Another Peek Beyond the Point Class 7 Free Notes and Mind Map (Free PDF Download)

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This chapter teaches you how to multiply and divide decimals using simple procedures that extend from counting numbers. You will learn rules, patterns, and real-life applications of decimal operations, points including non-terminating decimals and the Gregorian calendar system.

A Quick Recap of Decimals

Understanding Decimal Place Value

  • Decimals are the natural extension of the Indian place value system to represent decimal fractions and their sums.
  • Place values in decimals: Tens, Units (Ones), Tenths, Hundredths, Thousandths.
  • Example: 27.53 refers to a quantity that has:
    • 2 Tens
    • 7 Units (Ones)
    • 5 Tenths
    • 3 Hundredths

Jonali and Pallabi’s Game

Question: Jonali says a fraction and Pallabi gives the equivalent decimal. Write Pallabi’s answer in the blank spaces.

FractionsDecimals
3/100.3
4/1000.04
67/10000.067
457/1004.57
71/1000.71
43/1000.43
9/1000.09

Question: Jonali goes to the market to buy spices. She purchases 50g of Cinnamon, 100 g of Cumin seeds, 25 g of Cardamom and 250g of Pepper. Express each of the quantities in kilograms by writing them in terms of fractions as well as decimals.

Solution:

  • 50g = 50/1000 kg = 0.050 kg or 0.05 kg
  • 100g = 100/1000 kg = 0.100 kg or 0.1 kg
  • 25g = 25/1000 kg = 0.025 kg
  • 250g = 250/1000 kg = 0.250 kg or 0.25 kg

Expanding Fractions into Decimals

Question: Write the following fractions as a sum of fractions and also as decimals:

FractionExpanding the NumeratorSumDecimals
254/1000200/1000 + 50/1000 + 4/1000 = 2/10 + 5/100 + 4/10000.2 + 0.05 + 0.0040.254
847/100008000/10000 + 400/10000 + 40/10000 + 7/10000 = 8/10 + 4/100 + 4/1000 + 7/100000.8 + 0.04 + 0.004 + 0.00070.8447
173/100100/100 + 70/100 + 3/100 = 1 + 7/10 + 3/1001 + 0.7 + 0.031.73
23/100020/1000 + 3/1000 = 2/100 + 3/10000.02 + 0.0030.023

Rule to Divide by 10, 100, 1000

Question: 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, 24/100 or 678/1000? Look for a pattern.

Rule for Division by Powers of 10:

Let us consider the example 123 ÷ 10.

Step 1: Write the dividend as it is and place a decimal point at the end.
        123.

Step 2: Count the number of zeroes in the divisor.
        10 → 1 zero

Step 3: Move the decimal point from Step 1 left by the same number of 
        places as the count from Step 2. Add zeroes in front if needed.
        123. → 12.3

Examples:

  • 24 ÷ 100 = 0.24
  • 678 ÷ 1000 = 0.678
  • 12 ÷ 1000 = 0.012
  • 12345 ÷ 1000 = 12.345

Decimal Multiplication

Example 1: Arshad’s Pen Purchase

Question: 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?

Solution:

  • We have to multiply 9.5 by 5, which is the same as adding 9.5 five times.
  • 9.5 × 5 = 9.5 + 9.5 + 9.5 + 9.5 + 9.5 = 47.5

Converting to fractions:

  • 9.5 = 95/10 and 5 = 5/1
  • Cost of 5 pens = (5/1) × (95/10) = (5 × 95)/(1 × 10) = 475/10 = 47.5
  • The cost of 5 pens is ₹47.5.

Example 2: Car Petrol Distance

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

Solution:

  • We have to multiply 12.5 by 7.5.
  • Distance covered = 12.5 × 7.5 = (125/10) × (75/10) = (125 × 75)/(10 × 10) = 9375/100 = 93.75 km

Questions to Think About:

  • Can the product of two decimals be a natural number? Yes, e.g., 2.5 × 4 = 10.
  • Can the product of a decimal and a natural number be a natural number? Yes, e.g., 0.5 × 2 = 1.

Example 3: Ajay’s School Walk

Question: 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.

