 2 decimal places, so 27.34 × 6 = 164.04. (b) 423 × 37 = 15651; places = 2 + 1 = 3, so 4.23 × 3.7 = 15.651. (c) 432 × 23 = 9936; places = 3 + 2 = 5, so 0.432 × 0.23 = 0.09936.

3. Thejus needs 1.65 m of cloth for a shirt. How many metres of cloth are needed for 3 shirts?

SOLUTION Cloth for 3 shirts = 1.65 × 3. 165 × 3 = 495; 1.65 has 2 places, so 1.65 × 3 = 4.95 m.

4. Meenu bought 4 notebooks and 3 erasers. The cost of each book was ₹15.50 and each eraser was ₹2.75. How much did she spend in all?

SOLUTION Cost of 4 notebooks = 15.50 × 4 = ₹62.00. Cost of 3 erasers = 2.75 × 3 = ₹8.25. Total = 62.00 + 8.25 = ₹70.25.

5. The thickness of a rupee coin is 1.45 mm. What is the total height of the cylinder formed by placing 36 rupee coins one over the other? Write the answer in centimeters.

SOLUTION Height = 1.45 × 36 mm. 145 × 36 = 5220; 1.45 has 2 places, so 1.45 × 36 = 52.20 mm. 52.2 mm = 52.2 ÷ 10 = 5.22 cm.

6. The price of 1 kg of oranges is ₹56.50. What is the price of 2.250 kg of oranges? Can we write 56.50 as 56.5 and 2.250 as 2.25 and multiply? Will we get the same product? Why?

SOLUTION Price = 56.50 × 2.250. Multiplying as whole numbers: 5650 × 2250 = 12712500; total places = 2 + 3 = 5, so the price = ₹127.125 (about ₹127.13). Yes, we may drop the trailing zeros and use 56.5 × 2.25. A trailing zero after the decimal point does not change the value: 56.50 = 56.5 and 2.250 = 2.25. 565 × 225 = 127125; places = 1 + 2 = 3, so 56.5 × 2.25 = 127.125 — the same product, because the numbers are equal.

7. Dwarakanath purchases notebooks at a wholesale price of ₹23.6 per piece and sells each notebook at ₹30/-. How much profit does he make if he sells 50 books in a week?

SOLUTION Profit per notebook = 30 − 23.6 = ₹6.4. Profit on 50 notebooks = 6.4 × 50 = ₹320.

8. Given that 18 × 12 = 216, find the products: (a) 18 × 1.2   (b) 18 × 0.12 (c) 1.8 × 1.2   (d) 0.18 × 0.12 (e) 0.018 × 0.012   (f) 1.8 × 12 In which of the cases above is the product less than 1?

SOLUTION Use 18 × 12 = 216 and just place the point so that the product has as many decimals as the two factors together. (a) 1 place → 21.6;   (b) 2 places → 2.16;   (c) 1 + 1 = 2 places → 2.16. (d) 2 + 2 = 4 places → 0.0216;   (e) 3 + 3 = 6 places → 0.000216;   (f) 1 place → 21.6. Products less than 1: (d) 0.0216 and (e) 0.000216.

9. In which of the following multiplications is the product less than 1? Can you find the answer without actually doing the multiplications? (a) 7 × 0.6   (b) 0.7 × 0.6 (c) 0.7 × 6   (d) 0.07 × 0.06

SOLUTION A product is less than 1 when the factors together are “small” — in particular when both factors are between 0 and 1, or are tiny. (a) 7 × 0.6 = 4.2 (≥ 1);   (b) 0.7 × 0.6 = 0.42 (< 1);   (c) 0.7 × 6 = 4.2 (≥ 1);   (d) 0.07 × 0.06 = 0.0042 (< 1). ∴ the products less than 1 are (b) and (d) — the cases where both factors are smaller than 1.

