A Peek Beyond The Point

In the following figure, screws are placed above a scale. Measure them and write their length in the space provided.

Which scale helped you measure the length of the screws accurately? Why?

Answer

The scale in which each centimetre is divided into 10 equal parts (the smallest divisions) helped us measure the lengths accurately. This is because the screws are not a whole number of centimetres long — each screw ends somewhere between two consecutive centimetre marks. The finer divisions (tenths of a centimetre) let us read this extra fractional length exactly.

Can you explain why the unit was divided into smaller parts to measure the screws?

Answer

The lengths of the screws do not end exactly at a whole-centimetre mark. Therefore, measuring them using only centimetres does not give the exact length.

So each unit (1 cm) is divided into 10 equal smaller parts, each of length $\dfrac{1}{10}$ cm. These smaller divisions allow us to measure the extra fractional length and write the measurement exactly.

Hence, the unit was divided into smaller parts so that lengths could be measured more accurately.

Write the measurements of the objects shown in the picture:

Answer

Measurements of the objects shown:

1. Eraser = $2\dfrac{4}{10}$ cm

2. Pencil = $4\dfrac{5}{10}$ cm

3. Chalk = $1\dfrac{4}{10}$ cm

For the objects shown below, write their lengths in two ways. An example is given for the USB cable.

Answer

Reading each object against its own scale (which starts at 0), the length can be written in two ways — as a mixed form (units and one-tenths) and as a number of one-tenths.

(i) USB cable (given example):

Length = $4\dfrac{8}{10}$ units = $\dfrac{48}{10}$ units

(ii) Pencil box:

Length = $2\dfrac{8}{10}$ units = $\dfrac{28}{10}$ units

(iii) Leaf:

Length = $10\dfrac{9}{10}$ units = $\dfrac{109}{10}$ units

(iv) Finger:

Length = $1\dfrac{3}{10}$ units = $\dfrac{13}{10}$ units

Hence, each length can be written both as a mixed form and as a number of one-tenths.

Arrange these lengths in increasing order:

(a) $\dfrac{9}{10}$

(b) $1\dfrac{7}{10}$

(c) $\dfrac{130}{10}$

(d) $13\dfrac{1}{10}$

(e) $10\dfrac{5}{10}$

(f) $7\dfrac{6}{10}$

(g) $6\dfrac{7}{10}$

(h) $\dfrac{4}{10}$

Answer

Let us express each length as a number of one-tenths (i.e., write each as a fraction with denominator 10) so that they can be compared easily:

(a) $\dfrac{9}{10} = \dfrac{9}{10}$

(b) $1\dfrac{7}{10} = \dfrac{17}{10}$

(c) $\dfrac{130}{10} = \dfrac{130}{10}$

(d) $13\dfrac{1}{10} = \dfrac{131}{10}$

(e) $10\dfrac{5}{10} = \dfrac{105}{10}$

(f) $7\dfrac{6}{10} = \dfrac{76}{10}$

(g) $6\dfrac{7}{10} = \dfrac{67}{10}$

(h) $\dfrac{4}{10} = \dfrac{4}{10}$

Now the numerators are 9, 17, 130, 131, 105, 76, 67, 4.

Arranging these numerators in increasing order:

4 < 9 < 17 < 67 < 76 < 105 < 130 < 131

Hence, the increasing order is $\boldsymbol{\dfrac{4}{10} \lt \dfrac{9}{10} \lt 1\dfrac{7}{10} \lt 6\dfrac{7}{10} \lt 7\dfrac{6}{10} \lt 10\dfrac{5}{10} \lt \dfrac{130}{10} \lt 13\dfrac{1}{10}}$.

Arrange the following lengths in increasing order: $4\dfrac{1}{10}$, $\dfrac{4}{10}$, $\dfrac{41}{10}$, $41\dfrac{1}{10}$.

Answer

Let us express each length as a number of one-tenths:

$4\dfrac{1}{10} = \dfrac{41}{10}$

$\dfrac{4}{10} = \dfrac{4}{10}$

$\dfrac{41}{10} = \dfrac{41}{10}$

$41\dfrac{1}{10} = \dfrac{411}{10}$

Now the numerators are 41, 4, 41, 411

Arranging these numerators in increasing order:

4 < 41 = 41 < 411

Hence, the increasing order is $\boldsymbol {\dfrac{4}{10} \lt 4\dfrac{1}{10} = \dfrac{41}{10} \lt 41\dfrac{1}{10}}$.

The lengths of the body parts of a honeybee are given. Find its total length.

Head: $2\dfrac{3}{10}$ units

Thorax: $5\dfrac{4}{10}$ units

Abdomen: $7\dfrac{5}{10}$ units

Answer

Total length = Head + Thorax + Abdomen

$= 2\dfrac{3}{10} + 5\dfrac{4}{10} + 7\dfrac{5}{10} \text{ units}$

Adding the whole units and the one-tenths separately:

$= (2 + 5 + 7) + \left(\dfrac{3}{10} + \dfrac{4}{10} + \dfrac{5}{10}\right) \text{ units} \\[1em] = 14 + \dfrac{12}{10} \text{ units} \\[1em] = 14 + 1\dfrac{2}{10} \text{ units} \quad \text{[Since } \dfrac{12}{10} = 1\dfrac{2}{10} \text{]} \\[1em] = 15\dfrac{2}{10} \text{ units}$

Hence, the total length of the honeybee is $\bold {15\dfrac{2}{10}}$ units.

Using the method of converting both lengths to tenths, find the difference:

$12\dfrac{4}{10} - 6\dfrac{7}{10}$

Answer

Converting both lengths to tenths:

$12\dfrac{4}{10} = \dfrac{124}{10}$

$6\dfrac{7}{10} = \dfrac{67}{10}$

Subtracting:

$= \dfrac{124}{10} - \dfrac{67}{10} \text{ units} \\[1em] = \dfrac{124 - 67}{10} \text{ units} \\[1em] = \dfrac{57}{10} \text{ units} \\[1em] = 5\dfrac{7}{10} \text{ units}$

Hence, the difference between the two lengths is $\bold {5\dfrac{7}{10}}$ units.

A Celestial Pearl Danio's length is $2\dfrac{4}{10}$ cm, and the length of a Philippine Goby is $\dfrac{9}{10}$ cm. What is the difference in their lengths?

Answer

Given:

Length of the Celestial Pearl Danio = $2\dfrac{4}{10}$ cm = $\dfrac{24}{10}$ cm

Length of the Philippine Goby = $\dfrac{9}{10}$ cm

Difference in their lengths = (Longer length) − (Shorter length)

$= \dfrac{24}{10} - \dfrac{9}{10} \text{ cm} \\[1em] = \dfrac{24 - 9}{10} \text{ cm} \\[1em] = \dfrac{15}{10} \text{ cm} \\[1em] = 1\dfrac{5}{10} \text{ cm}$

Hence, the difference in their lengths = $\bold {1\dfrac{5}{10}}$ cm.

Observe the given sequences of numbers. Identify the change after each term and extend the pattern:

(a) $4$, $4\dfrac{3}{10}$, $4\dfrac{6}{10}$, .........., .........., .........., ..........

(b) $8\dfrac{2}{10}$, $8\dfrac{7}{10}$, $9\dfrac{2}{10}$, .........., .........., .........., ..........

(c) $7\dfrac{6}{10}$, $8\dfrac{7}{10}$, .........., .........., .........., ..........

(d) $5\dfrac{7}{10}$, $5\dfrac{3}{10}$, .........., .........., .........., ..........

(e) $13\dfrac{5}{10}$, $13$, $12\dfrac{5}{10}$, .........., .........., .........., ..........

(f) $11\dfrac{5}{10}$, $10\dfrac{4}{10}$, $9\dfrac{3}{10}$, .........., .........., .........., ..........