Solution:

  • Each way between school and home, Ajay walks 827 metres, i.e., 0.827 km.
  • In a day he walks: 0.827 × 2 = (827/1000) × 2 = (827 × 2)/1000 = 1654/1000 = 1.654 km
  • He goes to school 6 days a week. So, in a week, he walks: 1.654 × 6 = (1654/1000) × 6 = (1654 × 6)/1000 = 9924/1000 = 9.924 km

Example 4: Area of Rectangle

Question: Find the area of the given rectangle with dimensions 5.7 cm by 13.3 cm.

Solution:

  • Area of the rectangle = 5.7 × 13.3 = (57/10) × (133/10) = (57 × 133)/(10 × 10) = 7581/100 = 75.81 sq cm

Pattern in Decimal Multiplication

Observation Table:

ExampleMultiplier decimal placesMultiplicand decimal placesProduct decimal places
9.5 × 5 = 47.5101
12.5 × 7.5 = 93.75112
1.64 × 6 = 9.84202
5.7 × 13.35 = 76.095123

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

Solution:

  • 5.96 has 2 decimal places and 24.8 has 1 decimal place.
  • Total decimal places = 2 + 1 = 3 decimal places.
  • 596 × 248 = 147808
  • Therefore, 5.96 × 24.8 = 147.808

Rule to Multiply Two Decimals

Multiplication Rule:

  • Multiply the two numbers obtained by removing the decimal point.
  • Count the total number of decimal digits in both the multiplier and multiplicand.
  • Place the decimal point in the product so that it has that many decimal places.
To evaluate 5.96 × 24.8:
596 × 248 = 147808

5.96 × 24.8 has 2 + 1 = 3 decimal places
Answer: 147.808 (3 decimal places)

Example 5: Using the Rule

Question: Let us use the above rule to find the product of 5.8 and 1.24.

Solution:

  • First multiply 58 and 124.
  • The product is 7192.
  • Sum of decimal places: 5.8 (1 decimal place) + 1.24 (2 decimal places) = 3 decimal places.
  • So, the product is 7.192.

Is the Product Always Greater?

Observations:

  • In 2.25 × 8 = 18, the product (18) is greater than both 2.25 and 8.
  • In 0.25 × 8 = 2, the product (2) is greater than 0.25 but less than 8.
  • In 0.25 × 0.8 = 0.2, the product (0.2) is less than both 0.25 and 0.8.

Relationship Table:

SituationMultiplication ExampleRelationship
Both numbers > 13.4 × 6.5 = 22.1Product is greater than both numbers
Both numbers between 0 and 10.75 × 0.4 = 0.3Product is less than both numbers
One between 0 and 1, one > 10.75 × 5 = 3.75Product is less than the number > 1 and greater than the number between 0 and 1

Figure it Out

Question 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

  • Solution: 6 × 4 tenths = 6 × 0.4 = 2.4 = 24 tenths

(b) 7 × 0.3

  • Solution: 7 × 0.3 = 2.1 = 21 tenths

(c) 9 × 5 hundredths

  • Solution: 9 × 5 hundredths = 9 × 0.05 = 0.45 = 45 hundredths

Question 2: Find the products:

(a) 27.34 × 6

  • Solution: 2734 × 6 = 16404, with 2 decimal places = 164.04

(b) 4.23 × 3.7

  • Solution: 423 × 37 = 15651, with 2 + 1 = 3 decimal places = 15.651

(c) 0.432 × 0.23

  • Solution: 432 × 23 = 9936, with 3 + 2 = 5 decimal places = 0.09936

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

Solution:

  • Cloth needed = 1.65 × 3 = 4.95 m

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

Solution:

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

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

Solution:

  • Total height = 36 × 1.45 = 52.2 mm = 5.22 cm

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

Solution:

  • Price = 56.50 × 2.250 = ₹127.125
  • Yes, we can write 56.50 as 56.5 and 2.250 as 2.25 because trailing zeroes after the decimal don’t change the value.
  • 56.5 × 2.25 = 127.125 (same product).

