NCERT Solutions Class 7 Maths Chapter 12 Another Peek Beyond the Point

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 7 · M AT H S

NCERT Solutions

Chapter 12: Another Peek
Beyond the Point

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

Part II, 68 – 95 21 88 English

Solutions, notes, sample papers & more at 73 pages

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

CLASS 7 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 12: Another Peek Beyond
the Point
Chapter 4 of Ganita Prakash Grade 7 Part II takes decimals one step further — multiplying and dividing
them. Every rule here comes from one idea: a decimal is only a fraction whose denominator is 1 followed by
zeroes, so the old fraction rules do all the work and the decimal point simply lands where the zeroes say it
must.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 7) Part II, 68 – 95

SECTIONS QUESTIONS

21 88

MEDIUM

English

In-text Questions — Pages 67 – 68
Section 4.1 A Quick Recap of Decimals

MATH TALK

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)

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

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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

FRACTION WHAT IT MEANS DECIMAL

3/10 3 tenths 0.3

4/100 0 tenths and 4 hundredths 0.04

67/1000 0 tenths, 6 hundredths, 7 thousandths 0.067

457/100 4 ones, 5 tenths, 7 hundredths 4.57

71/100 7 tenths and 1 hundredth 0.71

43/100 4 tenths and 3 hundredths 0.43

9/100 0 tenths and 9 hundredths 0.09

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.

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

SPICE WEIGHT FRACTION OF A KG IN LOWEST TERMS DECIMAL (KG)

Cinnamon 50 g 50/1000 5/100 0.05 kg

Cumin seeds 100 g 100/1000 1/10 0.1 kg

Cardamom 25 g 25/1000 25/1000 0.025 kg

Pepper 250 g 250/1000 25/100 0.25 kg

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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

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.

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

FRACTION EXPANDING THE SUM OF TENTHS, DECIMAL
NUMERATOR HUNDREDTHS, …

254/1000 200/1000 + 50/1000 + 4/1000 2/10 + 5/100 + 4/1000 0.254

847/10000 800/10000 + 40/10000 + 7/10000 8/100 + 4/1000 + 7/10000 0.0847

173/100 100/100 + 70/100 + 3/100 1 + 7/10 + 3/100 1.73

23/1000 20/1000 + 3/1000 2/100 + 3/1000 0.023

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.

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Class 7 Maths Chapter 12 Another Peek Beyond the Point
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Tip: count the zeroes in the denominator, then count the digits of the numerator. If

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the numerator is short, pad it in front with zeroes: 847 → 0847 → 0.0847.

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Q4 Math Talk: Can you give a simple rule to divide any number by a number of the form

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1 followed by zeroes — 10, 100, 1000, etc.? For example, 123/10, 24/100 or 678/1000?

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Look for a pattern in the previous problems.

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Rule: write the number with a decimal point at its end, then move the point left by as many
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places as there are zeroes in the divisor. Put in extra zeroes in front if you run out of digits.
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Step 1: 123 → 123.
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Step 3: move the point 1 place left → 12.3

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123 ÷ 10 1 1 place left 12.3

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24 ÷ 100 2 2 places left
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678 ÷ 1000 3 3 places left 0.678

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12 ÷ 1000 3 3 places left (pad one 0) 0.012

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12345 ÷ 1000 3 places left 12.345

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Why it happens: dividing by 10 makes every digit worth one-tenth of what it was. A

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digit in the Tens place becomes a Ones digit, a Ones digit becomes a Tenths digit,
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and so on. Sliding the point one place left is exactly this drop in value for all the

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digits at once. Each extra zero in the divisor repeats the drop once more.
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In-text Questions — Page 69
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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Section 4.2 Decimal Multiplication

Q1 Example 1: Arshad goes to a stationery shop and purchases 5 pens. If one pen costs
₹9.5 (9 rupees and 50 paisa), how much should he pay the shopkeeper?

Arshad should pay ₹47.5.
Five pens means the price 9.5 taken 5 times.

9.5 × 5 = 9.5 + 9.5 + 9.5 + 9.5 + 9.5 = 47.5

The same answer comes out of the fraction method.

9.5 = 95/10 and 5 = 5/1
Cost of 5 pens = 5/1 × 95/10

= (5 × 95) / (1 × 10) (multiply numerators, multiply denominators)

= 475/10

= ₹47.5

Why it happens: ₹9.5 is 9 rupees and 50 paisa. Five pens cost 45 rupees plus 5 × 50
= 250 paisa = ₹2.50. Together ₹47.50 — the very same number.

Q2 What operation must we use here?

Multiplication.
Each pen costs the same amount, and there are 5 of them. Adding the same number again and
again is multiplication.

9.5 + 9.5 + 9.5 + 9.5 + 9.5 = 9.5 × 5

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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Why it happens: repeated addition of equal amounts is exactly what multiplication
was invented for. This does not change just because the amount is a decimal — 9.5 is
a number like any other.

Q3 Example 2: A car travels 12.5 km per litre of petrol. What is the distance covered
with 7.5 litres of petrol?

The car covers 93.75 km.

Distance = 12.5 × 7.5

= 125/10 × 75/10

= (125 × 75) / (10 × 10)

= 9375/100

= 93.75 km

Why it happens: each denominator is 10, so their product is 100 — two zeroes,
hence two digits after the point in the answer. Notice that 12.5 and 7.5 have one
decimal place each, and 1 + 1 = 2. That is the rule the chapter is heading towards.

Check it yourself: 7.5 litres is a little less than 8 litres, and 8 × 12.5 = 100 km. So the
answer must be a little under 100 km. 93.75 fits.

In-text Questions — Pages 70 – 71
Section 4.2 Decimal Multiplication

MATH TALK

Q1 Math Talk: Can the product of two decimals be a natural number?

Yes, it can.

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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

0.5 × 4.0 = 5/10 × 40/10 = 200/100 = 2

2.5 × 0.4 = 25/10 × 4/10 = 100/100 = 1

1.25 × 0.8 = 125/100 × 8/10 = 1000/1000 = 1

Why it happens: the product of the two fractions has a denominator made only of
tens. If the product of the numerators is a multiple of that denominator, the fraction
reduces to a whole number. In 2.5 × 0.4 the numerators give 25 × 4 = 100, and the
denominator is also 100, so the answer is exactly 1.

Tip: pair a decimal ending in 5 or 25 with one ending in 2, 4 or 8. Those pairs often
finish on a whole number, because 5 × 2 = 10.

Q2 Math Talk: Can the product of a decimal and a natural number be a natural
number?

Yes.

0.25 × 8 = 25/100 × 8 = 200/100 = 2

0.2 × 5 = 2/10 × 5 = 10/10 = 1

1.5 × 6 = 15/10 × 6 = 90/10 = 9

Why it happens: multiplying only changes the numerator. So the answer is a whole
number whenever the natural number cancels the denominator. 0.25 has
denominator 100, and 8 × 25 = 200, which 100 divides exactly.

Did you know? This is why shopkeepers like quarters. Four items at ₹0.25 come to
exactly ₹1 — no paisa left over.

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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Q3 Example 3: The distance between Ajay's school and his home is 827 m. He walks to
school in the morning and then walks back home in the evening, 6 days a week.
How much does he walk in a week? Answer in kilometres.

Ajay walks 9.924 km in a week.
First put the distance in kilometres. 827 m = 827/1000 km = 0.827 km.

In one day (to school and back):

0.827 × 2 = 827/1000 × 2 = 1654/1000 = 1.654 km

In a week (6 days):

1.654 × 6 = 1654/1000 × 6 = 9924/1000 = 9.924 km

Why it happens: multiplying by 2 and then by 6 only touches the numerator,
because 2 and 6 are whole numbers. The denominator stays 1000 throughout, so
the answer keeps exactly three digits after the point.

Check it yourself: 827 × 12 = 9924 metres in a week, and 9924 m = 9.924 km. Same
answer, reached the other way round.

Q4 Example 4: Find the area of the given rectangle. (Length 13.3 cm, breadth 5.7 cm)

The area is 75.81 sq cm.

Area = length × breadth = 5.7 × 13.3

= 57/10 × 133/10

= (57 × 133) / (10 × 10)

= 7581/100

= 75.81 sq cm

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Class 7 Maths Chapter 12 Another Peek Beyond the Point
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13.3 cm
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5.7 Area

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The rectangle from page 70. Multiply
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Why it happens: 57 × 133 = 7581 is an ordinary multiplication. Dividing by 10 twice
— once for each factor — is the same as dividing by 100, which slides the point two
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places left: 7581 → 75.81.
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Q5 Observe the number of digits after the decimal point in the multiplier, the
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multiplicand and the product. Also note the number of zeroes in the denominator.
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(Table: 9.5 × 5, 12.5 × 7.5, 1.64 × 6, 5.7 × 13.35)

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In every row, the decimal places in the product are the sum of the decimal places in the two

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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

EXAMPLE AS MULTIPLIER MULTIPLICAND PRODUCT ZEROES IN
FRACTIONS DENOMINATOR

9.5 × 5 = 95/10 × 5/1 = 1 0 1 1
47.5 475/10

12.5 × 7.5 = 125/10 × 75/10 1 1 2 2
93.75 = 9375/100

1.64 × 6 = 164/100 × 6/1 2 0 2 2
9.84 = 984/100

5.7 × 13.35 57/10 × 1 2 3 3
= 76.095 1335/100 =
76095/1000

Why it happens: a number with 1 decimal place is a fraction over 10, one with 2
decimal places is over 100. Multiplying the fractions multiplies the denominators, so
10 × 100 = 1000 — the zeroes simply add up. And the number of zeroes in the
denominator is the number of digits after the point. That is the whole reason the
decimal places add.

