Excercise 2.6 Fractions and Decimals- NCERT Solutions Class 7

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Chapter 2 Ex.2.6 Question 1

 Find

  i) \(0.2\times 6\)

 ii) \( 8 \times4.6 \)  

iii) \(2.71\times 5\)   

iv) \(20.1 \times 4\)

 v) \(0.05 \times 7\) 

vi) \(211.02 \times 4\) 

vii) \(2 \times 0.86\)

Solution

Video Solution

What is known?

One decimal number and one whole number.

What is unknown?

Product of these two numbers.

Reasoning:

Covert decimal number into fraction and then find the product.

Steps:

i) \(0.2\times 6\)

\[\begin{align}0.2 \times 6 &= \frac{2}{{10}} \times 6\\\,\,\,0.2 \times 6 &= \frac{{2 \times 6}}{{10}}\\\,\,\,\,0.2 \times 6 &= \frac{{12}}{{10}}\\\,\,\,\,0.2 \times 6 &= 1.2\\\end{align}\]

 ii) \( 8 \times4.6 \)  

\[\begin{align}8 \times 4.6 &= 8 \times \frac{{46}}{{10}}\\\,\,\,\,\,\,\,8 \times 4.6 &= \frac{{8 \times 46}}{{10}}\\\,\,\,\,\,\,\,\,8 \times 4.6 &= \frac{{368}}{{10}}\\\,\,\,\,\,\,\,\,8 \times 4.6 &= 3.68\end{align}\]

iii) \(2.71\times 5\)   

\[\begin{align}2.71 \times 5 &= \frac{{271}}{{100}} \times 5\\\,\,\,\,\,\,\,\,2.71 \times 5 &= \frac{{271 \times 5}}{{100}}\\\,\,\,\,\,\,\,\,2.71 \times 5 &= \frac{{1355}}{{100}}\\\,\,\,\,\,\,\,\,2.71 \times 5 &= 13.55\end{align}\]

iv) \(20.1 \times 4\)

\[\begin{align}20.1 \times 4 &= \frac{{201}}{{10}} \times 4\\\,\,\,\,\,\,\,20.1 \times 4 &= \frac{{201 \times 4}}{{10}}\\\,\,\,\,\,\,\,20.1 \times 4 &= \frac{{804}}{{10}}\\\,\,\,\,\,20.1 \times 4 &= 80.4\end{align}\]

v) \(0.05 \times 7\)

\[\begin{align}0.05 \times 7 &= \frac{5}{{100}} \times \frac{7}{1}\\\,\,\,\,\,\,0.05 \times 7 &= \frac{{5 \times 7}}{{100}}\\\,\,\,\,\,\,0.05 \times 7 &= \frac{{35}}{{100}}\\\,\,\,\,\,\,0.05 \times 7 &= 0.35\end{align}\]

vi) \(211.02 \times 4\) 

\[\begin{align}211.02 \times 4 &= \frac{{21102}}{{100}} \times \frac{4}{1}\\\,\,\,\,\,\,\,\,\,211.02 \times 4 &= \frac{{21102 \times 4}}{{100}}\\\,\,\,\,\,\,\,\,\,211.02 \times 4 &= \frac{{84408}}{{100}}\\\,\,\,\,\,\,\,\,\,211.02 \times 4 &= 844.08\end{align}\]

vii) \(2 \times 0.86\)

\[\begin{align}2 \times 0.86& = \frac{2}{1} \times \frac{{86}}{{100}}\\\,\,\,\,\,\,\,2 \times 0.86 &= \frac{{2 \times 86}}{{100}}\\\,\,\,\,\,\,\,2 \times 0.86 &= \frac{{172}}{{100}}\\\,\,\,\,\,\,2 \times 0.86 &= 1.72\end{align}\]

Chapter 2 Ex.2.6 Question 2

Find the area of rectangle whose length is \(\rm{}5.7\,cm\) and breadth is \(\rm{}3\, cm.\)

Solution

Video Solution

What is known?

Dimensions of rectangle.