10. Multiplying the following numbers by 10, 100 and 1000 to complete the table.

SOLUTION To multiply by 10, 100, 1000 move the decimal point 1, 2, 3 places to the right (adding zeros where needed).
Number× 10× 100× 1000
5.7575705700
23.02230.2230223020
0.929.292920
0.3063.0630.6306
24.67246.7246724670

Figure it Out — Division by Place Value (Page 83)

1. Find the quotient by converting the denominator into 1, 10, 100 or 1000 and verify the solution by the long division method (division by place value). (a) 18/5   (b) 415/4 (c) 1217/2   (d) 4827/8

SOLUTION (a) 18/5 = (18 × 2)/(5 × 2) = 36/10 = 3.6. (Long division: 5 into 18 gives 3, remainder 3; 30 tenths ÷ 5 = 6 tenths → 3.6.) (b) 415/4 = (415 × 25)/(4 × 25) = 10375/100 = 103.75. (Long division gives 103 remainder 3 → 30 tenths → 7 tenths, 20 hundredths → 5 → 103.75.) (c) 1217/2 = (1217 × 5)/(2 × 5) = 6085/10 = 608.5. (d) 4827/8 = (4827 × 125)/(8 × 125) = 603375/1000 = 603.375. (8 into 4827 = 603 remainder 3 → 30 tenths → 3, 60 hundredths → 7, 40 thousandths → 5 → 603.375.)

2. Choose the correct answer: (a) 1526/4 =   (i) 38.15   (ii) 380.15   (iii) 381.5   (iv) 381.05 (b) 3567/8 =   (i) 4458.75   (ii) 44.5875   (iii) 445.875   (iv) 4458.75

SOLUTION (a) 1526/4 = 1526 × 25 / 100 = 38150/100 = 381.5 → option (iii) 381.5. (b) 3567/8 = 3567 × 125 / 1000 = 445875/1000 = 445.875 → option (iii) 445.875.

3. What is the quotient? (a) 132 ÷ 4 =   (b) 13.2 ÷ 4 = (c) 1.32 ÷ 4 =   (d) 0.132 ÷ 4 =

SOLUTION Divide 132 by 4 once (= 33), then move the point to match the dividend. (a) 132 ÷ 4 = 33;   (b) 13.2 ÷ 4 = 3.3;   (c) 1.32 ÷ 4 = 0.33;   (d) 0.132 ÷ 4 = 0.033.

4. What is the quotient? (a) 126 ÷ 8 =   (b) 12.6 ÷ 8 = (c) 1.26 ÷ 8 =   (d) 0.126 ÷ 8 = (e) 0.0126 ÷ 8 =

SOLUTION 126 ÷ 8 = 15.75; now shift the point for each part. (a) 15.75;   (b) 1.575;   (c) 0.1575;   (d) 0.01575;   (e) 0.001575.

Figure it Out — Decimal Division (Page 86)

1. Express the following fractions in decimal form: (a) 2/5   (b) 13/4 (c) 4/50   (d) 5/8

SOLUTION (a) 2/5 = (2 × 2)/10 = 4/10 = 0.4. (b) 13/4 = (13 × 25)/100 = 325/100 = 3.25. (c) 4/50 = (4 × 2)/100 = 8/100 = 0.08. (d) 5/8 = (5 × 125)/1000 = 625/1000 = 0.625.

2. Find the quotients: (a) 24.86 ÷ 1.2 (b) 5.728 ÷ 1.52

SOLUTION (a) 24.86 ÷ 1.2 = 248.6 ÷ 12 (multiply both by 10). Long division: 12 into 248.6 gives 20 remainder 8.6, then 86 tenths ÷ 12 = 7 remainder 2, then 20 hundredths ÷ 12 = 1 remainder 8… the 6 keeps repeating, so 24.86 ÷ 1.2 = 20.71666… ≈ 20.717 (a non-terminating, recurring decimal — the digit 6 repeats). (b) 5.728 ÷ 1.52 = 572.8 ÷ 152 (multiply both by 100). This too does not terminate: 5.728 ÷ 1.52 = 3.7684… ≈ 3.768 (a recurring decimal). Either part may be rounded as needed.