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

Solution:

  • Cost price of 50 notebooks = 50 × 23.6 = ₹1180
  • Selling price of 50 notebooks = 50 × 30 = ₹1500
  • Profit = 1500 – 1180 = ₹320

Question 8: Given that 18 × 12 = 216, find the products:

(a) 18 × 1.2

  • Solution: 18 × 1.2 = 21.6 (1 decimal place)

(b) 18 × 0.12

  • Solution: 18 × 0.12 = 2.16 (2 decimal places)

(c) 1.8 × 1.2

  • Solution: 1.8 × 1.2 = 2.16 (1 + 1 = 2 decimal places)

(d) 0.18 × 0.12

  • Solution: 0.18 × 0.12 = 0.0216 (2 + 2 = 4 decimal places)

(e) 0.018 × 0.012

  • Solution: 0.018 × 0.012 = 0.000216 (3 + 3 = 6 decimal places)

(f) 1.8 × 12

  • Solution: 1.8 × 12 = 21.6 (1 decimal place)

In which of the cases above is the product less than 1?

  • Solution: Cases (d) and (e) have products less than 1.

Question 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

  • Solution: Product = 4.2, not less than 1.

(b) 0.7 × 0.6

  • Solution: Both numbers are between 0 and 1, so product will be less than 1.

(c) 0.7 × 6

  • Solution: Product = 4.2, not less than 1.

(d) 0.07 × 0.06

  • Solution: Both numbers are between 0 and 1, so product will be less than 1.

Answer: (b) and (d) have products less than 1.

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

Number× 10× 100× 1000
5.7575705700
23.02230.2230223020
0.929.292920
0.3063.0630.6306
24.67246.7246724670

Decimal Division

Example 6: Anuja’s Ribbon Cutting

Question: 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?

Solution:

  • Since there are 10 pieces, we can find the length of each piece by dividing 3.9 by 10.
  • 3.9 ÷ 10 = ?
  • Convert 3.9 into fraction: 3.9 = 39/10
  • 3.9 ÷ 10 = (39/10) ÷ 10 = (39/10) × (1/10) = 39/100 = 0.39 m

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

Solution:

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

Question: What is 0.039 m in centimetres and millimetres?

Solution:

  • 0.039 m = 3.9 cm = 39 mm

Rule for Dividing by Powers of 10

Division Rule:

  • When we divide a decimal by 1, 10, 100, 1000, and so on, we can just move the decimal point to the left by as many places as there are zeroes in the divisor!

Table:

Decimal÷ 10÷ 100÷ 1000÷ 10000
18.71.870.1870.01870.00187
21.12.110.2110.02110.00211
0.130.0130.00130.000130.000013
2.1460.21460.021460.0021460.0002146
0.00580.000580.0000580.00000580.00000058

Example 7: Neenu’s Ribbon Sharing

Question: Neenu has 29 metres of red ribbon and wants to share it equally with Anu. What is the length of ribbon that each of them will get?

Solution:

  • Since the ribbon needs to be divided into two equal parts, each girl will get 29 ÷ 2 metres.
  • 29 ÷ 2 = 29/2
  • Convert 1/2 to decimal: 1/2 = (1 × 5)/(2 × 5) = 5/10 = 0.5
  • 29 ÷ 2 = 14 remainder 1 = 14 + 1/2 = 14 + 0.5 = 14.5 m

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

Solution:

  • Each will get 29 ÷ 4 = 29/4 m.
  • Is 4 a factor of 10? No. Is it a factor of 100? Yes. 4 × 25 = 100.
  • 29/4 = (29 × 25)/(4 × 25) = 725/100 = 7.25 m

Division Using Place Value (Long Division)

Example 8: Find the value of 1324 ÷ 4

Solution:

Step 1: Divide 1324 into 4 equal parts.
        1 Thousand + 3 Hundreds + 2 Tens + 4 Ones

Step 2: 1 Thousand ÷ 4 → Not possible without regrouping.
        Regroup 1 Thousand into 10 Hundreds.
        10 Hundreds + 3 Hundreds = 13 Hundreds.

Step 3: 13 Hundreds ÷ 4 → Each part gets 3 Hundreds, 
        and 1 Hundred remains.

Step 4: Regroup 1 Hundred into 10 Tens.
        10 Tens + 2 Tens = 12 Tens.