What is unknown?

Area of rectangle.

Reasoning:

Area of rectangle \(=\) Length \(×\) Breadth

Steps:

Given:

Length of rectangle \(=\rm{} 5.7\, cm\)

Breadth of rectangle \(= \rm{}3\, cm \)

Therefore,

Area of rectangle \(=\) Length \(×\) Breadth

\[\begin{align}&= 5.7 \times 3\\&= 17.1\end{align}\]

Thus, the area of rectangle is \(\rm{}17.1\, cm\)

Chapter 2 Ex.2.6 Question 3

 Find:

   i) \(1.3 × 10 \)

  ii) \(\rm{}36.8 × 10 \)

 iii) \(\rm{}153.7× 10  \) 

 iv) \(\rm{}168.07 × 10\)

  v) \(\rm{}31.1×100\) 

 vi) \(\rm{}156.1 × 100\) 

 vii) \(\rm{}3.62 × 100  \) 

viii) \(\rm{}43.07 × 100\)

  ix) \(\rm{}0.5 × 10 \) 

  x) \(\rm{}0.08 × 10\) 

 xi) \(\rm{}0.9 ×100\) 

xii) \(\rm{}0.03 × 1000\)

Solution

Video Solution

What is known?

One decimal number and one whole number.

What is unknown?

Product of these numbers.

Reasoning:

We know that when a number is multiplied by \(10, 100\) or \(1000\) the digits in the products are same as in the decimal number but, the decimal point in the product is shifted to the right by as many places as there are zeros.

Steps:

   i) \(1.3 × 10 =13 .0 \)

  ii) \(\rm{}36.8 × 10 = 368 .0\)

 iii) \(\rm{}153.7× 10 = \rm{}1537.0\)

 iv) \(\rm{}168.07 × 10 = \rm{}1680.7\)

  v) \(\rm{}31.1×100 = \rm{}3110.0 \)

 vi) \(\rm{}156.1 × 100 = \rm{}15610.0\)

 vii) \(\rm{}3.62 × 100 = \rm{}362.0\)

viii) \(\rm{}43.07 × 100 = \rm{}4307.0\)

 ix) \(\rm{}0.5 × 10 =\rm{} 5.0 \)

  x) \(\rm{}0.08 × 10 = \rm{}0.8\)

 xi) \(\rm{}0.9 ×100 = \rm{}90.0 \)

xii) \(\rm{}0.03 × 1000 =\rm{} 30.0\)

Chapter 2 Ex.2.6 Question 4

A two wheeler covers a distance of \(\rm{}55.3\, km\) in \(\rm{}1\) liter of petrol. How much it will cover in \(\rm{}10\) liters of petrol?

Solution

Video Solution

What is known?

A two wheeler covers a distance of \(55.3\rm{ km}\) in \(1\) liter of petrol.

What is unknown?

How much it will cover in \(10\) liters of petrol.

Reasoning:

By using Unitary Method we can simply multiply \(55.3\) by \(10\) to get answer.

Steps:

Distance covered by two wheeler in \(1\) liter of petrol \(=\rm{}55.3\, km \)

Distance covered by \(10\) liter of petrol \(=\rm{} 55.3 × 10 =\rm{}553.0\, km \)

Therefore, it will cover a distance of \(\rm{}553 \,km\) in \(10\) liter of petrol.

Chapter 2 Ex.2.6 Question 5

 Find:

    i) \(2.5 × 0.3  \) 

   ii) \(\rm{}0.1 × 51.7  \) 

  iii) \(\rm{}0.2 × 316.8 \)

  iv) \(\rm{}1.3 × \rm{}3.1\)

   v) \(\rm{}0.5 ×0.05  \)

  vi) \(\rm{}11.2 × 0.15\)

 vii) \(\rm{}1.07× 0.02 \) 

viii) \(\rm{}10.05 × 1.05 \)

  ix) \(\rm{}101.01× 0.01 \)

   x) \(\rm{}100.01 × 1.1 \)

Solution

Video Solution

What is known?

