# Excercise 2.5 Fractions and Decimals - NCERT Solutions Class 7

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## Chapter 2 Ex.2.5 Question 1

What is greater?

i) $$0.5 \rm \,or\,0.05$$

ii) $$0.7 \rm \,or\,0.5$$

iii) $$7 \rm \,or\,0.7$$

iv) $$1.37 \rm \,or\,1.49$$

v) $$2.03 \rm \,or\,2.30$$

vi) $$0.8 \rm \,or\,0.88$$

### Solution

What is known?

Decimal numbers

What is unknown?

Which decimal number is greater.

Reasoning:

First convert these decimals into fractions then convert them into like fraction, now we can simply find out which fraction/decimal is greater.

Steps:

(i) $$0.5$$ or $$0.05$$

\begin{align}0.5 \quad&\boxed{\;\;}\quad 0.05\\\\\frac{5}{10} \quad &\boxed{\;\;} \quad \frac{5}{100}\end{align}

Converting them into like fractions, we get

\begin{align}\frac{5\times 10}{10\times 10} &\quad\boxed{\;\;}\quad \frac{5\times 1}{100\times 1}\\\\\frac{50}{100} &\quad\boxed{\;\;}\quad \frac{50}{100}\\\\\frac{50}{100} &\quad\boxed{\gt}\quad \frac{5}{100}\end{align}

Therefore$$,0.5 > 0.05.$$

ii) $$0.7$$ or $$0.5$$

\begin{align}~0.7\, \quad\boxed{\;\;}\quad 0.5\\\\\frac{7}{10} \quad\boxed{\;\;}\quad \frac{5}{10} \\\\ \frac{7}{10} \quad\boxed{\gt}\quad \frac{5}{10}\end{align}

Therefore,$$0.7$$ $$>$$ $$0.5$$.

iii) $$7$$ or $$0.7$$

\begin{align}\text{7} \quad\boxed{\;\;}\quad \frac{7}{10}\\\\=\frac{7\times 10}{1\times 10} \quad\boxed{\;\;}\quad \frac{7}{10}\\\\\frac{70}{10} \quad\boxed{\gt}\quad \frac{7}{10}\end{align}

Therefore, $$7 > 0.7.$$

$$7$$ is greater.

iv) $$1.37$$ or $$1.49$$

\begin{align}&=1\text{.37} \quad\boxed{\;\;}\quad \text{1}\text{.49}\\\\&=\frac{137}{100} \quad\boxed{\;\;}\quad \frac{149}{100}\\\\&=\frac{137}{100} \quad\boxed{\lt}\quad \frac{149}{100}\end{align}

Therefore, $$1.37$$ $$<$$$$1.49$$

$$1.49$$ is greater.

v) $$2.03$$ or $$2.30$$

\begin{align}\text{2}\text{.03} \quad\boxed{\;\;}\quad \text{2}\text{.30} \\\\\frac{203}{100} \quad\boxed{\;\;}\quad \frac{230}{100}\\\\\frac{203}{100} \quad\boxed{\lt}\quad \frac{230}{100}\end{align}

Therefore, $$2.03$$ $$<$$ $$2.30$$

$$2.30$$ is greater.

vi) $$0.8$$ or $$0.88$$

\begin{align}\frac{08}{10} \quad\boxed{\;\;}\quad \frac{088}{100}\end{align}

Converting them into like fractions, we get

\begin{align}\frac{8\times 10}{10\times 10} \quad\boxed{\;\;}\quad \frac{88}{100}\\\\\frac{80}{100} \quad\boxed{\lt}\quad \frac{88}{100} \end{align}

Therefore, $$0.8 < 0.88.$$

$$0.88$$ is greater.

## Chapter 2 Ex.2.5 Question 2

Express as rupees using decimals:

i) $$7$$ rupees $$7$$ paise

ii) $$7$$ rupees $$7$$ paise

iii) $$7$$ rupees $$7$$ paise

iv) $$7$$ rupees $$7$$ paise

v) $$7$$ rupees $$7$$ paise

### Solution

What is known?