Answer

(a) Change after each term: $4\dfrac{3}{10} - 4 = \dfrac{3}{10}$

Add $\dfrac{3}{10}$ each time:

$4,\ 4\dfrac{3}{10},\ 4\dfrac{6}{10},\ \boxed{4\dfrac{9}{10}},\ \boxed{5\dfrac{2}{10}},\ \boxed{5\dfrac{5}{10}},\ \boxed{5\dfrac{8}{10}}$

(b) Change after each term: $8\dfrac{7}{10} - 8\dfrac{2}{10} = \dfrac{5}{10}$

Add $\dfrac{5}{10}$ each time:

$8\dfrac{2}{10},\ 8\dfrac{7}{10},\ 9\dfrac{2}{10},\ \boxed{9\dfrac{7}{10}},\ \boxed{10\dfrac{2}{10}},\ \boxed{10\dfrac{7}{10}},\ \boxed{11\dfrac{2}{10}}$

(c) Change after each term: $8\dfrac{7}{10} - 7\dfrac{6}{10} = 1\dfrac{1}{10}$

Add $1\dfrac{1}{10}$ each time:

$7\dfrac{6}{10},\ 8\dfrac{7}{10},\ \boxed{9\dfrac{8}{10}},\ \boxed{10\dfrac{9}{10}},\ \boxed{12},\ \boxed{13\dfrac{1}{10}}$

(d) Change after each term: $5\dfrac{3}{10} - 5\dfrac{7}{10} = -\dfrac{4}{10}$

Subtract $\dfrac{4}{10}$ each time:

$5\dfrac{7}{10},\ 5\dfrac{3}{10},\ \boxed{4\dfrac{9}{10}},\ \boxed{4\dfrac{5}{10}},\ \boxed{4\dfrac{1}{10}},\ \boxed{3\dfrac{7}{10}}$

(e) Change after each term: $13 - 13\dfrac{5}{10} = -\dfrac{5}{10}$

Subtract $\dfrac{5}{10}$ each time:

$13\dfrac{5}{10},\ 13,\ 12\dfrac{5}{10},\ \boxed{12},\ \boxed{11\dfrac{5}{10}},\ \boxed{11},\ \boxed{10\dfrac{5}{10}}$

(f) Change after each term: $10\dfrac{4}{10} - 11\dfrac{5}{10} = -1\dfrac{1}{10}$

Subtract $1\dfrac{1}{10}$ each time:

$11\dfrac{5}{10},\ 10\dfrac{4}{10},\ 9\dfrac{3}{10},\ \boxed{8\dfrac{2}{10}},\ \boxed{7\dfrac{1}{10}},\ \boxed{6},\ \boxed{4\dfrac{9}{10}}$

How many one-hundredths make one-tenth? Can we also say that the length is 4 units and 4 one-tenths and 5 one-hundredths as 4 units and 45 one-hundredths?

Answer

We know that one-tenth is split into 10 equal parts, and each such part is one-hundredth. So:

$\dfrac{1}{10} = \dfrac{10}{100}$

∴ 10 one-hundredths make one-tenth.

For the second part, the given length is 4 units, 4 one-tenths and 5 one-hundredths, i.e.,

$4 + \dfrac{4}{10} + \dfrac{5}{100}$

Converting the tenths into hundredths:

$\dfrac{4}{10} + \dfrac{5}{100} = \dfrac{40}{100} + \dfrac{5}{100} \quad \text{[Since } \dfrac{4}{10} = \dfrac{40}{100} \text{]} \\[1em] = \dfrac{45}{100}$

So the length is $4 + \dfrac{45}{100}$, which is indeed 4 units and 45 one-hundredths.

Hence, 10 one-hundredths make one-tenth, and yes — the length can also be written as 4 units and 45 one-hundredths.

Observe the figure below. Notice the markings and the corresponding lengths written in the boxes when measured from 0. Fill the lengths in the empty boxes.

Answer

For the lengths shown below write the measurements and write the measures in words.

Answer

Reading each length from 0 up to the marked point (units, one-tenths and one-hundredths):

Measurement = $5\dfrac{37}{100}$

In words: Five and thirty-seven hundredths

Measurement = $15\dfrac{3}{100}$

In words: Fifteen and three-hundredths

Measurement = $7\dfrac{52}{100}$

In words: Seven and fifty-two hundredths

Measurement = $9\dfrac{80}{100}$

In words: Nine and eighty-hundredths

In each group, identify the longest and the shortest lengths. Mark each length on the scale.

(a) $\dfrac{3}{10}$, $\dfrac{3}{100}$, $\dfrac{33}{100}$

(b) $3\dfrac{1}{10}$, $\dfrac{30}{10}$, $1\dfrac{3}{10}$

(c) $\dfrac{45}{100}$, $\dfrac{54}{100}$, $\dfrac{5}{10}$, $\dfrac{4}{10}$

(d) $3\dfrac{6}{10}$, $3\dfrac{6}{100}$, $3\dfrac{6}{10}\dfrac{6}{100}$

(e) $\dfrac{8}{10}\dfrac{2}{100}$, $\dfrac{9}{100}$, $1\dfrac{8}{100}$

(f) $7\dfrac{3}{10}\dfrac{5}{100}$, $7\dfrac{5}{10}$, $7\dfrac{41}{100}$

(g) $\dfrac{65}{10}\dfrac{15}{100}$, $5\dfrac{87}{100}$, $5\dfrac{7}{100}$

Answer

To compare, we express each length as a number of hundredths (i.e. its value in units):

(a)

Given:

$\dfrac{3}{10}$, $\dfrac{3}{100}$, $\dfrac{33}{100}$

Explanation — The interval from 0 to 1 is divided into 100 equal parts. Therefore, each small division represents $\dfrac{1}{100}$. $\dfrac{3}{100}$ is marked at the 3rd division from 0. $\dfrac{3}{10} = \dfrac{30}{100}$ is marked at the 30th division from 0. $\dfrac{33}{100}$ is marked at the 33rd division from 0.

∴ Longest = $\dfrac{33}{100}$, Shortest = $\dfrac{3}{100}$.

(b)

Given:

$3\dfrac{1}{10}$, $\dfrac{30}{10}$, $1\dfrac{3}{10}$

Explanation — The interval from 0 to 10 is divided into 100 equal parts. Therefore, each small division represents $\dfrac{1}{10}$. $1\dfrac{3}{10}$ is marked 3 small divisions after 1. $\dfrac{30}{10}$ is marked at 3 units. $3\dfrac{1}{10}$ is marked 1 small division after 3.

∴ Longest = $3\dfrac{1}{10}$, Shortest = $1\dfrac{3}{10}$.

(c)

Given:

$\dfrac{45}{100}$, $\dfrac{54}{100}$, $\dfrac{5}{10}$, $\dfrac{4}{10}$

Explanation — The interval from 0 to 1 is divided into 100 equal parts. Therefore, each small division represents $\dfrac{1}{100}$. $\dfrac{4}{10} = \dfrac{40}{100}$ is marked at the 40th division from 0. $\dfrac{45}{100}$ is marked at the 45th division from 0. $\dfrac{5}{10} = \dfrac{50}{100}$ is marked at the 50th division from 0. $\dfrac{54}{100}$ is marked at the 54th division from 0.

∴ Longest = $\dfrac{54}{100}$, Shortest = $\dfrac{4}{10}$.

(d)

Given:

$3\dfrac{6}{10}$, $3\dfrac{6}{100}$, $3\dfrac{6}{10}\dfrac{6}{100}$

Explanation — The interval from 3 to 4 is divided into 10 equal parts, and each of these parts is further divided into 10 equal parts. So each small division represents $\dfrac{1}{100}$. $3\dfrac{6}{100}$ is marked 6 small divisions after 3. $3\dfrac{6}{10}$ is marked at 3 units and 6 tenths. $3\dfrac{6}{10}\dfrac{6}{100}$ is marked 6 small divisions after $3\dfrac{6}{10}$.

∴ Longest = $3\dfrac{6}{10}\dfrac{6}{100}$, Shortest = $3\dfrac{6}{100}$.

(e)

Given:

$\dfrac{8}{10}\dfrac{2}{100}$, $\dfrac{9}{100}$, $1\dfrac{8}{100}$

Explanation — The interval from 0 to 1 is divided into 100 equal parts. Therefore, each small division represents $\dfrac{1}{100}$. $\dfrac{9}{100}$ is marked at the 9th division after 0. $\dfrac{8}{10}\dfrac{2}{100}$ is marked at 8 tenths and 2 hundredths, i.e., $\dfrac{82}{100}$. $1\dfrac{8}{100}$ is marked at 1 unit and 8 hundredths, i.e., 8 hundredths after 1.

∴ Longest = $1\dfrac{8}{100}$, Shortest = $\dfrac{9}{100}$.

(f)

Given:

$7\dfrac{3}{10}\dfrac{5}{100}$, $7\dfrac{5}{10}$, $7\dfrac{41}{100}$

Explanation — The interval from 6.8 to 7.8 is divided into 100 equal parts. Therefore, each small division represents $\dfrac{1}{100}$. $7\dfrac{3}{10}\dfrac{5}{100}$ is marked at 7 units, 3 tenths and 5 hundredths. $7\dfrac{41}{100}$ is marked at 7 units and 41 hundredths. $7\dfrac{5}{10}$ is marked at 7 units and 5 tenths.

∴ Longest = $7\dfrac{5}{10}$, Shortest = $7\dfrac{3}{10}\dfrac{5}{100}$.

(g)

Given:

$\dfrac{65}{10}\dfrac{15}{100}$, $5\dfrac{87}{100}$, $5\dfrac{7}{100}$

Explanation — The interval shown on the scale is from 5.7 to 6.7, divided into hundredths. $5\dfrac{87}{100}$ is marked at 5 units and 87 hundredths. $\dfrac{65}{10}\dfrac{15}{100}$ is marked at 6 units, 5 tenths and 15 hundredths. $5\dfrac{7}{100}$ is less than 5.8, so it cannot be shown on this scale.

