[5 marks sum]

Question 15 Marks

If a boy works for six consecutive days for 8 hours, $7 \frac{1}{2}$ hours, $8 \frac{1}{4}$ hours, $6 \frac{1}{4}$ hours $+6 \frac{3}{4}$ hrs and 7 hours respectively. How much money will he earn at the rate of Rs. 36 per hour?

Answer

No. of hours, a boy worked in 6 days
$=8 \mathrm{hrs}+7 \frac{1}{2} \mathrm{hrs}+8 \frac{1}{4} \mathrm{hrs}+6 \frac{1}{4} \mathrm{hrs}+6 \frac{3}{4} \mathrm{hrs}+7 \mathrm{hrs}$
$=\left(8+\frac{15}{2}+\frac{33}{4}+\frac{25}{4}+\frac{27}{4}+7\right)$ hours
$=\frac{32+30+33+25+27+28}{4}$ hours
$=\frac{175}{4}$ hours $=43 \frac{3}{4}$ hours
Earning per hour $=$ Rs. 36
$\therefore$ Total earnings $=$ Rs. $\frac{175}{4} \times 36$
$=$ Rs. $175 \times 9=$ Rs, 1575

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Question 25 Marks

From a sack of potatoes weighing $120 \mathrm{~kg}$, a merchant sells portions weighing $6 \mathrm{~kg}, 5 \frac{1}{4} \mathrm{~kg}, 9 \frac{1}{2} \mathrm{~kg}$, and $9 \frac{3}{4} \mathrm{~kg}$ respectively.
(i) How many kg did he sell?
(ii) How many kg is still left in the sack?

Answer

Total quantities of potatoes $=120 \mathrm{~kg}$
(i) Quantity of potatoes he sold
$ \begin{aligned} & =6 \mathrm{~kg}+5 \frac{1}{4} \mathrm{~kg}+9 \frac{1}{2} \mathrm{~kg}+9 \frac{3}{4} \mathrm{~kg} \\ & =\left(6+\frac{21}{4}+\frac{19}{2}+\frac{39}{4}\right) \mathrm{kg} \\ & =\frac{24+21+38+39}{4} \mathrm{~kg} \\ & =\frac{122}{4}=\frac{61}{2} \mathrm{~kg}=30 \frac{1}{2} \mathrm{~kg} \end{aligned} $
(ii) Quantity of potatoes left
$ =120 \mathrm{~kg}-30 \frac{1}{2} \mathrm{~kg}=\left(\frac{120}{1}-\frac{61}{2}\right) \mathrm{kg} $
$ =\frac{240-61}{2}=\frac{179}{2} \mathrm{~kg}=89 \frac{1}{2} \mathrm{~kg} $

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Question 35 Marks

Simplify: $\frac{6}{5}$ of $\left(3 \frac{1}{3}-2 \frac{1}{2}\right)+\left(2 \frac{5}{21}-2\right)$

Answer

$\begin{aligned} & \frac{6}{5} \text { of }\left(3 \frac{1}{3}-2 \frac{1}{2}\right) \div\left(2 \frac{5}{21}-2\right) \\ & =\frac{6}{5} \text { of }\left(\frac{10}{3}-\frac{5}{2}\right) \div\left(\frac{47}{21}-\frac{2}{1}\right) \ldots(\text { Using BODMAS }) \\ & =\frac{6}{5} \text { of }\left(\frac{20-15}{6}\right) \div\left(\frac{47-42}{21}\right) \\ & =\frac{6}{5} \text { of } \frac{5}{6} \div \frac{5}{21} \\ & =1 \div \frac{5}{21} \\ & =1 \times \frac{21}{5}=\frac{21}{5}=4 \frac{1}{5}\end{aligned}$

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Question 45 Marks

Simplify: $1 \frac{1}{5}+\left\{2 \frac{1}{3}-(5+\overline{2-3})\right\}-3 \frac{1}{2}$

Answer

$\begin{aligned} & 1 \frac{1}{5}+\left\{2 \frac{1}{3}-(5+\overline{2-3})\right\}-3 \frac{1}{2} \\ & =\frac{6}{5} \div\left\{\frac{7}{3}-(5-1)\right\}-\frac{7}{2} \\ & =\frac{6}{5} \div\left\{\frac{7}{3} \div 4\right\}-\frac{7}{2}=\frac{6}{5} \div\left\{\frac{7-12}{3}\right\}-\frac{7}{2} \\ & =\frac{6}{5} \div \frac{-5}{3}-\frac{7}{2}=\frac{6}{5} \times \frac{3}{-5}-\frac{7}{2} \\ & =-\frac{18}{25}-\frac{7}{2}=\frac{-36-175}{50}=\frac{-211}{50} \\ & =-4 \frac{11}{50}\end{aligned}$