Tip: a whole number like 5 or 6 counts as 0 decimal places, because 5 = 5/1 and 1
has no zeroes at all.

In-text Questions — Page 71
Section 4.2 Decimal Multiplication — framing the rule

MATH TALK

Q1 Math Talk: Suppose we know that 596 × 248 = 147808, can you immediately write
down the product of 5.96 × 24.8?

Yes — the answer is 147.808.

Page 10 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

596 × 248 = 147808

5.96 has 2 decimal places

24.8 has 1 decimal place
2 + 1 = 3 decimal places in the product

So 5.96 × 24.8 = 147.808

596 × 248 = 147808

5.96 × 24.8 = 147.808
2 places 1 place 3 places

2+1=3

The digits never change — only where the point sits.

Why it happens: 5.96 = 596/100 and 24.8 = 248/10. Multiplying gives (596 × 248) /
(100 × 10) = 147808/1000. The numerator is the counting-number product you
already know, and the denominator has 2 + 1 = 3 zeroes, so the point moves 3 places
left.

Q2 By looking at the above examples, can you frame a rule to multiply two decimals?

Rule: remove the decimal points, multiply the two numbers as counting numbers, then put a
point in the product so that it has as many decimal places as the two numbers had together.

Page 11 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Step 1: drop the points and multiply.

Step 2: count the decimal places in the multiplier and in the multiplicand.

Step 3: add those two counts.
Step 4: place the point in the product so it has that many digits after it (pad with zeroes in

front if needed).

Worked out for 4.23 × 3.7:

423 × 37 = 15651

Decimal places: 2 + 1 = 3

4.23 × 3.7 = 15.651

Why it happens (the book's own argument):

Multiplying decimals is the same as multiplying their fractions.
The product of the numerators is the product of the numbers with the points
removed.
Both denominators are 1 followed by zeroes, so their product is also 1 followed by
zeroes — and the zeroes add up.
The number of zeroes in the denominator is the number of digits after the point.
So the decimal places add up too.

Tip: if the product has fewer digits than the places you need, write zeroes in front.
0.018 × 0.012 needs 6 places, but 18 × 12 = 216 has only 3 digits, so the answer is
0.000216.

In-text Questions — Page 72
Section 4.2 — Is the Product Always Greater than the Numbers Multiplied?

Q1 Example 5: Let us use the above rule to find the product of 5.8 and 1.24. […] Verify
this by converting the multiplier and multiplicand into fractions.

The product is 7.192.

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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

By the rule:

58 × 124 = 7192

Decimal places: 1 + 2 = 3

5.8 × 1.24 = 7.192

Verification with fractions:

5.8 = 58/10, 1.24 = 124/100

58/10 × 124/100 = (58 × 124) / (10 × 100)

= 7192/1000

= 7.192 ✓

Why it happens: the two ways are the same calculation written differently. The rule
is only a short way of saying "the denominators 10 and 100 multiply to 1000".

Check it yourself: 1.24 is just over 1, so the answer must be just over 5.8. And 7.192
is a little more than 5.8 — sensible, because 5.8 × 1.24 is 5.8 plus about a quarter of
5.8.

Q2 When is the product of two decimals greater than both the numbers? When is it less
than both the numbers?

It depends only on whether each number is bigger or smaller than 1.

SITUATION MULTIPLICATION RELATIONSHIP

Situation 1 Both numbers are greater than 1 (3.4 × 6.5) The product (22.1) is greater than both the
numbers

Situation 2 Both numbers are between 0 and 1 (0.75 × The product (0.3) is less than both the
0.4) numbers

Situation 3 One is between 0 and 1, one is greater than The product (3.75) is less than 5 but
1 (0.75 × 5) greater than 0.75

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Class 7 Maths Chapter 12 Another Peek Beyond the Point
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The book's three examples show the same thing:
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m l as
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2.25 × 8 = 18 → 18 is bigger than both 2.25 and 8
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0.25 × 8 = 2 → 2 is bigger than 0.25 but smaller than 8
a0.25 × 0.8 = 0.2 → 0.2 is smaller than both

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Why it happens: multiplying by a number greater than 1 means taking more than
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one whole copy, so the number grows. Multiplying by a number between 0 and 1

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means taking only a part of it, so the number shrinks. In 0.25 × 8 you are taking a
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quarter of 8 — smaller than 8, but eight quarters is far more than one quarter. That

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is exactly Situation 3.
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a gl Tip: multiplying by 1 changes nothing. So 1 is the dividing line: above it products
grow, below it they shrink.
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Figure it Out — Pages 73 a– 74 g
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Section 4.2 Decimal Multiplication

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MATH TALK

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a s em Recall that a tenth is 0.1, a hundredth is 0.01, and so on. Find the following products
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Q1
a in tenths, hundredths and so on: (a) 6 × 4 tenths = 24 tenths (b) 7 × 0.3 (c) 9 × 5
hundredths
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Count the pieces first, then name the piece.

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(a) 6 × 4 tenths = 24 tenths = 24/10 = 2.4
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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Why it happens: "tenth" and "hundredth" behave like units. Six lots of 4 tenths is 24
tenths, just as six lots of 4 pencils is 24 pencils. Then 24 tenths is more than 2 whole
ones, because 10 tenths make 1, so 24 tenths = 2 wholes and 4 tenths = 2.4.

Tip: the unit tells you the denominator. Tenths → over 10, hundredths → over 100.

Q2 Find the products: (a) 27.34 × 6 (b) 4.23 × 3.7 (c) 0.432 × 0.23

Drop the points, multiply, then count the places.

(a) 2734 × 6 = 16404

Places: 2 + 0 = 2 → 164.04

(b) 423 × 37 = 15651

Places: 2 + 1 = 3 → 15.651

(c) 432 × 23 = 9936

Places: 3 + 2 = 5, but 9936 has only 4 digits, so pad one zero in front → 0.09936

Why it happens: in (c) the fractions are 432/1000 and 23/100, so the denominator is
100000 — five zeroes. The numerator 9936 must therefore be written as 09936
before the point goes in front of it.

Check it yourself: (c) multiplies two numbers that are both less than 1, so the
answer must be smaller than both. 0.09936 is indeed smaller than 0.432 and smaller
than 0.23.

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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Q3 Thejus needs 1.65 m of cloth for a shirt. How many metres of cloth are needed for 3
shirts?

4.95 m of cloth is needed.

Cloth for 3 shirts = 1.65 × 3

165 × 3 = 495

Places: 2 + 0 = 2

= 4.95 m

Why it happens: 1.65 m is 1 m 65 cm. Three shirts need 3 m and 195 cm; 195 cm is 1
m 95 cm, so the total is 4 m 95 cm = 4.95 m.

Q4 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?

Meenu spent ₹70.25.

Notebooks: 15.50 × 4

1550 × 4 = 6200, places 2 + 0 = 2 → ₹62.00

Erasers: 2.75 × 3

275 × 3 = 825, places 2 + 0 = 2 → ₹8.25

Total = 62.00 + 8.25 = ₹70.25

Why it happens: the two items cost different amounts, so each group is multiplied
on its own and only then added. Adding first would mix up rupees per notebook
with rupees per eraser.

Page 16 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Check it yourself: ₹15.50 is a little over ₹15, so 4 books are a little over ₹60. Erasers
add about ₹8. The total should be a little over ₹68 — and ₹70.25 is.

Q5 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.

The stack is 5.22 cm tall.

Height = 1.45 × 36 mm

145 × 36 = 5220

Places: 2 + 0 = 2 → 52.20 mm

1 cm = 10 mm, so divide by 10:

52.2 ÷ 10 = 5.22 cm

Why it happens: changing mm to cm means dividing by 10, and dividing by 10
slides the point one place left. So 52.2 mm becomes 5.22 cm.

Check it yourself: 145 × 36 = 145 × 30 + 145 × 6 = 4350 + 870 = 5220. ✓

Q6 Math Talk: 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?

The price is ₹127.125, and yes — dropping the last zeroes gives exactly the same product.

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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

As printed: 5650 × 2250 = 12712500

Places: 2 + 3 = 5 → 127.12500

Shortened: 565 × 225 = 127125

Places: 1 + 2 = 3 → 127.125

Both answers are the same number: 127.12500 = 127.125.

