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Question 13 Marks
Find the sum:
$3\frac{1}{3}+4\frac{1}{4}+6\frac{1}{6}$
Answer
We have,
$L.C.M$. of $3, 4$ and $6 = (2 \times 2 \times 3) = 12$
$\begin{array}{c|c}3&3,4,6\\\hline2&1,4,2\\\hline2&1,2,1\\\hline&1,1,1\end{array}$
Therefore,
$3\frac{1}{3}+4\frac{1}{4}+6\frac{1}{6}$
$=\frac{10}{3}+\frac{17}{4}+\frac{37}{6}$
$=\frac{(40+51+74)}{12}$
$\Big(\frac{12}{3}=4,4\times10=40\Big)$
$\Big(\frac{12}{4}=3,3\times17=51\Big)$
and $\Big(\frac{12}{6}=2,2\times37=74\Big)$
$=\frac{165}{12}$
$=\frac{55}{4}$
$=13\frac{3}{4}$
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Question 23 Marks
Convert $\frac{1}{4},\frac{5}{8},\frac{7}{12}$ and $\frac{13}{24}$ into like fractions.
Answer
The given fractions are $\frac{1}{4},\frac{5}{8},\frac{7}{12}$ and $\frac{13}{24}$
$L.C.M.$ of $4, 8, 12$ and $24 = (4 \times 2 \times 3) = 24$
So, we convert the given fractions into equivalent fractions with $24$ as the denominator.
(But, one of the fractions already has $24$ as its denominator.
So, there is no need to convert it into an equivalent fraction.)
Thus, we have: $\frac{1}{4}=\frac{1\times6}{4\times6}=\frac{6}{24}$,
$\frac{5}{8}=\frac{5\times3}{8\times3}=\frac{15}{24}$,
$\frac{7}{12}=\frac{7\times2}{12\times2}=\frac{14}{24}$
Hence, the required like fractions are $\frac{6}{24},\frac{15}{24},\frac{14}{24}$ and $\frac{13}{24}$.
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Question 33 Marks
Find the sum:
$3\frac{2}{3}+1\frac{5}{6}+2$
Answer
We have,
$L.C.M.$ of $3$ and $6 = (2 \times 3) = 6$
$\begin{array}{c|c}3&3,6\\\hline2&1,2\\\hline&1,1\end{array}$
Therefore,
$3\frac{2}{3}+1\frac{5}{6}+2$
$=\frac{11}{3}+\frac{11}{6}+\frac{2}{1}$
$=\frac{(22+11+12)}{6}$
$\Big(\frac{6}{3}=2,2\times11=22\Big)$
$\Big(\frac{6}{6}=1,1\times11=11\Big)$
and $\Big(\frac{6}{1}=6,6\times2=12\Big)$
$=\frac{45}{6}$
$=\frac{15}{2}$
$=7\frac{1}{2}$
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Question 43 Marks
Lalita read $30$ pages of a book containing $100$ pages while Sarita read $\frac{2}{5}$ of the book. Who read more?
Answer
Lalita read $30$ pages of a book having $100$ pages Sarita read $\frac{2}{5}$ of the same book $\frac{2}{5}$ of $100$ pages = $\frac{2}{5}\times100$
$=\frac{200}{5}=40 = 40$ pages
Hence, Sarita read more pages than Lalita as $40$ is greater than $30.$
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Question 53 Marks
In a school $20$ students out of $25$ passed in $VI A$, while $24$ out of $30$ passed in $VI B$. Which section gave better result?
Answer
Fraction of students who passed in $ VI A =\frac{20}{25}$
$=\frac{20\div5}{25\div5}=\frac{4}{5}$
Fraction of students who passed in $VI B =\frac{24}{30}$
$=\frac{24\div6}{30\div6}=\frac{4}{5}$
In both the sections, the fraction of students who passed is the same, so both the sections have the same result.
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Question 63 Marks
Reduce the following fractions into its simplest form: $\frac{150}{60}$
Answer
Here, numerator $= 150$ and denominator $= 60$ Factors of $150$ are $1, 2, 3, 5, 6, 10, 15, 25, 30, 75$ and $150$
Factors of $60$ are $1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30$ and $60$
Common factors of $150$ and $60$ are $1, 2, 3, 5, 6, 10, 15$ and $30 H.C.F$. of $150$ and $60$ is $30.$
$\therefore\frac{150}{60}=\frac{150\div30}{60\div30}=\frac{5}{2}$
Hence, the simplest form of $\frac{150}{60}$ is $\frac{5}{2}$.
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Question 83 Marks
Subtract the sum of $3\frac{5}{9}$ and $3\frac{1}{3}$ from the sum of $5\frac{5}{6}$ and $4\frac{1}{9}$.
Answer
$\Big(5\frac{5}{6}+4\frac{1}{9}\Big)-\Big(3\frac{5}{9}+3\frac{1}{3}\Big)$$\Big(\frac{35}{6}+\frac{37}{9}\Big)-\Big(\frac{32}{9}+\frac{10}{3}\Big)$
$\begin{array}{c|c}2&6,9,3\\\hline3&3,9,3\\\hline3&1,3,1\\\hline&1,1,1\end{array}$
$L.C.M$. of $3, 6$ and $9 = (2 \times 2 \times 3) = 18$
$=\frac{[105+74]-[64+60]}{18}$
$\Big(\frac{18}{6}=3,3\times35=105\Big)$
and $\Big(\frac{18}{9}=2,2\times37=74\Big)$
$\Big(\frac{18}{9}=2,2\times32=64\Big)$
and $\Big(\frac{18}{3}=6,6\times10=60\Big)$
$=\frac{[179]-[124]}{18}$
$=\frac{55}{18}$
$=3\frac{1}{18}$
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Question 93 Marks
Find the sum: $3\frac{1}{8}+1\frac{5}{12}$
Answer
$L.C.M.$ of $8$ and $12 = (2 × 2 × 2 × 3) = 34$
$\begin{array}{c|c}2&8,12\\\hline2&4,6\\\hline2&2,3\\\hline3&1,3\\\hline&1,1\end{array} $
Therefore, $3\frac{1}{8}+1\frac{5}{12}$
$=\frac{25}{8}+\frac{17}{12}$
$=\frac{(75+34)}{24}$
$\Big(\frac{24}{8}=3,3\times25=75\Big)$ and $\Big(\frac{24}{12}=2,2\times17=34\Big)$
$=\frac{109}{24}$
$=4\frac{13}{24}$
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Question 103 Marks
The weight of an empty gas cylinder is $16\frac{4}{5}\text{kg}$ and it contains $14\frac{2}{3}\text{kg}$ of gas. What is the weight of the cylinder filled with gas?