Amount in paise.

What is unknown?

Amount in rupees.

Reasoning:

$$1$$ paise $$=$$ \begin{align}\frac{1}{{100}}\end{align}rupees

Steps:

i) $$7$$ paise

$$100$$ paise $$=$$ $$1$$ rupees

$$1$$ paise $$=$$ \begin{align}\frac{1}{{100}}\end{align} rupees

$$7$$ paise $$=$$ $$7$$ $$×$$\begin{align}\frac{1}{{100}}\end{align}rupees $$= 0.07$$ rupees

ii) $$7$$ rupees $$7$$ paise

$$100$$ paise $$=$$ $$1$$ rupees

$$1$$ paise $$=$$ \begin{align}\frac{1}{{100}}\end{align}rupees

$$7$$ rupees $$7$$ paise $$=$$  $$7$$ rupees $$+$$ $$7$$ paise

$$=$$ $$7$$ rupees $$+$$ $$7$$ $$×$$\begin{align}\frac{1}{{100}}\end{align}rupees

$$=$$ $$7$$ rupees $$+$$ $$0.07$$ rupees

$$=$$ $$7.07$$ rupees

iii) $$77$$ rupees $$77$$ paise

$$77$$ rupees $$77$$ paise $$=$$  $$77$$ rupees $$+$$ $$77$$ paise

$$=$$ $$77$$ rupees + $$77$$ $$×$$ \begin{align}\frac{1}{{100}}\end{align}rupees

$$\because$$$$100$$ paise $$=$$ $$1$$ rupees)

($$\therefore$$ $$1$$ paise $$=$$ \begin{align}\frac{1}{{100}}\end{align}rupees)

$$=$$ $$77$$ rupees $$+$$ $$0.77$$ rupees

$$=$$ $$77.77$$ rupees

iv) $$50$$ paise

$$\therefore$$ $$100$$ paise $$=$$ $$1$$ rupees

$$\therefore$$ $$1$$ paise $$=$$ \begin{align}\frac{1}{{100}}\end{align} rupees

$$50$$ paise $$=$$ $$50$$ $$×$$\begin{align}\frac{1}{{100}}\end{align}rupees

=> $$50$$ paise $$=$$ \begin{align}\frac{{50}}{{100}}\end{align}rupees

$$50$$ paise $$=$$ $$0.50$$ rupees

v) $$235$$ paise

$$\therefore$$ $$100$$ paise $$=$$ $$1$$ rupees

$$\therefore$$ $$1$$ paise $$=$$ \begin{align}\frac{1}{{100}}\end{align} rupees

$$235$$ paise $$=$$ $$235$$ $$×$$ \begin{align}\frac{1}{{100}}\end{align} rupees

$$235$$ paise $$=$$ \begin{align}\frac{{235}}{{100}}\end{align}rupees

$$235$$paise $$=$$ $$2.35$$ rupees

## Chapter 2 Ex.2.5 Question 3

i) Express $$5\,\rm{cm}$$ in meter and kilometer

ii) Express $$35\,\rm{mm}$$ in $$\,\rm{cm}$$,$$\,\rm{m}$$ and $$\,\rm{km}$$.

### Solution

What is known?

Lengths in centimeter and millimeter

What is unknown?

Length in different units.

Reasoning:

\begin{align}1\, \rm{cm} =\frac{1}{{100}}\rm{m}\end{align}

\begin{align}1\rm{m} = \frac{1}{{1000}}\,\rm{km}\end{align}

\begin{align}1\rm{mm} = \frac{1}{{10}}\rm{cm}\end{align}

Steps:

(i) $$5\,\rm{cm}$$

$$\because100\,\rm{cm}$$ $$=$$$$1\,\rm{m}$$

$$\therefore1\,\rm{cm}$$ $$=$$\begin{align}\frac{1}{{100}}\end{align}m

$$5\,\rm{cm}$$  $$=$$ $$5$$ $$×$$  \begin{align}\frac{{1}}{100}\rm{m}\end{align}

$$5\,\rm{cm}$$ $$=$$ $$0.05\,\rm{m}$$

Also, $$\because1000\,\rm{m}$$ $$=$$ $$1\,\rm{km}$$

$$\because1\,\rm{m}$$ $$=$$ \begin{align}\frac{1}{{1000}}\end{align}$$\rm{km}$$

Thus, $$0.05\,\rm{m}$$ $$=$$ $$0.05$$ $$×$$ \begin{align}\frac{1}{{1000}}\end{align}$$\rm{km}$$

\begin{align} & = \frac{{0.05}}{{1000}}{\text{km}} \\ & = 0.00005{\text{km}}\end{align}

(ii) Express $$35\,\rm{mm}$$ in $$\rm{cm}$$,$$\rm{m}$$ and $$\rm{km}$$.

$$\because10\,\rm{mm}$$ $$=$$ $$1\,\rm{cm}$$

$$\because1\,\rm{mm}$$ $$=$$ \begin{align}\frac{1}{{10}}\end{align} $$\rm{cm}$$

\begin{align}35{\text{mm}} &= 35 \times \frac{1}{{10}}{\text{cm}} \\ 35{\text{mm}} &= \frac{{35}}{{10}}{\text{cm}} \\ 35{\text{mm}} &= 3.5{\text{cm}} \end{align}

Now, converting $$3.5\,\rm{cm}$$ in $$\rm{m}$$, we get

$$\because100\,\rm{cm} = 1 \,\rm{m}$$

\because 1\,\rm{cm} = \begin{align}\frac{1}{{100}}\end{align}\rm{m}

3.5\,\rm{cm} = 3.5 × \begin{align}\frac{1}{{100}}\end{align}\,\rm{m}

\begin{align} 3.5{\text{cm}} &= \frac{{3.5}}{{100}}{\text{m}} \\ 3.5{\text{cm}} &= 0.035{\text{m}} \end{align}

Again, converting $$0.035\,\rm{m}$$ into $$\rm{km}$$

$$\because1000\,\rm{m}$$ $$=$$ $$1\,\rm{km}$$

\because 1\,\rm{m} = \begin{align}\frac{1}{{1000}}\end{align}\,\rm{km}

0.035\,\rm{m} = 0.035 × \begin{align}\frac{1}{{1000}}\end{align}\,\rm{km}

0.035 \,\rm{m} = \begin{align}\frac{{0.035}}{{1000}}\end{align}\,\rm{km}

$$0.035\,\rm{m} = 0.000035\,\rm{km}$$

## Chapter 2 Ex.2.5 Question 4

Express in $$\rm{kg}$$:

i) $$200$$ kg

ii) $$3470$$ g

iii) $$4\,\rm{kg} \ 8\,\rm{g}$$

### Solution

What is known?

Weight in grams

What is unknown?

Weight in kilograms.

Reasoning:

\begin{align}1\rm{g} = \frac{1}{{1000}}\rm{kg}\end{align}

Steps:

(i) $$200\,\rm{g}$$

$$\because1000\,\rm{g} = 1\,\rm{kg}$$

\because 1\,\rm{g} = \begin{align}\frac{1}{{1000}}\end{align}\,\rm{kg}

So,

\begin{align}&200\,\rm{g} = 200 ×\frac{1}{{1000}}\,\rm{kg}\\&=\frac{{200}}{{1000}}\,\rm{kg}\\&= 0.200\,\rm{kg}\end{align}

or

\begin{align} &=\frac{{200}}{{1000}}\\\\&=\frac{2}{{10}}\\\\&= 0.2\,\rm{kg}\end{align}

(ii) $$3470\,\rm{g}$$

$$\because 1000\,\rm{g} =1\,\rm{kg}$$

\because1\,\rm{g} =\begin{align}\frac{1}{{1000}}\end{align}\,\rm{kg}

\begin{align}3470\,\rm{g}& = 3470 \times (\frac{1}{{1000}})\,\rm{kg}\\3470\,\rm{g} &= (\frac{{3470}}{{1000}})\,\rm{kg}\\3470{\,\rm{g}} &= \frac{{347}}{{100}} \\ 3470{\,\rm{g}} &= 3.47{\,\rm{kg}} \end{align}