Why it happens: a zero at the end of a decimal adds no value. 56.50 = 5650/100 =
565/10, and 2.250 = 2250/1000 = 225/100. The fractions are equal, so the products
must be equal. Each zero you drop from a number removes one zero from its
denominator — and so removes one decimal place from the product. The two
changes cancel out exactly.

Tip: a zero before the first significant digit is different. In 0.05 the zero is holding the
tenths place, so it cannot be dropped.

Q7 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?

His profit is ₹320.

Profit on one notebook = 30 − 23.6 = ₹6.4

Profit on 50 notebooks = 6.4 × 50

64 × 50 = 3200, places 1 + 0 = 1 → ₹320

The longer route gives the same answer:

Cost price of 50 books = 23.6 × 50 = ₹1180
Selling price of 50 books = 30 × 50 = ₹1500

Profit = 1500 − 1180 = ₹320 ✓

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Class 7 Maths Chapter 12 Another Peek Beyond the Point
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Why it happens: profit per book multiplied by the number of books gives the same

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total as total selling price minus total cost price. Both are the same subtraction, done
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Given that 18 × 12 = 216, find the products: (a) 18 × 1.2 (b) 18 × 0.12 (c) 1.8 × 1.2 (d)
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Q8

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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?
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The digits are always 216. Only the position of the point changes.
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Why it happens: in (d) and (e) both numbers are smaller than 1, so each one shrinks
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the other. Everywhere else at least one number is bigger than 1. In (e) the answer

agl
needs 6 decimal places but 216 has only 3 digits, so three zeroes are written in front:
0.000216.

co m
m .
o m l a se
Tip: (b) and (c) give the same answer. Moving a point one place left in one factor and
ag because you have
.cone place right in the other leaves the product unchanged,
m divided by 10 and multiplied by 10.
l a se
ag
.c
s e m
m a
e m . co agl
g l as
a

co m
m .
m ase
.co


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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Q9 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

The product is less than 1 in (b) and (d).
You can decide without multiplying: a product is less than 1 only when both numbers are less
than 1.

MULTIPLICATION BOTH LESS THAN 1? PRODUCT LESS THAN 1?

(a) 7 × 0.6 No (7 > 1) 4.2 No

(b) 0.7 × 0.6 Yes 0.42 Yes

(c) 0.7 × 6 No (6 > 1) 4.2 No

(d) 0.07 × 0.06 Yes 0.0042 Yes

Why it happens: if both numbers are below 1, then taking a part of a part gives
something even smaller — smaller than each of them, and so certainly below 1. But
be careful: "one number below 1" is not enough. In (a), 0.6 of 7 is still 4.2, because 7
is large enough to survive being cut down.

Q10 Multiplying the following numbers by 10, 100 and 1000 to complete the table.

× 10 × 100 × 1000

5.7

23.02

0.92

0.306

24.67

Multiplying by 10, 100 or 1000 moves the point right by 1, 2 or 3 places.

Page 20 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

NUMBER × 10 × 100 × 1000

5.7 57 570 5700

23.02 230.2 2302 23020

0.92 9.2 92 920

0.306 3.06 30.6 306

24.67 246.7 2467 24670

Why it happens: multiplying by 10 makes every digit worth ten times as much. A
Tenths digit becomes a Ones digit, a Ones digit becomes a Tens digit. Sliding the
point one place right does that to all the digits at once. Once the point reaches the
end of the digits, extra zeroes appear to hold the empty places — that is why 0.92 ×
1000 is 920 and not 92.

Tip: this is the exact opposite of the division rule on page 68. Division moves the
point left, multiplication moves it right, and the number of zeroes tells you how far.

In-text Questions — Pages 74 – 75
Section 4.3 Decimal Division

Q1 Example 6: Anuja has a 3.9 m length of ribbon and she wants to cut it into 10 equal
pieces. What is the length of each piece in decimal?

Each piece is 0.39 m long.

3.9 ÷ 10

3.9 = 39/10

Dividing by 10 is the same as multiplying by 1/10:

39/10 ÷ 10 = 39/10 × 1/10 = 39/100

= 0.39 m

Page 21 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Why it happens: the reciprocal of 10 is 1/10, so dividing by 10 multiplies the
denominator by 10. The denominator grows from 10 to 100, which is one more zero
— so the point slides one more place left, from 3.9 to 0.39.

Check it yourself: ten pieces of 0.39 m give 3.9 m. ✓

Q2 What is the length of each piece if the ribbon is cut into 100 equal pieces?

Each piece is 0.039 m long.

3.9 ÷ 100 = 39/10 × 1/100 = 39/1000 = 0.039 m

Why it happens: 100 has two zeroes, so the point moves two places left instead of
one. Since 3.9 has only one digit before the point, a zero must be written in to hold
the tenths place: 0.039.

Q3 What is 0.039 m in centimetres and millimetres?

3.9 cm, which is the same as 39 mm.

1 m = 100 cm, so multiply by 100:

0.039 × 100 = 3.9 cm

1 m = 1000 mm, so multiply by 1000:

0.039 × 1000 = 39 mm

Why it happens: a centimetre is a smaller unit than a metre, so the same length
needs a bigger number. Multiplying by 100 moves the point two places right;
multiplying by 1000 moves it three places right and the point disappears from view,
leaving the whole number 39.

Page 22 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Check it yourself: 39 mm is about the width of three fingers — a believable piece of
ribbon after cutting 3.9 m into 100 parts.

Q4 Complete the table:

DECIMAL ÷ 10 ÷ 100 ÷ 1000 ÷ 10000

18.7 1.87 0.187 0.0187 0.00187

21.1

0.13

2.146

0.0058

Move the point left by 1, 2, 3 or 4 places. For the last two rows, work backwards — move the
point right.

DECIMAL ÷ 10 ÷ 100 ÷ 1000 ÷ 10000

18.7 1.87 0.187 0.0187 0.00187

21.1 2.11 0.211 0.0211 0.00211

0.13 0.013 0.0013 0.00013 0.000013

214.6 21.46 2.146 0.2146 0.02146

58 5.8 0.58 0.058 0.0058

Row 4: the number ÷ 100 = 2.146.

So the number = 2.146 × 100 = move the point 2 places right = 214.6

Row 5: the number ÷ 10000 = 0.0058.

So the number = 0.0058 × 10000 = move the point 4 places right = 58

Page 23 of 73

ase
Class 7 Maths Chapter 12 Another Peek Beyond the Point
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: multiplication undoes division. If dividing by 100 moved the point

m l a se
two places left, then getting back to the original number moves it two places right.
o
Once you.c g point sliding one
have the first cell of the row, the rest of the row is justathe
m
sefurther left each time.
l a
ag
place

o m
Example 7: Neenu has 29 metres ofm
e
. c ag
s
red ribbon and wants to share it equally with
a that each of them will get?
Q5

agl
Anu. What is the length of ribbon

co m
em.
as
Each girl gets 14.5 m.
m l
.co a g
a s e÷m2
gl
29
a Each gets 14 m, and 1 m is left over.

m a s
.co agl
That 1 m shared between two gives each another 1/2 m.

se m
g l a
1/2 = (1 × 5)/(2 × 5) = 5/10 = 0.5 a

co m
m .
m as e
l
Each girl gets 14 + 0.5 = 14.5 m
.co a g
a s em
a gl Why it happens: a fraction becomes a decimal easily only when its denominator is
10, 100, 1000 and so on. 2 is not one of these, but 2 × 5 = 10, so multiplying the top
se m
com g l a
.
and bottom by 5 turns 1/2 into 5/10 — the same value, now in a form we can read
m a
ase
off the place value chart.

agl

co m
Q6 How do we convert 1/2 into a decimal?
m .
as e
. com a g l
sem
a
agl Turn the denominator into 10, 100 or 1000 by multiplying the top and the bottom by the same
c
number.
m .
a s e
. c om agl
s e m2 × 5 = 10.
gla = 0.5
Is 2 a factor of 10? Yes, because

1/2 = (1 × 5)/(2 × 5) =a5/10

co m
m .
m ase
.co


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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

The same idea handles other denominators:

1/4: 4 × 25 = 100 → 25/100 = 0.25

1/5: 5 × 2 = 10 → 2/10 = 0.2

1/8: 8 × 125 = 1000 → 125/1000 = 0.125

Why it happens: the decimal system only has places worth tenths, hundredths,
thousandths … So a fraction can be written as a decimal in the direct way only when
it can be re-dressed with one of those denominators. Multiplying top and bottom by
the same number never changes the value — it only changes the clothes.

Tip: denominators like 2, 4, 5, 8, 20, 25 and 50 all work, because they are built only
from 2s and 5s — and 10 = 2 × 5.

In-text Questions — Page 76
Section 4.3 — Division Using Place Value

Q1 Now, what if the ribbon was shared between four friends instead of 2?