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Question 55 Marks

Simplify: $12 \frac{1}{2}-\left[8 \frac{1}{2}+\{9-(5-\overline{3-2})\}\right]$

Answer

$\begin{aligned} & 12 \frac{1}{2}-\left[8 \frac{1}{2}+\{9-(5-\overline{3-2})\}\right] \\ & =\frac{25}{2}-\left[\frac{17}{2}+\{9-(5-1)\}\right] \\ & =\frac{25}{2}-\left[\frac{17}{2}+\{9-4\}\right]=\frac{25}{2}-\left[\frac{17}{2}+5\right] \\ & =\frac{25}{2}-\frac{17}{2}-\frac{5}{1}=\frac{25-17-10}{2} \\ & =\frac{25-27}{2}=-\frac{2}{2}=-1\end{aligned}$

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Question 65 Marks

Simplify: $2 \frac{3}{4}-\left[3 \frac{1}{8} \div\left\{5-\left(4 \frac{2}{3}-\frac{11}{12}\right)\right\}\right]$

Answer

$\begin{aligned} & 2 \frac{3}{4}-\left[3 \frac{1}{8} \div\left\{5-\left(4 \frac{2}{3}-\frac{11}{12}\right)\right\}\right] \\ & =\frac{11}{4}-\left[\frac{25}{8} \div\left\{5-\left(\frac{14}{3}-\frac{11}{12}\right)\right\}\right] \\ & =\frac{11}{4}-\left[\frac{25}{8} \div\left\{5-\left(\frac{56-11}{12}\right)\right\}\right] \\ & =\frac{11}{4}-\left[\frac{25}{8} \div\left\{5-\frac{45}{12}\right\}\right] \\ & =\frac{11}{4}-\left[\frac{25}{8} \div\left\{\frac{60-45}{12}\right\}\right] \\ & =\frac{11}{4}-\left[\frac{25}{8} \div \frac{15}{12}\right]=\frac{11}{4}-\left[\frac{25}{8} \times \frac{12}{15}\right] \\ & =\frac{11}{4}-\frac{5}{2}=\frac{11-10}{4}=\frac{1}{4}\end{aligned}$

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Question 75 Marks

Reduce to a single fraction:
$\frac{2}{3}-\frac{3}{5}+3-\frac{1}{5}$

Answer

$\begin{aligned} & \frac{2}{3}-\frac{3}{5}+3-\frac{1}{5} \\ & \left.=\frac{2 \times 5}{3 \times 5}-\frac{3 \times 3}{5 \times 3}+\frac{3 \times 15}{15}-\frac{1 \times 3}{5 \times 3} \quad \text { (LCM of } 3 \text { and } 5=15\right.) \\ & =\frac{10}{15}-\frac{9}{15}+\frac{45}{15}-\frac{3}{15} \\ & =\frac{10-9+45-3}{15}=\frac{55-12}{15} \\ & =\frac{43}{15}=2 \frac{13}{15}\end{aligned}$

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Question 85 Marks

Arrange the given fraction in descending order by making numerator equal:
$\frac{1}{10}, \frac{6}{11}, \frac{8}{11} \text { and } \frac{3}{5}$

Answer

$ \frac{1}{10}, \frac{6}{11}, \frac{8}{11} \text { and } \frac{3}{5} $
L.C.M. of numerators $1,6,8$ and $3=24$
$ \begin{aligned} & \therefore \frac{1}{10}=\frac{1 \times 24}{10 \times 24}=\frac{24}{240} \\ & \frac{6}{11}=\frac{6 \times 4}{11 \times 4}=\frac{24}{44} \\ & \frac{8}{11}=\frac{8 \times 3}{11 \times 3}=\frac{24}{33} \\ & \frac{3}{5}=\frac{3 \times 8}{5 \times 8}=\frac{24}{40} \end{aligned} $
It is clear from the above that
$ \begin{aligned} & \frac{24}{33}>\frac{24}{40}>\frac{24}{44}>\frac{24}{240} \\ & \text { or } \frac{8}{11}>\frac{3}{5}>\frac{6}{11}>\frac{1}{10} \end{aligned} $
Hence, $\frac{8}{11}, \frac{3}{5}, \frac{6}{11}, \frac{1}{10}$ are in descending order.

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Question 95 Marks

Arrange the given fraction in descending order by making numerator equal:
$\frac{3}{7}, \frac{4}{8}, \frac{5}{7} \text { and } \frac{8}{11}$

Answer

$ \frac{3}{7}, \frac{4}{8}, \frac{5}{7} \text { and } \frac{8}{11} $
L.C.M. of numerators $3,4,5$ and $8=120$
$ \begin{aligned} & \therefore \frac{3}{7}=\frac{3 \times 40}{7 \times 40}=\frac{120}{280} \\ & \frac{4}{9}=\frac{4 \times 30}{9 \times 30}=\frac{120}{270} \\ & \frac{5}{7}=\frac{5 \times 24}{7 \times 24}=\frac{120}{168} \\ & \frac{8}{11}=\frac{8 \times 15}{11 \times 15}=\frac{120}{165} \end{aligned} $
It is clear from the above that
$ \begin{aligned} & \frac{120}{165}>\frac{120}{168}>\frac{120}{270}>\frac{120}{280} \\ & \text { or } \frac{8}{11}>\frac{5}{7}>\frac{4}{9}>\frac{3}{7} \end{aligned} $
Hence, $\frac{8}{11}, \frac{5}{7}, \frac{4}{9}, \frac{3}{7}$ are in descending order.