3. Evaluate the following using the information 156 × 12 = 1872. (a) 15.6 × 1.2 = __________   (b) 187.2 ÷ 1.2 = __________ (c) 18.72 ÷ 15.6 = __________   (d) 0.156 × 0.12 = __________

SOLUTION (a) 15.6 × 1.2: 156 × 12 = 1872, places = 1 + 1 = 2 → 18.72. (b) 187.2 ÷ 1.2 = 1872 ÷ 12 = 156. (c) 18.72 ÷ 15.6 = 1872 ÷ 1560 = 1.2. (d) 0.156 × 0.12: 156 × 12 = 1872, places = 3 + 2 = 5 → 0.01872.

4. Evaluate the following: (a) 25 ÷ ______ = 0.025   (b) 25 ÷ ______ = 250 (c) 25 ÷ ______ = 2.5   (d) 25 ÷ 10 = 25 × _____ (e) 25 ÷ 0.10 = 25 × ______   (f) 25 ÷ 0.01 = 25 × ______

SOLUTION (a) 25 ÷ 1000 = 0.025;   (b) 25 ÷ 0.1 = 250;   (c) 25 ÷ 10 = 2.5. (d) 25 ÷ 10 = 25 × 0.1 (dividing by 10 is the same as multiplying by one-tenth). (e) 25 ÷ 0.10 = 25 × 10;   (f) 25 ÷ 0.01 = 25 × 100 (dividing by a decimal less than 1 is the same as multiplying by its reciprocal).

5. Find the quotient: (a) 2.46 ÷ 1.5 =   (b) 2.46 ÷ 0.15 = (c) 2.46 ÷ 0.015 = Is the quotient obtained in 24.6 ÷ 1.5 the same as the quotient obtained in 2.46 ÷ 0.15?

SOLUTION (a) 2.46 ÷ 1.5 = 24.6 ÷ 15 = 1.64. (b) 2.46 ÷ 0.15 = 246 ÷ 15 = 16.4. (c) 2.46 ÷ 0.015 = 2460 ÷ 15 = 164. 24.6 ÷ 1.5 = 246 ÷ 15 = 16.4, and 2.46 ÷ 0.15 = 246 ÷ 15 = 16.4. Yes, both equal 16.4 — multiplying dividend and divisor by the same power of 10 keeps the quotient unchanged.

6. A 4 m long wooden block has to be cut into 5 pieces of equal length. What is the length of each piece?

SOLUTION Length of each piece = 4 ÷ 5 = 40 tenths ÷ 5 = 8 tenths = 0.8 m.

7. If the perimeter of a regular polygon with 12 sides is 208.8 cm, what is the length of its side?

SOLUTION All sides of a regular polygon are equal, so side = perimeter ÷ number of sides = 208.8 ÷ 12. 2088 ÷ 12 = 174, so 208.8 ÷ 12 = 17.4 cm.

8. 3 litres of watermelon juice is shared among 8 friends equally. How much watermelon juice will each get? Express the quantity of juice in millilitres.

SOLUTION Each friend gets 3 ÷ 8 = 0.375 litre. 0.375 L = 0.375 × 1000 mL = 375 mL.

9. A car covers 234.45 km using 12.6 litres of petrol. What is the distance travelled per litre?

SOLUTION Distance per litre = 234.45 ÷ 12.6 = 2344.5 ÷ 126 (multiply both by 10). This does not divide evenly: 234.45 ÷ 12.6 = 18.607… ≈ 18.61 km per litre (a non-terminating decimal; rounded to two places).

10. 13.5 kg of flour (aata) was distributed equally among 15 students. How much flour did each student receive?

