Class 7 Maths Chapter 4 Another Peek Beyond the Point notes and book solutions from Ganita Prakash Part 2 with easy explanations, important concepts, examples and textbook question answers for revision and exam preparation.
ANOTHER PEEK BEYOND THE POINT
Class 7 Mathematics | Ganita Prakash | Part II | Chapter 4
Easy Notes, Concepts and Complete Book Solutions
Chapter at a Glance
This chapter extends our understanding of decimals. We learn how to multiply and divide decimal numbers using the same place-value ideas that we already know for whole numbers and fractions.
Main ideas of the chapter:
- Decimal place value
- Converting fractions into decimals
- Multiplication of decimals
- Division by 10, 100, 1000 and so on
- Long division with decimal quotients
- Division by decimal divisors
- Terminating and repeating decimals
- Relationship between dividend, divisor and quotient
- Decimals in real-life situations
- Leap years and calendar calculations
4.1 A Quick Recap of Decimals
Recall that decimals are the natural extension of the Indian place value system to represent decimal fractions.
1 tenth = 1/10 = 0.1
1 hundredth = 1/100 = 0.01
1 thousandth = 1/1000 = 0.001
1 hundredth = 1/100 = 0.01
1 thousandth = 1/1000 = 0.001
Example: 27.53
| Tens | Ones | Tenths | Hundredths |
|---|---|---|---|
| 2 | 7 | 5 | 3 |
Therefore,
27.53 = 20 + 7 + 0.5 + 0.03
Decimal Place Value
Converting grams into kilograms
| Quantity | Fraction of 1 kg | Decimal in kg |
|---|---|---|
| 50 g Cinnamon | 50/1000 = 1/20 | 0.05 kg |
| 100 g Cumin seeds | 100/1000 = 1/10 | 0.1 kg |
| 25 g Cardamom | 25/1000 = 1/40 | 0.025 kg |
| 250 g Pepper | 250/1000 = 1/4 | 0.25 kg |
Easy rule: Dividing by 10, 100, 1000
When dividing by 10, move the decimal point 1 place to the left.
When dividing by 100, move it 2 places to the left.
When dividing by 1000, move it 3 places to the left.
When dividing by 100, move it 2 places to the left.
When dividing by 1000, move it 3 places to the left.
123 ÷ 10 = 12.3
24 ÷ 100 = 0.24
678 ÷ 1000 = 0.678
12 ÷ 1000 = 0.012
12345 ÷ 1000 = 12.345
24 ÷ 100 = 0.24
678 ÷ 1000 = 0.678
12 ÷ 1000 = 0.012
12345 ÷ 1000 = 12.345
Page 67
Book Question
Jonali and Pallabi play a game. Jonali says a fraction and Pallabi gives the equivalent decimal. Write Pallabi’s answer in the blank spaces.
Easy Solution
Convert each fraction into a decimal by dividing the numerator by the denominator. Fractions with denominators 10, 100, 1000 and so on can be written directly as decimals.
For example:
3/10 = 0.3
25/100 = 0.25
125/1000 = 0.125
3/10 = 0.3
25/100 = 0.25
125/1000 = 0.125
The individual fractions used in the game are shown in the textbook image. The supplied text does not contain those blank entries, so the method above is the required method for every blank.
Page 67
Book Question
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.
Easy Solution
1 kg = 1000 g.
50 g = 50/1000 kg = 1/20 kg = 0.05 kg
100 g = 100/1000 kg = 1/10 kg = 0.1 kg
25 g = 25/1000 kg = 1/40 kg = 0.025 kg
250 g = 250/1000 kg = 1/4 kg = 0.25 kg
100 g = 100/1000 kg = 1/10 kg = 0.1 kg
25 g = 25/1000 kg = 1/40 kg = 0.025 kg
250 g = 250/1000 kg = 1/4 kg = 0.25 kg
Page 67
Book Question
Write the following fractions as a sum of fractions and also as decimals:
| Fraction | Expanding the Numerator | Sum of one-tenths, one-hundredths, one-thousands,… | Decimals |
|---|---|---|---|
| 254/1000 | 200/1000 + 50/1000 + 4/1000 | 2/10 + 5/100 + 4/1000 | 0.254 |
| 847/10000 | |||
| 173/100 | |||
| 23/1000 |
Easy Solution
| Fraction | Sum of fractions | Decimal |
|---|---|---|
| 847/10000 | 8/100 + 4/1000 + 7/10000 | 0.0847 |
| 173/100 | 1 + 7/10 + 3/100 | 1.73 |
| 23/1000 | 2/100 + 3/1000 | 0.023 |
Page 68
Math Talk
Can you give a simple rule to divide any number by a number of the form 1 followed by zeroes — 10, 100, 1000, etc.? For example, 123 ÷ 10, 24 ÷ 100 or 678 ÷ 1000? Look for a pattern in the previous problems.
Easy Rule
Count the zeroes in the divisor and move the decimal point to the left by the same number of places.
÷ 10 → move 1 place left
÷ 100 → move 2 places left
÷ 1000 → move 3 places left
÷ 100 → move 2 places left
÷ 1000 → move 3 places left
Example:
123 ÷ 10 = 12.3
24 ÷ 100 = 0.24
678 ÷ 1000 = 0.678
24 ÷ 100 = 0.24
678 ÷ 1000 = 0.678
4.2 Decimal Multiplication
Easy Concept
Multiplying decimals is closely connected with multiplying fractions. First multiply the numbers as if the decimal points were not there. Then put the decimal point back by counting the total number of decimal places in the two numbers.
Rule:
Total decimal places in the answer = decimal places in first number + decimal places in second number.
Total decimal places in the answer = decimal places in first number + decimal places in second number.
Decimal Multiplication Idea
Important examples
9.5 × 5 = 47.5
12.5 × 7.5 = 93.75
0.827 × 2 = 1.654
1.654 × 6 = 9.924 km
5.7 × 13.3 = 75.81 cm²
5.8 × 1.24 = 7.192
12.5 × 7.5 = 93.75
0.827 × 2 = 1.654
1.654 × 6 = 9.924 km
5.7 × 13.3 = 75.81 cm²
5.8 × 1.24 = 7.192
Page 69
Math Talk
Can the product of two decimals be a natural number?
Yes.Example:
0.5 × 2 = 1, which is a natural number.