(iii) $$4\,\rm{kg} \ 8\,\rm{g}$$

\begin{align}4\,\rm{kg} \ 8\,\rm{g} &= 4\,\rm{kg} + 8\,\rm{g}\\4\,\rm{kg} \ 8 \,\rm{g} &=4\,\rm{kg} + \frac{8}{{1000}}\,\rm{kg} \end{align}

$$\because 1000\,\rm{g}= 1\,\rm{kg}$$

\because 1\,\rm{g} = \begin{align}\frac{1}{{1000}}\end{align}\,\rm{kg}

\begin{align}4 \,\rm{kg} \ 8\,\rm{g} &= 4 \,\rm{kg} + 0.008 \,\rm{kg}\\4\,\rm{kg} \ 8\,\rm{kg}&=4.008\,\rm{kg}\end{align}

## Chapter 2 Ex.2.5 Question 5

Write the following decimal numbers in expanded form:-

i)   $$20.03$$

ii)  $$2.03$$

iii) $$200.03$$

iv) $$2.034$$

### Solution

What is known?

Decimal numbers

What is unknown?

Decimal numbers in expanded form.

Reasoning:

 Hundreds Tens Ones Decimal Tenths Hundredths 100 10 1 --------- $\frac{1}{{10}}$ $\frac{1}{{100}}$

Steps:

i) $$20.03$$

\begin{align}20.03 = &2 \times 10 + 0 \times 1 +\\&0 \times\frac{1}{{10}} + 3 \times \frac{1}{{100}}\end{align}

ii) $$2.03$$

\begin{align}2.03 = &2 \times 1 + 0 \times\frac{1}{10} + \\&3 \times \frac{1}{100} \end{align}

iii) $$200.03$$

\begin{align} 200.03 = &2 \times 100 + 0 \times 10 + 0 \times 1 \\&+ 0 \times \frac{1}{10} + 3 \times \frac{1}{100} \end{align}

iv) $$2.034$$

\begin{align} 2.034 = &2 × 1 + 0 × \frac{1}{10} + \\& 3 \times \frac{1}{100} + 4 \times \frac{1}{1000} \end{align}

## Chapter 2 Ex.2.5 Question 6

Write the place value of $$2$$ in the following decimal numbers:-

i) $$2.56$$

ii) $$21.37$$

iii) $$10.25$$

iv) $$9.42$$

v) $$63.352$$

### Solution

What is known?

Decimal numbers

What is unknown?

Place value of $$2$$.

Reasoning:

 Hundreds Tens Ones Decimal Tenths Hundredths 100 10 1 ---- $\frac{1}{{10}}$ $\frac{1}{{100}}$

Steps:

i) $$2.56$$

$$\underline{2}.56$$

$$2$$ is at ones place.

ii) $$21.37$$

$$\underline{2}1.37$$

$$2$$ is at ten's place.

iii) $$10.25$$

$$10. \underline{2} 5$$

$$2$$ is at tenths place.

iv) $$9 .4 2$$

$$9 .4 \underline{2}$$

$$2$$ is at hundredth place.

V) $$63.352$$

$$63 .3 5\underline{2}$$

$$2$$ is at thousandth place.

## Chapter 2 Ex.2.5 Question 7

Dinesh went from place $$A$$ to place $$B$$ and from there to place $$C$$. $$A$$ is $$7.5\,\rm{km}$$ from $$B$$ and $$B$$ is $$12.7\,\rm{km}$$ from $$C$$. Ayub went from place $$A$$ to place $$D$$ and from there to place $$C$$. $$D$$ is $$9.3\,\rm{km}$$ from $$A$$ and $$C$$ is $$11.8\,\rm{km}$$ from $$D$$.Who travelled more and how much ? ### Solution

What is known?

Distance between points. Dinesh went from place $$A$$ to $$B$$ and then $$B$$ to $$C$$. Ayub went from place $$A$$ to $$D$$ then $$D$$ to $$C$$.