Each friend gets 7.25 m.

Each gets 29 ÷ 4 m, that is 29/4 m.

Is 4 a factor of 10? No.

Is 4 a factor of 100? Yes — 4 × 25 = 100.

29/4 = (29 × 25)/(4 × 25) = 725/100 = 7.25 m

Why it happens: 4 does not divide 10, so a single decimal place is not enough. But 4
does divide 100, so two decimal places are exactly right. This is the reason quarters
always end in .25 or .75.

Page 25 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Check it yourself: 7.25 × 4 = 29. ✓ And it makes sense that four friends get about
half of what two friends got (14.5 m).

Q2 Suppose we want to write the quotient 10/3 as a decimal. Can we convert this
fraction to an equivalent fraction with a denominator such as 1, 10, 100, 1000, etc.?

No, it is not possible.

10, 100, 1000 … are built only from 2s and 5s:

10 = 2 × 5, 100 = 2 × 2 × 5 × 5, 1000 = 2 × 2 × 2 × 5 × 5 × 5

The denominator 3 is neither a 2 nor a 5, and 3 never divides any of them.

Why it happens: to change 10/3 into a fraction over 100, we would need a whole
number k with 3 × k = 100. But 100 is not a multiple of 3 (1 + 0 + 0 = 1 is not divisible
by 3). The same test fails for 10, 1000, 10000 and every power of ten. So no
equivalent fraction of this kind exists, and we need a more general method — long
division using place value.

Did you know? This is exactly why 10 ÷ 3 turns out to be 3.333… and never stops.
The chapter comes back to it on page 84.

In-text Questions — Pages 77 – 80
Section 4.3 — Division Using Place Value · Division with a Decimal Quotient

Q1 Example 8: Find the value of 1324 ÷ 4.

1324 ÷ 4 = 331
Divide place by place, regrouping whenever a place is too small to share.

Page 26 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

1324 = 1 Thousand + 3 Hundreds + 2 Tens + 4 Ones

1 Thousand ÷ 4 → not possible. Regroup 1 Thousand as 10 Hundreds.

10 Hundreds + 3 Hundreds = 13 Hundreds

13 Hundreds ÷ 4 → each part gets 3 Hundreds, 1 Hundred remains

Regroup 1 Hundred as 10 Tens. 10 Tens + 2 Tens = 12 Tens

12 Tens ÷ 4 → each part gets 3 Tens, nothing remains

4 Ones ÷ 4 → each part gets 1 One

1324 ÷ 4 = 0 Thousands + 3 Hundreds + 3 Tens + 1 One = 331

Why it happens: this is ordinary long division, said out loud in place-value language.
"Bring down the next digit" is really "regroup what is left into ten of the next smaller
unit and add the digit already there".

Q2 Now, let us use this understanding of long division to find the value of 1325 ÷ 4.

1325 ÷ 4 = 331.25
The first three steps are the same as for 1324. The difference comes at the Ones.

Page 27 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

13 Hundreds ÷ 4 → 3 Hundreds each, 1 Hundred left

12 Tens ÷ 4 → 3 Tens each

5 Ones ÷ 4 → 1 One each, 1 One remains

1 One cannot be split into 4 equal whole parts. Regroup 1 One as 10 Tenths.

This is the moment the decimal point is placed in the quotient.

10 Tenths ÷ 4 → 2 Tenths each, 2 Tenths remain

Regroup 2 Tenths as 20 Hundredths.

20 Hundredths ÷ 4 → 5 Hundredths each, nothing remains

1325 ÷ 4 = 3 Hundreds + 3 Tens + 1 One + 2 Tenths + 5 Hundredths = 331.25

Why it happens: the place value chart does not stop at the Ones. To the right of the
Ones sit the Tenths, then the Hundredths. Regrouping 1 One into 10 Tenths is the
same move as regrouping 1 Hundred into 10 Tens — only one step further right. The
decimal point simply marks where the whole ones end, so it goes in the quotient
exactly when we cross that boundary.

Q3 Can we verify this by finding an equivalent fraction for 1325/4?

Yes — and it gives the same answer.

4 × 25 = 100, so multiply the top and the bottom by 25:

1325/4 = (1325 × 25)/(4 × 25) = 33125/100

= 331.25 ✓

Page 28 of 73

as e
Class 7 Maths Chapter 12 Another Peek Beyond the Point
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: the long division and the equivalent fraction are two roads to one

m l a se
place. The long division builds the answer digit by digit; the equivalent fraction
o g the two digits
.c whole thing at once with a denominator of 100, andathen
m
rewrites the
a e point can be read straight off.
sthe
ag l
after

com
ag
Check it yourself: 331.25 × 4 = 1325. ✓
m .
as e
a g l
In-text Questions — Pages 81 – 83
co m
m.
Section 4.3 — Division with a Decimal Dividend

m as e
. c o a g l
Q1 m
a s e Example 9: Find the value of 237 ÷ 8.

agl
s

m a
.co agl
237 ÷ 8 = 29.625

se m
g l a
a

co m
m .
m as e
.co a g l
se m
g l a
a
se m
com g l a
m . a
ase
agl

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

237 = 2 Hundreds + 3 Tens + 7 Ones

2 Hundreds ÷ 8 → not possible. Regroup as 20 Tens.
20 Tens + 3 Tens = 23 Tens

23 Tens ÷ 8 → 2 Tens each, 7 Tens remain

Regroup 7 Tens as 70 Ones. 70 + 7 = 77 Ones

77 Ones ÷ 8 → 9 Ones each, 5 Ones remain

Regroup 5 Ones as 50 Tenths — place the decimal point now.

50 Tenths ÷ 8 → 6 Tenths each, 2 Tenths remain

Regroup 2 Tenths as 20 Hundredths.

20 Hundredths ÷ 8 → 2 Hundredths each, 4 Hundredths remain

Regroup 4 Hundredths as 40 Thousandths.

40 Thousandths ÷ 8 → 5 Thousandths each, nothing remains

237 ÷ 8 = 29.625

Why it happens: the division ends because 8 = 2 × 2 × 2 is built only from 2s, so an
equivalent fraction over 1000 exists: 237/8 = (237 × 125)/1000 = 29625/1000 = 29.625.
Three 2s in the denominator need three decimal places.

Q2 Example 10: A shopkeeper has 9.5 kg of sugar and he wants to pack it equally in 4
bags. What is the weight of each bag of sugar?

Each bag weighs 2.375 kg.

Page 30 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

9.5 ÷ 4

9 Ones ÷ 4 → 2 Ones each, 1 One remains

Regroup 1 One as 10 Tenths; with the 5 Tenths already there → 15 Tenths.

Place the decimal point in the quotient.

15 Tenths ÷ 4 → 3 Tenths each, 3 Tenths remain

Regroup 3 Tenths as 30 Hundredths

30 Hundredths ÷ 4 → 7 Hundredths each, 2 Hundredths remain

Regroup 2 Hundredths as 20 Thousandths

20 Thousandths ÷ 4 → 5 Thousandths each, nothing remains

Each bag = 2.375 kg

Why it happens: a decimal dividend changes nothing about the method. The 5
Tenths of 9.5 simply join the 10 Tenths made by regrouping the leftover One, giving
15 Tenths to share. The point still goes in the quotient at the moment we move from
Ones to Tenths.

Check it yourself: 2.375 × 4 = 9.5. ✓ Four bags of a little under 2.5 kg make a little
under 10 kg.

Q3 Example 11: What is the value of 0.06 ÷ 5?

0.06 ÷ 5 = 0.012

Page 31 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

0.06 = 0 Ones + 0 Tenths + 6 Hundredths

0 Ones ÷ 5 → 0 Ones. Moving from Ones to Tenths, place the decimal point.
0 Tenths ÷ 5 → 0 Tenths

6 Hundredths ÷ 5 → 1 Hundredth each, 1 Hundredth remains

Regroup 1 Hundredth as 10 Thousandths

10 Thousandths ÷ 5 → 2 Thousandths

Quotient = 0.012

Why it happens: the zeros are not wasted steps — each one holds a place. Writing 0
in the Tenths place of the quotient is what keeps the 1 in the Hundredths place,
where it belongs. Without it the answer would read 0.12, which is ten times too big.

Check it yourself: 0.012 × 5 = 0.06. ✓

Figure it Out — Page 83
Section 4.3 Decimal Division

Q1 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

Find the number that turns the denominator into 10, 100 or 1000, then multiply the top by the
same number.