0.5 × 2 = 1, which is a natural number.
Page 69
Math Talk
Can the product of a decimal and a natural number be a natural number?
Yes.Example:
0.25 × 4 = 1.
0.25 × 4 = 1.
Math Talk
Observe the number of digits after the decimal point in the multiplier, the multiplicand and the product. Also note the number of zeroes in the denominator.
The number of decimal places in the product is equal to the total number
of decimal places in the two numbers being multiplied.
For example:
5.7 has 1 decimal place and 13.35 has 2 decimal places. Therefore the product has 3 decimal places:
5.7 × 13.35 = 76.095.
For example:
5.7 has 1 decimal place and 13.35 has 2 decimal places. Therefore the product has 3 decimal places:
5.7 × 13.35 = 76.095.
Page 71
Math Talk
Suppose we know that 596 × 248 = 147808, can you immediately write down the product of 5.96 × 24.8?
Yes.5.96 has 2 decimal places and 24.8 has 1 decimal place.Total = 2 + 1 = 3 decimal places.Therefore:
596 × 248 = 147808
5.96 × 24.8 = 147.808
596 × 248 = 147808
5.96 × 24.8 = 147.808
Page 71
Math Talk
By looking at the above examples, can you frame a rule to multiply two decimals?
Rule
- Ignore the decimal points temporarily.
- Multiply the numbers as whole numbers.
- Count the total number of decimal places in both factors.
- Put the decimal point in the product so that it has that many decimal places.
Page 72
Math Talk
When is the product of two decimals greater than both the numbers?
When both positive numbers are greater than 1.Example:
3.4 × 6.5 = 22.1
22.1 is greater than both 3.4 and 6.5.
3.4 × 6.5 = 22.1
22.1 is greater than both 3.4 and 6.5.
Page 72
Math Talk
When is it less than both the numbers?
When both positive numbers lie between 0 and 1.Example:
0.75 × 0.4 = 0.3
0.3 is less than both 0.75 and 0.4.
0.75 × 0.4 = 0.3
0.3 is less than both 0.75 and 0.4.
Page 73
Figure it Out — 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
(a) 6 × 4 tenths = 24 tenths
(b) 7 × 0.3
(c) 9 × 5 hundredths
(a) 24 tenths = 2.4
(b) 7 × 0.3 = 21 tenths = 2.1
(c) 9 × 5 hundredths = 45 hundredths = 0.45
(b) 7 × 0.3 = 21 tenths = 2.1
(c) 9 × 5 hundredths = 45 hundredths = 0.45
Page 73
Figure it Out — 2
Find the products:
(a) 27.34 × 6
(b) 4.23 × 3.7
(c) 0.432 × 0.23
(a) 27.34 × 6
(b) 4.23 × 3.7
(c) 0.432 × 0.23
(a)
27.34 × 6 = 164.04
(b) 4.23 × 3.7 = 4.23 × 37 ÷ 10 = 156.51 ÷ 10 = 15.651
(c) 0.432 × 0.23 = 432 × 23 with 5 decimal places = 9936/100000 = 0.09936
(b) 4.23 × 3.7 = 4.23 × 37 ÷ 10 = 156.51 ÷ 10 = 15.651
(c) 0.432 × 0.23 = 432 × 23 with 5 decimal places = 9936/100000 = 0.09936
Page 73
Figure it Out — 3
Thejus needs 1.65 m of cloth for a shirt. How many metres of cloth are needed for 3 shirts?
1.65 × 3 = 4.95 m.
Therefore, 4.95 metres of cloth are needed.
Therefore, 4.95 metres of cloth are needed.
Page 73
Figure it Out — 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?
Cost of 4 notebooks:
4 × ₹15.50 = ₹62.00
Cost of 3 erasers:
3 × ₹2.75 = ₹8.25
Total:
₹62.00 + ₹8.25 = ₹70.25
4 × ₹15.50 = ₹62.00
Cost of 3 erasers:
3 × ₹2.75 = ₹8.25
Total:
₹62.00 + ₹8.25 = ₹70.25
Page 73
Figure it Out — 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.
Height = 1.45 × 36
= 52.2 mm
Since 10 mm = 1 cm:
52.2 mm = 5.22 cm.
= 52.2 mm
Since 10 mm = 1 cm:
52.2 mm = 5.22 cm.
Page 73
Figure it Out — 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?
Price = 56.50 × 2.250
= ₹127.125
Yes, we can write:
56.50 = 56.5
2.250 = 2.25
Removing zeroes at the end of a decimal does not change its value. Therefore:
56.5 × 2.25 = 127.125.
= ₹127.125
Yes, we can write:
56.50 = 56.5
2.250 = 2.25
Removing zeroes at the end of a decimal does not change its value. Therefore:
56.5 × 2.25 = 127.125.
Page 73
Figure it Out — 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?
Profit per notebook:
₹30 − ₹23.6 = ₹6.4
Profit on 50 notebooks:
₹6.4 × 50 = ₹320
₹30 − ₹23.6 = ₹6.4
Profit on 50 notebooks:
₹6.4 × 50 = ₹320
Page 73
Figure it Out — 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?
(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?
(a) 18 × 1.2 = 21.6
(b) 18 × 0.12 = 2.16
(c) 1.8 × 1.2 = 2.16
(d) 0.18 × 0.12 = 0.0216
(e) 0.018 × 0.012 = 0.000216
(f) 1.8 × 12 = 21.6
Products less than 1: (d) and (e).
(b) 18 × 0.12 = 2.16
(c) 1.8 × 1.2 = 2.16
(d) 0.18 × 0.12 = 0.0216
(e) 0.018 × 0.012 = 0.000216
(f) 1.8 × 12 = 21.6
Products less than 1: (d) and (e).
Page 73
Figure it Out — 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
(a) 7 × 0.6
(b) 0.7 × 0.6
(c) 0.7 × 6
(d) 0.07 × 0.06
A product of two positive numbers less than 1 is less than 1.
Therefore:
(b) 0.7 × 0.6
(d) 0.07 × 0.06
Both products are less than 1.
Therefore:
(b) 0.7 × 0.6
(d) 0.07 × 0.06
Both products are less than 1.
