Another Peek Beyond The Point

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

The fractions are $\dfrac{3}{10}$, $\dfrac{4}{100}$, $\dfrac{67}{1000}$, $\dfrac{457}{100}$, $\dfrac{71}{100}$, $\dfrac{43}{100}$ and $\dfrac{9}{100}$ (for example, $\dfrac{3}{10}$ = 0.3).

Answer

Pallabi converts each fraction into its equivalent decimal. Since the denominators are 10, 100 and 1000, we write the numerator and place the decimal point so that the number of digits after the point equals the number of zeroes in the denominator:

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.

Answer

We know that 1 kg = 1000 g. So, to convert a weight from grams to kilograms, we divide it by 1000.

Cinnamon: $50 \text{ g} = \dfrac{50}{1000} \text{ kg} = 0.05 \text{ kg}$

Cumin seeds: $100 \text{ g} = \dfrac{100}{1000} \text{ kg} = 0.1 \text{ kg}$

Cardamom: $25 \text{ g} = \dfrac{25}{1000} \text{ kg} = 0.025 \text{ kg}$

Pepper: $250 \text{ g} = \dfrac{250}{1000} \text{ kg} = 0.25 \text{ kg}$

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

Answer

Hence, the decimals are 0.254, 0.0847, 1.73 and 0.023 respectively.

Can the product of two decimals be a natural number?

Answer

Yes, the product of two decimals can be a natural number.

For example, consider 2.5 × 0.4.

Converting the decimals into fractions:

$2.5 \times 0.4 = \dfrac{25}{10} \times \dfrac{4}{10} \\[1em] = \dfrac{25 \times 4}{10 \times 10} \\[1em] = \dfrac{100}{100}\\[1em] = 1$

Here the product 1 is a natural number.

Hence, the product of two decimals can be a natural number.

Can the product of a decimal and a natural number be a natural number?

Answer

Yes, the product of a decimal and a natural number can be a natural number.

For example, consider 0.25 × 8.

Converting the decimal into a fraction:

$0.25 \times 8 = \dfrac{25}{100} \times 8\\[1em] = \dfrac{25 \times 8}{100}\\[1em] = \dfrac{200}{100}\\[1em] = 2$

Here the product 2 is a natural number.

Hence, the product of a decimal and a natural number can be a natural number.

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.

Answer

Let us look at each multiplication and count the digits after the decimal point in the multiplier, the multiplicand and the product, along with the number of zeroes in the denominator.

$9.5 \times 5 = \dfrac{95 \times 5}{10} = \dfrac{475}{10} = 47.5$ — the multiplier has 1 digit and the multiplicand has 0 digits after the decimal point, while the product has 1 digit; the denominator has 1 zero.

$12.5 \times 7.5 = \dfrac{125}{10} \times \dfrac{75}{10} = \dfrac{9375}{100} = 93.75$ — the multiplier has 1 digit and the multiplicand has 1 digit, while the product has 2 digits; the denominator has 2 zeroes.

$1.64 \times 6 = \dfrac{164}{100} \times 6 = \dfrac{984}{100} = 9.84$ — the multiplier has 2 digits and the multiplicand has 0 digits, while the product has 2 digits; the denominator has 2 zeroes.

$5.7 \times 13.35 = \dfrac{57}{10} \times \dfrac{1335}{100} = \dfrac{57 \times 1335}{10 \times 100} = \dfrac{76095}{1000} = 76.095$ — the multiplier has 1 digit and the multiplicand has 2 digits, while the product has 3 digits; the denominator has 3 zeroes.

In every case, the number of digits after the decimal point in the product is equal to the sum of the number of digits after the decimal point in the multiplier and the multiplicand. Moreover, the number of zeroes in the denominator is the same as the number of digits after the decimal point in the product.

Hence, the number of decimal digits in the product = (number of decimal digits in the multiplier) + (number of decimal digits in the multiplicand), which is also equal to the number of zeroes in the denominator.

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

Answer

Yes. On removing the decimal points, 5.96 becomes 596 and 24.8 becomes 248, and we are given that 596 × 248 = 147808.

The number 5.96 has 2 digits after the decimal point and 24.8 has 1 digit after the decimal point, so the product must have 2 + 1 = 3 digits after the decimal point.

Placing the decimal point in 147808 so that there are 3 digits after it:

5.96 × 24.8 = 147.808

Hence, 5.96 × 24.8 = 147.808.

Find the product of 5.8 and 1.24 using the above rule. Verify this by converting the multiplier and multiplicand into fractions.

Answer

Using the rule:

First, we multiply the numbers obtained by removing the decimal points, i.e., 58 and 124:

58 × 124 = 7192

The number 5.8 has 1 digit after the decimal point and 1.24 has 2 digits after the decimal point, so the product has 1 + 2 = 3 digits after the decimal point:

5.8 × 1.24 = 7.192

Verification by converting into fractions:

$5.8 \times 1.24 = \dfrac{58}{10} \times \dfrac{124}{100}\\[1em] = \dfrac{58 \times 124}{10 \times 100}\\[1em] = \dfrac{7192}{1000}\\[1em] = 7.192$

Both methods give the same result.

Hence, 5.8 × 1.24 = 7.192.

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

Answer

(a) 6 × 4 tenths = 24 tenths = 2.4

(b) 7 × 0.3

Here 0.3 = 3 tenths, so:

7 × 3 tenths = 21 tenths = 2.1

(c) 9 × 5 hundredths

9 × 5 hundredths = 45 hundredths = 0.45

Hence, 7 × 0.3 = 2.1 and 9 × 5 hundredths = 0.45.

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

Answer

Converting each decimal into a fraction and multiplying:

(a) 27.34 × 6

$= \dfrac{2734}{100} \times 6 \\[1em] = \dfrac{2734 \times 6}{100} \\[1em] = \dfrac{16404}{100} \\[1em] = 164.04$

Hence, 27.34 × 6 = 164.04

(b) 4.23 × 3.7

$= \dfrac{423}{100} \times \dfrac{37}{10} \\[1em] = \dfrac{423 \times 37}{100 \times 10} \\[1em] = \dfrac{15651}{1000} \\[1em] = 15.651$

Hence, 4.23 × 3.7 = 15.651.