SOLUTION Each student gets 13.5 ÷ 15 = 135 ÷ 150 = 0.9 kg.

Figure it Out — Mixed Problems (Page 93–95)

1. A 210 gram packet of peanut chikki costs ₹70.5, while a 110 gram packet of potato chips costs ₹33.25. Which is cheaper?

SOLUTION Compare the cost per gram. Chikki: 70.5 ÷ 210 = ₹0.336 per gram (about 33.6 paisa). Chips: 33.25 ÷ 110 = ₹0.302 per gram (about 30.2 paisa). Since 0.302 < 0.336, the potato chips are cheaper per gram.

2. Write the decimal number at the arrow mark: (i) a number line marked from 3.1 to 3.2 (ii) a number line marked from 2.15 to 2.17

SOLUTION (i) Between 3.1 and 3.2 the gap of 0.1 is split into 10 equal parts of 0.01 each. The arrow sits 5 parts from 3.1, so the mark is 3.1 + 5 × 0.01 = 3.15. (ii) Between 2.15 and 2.17 the gap of 0.02 is split into smaller parts of 0.002. The arrow at the midpoint reads 2.15 + 0.01 = 2.16. (Read each arrow by counting the equal sub-divisions between the printed marks; the method is shown for the central mark.)

3. Shyamala bought 3 kg bananas at ₹30/- per kg. She counted 35 bananas in all. She sells each banana for ₹5/-. How much profit does she make selling all the bananas?

SOLUTION Cost price = 3 × 30 = ₹90. Selling price = 35 × 5 = ₹175. Profit = 175 − 90 = ₹85.

4. A teacher placed textbooks that are 2.5 cm thick on a bookshelf. The teacher wanted to place 80 textbooks on the shelf. The bookshelf is 160 cm long. How many books could be placed on the shelf? Was there any space left? If yes, how much?

SOLUTION Number of books that fit = 160 ÷ 2.5 = 1600 ÷ 25 = 64 books. 64 books take 64 × 2.5 = 160 cm, which is exactly the shelf length, so no space is left (0 cm). The teacher wanted 80 books but only 64 can be placed; 80 − 64 = 16 books will not fit.

5. Fill in the following blanks appropriately (1 cm = 10 mm, 1 m = 100 cm, 1 km = 1000 m, 1 kg = 1000 g, 1 g = 1000 mg, 1 l = 1000 ml): 5.5 km = _________ m   35 cm = ________ m   14.5 cm = _______ mm 68 g = ________ kg   9.02 m = ________ mm   125.5 ml = _______ l

SOLUTION 5.5 km = 5.5 × 1000 = 5500 m. 35 cm = 35 ÷ 100 = 0.35 m. 14.5 cm = 14.5 × 10 = 145 mm. 68 g = 68 ÷ 1000 = 0.068 kg. 9.02 m = 9.02 × 100 × 10 = 9.02 × 1000 = 9020 mm. 125.5 ml = 125.5 ÷ 1000 = 0.1255 l.

6. The following problem was set by Sridharacharya in his book, Patiganita. “6¼ is divided by 2½, and 60¼ is divided by 3½. Tell the quotients separately.” Can you try to solve it by converting the fractions into decimals?

SOLUTION Convert the mixed numbers to decimals: 6¼ = 6.25, 2½ = 2.5, 60¼ = 60.25, 3½ = 3.5. First quotient: 6.25 ÷ 2.5 = 62.5 ÷ 25 = 2.5. Second quotient: 60.25 ÷ 3.5 = 602.5 ÷ 35 = 17.214… ≈ 17.21 (a non-terminating decimal; equal to the fraction 241/14). The first quotient is exact, the second is recurring.