Page 73
Figure it Out — 10
Multiplying the following numbers by 10, 100 and 1000 to complete the table.
| Number | × 10 | × 100 | × 1000 |
|---|---|---|---|
| 5.7 | |||
| 23.02 | |||
| 0.92 | |||
| 0.306 | |||
| 24.67 |
| 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 |
4.3 Decimal Division
Dividing by 10, 100 and 1000
Decimal Point Moves Left
When dividing by 10, 100, 1000, 10000, move the decimal point left
by 1, 2, 3, 4 places respectively.
Decimal division by long division
In long division, if there is a remainder after the Ones place, regroup the remaining Ones into Tenths. At this point put the decimal point in the quotient. Continue with Hundredths, Thousandths and so on.
1325 ÷ 4 = 331.25
237 ÷ 8 = 29.625
9.5 ÷ 4 = 2.375
0.06 ÷ 5 = 0.012
237 ÷ 8 = 29.625
9.5 ÷ 4 = 2.375
0.06 ÷ 5 = 0.012
Dividing by a decimal divisor
If the divisor is a decimal, multiply both the dividend and divisor by
the same power of 10 until the divisor becomes a whole number.
4.68 ÷ 0.13
= 468 ÷ 13
= 36
= 468 ÷ 13
= 36
Page 74
Book Question
What is the length of each piece if the ribbon is cut into 100 equal pieces?
3.9 ÷ 100 = 0.039 m.
Page 74
Book Question
What is 0.039 m in centimetres and millimetres?
0.039 m = 3.9 cm = 39 mm.
Page 74
Book Question
By looking at these divisions, we can frame a simple rule for dividing decimals by 1, 10, 100, 1000, and so on.
The decimal point moves to the left by as many places as there are
zeroes in the divisor.
Page 74
Book Question
Decimal ÷ 10 ÷ 100 ÷ 1000 ÷ 10000
18.7 1.87 0.187 0.0187 0.00187
21.1
0.13
2.146
0.0058
18.7 1.87 0.187 0.0187 0.00187
21.1
0.13
2.146
0.0058
| 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 |
| 2.146 | 0.2146 | 0.02146 | 0.002146 | 0.0002146 |
| 0.0058 | 0.00058 | 0.000058 | 0.0000058 | 0.00000058 |
Page 75
Book Question
So, can we find a fraction equivalent to 1/2 with such a denominator?
Yes.
1/2 = 5/10 = 0.5.
1/2 = 5/10 = 0.5.
Page 75
Book Question
Now, what if the ribbon was shared between four friends instead of 2?
Each friend gets:
29 ÷ 4 = 29/4
= 725/100
= 7.25 m.
29 ÷ 4 = 29/4
= 725/100
= 7.25 m.
Page 75
Book Question
Can we find a fraction equivalent to 29/4 with such a denominator?
Yes. Since 4 × 25 = 100:
29/4 = (29 × 25)/(4 × 25) = 725/100 = 7.25.
29/4 = (29 × 25)/(4 × 25) = 725/100 = 7.25.
Page 76
Example 8
Find the value of 1324 ÷ 4.
1324 ÷ 4 = 331.
Place-value division:
13 hundreds ÷ 4 = 3 hundreds, remainder 1 hundred.
Regroup 1 hundred as 10 tens.
12 tens ÷ 4 = 3 tens.
4 ones ÷ 4 = 1 one.
Therefore, 1324 ÷ 4 = 331.
Place-value division:
13 hundreds ÷ 4 = 3 hundreds, remainder 1 hundred.
Regroup 1 hundred as 10 tens.
12 tens ÷ 4 = 3 tens.
4 ones ÷ 4 = 1 one.
Therefore, 1324 ÷ 4 = 331.
Page 78–80
Example
1325 ÷ 4
1325 ÷ 4 = 331.25.
After dividing the Ones, 1 One remains. Regroup it as 10 Tenths:
10 Tenths ÷ 4 = 2 Tenths, remainder 2 Tenths.
Regroup 2 Tenths as 20 Hundredths:
20 Hundredths ÷ 4 = 5 Hundredths.
Therefore:
1325 ÷ 4 = 331.25.
After dividing the Ones, 1 One remains. Regroup it as 10 Tenths:
10 Tenths ÷ 4 = 2 Tenths, remainder 2 Tenths.
Regroup 2 Tenths as 20 Hundredths:
20 Hundredths ÷ 4 = 5 Hundredths.
Therefore:
1325 ÷ 4 = 331.25.
Page 80–81
Example 9
Find the value of 237 ÷ 8.
237 ÷ 8 = 29.625.
2 hundreds are regrouped into 20 tens.
20 tens + 3 tens = 23 tens.
23 tens ÷ 8 = 2 tens, remainder 7 tens.
7 tens = 70 ones.
70 + 7 = 77 ones.
77 ÷ 8 = 9 ones, remainder 5 ones.
5 ones = 50 tenths.
50 ÷ 8 = 6 tenths, remainder 2 tenths.
2 tenths = 20 hundredths.
20 ÷ 8 = 2 hundredths, remainder 4 hundredths.
4 hundredths = 40 thousandths.
40 ÷ 8 = 5 thousandths.
Therefore:
237 ÷ 8 = 29.625.
2 hundreds are regrouped into 20 tens.
20 tens + 3 tens = 23 tens.
23 tens ÷ 8 = 2 tens, remainder 7 tens.
7 tens = 70 ones.
70 + 7 = 77 ones.
77 ÷ 8 = 9 ones, remainder 5 ones.
5 ones = 50 tenths.
50 ÷ 8 = 6 tenths, remainder 2 tenths.
2 tenths = 20 hundredths.
20 ÷ 8 = 2 hundredths, remainder 4 hundredths.
4 hundredths = 40 thousandths.
40 ÷ 8 = 5 thousandths.
Therefore:
237 ÷ 8 = 29.625.
Page 82
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?
9.5 ÷ 4 = 2.375 kg.
Therefore, each bag contains 2.375 kg of sugar.
Therefore, each bag contains 2.375 kg of sugar.
Page 82
Example 11
What is the value of 0.06 ÷ 5?
0.06 ÷ 5 = 0.012.
6 hundredths = 60 thousandths.
60 ÷ 5 = 12 thousandths.
Therefore, the answer is 0.012.
6 hundredths = 60 thousandths.
60 ÷ 5 = 12 thousandths.
Therefore, the answer is 0.012.