∴ Longest = $\dfrac{65}{10}\dfrac{15}{100}$, Shortest = $5\dfrac{7}{100}$.

Hence, the longest and shortest lengths in each group are as identified above.

Observe the addition done below for 483 + 268. Do you see any similarities between the methods shown above?

(400 + 80 + 3) + (200 + 60 + 8)
= (400 + 200) + (80 + 60) + (3 + 8)
= 600 + 140 + 11
= 600 + 150 + 1
= 700 + 50 + 1
= 751

Answer

Yes. In both cases, numbers are separated according to their place values and the corresponding places are added together.

For 483 + 268:

• Hundreds are added to hundreds.
• Tens are added to tens.
• Ones are added to ones.

When a place value becomes 10 or more, it is regrouped into the next higher place. For example, 11 ones = 1 ten and 1 one, and 14 tens = 1 hundred and 4 tens.

Similarly, in decimal addition, hundredths are added to hundredths, tenths to tenths, and units to units. Also, 10 hundredths = 1 tenth and 10 tenths = 1 unit.

Hence, both methods use the same place-value grouping and regrouping (carrying) process.

Solve this by converting to hundredths. What is the difference $15\dfrac{3}{10}\dfrac{4}{100} - 2\dfrac{6}{10}\dfrac{8}{100}$?

Answer

We have:

$15\dfrac{3}{10}\dfrac{4}{100} - 2\dfrac{6}{10}\dfrac{8}{100}$

Converting each quantity completely into hundredths:

$15\dfrac{3}{10}\dfrac{4}{100} = \dfrac{1534}{100} \\[1em] \text{[15 units, 3 tenths, 4 hundredths} = 1534 \text{ hundredths]} \\[1em] 2\dfrac{6}{10}\dfrac{8}{100} = \dfrac{268}{100} \\[1em] \text{[2 units, 6 tenths, 8 hundredths} = 268 \text{ hundredths]}$

Now subtracting:

$\phantom{=} \dfrac{1534}{100} - \dfrac{268}{100} \\[1em] = \dfrac{1534 - 268}{100} \\[1em] = \dfrac{1266}{100} \\[1em] = \dfrac{1200}{100} + \dfrac{66}{100} \\[1em] = 12 + \dfrac{66}{100} \\[1em] = 12\dfrac{6}{10}\dfrac{6}{100} \quad \text{[∵ } \dfrac{66}{100} = \dfrac{6}{10} + \dfrac{6}{100} \text{]}$

Hence, the difference is $\bold {12\dfrac{6}{10}\dfrac{6}{100}}$ (that is, 12.66).

Observe the subtraction done below for 653 – 268. Do you see any similarities with the methods shown above?

(600 + 50 + 3) – (200 + 60 + 8)
= (600 – 200) + (50 – 60) + (3 – 8)
= (600 – 200) + (40 – 60) + (13 – 8)
= (600 – 200) + (40 – 60) + 5
= (500 – 200) + (140 – 60) + 5
= 300 + 80 + 5
= 385

Answer

Yes. In both methods, the numbers are separated according to place value, and corresponding place values are subtracted.

When the digit in one place is smaller than the digit being subtracted, 1 unit from the next higher place is regrouped:

• 1 ten is regrouped as 10 ones, so 3 ones become 13 ones.
• 1 hundred is regrouped as 10 tens, so 4 tens become 14 tens; equivalently, 40 becomes 140.

Similarly, in decimal subtraction, 1 unit can be regrouped as 10 tenths and 1 tenth can be regrouped as 10 hundredths.

Hence, both methods use the same place-value regrouping process.

Find the sums and differences:

(a) $\dfrac{3}{10} + 3\dfrac{4}{100}$

(b) $9\dfrac{5}{10}\dfrac{7}{100} + 2\dfrac{1}{10}\dfrac{3}{100}$

(c) $15\dfrac{6}{10}\dfrac{4}{100} + 14\dfrac{3}{10}\dfrac{6}{100}$

(d) $7\dfrac{7}{100} - 4\dfrac{4}{100}$

(e) $8\dfrac{6}{100} - 5\dfrac{3}{100}$

(f) $12\dfrac{6}{10}\dfrac{2}{100} - \dfrac{9}{10}\dfrac{9}{100}$

Answer

(a)

We have:

$\phantom{=} \dfrac{3}{10} + 3\dfrac{4}{100} \\[1em] = (0 + 3) + \dfrac{3}{10} + \dfrac{4}{100} \\[1em] \text{[Grouping ones, tenths and hundredths]} \\[1em] = 3 + \dfrac{30}{100} + \dfrac{4}{100} \quad \Big[\dfrac{3}{10} = \dfrac{30}{100}\Big] \\[1em] = 3 + \dfrac{34}{100} \\[1em] = 3\dfrac{34}{100}$

∴ The answer is $\bold {3\dfrac{34}{100}}$

(b)

We have:

$\phantom{=} 9\dfrac{5}{10}\dfrac{7}{100} + 2\dfrac{1}{10}\dfrac{3}{100} \\[1em] = (9 + 2) + \Big(\dfrac{5}{10} + \dfrac{1}{10}\Big) + \Big(\dfrac{7}{100} + \dfrac{3}{100}\Big) \\[1em] \text{[Grouping ones, tenths and hundredths]} \\[1em] = 11 + \dfrac{6}{10} + \dfrac{10}{100} \\[1em] = 11 + \dfrac{6}{10} + \dfrac{1}{10} \quad \Big[\dfrac{10}{100} = \dfrac{1}{10}\Big] \\[1em] = 11 + \dfrac{7}{10} \\[1em] = 11\dfrac{7}{10}$

∴ The answer is $\mathbf{11\dfrac{7}{10}}$

(c)

We have:

$\phantom{=} 15\dfrac{6}{10}\dfrac{4}{100} + 14\dfrac{3}{10}\dfrac{6}{100} \\[1em] = (15 + 14) + \Big(\dfrac{6}{10} + \dfrac{3}{10}\Big) + \Big(\dfrac{4}{100} + \dfrac{6}{100}\Big) \\[1em] \text{[Grouping ones, tenths and hundredths]} \\[1em] = 29 + \dfrac{9}{10} + \dfrac{10}{100} \\[1em] = 29 + \dfrac{9}{10} + \dfrac{1}{10} \quad \Big[\dfrac{10}{100} = \dfrac{1}{10}\Big] \\[1em] = 29 + \dfrac{10}{10} \\[1em] = 29 + 1 \\[1em] = 30$

∴ The answer is 30

(d)

We have:

$\phantom{=} 7\dfrac{7}{100} - 4\dfrac{4}{100} \\[1em] = (7 - 4) + \Big(\dfrac{7}{100} - \dfrac{4}{100}\Big) \\[1em] \text{[Grouping ones and hundredths]} \\[1em] = 3 + \dfrac{3}{100} \\[1em] = 3\dfrac{3}{100}$

∴ The answer is $\mathbf {3\dfrac{3}{100}}$

(e)

We have:

$\phantom{=} 8\dfrac{6}{100} - 5\dfrac{3}{100} \\[1em] = (8 - 5) + \Big(\dfrac{6}{100} - \dfrac{3}{100}\Big) \\[1em] \text{[Grouping ones and hundredths]} \\[1em] = 3 + \dfrac{3}{100} \\[1em] = 3\dfrac{3}{100}$

∴ The answer is $\mathbf{ 3\dfrac{3}{100}}$

(f) We have:

$\phantom{=} 12\dfrac{6}{10}\dfrac{2}{100} - \dfrac{9}{10}\dfrac{9}{100} \\[1em] = 12\dfrac{62}{100} - \dfrac{99}{100} \\[1em] \text{[Writing tenths and hundredths together as hundredths]} \\[1em] = 11 + \dfrac{100}{100} + \dfrac{62}{100} - \dfrac{99}{100} \\[1em] \Big[\text{Regrouping 1 unit as } \dfrac{100}{100}\Big] \\[1em] = 11 + \dfrac{162}{100} - \dfrac{99}{100} \\[1em] = 11 + \dfrac{63}{100} \\[1em] = 11\dfrac{63}{100}$

∴ The answer is $\mathbf {11\dfrac{63}{100}}$

Can we extend this further?

Answer

Yes. The place value system can be extended further to the right.

When $\dfrac{1}{100}$ is divided into 10 equal parts, each part is $\dfrac{1}{1000}$. Similarly,

$\dfrac{1}{1000} \div 10 = \dfrac{1}{10000}$

Thus, each place value is 10 times smaller than the place value immediately to its left. This process can continue indefinitely.

Hence, we can extend the place value system further to obtain $\mathbf {\dfrac{1}{1000}, \dfrac{1}{10000}, \dots}$ (each 10 times smaller than the previous one).