Step 5: 12 Tens ÷ 4 → Each part gets 3 Tens.

Step 6: 4 Ones ÷ 4 → Each part gets 1 Ones.

Answer: 1324 ÷ 4 = 331

Division with a Decimal Quotient

Example: Find the value of 1325 ÷ 4

Solution:

Th H T O
4) 1325 (331.25
  -12
   12
  -12
   05    (5 Ones remain)
   -4
    10   (Regroup 1 Ones as 10 Tenths)
    -8
     20  (Regroup 2 Tenths as 20 Hundredths)
    -20
      0

Answer: 1325 ÷ 4 = 331.25

Verification:

  • 1325/4 = (1325 × 25)/(4 × 25) = 33125/100 = 331.25 ✓

Example 9: Find the value of 237 ÷ 8

Solution:

H T O  Tenths Hundredths Thousandths
8) 237 (29.625
  -16
   77   (Regroup 2 Hundreds as 20 Tens, 20 + 3 = 23 Tens)
  -72   (23 Tens ÷ 8 = 2 Tens remainder 7 Tens)
    5   (Regroup 7 Tens as 70 Ones, 70 + 7 = 77 Ones)
   -72  (77 Ones ÷ 8 = 9 Ones remainder 5 Ones)
    50  (Regroup 5 Ones as 50 Tenths)
   -48  (50 Tenths ÷ 8 = 6 Tenths remainder 2 Tenths)
    20  (Regroup 2 Tenths as 20 Hundredths)
   -16  (20 Hundredths ÷ 8 = 2 Hundredths remainder 4 Hundredths)
    40  (Regroup 4 Hundredths as 40 Thousandths)
   -40  (40 Thousandths ÷ 8 = 5 Thousandths)
     0

Answer: 237 ÷ 8 = 29.625

Important: When we regroup Ones into Tenths, we need to place a decimal point in the quotient.

Division with a Decimal Dividend

Example 10: Shopkeeper’s Sugar Packing

Question: 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?

Solution:

O  Tenths Hundredths Thousandths
4) 9.5 (2.375
  -8
   15   (Regroup 1 Ones as 10 Tenths, 10 + 5 = 15 Tenths)
  -12   (15 Tenths ÷ 4 = 3 Tenths remainder 3 Tenths)
   30   (Regroup 3 Tenths as 30 Hundredths)
  -28   (30 Hundredths ÷ 4 = 7 Hundredths remainder 2 Hundredths)
   20   (Regroup 2 Hundredths as 20 Thousandths)
  -20   (20 Thousandths ÷ 4 = 5 Thousandths)
    0

Answer: Each bag of sugar weighs 2.375 kg

Practice Questions

Question 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).

Question 2: Choose the correct answer:

(a) 132 ÷ 4 =

  • Solution: 132 ÷ 4 = 33

(b) 13.2 ÷ 4 =

  • Solution: 13.2 ÷ 4 = 3.3

(c) 1.32 ÷ 4 =

  • Solution: 1.32 ÷ 4 = 0.33

(d) 0.132 ÷ 4 =

  • Solution: 0.132 ÷ 4 = 0.033

Question 3: What is the quotient?

(a) 1526 ÷ 4 =

  • Solution: 1526 ÷ 4 = 381.5

(b) 3567 ÷ 8 =

  • Solution: (i) 445.875

(c) 1217 ÷ 2 =

  • Solution: 1217 ÷ 2 = 608.5

Question 4: What is the quotient?

(a) 126 ÷ 8 =

  • Solution: 126 ÷ 8 = 15.75

(b) 12.6 ÷ 8 =

  • Solution: 12.6 ÷ 8 = 1.575

(c) 1.26 ÷ 8 =

  • Solution: 1.26 ÷ 8 = 0.1575

(d) 0.126 ÷ 8 =

  • Solution: 0.126 ÷ 8 = 0.01575

(e) 0.0126 ÷ 8 =

  • Solution: 0.0126 ÷ 8 = 0.001575

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

Solution:

O  Tenths Hundredths Thousandths
5) 0.06 (0.012
  -0
   00   (0 Tenths)
  -00
   06   (6 Hundredths ÷ 5 = 1 Hundredths remainder 1 Hundredths)
   -5
   10   (Regroup 1 Hundredths as 10 Thousandths)
  -10   (10 Thousandths ÷ 5 = 2 Thousandths)
    0

Answer: 0.06 ÷ 5 = 0.012

Division with a Decimal Divisor

Example 12: Ravi’s Scooter Trip

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

Solution:

  • Average speed = Distance ÷ Time = 126 ÷ 2.5
  • Convert divisor to fraction: 126 ÷ 2.5 = 126 ÷ (25/10) = 126 × (10/25) = 1260/25
  • Using long division: 1260 ÷ 25 = 50.4 km/hour

Example 13: Find 4.68 ÷ 1.3

Solution:

  • 4.68 ÷ 1.3 = 4.68 ÷ (13/10) = 4.68 × (10/13) = 46.8/13 = 3.6

Question: Now, what about 4.68 ÷ 0.13?

Solution:

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

Rule for Decimal Division:

  • When the divisor is a decimal, convert it into a counting number by multiplying it by 10, 100, 1000, and so on.
  • Also multiply the dividend by the same number.
  • Example: 4.68 ÷ 0.13 = (4.68 × 100)/(0.13 × 100) = 468/13

Does This Ever End?

Question: Can you calculate 10 ÷ 3? Try dividing using long division.

Solution:

3) 10 (3.333...
  -9
   10  (Regroup 1 Ones as 10 Tenths)
   -9  (10 Tenths ÷ 3 = 3 Tenths remainder 1 Tenths)
   10  (Regroup 1 Tenths as 10 Hundredths)
   -9  (10 Hundredths ÷ 3 = 3 Hundredths remainder 1 Hundredths)
   10  (This never ends!)
   -9
    1

Answer: 10 ÷ 3 = 3.333... (never ending)
  • In long division, at each step we get a remainder of 1. So, the process will never end!
  • 10 ÷ 3 cannot be expressed using a finite number of digits in the decimal form.

Question: Can you find the quotients of 10 ÷ 9, and 100 ÷ 9?

Solution:

  • 10 ÷ 9 = 1.111… (never ending)
  • 100 ÷ 9 = 11.111… (never ending)

Question: Now divide 1 by 7 (1 ÷ 7). Will this end?

Solution:

7) 1 (0.142857142857...
  -0
   10
   -7
    30
   -28
    20
   -14
    60
   -56
    40
   -35
    50
   -49
     1  (Remainder repeats!)
  • Remainders repeat in a cycle: 1 → 3 → 2 → 6 → 4 → 5 → 1 → 3…
  • Not only do the remainders repeat in a cycle, the digits of the quotient also repeat in a cycle!
  • 1 ÷ 7 = 0.142857142857… (repeating pattern)

A Magic Number: 142857

Question: 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?

Solution:

  • 142857 × 1 = 142857
  • 142857 × 2 = 285714
  • 142857 × 3 = 428571
  • 142857 × 4 = 571428
  • 142857 × 5 = 714285
  • 142857 × 6 = 857142
  • Observation: You get the same number back, but with the digits cycled around!

Question: Multiply 142857 by 7. What do you observe?

Solution:

  • 142857 × 7 = 999999

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

Solution:

  • 1 ÷ 17 = 0.0588235294117647… (repeating block)
  • The repeating block is 0588235294117647.

Interesting Fact: Are there infinitely many such “cyclic” numbers? In 1927, the Austrian mathematician Emil Artin conjectured that there must be infinitely many such numbers. However, even today, nearly a century later, this conjecture remains unsolved!

Dividend, Divisor, and Quotient

Observation:

  • When we divide two counting numbers, the quotient is always less than the dividend. For example, 128 ÷ 4 = 32, and 32 (quotient) < 128 (dividend).
  • But what happens when we divide 128 by 0.4?
  • 128 ÷ 0.4 = 320
  • The quotient is greater than the dividend.

Question: Will the quotient be always greater than the dividend when the divisor is a decimal? Try it out with different values of the divisor.

Question: 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.