(c) 0.432 × 0.23

$= \dfrac{432}{1000} \times \dfrac{23}{100} \\[1em] = \dfrac{432 \times 23}{1000 \times 100} \\[1em] = \dfrac{9936}{100000} \\[1em] = 0.09936$

Hence, 0.432 × 0.23 = 0.09936.

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

Answer

Given:

Cloth needed for 1 shirt = 1.65 m

Cloth needed for 3 shirts = 1.65 × 3 metres.

$= \dfrac{165}{100} \times 3 \\[1em] = \dfrac{165 \times 3}{100} \\[1em] = \dfrac{495}{100} \\[1em] = 4.95 \text{ m}$

∴ The cloth needed for 3 shirts is 4.95 m.

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?

Answer

Given:

The cost of one notebook = ₹15.50

The cost of one eraser = ₹2.75

Cost of 4 notebooks = 4 × 15.50 = 62.00

Cost of 3 erasers = 3 × 2.75 = 8.25

Total amount spent = 62.00 + 8.25 = 70.25

∴ Meenu spent ₹70.25 in all.

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.

Answer

Given:

The thickness of a rupee coin = 1.45 mm

Total height of 36 coins = 36 × 1.45 mm.

$= \dfrac{145}{100} \times 36 \\[1em] = \dfrac{145 \times 36}{100} \\[1em] = \dfrac{5220}{100} \\[1em] = 52.20 \text{ mm}$

Converting millimetres into centimetres (since 1 cm = 10 mm):

$52.20 \text{ mm} = \dfrac{52.20}{10} \text{ cm} = 5.22 \text{ cm}$

∴ The total height of the cylinder of 36 coins is 5.22 cm.

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?

Answer

Given:

The price of 1 kg of oranges = ₹56.50

Price of 2.250 kg of oranges = 56.50 × 2.250

$= \dfrac{5650}{100} \times \dfrac{2250}{1000} \\[1em] = \dfrac{5650 \times 2250}{100 \times 1000} \\[1em] = \dfrac{12712500}{100000} \\[1em] = 127.125$

So, the price of 2.250 kg of oranges is ₹127.125.

Yes, we can write 56.50 as 56.5 and 2.250 as 2.25 because trailing zeroes after the decimal point do not change the value of a decimal number. Therefore, we get the same product.

So 56.50 = 56.5 and 2.250 = 2.25, and therefore:

$56.5 \times 2.25 = \dfrac{565}{10} \times \dfrac{225}{100} = \dfrac{127125}{1000} = 127.125$

Hence, the price of 2.250 kg of oranges is ₹127.125.

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?

Answer

Given:

The cost price of one notebook = ₹23.6

The selling price of one notebook = ₹30

Profit on 50 notebooks = ?

Profit on one notebook = Selling price − Cost price.

= 30 − 23.6

= 6.40

So, the profit on one notebook is ₹6.40.

Profit on 50 notebooks = 6.40 × 50.

$= \dfrac{640}{100} \times 50 \\[1em] = \dfrac{640 \times 50}{100} \\[1em] = \dfrac{32000}{100} \\[1em] = 320$

∴ Dwarakanath makes a profit of ₹320 by selling 50 books in a week.

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?

Answer

We are given that 18 × 12 = 216.

For each multiplication, remove the decimal points, use 18 × 12 = 216, and then place the decimal point according to the total number of decimal places in the factors.

(a) 18 × 1.2 — total decimal digits = 0 + 1 = 1

18 × 1.2 = 21.6

(b) 18 × 0.12 — total decimal digits = 0 + 2 = 2

18 × 0.12 = 2.16

(c) 1.8 × 1.2 — total decimal digits = 1 + 1 = 2

1.8 × 1.2 = 2.16

(d) 0.18 × 0.12 — total decimal digits = 2 + 2 = 4

0.18 × 0.12 = 0.0216

(e) 0.018 × 0.012 — total decimal digits = 3 + 3 = 6

0.018 × 0.012 = 0.000216

(f) 1.8 × 12 — total decimal digits = 1 + 0 = 1

1.8 × 12 = 21.6

The products are 0.0216 in (d) and 0.000216 in (e), both of which are less than 1.

Hence, the product is less than 1 in cases (d) and (e).

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

Answer

We can decide this without multiplying, by looking at the two numbers being multiplied:

• When both factors lie between 0 and 1, their product is less than each factor and therefore less than 1.

• When one factor is greater than 1 and the other lies between 0 and 1, the product is not always greater than 1; the actual values must be considered.

(a) 7 × 0.6

Since $0.6 \gt \dfrac{1}{7}$, multiplying 7 by 0.6 gives a number greater than 1.

∴ The product is greater than 1.

(b) 0.7 × 0.6

Both factors lie between 0 and 1. Their product is therefore less than each factor.

∴ The product is less than 1.

(c) 0.7 × 6

Since $0.7 \gt \dfrac{1}{6}$, multiplying 6 by 0.7 gives a number greater than 1.

∴ The product is greater than 1.

(d) 0.07 × 0.06

Both factors lie between 0 and 1.

∴ The product is less than 1.

Hence, the product is less than 1 in cases (b) and (d).

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

Answer

To multiply a decimal by 10, 100 or 1000, we move the decimal point to the right by as many places as there are zeroes in the multiplier (adding zeroes at the end when needed).

Hence, the completed table is as shown above.

What is 0.039 m in centimetres and millimetres?

Answer

We know that 1 m = 100 cm and 1 m = 1000 mm.

Converting into centimetres:

0.039 m = 0.039 × 100 cm = 3.9 cm

Converting into millimetres:

0.039 m = 0.039 × 1000 mm = 39 mm

∴ 0.039 m = 3.9 cm = 39 mm.

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

Answer

To write a fraction as a terminating decimal by this method, we need an equivalent fraction whose denominator is 1, 10, 100, 1000, and so on.

