Find:
$(i).\ 2.5 \times 0.3$
$(ii).\ 0.1 \times 51.7$
$(iii).\ 0.2 \times 316.8$
$(iv).\ 1.3 \times 3.1$
$(v).\ 0.5 \times 0.05$
$(vi).\ 11.2 \times 0.15$
$(vii).\ 1.07 \times 0.02$
$(viii).\ 10.05 \times 1.05$
$(ix).\ 101.01 \times 0.01$
$(x).\ 100.01 \times 1.1$

AcademicMathematicsNCERTClass 7

To do:

We have to find 

(i) $2.5 \times 0.3$

(ii) $0.1 \times 51.7$

(iii) $0.2 \times 316.8$

(iv) $1.3 \times 3.1$

(v) $0.5 \times 0.05$

(vi) $11.2 \times 0.15$

(vii) $1.07 \times 0.02$

(viii) $10.05 \times 1.05$

(ix) $101.01 \times 0.01$

(x) $100.01 \times 1.1$

Solution:

We know that,

On dividing a decimal by $10^n$, the decimal point shifts to the left by $n$ places.

(i) $2.5\times0.3=\frac{25}{10}\times\frac{3}{10}$

$=\frac{25\times3}{10\times10}$

$=\frac{75}{100}$

$=0.75$

(ii) $0.1\times51.7=\frac{1}{10}\times\frac{517}{10}$

$=\frac{1\times517}{10\times10}$

$=\frac{517}{100}$

$=5.17$

(iii) $0.2\times316.8=\frac{2}{10}\times\frac{3168}{10}$

$=\frac{2\times3168}{10\times10}$

$=\frac{6336}{100}$

$=63.36$

(iv) $1.3\times3.1=\frac{13}{10}\times\frac{31}{10}$

$=\frac{13\times31}{10\times10}$

$=\frac{403}{100}$

$=4.03$

(v) $0.5\times0.05=\frac{5}{10}\times\frac{5}{100}$

$=\frac{5\times5}{10\times100}$

$=\frac{25}{1000}$

$=0.025$

(vi) $11.2\times0.15=\frac{112}{10}\times\frac{15}{100}$

$=\frac{112\times15}{10\times100}$

$=\frac{1680}{1000}$

$=1.68$

(vii) $1.07\times0.02=\frac{107}{100}\times\frac{2}{100}$

$=\frac{107\times2}{100\times100}$

$=\frac{214}{10000}$

$=0.0214$

(viii) $10.05\times1.05=\frac{1005}{100}\times\frac{105}{100}$

$=\frac{1005\times105}{100\times100}$

$=\frac{105525}{10000}$

$=10.5525$

(ix) $101.01\times0.01=\frac{10101}{100}\times\frac{1}{100}$

$=\frac{10101\times1}{100\times100}$

$=\frac{10101}{10000}$

$=1.0101$

(x) $100.01\times1.1=\frac{10001}{100}\times\frac{11}{10}$

$=\frac{10001\times11}{100\times10}$

$=\frac{110011}{1000}$

$=110.011$

raja
Updated on 10-Oct-2022 13:32:37

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