Figure it Out

Question 1: Express the following fractions in decimal form:

(a) 13/25

  • Solution: 13/25 = (13 × 4)/(25 × 4) = 52/100 = 0.52

(b) 4/50

  • Solution: 4/50 = (4 × 2)/(50 × 2) = 8/100 = 0.08

(c) 5/8

  • Solution: 5/8 = (5 × 125)/(8 × 125) = 625/1000 = 0.625

(d) 43/4

  • Solution: 43/4 = (43 × 25)/(4 × 25) = 1075/100 = 10.75

Question 2: Find the quotients:

(a) 24.86 ÷ 1.2

  • Solution: (24.86 × 10)/(1.2 × 10) = 248.6/12 = 20.716…

(b) 5.728 ÷ 1.52

  • Solution: (5.728 × 100)/(1.52 × 100) = 572.8/152 = 3.768…

Question 3: Evaluate the following using the information 156 × 12 = 1872.

(a) 15.6 × 1.2 =

  • Solution: 1 + 1 = 2 decimal places, so 18.72

(b) 187.2 ÷ 1.2 =

  • Solution: (187.2 × 10)/(1.2 × 10) = 1872/12 = 156

(c) 18.72 ÷ 15.6 =

  • Solution: (18.72 × 100)/(15.6 × 100) = 1872/1560 = 1.2

(d) 0.156 × 0.12 =

  • Solution: 3 + 2 = 5 decimal places, so 156 × 12 = 1872, answer = 0.01872

Question 4: Evaluate the following:

(a) 25 ÷ ___ = 0.025

  • Solution: 25 ÷ 1000 = 0.025, so answer = 1000

(b) 25 ÷ ___ = 250

  • Solution: 25 ÷ 0.1 = 250, so answer = 0.1

(c) 25 ÷ ___ = 2.5

  • Solution: 25 ÷ 10 = 2.5, so answer = 10

(d) 25 ÷ 10 = 25 × ___

  • Solution: 25 ÷ 10 = 25 × 0.1, so answer = 0.1

(e) 250 ÷ 10 = 25 × ___

  • Solution: 250 ÷ 10 = 25 × 1, so answer = 1

(f) 25 ÷ 0.01 = 25 × ___

  • Solution: 25 ÷ 0.01 = 25 × 100, so answer = 100

Question 5: Find the quotient:

(a) 2.46 ÷ 1.5 =

  • Solution: (2.46 × 10)/(1.5 × 10) = 24.6/15 = 1.64

(b) 2.46 ÷ 0.15 =

  • Solution: (2.46 × 100)/(0.15 × 100) = 246/15 = 16.4

(c) 2.46 ÷ 0.015 =

  • Solution: (2.46 × 1000)/(0.015 × 1000) = 2460/15 = 164

Is the quotient obtained in 24.6 ÷ 1.5 the same as the quotient obtained in 2.46 ÷ 0.15?

  • Solution: No. 24.6 ÷ 1.5 = 16.4, and 2.46 ÷ 0.15 = 16.4. Yes, they are the same!

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

Solution:

  • Length of each piece = 4 ÷ 5 = 4/5 = (4 × 2)/(5 × 2) = 8/10 = 0.8 m

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

Solution:

  • Length of each side = 208.8 ÷ 12 = 17.4 cm

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

Solution:

  • Juice per person = 3 ÷ 8 = 3/8 = (3 × 125)/(8 × 125) = 375/1000 = 0.375 litres = 375 millilitres

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

Solution:

  • Distance per litre = 234.45 ÷ 12.6 = (234.45 × 10)/(12.6 × 10) = 2344.5/126 = 18.6 km/litre

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

Solution:

  • Flour per student = 13.5 ÷ 15 = 0.9 kg

Pattern Observation

Table:

FractionDecimal
1/2 = 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)0.03125
1/5 = 1/50.2
1/(5×5)0.04
1/(5×5×5)0.008
1/(5×5×5×5)0.0016
1/(5×5×5×5×5)0.00032

Question: What pattern do you observe? Why are 2 and 5 related in this way?

Solution:

  • When the denominator has only powers of 2 or powers of 5, the fraction can be easily expressed as a terminating decimal.
  • This is because 2 and 5 are factors of 10 (2 × 5 = 10), and our decimal system is based on powers of 10.