Page 83
Figure it Out — 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
(a) 18/5 (b) 415/4
(c) 1217/2 (d) 4827/8
(a) 18/5 = 36/10 = 3.6
(b) 415/4 = 10375/100 = 103.75
(c) 1217/2 = 6085/10 = 608.5
(d) 4827/8 = 603375/1000 = 603.375
(b) 415/4 = 10375/100 = 103.75
(c) 1217/2 = 6085/10 = 608.5
(d) 4827/8 = 603375/1000 = 603.375
Page 83
Figure it Out — 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
(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)
1526 ÷ 4 = 381.5.
Correct answer: (iii).
(b) 3567 ÷ 8 = 445.875.
Correct answer: (iii).
Correct answer: (iii).
(b) 3567 ÷ 8 = 445.875.
Correct answer: (iii).
Page 83
Figure it Out — 3
What is the quotient?
(a) 132 ÷ 4 = (b) 13.2 ÷ 4 =
(c) 1.32 ÷ 4 = (d) 0.132 ÷ 4 =
(a) 132 ÷ 4 = (b) 13.2 ÷ 4 =
(c) 1.32 ÷ 4 = (d) 0.132 ÷ 4 =
(a) 132 ÷ 4 = 33
(b) 13.2 ÷ 4 = 3.3
(c) 1.32 ÷ 4 = 0.33
(d) 0.132 ÷ 4 = 0.033
(b) 13.2 ÷ 4 = 3.3
(c) 1.32 ÷ 4 = 0.33
(d) 0.132 ÷ 4 = 0.033
Page 83
Figure it Out — 4
What is the quotient?
(a) 126 ÷ 8 = (b) 12.6 ÷ 8 =
(c) 1.26 ÷ 8 = (d) 0.126 ÷ 8 =
(e) 0.0126 ÷ 8 =
(a) 126 ÷ 8 = (b) 12.6 ÷ 8 =
(c) 1.26 ÷ 8 = (d) 0.126 ÷ 8 =
(e) 0.0126 ÷ 8 =
(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
(b) 12.6 ÷ 8 = 1.575
(c) 1.26 ÷ 8 = 0.1575
(d) 0.126 ÷ 8 = 0.01575
(e) 0.0126 ÷ 8 = 0.001575
Page 84
Example 12
Ravi went from Pune to Matheran by scooter in 2.5 hours. The distance was 126 km. What was his average speed?
Average speed = Distance ÷ Time
= 126 ÷ 2.5
= 1260 ÷ 25
= 50.4 km per hour.
= 126 ÷ 2.5
= 1260 ÷ 25
= 50.4 km per hour.
Page 84
Example 13
Find 4.68 ÷ 1.3.
Multiply both numbers by 10:
4.68 ÷ 1.3 = 46.8 ÷ 13
= 3.6.
4.68 ÷ 1.3 = 46.8 ÷ 13
= 3.6.
Page 84
Book Question
Now, what about 4.68 ÷ 0.13?
Multiply both numbers by 100:
4.68 ÷ 0.13 = 468 ÷ 13
= 36.
4.68 ÷ 0.13 = 468 ÷ 13
= 36.
Page 84
Book Question
What do you notice in these cases?
The divisor can be changed into a whole number by multiplying it by
10, 100, 1000 and so on. We must multiply the dividend by the same
number.
For example:
4.68 ÷ 0.13 = 468 ÷ 13.
For example:
4.68 ÷ 0.13 = 468 ÷ 13.
Page 84–85
Book Question
Will this process end?
No.
In 10 ÷ 3, every step leaves remainder 1. Therefore the same process starts again and again.
10 ÷ 3 = 3.333…
The decimal never terminates.
In 10 ÷ 3, every step leaves remainder 1. Therefore the same process starts again and again.
10 ÷ 3 = 3.333…
The decimal never terminates.
Page 85
Book Question
Can you find the quotients of 10 ÷ 9, and 100 ÷ 11?
10 ÷ 9 = 1.111…
100 ÷ 11 = 9.090909…
100 ÷ 11 = 9.090909…
Page 85
Book Question
Now divide 1 by 7 (1 ÷ 7). Will this end?
1 ÷ 7 = 0.142857142857…
It does not end. The digits 142857 repeat.
It does not end. The digits 142857 repeat.
Page 85
Book Question
What do you observe? Can you explain why this division never ends?
The remainders repeat in a cycle:
1, 3, 2, 6, 4, 5, 1, 3, 2, …
Because the same remainder appears again, the same digits of the quotient start repeating.
Therefore:
1 ÷ 7 = 0.142857142857….
1, 3, 2, 6, 4, 5, 1, 3, 2, …
Because the same remainder appears again, the same digits of the quotient start repeating.
Therefore:
1 ÷ 7 = 0.142857142857….
Page 85
Book Question
What are the products? What do you notice?
For 142857:
142857 × 1 = 142857
142857 × 2 = 285714
142857 × 3 = 428571
142857 × 4 = 571428
142857 × 5 = 714285
142857 × 6 = 857142
The same six digits appear again, but in a different cyclic order.
142857 × 1 = 142857
142857 × 2 = 285714
142857 × 3 = 428571
142857 × 4 = 571428
142857 × 5 = 714285
142857 × 6 = 857142
The same six digits appear again, but in a different cyclic order.
Page 85
Book Question
Multiply 142857 by 7. What do you observe?
142857 × 7 = 999999.
This is special because multiplying the cyclic number by 7 gives six 9s.
This is special because multiplying the cyclic number by 7 gives six 9s.
Page 85
Book Question
Are there other such numbers?
Yes. The chapter gives 1 ÷ 17 as another way to find a repeating block
of digits. Such numbers can produce interesting cyclic patterns.
Page 85
Book Question
Are there infinitely many such “cyclic” numbers? That is, can we keep finding more cyclic numbers, or do they eventually stop?
This is an interesting mathematical question. The chapter explains that
Emil Artin conjectured that there should be infinitely many such numbers,
but this conjecture remains unsolved.
So, we do not yet have a final proof answering the question.
So, we do not yet have a final proof answering the question.
Page 86
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.