Answer the following:

(a) How many thousandths make one unit?
(b) How many thousandths make one tenth?
(c) How many thousandths make one hundredth?
(d) How many tenths make one ten?
(e) How many hundredths make one ten?

Answer

(a) One unit = $1 = \dfrac{1000}{1000} = 1000 \times \dfrac{1}{1000}$

∴ 1000 thousandths make one unit.

(b) One-tenth = $\dfrac{1}{10} = \dfrac{100}{1000} = 100 \times \dfrac{1}{1000}$

∴ 100 thousandths make one-tenth.

(c) One-hundredth = $\dfrac{1}{100} = \dfrac{10}{1000} = 10 \times \dfrac{1}{1000}$

∴ 10 thousandths make one-hundredth.

(d) One ten = $10 = 100 \times \dfrac{1}{10} = 100 \times \dfrac{1}{10}$

That is, 100 tenths = $100 \times \dfrac{1}{10} = \dfrac{100}{10} = 10$.

∴ 100 tenths make one ten.

(e) One ten = $10 = 1000 \times \dfrac{1}{100}$

That is, 1000 hundredths = $1000 \times \dfrac{1}{100} = \dfrac{1000}{100} = 10$.

∴ 1000 hundredths make one ten.

Make a few more questions of this kind and answer them.

Answer

Sample 1. How many hundredths make one unit?

One unit = $1 = \dfrac{100}{100} = 100 \times \dfrac{1}{100}$

Hence, 100 hundredths make one unit.

Sample 2. How many tenths make one hundred?

One hundred = $100 = 1000 \times \dfrac{1}{10}$

Hence, 1000 tenths make one hundred.

Sample 3. How many thousandths make one ten?

One ten = $10 = 10000 \times \dfrac{1}{1000}$

Hence, 10000 thousandths make one ten.

Can the quantity $4\dfrac{2}{10}$ be written as 42 (skipping the $\dfrac{1}{10}$ in $2 \times \dfrac{1}{10}$)?

Answer

No. The symbols 4 and 2 alone would normally form the whole number 42:

$42 = 4 \text{ tens} + 2 \text{ ones} = 40 + 2$

However,

$4\dfrac{2}{10} = 4 + \dfrac{2}{10}$

These are different quantities. A decimal point is used to show where the whole-number part ends and the fractional part begins:

$4 + \dfrac{2}{10} = 4.2$

Hence, $4\dfrac{2}{10}$ cannot be written as 42; it is written as 4.2.

Make a place value table similar to the one above. Write each quantity in decimal form and in terms of place value, and read the number:

(a) 2 ones, 3 tenths and 5 hundredths

(b) 1 ten and 5 tenths

(c) 4 ones and 6 hundredths

(d) 1 hundred, 1 one and 1 hundredth

(e) $\dfrac{8}{100}$ and $\dfrac{9}{10}$

(f) $\dfrac{5}{100}$

(g) $\dfrac{1}{10}$

(h) $2\dfrac{1}{100}$, $4\dfrac{1}{10}$ and $7\dfrac{7}{1000}$

Answer

The place value table for each quantity is given below.

For part (h), the three quantities are written separately:

∴ The decimal forms are (a) 2.35, (b) 10.5, (c) 4.06, (d) 101.01, (e) 0.98, (f) 0.05, (g) 0.1, and (h) 2.01, 4.1, 7.007.

Write these quantities in decimal form:
(a) 234 hundredths,
(b) 105 tenths.

Answer

(a) 234 hundredths

$\phantom{=} 234 \text{ hundredths} = \dfrac{234}{100} \\[1em] = \dfrac{200}{100} + \dfrac{30}{100} + \dfrac{4}{100} \\[1em] = 2 + \dfrac{3}{10} + \dfrac{4}{100} \\[1em] = 2.34$

∴ 234 hundredths = 2.34

(b) 105 tenths

$\phantom{=} 105 \text{ tenths} = \dfrac{105}{10} \\[1em] = \dfrac{100}{10} + \dfrac{5}{10} \\[1em] = 10 + \dfrac{5}{10} \\[1em] = 10.5$

∴ 105 tenths = 10.5

Fill in the blanks below (mm <–> cm):

Answer

We know that 1 mm = $\dfrac{1}{10}$ cm = 0.1 cm, and 1 cm = 10 mm.

$70 \text{ mm} = \dfrac{70}{10}\text{ cm} = 7.0\text{ cm} \\[1em] 0.9 \text{ cm} = (0.9 \times 10)\text{ mm} = 9\text{ mm} \\[1em] 134 \text{ mm} = \dfrac{134}{10}\text{ cm} = 13.4\text{ cm} \\[1em] 203.6 \text{ cm} = 203.6 \times 10 \text{ mm} = 2036 \text{ mm}$

The completed table is:

Fill in the blanks below (cm <–> m):

Answer

We know that 1 cm = $\dfrac{1}{100}\text{ m}$ = 0.01 m, and 1 m = 100 cm.

$36 \text{ cm} = \dfrac{36}{100}\text{ m} = 0.36\text{ m} \\[1em] 50 \text{ cm} = \dfrac{50}{100}\text{ m} = 0.5\text{ m} \\[1em] 0.89 \text{ m} = (0.89 \times 100)\text{ cm} = 89\text{ cm} \\[1em] 4 \text{ cm} = \dfrac{4}{100}\text{ m} = 0.04\text{ m} \\[1em] 325 \text{ cm} = \dfrac{325}{100}\text{ m} = 3.25\text{ m} \\[1em] 2.07 \text{ m} = (2.07 \times 100)\text{ cm} = 207\text{ cm}$

The completed table is:

How many mm does 1 meter have?

Answer

We know that:

1 m = 100 cm
1 cm = 10 mm
∴ 1 m = (100 × 10) mm
= 1000 mm

∴ 1 meter has 1000 mm.

Can we write 1 mm = $\dfrac{1}{1000}$ m?

Answer

We know the relation between metre and millimetre:

1 m = 100 cm

1 cm = 10 mm

⇒ 1 m = 100 × 10 mm = 1000 mm

Since 1 metre is made up of 1000 equal millimetres, each millimetre is one-thousandth of a metre:

$\phantom{=} 1 \text{ mm} = \dfrac{1}{1000} \text{ m} \\[1em] = 0.001 \text{ m}$

This is exactly the same pattern seen for weight, where 1 g = $\dfrac{1}{1000}\text{ kg}$ and 1 mg = $\dfrac{1}{1000}\text{ g}$.

∴ Yes, we can write $\mathbf {1 \\ mm = \dfrac{1}{1000}\ m = 0.001\ m}$.

Fill in the blanks below (g <–> kg):

Answer

Since 1 kg = 1000 g, each gram is one-thousandth of a kilogram:

$1 \text{ g} = \dfrac{1}{1000} \text{ kg} = 0.001 \text{ kg}$

So, to change g into kg we divide by 1000 (move the decimal point 3 places to the left), and to change kg into g we multiply by 1000 (move the decimal point 3 places to the right).

Working out each blank:

$465 \text{ g} = \dfrac{465}{1000}\text{ kg} = 0.465 \text{ kg} \\[1em] 68 \text{ g} = \dfrac{68}{1000}\text{ kg} = 0.068 \text{ kg} \\[1em] 1560 \text{ g} = \dfrac{1560}{1000}\text{ kg} = 1.560 \text{ kg} \\[1em] 704 \text{ g} = \dfrac{704}{1000}\text{ kg} = 0.704 \text{ kg} \\[1em] 0.56 \text{ kg} = (0.56 \times 1000)\text{ g} = 560 \text{ g} \\[1em] 2.5 \text{ kg} = (2.5 \times 1000)\text{ g} = 2500 \text{ g}$

The completed table:

Fill in the blanks below (rupee <–> paise):

Answer

Since 1 rupee = 100 paise, each paisa is one-hundredth of a rupee:

$1 \text{ paisa} = ₹\dfrac{1}{100} = ₹0.01$

So, to change paise into rupees we divide by 100 (move the decimal point 2 places to the left), and to change rupees into paise we multiply by 100 (move the decimal point 2 places to the right).

Working out each blank:

$10 \text{ p} = ₹\dfrac{10}{100} = ₹0.10 \\[1em] ₹0.05 = (0.05 \times 100)\text{ p} = 5 \text{ p} \\[1em] ₹0.36 = (0.36 \times 100)\text{ p} = 36 \text{ p} \\[1em] ₹0.50 = (0.50 \times 100)\text{ p} = 50 \text{ p} \\[1em] 99 \text{ p} = ₹\dfrac{99}{100} = ₹0.99 \\[1em] 250 \text{ p} = ₹\dfrac{250}{100} = ₹2.50$

The completed table:

Name all the divisions between 1 and 1.1 on the number line.