Answer
Weight of the cylinder filled with gas = Weight of the empty cylinder $+$ Weight of the gas inside the cylinder
Thus, we have: $(L.C.M .of 5$ and $3 = (5 \times 3) = 15$
$\Big(16\frac{4}{5}+14\frac{2}{3}\Big)\text{kg}$
$=\Big(\frac{84}{5}+\frac{44}{3}\Big)\text{kg}$
$=\frac{(252+220)}{15}\text{kg}$
$=\frac{472}{15}\text{kg}$
$=31\frac{7}{15}\text{kg}$
Hence, the weight of the cylinder filled with gas is $31\frac{7}{15}\text{kg}$.
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Question 113 Marks
Simplify:$\frac{5}{8}+\frac{3}{4}-\frac{7}{12}$
Answer
We have:$\frac{5}{8}+\frac{3}{4}-\frac{7}{12}$
$\begin{array}{c|c}2&4,8,12\\\hline2&2,4,6\\\hline2&1,2,3\\\hline3&1,1,3\\\hline&1,1,1\end{array}$
$L.C.M.$ of $4, 8$ and $12 = (2 \times 2 \times 2 \times 3) = 24$
$=\frac{(15+18-14)}{24}$
$\Big(\frac{24}{8}=3,3\times5=15\Big)$
$\Big(\frac{24}{4}=6,6\times3=18\Big)$ and $\Big(\frac{24}{12}=2,2\times7=14\Big)$
$=\frac{(33-14)}{24}$
$=\frac{19}{24}$
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Question 123 Marks
Mrs Soni bought $7\frac{1}{2}$ litres of milk. Out of this milk, $5\frac{3}{4}$ litres was consumed. How much milk is left with her?
Answer
Amount of milk left with Mrs. Soni = Total amount of milk bought by her - Amount of milk consumed
$\therefore$ Amount of milk left with Mrs. Soni $=7\frac{1}{2}-5\frac{3}{4}$
$=\frac{15}{2}-\frac{23}{4}$
$L.C.M.$ of $2$ and $4 = (2 \times 2) = 4$
$=\frac{30-23}{4}$
$\Big(\frac{4}{2}=2,2\times15=30\Big)$ and $\Big(\frac{4}{4}=1,1\times23=23\Big)$
$=\frac{7}{4}$
$=1\frac{3}{4}\ \text{litres}$
Therefore, Milk left with Mrs. Soni $=1\frac{3}{4}\ \text{litres}$
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Question 133 Marks
What should be added to $9\frac{2}{3}$ to get $19?$
Answer
Let x be added to $9\frac{2}{3}$ to get $19$
Therefore, $9\frac{2}{3}+\text{x}=19$ Thus,
we have: $\text{x}=19-9\frac{2}{3}$ $=\frac{19}{1}-\frac{29}{3}$
$L.C.M.$ of $1$ and $3$ is 3$=\frac{(57-29)}{3}$
$\Big(\frac{3}{1}=3,3\times19=57\Big)$
and $\Big(\frac{3}{3}=1,1\times29=29\Big)$
$=\frac{28}{3}$
$=9\frac{1}{3}$
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Question 143 Marks
Show that the following fractions is in the simplest form: $\frac{9}{14}$
Answer
Here, numerator $= 9$ and denominator $= 14$
Factors of $9$ are $1, 3$ and $9$
Factors of $14$ are $1, 2, 7$ and $14$
Common factor of $9$ and $14$ is $1$
Thus, $H.C.F.$ of $9$ and $14$ is $1$
​​​​​​​Hence, $\frac{9}{14}$ is the simplest form.
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Question 153 Marks
Sohini bought $4\frac{1}{2}\text{m}$ of cloth for her kurta and $2\frac{2}{3}\text{m}$ of cloth for her pyjamas. Ho much cloth did she purchase in all?
Answer
Total cloth purchased by Sohini = Cloth for kurta + Cloth for pyjamas
Thus, we have:
$\Big(4\frac{1}{2}+2\frac{2}{3}\Big)\ \text{m}$
$=\Big(\frac{9}{2}+\frac{8}{3}\Big)\ \text{m}$
$(L.C.M.$ of $2$ and $3 = (2 \times 3) = 6)$
$=\Big(\frac{(27+16)}{6}\Big)\ \text{m}$
$\Big(\frac{16}{2}=3,3\times9=27\Big)$
and $\Big(\frac{6}{3}=2,2\times8=16\Big)$
$=\frac{43}{6}\ \text{m}$
$=7\frac{1}{6}\ \text{m}$
$\therefore$ Total length of cloth purchased $=7\frac{1}{6}\ \text{m}$
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Question 163 Marks
Simplify:$5\frac{3}{4}-4\frac{5}{12}+3\frac{1}{6}$
Answer
We have:$5\frac{3}{4}-4\frac{5}{12}+3\frac{1}{6}$
$\frac{23}{4}-\frac{53}{12}+\frac{19}{6}$
$\begin{array}{c|c}2&4,12,6\\\hline2&2,6,3\\\hline3&1,2,3\\\hline2&1,2,1\\\hline&1,1,1\end{array}$
$L.C.M$. of $4, 12$ and $6 = (2 \times 2 \times 3) = 12$
$=\frac{(69-53+38)}{12}$
$\Big(\frac{12}{4}=3,3\times23=69\Big)$
$\Big(\frac{12}{12}=1,1\times53=53\Big)$ and
$\Big(\frac{12}{6}=2,2\times19=38\Big)$
$=\frac{(107-53)}{12}$
$=\frac{54}{12}$
$=\frac{9}{2}$
$=4\frac{1}{2}$
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Question 173 Marks
Neelam has $25$ pencils. She gives $\frac{4}{5}$ of them to Meena. How many pencils does Meena get? How many pencils are left with Neelam?