For positive numbers:
| Divisor | Effect on quotient | Example |
|---|---|---|
| Greater than 1 | Quotient is smaller than dividend | 128 ÷ 4 = 32 |
| Equal to 1 | Quotient equals dividend | 128 ÷ 1 = 128 |
| Between 0 and 1 | Quotient is greater than dividend | 128 ÷ 0.4 = 320 |
Page 86
Figure it Out — 1
Express the following fractions in decimal form:
(a) 2/5 (b) 13/4
(c) 4/50 (d) 5/8
(a) 2/5 (b) 13/4
(c) 4/50 (d) 5/8
(a) 2/5 = 4/10 = 0.4
(b) 13/4 = 325/100 = 3.25
(c) 4/50 = 8/100 = 0.08
(d) 5/8 = 625/1000 = 0.625
(b) 13/4 = 325/100 = 3.25
(c) 4/50 = 8/100 = 0.08
(d) 5/8 = 625/1000 = 0.625
Page 86
Figure it Out — 2
Find the quotients:
(a) 24.86 ÷ 1.2
(b) 5.728 ÷ 1.52
(a) 24.86 ÷ 1.2
(b) 5.728 ÷ 1.52
(a)
24.86 ÷ 1.2
= 248.6 ÷ 12
= 20.716666…
(b) 5.728 ÷ 1.52
= 5728 ÷ 152
= 716/19
= 37.684210526…
= 248.6 ÷ 12
= 20.716666…
(b) 5.728 ÷ 1.52
= 5728 ÷ 152
= 716/19
= 37.684210526…
Page 86
Figure it Out — 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 = __________
(a) 15.6 × 1.2 = __________ (b) 187.2 ÷ 1.2 = __________
(c) 18.72 ÷ 15.6 = __________ (d) 0.156 × 0.12 = __________
(a) 15.6 × 1.2 = 18.72
(b) 187.2 ÷ 1.2 = 156
(c) 18.72 ÷ 15.6 = 1.2
(d) 0.156 × 0.12 = 0.01872
(b) 187.2 ÷ 1.2 = 156
(c) 18.72 ÷ 15.6 = 1.2
(d) 0.156 × 0.12 = 0.01872
Page 86
Figure it Out — 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 × ______
(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) 1000
(b) 0.1
(c) 10
(d) 0.1
(e) 10
(f) 100
(b) 0.1
(c) 10
(d) 0.1
(e) 10
(f) 100
Page 87
Figure it Out — 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?
(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 = 1.64
(b) 2.46 ÷ 0.15 = 16.4
(c) 2.46 ÷ 0.015 = 164
24.6 ÷ 1.5 = 16.4.
2.46 ÷ 0.15 = 16.4.
Yes, the quotients are the same because both dividend and divisor have been multiplied by 10.
(b) 2.46 ÷ 0.15 = 16.4
(c) 2.46 ÷ 0.015 = 164
24.6 ÷ 1.5 = 16.4.
2.46 ÷ 0.15 = 16.4.
Yes, the quotients are the same because both dividend and divisor have been multiplied by 10.
Page 87
Figure it Out — 6
A 4 m long wooden block has to be cut into 5 pieces of equal length. What is the length of each piece?
4 ÷ 5 = 0.8 m.
Page 87
Figure it Out — 7
If the perimeter of a regular polygon with 12 sides is 208.8 cm, what is the length of its side?
Side length = Perimeter ÷ Number of sides
= 208.8 ÷ 12
= 17.4 cm.
= 208.8 ÷ 12
= 17.4 cm.
Page 87
Figure it Out — 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.
3 litres = 3000 ml.
3000 ÷ 8 = 375 ml.
Each friend gets 375 ml.
3000 ÷ 8 = 375 ml.
Each friend gets 375 ml.
Page 87
Figure it Out — 9
A car covers 234.45 km using 12.6 litres of petrol. What is the distance travelled per litre?
234.45 ÷ 12.6
= 2344.5 ÷ 126
= 18.607142857… km per litre.
Approximately 18.61 km per litre.
= 2344.5 ÷ 126
= 18.607142857… km per litre.
Approximately 18.61 km per litre.
Page 87
Figure it Out — 10
13.5 kg of flour (aata) was distributed equally among 15 students. How much flour did each student receive?
13.5 ÷ 15 = 0.9 kg.
Each student received 0.9 kg of flour.
Each student received 0.9 kg of flour.
Page 87
Math Talk
What pattern do you observe? Why are 2 and 5 related in this way?
The denominators are powers of 2 and powers of 5.
2 × 5 = 10.
Therefore, powers of 2 and powers of 5 can be combined to make powers of 10. This is why fractions with denominators made only from 2s and 5s can often be written as terminating decimals.
For example:
1/2 = 0.5
1/4 = 0.25
1/8 = 0.125
1/5 = 0.2
1/25 = 0.04
2 × 5 = 10.
Therefore, powers of 2 and powers of 5 can be combined to make powers of 10. This is why fractions with denominators made only from 2s and 5s can often be written as terminating decimals.
For example:
1/2 = 0.5
1/4 = 0.25
1/8 = 0.125
1/5 = 0.2
1/25 = 0.04
Page 87
Book Question
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) = ?
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?
1/(2 × 2) = 0.25
1/(2 × 2 × 2) = 0.125
1/(2 × 2 × 2 × 2) = 0.0625
1/(2 × 2 × 2 × 2 × 2) = ?
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?
For powers of 2:
1/32 = 0.03125.
For powers of 5:
1/3125 = 0.00032.
The decimals move towards smaller values as the denominator becomes larger.
2 and 5 are related because 2 × 5 = 10.
1/32 = 0.03125.
For powers of 5:
1/3125 = 0.00032.
The decimals move towards smaller values as the denominator becomes larger.
2 and 5 are related because 2 × 5 = 10.
4.4 Look Before You Leap!
Leap Year Concept
The chapter uses 365.2422 days as the approximate time taken by Earth to complete one revolution around the Sun.
365.2422 − 365 = 0.2422 day
After 100 years:
0.2422 × 100 = 24.22 days
To keep the calendar closer to the Earth’s revolution, an extra day is added in leap years.
Leap Year Decision Idea
Simple classroom rule:
A year divisible by 4 is normally a leap year.
A century year must also be divisible by 400 to be a leap year.
Examples:
2000 → leap year
1900 → not a leap year
2024 → leap year
A year divisible by 4 is normally a leap year.
A century year must also be divisible by 400 to be a leap year.