Answer

Between 1 and 1.1 the number line is split into 10 equal parts. Each small part is one-hundredth $\dfrac{1}{100}$.

Hence, the divisions between 1 and 1.1 are:

1.01, 1.02, 1.03, 1.04, 1.05, 1.06, 1.07, 1.08, 1.09

∴ The divisions between 1 and 1.1 are 1.01, 1.02, 1.03, 1.04, 1.05, 1.06, 1.07, 1.08 and 1.09.

Identify and write the decimal numbers against the letters.

Answer

The number line shown is from 5 to 5.4. Each interval of one-tenth (such as 5.1 to 5.2) is divided into 10 equal parts. Therefore, each small division represents one-hundredth (0.01).

Reading the position of each letter against these hundredth marks:

A = 5.09
B = 5.13
C = 5.20
D = 5.31

Hence, A = 5.09, B = 5.13, C = 5.20 and D = 5.31.

Sonu says that 0.2 can also be written as 0.20, 0.200; Zara thinks that putting zeros on the right side may alter the value of the decimal number. What do you think?

Answer

Sonu is correct, and Zara's worry is not needed. Writing extra zeros to the right end of the decimal part does not change the value, because those zeros do not add any new quantity:

$0.2 = \dfrac{2}{10} \\[1em] 0.20 = \dfrac{20}{100} = \dfrac{2}{10} \\[1em] 0.200 = \dfrac{200}{1000} = \dfrac{2}{10}$

All three represent 2 tenths, so 0.2 = 0.20 = 0.200.

However, this is only true for zeros placed at the far right. A zero placed between the decimal point and a digit does change the value, since it shifts the digit to a smaller place. For example, 0.2 (2 tenths), 0.02 (2 hundredths) and 0.002 (2 thousandths) are all different.

Hence, Sonu is right: adding zeros on the right does not change the value, so 0.2 = 0.20 = 0.200.

Can you tell which of these is the smallest and which is the largest? (0.2, 0.02, 0.002)

Answer

Writing each number with the same number of decimal places (as thousandths) makes the comparison easy:

$0.2 = \dfrac{2}{10} = \dfrac{200}{1000} = 0.200 \\[1em] 0.02 = \dfrac{2}{100} = \dfrac{20}{1000} = 0.020 \\[1em] 0.002 = \dfrac{2}{1000} = 0.002$

Comparing the numerators over 1000:

200 > 20 > 2 ⇒ 0.2 > 0.02 > 0.002

Hence, the largest number is 0.2 and the smallest number is 0.002.

Which of these are the same: 4.5, 4.05, 0.405, 4.050, 4.50, 4.005, 04.50?

Answer

Two decimal numbers are equal only if they have the same digits in the same place values. Zeros written at the far right of the decimal part, or a zero written before the whole-number part, do not change the value.

Comparing each number:

$4.5 = 4.50 = 04.50 = \dfrac{45}{10} \quad \text{(all equal to 4.5)} \\[1em] 4.05 = 4.050 = \dfrac{405}{100} \quad \text{(both equal to 4.05)} \\[1em] 0.405 = \dfrac{405}{1000} \quad \text{(different)} \\[1em] 4.005 = \dfrac{4005}{1000} \quad \text{(different)}$

So there are two groups of equal numbers, while 0.405 and 4.005 stand on their own.

Hence, 4.5, 4.50 and 04.50 are the same (each equal to 4.5); and 4.05 and 4.050 are the same (each equal to 4.05).

Identify the decimal number in the last number line in Figure (b) denoted by '?'.

Answer

Explanation

• On the first number line (0 to 10), the marked point lies between 3 and 4.
• The next enlarged number line divides the interval 3 to 4 into 10 equal parts. The marked point lies between 3.0 and 3.1.
• The third enlarged number line divides the interval 3.0 to 3.1 into 10 equal parts. The marked point lies between 3.05 and 3.06.
• The final enlarged number line divides the interval 3.05 to 3.06 into 10 equal parts. The marked point is at the 9th division after 3.05.

Therefore, 3.05 + 0.009 = 3.059

Hence, the decimal number denoted by '?' is 3.059.

Make such number lines for the decimal numbers:
(a) 9.876
(b) 0.407.

Answer

To locate a decimal number, we begin with the whole-number segment on the number line and then magnify the relevant part step by step, dividing each chosen unit into 10 equal parts at every level.

(a) 9.876

We locate 9.876 through successive magnifications:

Level 1: Between 0 and 10, the number lies between 9 and 10.

Level 2: Divide 9 to 10 into 10 parts; 9.876 lies between 9.8 and 9.9.

Level 3: Divide 9.8 to 9.9 into 10 parts; 9.876 lies between 9.87 and 9.88.

Level 4: Divide 9.87 to 9.88 into 10 parts; 9.876 is the 6^th mark.

(b) 0.407

We locate 0.407 through successive magnifications:

Level 1: Between 0 and 1, the number lies between 0.4 and 0.5.

Level 2: Divide 0.4 to 0.5 into 10 parts; 0.407 lies between 0.40 and 0.41.

Level 3: Divide 0.40 to 0.41 into 10 parts; 0.407 is the 7^th mark.

In the number line shown below, what decimal numbers do the boxes labelled 'a', 'b', and 'c' denote?

Answer

The interval from 5 to 10 has length 5 units and is divided into 10 equal parts.

Value of each division = $\dfrac{5}{10} = \dfrac{1}{2} = 0.5$ $\quad \text{[5 units split into 10 parts]}$

So, starting from 5, the successive tick marks are:

5, 5.5, 6, 6.5, 7, 7.5, 8, 8.5, 9, 9.5, 10

Reading off the marks pointed to by the boxes:

(a) Box 'a' is on the 2^nd division after 5 ⇒ a = 6

(b) Box 'b' is on the 5^th division after 5 ⇒ b = 7.5

(c) Box 'c' is on the 9^th division after 5 ⇒ c = 9.5

Hence, a = 6, b = 7.5, and c = 9.5.

Using similar reasoning find out the decimal numbers in the boxes below.

Answer

We read each number line by first finding the value of one division.

First number line (from 8 to 8.1):

The interval from 8 to 8.1 is divided into 10 equal parts.

Value of each division = $\dfrac{8.1 - 8}{10} = \dfrac{0.1}{10} = 0.01$

So the marks are 8.00, 8.01, 8.02, ...., 8.10.

(i) Box 'd' is on the 1^st division after 8 ⇒ d = 8.01

(ii) Box 'e' is on the 5^th division after 8 ⇒ e = 8.05

Second number line (from 4.3 to 4.8):

The interval from 4.3 to 4.8 is divided into 10 equal parts.

Value of each division = $\dfrac{4.8 - 4.3}{10} = \dfrac{0.5}{10} = 0.05$

So the marks are 4.30, 4.35, 4.40, ...., 4.80, 4.85

(iii) Box 'f' is on the 1^st division after 4.3 ⇒ f = 4.35

(iv) Box 'g' is on the 4^th division after 4.3 ⇒ g = 4.50

(v) Box 'h' is on the 11^th division after 4.3 ⇒ h = 4.85

Hence, d = 8.01, e = 8.05, f = 4.35, g = 4.50, and h = 4.85.

Which decimal number is greater?

(a) 1.23 or 1.32
(b) 3.81 or 13.800
(c) 1.009 or 1.090

Answer

We compare the numbers place value by place value, starting from the highest place value.

(a) 1.23 or 1.32

Ones: 1 = 1 $\quad$ [same]

Tenths: 2 < 3 $\quad$ [first differing place]

⇒ 1.23 < 1.32

Hence, 1.32 is greater.

(b) 3.81 or 13.800

Whole part: 3 < 13 $\quad$ [3 is a 1-digit number, 13 is a 2-digit number]

⇒ 3.81 < 13.800

Hence, 13.800 is greater.

(c) 1.009 or 1.090

Ones: 1 = 1 $\quad$ [same]

Tenths: 0 < 9 $\quad$ [first differing place]

⇒ 1.009 < 1.090

Hence, 1.090 is greater.

Which of the above is closest to 1.09? (0.9, 1.1, 1.01, 1.11)

Answer

Find the difference between 1.09 and each number:

|1.09 - 0.9| = 0.19

|1.1 - 1.09| = 0.01

|1.09 - 1.01| = 0.08

|1.11 - 1.09| = 0.02

0.01 < 0.02 < 0.08 < 0.19

The smallest distance is 0.01.

Hence, 1.1 is closest to 1.09.

Which among these is closest to 4: 3.56, 3.65, 3.099?

Answer

We find how far each number is from 4 by taking the difference:

|4 - 3.56| = 0.44

|4 - 3.65| = 0.35

|4 - 3.099| = 0.901

The smallest distance is 0.35.

Hence, 3.65 is closest to 4.

Which among these is closest to 1: 0.8, 0.69, 1.08?