Answer
Neelam gives $\frac{4}{5}$ of $25$
pencils to Meena $\Big(\frac{4}{5}\times25\Big) = 20$
Pencils Thus, Meena gets $20$ pencils.
Therefore, Number of pencils left with Neelam $= 25 - 20 = 5$ pencils
Thus, $5$ pencils are left with Neelam
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Question 183 Marks
Rohit bought a pencil for Rs. $3\frac{2}{5}$ and an eraser for Rs. $2\frac{7}{10}$. What is the total cost of both the articles?
Answer
Total cost of both articles $=$ Cost of pencil $+$ Cost of eraser
Thus, we have: $\text{Rs.}\ 3\frac{2}{5}+\text{Rs.}\ 2\frac{7}{10}$
$=\frac{17}{5}+\frac{27}{10}$
$=\frac{(34+27)}{10}$
$(L.C.M.$ of $5$ and $10 = (5 \times 2) = 10)$
$=\frac{61}{10}$
$=\text{Rs.}\ 6\frac{1}{10}$
Hence, the total cost of both the articles is $\text{Rs.}\ 6\frac{1}{10}$.
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Question 193 Marks
Show that the following fractions is in the simplest form: $\frac{8}{15}$
Answer
Here, numerator $= 8$ and denominator $= 15$
Factors of 8 are $1, 2, 4$ and $8$ Factors of $15$ are $1, 3, 5$ and $15$
Common factor of $8$ and $15$ is $1$
Thus, $H.C.F$. of $8$ and $15$ is $1$
Hence, $\frac{8}{15}$ is the simplest form.
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Question 203 Marks
Reduce the following fractions into its simplest form:
$\frac{48}{60}$
Answer
Here, numerator $= 84$ and denominator $= 98$
Factors of $84$ are $1, 2, 3, 4, 6, 7, 12, 14, 21, 42$ and $84$
Factors of $98$ are $1, 2, 7, 14, 49$ and $98$
Common factors of $84$ and $98$ are $1, 2, 7$ and $14$
$H.C.F.$ of $84$ and $98$ is $14$
$\therefore\frac{48}{60}=\frac{48\div12}{60\div12}=\frac{4}{5}$
Hence, the simplest form of $\frac{48}{60}$ is $\frac{4}{5}$.
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Question 213 Marks
What should be added to $6\frac{7}{15}$ to get $8\frac{1}{5}$?
Answer
Let $x$ be added to $6\frac{7}{15}$ to get $8\frac{1}{5}$
Therefore, $6\frac{7}{15}+\text{x}=8\frac{1}{5}$
Thus, we have: $\text{x}=8\frac{1}{5}-6\frac{7}{15}$
$=\frac{41}{5}-\frac{97}{15}$ $L.C.M$. of $5$ and $15 = (5 \times 3) = 15 =\frac{(123-97)}{15}$
$\Big(\frac{15}{5}=3,3\times41=123\Big)$
and $\Big(\frac{15}{15}=1,1\times97=97\Big)$
$=\frac{26}{15}$
$=1\frac{11}{15}$
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Question 223 Marks
Simplify:$3+1\frac{1}{5}+\frac{2}{3}-\frac{7}{15}$
Answer
We have:$3+1\frac{1}{5}+\frac{2}{3}-\frac{7}{15}$
$=\frac{3}{1}+\frac{6}{5}+\frac{2}{3}-\frac{7}{15}$
$\begin{array}{c|c}5&5,3,15\\\hline3&1,3,3\ \\\hline&1,1,1\ \end{array}$
$L.C.M$. of $5, 3$ and $15 = (5 \times 3) = 15$
 $=\frac{(45+18+10-7)}{15}$
$\Big(\frac{15}{1}=15,15\times3=45\Big)$
$\Big(\frac{15}{5}=3,3\times6=18\Big)$
$\Big(\frac{15}{3}=5,5\times2=10\Big)$ and
$\Big(\frac{15}{15}=1,1\times7=7\Big)$
$=\frac{(73-7)}{15}$
$=\frac{66}{15}$
$=\frac{22}{5}$
$=4\frac{2}{5}$
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Question 233 Marks
Show that the following fractions is in the simplest form:
$\frac{21}{10}$
Answer
Here, numerator $= 21$ and denominator $= 10$
Factors of $21$ are $1, 3, 7$ and $21$
Factors of $10$ are $1, 2, 5$ and $10$
Common factor of $21$ and $10$ is $1$
Thus, $H.C.F.$ of $21$ and $10$ is $1.$
Hence, $\frac{21}{10}$ is the simplest form.
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Question 243 Marks
Rafiq exercised for $\frac{2}{3}$ hour, while Rohit exercise for $\frac{3}{4}$ hour. Who exercised for a longer time?
Answer
To know who exercised for a longer time,
we have to compare $\frac{2}{3}$ hour with $\frac{3}{4}$
hour On cross multiplying: $4 \times 2 = 8$ and $3 \times 3 = 9$
Clearly, $8 < 9$
$\therefore\frac{2}{3}$
hour $< \frac{3}{4}$ hour
 Hence, Rohit exercised for a longer time.
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Question 253 Marks
Convert $\frac{3}{5},\frac{7}{10},\frac{8}{15}$ and $\frac{11}{30}$ into like fractions.