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Question 105 Marks

Arrange the given fraction in descending order by making numerator equal:
$\frac{5}{6}, \frac{4}{15}, \frac{8}{9} \text { and } \frac{1}{3}$

Answer

$ \frac{5}{6}, \frac{4}{15}, \frac{8}{9} \text { and } \frac{1}{3} $
L.C.M. of numerators 5, 4, 8 and $1=40$
$ \begin{aligned} & \therefore \frac{5}{6}=\frac{5 \times 8}{6 \times 8}=\frac{40}{48} \\ & \frac{4}{15}=\frac{4 \times 10}{15 \times 10}=\frac{40}{150} \end{aligned} $
$ \frac{8}{9}=\frac{8 \times 5}{9 \times 5}=\frac{40}{45} $
$ \frac{1}{3}=\frac{1 \times 40}{3 \times 40}=\frac{40}{120} $
From above we see that
$ \begin{aligned} & \frac{40}{45}>\frac{40}{48}>\frac{40}{120}>\frac{40}{150} \\ & \therefore \frac{8}{9}>\frac{5}{6}>\frac{1}{3}>\frac{4}{15} \end{aligned} $
Hence, $\frac{8}{9}, \frac{5}{6}, \frac{1}{3}, \frac{4}{15}$ are in descending order.

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Question 115 Marks

Put the given fraction in ascending order by making denominator equal:
$\frac{5}{7}, \frac{3}{8}, \frac{9}{14} \text { and } \frac{20}{21}$

Answer

$ \frac{5}{7}, \frac{3}{8}, \frac{9}{14} \text { and } \frac{20}{21} $
L.C.M. of denominators 7, 8, 14 and $21=168$
$ \begin{aligned} & \therefore \frac{5}{7}=\frac{5 \times 24}{7 \times 24}=\frac{120}{168} \\ & \frac{3}{8}=\frac{3 \times 21}{8 \times 21}=\frac{63}{168} \\ & \frac{9}{14}=\frac{9 \times 12}{14 \times 12}=\frac{108}{168} \\ & \frac{20}{21}=\frac{20 \times 8}{21 \times 8}=\frac{160}{168} \end{aligned} $
It is clear from the above that
$ \begin{aligned} & \frac{63}{168}<\frac{108}{168}<\frac{120}{168}<\frac{160}{168} \\ & \text { or } \frac{3}{8}<\frac{9}{14}<\frac{5}{7}<\frac{20}{21} \end{aligned} $
Hence, $\frac{3}{8}, \frac{9}{14}, \frac{5}{7}$ and $\frac{20}{21}$ are in ascending order.

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Question 125 Marks

Put the given fraction in ascending order by making denominator equal:
$\frac{5}{6}, \frac{7}{8}, \frac{11}{12} \text { and } \frac{3}{10}$

Answer

$ \frac{5}{6}, \frac{7}{8}, \frac{11}{12} \text { and } \frac{3}{10} $
L.C.M. of denominators $6,8,12$ and $10=240$
$ \begin{aligned} & \therefore \frac{5}{6}=\frac{5 \times 40}{6 \times 40}=\frac{200}{240} \\ & \frac{7}{8}=\frac{7 \times 30}{8 \times 30}=\frac{210}{240} \\ & \frac{11}{12}=\frac{11 \times 20}{12 \times 20}=\frac{220}{240} \\ & \frac{3}{10}=\frac{3 \times 24}{10 \times 24}=\frac{72}{240} \end{aligned} $
It is clear from the above that
$ \begin{aligned} & \frac{72}{240}<\frac{200}{240}<\frac{210}{240}<\frac{220}{240} \\ & \text { or } \frac{3}{10}<\frac{5}{6}<\frac{7}{8}<\frac{11}{12} \end{aligned} $
Hence, $\frac{3}{10}, \frac{5}{6}, \frac{7}{8}$ and $\frac{11}{12}$ are in ascending order.

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Question 135 Marks

Convert given fraction into a fraction with equal denominator:
$\frac{4}{5}, \frac{17}{20}, \frac{23}{40} \text { and } \frac{11}{16}$

Answer

$ \ln \frac{4}{5}, \frac{17}{20}, \frac{23}{40} \text { and } \frac{11}{16} $
L.C.M of $5,20,40$ and $16=80$
$ \therefore \frac{4}{5}=\frac{4 \times 16}{5 \times 16}=\frac{64}{80} $
$ \frac{17}{20}=\frac{17 \times 4}{20 \times 4}=\frac{68}{80} $
$ \frac{23}{40}=\frac{23 \times 2}{40 \times 2}=\frac{46}{80} $
and $\frac{11}{16}=\frac{11 \times 5}{16 \times 5}=\frac{55}{80}$
Hence, the required fractions are $\frac{64}{80}, \frac{68}{80}, \frac{46}{80}$ and $\frac{55}{80}$.

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