The denominators 10,100,1000,… are made up only of the prime factors 2 and 5 (for example, 10 = 2 × 5, 100 = 2 × 2 × 5 × 5).

Here the denominator is 3, which is neither 2 nor 5, and 3 is not a factor of 10, 100, 1000, or any such number. So we cannot find an equivalent fraction of $\dfrac{10}{3}$ with a denominator of the form 1, 10, 100, 1000, etc.

Hence, we need a more general method — division using place value (long division). On dividing, we find:

$\dfrac{10}{3} = 3.333\ldots$

which is a non-terminating decimal.

∴ No, $\mathbf{\dfrac{10}{3}}$ cannot be converted to an equivalent fraction with denominator 1, 10, 100, 1000, etc., so it cannot be written using a finite number of decimal digits.

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) $\dfrac{18}{5}$

(b) $\dfrac{415}{4}$

(c) $\dfrac{1217}{2}$

(d) $\dfrac{4827}{8}$

Answer

(a) $\dfrac{18}{5}$

Since 5 × 2 = 10, we multiply the numerator and the denominator by 2:

$\dfrac{18}{5} = \dfrac{18 \times 2}{5 \times 2} = \dfrac{36}{10} = 3.6$

Verifying by long division:

$\begin{array}{r} 3.6 \\ 5 \overline{\smash{)} 18.0 } \\ \underline{15}\phantom{.0} \\ 30 \phantom{.} \\ \underline{30} \phantom{.} \\ 0 \phantom{.} \end{array}$

∴ $\mathbf {\dfrac{18}{5} = 3.6}$.

(b) $\dfrac{415}{4}$

Since 4 × 25 = 100, we multiply the numerator and the denominator by 25:

$\dfrac{415}{4} = \dfrac{415 \times 25}{4 \times 25} = \dfrac{10375}{100} = 103.75$

Verifying by long division:

$\begin{array}{r} 103.75 \\ 4 \overline{\smash{)} 415.00 } \\ \underline{4}\phantom{15.00} \\ 01\phantom{5.00} \\ \underline{0}\phantom{5.00} \\ 15\phantom{.00} \\ \underline{12}\phantom{.00} \\ 30\phantom{0} \\ \underline{28}\phantom{0} \\ 20 \\ \underline{20} \\ 0 \end{array}$

∴ $\mathbf{\dfrac{415}{4} = 103.75}$.

(c) $\dfrac{1217}{2}$

Since 2 × 5 = 10, we multiply the numerator and the denominator by 5:

$\dfrac{1217}{2} = \dfrac{1217 \times 5}{2 \times 5} = \dfrac{6085}{10} = 608.5$

Verifying by long division:

$\begin{array}{r} 608.5 \\ 2 \overline{\smash{)} 1217.0 } \\ \underline{12}\phantom{17.0} \\ 01\phantom{7.0} \\ \underline{0}\phantom{7.0} \\ 17\phantom{.0} \\ \underline{16}\phantom{.0} \\ 10 \\ \underline{10} \\ 0 \end{array}$

∴ $\mathbf{\dfrac{1217}{2} = 608.5}$.

(d) $\dfrac{4827}{8}$

Since $8 \times 125 = 1000$, we multiply the numerator and the denominator by 125:

$\dfrac{4827}{8} = \dfrac{4827 \times 125}{8 \times 125} = \dfrac{603375}{1000} = 603.375$

Verifying by long division:

$\begin{array}{r} 603.375 \\ 8 \overline{\smash{)} 4827.000 } \\ \underline{48}\phantom{27.000} \\ 02\phantom{7.000} \\ \underline{0}\phantom{7.000} \\ 27\phantom{.000} \\ \underline{24}\phantom{.000} \\ 30\phantom{00} \\ \underline{24}\phantom{00} \\ 60\phantom{0} \\ \underline{56}\phantom{0} \\ 40 \\ \underline{40} \\ 0 \end{array}$

∴ $\mathbf {\dfrac{4827}{8} = 603.375}$.

Choose the correct answer:

$\dfrac{1526}{4}$ =

1. 38.15
2. 380.15
3. 381.5
4. 381.05

Answer

Since 4 × 25 = 100, we multiply the numerator and the denominator by 25:

$\dfrac{1526}{4} = \dfrac{1526 \times 25}{4 \times 25} = \dfrac{38150}{100} = 381.5$

Hence, option 3 is the correct option.

Choose the correct answer:

$\dfrac{3567}{8}$ =

1. 4458.75
2. 44.5875
3. 445.875
4. 4458.75

Answer

Since 8 × 125 = 1000, we multiply the numerator and the denominator by 125:

$\dfrac{3567}{8} = \dfrac{3567 \times 125}{8 \times 125} = \dfrac{445875}{1000} = 445.875$

Hence, option 3 is the correct option.

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

Answer

First, let us divide 132 by 4 using long division:

$\begin{array}{r} 33 \\ 4 \overline{\smash{)} 132 } \\ \underline{12}\phantom{2} \\ 12 \\ \underline{12} \\ 0 \end{array}$

So, 132 ÷ 4 = 33.

Dividing by 4 always gives the same digits (33); only the position of the decimal point changes, moving one place to the left each time the decimal point in the dividend moves one place to the left.

(a) 132 ÷ 4 = 33

(b) 13.2 ÷ 4 = 3.3

(c) 1.32 ÷ 4 = 0.33

(d) 0.132 ÷ 4 = 0.033

Hence, the quotients are 33, 3.3, 0.33 and 0.033 respectively.

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

Answer

First, let us divide 126 by 8 using long division. When we regroup the Ones into Tenths, we place a decimal point in the quotient.

$\begin{array}{r} 15.75 \\ 8 \overline{\smash{)} 126.00 } \\ \underline{8}\phantom{26.00} \\ 46\phantom{.00} \\ \underline{40}\phantom{.00} \\ 60\phantom{0} \\ \underline{56}\phantom{0} \\ 40 \\ \underline{40} \\ 0 \end{array}$

So, 126 ÷ 8 = 15.75

Dividing by 8 always gives the same digits (1575); only the position of the decimal point changes, moving one place to the left each time the decimal point in the dividend moves one place to the left.