Answer
The given fractions are $\frac{3}{5},\frac{7}{10},\frac{8}{15}$ and $\frac{11}{30}$
$\begin{array}{c|c}5&5,10,15,30\\\hline2&1,2,3,6\ \ \ \ \ \ \\\hline3&1,1,3,3\ \ \ \ \ \\\hline&1,1,1,1\ \ \ \ \ \end{array}$
$L.C.M$. of $5, 10, 15$ and $30 = (5 \times 2 \times 3) = 30$
So, we convert the given fractions into equivalent fractions with $30$ as the denominator.
(But, one of the fractions already has $30$ as its denominator.
So, there is no need to convert it into an equivalent fraction.)
Thus, we have: $\frac{3}{5}=\frac{3\times6}{5\times6}=\frac{18}{30}$,
$\frac{7}{10}=\frac{7\times3}{10\times3}=\frac{21}{30}$,
$\frac{8}{15}=\frac{8\times2}{15\times2}=\frac{16}{30}$
Hence, the required like fractions are $\frac{18}{30},\frac{21}{30},\frac{16}{30}$ and $\frac{11}{30}$.
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Question 263 Marks
Find the sum: $2\frac{1}{3}+1\frac{1}{4}+2\frac{5}{6}+3\frac{7}{12}$
Answer
We have, $L.C.M$. of $3, 4, 6$ and $12 = (2 \times 2 \times 3) = 12$
$\begin{array}{c|c}2&3,4,6,12\\\hline2&3,2,3,6\ \ \\\hline3&3,1,3,3\ \ \\\hline&1,1,1,1\ \ \end{array}$
Therefore, $2\frac{1}{3}+1\frac{1}{4}+2\frac{5}{6}+3\frac{7}{12}$
$=\frac{7}{3}+\frac{5}{4}+\frac{17}{6}+\frac{43}{12}$
$=\frac{(28+15+34+43)}{12}$
$\Big(\frac{12}{3}=4,4\times7=28\Big)$
$\Big(\frac{12}{4}=3,3\times5=15\Big)$
$\Big(\frac{12}{6}=2,2\times17=34\Big)$ and
$\Big(\frac{12}{12}=1,1\times43=43\Big)$
$=\frac{120}{12}$
$=10$
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Question 273 Marks
Simplify:$2+5\frac{7}{10}-3\frac{14}{15}$
Answer
We have:$2+5\frac{7}{10}-3\frac{14}{15}$
$=\frac{2}{1}+\frac{57}{10}-\frac{59}{15}$
$\begin{array}{c|c}5&4,10,15\\\hline2&1,2,3\ \ \\\hline3&1,1,3\ \ \\\hline&1,1,1\ \ \end{array}$
$L.C.M.$ of $10$ and $15 = (2 \times 5 \times 3) = 30$
$=\frac{(69+171-118)}{30}$
$\Big(\frac{30}{1}=30,30\times2=60\Big)$
$\Big(\frac{30}{10}=3,3\times57=171\Big)$ and $\Big(\frac{30}{15}=2,2\times59=118\Big)$
$=\frac{(231-118)}{30}$
$=\frac{113}{30}$
$=3\frac{23}{30}$
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Question 283 Marks
Reduce the following fractions into its simplest form: $\frac{84}{98}$
Answer
Here, numerator $= 84$ and
denominator $= 98$
Factors of $84$ are $1, 2, 3, 4, 6, 7, 12, 14, 21, 42$ and $84$
Factors of $98$ are $1, 2, 7, 14, 49$ and $98$
Common factors of $84$ and $98$ are $1, 2, 7$ and $14 H.C.F$. of $84$ and $98$ is $14$
$\therefore\frac{84}{98}=\frac{84\div14}{98\div14}=\frac{6}{7}$
Hence, the simplest form of $\frac{84}{98}$ is $\frac{6}{7}$.
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Question 293 Marks
Simplify:
$\frac{5}{6}-\frac{4}{9}+\frac{2}{3}$
Answer
We have, $\frac{5}{6}-\frac{4}{9}+\frac{2}{3}$
$\begin{array}{c|c}3&3,6,9\\\hline3&1,2,3\\\hline2&1,2,1\\\hline&1,1,1\end{array}$
$L.C.M.$ of $3, 6$ and $9 = (2 \times 3 \times 3) = 18$
$=\frac{(15+8+12)}{18}$
$\Big(\frac{18}{6}=3,3\times5=15\Big)$
$\Big(\frac{18}{9}=2,2\times4=8\Big)$ and
$\Big(\frac{18}{3}=6,6\times2=12\Big)$
$=\frac{(27-8)}{18}$
$=\frac{19}{18}$
$=1\frac{1}{18}$
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Question 303 Marks
Find the difference:
$2\frac{7}{9}-1\frac{8}{15}$
Answer
$\begin{array}{c|c}3&9,15\\\hline3&3,5\ \ \\\hline5&1,5\ \ \\\hline&1,1\ \ \end{array}$
$L.C.M$ of $9$ and $15 = (3 \times 3 \times 5) = 45$
$2\frac{7}{9}-1\frac{8}{15}$
$=\frac{25}{9}-\frac{23}{15}$
$=\frac{(125-69)}{45}$
$=\frac{56}{45}$
$=1\frac{11}{45}$
$\Big(\frac{45}{9}=5,5\times25=125\Big)$
and $\Big(\frac{45}{15}=3,3\times23=69\Big)$
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Question 313 Marks
Simplify:$8\frac{5}{6}-3\frac{3}{8}+2\frac{7}{12}$
Answer
We have:$8\frac{5}{6}-3\frac{3}{8}+2\frac{7}{12}$
$=\frac{53}{6}-\frac{27}{8}+\frac{31}{12}$
$\begin{array}{c|c}2&6,8,12\\\hline2&3,4,6\ \\\hline3&3,2,3\ \\\hline2&1,2,1\ \\\hline&1,1,1\ \end{array}$
$L.C.M.$ of $6, 8$ and $12 = (2 \times 2 \times 2 \times 3) = 24$
$=\frac{(212-81+62)}{24}$
$\Big(\frac{24}{6}=4,4\times53=212\Big)$
$\Big(\frac{24}{8}=23,3\times7=81\Big)$ and
$\Big(\frac{24}{12}=2,2\times31=62\Big)$$=\frac{(274-81)}{24}$
$=\frac{193}{24}$
$=8\frac{1}{24}$
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Question 323 Marks
Show that the following fractions is in the simplest form: $\frac{25}{36}$
Answer
Here, numerator $= 25$ and denominator $= 36$
Factors of $25$ are $1, 5$ and $25$
Factors of $36$ are $1, 2, 3, 4, 6, 9, 12, 18$ and $36$
Common factor of $25$ and $36$ is $1$ Thus,
$H.C.F$. of $25$ and $36$ is $1$
Hence, $\frac{25}{36}$ is the simplest form.