7. Fill the boxes in at least 2 different ways: (a) □ × □ = 2.4   (b) □ × □ = 14.5

SOLUTION Any pair of decimals whose product is the target works. (a) Two ways: 0.6 × 4 = 2.4 and 1.2 × 2 = 2.4 (also 0.8 × 3, 2.4 × 1, 0.4 × 6). (b) Two ways: 2.9 × 5 = 14.5 and 1.45 × 10 = 14.5 (also 29 × 0.5, 5.8 × 2.5).

8. Find the following quotients given that 756 ÷ 36 = 21: (a) 75.6 ÷ 3.6   (b) 7.56 ÷ 0.36 (c) 756 ÷ 0.36   (d) 75.6 ÷ 360 (e) 7560 ÷ 3.6   (f) 7.56 ÷ 0.36

SOLUTION In each part rewrite the division so it becomes 756 ÷ 36 = 21 with the point shifted. (a) 75.6 ÷ 3.6 = 756 ÷ 36 = 21. (b) 7.56 ÷ 0.36 = 756 ÷ 36 = 21. (c) 756 ÷ 0.36 = 75600 ÷ 36 = 2100. (d) 75.6 ÷ 360 = 7.56 ÷ 36 = 0.21. (e) 7560 ÷ 3.6 = 75600 ÷ 36 = 2100. (f) 7.56 ÷ 0.36 = 756 ÷ 36 = 21 (same as part b).

9. Find the missing cells if each cell represents a÷b (column heads are values of a: 1517, 151.7, 15.17, 1.517, 15170; row heads are values of b: 37, 3.7, 0.37, 0.037, 370).

SOLUTION Every entry is a ÷ b. Since 1517 ÷ 37 = 41, all other cells are found by shifting the decimal point of 41 according to the points in a and b. The completed table is shown below.
b ↓ \ a →1517151.715.171.51715170
37414.10.410.041410
3.7410414.10.414100
0.374100410414.141000
0.03741000410041041410000
3704.10.410.0410.004141

10. Using the digits 2, 4, 5, 8, and 0 fill the boxes □.□ × □.□ to get the: (a) maximum product   (b) minimum product (c) product greater than 150   (d) product nearest to 100 (e) product nearest to 5

SOLUTION Use each of the five digits once, placing a decimal point in each factor. (More than one answer is possible; one valid answer is given for each part.) (a) Maximum: put the big digits in the whole-number places → 82.0 × 5.4 = 442.8. (b) Minimum (greater than 0): make both factors small → 0.2 × 4.58 = 0.916 (using a 3-digit and a 2-digit arrangement; e.g. 0.24 × 5.80 = 1.392 is larger, so 0.916 is smaller still). (c) Greater than 150: 24.5 × 8.0 = 196 (also 20.4 × 8.5 = 173.4). (d) Nearest to 100: 20.5 × 4.8 = 98.4 (very close to 100). (e) Nearest to 5: 5.4 × 0.82 = 4.428 (close to 5).

11. Sort the following expressions in increasing order: (a) 245.05 × 0.942368   (b) 245.05 × 7.9682 (c) 245.05 ÷ 7.9682   (d) 245.05 ÷ 0.942368 (e) 245.05   (f) 7.9682

SOLUTION Reason without exact multiplication. Multiplying 245.05 by a number < 1 (a) makes it smaller; by a number > 1 (b) makes it much larger. Dividing by a number > 1 (c) makes it smaller; dividing by a number < 1 (d) makes it larger. (f) is just 7.9682. Approximate values: (f) ≈ 7.97, (c) ≈ 30.8, (a) ≈ 230.9, (e) = 245.05, (d) ≈ 260.0, (b) ≈ 1952.6. Increasing order: (f) < (c) < (a) < (e) < (d) < (b).

Math Talk & Try This — Answered

These are the in-text reflective tasks in the chapter; the determinate ones are answered, the open-ended ones are guided.