(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

Hence, the quotients are 15.75, 1.575, 0.1575, 0.01575 and 0.001575 respectively.

Find 4.68 ÷ 1.3.

Now, what about 4.68 ÷ 0.13?

Answer

When the divisor is a decimal, we convert it into a fraction and then multiply the dividend by its reciprocal.

4.68 ÷ 1.3:

Here $1.3 = \dfrac{13}{10}$, so

$\phantom{=} 4.68 \div 1.3 = 4.68 \div \dfrac{13}{10}\\[1em] = 4.68 \times \dfrac{10}{13}\\[1em] = \dfrac{46.8}{13}\\[1em] = 3.6$

Hence, 4.68 ÷ 1.3 = 3.6

4.68 ÷ 0.13:

Here $0.13 = \dfrac{13}{100}$, so

$\phantom{=} 4.68 \div 0.13 = 4.68 \div \dfrac{13}{100}\\[1em] = 4.68 \times \dfrac{100}{13}\\[1em] = \dfrac{468}{13}\\[1em] = 36$

Hence, 4.68 ÷ 0.13 = 36.

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

Answer

Both of these are carried out by long division, and in each case the remainder keeps repeating, so the quotient never ends.

10 ÷ 9:

$\begin{array}{r} 1.111\ldots \\ 9 \overline{\smash{)} 10.000 }\phantom{\ldots} \\ \underline{9}\phantom{.000\ldots} \\ 10\phantom{00\ldots} \\ \underline{9}\phantom{00\ldots} \\ 10\phantom{0\ldots} \\ \underline{9}\phantom{0\ldots} \\ 1\phantom{0}\ldots \end{array}$

At each step the remainder is 1, so

10 ÷ 9 = 1.111 …

100 ÷ 11:

$\begin{array}{r} 9.0909\ldots \\ 11 \overline{\smash{)} 100.0000 }\phantom{\ldots} \\ \underline{99}\phantom{.0000\ldots} \\ 10\phantom{000\ldots} \\ \underline{0}\phantom{000\ldots} \\ 100\phantom{00\ldots} \\ \underline{99}\phantom{00\ldots} \\ 1\phantom{00}\ldots \end{array}$

Here the remainders repeat in the cycle 1, 10, 1, 10, …, so

100 ÷ 11 = 9.0909 …

Hence, 10 ÷ 9 = 1.111 … and 100 ÷ 11 = 9.0909 …, both non-terminating decimals.

Multiply 142857 by numbers from 1 to 6. What are the products? What do you notice?

(i) Multiply 142857 by 7. What do you observe?

(ii) Are there other such numbers? To find one such number, you can find 1 ÷ 17 in decimal, and use the repeating block of digits.

Answer

Multiplying 142857 by each number from 1 to 6:

142857 × 1 = 142857

142857 × 2 = 285714

142857 × 3 = 428571

142857 × 4 = 571428

142857 × 5 = 714285

142857 × 6 = 857142

We notice that every product uses exactly the same six digits 1, 4, 2, 8, 5, 7 in the same circular order — only the starting point changes. In each case the digits of 142857 appear "cycled around."

Hence, multiplying 142857 by 1, 2, 3, 4, 5 and 6 gives back the same digits, merely cyclically rotated.

(i)

142857 × 7 = 999999

Unlike the products for 1 to 6, multiplying by 7 does not give a cyclic rearrangement of the digits. Instead, all the digits become 9.

This is because 142857 is the repeating block obtained from $\dfrac{1}{7} = 0.142857142857…$, and $7 \times \dfrac{1}{7} = 1 = 0.999999\ldots$, so 142857 × 7 = 999999.

Hence, 142857 × 7 = 999999.

(ii)

Dividing 1 by 17 by long division gives a non-terminating, repeating decimal:

$\dfrac{1}{17} = 0.\overline{0588235294117647}$

The repeating block has 16 digits: 0588235294117647.

This block itself is a "cyclic" number. Multiplying it by 1, 2, 3, $\ldots$, 16 gives back the same 16 digits, each time cycled around (writing the results as 16-digit numbers, with leading zeroes kept):

0588235294117647 × 1 = 0588235294117647

0588235294117647 × 2 = 1176470588235294

0588235294117647 × 3 = 1764705882352941

0588235294117647 × 4 = 2352941176470588
⋮
0588235294117647 × 16 = 9411764705882352

(And 0588235294117647 × 17 = 9999999999999999, just as 142857 × 7 = 999999.)

Hence, from $\mathbf {\dfrac{1}{17} = 0.\overline{0588235294117647}}$, the repeating block 0588235294117647 is another cyclic number.

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.

Answer

Let us keep the dividend fixed at 12 and divide by different divisors:

12 ÷ 3 = 4

12 ÷ 1.5 = 8

12 ÷ 1 = 12

12 ÷ 0.5 = 24

12 ÷ 0.4 = 30

12 ÷ 0.1 = 120

So the quotient is not always greater than the dividend when the divisor is a decimal. For example, 12 ÷ 1.5 = 8, which is less than 12. What matters is whether the divisor is greater than 1, equal to 1, or between 0 and 1.

Capturing this in a table:

Hence, the quotient is greater than the dividend only when the divisor lies between 0 and 1; it equals the dividend when the divisor is 1, and it is less than the dividend when the divisor is greater than 1.

Express the following fractions in decimal form:

(a) $\dfrac{2}{5}$

(b) $\dfrac{13}{4}$

(c) $\dfrac{4}{50}$

(d) $\dfrac{5}{8}$

Answer

To express a fraction as a decimal, we find an equivalent fraction whose denominator is of the form 10, 100, 1000, and so on, and then write it as a decimal.