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Question 333 Marks
Find the sum: $2+\frac{3}{4}+1\frac{5}{8}+3\frac{7}{16}$
Answer
We have, $L.C.M$. of $4, 8$ and $16 = (2 \times 2 \times 2 \times 2) = 16$
$\begin{array}{c|c}2&4,8,16\\\hline2&2,4,8\\\hline2&1,2,4\\\hline2&1,1,2\\\hline&1,1,1\end{array}$
Therefore, $2+\frac{3}{4}+1\frac{5}{8}+3\frac{7}{16}$
$=\frac{2}{1}+\frac{3}{4}+\frac{13}{8}+\frac{55}{16}$
$=\frac{(32+12+26+55)}{16}$
$\Big(\frac{16}{1}=16,6\times2=32\Big)$
$\Big(\frac{16}{4}=4,4\times3=12\Big)$
$\Big(\frac{16}{8}=2,2\times13=26\Big)$ and
$\Big(\frac{16}{16}=1,1\times55=55\Big)$
$=\frac{125}{16}$
$=7\frac{13}{16}$
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Question 343 Marks
Compare the fractions given below: $\frac{7}{8},\frac{9}{10}$
Answer
$L.C.M$. of $8$ and $10 = (2 \times 5 \times 2 \times 2) = 40$
Now, we convert $\frac{7}{8}$ and $\frac{9}{10}$
into equivalent fractions having $40$ as the denominator.
$\therefore\frac{7}{8}=\frac{7\times5}{8\times5}=\frac{35}{40}$ and
$\frac{9}{10}=\frac{9\times4}{10\times4}=\frac{36}{40}$
Clearly, $\frac{35}{40}<\frac{36}{40}$
$\therefore\frac{7}{8}<\frac{9}{10}$
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Question 353 Marks
Simplify:$8-3\frac{1}{2}-2\frac{1}{4}$
Answer
We have:$8-3\frac{1}{2}-2\frac{1}{4}$
$=\frac{8}{1}-\frac{7}{2}-\frac{9}{4}$
$\begin{array}{c|c}2&1,2,4\\\hline2&1,1,2\\\hline&1,1,1\end{array}$
$L.C.M.$ of $1, 2$ and $4 = (2 \times 2) = 4$
$=\frac{(32-14-9)}{4}$
$\Big(\frac{4}{1}=4,4\times8=32\Big)$
$\Big(\frac{4}{2}=2,2\times7=14\Big)$ and
$\Big(\frac{4}{4}=1,1\times9=9\Big)$
$=\frac{(32-23)}{34}$
$=\frac{9}{4}$
$=2\frac{1}{4}$
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Question 363 Marks
In one day, a rickshaw puller earned $\text{Rs.}\ 137\frac{1}{2}$. Out of this money, he spent $\text{Rs.}\ 56\frac{3}{4}$ on food. How much money is left with him?
Answer
Money left with the rickshaw puller $=$ Money earned by him in a day $-$ Money spent by him on food
$=\text{Rs.}\ \Big(137\frac{1}{2}-56\frac{3}{4}\Big)$
$L.C.M.$ of $2$ and $4 = (2 \times 2) = 4$
$=\text{Rs.}\ \Big(\frac{275}{2}-\frac{227}{4}\Big)$
$\Big(\frac{4}{2}=2,2\times275=550\Big)$ and $\Big(\frac{4}{4}=1,1\times227=227\Big)$
$=\text{Rs.}\ \Big(\frac{550-227}{4}\Big)$
$=\text{Rs.}\ \Big(\frac{323}{4}\Big)$
$=\text{Rs.}\ 80\frac{3}{4}$ Hence, $\text{Rs.}\ 80\frac{3}{4}$ is left with the rickshaw puller.
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Question 373 Marks
Find the sum:
$\frac{2}{3}+3\frac{1}{6}+4\frac{2}{9}+2\frac{5}{18}$
Answer
We have,
$L.C.M.$ of $3, 6, 9$ and $18 = (3 \times 3 \times 2) = 18$
$\begin{array}{c|c}3&3,6,9,18\\\hline3&1,2,3,6\ \ \\\hline2&1,2,1,2\ \ \\\hline&1,1,1,1\ \end{array}$
Therefore,
$\frac{2}{3}+3\frac{1}{6}+4\frac{2}{9}+2\frac{5}{18}$
$=\frac{2}{3}+\frac{19}{6}+\frac{38}{9}+\frac{41}{18}$
$=\frac{(12+57+76+41)}{18}$
$\Big(\frac{18}{3}=6,6\times2=12\Big)$
$\Big(\frac{18}{6}=3,3\times19=57\Big)$
$\Big(\frac{18}{9}=2,2\times38=76\Big)$
and $\Big(\frac{18}{18}=1,1\times41=41\Big)$
$=\frac{186}{18}$
$=\frac{31}{3}$
$=10\frac{1}{3}$
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Question 383 Marks
Find the difference: $\frac{5}{6}-\frac{4}{9}$
Answer
$\begin{array}{c|c}3&6,9\\\hline2&2,3\\\hline3&1,3\\\hline&1,1\end{array}$
$L.C.M$ of $6$ and $9 = (3 \times 2 \times 3) = 18$
Now, we have:
$\frac{5}{6}=\frac{5\times3}{6\times3}=\frac{15}{18}$
$\frac{4}{9}=\frac{4\times2}{9\times2}=\frac{8}{18}$
Therefore,
$\frac{5}{6}-\frac{4}{9}$
$=\frac{15}{18}-\frac{8}{18}$
$=\frac{(15-8)}{18}$
$=\frac{7}{18}$
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Question 393 Marks
Convert the fractions $\frac{1}{2},\frac{ 2}{3}, \frac{4}{9}$ and $\frac{5}{6}$ into like fractions.
