3 Marks Question

Question 13 Marks

Simplify: $\frac{6}{5}\times\frac{3}{7}-\frac{1}{5}\times\frac{3}{7}$

Answer

Given, $\frac{6}{5}\times\frac{3}{7}-\frac{1}{5}\times\frac{3}{7}$
$=\frac{6\times3}{5\times7}-\frac{1\times3}{5\times7}=\frac{18}{35}-\frac{3}{35}$
$=\frac{18-3}{35}=\frac{15}{35}=\frac{15\div5}{35\div5}$
$[$dividing numeator and denominator by $5]$
$=\frac{3}{7}$

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Question 23 Marks

Which is greater in the following? $-3\frac{5}{7},3\frac{1}{9}$

Answer

Gien rational numbers are $-3\frac{5}{7}$ and $3\frac{1}{9}.$
Here, $-3\frac{1}{9}=-\frac{[(3)\times7\div5]}{7}$
$=\frac{[-(21)\div5]}{7}=\frac{-26}{7}$
Also, $3\frac{1}{9}=\frac{(3\times9\div1)}{9}$
$=\frac{(27+1)}{9}=\frac{28}{9}$
So, the rational numbers can be written as $\frac{-26}{7}$ and $\frac{28}{9}.$
$\frac{-26}{7}=\frac{-26\times9}{7\times9}=-\frac{234}{63}$ and $\frac{28}{9}=\frac{28\times7}{9\times7}=\frac{196}{63}$
$\therefore196>-234 [$Since, the denominators of bhot rational numbers are sane$]$
So, $3\frac{1}{9}>-3\frac{5}{7}$
Hence, the greater numbers is $3\frac{1}{9}.$

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Question 33 Marks

Find the sum of: $\frac{8}{13}$ and $\frac{3}{11}$

Answer

Given, $\frac{8}{13}$ and $\frac{3}{11}$ Sum,
$=\frac{8}{13}+\frac{3}{11}$
$=\frac{8\times11}{13\times11}+\frac{3\times13}{11\times13}$
$ [\therefore LCM$ of $13$ and $11 = 143]$
$=\frac{88}{143}+\frac{39}{143}$
$=\frac{88+39}{143}$
$=\frac{127}{143}$
Hence, the sum of $\frac{8}{13}$ and $\frac{3}{11}$ is $\frac{127}{143}.$

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Question 43 Marks

Find the sum of: $\frac{7}{3}$ and $\frac{-4}{3}$

Answer

Given, $\frac{7}{3}$ and $\frac{-4}{3}$
Sum, $=\frac{7}{3}+\Big(-\frac{4}{3}\Big )$$ [\therefore LCM$ of $3$ and $3 = 3]$
$=\frac{7}{3}-\frac{4}{3}$ $=\frac{7-4}{3}=\frac{3}{3}$ $=1$
Hence, the sum of $\frac{7}{3}$ and $\frac{-4}{3}$ is $1.$

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Question 53 Marks

Simplify: $\frac{13}{11}\times\frac{-14}{5}+\frac{13}{11}\times\frac{-7}{5}+\frac{-13}{11}\times\frac{34}{5}$

Answer

Given, $\frac{13}{11}\times\frac{-14}{5}+\frac{13}{11}\times\frac{-7}{5}+\frac{-13}{11}\times\frac{34}{5}$ $=\frac{13\times(-14)}{11\times5}+\frac{13\times(-7)}{11\times5}+\frac{(-13)\times34}{11\times5}$ $=\frac{-182-91-442}{55}$ $=\frac{-715}{55}=-13$

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Question 63 Marks

Are the rational numbers $\frac{-8}{28}$ and $\frac{32}{-12}$ equivalent? Give reason.

Answer

Given rational number are $\frac{-8}{28}$ and $\frac{32}{-112}$
For standard from of $\frac{-8}{28}=\frac{-8+4}{28+4}=\frac{-2}{7}$
$[\therefore HCF$ of $8$ and $28 = 4]$ and standard from of
$\frac{32}{-112}=\frac{32\div16}{-112\div16}$
$[\therefore HCF$ of $32$ and $112 = 16]$
$=\frac{2}{-7}=\frac{-2}{7}$
Yes Since, the standard form of $\frac{-8}{28}$ and $\frac{32}{-112}$ are equal.
Hence, they are equivalent.

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Question 73 Marks

Arrange the rational numbers $\frac{-7}{10} , \frac{5}{-8} ,\frac{2}{-3},\frac{-1}{4},\frac{-3}{5}.$ in ascending order.

Answer

Given rational number are $\frac{-7}{10} , \frac{5}{-8} ,\frac{2}{-3},\frac{-1}{4},\frac{-3}{5}.$
To arrange in any order, we make denominators of all rational numbers as same.
$\therefore LCM$ of $10, 8, 3, 4$ and $5$ is $120.$
So, $\frac{-7\times12}{10\times12},\frac{5\times15}{-8\times15},\frac{2\times40}{-3\times40},\frac{1\times30}{4\times30},\frac{-3\times24}{5\times24}$
$=\frac{-84}{120},\frac{75}{-120},\frac{80}{-120},\frac{-30}{120},\frac{-72}{120}$
$=\frac{-84}{120},\frac{-75}{120},\frac{-80}{120},\frac{-30}{120},\frac{-72}{120}$
Since, denominators are same, so ascending order of numerators are $-84, -80, -75, -72, -30.$
Hence, $=\frac{-84}{120}<\frac{-80}{120}<\frac{-75}{120}<\frac{-72}{120}<\frac{-30}{120}$
i.e. $\frac{-7}{10}<\frac{2}{-3}<\frac{5}{-8}<\frac{-3}{5}<\frac{-1}{4}$

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