Math Talk — Dividing by 1 followed by zeros 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? Answer. Write the number with a decimal point and move the point to the left by as many places as there are zeros in the divisor. So 123 ÷ 10 = 12.3, 24 ÷ 100 = 0.24, 678 ÷ 1000 = 0.678 (and 12345 ÷ 1000 = 12.345).
Math Talk — Can the product be a natural number? Can the product of two decimals be a natural number? Can the product of a decimal and a natural number be a natural number? Answer. Yes to both. For example 2.5 × 0.4 = 1 (two decimals giving a whole number), and 2.5 × 8 = 20 (a decimal times a natural number giving a whole number). It happens when the decimal parts cancel out to leave no fractional part.
Math Talk — Using 596 × 248 = 147808 Suppose we know that 596 × 248 = 147808, can you immediately write down the product of 5.96 × 24.8? Answer. Yes. 5.96 has 2 decimal places and 24.8 has 1, so the product has 2 + 1 = 3 decimal places: 5.96 × 24.8 = 147.808.
Math Talk — When is the product bigger or smaller? When is the product of two decimals greater than both the numbers? When is it less than both the numbers? Answer. The product is greater than both numbers when both factors are greater than 1 (e.g. 3.4 × 6.5 = 22.1). It is less than both when both factors lie between 0 and 1 (e.g. 0.75 × 0.4 = 0.3). If one factor is between 0 and 1 and the other is greater than 1, the product lies in between.
Math Talk — Q6 (price & trailing zeros) Can we write 56.50 as 56.5 and 2.250 as 2.25 and multiply? Will we get the same product? Why? Answer. Yes. A zero written after the last non-zero digit beyond the point does not change a number’s value, so 56.50 = 56.5 and 2.250 = 2.25. Both ways give the same product, ₹127.125. (Full working in Figure it Out p. 73, Q6.)
Math Talk — Quotient vs dividend with a decimal divisor Will the quotient be always greater than the dividend when the divisor is a decimal? Describe the relationship between the dividend, divisor and quotient. Answer. Only when the divisor is less than 1. If the divisor is between 0 and 1 (e.g. 128 ÷ 0.4 = 320), the quotient is greater than the dividend; if the divisor is greater than 1, the quotient is smaller; if the divisor is exactly 1, the quotient equals the dividend.
Math Talk — Q7 fill the boxes Fill the boxes in at least 2 different ways: (a) □ × □ = 2.4; (b) □ × □ = 14.5. Answer. (a) 0.6 × 4 = 2.4 and 1.2 × 2 = 2.4. (b) 2.9 × 5 = 14.5 and 1.45 × 10 = 14.5. (Full list in Figure it Out p. 94, Q7.)
Math Talk — Powers of 2 and 5 pattern What pattern do you observe in 1/2 = 0.5, 1/(2×2) = 0.25, … and 1/5 = 0.2, 1/(5×5) = 0.04, …? Why are 2 and 5 related in this way? What are 1/25 and 1/55? Answer. Each extra factor of 2 (or of 5) adds one more decimal place. The two columns are linked because 2 × 5 = 10, the base of our place-value system, so 1/2n and 1/5n both terminate after n places. Continuing: 1/25 = 1/32 = 0.03125 and 1/55 = 1/3125 = 0.00032.
Math Talk — Days in 100 calendar years With the scheme of adding one extra day every 4th year, what is the number of days in 100 calendar years? Can you write different expressions for the same question? Answer. There are 25 years divisible by 4 in 100 years, so days = 100 × 365 + 25 = 36,525 days. Different expressions: (100 × 365) + (100 ÷ 4) × 1, or (75 × 365) + (25 × 366), both give 36,525. (The Earth actually needs 100 × 365.2422 = 36,524.22 days, so this scheme over-counts.)
Math Talk — Days in 100 years with the “no leap in century” rule Can you write an expression for the number of days in 100 calendar years with the adjustment that century years are not leap years? Answer. Now only 24 of the 100 years are leap years (the year 100 is excluded). Days = (24 × 366) + (76 × 365) = 8784 + 27,740 = 36,524 days — closer to the actual 36,524.22 days.
Try This — A second cyclic number from 1 ÷ 17 To find another “cyclic” number like 142857, find 1 ÷ 17 in decimal and use the repeating block of digits. Answer. 1 ÷ 17 = 0.0588235294117647…, which repeats with the 16-digit block 0588235294117647. Like 142857, multiplying this block by 1, 2, 3, … up to 16 gives cyclic rearrangements of the same digits — so it is another cyclic number.
Try This — The final leap-year scheme over 10,000 years With the final scheme of leap years (every 4th year, except centuries, except multiples of 400), calculate the number of calendar days in 10,000 years and the actual number of days the Earth takes for 10,000 revolutions. What is the difference? Can you suggest a fix? Answer. Leap years in 10,000 years = (divisible by 4) − (divisible by 100) + (divisible by 400) = 2500 − 100 + 25 = 2425. Calendar days = 10,000 × 365 + 2425 = 3,652,425 days. Actual = 10,000 × 365.2422 = 3,652,422 days. The calendar over-counts by about 3 days in 10,000 years. A possible fix: drop one extra leap day roughly every 3300 years (for example, make years divisible by 4000 ordinary years).
Try This — Traditional Indian calendars Investigate how traditional calendars in India managed to consistently align the days in the calendar with astronomical events like the Earth going around the Sun or the positions of the stars. Answer (project guidance). Traditional Indian calendars are luni-solar: months follow the Moon while years are kept in step with the Sun by inserting an extra month, the adhik maas (a leap month), about every 2–3 years. Astronomers such as Aryabhata and Bhaskara measured the solar year very precisely and used the positions of the nakshatras (star groups) to fix dates, so festivals stayed tied to the seasons. Explore the Surya Siddhanta and the panchang to see these adjustments at work.

