NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals (EX 2.5) Exercise 2.5

Q.1

\begin{array}{l}Find:\\ \left(i\right)0.2\times 6\\ \left(ii\right)8\times 4.6\\ \left(iii\right)2.71\times 5\\ \left(iv\right)20.1\times 4\\ \left(v\right)0.05\times 7\\ \left(vi\right)211.02\times 4\\ \left(vii\right)2\times 0.86\end{array}

Ans.

\begin{array}{l}\left(\text{i}\right)0.\text{2}\times \text{6 =}\boxed{1.2}\\ \left(\text{ii}\right)\text{8}\times \text{4}.\text{6}=\boxed{36.8}\\ \left(\text{iii}\right)\text{2}.\text{71}\times \text{5}=\boxed{13.55}\\ \left(\text{iv}\right)\text{2}0.\text{1}\times \text{4}\text{=}\boxed{\text{80}\text{.4}}\\ \left(\text{v}\right)0.0\text{5}\times \text{7}\text{=}\boxed{\text{0}\text{.35}}\\ \left(\text{vi}\right)\text{211}.0\text{2}\times \text{4}=\boxed{844.08}\\ \left(\text{vii}\right)\text{2}\times 0.\text{86}\text{=}\boxed{\text{1}\text{.72}}\end{array}

Q.2

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{are}\mathrm{a}\mathrm{of}\mathrm{rectangle}\mathrm{whose}\mathrm{length}\mathrm{is}5.7\mathrm{cm}\\ \mathrm{and}\mathrm{breadth}\mathrm{is}3\mathrm{cm}.\end{array}$

Ans.

$\begin{array}{l}\text{The area of a rectangle is given by length}\times \text{breadth.}\\ \text{So, the area of given rectangle would be 5.7 cm}\times 3\text{\hspace{0.17em}cm}\\ \text{=}\boxed{{\text{17.1 cm}}^2}\end{array}$

Q.3

\begin{array}{l}Find:\\ \left(i\right)1.3\times 10\left(ii\right)36.8\times 10\left(iii\right)153.7\times 10\\ \left(iv\right)168.07\times 10\left(v\right)31.1\times 100\left(vi\right)156.1\times 100\\ \left(vii\right)3.62\times 100\left(viii\right)43.07\times 100\left(ix\right)0.5\times 10\\ \left(x\right)0.08\times 10\left(xi\right)0.9\times 100\left(xii\right)0.03\times 1000\end{array}

Ans.

\begin{array}{l}\left(\text{i}\right)\text{1}.\text{3}\times \text{1}0=\boxed{13}\\ \left(\text{ii}\right)\text{36}.\text{8}\times \text{1}0=\boxed{368}\\ \left(\text{iii}\right)\text{153}.\text{7}\times \text{1}0=\boxed{1537}\\ \left(\text{iv}\right)\text{168}.0\text{7}\times \text{1}0=\boxed{1680.7}\\ \left(\text{v}\right)\text{31}.\text{1}\times \text{1}00=\boxed{3110}\\ \left(\text{vi}\right)\text{156}.\text{1}\times \text{1}00=\boxed{15610}\\ \left(\text{vii}\right)\text{3}.\text{62}\times \text{1}00=\boxed{362}\\ \left(\text{viii}\right)\text{43}.0\text{7}\times \text{1}00=\boxed{4307}\\ \left(\text{ix}\right)0.\text{5}\times \text{1}0=\boxed{5}\\ \left(\text{x}\right)0.0\text{8}\times \text{1}0=\boxed{0.8}\\ \left(\text{xi}\right)0.\text{9}\times \text{1}00=\boxed{90}\\ \left(\text{xii}\right)0.0\text{3}\times \text{1}000=\boxed{30}\end{array}

Q.4

$\begin{array}{l}\mathrm{A}\mathrm{two}–\mathrm{wheeler}\mathrm{covers}\mathrm{a}\mathrm{distance}\mathrm{of}55.3\mathrm{km}\mathrm{in}\mathrm{one}\\ \mathrm{litre}\mathrm{of}\mathrm{petrol}.\mathrm{How}\mathrm{much}\mathrm{distance}\mathrm{will}\mathrm{it}\mathrm{cover}\mathrm{in}10\mathrm{litres}\\ \mathrm{of}\mathrm{petrol}?\end{array}$

Ans.

$\begin{array}{l}\text{Distance covered by a two-wheeler in 1 litres}\\ \text{= 55.3 km}\\ \text{So, distance covered by a two-wheeler in 10 litres}\\ \text{=55.3 km}\times \text{10}\\ \text{=}\boxed{\text{553 km}}\end{array}$

Q.5

\begin{array}{l}Find:\\ \left(i\right)2.5\times 0.3\left(ii\right)0.1\times 51.7\left(iii\right)0.2\times 316.8\\ \left(iv\right)1.3\times 3.1\left(v\right)0.5\times 0.05\left(vi\right)11.2\times 0.15\\ \left(vii\right)1.07\times 0.02\left(viii\right)10.05\times 1.05\left(ix\right)101.01\times 0.01\\ \left(x\right)100.01\times 1.1\end{array}

Ans.

\begin{array}{l}\left(\text{i}\right)\text{2}.\text{5}\times 0.\text{3 =}\boxed{\text{0}\text{.75}}\\ \left(\text{ii}\right)0.\text{1}\times \text{51}.\text{7}=\boxed{5.17}\\ \left(\text{iii}\right)0.\text{2}\times \text{316}.\text{8}=\boxed{63.36}\\ \left(\text{iv}\right)\text{1}.\text{3}\times \text{3}.\text{1=}\boxed{\text{4}\text{.03}}\\ \left(\text{v}\right)0.\text{5}\times 0.0\text{5}=\boxed{0.025}\\ \left(\text{vi}\right)\text{11}.\text{2}\times 0.\text{15=}\boxed{\text{1}\text{.68}}\\ \left(\text{vii}\right)\text{1}.0\text{7}\times 0.0\text{2=}\boxed{\text{0}\text{.0214}}\\ \left(\text{viii}\right)\text{1}0.0\text{5}\times \text{1}.0\text{5=}\boxed{\text{10}\text{.5525}}\\ \left(\text{ix}\right)\text{1}0\text{1}.0\text{1}\times 0.0\text{1=}\boxed{\text{1}\text{.0101}}\\ \left(\text{x}\right)\text{1}00.0\text{1}\times \text{1}.\text{1=}\boxed{\text{110}\text{.011}}\end{array}