Examples:
2000 → leap year
1900 → not a leap year
2024 → leap year
Page 88
Book Question
Do you know which month has this extra day?
The extra day is added to February.
February has 29 days in a leap year instead of 28.
February has 29 days in a leap year instead of 28.
Page 89
Book Question
What is the number of days that the Earth needs to make 4 full revolutions around the Sun?
4 × 365.2422 = 1460.9688 days.
Page 89
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?
There are 25 years divisible by 4 in 100 years.
Days = 100 × 365 + 25
= 36,500 + 25
= 36,525 days.
Days = 100 × 365 + 25
= 36,500 + 25
= 36,525 days.
Page 89
Math Talk
How many years are divisible by 4 in 100 years?
100 ÷ 4 = 25 years.
Page 89
Math Talk
Can you form different expressions for the same question?
One expression is:
(100 × 365) + (100 ÷ 4 × 1)
= 36,500 + 25
= 36,525.
Another way:
25 × 366 + 75 × 365
= 9,150 + 27,375
= 36,525.
(100 × 365) + (100 ÷ 4 × 1)
= 36,500 + 25
= 36,525.
Another way:
25 × 366 + 75 × 365
= 9,150 + 27,375
= 36,525.
Page 89
Book Question
The actual number of days that the Earth takes to go around the Sun 100 times is,
100 × 365.2422 = 36,524.22 days.
Thus, by adding a day every fourth year, after 100 years, the calendar days are more than the actual number of days taken by the Earth to go around the Sun. We have overcompensated.
100 × 365.2422 = 36,524.22 days.
Thus, by adding a day every fourth year, after 100 years, the calendar days are more than the actual number of days taken by the Earth to go around the Sun. We have overcompensated.
The calendar gives:
36,525 days.
Actual approximate time:
36,524.22 days.
Difference:
36,525 − 36,524.22 = 0.78 day.
So the four-year leap rule adds slightly too much over 100 years.
36,525 days.
Actual approximate time:
36,524.22 days.
Difference:
36,525 − 36,524.22 = 0.78 day.
So the four-year leap rule adds slightly too much over 100 years.
Page 90
Math Talk
Can you write an expression for the number of days in 100 calendar years with this new adjustment?
There are 25 years divisible by 4, but the 100th year is excluded.
So there are 24 leap years.
The remaining 76 years have 365 days.
Expression:
(24 × 366) + (76 × 365)
= 36,524 days.
So there are 24 leap years.
The remaining 76 years have 365 days.
Expression:
(24 × 366) + (76 × 365)
= 36,524 days.
Page 90
Math Talk
This is close to 36524.22 days but is it close enough? What happens after 1000 years with this adjustment?
For 100 years the difference is:
36,524.22 − 36,524 = 0.22 day.
For 1000 years:
36,524 × 10 = 365,240 days.
Actual:
1000 × 365.2422 = 365,242.2 days.
Difference:
365,242.2 − 365,240 = 2.2 days.
36,524.22 − 36,524 = 0.22 day.
For 1000 years:
36,524 × 10 = 365,240 days.
Actual:
1000 × 365.2422 = 365,242.2 days.
Difference:
365,242.2 − 365,240 = 2.2 days.
Page 91
Book Question
In 1000 calendar years, how many years are divisible by 400?
1000 ÷ 400 = 2.5.
The multiples of 400 within 1000 are 400 and 800.
Therefore, 2 years.
The multiples of 400 within 1000 are 400 and 800.
Therefore, 2 years.
Page 91
Book Question
In 1000 calendar years, how many years are divisible by 100 but not divisible by 400?
There are 10 multiples of 100 in 1000 years.
Two of them are also divisible by 400.
10 − 2 = 8 years.
Two of them are also divisible by 400.
10 − 2 = 8 years.
Page 91
Book Question
In 1000 calendar years, how many years are divisible by 4 but not divisible by 100 and 400?
There are:
1000 ÷ 4 = 250 multiples of 4.
10 multiples of 100 must be removed.
250 − 10 = 240 years.
1000 ÷ 4 = 250 multiples of 4.
10 multiples of 100 must be removed.
250 − 10 = 240 years.
Page 91
Book Question
The rest of the years are 1000 – (2 + 8 + 240) = 750.
2 + 8 + 240 = 250.
1000 − 250 = 750 years.
1000 − 250 = 750 years.
Page 92
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?
Step 1: Count leap years.
Multiples of 4:
10000 ÷ 4 = 2500
Multiples of 100:
10000 ÷ 100 = 100
Multiples of 400:
10000 ÷ 400 = 25
Therefore:
2500 − 100 + 25 = 2425 leap years.
Step 2: Calendar days.
10000 × 365 + 2425
= 3,650,000 + 2,425
= 3,652,425 days.
Step 3: Actual days.
10000 × 365.2422
= 3,652,422 days.
Step 4: Difference.
3,652,425 − 3,652,422
= 3 days.
The difference is small, but over a very long period it accumulates. A mathematical correction could omit an occasional leap day in a much longer cycle. The exact calendar system used in practice is decided by calendar rules rather than by this classroom calculation.
Multiples of 4:
10000 ÷ 4 = 2500
Multiples of 100:
10000 ÷ 100 = 100
Multiples of 400:
10000 ÷ 400 = 25
Therefore:
2500 − 100 + 25 = 2425 leap years.
Step 2: Calendar days.
10000 × 365 + 2425
= 3,650,000 + 2,425
= 3,652,425 days.
Step 3: Actual days.
10000 × 365.2422
= 3,652,422 days.
Step 4: Difference.
3,652,425 − 3,652,422
= 3 days.
The difference is small, but over a very long period it accumulates. A mathematical correction could omit an occasional leap day in a much longer cycle. The exact calendar system used in practice is decided by calendar rules rather than by this classroom calculation.
Page 92
Try This
Do you wonder how people figured out that the Earth completes one revolution around the Sun in exactly 364.2422 days?
Astronomers observed the repeated positions of the Sun and stars,
seasonal changes and the length of the year over very long periods.
By recording these observations and comparing them over many years,
they could make increasingly accurate estimates of the length of the
solar year.
The value 365.2422 days is an approximate value used in this chapter for mathematical calculations.
The value 365.2422 days is an approximate value used in this chapter for mathematical calculations.