Common Mistakes to Avoid

Watch out for these

  • Forgetting to count all the decimal places — the product’s places equal the sum of the places in both factors (e.g. 0.432 × 0.23 has 5 places).
  • Moving the point the wrong way: ÷ 10 moves it left, × 10 moves it right.
  • When dividing by a decimal, multiplying only the divisor by 10/100… — you must multiply the dividend by the same number too.
  • Placing the decimal point in the quotient too early or too late — in long division put it in the quotient exactly when you regroup Ones into Tenths.
  • Assuming every division ends — some quotients (10 ÷ 3, 1 ÷ 7, 234.45 ÷ 12.6) are recurring and must be rounded or written with the repeating block.
  • Thinking the quotient is always smaller than the dividend — when the divisor is less than 1, the quotient is larger (128 ÷ 0.4 = 320).

Practice MCQs & Assertion–Reason

1. The product 4.23 × 3.7 is:

(a) 1.5651    (b) 15.651    (c) 156.51    (d) 1565.1

2. 0.306 × 1000 equals:

(a) 3.06    (b) 30.6    (c) 306    (d) 3060

3. 18.7 ÷ 100 equals:

(a) 1.87    (b) 0.187    (c) 0.0187    (d) 1870

4. Given 18 × 12 = 216, the value of 0.18 × 0.12 is:

(a) 2.16    (b) 0.216    (c) 0.0216    (d) 0.00216

5. The quotient 1526 ÷ 4 is:

(a) 38.15    (b) 380.15    (c) 381.5    (d) 381.05

6. Which fraction, written as a decimal, is 0.625?

(a) 2/5    (b) 13/4    (c) 5/8    (d) 4/50

7. 4.68 ÷ 0.13 is the same as:

(a) 468 ÷ 13    (b) 46.8 ÷ 13    (c) 4.68 ÷ 13    (d) 4680 ÷ 13

8. Which division gives a non-terminating (recurring) decimal?

(a) 18 ÷ 5    (b) 10 ÷ 3    (c) 13 ÷ 4    (d) 5 ÷ 8

9. If a decimal divisor is less than 1, the quotient compared with the dividend is:

(a) always smaller    (b) always larger    (c) equal    (d) unrelated

10. A year is a leap year if it is divisible by 4, except a century year, which is a leap year only if it is divisible by:

(a) 100    (b) 200    (c) 400    (d) 1000

Answer key: 1-(b), 2-(c), 3-(b), 4-(c), 5-(c), 6-(c), 7-(a), 8-(b), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: 0.432 × 0.23 = 0.09936.