(a) $\dfrac{2}{5}$:

$\dfrac{2}{5} = \dfrac{2 \times 2}{5 \times 2} = \dfrac{4}{10} = 0.4$

Hence, $\mathbf {\dfrac{2}{5} = 0.4}$

(b) $\dfrac{13}{4}$:

Since $4 \times 25 = 100$,

$\dfrac{13}{4} = \dfrac{13 \times 25}{4 \times 25} = \dfrac{325}{100} = 3.25$

Hence, $\mathbf {\dfrac{13}{4} = 3.25}$

(c) $\dfrac{4}{50}$:

Since $50 \times 2 = 100$,

$\dfrac{4}{50} = \dfrac{4 \times 2}{50 \times 2} = \dfrac{8}{100} = 0.08$

Hence, $\mathbf {\dfrac{4}{50} = 0.08}$

(d) $\dfrac{5}{8}$:

Since $8 \times 125 = 1000$,

$\dfrac{5}{8} = \dfrac{5 \times 125}{8 \times 125} = \dfrac{625}{1000} = 0.625$

Hence, $\mathbf {\dfrac{5}{8} = 0.625}$

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

Answer

When the divisor is a decimal, we first convert it into a whole number by multiplying both the dividend and the divisor by the same power of 10, and then carry out the division.

(a) 24.86 ÷ 1.2

Multiplying both the dividend and the divisor by 10 (to make the divisor a whole number):

$24.86 \div 1.2 = \dfrac{24.86 \times 10}{1.2 \times 10} = \dfrac{248.6}{12}$

Now dividing 248.6 by 12:

248.6 ÷ 12 = 20.7166 …

Here the digit 6 keeps repeating, so the division never ends.

∴ 24.86 ÷ 1.2 = 20.7166… (a non-terminating, recurring decimal in which the digit 6 repeats).

(b) 5.728 ÷ 1.52

Multiplying both the dividend and the divisor by 100 (to make the divisor a whole number):

$5.728 \div 1.52 = \dfrac{5.728 \times 100}{1.52 \times 100} = \dfrac{572.8}{152}$

Now dividing 572.8 by 152:

572.8 ÷ 152 = 3.768421052631578947…

This division does not end; the quotient is a non-terminating recurring decimal.

∴ 5.728 ÷ 1.52 = 3.768421052631578947… (a non-terminating, recurring decimal).

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 = ...............

Answer

In each part, the digits (ignoring the decimal points) come from 156 × 12 = 1872. We only need to place the decimal point correctly.

(a) 15.6 × 1.2

15.6 has 1 digit after the decimal point and 1.2 has 1 digit after the decimal point, so the product has 1 + 1 = 2 digits after the decimal point.

15.6 × 1.2 = 18.72

(b) 187.2 ÷ 1.2

$187.2 \div 1.2 = \dfrac{187.2 \times 10}{1.2 \times 10} = \dfrac{1872}{12} = 156$

187.2 ÷ 1.2 = 156

(c) 18.72 ÷ 15.6

$18.72 \div 15.6 = \dfrac{18.72 \times 10}{15.6 \times 10} = \dfrac{187.2}{156} = 1.2$

18.72 ÷ 15.6 = 1.2

(d) 0.156 × 0.12

0.156 has 3 digits after the decimal point and 0.12 has 2 digits after the decimal point, so the product has 3 + 2 = 5 digits after the decimal point.

0.156 × 0.12 = 0.01872

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 × ..........

Answer

(a) The number by which 25 is divided is 25 ÷ 0.025:

$\phantom{=} \dfrac{25}{0.025} = \dfrac{25 \times 1000}{0.025 \times 1000} \\[1em] = \dfrac{25000}{25} \\[1em] = 1000$

So, 25 ÷ 1000 = 0.025.

(b) The number by which 25 is divided is 25 ÷ 250:

$\phantom{=} \dfrac{25}{250} = \dfrac{25}{250} \\[1em] = \dfrac{1}{10} \\[1em] = 0.1$

So, 25 ÷ 0.1 = 250.

(c) The number by which 25 is divided is 25 ÷ 2.5:

$\phantom{=} \dfrac{25}{2.5} = \dfrac{25 \times 10}{2.5 \times 10} \\[1em] = \dfrac{250}{25} \\[1em] = 10$

So, 25 ÷ 10 = 2.5.

(d) Dividing by 10 is the same as multiplying by $\dfrac{1}{10}$, i.e., by 0.1:

$\phantom{=} \dfrac{25}{10} = 2.5 = 25 \times 0.1$

So, 25 ÷ 10 = 25 × 0.1.

(e) Dividing by 0.10 is the same as multiplying by 10:

$\phantom{=} \dfrac{25}{0.10} = 250 = 25 \times 10$

So, 25 ÷ 0.10 = 25 × 10.

(f) Dividing by 0.01 is the same as multiplying by 100:

$\phantom{=} \dfrac{25}{0.01} = 2500 = 25 \times 100$

So, 25 ÷ 0.01 = 25 × 100.

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?

Answer

We convert each divisor into a whole number by multiplying both the dividend and the divisor by the same power of 10.

(a) 2.46 ÷ 1.5

$\phantom{=} \dfrac{2.46}{1.5} = \dfrac{2.46 \times 10}{1.5 \times 10} \\[1em] = \dfrac{24.6}{15} \\[1em] = 1.64$

Hence, 2.46 ÷ 1.5 = 1.64

(b) 2.46 ÷ 0.15

$\phantom{=} \dfrac{2.46}{0.15} = \dfrac{2.46 \times 100}{0.15 \times 100} \\[1em] = \dfrac{246}{15} \\[1em] = 16.4$

Hence, 2.46 ÷ 0.15 = 16.4.

(c) 2.46 ÷ 0.015

$\phantom{=} \dfrac{2.46}{0.015} = \dfrac{2.46 \times 1000}{0.015 \times 1000} \\[1em] = \dfrac{2460}{15} \\[1em] = 164$

Hence, 2.46 ÷ 0.015 = 164

Comparison of 24.6 ÷ 1.5 and 2.46 ÷ 0.15:

$24.6 \div 1.5 = \dfrac{24.6 \times 10}{1.5 \times 10} = \dfrac{246}{15} = 16.4$

$2.46 \div 0.15 = \dfrac{2.46 \times 100}{0.15 \times 100} = \dfrac{246}{15} = 16.4$

Both divisions reduce to 246 ÷ 15, so they give the same quotient.