Page 92
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.
Traditional Indian calendar systems used careful observations and
astronomical calculations of solar and lunar cycles. Different regional
calendars used solar, lunar or lunisolar methods.
Astronomical texts and traditions used observations of the Sun, Moon, stars and seasons to keep calendar dates connected with astronomical events.
For a school project, students can investigate the Saka calendar, regional solar calendars and traditional astronomical works such as the Surya Siddhanta.
Astronomical texts and traditions used observations of the Sun, Moon, stars and seasons to keep calendar dates connected with astronomical events.
For a school project, students can investigate the Saka calendar, regional solar calendars and traditional astronomical works such as the Surya Siddhanta.
Page 93
Figure it Out — 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?
If we compare price per gram:
Chikki:
₹70.5 ÷ 210 ≈ ₹0.336 per gram
Chips:
₹33.25 ÷ 110 ≈ ₹0.302 per gram
Therefore, the potato chips are cheaper per gram.
Chikki:
₹70.5 ÷ 210 ≈ ₹0.336 per gram
Chips:
₹33.25 ÷ 110 ≈ ₹0.302 per gram
Therefore, the potato chips are cheaper per gram.
Page 93
Figure it Out — 2
Write the decimal number at the arrow mark:
3.1 3.2
2.15 2.17
3.1 3.2
2.15 2.17
First Number Line
The interval 3.1 to 3.2 is divided into 10 equal parts.
Each part is 0.01.
The arrow is at the sixth mark after 3.1.
3.1 + 0.06 = 3.16.
Second Number Line
The interval 2.15 to 2.17 is 0.02.
Each of 10 equal divisions is:
0.02 ÷ 10 = 0.002.
Six divisions after 2.15:
2.15 + 6 × 0.002 = 2.162.
Page 93
Figure it Out — 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?
Cost price:
3 × ₹30 = ₹90
Selling price:
35 × ₹5 = ₹175
Profit:
₹175 − ₹90 = ₹85.
3 × ₹30 = ₹90
Selling price:
35 × ₹5 = ₹175
Profit:
₹175 − ₹90 = ₹85.
Page 93
Figure it Out — 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?
80 books would require:
80 × 2.5 = 200 cm.
But the shelf is only 160 cm.
Number of books that fit:
160 ÷ 2.5 = 64 books.
Space used:
64 × 2.5 = 160 cm.
Space left = 0 cm.
So, 64 books can be placed and 16 books remain outside.
80 × 2.5 = 200 cm.
But the shelf is only 160 cm.
Number of books that fit:
160 ÷ 2.5 = 64 books.
Space used:
64 × 2.5 = 160 cm.
Space left = 0 cm.
So, 64 books can be placed and 16 books remain outside.
Page 94
Figure it Out — 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
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
5.5 km = 5500 m
35 cm = 0.35 m
14.5 cm = 145 mm
68 g = 0.068 kg
9.02 m = 9020 mm
125.5 ml = 0.1255 l
35 cm = 0.35 m
14.5 cm = 145 mm
68 g = 0.068 kg
9.02 m = 9020 mm
125.5 ml = 0.1255 l
Page 94
Figure it Out — 6
The following problem was set by Sridharacharya in his book, Patiganita. “6 1/4 is divided by 2 1/2, and 60 1/4 is divided by 3 1/2. Tell the quotients separately.” Can you try to solve it by converting the fractions into decimals?
Convert mixed numbers into decimals:
6 1/4 = 6.25
2 1/2 = 2.5
6.25 ÷ 2.5 = 2.5.
Next:
60 1/4 = 60.25
3 1/2 = 3.5
60.25 ÷ 3.5
= 6025 ÷ 350
= 241/14
= 17.214285714…
= 17 3/14.
Therefore the two quotients are:
2.5 and 17 3/14.
6 1/4 = 6.25
2 1/2 = 2.5
6.25 ÷ 2.5 = 2.5.
Next:
60 1/4 = 60.25
3 1/2 = 3.5
60.25 ÷ 3.5
= 6025 ÷ 350
= 241/14
= 17.214285714…
= 17 3/14.
Therefore the two quotients are:
2.5 and 17 3/14.
Page 94
Figure it Out — 7
Fill the boxes in at least 2 different ways:
(a) □ × □ = 2.4
(b) □ × □ = 14.5
(a) □ × □ = 2.4
(b) □ × □ = 14.5
(a)
1.2 × 2 = 2.4
0.6 × 4 = 2.4
(b)
2.9 × 5 = 14.5
1.45 × 10 = 14.5
Other correct answers are also possible.
1.2 × 2 = 2.4
0.6 × 4 = 2.4
(b)
2.9 × 5 = 14.5
1.45 × 10 = 14.5
Other correct answers are also possible.
Page 94
Figure it Out — 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
(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
(a) 75.6 ÷ 3.6 = 21
(b) 7.56 ÷ 0.36 = 21
(c) 756 ÷ 0.36 = 2100
(d) 75.6 ÷ 360 = 0.21
(e) 7560 ÷ 3.6 = 2100
(f) 7.56 ÷ 0.36 = 21
(b) 7.56 ÷ 0.36 = 21
(c) 756 ÷ 0.36 = 2100
(d) 75.6 ÷ 360 = 0.21
(e) 7560 ÷ 3.6 = 2100
(f) 7.56 ÷ 0.36 = 21
Page 94
Figure it Out — 9
Find the missing cells if each cell represents a÷b:
b ↓ a → 1517, 151.7, 15.17, 1.517, 15170
37
3.7
0.37
0.037
370
b ↓ a → 1517, 151.7, 15.17, 1.517, 15170
37
3.7
0.37
0.037
370
Each cell = a ÷ b. The completed table is:
| b ↓ a → | 1517 | 151.7 | 15.17 | 1.517 | 15170 |
|---|---|---|---|---|---|
| 37 | 41 | 4.1 | 0.41 | 0.041 | 410 |
| 3.7 | 410 | 41 | 4.1 | 0.41 | 4100 |
| 0.37 | 4100 | 410 | 41 | 4.1 | 41000 |
| 0.037 | 41000 | 4100 | 410 | 41 | 410000 |
| 370 | 4.1 | 0.41 | 0.041 | 0.0041 | 41 |
Example:
1517 ÷ 37 = 41.