Answer
The given fractions are $\frac{1}{2},\frac{ 2}{3}, \frac{4}{9},\frac{5}{6}$
$L.C.M$. of $2, 3, 9$ and $6 = (2 ⨯ 3 ​⨯ 3) = 18$
Now, we have: $\frac{1}{2}=\frac{1\times9}{2\times9}=\frac{9}{18}$
$\frac{2}{3}=\frac{2\times6}{3\times6}=\frac{12}{18}$
$\frac{4}{9}=\frac{4\times2}{9\times2}=\frac{8}{18}$
$\frac{5}{6}=\frac{5\times3}{6\times3}=\frac{15}{18}$
Hence, $\frac{9}{18},\frac{12}{18}, \frac{8}{18}$ and $\frac{15}{18}$ are like fractions.
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Question 403 Marks
Find the sum:
$3+1\frac{4}{15}+1\frac{3}{20}$
Answer
We have,
$L.C.M$. of $15$ and $20 = (2 \times 2 \times 3 \times 5) = 60$
$\begin{array}{c|c}5&15,20\\\hline3&3,4\ \ \ \ \\\hline2&1,4\ \ \ \ \\\hline2&1,2\ \ \ \ \ \\\hline&1,1\ \ \ \ \ \end{array} $
Therefore,
$3+1\frac{4}{15}+1\frac{3}{20}$
$=\frac{3}{1}+\frac{19}{15}+\frac{23}{20}$
$=\frac{(180+76+69)}{60}$
$\Big(\frac{60}{1}=60,60\times3=180\Big)$
$\Big(\frac{60}{15}=4,4\times19=76\Big)$
and $\Big(\frac{60}{20}=3,3\times23=69\Big)$
$=\frac{325}{60}$
$=\frac{65}{12}$
$=5\frac{5}{12}$
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Question 413 Marks
Simplify:$6\frac{1}{6}-5\frac{1}{5}+3\frac{1}{3}$
Answer
We have:$6\frac{1}{6}-5\frac{1}{5}+3\frac{1}{3}$
$=\frac{37}{6}-\frac{26}{5}+\frac{10}{3}$
$\begin{array}{c|c}2&6,5,3\\\hline3&3,5,3\\\hline5&1,5,1\\\hline&1,1,1\end{array}$
$L.C.M$. of $6, 5$ and $3 = (2 \times 5 \times 3) = 30$
$=\frac{(185-156+100)}{30}$
$\Big(\frac{30}{6}=5,5\times37=185\Big)$
$\Big(\frac{30}{5}=6,6\times26=156\Big)$ and
$\Big(\frac{30}{3}=10,10\times10=100\Big)$
​​​​​​​$=\frac{(285-156)}{30}$
$=\frac{129}{30}$
$=\frac{43}{10}$
$=4\frac{3}{10}$
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Question 423 Marks
Find the sum: $2\frac{7}{10}+3\frac{8}{15}$
Answer
We have, $L.C.M$. of $10$ and $15 = (2 \times 3 \times 5) = 30$
$\begin{array}{c|c}5&10,15\\\hline2&2,3\\\hline3&1,3\\\hline&1,1\end{array} $
Therefore, $2\frac{7}{10}+3\frac{8}{15}$
$=\frac{27}{10}+\frac{53}{15}$
$=\frac{(81+106)}{30}$
$\Big(\frac{30}{10}=3,3\times27=81\Big)$ and
$\Big(\frac{30}{15}=2,2\times53=106\Big)$
$=\frac{187}{30}$
$=6\frac{7}{30}$
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Question 433 Marks
Reduce $\frac{84}{98}$ to the simplest form.
Answer
The factors of $84$ are $1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84$
The factors of $98$ are $1, 2, 7, 14, 49, 98$
The common factors of $84$ and $98$ are $1, 2, 7, 14$
The $H.C.F.$ of $84$ and $98$ is $14$
Dividing both the numerator and the denominator by the
$H.C.F \frac{84}{98}=\frac{84\div14}{98\div14}=\frac{6}{7}$
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Question 443 Marks
Compare the fractions given below:
$\frac{4}{5},\frac{7}{10}$
Answer
$L.C.M.$ of $5$ and $10 = (5 \times 2) = 10$
Now, we convert $\frac{4}{5}$ into equivalent fractions having $10$ as the denominator.
$\therefore\frac{4}{5}=\frac{4\times2}{5\times2}=\frac{8}{10}$
Clearly, $\frac{8}{10}<\frac{7}{10}$
$\therefore\frac{4}{5}<\frac{7}{10}$
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Question 453 Marks
Simplify:$2+\frac{11}{15}-\frac{5}{9}$
Answer
We have:$2+\frac{11}{15}-\frac{5}{9}$
$\begin{array}{c|c}3&1,15,9\\\hline3&1,5,3\ \\\hline5&1,5,1\ \\\hline&1,1,1\ \end{array}$
$L.C.M.$ of $15$ and $9 = (3 \times 3 \times 5) = 45$
$=\frac{(90+33-25)}{45}$
$\Big(\frac{45}{1}=45,45\times2=90\Big)$
$\Big(\frac{45}{15}=3,3\times11=33\Big)$ and
$\Big(\frac{45}{9}=5,5\times5=25\Big)$
$=\frac{(98+8)}{45}$
$=\frac{98}{45}$
$=2\frac{8}{45}$
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Question 463 Marks
Compare the fractions given below: $\frac{4}{9},\frac{5}{6}$
Answer
$L.C.M.$ of $9$ and $6 = (3 \times 3 \times 2) = 18$
Now, we convert $\frac{4}{9}$ and $\frac{5}{6}$ into equivalent fractions having $18$ as the denominator.