Answer

We find how far each number is from 1 by taking the difference:

|1 - 0.8| = 0.2

|1 - 0.69| = 0.31

|1.08 - 1| = 0.08

The smallest distance is 0.08.

Hence, 1.08 is closest to 1.

In each case below use the digits 4, 1, 8, 2, and 5 exactly once and try to make a decimal number as close as possible to 25.

Answer

1. Green box:

To be as close as possible to 25, use 25 as the whole-number part.

The remaining digits are 1, 4, 8. Since the number is already equal to 25 before the decimal point, we must make the decimal part as small as possible.

148 < 184 < 418 < 481 < 814 < 841

So the smallest decimal part is .148.

⇒ 25.148

2. Yellow box:

Only one digit can be placed before the decimal point. To get as close as possible to 25, choose the largest digit, 8.

Now the remaining digits are 1, 2, 4, 5. Since every number will still be much smaller than 25, we make the decimal part as large as possible.

542 > 541 > 524 >⋯

So the largest decimal part is .542.

⇒ 8.542

3. Pink box:

We need a 3-digit whole-number part. To get closest to 25, the whole-number part should be as close to 25 as possible.

Since we need three digits before the decimal, the smallest possible 3-digit part is obtained by using the smallest available digits first:

124, 125, 128, 142,…

Among all possible numbers, 124 is the closest 3-digit whole-number part to 25.

The remaining digits are 5 and 8. To make the number as close as possible to 25, make the decimal part as small as possible:

.58 < .85

⇒ 124.58

Hence, the decimal numbers closest to 25 are 25.148, 8.5421 and 124.58.

Write the detailed place value computation for 84.691 – 77.345, and its compact form.

Answer

We have:

84.691 - 77.345

Detailed place value computation:

Writing each number in expanded place value form:

$84.691 = 8 \times 10 + 4 \times 1 + 6 \times \dfrac{1}{10} + 9 \times \dfrac{1}{100} + 1 \times \dfrac{1}{1000} \\[1em] 77.345 = 7 \times 10 + 7 \times 1 + 3 \times \dfrac{1}{10} + 4 \times \dfrac{1}{100} + 5 \times \dfrac{1}{1000}$

Subtracting place by place (from the smallest place value), with borrowing where needed:

$\text{Thousandths: } (1 - 5) \Rightarrow (11 - 5)\times\dfrac{1}{1000} = 6 \times \dfrac{1}{1000} \\[1em] \text{[borrow 1 hundredth} = 10 \text{ thousandths]} \\[1em] \text{Hundredths: } (8 - 4)\times\dfrac{1}{100} = 4 \times \dfrac{1}{100} \\[1em] \text{[9 became 8 after lending]} \\[1em] \text{Tenths: } (6 - 3)\times\dfrac{1}{10} = 3 \times \dfrac{1}{10} \\[1em] \text{Ones: } (4 - 7) \Rightarrow (14 - 7)\times 1 = 7 \times 1 \\[1em] \text{[borrow 1 ten} = 10 \text{ ones]} \\[1em] \text{Tens: } (7 - 7)\times 10 = 0 \times 10 \\[1em] \text{[8 became 7 after lending]}$

Putting the place values together:

$0 \times 10 + 7 \times 1 + 3 \times \dfrac{1}{10} + 4 \times \dfrac{1}{100} + 6 \times \dfrac{1}{1000} = 7.346$

Compact form:

In the compact form, the same borrowing is carried out directly in columns (a ten is borrowed for the ones place, and a hundredth is borrowed for the thousandths place):

$\begin{array}{r} 84.691 \\ -77.345 \\ \hline \phantom{0}7.346 \\ \hline \end{array}$

That is:

84.691 - 77.345 = 7.346

Hence, 84.691 – 77.345 = 7.346.

Find the sums:

(a) 5.3 + 2.6
(b) 18 + 8.8
(c) 2.15 + 5.26
(d) 9.01 + 9.10
(e) 29.19 + 9.91
(f) 0.934 + 0.6
(g) 0.75 + 0.03
(h) 6.236 + 0.487

Answer

In each case we write the numbers one below the other so that the decimal points (and hence the place values) line up, putting extra zeros where needed, and then add just as we do for whole numbers.

(a) 5.3 + 2.6

$\begin{array}{r} 5.3 \\ +\phantom{0}2.6 \\ \hline 7.9 \\ \hline \end{array}$

Hence, the sum is 7.9.

(b) 18 + 8.8 = 18.0 + 8.8

$\begin{array}{r} 18.0 \\ +\phantom{0}8.8 \\ \hline 26.8 \\ \hline \end{array}$

Hence, the sum is 26.8.

(c) 2.15 + 5.26

$\begin{array}{r} 2.15 \\ +\phantom{0}5.26 \\ \hline 7.41 \\ \hline \end{array}$

Hence, the sum is 7.41.

(d) 9.01 + 9.10

$\begin{array}{r} 9.01 \\ +\phantom{0}9.10 \\ \hline 18.11 \\ \hline \end{array}$

Hence, the sum is 18.11.

(e) 29.19 + 9.91

$\begin{array}{r} 29.19 \\ +\phantom{0}9.91 \\ \hline 39.10 \\ \hline \end{array}$

Hence, the sum is 39.1.

(f) 0.934 + 0.6 = 0.934 + 0.600

$\begin{array}{r} 0.934 \\ +\phantom{0}0.600 \\ \hline 1.534 \\ \hline \end{array}$

Hence, the sum is 1.534.

(g) 0.75 + 0.03

$\begin{array}{r} 0.75 \\ +\phantom{0}0.03 \\ \hline 0.78 \\ \hline \end{array}$

Hence, the sum is 0.78.

(h) 6.236 + 0.487

$\begin{array}{r} 6.236 \\ +\phantom{0}0.487 \\ \hline 6.723 \\ \hline \end{array}$

Hence, the sum is 6.723.

Find the differences:

(a) 5.6 – 2.3
(b) 18 – 8.8
(c) 10.4 – 4.5
(d) 17 – 16.198
(e) 17 – 0.05
(f) 34.505 – 18.1
(g) 9.9 – 9.09
(h) 6.236 – 0.487

Answer

As with addition, we line up the decimal points, add zeros so that both numbers have the same number of decimal places, and then subtract just as we do for whole numbers.

(a) 5.6 – 2.3

$\begin{array}{r} 5.6 \\ -\phantom{0}2.3 \\ \hline 3.3 \\ \hline \end{array}$

Hence, the difference is 3.3.

(b) 18 – 8.8 = 18.0 – 8.8

$\begin{array}{r} 18.0 \\ -\phantom{0}8.8 \\ \hline \phantom{0}9.2 \\ \hline \end{array}$

Hence, the difference is 9.2.

(c) 10.4 – 4.5

$\begin{array}{r} 10.4 \\ -\phantom{0}4.5 \\ \hline \phantom{0}5.9 \\ \hline \end{array}$

Hence, the difference is 5.9.

(d) 17 – 16.198 = 17.000 – 16.198

$\begin{array}{r} 17.000 \\ -\phantom{0}16.198 \\ \hline \phantom{0}0.802 \\ \hline \end{array}$

Hence, the difference is 0.802.

(e) 17 – 0.05 = 17.00 – 0.05

$\begin{array}{r} 17.00 \\ -\phantom{0}0.05 \\ \hline 16.95 \\ \hline \end{array}$

Hence, the difference is 16.95.

(f) 34.505 – 18.1 = 34.505 – 18.100

$\begin{array}{r} 34.505 \\ -\phantom{0}18.100 \\ \hline 16.405 \\ \hline \end{array}$

Hence, the difference is 16.405.

(g) 9.9 – 9.09 = 9.90 – 9.09

$\begin{array}{r} 9.90 \\ -\phantom{0}9.09 \\ \hline 0.81 \\ \hline \end{array}$

Hence, the difference is 0.81.

(h) 6.236 – 0.487

$\begin{array}{r} 6.236 \\ -\phantom{0}0.487 \\ \hline 5.749 \\ \hline \end{array}$

Hence, the difference is 5.749.

Continue this sequence and write the next 3 terms.

4.4, 4.8, 5.2, 5.6, 6.0, …

Answer

We first find the change from one term to the next:

4.8 - 4.4 = 0.4

So each term is obtained by adding 0.4 to the previous term. Continuing the pattern:

6.0 + 0.4 = 6.4

6.4 + 0.4 = 6.8

6.8 + 0.4 = 7.2

Hence, the next 3 terms are 6.4, 6.8, 7.2.

Identify the change and write the next 3 terms for each sequence given below.