Page 32 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

(a) 5 × 2 = 10

18/5 = (18 × 2)/(5 × 2) = 36/10 = 3.6

(b) 4 × 25 = 100

415/4 = (415 × 25)/(4 × 25) = 10375/100 = 103.75

(c) 2 × 5 = 10

1217/2 = (1217 × 5)/(2 × 5) = 6085/10 = 608.5

(d) 8 × 125 = 1000

4827/8 = (4827 × 125)/(8 × 125) = 603375/1000 = 603.375

Verification by long division (the last steps only):

DIVISION WHOLE PART REGROUPING QUOTIENT

18 ÷ 5 3, remainder 3 3 Ones → 30 Tenths; 30 ÷ 5 = 6 Tenths 3.6

415 ÷ 4 103, remainder 3 30 Tenths → 7 Tenths r 2; 20 Hundredths → 5 103.75

1217 ÷ 2 608, remainder 1 10 Tenths ÷ 2 = 5 Tenths 608.5

4827 ÷ 8 603, remainder 3 30 Tenths → 3 r 6; 60 Hund. → 7 r 4; 40 Thou. → 5 603.375

Why it happens: 5, 4, 2 and 8 are all built only from 2s and 5s, so each of them
divides some power of ten exactly. That is why all four quotients stop after a few
decimal places.

Q2 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

(a) (iii) 381.5 (b) (iii) 445.875

Page 33 of 73

as e
Class 7 Maths Chapter 12 Another Peek Beyond the Point
a g l AglaSem · NCERT Solutions

co m
em.
(a) 1526/4 = (1526 × 25)/100 = 38150/100 = 381.5
m l as
m .co a g
l a se
g
(b) 3567/8 = (3567 × 125)/1000 = 445875/1000 = 445.875
a

. com
Why it happens: you can pick the answer without dividing fully, by estimating. For
ag
a s
(a), 4 × 400 = 1600, which is a bit moreemthan 1526, so the quotient is a little under 400
agl= 3600, a little more than 3567, so the quotient is a
— only 381.5 fits. For (b), 8 × 450
little under 450 — only 445.875 fits.

co m
e m.
c o m l as
Note: in the book, options (i) and (iv) of part (b) are both printed as 4458.75. It is a
g
m . slip; neither is correct, since both are ten times too big.
a
e
printing
s
a gla
m a s
.co agl
Q3 What is the quotient? (a) 132 ÷ 4 = (b) 13.2 ÷ 4 = (c) 1.32 ÷ 4 = (d) 0.132 ÷ 4 =

se m
g l a
a
Only the dividend's point moves, so only the quotient's point moves.

co m
m .
as e
com
DIVISION QUOTIENT

. a g l
s e m
a
(a) 132 ÷ 4 33

agl (b) 13.2 ÷ 4 3.3

se m
com g l a
. a
(c) 1.32 ÷ 4 0.33

m
ase
agl
(d) 0.132 ÷ 4 0.033

Why it happens: each dividend is one-tenth of the one above it, and the divisor
co m
m .
never changes. Sharing one-tenth as much among the same 4 people gives each
m ase
.co a g l
person one-tenth as much. So the quotient's point slides one place left each time,

s e m exactly as the dividend's did.
agla
.c
s e m
m a
e m . co agl
g l as
a

co m
m .
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.co


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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Q4 What is the quotient? (a) 126 ÷ 8 = (b) 12.6 ÷ 8 = (c) 1.26 ÷ 8 = (d) 0.126 ÷ 8 = (e) 0.0126 ÷
8=

Divide 126 by 8 once, then slide the point.

126 ÷ 8: 8 × 15 = 120, remainder 6

60 Tenths ÷ 8 → 7 Tenths, 4 Tenths remain

40 Hundredths ÷ 8 → 5 Hundredths, nothing remains

So 126 ÷ 8 = 15.75

DIVISION QUOTIENT

(a) 126 ÷ 8 15.75

(b) 12.6 ÷ 8 1.575

(c) 1.26 ÷ 8 0.1575

(d) 0.126 ÷ 8 0.01575

(e) 0.0126 ÷ 8 0.001575

Why it happens: dividing the dividend by 10 divides the quotient by 10 as well,
because the divisor is untouched. The digits 1575 stay put; only their place value
drops one step each row.

Check it yourself: 0.001575 × 8 = 0.0126. ✓

In-text Questions — Pages 84 – 85
Section 4.3 — Division with a Decimal Divisor · Does This Ever End?

Q1 Example 12: Ravi went from Pune to Matheran by scooter in 2.5 hours. The distance
was 126 km. What was his average speed?

His average speed was 50.4 km per hour.

Page 35 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Average speed = distance ÷ time = 126 ÷ 2.5

The divisor is a decimal, so turn it into a fraction:
126 ÷ 25/10 = 126 × 10/25 = 1260/25

Long division: 25 × 50 = 1250, remainder 10

100 Tenths ÷ 25 = 4 Tenths

= 50.4 km/h

Why it happens: dividing by a fraction is multiplying by its reciprocal. The reciprocal
of 25/10 is 10/25, so the dividend gets multiplied by 10 while the divisor becomes
the counting number 25. The value of the quotient does not change, because
1260/25 and 126/2.5 are the same fraction written differently.

Check it yourself: 50.4 × 2.5 = 126. ✓ A scooter at about 50 km/h for two and a half
hours covering 126 km sounds right.

Q2 Example 13: Find 4.68 ÷ 1.3. Now, what about 4.68 ÷ 0.13?

4.68 ÷ 1.3 = 3.6 and 4.68 ÷ 0.13 = 36.

First: 4.68 ÷ 13/10 = 4.68 × 10/13 = 46.8/13

13 × 3 = 39, remainder 7.8 → 78 Tenths ÷ 13 = 6 Tenths

= 3.6

Second: 4.68 ÷ 13/100 = 4.68 × 100/13 = 468/13

13 × 36 = 468

= 36

Page 36 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Why it happens: 0.13 is one-tenth of 1.3. Sharing the same 4.68 into pieces that are
ten times smaller gives ten times as many pieces. So the second quotient is ten
times the first.

Q3 What do you notice in these cases?

When the divisor is a decimal, we can make it a counting number — as long as we do the same
to the dividend.

4.68/0.13 = (4.68 × 100)/(0.13 × 100) = 468/13 = 36

The rule: multiply the divisor by 10, 100, 1000 … until it becomes a whole number, multiply the
dividend by the same number, and then do ordinary long division.

DIVISION MULTIPLY BOTH BY BECOMES QUOTIENT

126 ÷ 2.5 10 1260 ÷ 25 50.4

4.68 ÷ 1.3 10 46.8 ÷ 13 3.6

4.68 ÷ 0.13 100 468 ÷ 13 36

Why it happens: a division is a fraction, and multiplying the numerator and the
denominator of a fraction by the same number gives an equivalent fraction — the
same value. So the quotient is untouched, while the divisor becomes friendly
enough for long division.

Q4 Can you calculate 10 ÷ 3? Try dividing using long division. Will this process end?

No, it never ends. 10 ÷ 3 = 3.333…

Page 37 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Step 1: regroup 1 Ten as 10 Ones. 10 Ones ÷ 3 → 3 Ones, 1 One remains

Step 2: regroup 1 One as 10 Tenths. 10 Tenths ÷ 3 → 3 Tenths, 1 Tenth remains

Step 3: regroup 1 Tenth as 10 Hundredths. 10 Hundredths ÷ 3 → 3 Hundredths, 1
Hundredth remains

Step 4: regroup 1 Hundredth as 10 Thousandths. 10 Thousandths ÷ 3 → 3 Thousandths, 1

Thousandth remains

… and so on for ever.

10 ÷ 3 = 3.333…

Why it happens: every single step ends with a remainder of 1 in the next smaller
place. Regrouping that 1 always gives 10 of the next unit, and 10 ÷ 3 always leaves 1
again. Since the step repeats exactly, it can never stop. So 10 ÷ 3 cannot be written
with a finite number of decimal digits.

Did you know? This links back to page 76 — 3 is not made of 2s and 5s, so no power
of ten is a multiple of 3, and no equivalent fraction over 10, 100 or 1000 exists.

In-text Questions — Pages 85 – 86
Section 4.3 — Does This Ever End? · A Magic Number: 142857

TRY THIS

Q1 Can you find the quotients of 10 ÷ 9, and 100 ÷ 11?

Both quotients repeat for ever.