If the divisor becomes 10 times smaller, the quotient becomes 10 times
larger.
Page 94
Figure it Out — 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
(a) maximum product
(b) minimum product
(c) product greater than 150
(d) product nearest to 100
(e) product nearest to 5
The box pattern is:
□ □ . □ × □ . □
Each digit 2, 4, 5, 8 and 0 is used once.
(a) maximum product
82.0 × 5.4 = 442.8
(b) minimum product
45.8 × 0.2 = 9.16
(c) product greater than 150
54.0 × 2.8 = 151.2
151.2 > 150
151.2 > 150
Other arrangements greater than 150 are also possible.
(d) product nearest to 100
20.5 × 4.8 = 98.4
Difference from 100 = 1.6
Difference from 100 = 1.6
(e) product nearest to 5
45.8 × 0.2 = 9.16
Difference from 5 = 4.16
Difference from 5 = 4.16
The smallest possible product is 9.16, so it is also the product closest to 5.
Page 94
Figure it Out — 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
(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
Approximate values:
| Expression | Approximate value |
|---|---|
| (a) 245.05 × 0.942368 | 230.9273 |
| (b) 245.05 × 7.9682 | 1952.6074 |
| (c) 245.05 ÷ 7.9682 | 30.7535 |
| (d) 245.05 ÷ 0.942368 | 260.0364 |
| (e) 245.05 | 245.05 |
| (f) 7.9682 | 7.9682 |
Therefore, increasing order is:
(f) < (c) < (a) < (e) < (d) < (b)
SUMMARY
- In this chapter, we learnt procedures to perform decimal multiplication and division.
- For decimal multiplication, we first multiply the multiplier and multiplicand as counting numbers. The number of decimal digits in the product is the total number of decimal digits in the multiplier and multiplicand.
- Division of decimals uses the same procedure, i.e., division using place value (long division), as with counting numbers. The regrouping continues after the Ones place to Tenths, Hundredths, Thousandths, and so on. When the Ones are regrouped to Tenths, a decimal point is placed in the quotient.
- There are decimal divisions where the quotient never ends. After each regrouping and dividing there is always a remainder.
Quick Revision:
Decimal multiplication:
Ignore decimal points → multiply → count decimal places → put decimal point back.
Division by 10, 100, 1000:
Move decimal point left by 1, 2, 3 places.
Decimal divisor:
Multiply dividend and divisor by the same power of 10.
Long division:
Continue regrouping Ones → Tenths → Hundredths → Thousandths.
Decimal multiplication:
Ignore decimal points → multiply → count decimal places → put decimal point back.
Division by 10, 100, 1000:
Move decimal point left by 1, 2, 3 places.
Decimal divisor:
Multiply dividend and divisor by the same power of 10.
Long division:
Continue regrouping Ones → Tenths → Hundredths → Thousandths.
Puzzle: Hidato
In Hidato, a grid of cells is given. It is usually square-shaped, like Sudoku or Kakuro, but it can also include hexagons or any shape that forms a tessellation. It can have inner holes (like a disc), but it is made of only one piece.
Usually, in every Hidato puzzle the lowest and the highest numbers are given on the grid.
Your task is to fill the grid such that there is a continuous path of consecutive numbers from the lowest to the highest number. The next number must be in any one of the adjacent cells, including diagonally adjacent cells.
The grid comes pre-filled with some numbers (with values between the smallest and the highest) to ensure that these puzzles have a single solution.
Try solving the following Hidato puzzles.
Usually, in every Hidato puzzle the lowest and the highest numbers are given on the grid.
Your task is to fill the grid such that there is a continuous path of consecutive numbers from the lowest to the highest number. The next number must be in any one of the adjacent cells, including diagonally adjacent cells.
The grid comes pre-filled with some numbers (with values between the smallest and the highest) to ensure that these puzzles have a single solution.
Try solving the following Hidato puzzles.
Easy Explanation
In Hidato, every consecutive number must touch the next number. Touching can be horizontal, vertical or diagonal.
For example, 1 can be connected to 2, then 2 to 3, then 3 to 4,
and so on. The path must continue without breaking until the highest
number.
Textbook Puzzle Solution
The supplied textbook page gives the completed solution path. The solution numbers are:
1 → 2 → 3 → 4 → 5 → 6 → 7 → 8 → 9 → 10 → 11 → 12 → 13 → 14 → 15
→ 16 → 17 → 18 → 19 → 20 → 21 → 22 → 23 → 24 → 25 → 26 → 27
→ 28 → 29 → 30 → 31 → 32 → 33 → 34 → 35 → 36 → 37 → 38 → 39 → 40
The important idea is that every number must be adjacent to the next number, including diagonal adjacency.
Try solving the following Hidato puzzles.
How to solve
- Find the smallest number.
- Find the highest number.
- Join consecutive numbers.
- Remember that diagonal cells also count as adjacent.
- Use the already-filled numbers as clues.
- If one path becomes impossible, go back and try another adjacent cell.
The later practice grids are graphical puzzles in the textbook. Their exact cell positions are needed to reproduce a complete filled grid. The method above is the correct method for solving them.
Chapter 4 Final Revision Sheet
| Situation | Remember |
|---|---|
| Multiply decimals | Multiply as whole numbers, then count total decimal places. |
| Divide by 10 | Move decimal 1 place left. |
| Divide by 100 | Move decimal 2 places left. |
| Divide by 1000 | Move decimal 3 places left. |
| Decimal divisor | Multiply both dividend and divisor by the same power of 10. |
| Long division | Regroup Ones into Tenths, then Hundredths, then Thousandths. |
| Repeating decimal | The quotient may continue forever with a repeating pattern. |
| Divisor > 1 | For positive numbers, quotient is smaller than dividend. |
| Divisor between 0 and 1 | For positive numbers, quotient is greater than dividend. |
| Leap year | Century years need the 400-year rule. |
Prepare for your exams with this easy and student-friendly worksheet based on Class 7 Maths Chapter 4 from Ganita Prakash Part 2. Practice all the questions carefully to revise the chapter, strengthen your concepts, and improve your mathematical understanding.
CLASS 7 • MATHEMATICS PART 2
Class 7 Maths Part 2 Chapter 4 Worksheet
Practice worksheet for Class 7 Mathematics Part 2 Chapter 4 with important questions, extra practice and answer key.
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