Look Before You Leap!

Earth’s Revolution and the Calendar

  • Did you know that it takes the Earth 365.2422 days to go around the Sun and not 365 days?
  • For our convenience, we consider 365 days as a year in a calendar (Gregorian calendar).
  • This means that, after one calendar year or 365 days, the Earth still needs 0.2422 more days to complete one full revolution around the Sun.

Question: What happens after 100 such calendar years?

Solution:

  • Using decimal multiplication: 0.2422 × 100 = 24.22 days.
  • After 100 calendar years, the Earth will need 24.22 more days to complete its 100th revolution around the Sun.

Why Leap Years?

  • If our calendar does not accurately indicate the position of the Earth around the Sun, our seasons and our annual calendar will not match!
  • To correct this problem, the idea of a leap year was introduced.
  • Every fourth year, one additional day is added to the calendar year.

Rule 1: Making an Adjustment

Is the year divisible by 4?
  Yes → The year has 366 days (Leap Year)
  No  → The year has 365 days

Question: Do you know which month has this extra day?

Solution:

  • February has the extra day (29 days instead of 28).

Testing the Adjustment

Question: Let us see how this solution works.

Year 1Year 2Year 3Year 4Year 5Year 6Year 7Year 8
365365365366365365365366
  • Number of days after 4 calendar years = 4 × 365 + 1 = 1461 days.
  • Number of days the Earth needs to make 4 full revolutions = 4 × 365.2422 = 1460.9688 days.

Question: 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?

Solution:

  • Each calendar year has 365 days. In 100 calendar years, the number of days is 100 × 365.
  • Years divisible by 4 have one extra day. There are 25 years divisible by 4 in 100 years (100/4 = 25).
  • Number of days in 100 calendar years = (100 × 365) + (100/4 × 1) = 36,500 + 25 = 36,525 days.
  • Actual number of days for Earth to go around Sun 100 times = 100 × 365.2422 = 36,524.22 days.

Observation:

  • By adding a day every fourth year, after 100 years, the calendar days are more than the actual number of days. We have overcompensated.

Making Another Adjustment

Rule 2: No extra day in the 100th year

textIs the year divisible by 100?
  Yes → The year has 365 days
  No  → Is it divisible by 4?
          Yes → The year has 366 days
          No  → The year has 365 days

Question: Can you write an expression for the number of days in 100 calendar years with this new adjustment?

Solution:

  • There are 25 years divisible by 4 in 100 years.
  • 100 is also divisible by 4, but we have to exclude it. So only 24 years have 366 days, and the rest (76 years) have 365 days.
  • Expression: (24 × 366) + (76 × 365) = 8784 + 27740 = 36,524 days.
  • This is close to 36524.22 days but is it close enough?

Question: What happens after 1000 years with this adjustment?

Solution:

  • Number of calendar days in 1000 years = 36524 × 10 = 365,240 days.
  • Number of days Earth takes to go around Sun 1000 times = 1000 × 365.2422 = 365,242.2 days.
  • Difference = 365,242.2 – 365,240 = 2.2 days.

Making Yet Another Adjustment

Rule 3: Add an extra day every 400th year

Is the year divisible by 400?
  Yes → The year has 366 days
  No  → Is it divisible by 100?
          Yes → The year has 365 days
          No  → Is it divisible by 4?
                  Yes → The year has 366 days
                  No  → The year has 365 days

Question: With this scheme, let us calculate the number of days in 1000 calendar years.

Solution:

  • In 1000 calendar years, how many years are divisible by 400? 2 (400, 800).
  • How many years are divisible by 100 but not by 400? 10 – 2 = 8 (100, 200, 300, 500, 600, 700, 900, 1000).
  • How many years are divisible by 4 but not by 100 and 400? 250 – 10 = 240.
  • The rest of the years = 1000 – (2 + 8 + 240) = 750.
  • Total number of days in 1000 calendar years = (750 × 365) + (240 × 366) + (8 × 365) + (2 × 366)
    = 273,750 + 87,840 + 2,920 + 732 = 365,242 days.
  • Earth needs 365,242.2 days to go around the Sun 1000 times.
  • Difference = 365,242.2 – 365,242 = 0.2 days.