Page 38 of 73

as e
Class 7 Maths Chapter 12 Another Peek Beyond the Point
a g l AglaSem · NCERT Solutions

co m
e m.
10 ÷ 9: 9 goes into 10 once, remainder 1.
m l as
.co
10 Tenths ÷ 9 → 1 Tenth, remainder 1 … the same step every time.
m a g
l a se
g
10 ÷ 9 = 1.111…
a

co m
. ag
100 ÷ 11: 11 × 9 = 99, remainder 1.
e m
10 Tenths ÷ 11 → 0 Tenths, remainder 10
g l as
a
100 Hundredths ÷ 11 → 9 Hundredths, remainder 1 … now it repeats.

co m
m.
100 ÷ 11 = 9.0909…

m as e
.co a g l
a s em
Why it happens: in each case a remainder comes back that has already appeared.

a gl Once a remainder repeats, every step after it must repeat too, because the same
remainder with the same divisor gives the same next digit. 9 and 11 are not built

m a s
.co agl
from 2s and 5s, so neither divides any power of ten and neither division can stop.

se m
g l a
a
Now divide 1 by 7 (1 ÷ 7). Will this end? Note all the remainders we get. It starts with
m
Q2
1, then 3, then 2, then 6, and so on. What do you observe? Can you explain why this
. co
e m
as
division never ends?
m l
.co a g
a s em
gl

a It never ends. 1 ÷ 7 = 0.142857 142857 14…

se m
com g l a
. a
STEP DIVIDE QUOTIENT DIGIT REMAINDER
m
a1 se
agl
1 10 ÷ 7 3

2 30 ÷ 7 4 2

co m
m .
as e
3 20 ÷ 7 2 6

m l
.4co a g
sem
60 ÷ 7 8 4

a
agl 5 40 ÷ 7 5 5
c
m .
m a s e
.co agl
6 50 ÷ 7 7 1 ← back to the start

se m
g l a
The remainders run 1 → 3 → 2 → 6 → 4 → 5 → 1 → 3 … — a closed chain of six.
a

com
m .
m ase
.co


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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Why it happens: when you divide by 7, the only possible remainders are 0, 1, 2, 3, 4,
5 and 6. A remainder of 0 would end the division, so at most six remainders can
occur. With only six choices, a remainder must sooner or later come round again —
here after exactly six steps. And the moment a remainder repeats, the whole block
of quotient digits repeats with it. That is why not only the remainders but also the
digits 142857 cycle for ever.

Tip: the same argument shows that every division of whole numbers either stops or
starts repeating — the remainders can never all be new for ever.

Q3 A Magic Number: 142857. Let us consider the number 142857 that arose when
dividing 1 by 7. Multiply 142857 by numbers from 1 to 6. What are the products?
What do you notice? Multiply 142857 by 7. What do you observe?

You get the same six digits back, only rolled around to a different starting point.

MULTIPLICATION PRODUCT WHAT HAPPENED

142857 × 1 142857 starts at 1

142857 × 2 285714 starts at 2

142857 × 3 428571 starts at 4

142857 × 4 571428 starts at 5

142857 × 5 714285 starts at 7

142857 × 6 857142 starts at 8

142857 × 7 999999 all nines!

Why it happens: 142857 is the repeating block of 1 ÷ 7. Dividing 2 by 7, 3 by 7 and
so on starts the very same chain of remainders at a different place, so the same six
digits appear in the same cyclic order. And 7 × 142857 = 999999 because 1/7 + 2/7 +
4/7 … in fact 7 × (1/7) = 1, and 0.142857142857… × 7 = 0.999999… = 1.

Page 40 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Did you know? Numbers like this are called cyclic numbers. 142857 is the smallest
one.

Q4 Try This: To find one such number, you can find 1 ÷ 17 in decimal, and use the
repeating block of digits.

Divide 1 by 17 by long division. The remainders take 16 steps to return to 1.

1 ÷ 17 = 0.0588235294117647 0588235294117647 …

So the repeating block is the 16-digit number 0588235294117647. It is cyclic, exactly like 142857:

588235294117647 × 2 = 1176470588235294

588235294117647 × 3 = 1764705882352941

Each product uses the same digits, started at a different point in the cycle.

Why it happens: dividing by 17 can leave only the remainders 1 to 16. Here all
sixteen actually occur before the chain closes, so the repeating block is as long as it
possibly can be. Primes that behave like this always produce cyclic numbers.

Tip: 7 and 17 work, but 11 does not — 1 ÷ 11 = 0.0909…, a block of only two digits,
far shorter than 10. Try 19 and 23 next; both give full-length cycles.

In-text Question — Page 86

Page 41 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Section 4.3 — Dividend, Divisor, and Quotient

MATH TALK

Q1 Math Talk: Will the quotient be always greater than the dividend when the divisor is
a decimal? Try it out with different values of the divisor. Describe the relationship
between the dividend, divisor, and the quotient. Create a table for capturing this
relationship in different situations, like we did for multiplication.

No. What matters is not whether the divisor is a decimal, but whether it is smaller or larger than
1.

128 ÷ 0.4 = 320 → quotient greater than 128 (divisor is below 1)

128 ÷ 2.5 = 51.2 → quotient less than 128 (divisor is above 1, though still a decimal)

128 ÷ 1.0 = 128 → quotient equal to the dividend

SITUATION DIVISOR EXAMPLE RELATIONSHIP

Situation 1 Greater than 1 128 ÷ 4 = 32; 128 ÷ 2.5 = 51.2 Quotient is less than the dividend

Situation 2 Equal to 1 128 ÷ 1 = 128 Quotient equals the dividend

Situation 3 Between 0 and 1 128 ÷ 0.4 = 320; 128 ÷ 0.5 = 256 Quotient is greater than the dividend

Why it happens: a quotient answers "how many of the divisor fit into the dividend?".
If the divisor is bigger than 1, fewer than 128 of them fit into 128. If the divisor is
smaller than 1, more than 128 of them fit — 0.4 fits into 128 exactly 320 times,
because each piece is less than half a unit. Dividing by a number below 1 is the
mirror image of multiplying by a number below 1: one makes things bigger, the
other makes them smaller.

Tip: compare the divisor with 1, not with the dividend. 128 ÷ 2.5 has a decimal
divisor, yet the quotient 51.2 is smaller than 128.

Figure it Out — Pages 86 – 87

Page 42 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Section 4.3 Decimal Division

Q1 Express the following fractions in decimal form: (a) 2/5 (b) 13/4 (c) 4/50 (d) 5/8

Build a denominator of 10, 100 or 1000 in each case.

(a) 5 × 2 = 10 → 2/5 = 4/10 = 0.4

(b) 4 × 25 = 100 → 13/4 = 325/100 = 3.25

(c) 50 × 2 = 100 → 4/50 = 8/100 = 0.08

(d) 8 × 125 = 1000 → 5/8 = 625/1000 = 0.625

Why it happens: every one of these denominators — 5, 4, 50, 8 — is made only of 2s
and 5s, so each of them divides some power of ten. The number of decimal places
you need is decided by how big that power of ten has to be: 10 for (a), 100 for (b) and
(c), 1000 for (d).

Check it yourself: 13/4 is 3 wholes and 1/4 left over, and a quarter is 0.25. So 3.25. ✓

Q2 Find the quotients: (a) 24.86 ÷ 1.2 (b) 5.728 ÷ 1.52

First clear the decimal from the divisor, then divide. Neither of these quotients ends.

Page 43 of 73

ase
Class 7 Maths Chapter 12 Another Peek Beyond the Point
a g l AglaSem · NCERT Solutions

co m
e m.
(a) 24.86 ÷ 1.2 = (24.86 × 10)/(1.2 × 10) = 248.6 ÷ 12
m l as
.co
12 × 20 = 240, remainder 8.6
m a g
l a se
g
86 Tenths ÷ 12 → 7 Tenths, 2 Tenths remain
a20 Hundredths ÷ 12 → 1 Hundredth, 8 Hundredths remain

com
. ag
80 Thousandths ÷ 12 → 6 Thousandths, 8 Thousandths remain … the 8 keeps coming back
e m
= 20.7166… (the 6 repeats for ever) ≈ 20.72
g l as
a

co m
m.
(b) 5.728 ÷ 1.52 = (5.728 × 100)/(1.52 × 100) = 572.8 ÷ 152

m as e
.co l
152 × 3 = 456, remainder 116.8
a g
a s em
Carrying on: 3.7684210526315789473…

a gl = 3.7684… ≈ 3.77

om a s
Why it happens: after clearing them
e
. c agl
s
divisors we are dividing by 12 = 2 × 2 × 3 and by

a ga
152 = 2 × 2 × 2 × 19. The extralfactors 3 and 19 are not 2s or 5s, so no power of ten is
a multiple of them, and the division cannot stop. As on page 85, a remainder repeats
and drags the digits round with it.
co m
m .
o m l a se
g Two decimal places
.cwhen a quotient does not end, say how far you have agone.
m
Tip:

aseis usually enough for a practical answer.
agl
se m
com g l a
m . a
e
Evaluate the following using the information 156 × 12 = 1872. (a) 15.6 × 1.2 = (b) 187.2
as 0.156 × 0.12 =
Q3

g l
÷ 1.2 = (c) 18.72 ÷ 15.6 = (d)
a

co m

m .
e
The digits are always 156, 12 and 1872. Only the point moves.
m l as
m .co a g
l a se
ag
.c
s e m
m a
e m . co agl
g l as
a

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m .
m ase
.co


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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

(a) 15.6 × 1.2: places 1 + 1 = 2 → 18.72

(b) 187.2 ÷ 1.2 = 1872 ÷ 12 = 156

(both multiplied by 10)

(c) 18.72 ÷ 15.6 = 1872 ÷ 1560 = 1.2

(both multiplied by 100)

(d) 0.156 × 0.12: places 3 + 2 = 5 → 0.01872

Why it happens: multiplication and division are opposites. Since 15.6 × 1.2 = 18.72,
we already know 18.72 ÷ 1.2 = 15.6 and 18.72 ÷ 15.6 = 1.2 without dividing at all. For
(b), 187.2 is ten times 18.72, so its quotient is ten times 15.6 — that is, 156.