What is unknown?

Who travelled more and how much.

Reasoning:

Calculate by all the distance travelled by Dinesh to how much he travelled then calculate distance travel by Ayub.

Steps:

Given Distance travelled by Dinesh from $$A$$ to $$B = 7. 5\,\rm{km}$$

And from place $$B$$ to place $$C = 12.7 \,\rm{km}$$

$$\therefore$$ Total distance travelled by Dinesh

\begin{align} &= AB+ BC\\&= 7.5\,\rm{km} + 12.7\,\rm{km}\\&= 20.2 \,\rm{km}\end{align}

Distance travelled by Ayub from place $$A$$ to place $$D = 9.3 \,\rm{km}$$

And from place $$D$$ to place $$C = 11.8\,\rm{km}$$

$$\therefore$$ Total distance travelled by Ayub

\begin{align}&= 9.3\,\rm{km} + 11.8 \,\rm{km}\\&= 21.1 \,\rm{km}\end{align}

On comparing the total distance travelled by Dinesh and Ayub, we get

$$21.1\,\rm{km}$$$$>$$ $$20.2\,\rm{km}$$

Distance travelled by Ayub > Distance travelled by Dinesh

$$\therefore$$ Ayub covered more distance by Dinesh is

\begin{align}&= 21.1 -20.2\\&= 0.9 \,\rm{km}\\&= 0.9 × 1000 \,\rm{m}\\&= 900\,\rm{m}\end{align}

## Chapter 2 Ex.2.5 Question 8

Shyama bought $$5\,\rm{kg}\,300\,\rm{g}$$. apples and $$3\,\rm{kg}\ 250\,\rm{g}$$ mangoes. Sarala bought $$4\,\rm{kg}\ 800\,\rm{g}$$ oranges and $$4\,\rm{kg}\ 150\,\rm{g}$$ bananas. Who bought more fruits?

### Solution

What is known?

Fruits bought by Shyama and Sarala.

What is unknown?

Who bought more fruits.

Reasoning:

Find out total fruits purchased by both Shyama and Sarala then we easily find out who bought more.

Steps:

Weight of apples bought by Shyama $$=5\,\rm{kg}\ 300\,\rm{g}$$

Weight of mangoes bought by Shyama $$=3\,\rm{kg}\ 250\,\rm{g}$$

$$\therefore$$ Total weight of fruits bought by Shyama

\begin{align}&= 5 \,\rm{kg}\ 300 \,\rm{g} + 3 \,\rm{kg}\ 250 \,\rm{g} \\&= 8 \,\rm{kg}\ 550 \,\rm{g}\end{align}

Also,

Weight of oranges bought by Sarala $$=4\,\rm{kg}\ 800\,\rm{g}$$

Weight of oranges bought by Sarala $$=4\,\rm{kg}\ 150\,\rm{g}$$

$$\therefore$$ Total weight of fruits bought by Sarala

\begin{align}&= 4\,\rm{kg}\ 800 \,\rm{g}\ + 4 \,\rm{kg}\ 150 \,\rm{g}\ \\&= 8\,\rm{kg}\ 950\,\rm{kg}\end{align}

On comparing the quantity of fruits, we get

$$8\,\rm{kg}\ 950\,\rm{g} > 8 \,\rm{kg}\ 550 \,\rm{kg}$$

Thus, Sarala bought more fruits.

## Chapter 2 Ex.2.5 Question 9

How much less is $$28\,\rm{km}$$ than $$42.6\,\rm{km}$$?

### Solution

What is known?

Two numbers.

What is unknown?

Difference between these two numbers.

Reasoning:

We can simply calculate it by subtracting smaller number from bigger number.

Steps:

Here,

We have to find out the difference between $$28\,\rm{km}$$ and $$42.6\,\rm{km}$$

\begin{align}\therefore\text{Difference}&= 42.6\ – 28\\&= 14. 6 \,\rm{km}\end{align}

Thus, $$14.6\,\rm{km}$$ less is $$28\,\rm{km}$$ than $$42.6$$.

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