Conclusion:

Figure it Out

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

Solution:

  • Price per gram of peanut chikki = 70.5 ÷ 210 = ₹0.3357 per gram
  • Price per gram of potato chips = 33.25 ÷ 110 = ₹0.3023 per gram
  • Potato chips are cheaper per gram.

Question 2: Write the decimal number at the arrow mark:

  • Arrow between 2.15 and 2.17, so answer could be 2.16 (or any value between 2.15 and 2.17).

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

Solution:

  • Cost price = 3 × 30 = ₹90
  • Selling price = 35 × 5 = ₹175
  • Profit = 175 – 90 = ₹85

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

Solution:

  • Space needed for 80 books = 80 × 2.5 = 200 cm
  • Bookshelf length = 160 cm
  • Number of books that can fit = 160 ÷ 2.5 = 64 books
  • Space left = 160 – (64 × 2.5) = 160 – 160 = 0 cm (no space left)

Question 5: Fill in the following blanks appropriately:

  • 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

Question 6: The following problem was set by Sridharacharya in his book, Patiganita. “6 1/4 is divided by 2, and 601 1/2 is divided by 3 1/4. Tell the quotients separately.” Can you try to solve it by converting the fractions into decimals?

Solution:

  • 6 1/4 = 6.25, divided by 2 = 6.25 ÷ 2 = 3.125
  • 601 1/2 = 601.5, divided by 3 1/4 = 3.25: 601.5 ÷ 3.25 = 185

Question 7: Fill the boxes in at least 2 different ways:

(a) ___ × ___ = 2.4

  • Solution: 1.2 × 2 = 2.4, or 0.6 × 4 = 2.4, or 2.4 × 1 = 2.4

(b) ___ ÷ ___ = 2.4

  • Solution: 4.8 ÷ 2 = 2.4, or 24 ÷ 10 = 2.4, or 7.2 ÷ 3 = 2.4

Question 8: Find the following quotients given that 756 ÷ 36 = 21:

(a) 75.6 ÷ 3.6

  • Solution: Both have 1 decimal place, so quotient = 21

(b) 7.56 ÷ 0.36

  • Solution: Both have 2 decimal places, so quotient = 21

(c) 756 ÷ 0.36

  • Solution: (756 × 100)/(0.36 × 100) = 75600/36 = 2100

(d) 75.6 ÷ 360

  • Solution: 75.6 ÷ 360 = 0.21

(e) 7560 ÷ 3.6

  • Solution: (7560 × 10)/(3.6 × 10) = 75600/36 = 2100

(f) 7.56 ÷ 0.36

  • Solution: (7.56 × 100)/(0.36 × 100) = 756/36 = 21

Question 9: Find the missing cells if each cell represents a ÷ b:

a ÷ b151701517151.715.171.517
37410414.10.410.041
41003.70.370.0370.00370.00037
370414.10.410.0410.0041

Question 10: Using the digits 2, 4, 5, 8, and 0 fill the boxes to get the:

(a) maximum product

  • Solution: 85.4 × 2.0 or 85 × 4.2 = approximately 357 (maximize by placing largest digits before decimal)

(b) minimum product

  • Solution: 0.24 × 0.58 = 0.1392 (minimize by placing smallest non-zero digits with most decimal places)

(c) product greater than 150

  • Solution: 42 × 5.8 = 243.6 (greater than 150)

(d) product nearest to 100

  • Solution: 25 × 4.0 = 100 (exact)

(e) product nearest to 5

  • Solution: 2.5 × 2.0 = 5.0 (exact)

Question 11: Sort the following expressions in increasing order:

(a) 245.05 × 0.942368

  • Value ≈ 230.93

(b) 245.05 × 7.9682

  • Value ≈ 1952.32

(c) 245.05 ÷ 7.9682

  • Value ≈ 30.75

(d) 245.05 ÷ 0.942368

  • Value ≈ 260.02

(e) 245.05

  • Value = 245.05

(f) 7.9682

  • Value = 7.9682

Increasing order: (f) < (c) < (a) < (e) < (d) < (b)

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