Check it yourself: 1.2 × 156 = 187.2. ✓

Page 45 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Q4 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 × ______

(a) 25 → 0.025: the point moved 3 places left, so divide by 1000

(b) 25 → 250: the answer is bigger, so the divisor is below 1. 25 ÷ 0.1 = 250

(c) 25 → 2.5: the point moved 1 place left, so divide by 10

(d) 25 ÷ 10 = 25 × 0.1 (that is, 1/10)

(e) 25 ÷ 0.10 = 25 × 10

(f) 25 ÷ 0.01 = 25 × 100

Why it happens: dividing by a number is the same as multiplying by its reciprocal.
The reciprocal of 10 is 0.1, and the reciprocal of 0.1 is 10. This is why (d) and (e) point
in opposite directions even though both involve a 10 and a 0.1. In (f), 0.01 is one
hundredth, and there are 100 hundredths in each whole — so 25 wholes hold 2500
of them.

Check it yourself: 25 ÷ 0.01 = 2500, and 25 × 100 = 2500. ✓

Page 46 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Q5 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?

(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

Yes, 24.6 ÷ 1.5 gives the same quotient as 2.46 ÷ 0.15.

24.6 ÷ 1.5 = 246 ÷ 15 = 16.4

2.46 ÷ 0.15 = 246 ÷ 15 = 16.4 ✓

Why it happens: both divisions become 246 ÷ 15 once the divisor is cleared. Going
from 24.6 ÷ 1.5 to 2.46 ÷ 0.15 divides the dividend by 10 and the divisor by 10.
Making both ten times smaller leaves the quotient exactly where it was — the pieces
shrank, but so did the whole.

Tip: in (a), (b), (c) only the divisor shrinks. Ten times smaller pieces means ten times
as many of them, so each quotient is ten times the one before.

Q6 A 4 m long wooden block has to be cut into 5 pieces of equal length. What is the
length of each piece?

Each piece is 0.8 m long, that is 80 cm.

4 ÷ 5 = 4/5 = (4 × 2)/(5 × 2) = 8/10 = 0.8 m

Page 47 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Why it happens: 4 whole metres cannot be shared into 5 equal whole metres, so we
regroup: 4 Ones become 40 Tenths, and 40 Tenths ÷ 5 = 8 Tenths each. Eight tenths
of a metre is 0.8 m.

Check it yourself: 0.8 × 5 = 4. ✓

Q7 If the perimeter of a regular polygon with 12 sides is 208.8 cm, what is the length of
its side?

Each side is 17.4 cm.

In a regular polygon all sides are equal.

Side = perimeter ÷ number of sides = 208.8 ÷ 12

12 × 17 = 204, remainder 4.8

48 Tenths ÷ 12 = 4 Tenths

= 17.4 cm

Why it happens: the perimeter is the total distance round the shape. Since the 12
sides are all the same length, sharing the perimeter equally among them gives one
side.

Check it yourself: 17.4 × 12 = 208.8. ✓

Q8 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.

Each friend gets 0.375 litres, which is 375 ml.

Page 48 of 73

as e
Class 7 Maths Chapter 12 Another Peek Beyond the Point
a g l AglaSem · NCERT Solutions

co m
e m.
3 ÷ 8 = 3/8
m l as
.co
8 × 125 = 1000, so 3/8 = 375/1000 = 0.375 l
m a g
l a se
g
a1 l = 1000 ml, so multiply by 1000:

com
. ag
0.375 × 1000 = 375 ml
e m
g l as
a
Why it happens: 3 whole litres split among 8 gives less than one litre each, so the
answer must be below 1. Regrouping 3 Ones as 30 Tenths gives 3 Tenths each with 6
co m
Tenths left, then 60 Hundredths gives 7 each with 4 left, then 40 Thousandths gives 5
em.
m l as
.co g
each — 0.375, exactly what the fraction method gave.
m a
l a se
a g Check it yourself: 375 × 8 = 3000 ml = 3 l. ✓

m a s
m .co agl
l a se
g
A car covers 234.45 km using 12.6 litres of petrol. What is the distance travelled per
a
Q9
litre?

co m
m .
e

m l as
.co g
The car travels about 18.61 km per litre.

em a
a s
agl Distance per litre = 234.45 ÷ 12.6
se m
com a
Clear the divisor — multiply both by 10:
. a g l
m
ase
= 2344.5 ÷ 126

agl
m
126 × 18 = 2268, remainder 76.5

. co
765 Tenths ÷ 126 → 6 Tenths, 9 Tenths remain
em
m l as
.co
m 900 Thousandths ÷ 126 → 7, 18 remain …
90 Hundredths ÷ 126 → 0, 90 remain
a g
l a se
ag
.c
e m
= 18.6071… ≈ 18.61 km per litre
m a s
e m . co agl
g l as
a

co m
m .
m ase
.co


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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Why it happens: this quotient does not end, because after clearing the divisor we
are dividing by 126 = 2 × 3 × 3 × 7, and the 3s and the 7 are not factors of any power
of ten. For a real answer about petrol, two decimal places is plenty.

Check it yourself: 18.6 × 12.6 = 234.36 km, just short of 234.45 — so a shade more
than 18.6 is right.

Q10 13.5 kg of flour (aata) was distributed equally among 15 students. How much flour
did each student receive?

Each student received 0.9 kg, that is 900 g.

13.5 ÷ 15

13 Ones ÷ 15 → 0 Ones. Place the point.

Regroup 13 Ones as 130 Tenths; with the 5 Tenths → 135 Tenths

135 Tenths ÷ 15 = 9 Tenths each

= 0.9 kg

Why it happens: 13.5 is less than 15, so each share must be less than 1 kg. The
quotient starts with 0 in the Ones place, and the first real digit appears in the Tenths
place.

Check it yourself: 0.9 × 15 = 13.5. ✓

In-text Question — Page 87

Page 50 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Section 4.3 — the pattern box with powers of 2 and 5

MATH TALK

Q1 Math Talk: Complete the pattern — 1/2 = 0.5, 1/(2×2) = 0.25, 1/(2×2×2) = 0.125,
1/(2×2×2×2) = 0.0625, 1/(2×2×2×2×2) = ? and 1/5 = 0.2, 1/(5×5) = 0.04, 1/(5×5×5) = 0.008,
1/(5×5×5×5) = 0.0016, 1/(5×5×5×5×5) = ? What pattern do you observe? Why are 2 and
5 related in this way?

The two missing values are 0.03125 and 0.00032.

POWERS OF 2 DECIMAL POWERS OF 5 DECIMAL

1/2 0.5 1/5 0.2

1/(2×2) 0.25 1/(5×5) 0.04

1/(2×2×2) 0.125 1/(5×5×5) 0.008

1/(2×2×2×2) 0.0625 1/(5×5×5×5) 0.0016

1/(2×2×2×2×2) 0.03125 1/(5×5×5×5×5) 0.00032

The patterns:

Every one of these decimals ends — none of them goes on for ever.
At each step down the list the decimal is halved on the left and divided by 5 on the right.
Both columns need one more decimal place at each step: 1, 2, 3, 4, 5 places.
The digits in each row of the two columns multiply to a power of ten: 5 × 2 = 10, 25 × 4 = 100,
125 × 8 = 1000, 625 × 16 = 10000, 3125 × 32 = 100000.

1/32 = 1×3125 / 32×3125 = 3125/100000 = 0.03125

1/3125 = 1×32 / 3125×32 = 32/100000 = 0.00032

Why 2 and 5 are related this way: because 10 = 2 × 5. Every power of ten is made
of exactly as many 2s as 5s: 100000 = 2⁵ × 5⁵. So a denominator built only from 2s
can always be completed into a power of ten by supplying the missing 5s, and a
denominator built only from 5s by supplying the missing 2s. That is why fractions
with denominators built only from 2s and 5s always give decimals that stop — and
why every other denominator, like 3 or 7, gives a decimal that never stops.

Page 51 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Tip: the two decimals in a row are a matching pair. 0.03125 and 0.00032 both have 5
decimal places, and 3125 × 32 = 100000 shows exactly why.

In-text Questions — Pages 89 – 90
Section 4.4 Look Before You Leap! — Making an Adjustment

MATH TALK

Q1 Do you know which month has this extra day?

February.
In an ordinary year February has 28 days. In a leap year it has 29, and the year has 366 days
instead of 365.

Ordinary year: 365 days, February has 28 days

Leap year: 366 days, February has 29 days

Why it happens: the Earth takes 365.2422 days to go round the Sun, but a calendar
can only count whole days. The leftover 0.2422 of a day is saved up, and after four
years it has grown to about one whole day (4 × 0.2422 = 0.9688). That saved-up day
is added to the shortest month.