$\therefore\frac{4}{9}=\frac{4\times2}{9\times2}=\frac{8}{18}$ and
$\frac{5}{6}=\frac{5\times3}{6\times3}=\frac{15}{18}$
Clearly, $\frac{8}{18}<\frac{15}{18}$
$\therefore\frac{4}{9}<\frac{5}{6}$
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Question 473 Marks
Of $\frac{3}{4}$ and $\frac{5}{7}$, which is greater and by how much?
Answer
Let us compare $\frac{3}{4}$ and $\frac{5}{7}$ By
cross multiplying: $3 ⨯ 7 = 21$ and $​4 ⨯ 5 = 20$
​​​​​​​ Clearly, $21 > 20$
$\therefore\frac{3}{4}>\frac{5}{7}$
Their difference: $\frac{3}{4}-\frac{5}{7}$
$L.C.M.$ of $4$ and $7 = (2 \times 2 \times 7) = 28$
$(28\div4=7,7\times3=21)$ and $(28\div7=4,4\times5=20)$
$=\frac{21-20}{28}$
$=\frac{1}{28}$
Hence, $\frac{3}{4}$ is greater than $\frac{5}{7}$ by $\frac{1}{28}$.
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Question 483 Marks
Find the equivalent fraction of $\frac{3}{5}$ having denominator $30.$
Answer
Let $\frac{3}{5}=\frac{}{30}$
$30 = 5 ​⨯ 6$
So, we have to multiply the numerator by $6$ to get the equivalent fraction having denominator $30$
$\therefore\frac{3}{5}=\frac{3\times6}{5\times6}=\frac{18}{30}$
Thus, $\frac{18}{30}$ is the equivalent fraction of $\frac{3}{5}$.
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Question 493 Marks
Find the difference: $\frac{5}{8}-\frac{7}{12}$
Answer
$\begin{array}{c|c}2&8,12\\\hline2&4,6\\\hline2&2,3\\\hline3&1,3\\\hline&1,1\end{array}$
$L.C.M$ of $2$ and $8 = (2 \times 2 \times 2 \times 3) = 24$
Now, we have:
$\frac{5}{8}=\frac{5\times3}{8\times3}=\frac{15}{24}$
$\frac{7}{12}=\frac{7\times2}{12\times2}=\frac{14}{24}$
Therefore,
$\frac{5}{8}-\frac{7}{12}$
$=\frac{15}{24}-\frac{14}{24}$
$=\frac{(15-14)}{4}$
$=\frac{1}{24}$
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Question 503 Marks
While coming back home from his school, Kishan covered $4\frac{3}{4}\text{km}$ by rickshaw and $1\frac{1}{2}\text{km}$ on foot. What is the distance of his house from the school?
Answer
Distance from Kishan's house to school $=$ Distance covered by him by rickshaw $+$ Distance covered by him on foot
Thus, we have:$\Big(4\frac{3}{4}+1\frac{1}{2}\Big)\text{km}$
$=\frac{19}{4}+\frac{3}{2}\text{km}$
$=\frac{(19+6)}{4}\text{km}$
$=\frac{25}{4}\text{km}$
$=6\frac{1}{4}\text{km}$
$(L.C.M $.of $2$ and $4 = (2 × 2) = 4$
$\begin{array}{c|c}2&2,4\\\hline2&1,2\\\hline&1,1\end{array}$
Hence, the distance from Kishan's house to school is $6\frac{1}{4}\text{km}$.
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Question 513 Marks
Of $\frac{3}{4}$ and $\frac{5}{7}$, which is greater and by how much?
Answer
Let us compare $\frac{3}{4}$ and $\frac{5}{7}$
$3 \times 7 = 21$ and $4 \times 5 = 20$
Clearly, 21 > 20 Therefore, $\frac{3}{4}>\frac{5}{7}$ Required difference:$=\frac{3}{4}-\frac{5}{7}$
$L.C.M.$ of $4$ and $7 = (2 \times 2 \times 7) = 28$
$=\frac{21-20}{28}$
$\Big(\frac{28}{4}=7,7\times3=21\Big)$
and $\Big(\frac{28}{7}=4,4\times5=20\Big)$
$=\frac{1}{28}$
Hence, $\frac{3}{4}$ is greater than $\frac{5}{7}$ by $\frac{1}{28}$.
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Question 523 Marks
A piece of wire, $2\frac{3}{4}$ metres long, broke into two pieces. One piece is $\frac{5}{8}$ metre long. How long is the other piece?
Answer
The length of the other piece = (Length of the wire $-$ Length of one piece)
$=\Big(2\frac{3}{4}-\frac{5}{8}\Big)\ \text{m}$
$=\Big(\frac{11}{4}-\frac{5}{8}\Big)\ \text{m}$
$L.C.M.$ of $4$ and $8 = (2 \times 2 \times 2) = 8$
$=\Big(\frac{22-5}{8}\Big)\ \text{m}$
$\Big(\frac{8}{4}=2,2\times11=22\Big)$
and $\Big(\frac{8}{8}=1,1\times5=5\Big)$
$=\Big(\frac{17}{8}\Big)\ \text{m}$
$=2\frac{1}{8}\ \text{m}$
Hence, the other piece is $2\frac{1}{8}\ \text{m}$ long.
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Question 533 Marks
The weight of an empty gas cylinder is $16\frac{4}{5}\text{kg}$ and it contains $14\frac{2}{3}\text{kg}$ of gas. What is the weight of the cylinder filled with gas?