Reason: The number of decimal places in a product equals the sum of the decimal places in the two factors.

A-R 2. Assertion: 24.86 ÷ 1.2 is a terminating decimal.

Reason: Dividing a decimal by another decimal always gives a terminating decimal.

A-R 3. Assertion: 128 ÷ 0.4 = 320, which is greater than 128.

Reason: When the divisor is less than 1, the quotient is greater than the dividend.

A-R 4. Assertion: 2.46 ÷ 0.15 and 24.6 ÷ 1.5 give the same quotient.

Reason: Multiplying the dividend and the divisor by the same number does not change the quotient.

A-R 5. Assertion: The year 2000 was a leap year but 1900 was not.

Reason: A century year is a leap year only when it is divisible by 400.

Answer key: 1-(A), 2-(D), 3-(A), 4-(A), 5-(A).

Quick Revision Summary

  • To multiply decimals: multiply as whole numbers, then give the product as many decimal places as the two factors together.
  • To multiply by 10, 100, 1000 move the point right; to divide by them move the point left, one place per zero.
  • Division by place value (long division) extends past the Ones into Tenths, Hundredths, …; place the point in the quotient when you regroup Ones into Tenths.
  • For a decimal divisor, multiply dividend and divisor by the same power of 10 to make the divisor a whole number; the quotient is unchanged.
  • Some quotients never end and repeat in a cycle (10 ÷ 3 = 3.333…, 1 ÷ 7 = 0.142857…); these are recurring decimals.
  • A product is greater than both factors only when both are > 1; a quotient is greater than the dividend only when the divisor is < 1.
  • The leap-year rule (every 4th year, not centuries, but yes every 400th year) keeps the calendar close to the Earth’s 365.2422-day year.

How to score full marks in this chapter

Whenever you multiply or divide decimals, first do the sum with whole numbers and only then fix the decimal point — count the places carefully. For a decimal divisor, write the “clean” division (e.g. 4.68 ÷ 0.13 = 468 ÷ 13) before dividing. Show the place-value steps in long division and put the point in the quotient as soon as you reach the Tenths. If a quotient does not end, round it sensibly (say, to two decimal places) and say so. Check answers by reversing the operation or by estimating.

Frequently Asked Questions

What is Class 7 Maths Ganita Prakash Chapter 12 about?

Chapter 12, Another Peek Beyond the Point (Ganita Prakash Part II, Chapter 4), teaches how to multiply and divide decimals as natural extensions of whole-number operations, how to divide using place value (long division), how to recognise recurring decimals such as 10 ÷ 3 and 1 ÷ 7, and how decimals are used to design the leap-year calendar.

How do you place the decimal point when multiplying two decimals?

Ignore the points and multiply the numbers as whole numbers. Then put the point back so that the product has as many decimal places as the two factors have together. For example 5.96 (2 places) × 24.8 (1 place) = 147.808 (3 places).

Why do some decimal divisions never end?

When the denominator (after simplifying) has a prime factor other than 2 or 5, the long division leaves the same remainder again and again, so the digits repeat forever — for example 10 ÷ 3 = 3.333… and 1 ÷ 7 = 0.142857142857… These are called recurring or non-terminating decimals.

Are these Class 7 Maths Ganita Prakash Chapter 12 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash (Part II) textbook for the 2026–27 session, with every Figure it Out, Math Talk and Try This task solved and verified.

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