Did you know? 29 February is called an intercalary day. Someone born on it gets a
birthday only once every four years!

Q2 What is the number of days that the Earth needs to make 4 full revolutions around
the Sun?

1460.9688 days.

Page 52 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

4 × 365.2422

3652422 × 4 = 14609688

Decimal places: 4 + 0 = 4

= 1460.9688 days

The calendar, meanwhile, counts

4 × 365 + 1 = 1461 days

Why it happens: the calendar with one leap day every four years gives 1461 days,
but the Earth needs only 1460.9688. The calendar is ahead by 1461 − 1460.9688 =
0.0312 days every four years. Tiny — but it adds up, which is what the rest of the
section is about.

Q3 Math Talk: With this new scheme of adding one extra day every 4th year, what is the
number of days in 100 calendar years? Can you write an expression to calculate that
number?

36,525 days.

Each calendar year has 365 days → 100 × 365 = 36500 days

Years divisible by 4 get one extra day → 100/4 = 25 extra days

(100 × 365 + 100/4 × 1) = 36500 + 25 = 36,525 days

Why it happens: the expression counts in two layers. First give every year the basic
365 days. Then hand out one bonus day to each of the 25 leap years. Adding the
bonus separately is easier than sorting the years into two groups first.

Page 53 of 73

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Class 7 Maths Chapter 12 Another Peek Beyond the Point
a g l AglaSem · NCERT Solutions

co m
m.
How many years are divisible by 4 in 100 years?
e
Q4

m l as
m .co a g
ase

25 g
a l
years.

com
. ag
The multiples of 4 from 1 to 100 are 4, 8, 12, …, 100
e m
Count = 100 ÷ 4 = 25
g l as
a

. com
Why it happens: the multiples of 4 come one in every block of four consecutive
years. A hundred years hold 25 such blocks, so there are 25 of them.em
m a s Note that 100

. o
cdivisible a glpage, where year
emhas to be taken out again.
itself is by 4, so it is counted — this matters on the next

a s
gl
100
a
om a s
Math Talk: Can you form different.cexpressions for the same question?
em agl
s
Q5
a
agl

co m
.
Yes. Here are three that all give 36,525.

e m
comdays plus bonus days: g l as
. a
sem
1. Basic
a (100 × 365) + (100/4 × 1) = 36500 + 25 = 36,525
agl
se m
2. Sort the years into two groups first:.com g l a
m
e27375 a
a s
agl
(25 × 366) + (75 × 365) = 9150 + = 36,525

co m
3. Pretend every year is a leap year, then take back the extra days:
m .
m as e
.co
(100 × 366) − 75 = 36600 − 75 = 36,525
a g l
se m
g l a
a c
Why it happens: all three count the same collection of days, only grouped
m .
m
differently. The first adds a bonus, the second splits the years into leap and ordinary,
a s e
m . co
the third over-counts and then corrects. Being able to move between such
e agl
l as
expressions is what makes a calculation easy to check.
g
a

co m
m .
m ase
.co


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Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Compare: the Earth actually needs 100 × 365.2422 = 36,524.22 days for 100
revolutions. The calendar has 36,525 — it is ahead by 0.78 days. We have
overcompensated.

Q6 Math Talk: Can you write an expression for the number of days in 100 calendar
years with this new adjustment? (no extra day in the 100th year)

36,524 days.

Years divisible by 4 in 100 years = 100/4 = 25

But 100 itself must be left out → subtract 100/100 = 1

Leap years = 100/4 − 100/100 = 25 − 1 = 24

Ordinary years = 100 − 24 = 76

(100/4 − 100/100) × 366 + (100 − (100/4 − 100/100)) × 365

= (24 × 366) + (76 × 365)

= 8784 + 27740

= 36,524 days

Why it happens: the old scheme gave 36,525 days against the Earth's 36,524.22 —
too many by 0.78 of a day. Removing one leap day per century takes the calendar
down to 36,524, which is now short by only 0.22 of a day. Much closer.

Q7 This is close to 36524.22 days but is it close enough? What happens after 1000 years
with this adjustment?

Over 1000 years the small shortfall grows to 2.2 days — so it is not close enough.

Page 55 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Calendar days in 1000 years = 36524 × 10 = 3,65,240 days

Days the Earth needs = 1000 × 365.2422 = 3,65,242.2 days

Difference = 365242.2 − 365240 = 2.2 days

Why it happens: a shortfall of 0.22 days per century is repeated ten times in a
thousand years, and 10 × 0.22 = 2.2. Now the calendar is behind the Earth, the
opposite of the earlier problem. To bridge the gap the calendar makers put one leap
day back: every 400th year is a leap year after all.

Result: with that final rule, 1000 calendar years hold (750 × 365) + (240 × 366) + (8 ×
365) + (2 × 366) = 3,65,242 days, against the Earth's 3,65,242.2 — just 0.2 days apart
in a thousand years.

Try This — Page 92
Section 4.4 — Making Yet Another Adjustment

TRY THIS

Q1 Try This: With this final scheme of leap years can you calculate the number of
calendar days in 10,000 years and the number of actual days the Earth will take to
make 10,000 revolutions around the Sun? What is the difference? If there is a big
difference, can you suggest a way to fix this problem?

The calendar runs 3 days ahead in 10,000 years.

Page 56 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Step 1 — count the leap years in 10,000 years.

Divisible by 4: 10000/4 = 2500

Of these, drop the ones divisible by 100: 10000/100 = 100

Put back the ones divisible by 400: 10000/400 = 25

Leap years = 2500 − 100 + 25 = 2425

Step 2 — calendar days.

10000 × 365 + 2425 = 36,50,000 + 2425 = 36,52,425 days

Step 3 — actual days.

10000 × 365.2422 = 36,52,422 days

Difference = 3652425 − 3652422 = 3 days

A way to fix it: take away 3 leap days somewhere in the 10,000 years. One neat rule is:

A year divisible by 4000 shall not be a leap year.

Years 4000 and 8000 lose their extra day → 2 days saved.

That leaves the calendar only 1 day ahead in 10,000 years.

Why it happens: each rule is a smaller correction on top of the last one. Every 4
years adds too much, every 100 years takes back a little too much, every 400 years
puts back slightly too much again. The leftovers keep shrinking — 0.78 days in 100
years, 0.2 days in 1000 years, 3 days in 10,000 years — so each new rule fires less
often than the one before.

Did you know? A 3-day drift in 10,000 years is about 1 day in 3,300 years. The
calendar makers decided that was somebody else's problem!

In-text Questions — Page 93

Page 57 of 73

Class 7 Maths Chapter 12 Another Peek Beyond the Point AglaSem · NCERT Solutions

Section 4.4 — how calendars were worked out

TRY THIS

Q1 Do you wonder how people figured out that the Earth completes one revolution
around the Sun in exactly 364.2422 days?

By watching the sky patiently for a very long time and then dividing.
The method is simple to describe, though it needs great care.

Fix a moment that repeats every year — for example the day the shadow of a pole at noon is
longest (the winter solstice), or the day the Sun rises exactly due east (an equinox).
Count the whole days from one such moment to the same moment many years later.
Divide the total number of days by the number of years.

If 100 years are found to hold 36,524.22 days, then

one year = 36524.22 ÷ 100 = 365.2422 days

Why it happens: one year alone cannot be measured to four decimal places — no
one can note the exact instant of a solstice that finely. But spreading the error over a
hundred years divides it by a hundred as well. That is why the number is written with
four decimal places only after long records are collected.

Note: the book prints 364.2422 here, but on pages 88, 89, 91 and 92 the figure used
is 365.2422 days. The correct value is 365.2422; 364.2422 is a printing slip.

Q2 Try This: 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 even the positions of the stars in the sky accurately.

Indian calendars are luni-solar: the months follow the Moon, while the year is kept in step with
the Sun.
What to look for in your investigation:

A lunar month (new Moon to new Moon) is about 29.53 days, so twelve of them make about
354.37 days — roughly 11 days short of a solar year.

Page 58 of 73

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Class 7 Maths Chapter 12 Another Peek Beyond the Point
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Left alone, the festivals would slide backwards through the seasons, as they do in a purely
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19 solar years = 19 × 365.2422 ≈ 6939.6 days

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235 lunar months = 235 × 29.5306 ≈ 6939.7 days
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Sample answer: "The Sūrya Siddhānta and later the Pañcāṅga tradition track the Sun's position
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leap month. Both are the same idea — save up the leftover, then pay it back in one
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Ask at home which festivals shift by about 11 days each year, and which ones do not.

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Figure it Out — Pages 93 – 95
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End-of-chapter question set

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MATH TALK

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Q1 A 210 gram packet of peanut chikki costs ₹70.5, while a 110 gram packet of potato

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The potato chips are cheaper.

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The packets are different sizes, so compare the price of the same weight of each — say 100 g.

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a g l Page 59 of 73