Decimal numbers.

What is unknown?

Product of these two numbers.

Reasoning:

Covert decimal number into fraction and then find the product.

Steps:

i) \(2.5 × 0.3  \) 

\[\begin{align}&= \frac{{25}}{{10}} \times \frac{3}{{10}} \hfill \\\,\,\, &= \frac{{25 \times 3}}{{100}} \hfill \\\,\,\, &= \frac{{75}}{{100}} \hfill \\\,\,\, &= 0.75 \hfill \\\hfill \\\end{align}\] 

ii) \(\rm{}0.1 × 51.7  \) 

\[\begin{align}&= \frac{1}{{10}} \times \frac{{517}}{{10}} \hfill \\\,\,\, &= \frac{{1 \times 517}}{{100}} \hfill \\\,\,\, &= \frac{{517}}{{100}} \hfill \\\,\,\, &= 5.17 \hfill \\\end{align}\]

iii) \(\rm{}0.2 × 316.8 \)

\[\begin{align}&= \frac{2}{{10}} \times \frac{{3168}}{{10}} \hfill \\\,\,\,\,\, &= \frac{{2 \times 3168}}{{100}} \hfill \\\,\,\,\,\, &= \frac{{6336}}{{100}} \hfill \\\,\,\,\,\, &= 63.36 \hfill \\\end{align}\]

iv) \(\rm{}1.3 × \rm{}3.1\)

\[\begin{align}&= \frac{{13}}{{10}} \times \frac{{31}}{{10}} \hfill \\\,\,\,\,\,\,\, &= \frac{{13 \times 31}}{{100}} \hfill \\\,\,\,\,\,\,\, &= \frac{{403}}{{100}} \hfill \\\,\,\,\,\,\,\, &= 4.03 \hfill \\\end{align}\]

v) \(\rm{}0.5 ×0.05  \)

\[\begin{align}&= \frac{5}{{10}} \times \frac{5}{{100}} \hfill \\\,\,\,\, &= \frac{{5 \times 5}}{{1000}} \hfill \\\,\,\, &= \frac{{25}}{{1000}} \hfill \\\,\, &= 0.025 \hfill \\\hfill \\\end{align}\]

vi) \(\rm{}11.2 × 0.15\)

\[\begin{align}&= \frac{{112}}{{10}} \times \frac{{15}}{{100}} \hfill \\\,\,\,\,\, &= \frac{{112 \times 15}}{{1000}} \hfill \\\,\,\,\, &= \frac{{1680}}{{1000}} \hfill \\\,\,\, &= 1.68 \hfill \\\end{align}\]

vii) \(\rm{}1.07× 0.02 \) 

\[\begin{align}&= \frac{{107}}{{100}} \times \frac{2}{{100}} \hfill \\\,\,\,\,\,\, &= \frac{{107 \times 2}}{{10000}} \hfill \\\,\,\,\,\, &= \frac{{214}}{{10000}} \hfill \\\,\,\,\,\, &= 0.0214 \hfill \\\end{align}\]

viii) \(\rm{}10.05 × 1.05 \)

\[\begin{align}&= \frac{{1005}}{{100}} \times \frac{{105}}{{100}} \\{}& = \frac{{1005 \times 105}}{{10000}} \\{}& = \frac{{105525}}{{10000}} \\{}& = 10.5525 \\\end{align}\]

ix) \(\rm{}101.01× 0.01 \)

\[\begin{align}& = \frac{{10101}}{{100}} \times \frac{1}{{100}} \\{}& = \frac{{10101 \times 1}}{{10000}} \\{}& = \frac{{10101}}{{10000}} \\{}& = 1.0101 \\\end{align}\]

 x) \(\rm{}100.01 × 1.1 \)

\[\begin{align}& = \frac{{10001}}{{100}} \times \frac{{11}}{{10}} \\{}& = \frac{{10001 \times 11}}{{1000}} \\{}& = \frac{{110011}}{{1000}} \\{}& = 110.011 \\\end{align}\]

  
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