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

Answer

Given:

Total length of wooden block = 4 m

Number of pieces = 5

Length of each piece = Total length ÷ Number of pieces

$= 4 \div 5 \text{ m}\\[1em] = \dfrac{4}{5} \text{ m}\\[1em] = \dfrac{4 \times 2}{5 \times 2} \text{ m}\\[1em] = \dfrac{8}{10} \text{ m}\\[1em] = 0.8 \text{ m}$

Hence, the length of each piece is 0.8 m.

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

Answer

Given:

Perimeter of a regular polygon with 12 sides = 208.8 cm

A regular polygon has all sides equal in length. So, the length of one side is the perimeter divided by the number of sides.

Length of each side = 208.8 ÷ 12 cm

Dividing:

$\dfrac{208.8}{12}$ = 17.4 cm

Hence, the length of each side of the regular polygon is 17.4 cm.

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.

Answer

Given:

Total juice = 3 litres

Number of friends = 8

Quantity of juice each friend gets = Total juice ÷ Number of friends

$= 3 \div 8 \text{ litres}\\[1em] = \dfrac{3}{8} \text{ litres}\\[1em] = \dfrac{3 \times 125}{8 \times 125} \text{ litres} \quad [\text{since } 8 \times 125 = 1000]\\[1em] = \dfrac{375}{1000} \text{ litres}\\[1em] = 0.375 \text{ litres}$

Converting to millilitres, using 1 litre = 1000 ml:

0.375 litres = 0.375 × 1000 ml = 375 ml

Hence, each friend will get 0.375 litres, that is, 375 ml of watermelon juice.

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

Answer

Distance travelled per litre = Total distance ÷ Quantity of petrol

= 234.45 ÷ 12.6 km

Converting the divisor into a whole number by multiplying both the dividend and the divisor by 10:

$234.45 \div 12.6 = \dfrac{234.45 \times 10}{12.6 \times 10} = \dfrac{2344.5}{126}$

Carrying out the division:

2344.5 ÷ 126 = 18.6071428…

Here the block of digits 714285 keeps repeating, so the quotient is a non-terminating recurring decimal. For practical purposes, the distance travelled per litre is about 18.61 km.

Hence, the car travels 18.6071… km per litre of petrol (approximately 18.61 km per litre).

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

Answer

Flour received by each student = Total flour ÷ Number of students

= 13.5 ÷ 15 kg

Converting the divisor and dividend so that the division is straightforward:

$13.5 \div 15 = \dfrac{135}{150} = \dfrac{9}{10} = 0.9 \text{ kg}$

Hence, each student received 0.9 kg of flour.

Complete the following patterns and answer the questions below.

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

Answer

Completing the two patterns:

For the powers of 2, we multiply the numerator and denominator by the matching power of 5 so that the denominator becomes a power of 10:

$\dfrac{1}{2 \times 2 \times 2 \times 2 \times 2} = \dfrac{1}{32} = \dfrac{1 \times 3125}{32 \times 3125} = \dfrac{3125}{100000} = 0.03125$

For the powers of 5, we multiply the numerator and denominator by the matching power of 2:

$\dfrac{1}{5 \times 5 \times 5 \times 5 \times 5} = \dfrac{1}{3125} = \dfrac{1 \times 32}{3125 \times 32} = \dfrac{32}{100000} = 0.00032$

The pattern observed:

In every row, each of these fractions gives a decimal that terminates (ends), and the number of decimal places equals the number of 2's (or 5's) multiplied together.

Looking more closely:

$\dfrac{1}{2^n} = \dfrac{5^n}{10^n}$, so the significant digits of the first list are the powers of 5, namely 5, 25, 125, 625, 3125.

$\dfrac{1}{5^n} = \dfrac{2^n}{10^n}$, so the significant digits of the second list are the powers of 2, namely 2, 4, 8, 16, 32.

Also, multiplying the two fractions in the same row gives a power of 10 in the denominator:

$\dfrac{1}{2^n} \times \dfrac{1}{5^n} = \dfrac{1}{(2 \times 5)^n} = \dfrac{1}{10^n}$

Why 2 and 5 are related in this way:

The reason is that 2 × 5 = 10; that is, 2 and 5 are the two prime factors of 10. To write a fraction as a decimal, it helps to make the denominator a power of 10 (10, 100, 1000, …). A denominator that is a power of 2 can be turned into a power of 10 by multiplying by the matching power of 5, and a denominator that is a power of 5 can be turned into a power of 10 by multiplying by the matching power of 2. Because each such fraction can be rewritten with a power of 10 in the denominator, all of these fractions give terminating (finite) decimals.

Hence, $\mathbf {\dfrac{1}{32} = 0.03125}$ and $\mathbf {\dfrac{1}{3125} = 0.00032}$; the two patterns are linked because 2 × 5 = 10, so $\mathbf {\dfrac{1}{2^n} = \dfrac{5^n}{10^n}}$ and $\mathbf {\dfrac{1}{5^n} = \dfrac{2^n}{10^n}}$, and every such fraction gives a terminating decimal.

Do you know which month has this extra day?

Answer

An ordinary calendar year has 365 days, and in it the month of February has 28 days.

Every fourth year (a leap year) has one extra day, making the year 366 days long. This extra day is added to the month of February, giving it 29 days. So the extra day is 29 February.

Hence, the extra day of a leap year is added to the month of February (29 February).

How many years are divisible by 4 in 100 years?

Can you form different expressions for the same question?

Answer

The years divisible by 4 are 4, 8, 12, …, 100. Their count is:

100 ÷ 4 = 25

∴ 25 years are divisible by 4 in 100 years.

Among 100 calendar years, the 25 years that are divisible by 4 are leap years having 366 days each, and the remaining 100 - 25 = 75 years have 365 days each.

So the number of days in 100 calendar years can also be written as:

= (25 × 366) + (75 × 365)

= 9150 + 27375

= 36525 days

Hence, (25 × 366) + (75 × 365) = 36525 days is another expression that gives the same number of days.