Answer
Weight of the cylinder filled with gas = Weight of the empty cylinder $+$ Weight of the gas inside the cylinder Thus, we have: $(L.C.M$ .of $5$ and $3 = (5 \times 3) = 15$
$\Big(16\frac{4}{5}+14\frac{2}{3}\Big)\text{kg}$
$=\Big(\frac{84}{5}+\frac{44}{3}\Big)\text{kg}$
$=\frac{(252+220)}{15}\text{kg}$
$=\frac{472}{15}\text{kg}$
$=31\frac{7}{15}\text{kg}$
Hence, the weight of the cylinder filled with gas is $31\frac{7}{15}\text{kg}$.
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Question 543 Marks
Find the sum: $2\frac{3}{4}+5\frac{5}{6}$
Answer
We have, $L.C.M$. of $4$ and $6 = (2 \times 2 \times 3) = 12$
$\begin{array}{c|c}2&4,6\\\hline2&2,3\\\hline3&1,3\\\hline&1,1\end{array} $
Therefore, $2\frac{3}{4}+5\frac{5}{6}$
$=\frac{11}{4}+\frac{35}{6}$
$=\frac{(66+140)}{24}$
$\Big(\frac{24}{4}=6,6\times11=66\Big)$ and
$\Big(\frac{24}{6}=4,4\times35=140\Big)$
$=\frac{206}{24}$
$=\frac{103}{12}$
$=8\frac{7}{12}$
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Question 553 Marks
A film show lasted for $3\frac{1}{3}$ hours. Out of his time, $1\frac{3}{4}$ hours was spent on advertisements. What was the actual duration of the film?
Answer
Actual duration of the film = Total duration of the show - Time spent on advertisements
$=\Big(3\frac{1}{3}-1\frac{3}{4}\Big)\ \text{hours}$
$=\Big(\frac{10}{3}-\frac{7}{4}\Big)\ \text{hours}$
$L.C.M$. of $3$ and $4 = (2 \times 2 \times 3) = 12$
$=\Big(\frac{40-21}{12}\Big)\ \text{hours}$
$\Big(\frac{12}{3}=4,4\times10=40\Big)$
and $\Big(\frac{12}{4}=3,3\times7=21\Big)$
$=\Big(\frac{19}{12}\Big)\ \text{hours}$
$=1\frac{7}{12}\ \text{hours}$
Thus, the actual duration of the film was $1\frac{7}{12}\ \text{hours}$.
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Question 563 Marks
Reduce the following fractions into its simplest form: $\frac{9}{15}$
Answer
Here, numerator $= 9$ and denominator $= 15$
Factors of $9$ are $1, 3$ and $9$
 Factors of $15$ are $1, 3, 5$ and $15$
 Common factors of $9$ and $15$ are $1$ and $3 H.C.F$. of $9$ and $15$ is $3$
$\therefore\frac{9}{15}=\frac{9\div3}{15\div3}=\frac{3}{5}$
Hence, the simplest form of $\frac{9}{15}$ is $\frac{3}{5}$.
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Question 573 Marks
Reduce the following fractions into its simplest form: $\frac{72}{90}$
Answer
Here, numerator $= 72$ and denominator $= 90$
Factors of $72$ are $1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36$ and $72$
Factors of $90$ are $1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45$ and $90$
Common factors of $72$ and $90$ are $1, 2, 3, 6, 9$ and $18 H.C.F.$ of $72$ and $90$ is $18$
$\therefore\frac{72}{90}=\frac{72\div18}{90\div18}=\frac{4}{5}$
Hence, the simplest form of $\frac{72}{90}$ is $\frac{4}{5}$.
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Question 583 Marks
Arrange the following fractions in descending order: $\frac{3}{7},\frac{3}{11},\frac{3}{5},\frac{3}{13},\frac{3}{4},\frac{3}{17}$
Answer
The given fractions are $\frac{3}{7},\frac{3}{11},\frac{3}{5},\frac{3}{13},\frac{3}{4},\frac{3}{17}$
As the fractions have the same numerator,
we can follow the rule for the comparison of such fractions.
This rule states that when two fractions have the same numerator,
the fraction having the smaller denominator is the greater one.
Clearly, $\frac{3}{4} >\frac{3}{5} >\frac{3}{7} >\frac{3}{11}>\frac{3}{13}>\frac{3}{17}$
Hence, the given fractions can be arranged in the descending order as follows: $\frac{3}{4},\frac{3}{5},\frac{3}{7} ,\frac{3}{11},\frac{3}{13},\frac{3}{17}$
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Question 593 Marks
Arrange the following fractions in descending order:
$\frac{1}{12},\frac{1}{23},\frac{1}{7},\frac{1}{9},\frac{1}{17},\frac{1}{50}$
Answer
The given fractions are $\frac{1}{12},\frac{1}{23},\frac{1}{7},\frac{1}{9},\frac{1}{17},\frac{1}{50}$
As the fractions have the same numerator, we can follow the rule for the comparison of such fractions.
This rule states that when two fractions have the same numerator,
the fraction having the smaller denominator is the greater one.
Clearly, $\frac{1}{7} >\frac{1}{9} >\frac{1}{12} >\frac{1}{17}>\frac{1}{23}>\frac{1}{50}$
Hence, the given fractions can be arranged in the descending order as follows:
$\frac{1}{7},\frac{1}{9},\frac{1}{12},\frac{1}{17},\frac{1}{23},\frac{1}{50}$
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Question 603 Marks
Compare the fractions given below: $\frac{11}{12},\frac{13}{15}$
Answer
$L.C.M.$ of $12$ and $15 = (2 \times 2 \times 3 \times 5) = 60$
Now, we convert $\frac{11}{12}$ and $\frac{13}{15}$ into equivalent fractions having $40$ as the denominator.
$\therefore\frac{11}{12}=\frac{11\times5}{12\times5}=\frac{55}{60}$ and
$\frac{13}{15}=\frac{13\times4}{15\times4}=\frac{52}{60}$
Clearly, $\frac{55}{60}>\frac{52}{60}$
$\therefore\frac{11}{12}>\frac{13}{15}$
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