(a) 4.4, 4.45, 4.5, …
(b) 25.75, 26.25, 26.75, …
(c) 10.56, 10.67, 10.78, …
(d) 13.5, 16, 18.5, …
(e) 8.5, 9.4, 10.3, …
(f) 5, 4.95, 4.90, …
(g) 12.45, 11.95, 11.45, …
(h) 36.5, 33, 29.5, …

Answer

We first find the change from one term to the next:

(a) 4.4, 4.45, 4.5, …

Change = 4.45 - 4.4 = 0.05 (add 0.05).

4.5 + 0.05 = 4.55, $\quad$ 4.55 + 0.05 = 4.6, $\quad$ 4.6 + 0.05 = 4.65

Hence, the next 3 terms are 4.55, 4.6, 4.65.

(b) 25.75, 26.25, 26.75, …

Change = 26.25 - 25.75 = 0.5 (add 0.5).

26.75 + 0.5 = 27.25, $\quad$ 27.25 + 0.5 = 27.75, $\quad$ 27.75 + 0.5 = 28.25

Hence, the next 3 terms are 27.25, 27.75, 28.25.

(c) 10.56, 10.67, 10.78, …

Change = 10.67 - 10.56 = 0.11 (add 0.11).

10.78 + 0.11 = 10.89, $\quad$ 10.89 + 0.11 = 11.00, $\quad$ 11.00 + 0.11 = 11.11

Hence, the next 3 terms are 10.89, 11.00, 11.11.

(d) 13.5, 16, 18.5, …

Change = 16 - 13.5 = 2.5 (add 2.5).

18.5 + 2.5 = 21, $\quad$ 21 + 2.5 = 23.5, $\quad$ 23.5 + 2.5 = 26

Hence, the next 3 terms are 21, 23.5, 26.

(e) 8.5, 9.4, 10.3, …

Change = 9.4 - 8.5 = 0.9 (add 0.9).

10.3 + 0.9 = 11.2, $\quad$ 11.2 + 0.9 = 12.1, $\quad$ 12.1 + 0.9 = 13

Hence, the next 3 terms are 11.2, 12.1, 13.

(f) 5, 4.95, 4.90, …

Change = 4.95 - 5 = -0.05 (subtract 0.05).

4.90 - 0.05 = 4.85, $\quad$ 4.85 - 0.05 = 4.80, $\quad$ 4.80 - 0.05 = 4.75

Hence, the next 3 terms are 4.85, 4.80, 4.75.

(g) 12.45, 11.95, 11.45, …

Change = 11.95 - 12.45 = -0.5 (subtract 0.5).

11.45 - 0.5 = 10.95, $\quad$ 10.95 - 0.5 = 10.45, $\quad$ 10.45 - 0.5 = 9.95

Hence, the next 3 terms are 10.95, 10.45, 9.95.

(h) 36.5, 33, 29.5, …

Change = 33 - 36.5 = -3.5 (subtract 3.5).

29.5 - 3.5 = 26, $\quad$ 26 - 3.5 = 22.5, $\quad$ 22.5 - 3.5 = 19

Hence, the next 3 terms are 26, 22.5, 19.

Sonu observed sums and differences of decimal numbers and says, "If we add two decimal numbers, then the sum will always be greater than the sum of their whole number parts. Also, the sum will always be less than 2 more than the sum of their whole number parts." (For example, if the two numbers to be added are 25.936 and 8.202, the claim is that their sum will be greater than 25 + 8 and will be less than 25 + 1 + 8 + 1.) What do you think about this claim? Verify if this is true for these numbers. Will it work for any 2 decimal numbers?

Answer

Verifying for 25.936 and 8.202

First we find the actual sum:

$\begin{array}{r} 25.936 \\ +\phantom{0}8.202 \\ \hline 34.138 \\ \hline \end{array}$

The sum of the whole number parts is 25 + 8 = 33, and 2 more than this is 25 + 1 + 8 + 1 = 35.

33 < 34.138 < 35

So the sum lies between 33 and 35, and Sonu's claim is true for these numbers.

Will it work for any 2 decimal numbers?

Every decimal number can be written as its whole number part plus its fractional part, where the fractional part is always at least 0 and less than 1.

Let the two numbers be a = w₁ + f₁ and b = w₂ + f₂, where w₁, w₂ are the whole number parts and 0 ≤ f₁ < 1, 0 ≤ f₂ < 1 are the fractional parts. Then:

a + b = (w₁ + w₂) + (f₁ + f₂)

Since 0 ≤ f₁ + f₂ < 2, we get:

(w₁ + w₂) ≤ a + b < (w₁ + w₂) + 2

Thus, the sum is always at least the sum of the whole number parts and always less than 2 more than that sum.

Hence, Sonu's observation is true. For the given numbers, 33 < 34.138 < 35, and in general the sum of two decimal numbers lies between the sum of their whole number parts and 2 more than that sum.

What about for the sum of 25.93603259 and 8.202?

Answer

Here one number has many more decimal places, but the reasoning is exactly the same — only the whole number part and the fractional part matter.

Adding the numbers:

25.93603259 + 8.202 = 34.13803259

The sum of the whole number parts is 25 + 8 = 33, and 2 more than this is 35.

33 < 34.13803259 < 35

The fractional parts (0.93603259 and 0.202) still add up to a value between 0 and 2, so the sum again lies between the sum of the whole number parts and 2 more than it. Adding more decimal digits does not change this.

Hence, the claim holds here as well, since 33 < 34.13803259 < 35.

Similarly, come up with a way to narrow down the range of whole numbers within which the difference of two decimal numbers will lie.

Answer

Let the two numbers be a = w₁ + f₁ and b = w₂ + f₂, where w₁ and w₂ are the whole number parts, and f₁ and f₂ are the decimal parts.

Then,

a − b = (w₁ − w₂) + (f₁ − f₂)

Since each decimal part is less than 1,

−1 < f₁ − f₂ < 1

Therefore:

(w₁ − w₂) − 1 < a − b < (w₁ − w₂) + 1

So, the difference of two decimal numbers always lies between one less and one more than the difference of their whole number parts.

Example:

25.936 − 8.202 = 17.734

The difference of the whole number parts is 25 − 8 = 17

Since 16 < 17.734 < 18,

the difference lies between 17 − 1 and 17 + 1.

Hence, if w₁ and w₂ are the whole number parts of two decimal numbers, then their difference lies between

(w₁ − w₂) − 1 and (w₁ − w₂) + 1

Where else can we see'non-decimals' with a decimal-like notation?

Answer

There are several everyday situations where a number is written with a point but the part after the point is not counted in tenths, hundredths, etc. Some examples are:

(i) Cricket overs — "Overs left: 5.5" means 5 overs and 5 balls, not 5 and a half overs, because 1 over = 6 balls (base 6), so 5.5 overs means $5\dfrac{5}{6}$ overs.

(ii) Time written as hours — "The bus will reach 4.5 hours post noon." Here 0.5 hour means half of an hour = 30 minutes, so the time is 4:30, not 4:50.

(iii) Clock/time readings — a reading like "6.45" for the time means 6 hours and 45 minutes (base 60), not 6 and 45 hundredths.

(iv) Feet and inches — "2.5 ft" can be confused with 2 ft 5 inches, though 0.5 ft actually equals 6 inches (1 ft = 12 inches).

∴ Cricket overs, time expressed in hours, clock readings and feet–inches are common places where a decimal-like point is used but the value after the point is not measured in tenths.

Convert the following fractions into decimals:

(a) $\dfrac{5}{100}$

(b) $\dfrac{16}{1000}$

(c) $\dfrac{12}{10}$

(d) $\dfrac{254}{1000}$

Answer

(a) $\dfrac{5}{100}$

The denominator 100 has 2 zeros ⇒ 2 decimal places. Writing 5 with 2 digits after the point:

$\dfrac{5}{100} = 0.05$

Hence, the decimal form is 0.05.

(b) $\dfrac{16}{1000}$

The denominator 1000 has 3 zeros ⇒ 3 decimal places. Writing 16 with 3 digits after the point:

$\dfrac{16}{1000} = 0.016$

Hence, the decimal form is 0.016.

(c) $\dfrac{12}{10}$

The denominator 10 has 1 zero ⇒ 1 decimal place. Writing 12 with 1 digit after the point:

$\dfrac{12}{10} = 1.2$

Hence, the decimal form is 1.2.

(d) $\dfrac{254}{1000}$

The denominator 1000 has 3 zeros ⇒ 3 decimal places. Writing 254 with 3 digits after the point:

$\dfrac{254}{1000} = 0.254$

Hence, the decimal form is 0.254.

Convert the following decimals into a sum of tenths, hundredths and thousandths:

(a) 0.34
(b) 1.02
(c) 0.8
(d) 0.362

Answer

We know that the place after the decimal point are tenths $\left(\dfrac{1}{10}\right)$, hundredths $\left(\dfrac{1}{100}\right)$ and thousandths $\left(\dfrac{1}{1000}\right)$.