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?

Answer

In the final scheme, a year has 366 days if it is divisible by 400, or if it is divisible by 4 but not by 100; otherwise it has 365 days.

Let us count the years in 10,000 years:

Years divisible by 400 = 10000 ÷ 400 = 25

Years divisible by 100 but not by 400 = 100 - 25 = 75

Years divisible by 4 but not by 100 = 2500 - 100 = 2400

Remaining years = 10000 - (25 + 75 + 2400) = 7500

So the number of leap years (366 days) = 25 + 2400 = 2425, and the number of ordinary years (365 days) = 75 + 7500 = 7575.

Number of calendar days in 10,000 years:

= (2425 × 366) + (7575 × 365)

= 887550 + 2764875

= 36,52,425 days

The actual number of days the Earth takes to make 10,000 revolutions:

= 10000 × 365.2422

= 36,52,422 days

Difference = 36,52,425 − 36,52,422 = 3 days. So even with this scheme, in 10,000 years the calendar over-counts by 3 days.

Since these 3 extra days accumulate over 10,000 years, one way to fix the problem is to drop 3 leap days in every 10,000 years — for example, by declaring that 3 suitably chosen years in each 10,000-year cycle (roughly one every 3333 years) which would otherwise be leap years are treated as ordinary years. This keeps the calendar in step with the Earth's revolution around the Sun.

Hence, the calendar counts 36,52,425 days while the Earth actually takes 36,52,422 days for 10,000 revolutions — a difference of 3 days, which can be corrected by removing 3 leap days over every 10,000 years.

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

Answer

People worked out the length of the year through careful astronomical observation carried out over many centuries.

By recording the positions of the Sun and the stars — for example, noting the exact day of a solstice or equinox, or the day on which a particular star first became visible just before sunrise — and observing how these events slowly shifted with respect to the calendar, astronomers could measure how many days the seasons take to repeat. Averaging such observations over hundreds or even thousands of years cancels out small measurement errors and gives a very precise value, close to 365.2422 days, for one revolution of the Earth around the Sun.

Hence, the length of the year was determined by patiently observing and averaging the recurrence of astronomical events (like solstices, equinoxes, and star positions) over very long periods of time.

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

Answer

Given:

Cost of the 210-gram packet of peanut chikki = ₹70.5

Cost of the 110-gram packet of potato chips = ₹33.25

To decide which is cheaper, let us find the cost of 1 gram (the unit price) of each item.

Cost of 1 gram of peanut chikki:

= 70.5 ÷ 210

= 0.3357…

≈ ₹0.336

Cost of 1 gram of potato chips:

= 33.25 ÷ 110

= 0.3022…

≈ ₹0.302

Since ₹0.302 < ₹0.336, one gram of potato chips costs less than one gram of peanut chikki.

Hence, the packet of potato chips is cheaper because it costs less per gram.

Write the decimal number at the arrow mark:

Answer

The figure shows two number lines. Let us read the arrow mark on each.

First number line (from 3.1 to 3.2):

The gap from 3.1 to 3.2 is 0.1, and it is divided into 10 equal parts, so each small division = 0.1 ÷ 10 = 0.01.

Counting from 3.1, the arrow is at the 6^th mark:

= 3.1 + 6 × 0.01 = 3.1 + 0.06 = 3.16

Second number line (from 2.15 to 2.17):

The gap from 2.15 to 2.17 is 0.02, and it is divided into 10 equal parts, so each small division = 0.02 ÷ 10 = 0.002.

Counting from 2.15, the arrow is at the 6^th mark:

= 2.15 + 6 × 0.002 = 2.15 + 0.012 = 2.162

Hence, the decimal numbers at the arrow marks are 3.16 and 2.162.

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?

Answer

Given:

Total number of bananas = 35

Cost price of 1 kg bananas = ₹30

Cost of 3 kg of bananas = 3 × ₹30 = ₹90

Selling price of one banana = ₹5

Selling price of 35 bananas = 35 × ₹5 = ₹175

Profit = Selling price of 35 bananas − Cost price of 35 bananas:

= ₹175 - ₹90

= ₹85

Hence, Shyamala makes a profit of ₹85 on selling all the bananas.

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?

Answer

Given:

Thickness of each textbook = 2.5 cm

Length of the bookshelf = 160 cm

The number of books that can be placed = $\dfrac{\text{Length of the bookshelf}}{\text{Thickness of each textbook}}$

Number of books = 160 ÷ 2.5

= 1600 ÷ 25

= 64

So only 64 books could be placed on the shelf (not all 80 that the teacher wanted).

Length occupied by these 64 books = 64 × 2.5 = 160 cm, which is exactly the full length of the shelf.

Space left = 160 - 160 = 0 cm.

Hence, 64 books could be placed on the shelf, and no space was left over.

Fill in the following blanks appropriately:

5.5 km = ............... m
35 cm = ............... m
14.5 cm = ............... mm
68 g = ............... kg
9.02 m = ............... mm
125.5 ml = ............... l

Answer

To change to a smaller unit we multiply, and to change to a larger unit we divide, by the appropriate power of 10.

5.5 km = 5.5 × 1000 m = 5500 m

35 cm = 35 ÷ 100 m = 0.35 m

14.5 cm = 14.5 × 10 mm = 145 mm

68 g = 68 ÷ 1000 kg = 0.068 kg

9.02 m = 9.02 × 1000 mm = 9020 mm   [Since 1 m = 100 cm = 1000 mm]

125.5 ml = 125.5 ÷ 1000 l = 0.1255 l

Hence, the blanks are filled as 5500 m, 0.35 m, 145 mm, 0.068 kg, 9020 mm and 0.1255 l respectively.

The following problem was set by Sridharacharya in his book, Patiganita. “$6\dfrac{1}{4}$ is divided by $2\dfrac{1}{2}$, and $60\dfrac{1}{4}$ is divided by $3\dfrac{1}{2}$. Tell the quotients separately.” Can you try to solve it by converting the fractions into decimals?

