Excercise 2.5 Fractions and Decimals - NCERT Solutions Class 7

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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

Video 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.

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

Video 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

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

Video 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 \(\rm{km}\)

 \(\because1000\,\rm{m}\) \(=\) \(1\,\rm{km}\)

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

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

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

\(0.035\,\rm{m} = 0.000035\,\rm{km} \)

Question 4

Express in \(\rm{kg}\):

i) \(200\) kg

ii) \(3470\) g

iii) \(4\,\rm{kg} \ 8\,\rm{g}\)

Solution

Video 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) \(4\,\rm{kg} \ 8\,\rm{g}\)

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

\(\because 1000\,\rm{g}= 1\,\rm{kg}\)

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

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

Question 5

Write the following decimal numbers in expanded form:-

i)   \( 20.03\)

ii)  \(2.03\)

iii) \(200.03\)

iv) \(2.034\)

Solution

Video 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) \(2.034\)

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

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

Video 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.

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

Video 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

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

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

Video 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.

Question 9

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

Solution

Video 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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