(a) 0.34

$0.34 = \dfrac{3}{10} + \dfrac{4}{100} \quad \text{[3 tenths and 4 hundredths]}$

(b) 1.02

$\phantom{=} 1.02 = 1 + \dfrac{0}{10} + \dfrac{2}{100} \\[1em] = 1 + \dfrac{2}{100} \quad \text{[1 unit and 2 hundredths]}$

(c) 0.8

$0.8 = \dfrac{8}{10} \quad \text{[8 tenths]}$

(d) 0.362

$0.362 = \dfrac{3}{10} + \dfrac{6}{100} + \dfrac{2}{1000} \quad \text{[3 tenths, 6 hundredths and 2 thousandths]}$

What decimal number does each letter represent in the number line below?

Answer

The interval from 6.4 to 6.5 is divided into 4 equal parts.

$\text{Value of each small division} = \dfrac{6.5 - 6.4}{4} \\[1em] = \dfrac{0.1}{4} \\[1em] = 0.025$

Therefore the successive marks are:

6.4, 6.425, 6.45, 6.475, 6.5, 6.525, 6.55, 6.575, 6.6.

Now, reading each letter by counting the divisions:

(a) The letter a is 2 divisions to the right of 6.4.

⇒ a = 6.45

(c) The letter c is 1 division to the right of 6.5.

⇒ c = 6.525

(b) The letter b is 2 divisions to the right of 6.5.

⇒ b = 6.55

Hence, a = 6.45, c = 6.525 and b = 6.55.

Arrange the following quantities in descending order:

(a) 11.01, 1.011, 1.101, 11.10, 1.01
(b) 2.567, 2.675, 2.768, 2.499, 2.698
(c) 4.678 g, 4.595 g, 4.600 g, 4.656 g, 4.666 g
(d) 33.13 m, 33.31 m, 33.133 m, 33.331 m, 33.313 m

Answer

To arrange the numbers in descending order, we compare them digit by digit starting from the leftmost (highest) place value.

(a) 11.01, 1.011, 1.101, 11.10, 1.01

Comparing the whole number parts, 11 > 1.

So 11.10 and 11.01 come first, then the numbers with whole part 1.

Hence, 11.10 > 11.01 > 1.101 > 1.011 > 1.01.

(b) 2.567, 2.675, 2.768, 2.499, 2.698

All have the same whole part 2, so we compare the tenths, then hundredths.

Hence, 2.768 > 2.698 > 2.675 > 2.567 > 2.499.

(c) 4.678 g, 4.595 g, 4.600 g, 4.656 g, 4.666 g

Hence, 4.678 g > 4.666 g > 4.656 g > 4.600 g > 4.595 g.

(d) 33.13 m, 33.31 m, 33.133 m, 33.331 m, 33.313 m

Hence, 33.331 m > 33.313 m > 33.31 m > 33.133 m > 33.13 m.

Using the digits 1, 4, 0, 8, and 6 make:

(a) the decimal number closest to 30
(b) the smallest possible decimal number between 100 and 1000.

Answer

We have the digits 1, 4, 0, 8 and 6.

(a) The decimal number closest to 30

We must use the digits 1, 4, 0, 8, 6 exactly once.

Since 30 cannot be formed, we look for numbers nearest to 30.

The largest number less than 30 is obtained by making the whole-number part as large as possible below 30:

18.640

30 − 18.640 = 11.360

The smallest number greater than 30 is obtained by making the whole-number part as small as possible above 30 is 40.168

40.168 − 30 = 10.168

Since

10.168 < 11.360,

The decimal number closest to 30 is 40.168

(b) The smallest possible decimal number between 100 and 1000

A number between 100 and 1000 must have a 3-digit whole number part.

To make the number as small as possible:

• Choose the smallest non-zero digit for the hundreds place: 1
• Choose the smallest remaining digit for the tens place: 0
• Choose the smallest remaining digit for the ones place: 4

So the whole-number part is 104

The remaining digits are 6 and 8.

To make the decimal part as small as possible, arrange them in increasing order that is .68

Hence, the smallest number between 100 and 1000 is 104.68.

Will a decimal number with more digits be greater than a decimal number with fewer digits?

Answer

No, a decimal number with more digits is not always greater than a decimal number with fewer digits.

The value of a decimal number depends on the place values of its digits, not on the total number of digits.

For example, compare 2.5 and 2.05:

$2.5 = 2 + \dfrac{5}{10} \\[1em] 2.05 = 2 + \dfrac{0}{10} + \dfrac{5}{100}$

Here 2.05 has more digits than 2.5, but 2.5 > 2.05.

Hence, a decimal number with more digits is not always greater than one with fewer digits.

Mahi purchases 0.25 kg of beans, 0.3 kg of carrots, 0.5 kg of potatoes, 0.2 kg of capsicums, and 0.05 kg of ginger. Calculate the total weight of the items she bought.

Answer

Given:

Weight of beans = 0.25 kg

Weight of carrots = 0.3 kg

Weight of potatoes = 0.5 kg

Weight of capsicums = 0.2 kg

Weight of ginger = 0.05 kg

Total weight = Sum of the weights of all items

$\begin{array}{r} 0.25 \\ 0.30 \\ 0.50 \\ 0.20 \\ +\phantom{0}0.05 \\ \hline 1.30 \\ \hline \end{array}$

Hence, the total weight of the items Mahi bought = 1.3 kg.

Pinto supplies 3.79 L, 4.2 L, and 4.25 L of milk to a milk dairy in the first three days. In 6 days, he supplies 25 litres of milk. Find the total quantity of milk supplied to the dairy in the last three days.

Answer

Given:

Milk supplied in the first three days = 3.79 L, 4.2 L and 4.25 L

Total milk supplied in 6 days = 25 L

Total milk supplied in the first three days:

$\begin{array}{r} 3.79 \\ 4.20 \\ +\phantom{0}4.25 \\ \hline 12.24 \\ \hline \end{array}$

Milk supplied in the last three days = (Total milk in 6 days) − (Milk in first three days)

$\begin{array}{r} 25.00 \\ -\phantom{0}12.24 \\ \hline 12.76 \\ \hline \end{array}$

Hence, the total quantity of milk supplied in the last three days = 12.76 L.

Tinku weighed 35.75 kg in January and 34.50 kg in February. Has he gained or lost weight? How much is the change?

Answer

Given:

Weight in January = 35.75 kg

Weight in February = 34.50 kg

Since 34.50 kg < 35.75 kg, Tinku has lost weight.

Change in weight = (Weight in January) − (Weight in February)

$\begin{array}{r} 35.75 \\ -\phantom{0}34.50 \\ \hline \phantom{0}1.25 \\ \hline \end{array}$

Hence, Tinku has lost weight, and the change in weight = 1.25 kg.

Extend the pattern: 5.5, 6.4, 6.39, 7.29, 7.28, 8.18, 8.17, .........., ..........

Answer

Let us observe the pattern by finding the change at each step:

5.5 + 0.9 = 6.4 $\quad$ [Add 0.9]

6.4 - 0.01 = 6.39 $\quad$ [Subtract 0.01]

6.39 + 0.9 = 7.29 $\quad$ [Add 0.9]

7.29 - 0.01 = 7.28 $\quad$ [Subtract 0.01]

7.28 + 0.9 = 8.18 $\quad$ [Add 0.9]

8.18 - 0.01 = 8.17 $\quad$ [Subtract 0.01]

So the pattern is: add 0.9, then subtract 0.01, repeating.

Continuing after 8.17:

8.17 + 0.9 = 9.07 $\quad$ [Add 0.9]

9.07 - 0.01 = 9.06 $\quad$ [Subtract 0.01]

Hence, the next two numbers are 9.07 and 9.06.

How many millimeters make 1 kilometer?

Answer

We convert step by step using the metric relations.

1 km = 1000 m

1 m = 1000 mm

Therefore,

1 km = 1000 × 1000 mm

= 10,00,000 mm

Hence, 1 kilometre = 10,00,000 mm.

Indian Railways offers optional travel insurance for passengers who book e-tickets. It costs 45 paise per passenger. If 1 lakh people opt for insurance in a day, what is the total insurance fee paid?

Answer

Given:

Insurance cost for 1 passenger = 45 paise

Number of passengers = 1 lakh = 1,00,000

Total insurance fee = (Cost per passenger) × (Number of passengers)

= (45 × 1,00,000) paise

= 45,00,000 paise

Converting paise into rupees:

₹1 = 100 paise

45,00,000 paise = ?

= $₹\dfrac{45,00,000}{100}$

= ₹45,000

Hence, the total insurance fee paid = ₹45,000.

Which is greater?

(a) $\dfrac{10}{1000}$ or $\dfrac{1}{10}$?

(b) One-hundredth or 90 thousandths?

(c) One-thousandth or 90 hundredths?

Answer

We convert each quantity into decimal form and then compare.

(a) $\dfrac{10}{1000}$ or $\dfrac{1}{10}$