Answer

Let us first convert each mixed fraction into a decimal.

$6\dfrac{1}{4} = 6.25$  and  $2\dfrac{1}{2} = 2.5$

$60\dfrac{1}{4} = 60.25$  and  $3\dfrac{1}{2} = 3.5$

First quotient: 6.25 ÷ 2.5

When the divisor is a decimal, we multiply both the dividend and the divisor by the same power of 10 to make the divisor a whole number:

$6.25 \div 2.5 = \dfrac{6.25 \times 10}{2.5 \times 10} = \dfrac{62.5}{25} \\[1em] = 2.5$

Second quotient: 60.25 ÷ 3.5

$60.25 \div 3.5 = \dfrac{60.25 \times 10}{3.5 \times 10} = \dfrac{602.5}{35} \\[1em] = 17.2142857\ldots$

Here the division does not end — the block of digits 142857 keeps repeating, so the quotient is the non-terminating decimal 17.2142857… (equal to $17\dfrac{3}{14}$).

Hence, the two quotients are 2.5 and $\mathbf{17.2142857\ldots}$ (i.e. $\mathbf{17\dfrac{3}{14}}$) respectively.

Fill the boxes in at least 2 different ways:
(a) $\square$ × $\square$ = 2.4
(b) $\square$ × $\square$ = 14.5

Answer

(a) $\square \times \square = 2.4$

Way 1: $1.2 \times 2 = 2.4$

Way 2: $0.8 \times 3 = 2.4$

(b) $\square \times \square = 14.5$

Way 1: $2.9 \times 5 = 14.5$

Way 2: $5.8 \times 2.5 = 14.5$

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

Answer

In each case we multiply (or divide) both the dividend and the divisor by the same power of 10 so that the division becomes 756 ÷ 36 = 21, and then adjust the answer.

(a) 75.6 ÷ 3.6

$\phantom{=} \dfrac{75.6}{3.6} \\[1em] = \dfrac{75.6 \times 10}{3.6 \times 10} \\[1em] = \dfrac{756}{36} \\[1em] = 21$

Hence, 75.6 ÷ 3.6 = 21

(b) 7.56 ÷ 0.36

$\phantom{=} \dfrac{7.56}{0.36} \\[1em] = \dfrac{7.56 \times 100}{0.36 \times 100} \\[1em] = \dfrac{756}{36} \\[1em] = 21$

Hence, 7.56 ÷ 0.36 = 21

(c) 756 ÷ 0.36

$\phantom{=} \dfrac{756}{0.36} \\[1em] = \dfrac{756 \times 100}{0.36 \times 100} \\[1em] = \dfrac{75600}{36} \\[1em] = 2100$

Hence, 756 ÷ 0.36 = 2100

(d) 75.6 ÷ 360

$\phantom{=} \dfrac{75.6}{360} \\[1em] = \dfrac{756}{3600} \\[1em] = 0.21$

Hence, 75.6 ÷ 360 = 0.21

(e) 7560 ÷ 3.6

$\phantom{=} \dfrac{7560}{3.6} \\[1em] = \dfrac{7560 \times 10}{3.6 \times 10} \\[1em] = \dfrac{75600}{36} \\[1em] = 2100$

Hence, 7560 ÷ 3.6 = 2100

(f) 7.56 ÷ 0.36

$\phantom{=} \dfrac{7.56}{0.36} \\[1em] = \dfrac{7.56 \times 100}{0.36 \times 100} \\[1em] = \dfrac{756}{36} \\[1em] = 21$

Hence, 7.56 ÷ 0.36 = 21

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

Answer

Every cell is found from the single basic fact 1517 ÷ 37 = 41. Multiplying or dividing the dividend a or the divisor b by 10, 100, 1000, … only shifts the decimal point in the quotient. So each entry is 41 with its decimal point suitably shifted.

For example, $\dfrac{151.7}{37} = \dfrac{1517}{370} = 4.1$ and $\dfrac{1517}{0.37} = \dfrac{151700}{37} = 4100$.

Completing every cell as a ÷ b:

Hence, every missing cell is a power-of-ten shift of the value 41, as shown in the completed table above.

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

Answer

Each of the five digits 2, 4, 5, 8 and 0 is used exactly once to fill the five boxes in $\square\square.\square \times \square.\square$.

(a) Maximum product:

To make the product largest, the two leading digits 8 and 5 should occupy the highest places.

82.0 × 5.4 = 442.8

(The same maximum is also obtained from 54.0 × 8.2 = 442.8)

(b) Minimum product:

To make the product smallest, the second factor should be as small as possible, i.e. 0.2.

45.8 × 0.2 = 9.16

(c) Product greater than 150:

Many arrangements work, for example

54.2 × 8.0 = 433.6 or 82.4 × 5.0 = 412

(The smallest product that is still greater than 150 is 54.0 × 2.8 = 151.2.)

(d) Product nearest to 100:

20.5 × 4.8 = 98.4

(e) Product nearest to 5:

The smallest product that can be formed with these digits is 9.16, so this is also the product that comes closest to 5.

45.8 × 0.2 = 9.16

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

Answer

We can sort these without full computation by reasoning about how each operation changes 245.05, and then confirm with approximate values.

Multiplying 245.05 by a number less than 1 makes it smaller — (a) is a little less than 245.05.

Multiplying by 7.9682 makes it much larger — (b) is the biggest.

Dividing by 7.9682 makes it much smaller — (c).

Dividing by a number less than 1 makes it larger — (d) is a little more than 245.05.

The approximate values are:

(a) 245.05 × 0.942368 ≈ 230.93

(b) 245.05 × 7.9682 ≈ 1952.61

(c) 245.05 ÷ 7.9682 ≈ 30.75

(d) 245.05 ÷ 0.942368 ≈ 260.04

(e) 245.05

(f) 7.9682

Arranging these in increasing order:

7.9682 < 30.75 < 230.93 < 245.05 < 260.04 < 1952.61

Hence, the increasing order is (f) < (c) < (a) < (e) < (d) < (b).