5 Marks Questions

Question 15 Marks

Arrange the following fractions in descending order:$\frac{3}{4},\frac{5}{8},\frac{11}{12}\ \text{and}\ \frac{17}{24}$

Answer

The given fractions are $\frac{3}{4},\frac{5}{8},\frac{11}{12}\ \text{and}\ \frac{17}{24}$
$\begin{array}{c|c}2&4,8,12,24\ \ \ \ \ \ \\\hline2&2,4,6,12\ \ \ \ \ \ \ \\\hline3&1,2,3,6\ \ \ \ \ \ \ \ \ \\\hline2&1,2,1,2\ \ \ \ \ \ \ \ \ \\\hline&1,1,1,1\ \ \ \ \ \ \ \ \ \end{array}$
$L.C.M$. of $4, 8, 12$ and $24 = (2 \times 2 \times 2 \times 3) = 24$
So, we convert each of the fractions whose denominator is not
equal to $24$ into an equivalent fraction with denominator $24.$
Thus, we have: $\frac{3}{4}=\frac{3\times6}{4\times6}=\frac{18}{24}$,
$\frac{5}{8}=\frac{5\times3}{8\times3}=\frac{15}{24}$,
$\frac{11}{12}=\frac{11\times2}{12\times2}=\frac{22}{24}$
Clearly, $\frac{22}{24}>\frac{18}{24}>\frac{17}{24}>\frac{15}{24}$
$\therefore\frac{11}{12}>\frac{3}{4}>\frac{17}{24}>\frac{5}{8}$
Hence, the given fractions can be arranged in
the descending order as follows: $\frac{11}{12},\frac{3}{4},\frac{17}{24},\frac{5}{8}$

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Question 25 Marks

Arrange the following fractions in ascending order:$\frac{1}{2},\frac{3}{4},\frac{5}{6}\ \text{and}\ \frac{7}{8}$

Answer

The given fractions are $\frac{1}{2},\frac{3}{4},\frac{5}{6}\ \text{and}\ \frac{7}{8}$
$\begin{array}{c|c}2&2,4,6,8\ \ \ \ \ \ \\\hline2&1,2,3,4\ \ \ \ \ \ \\\hline2&1,1,3,2\ \ \ \ \ \\\hline3&1,1,3,1\ \ \ \ \ \\\hline&1,1,1,1\ \ \ \ \ \end{array}$
$L.C.M$. of $2, 4, 6$ and $8 = (2 \times 2 \times 2 \times 3) = 24$
We convert each of the given fractions into an equivalent fraction with denominator $24.$
Now, we have: $\frac{1}{2}=\frac{1\times12}{2\times12}=\frac{12}{24}$,
$\frac{3}{4}=\frac{3\times6}{4\times6}=\frac{18}{24}$,
$\frac{5}{6}=\frac{5\times4}{6\times4}=\frac{20}{24}$, $\frac{7}{8}=\frac{7\times3}{8\times3}=\frac{21}{24}$
Clearly, $\frac{12}{24}<\frac{18}{24}<\frac{20}{24}<\frac{21}{24}$
$\therefore\frac{1}{2}<\frac{3}{4}<\frac{5}{6}<\frac{7}{8}$
Hence, the given fractions can be arranged in
the ascending order as follows: $\frac{1}{2},\frac{3}{4},\frac{5}{6},\frac{7}{8}$

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Question 35 Marks

Arrange the following fractions in ascending order:$\frac{2}{5},\frac{7}{10},\frac{11}{15}\ \text{and}\ \frac{17}{30}$

Answer

The given fractions are $\frac{2}{5},\frac{7}{10},\frac{11}{15}\ \text{and}\ \frac{17}{30}$
$\begin{array}{c|c}5&5,10,15,30 \ \\\hline2&1,2,3,6\ \ \ \ \ \ \ \\\hline3&1,1,3,3\ \ \ \ \ \ \\\hline&1,1,1,1\ \ \ \ \ \ \end{array}$
$L.C.M.$ of $5, 10, 15$ and $30 = (2 \times 5 \times 3) = 30$
So, we convert each of the fractions whose denominator is not
equal to $30$ into an equivalent fraction with denominator $30$.
Now, we have: $\frac{2}{5}=\frac{2\times6}{5\times6}=\frac{12}{30}$,
$\frac{7}{10}=\frac{7\times3}{10\times3}=\frac{21}{30}$,
$\frac{11}{15}=\frac{11\times2}{15\times2}=\frac{22}{30}$
Clearly, $\frac{12}{30}<\frac{17}{30}<\frac{21}{30}<\frac{22}{30}$
$\therefore\frac{2}{5}<\frac{17}{30}<\frac{7}{10}<\frac{11}{15}$
Hence, the given fractions can be arranged in
the ascending order as follows: $\frac{2}{5},\frac{17}{30},\frac{7}{10},\frac{11}{15}$

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Question 45 Marks

Arrange the following fractions in descending order:$\frac{7}{9},\frac{5}{12},\frac{11}{18}\ \text{and}\ \frac{17}{36}$

Answer

The given fractions are $\frac{7}{9},\frac{5}{12},\frac{11}{18}\ \text{and}\ \frac{17}{36}$
$\begin{array}{c|c}3&9,12,18,36\ \ \ \\\hline3&3,4,6,12\ \ \ \ \ \ \ \\\hline2&1,4,2,4\ \ \ \ \ \ \ \ \ \\\hline2&1,2,1,2\ \ \ \ \ \ \ \ \ \\\hline&1,1,1,1\ \ \ \ \ \ \ \ \ \end{array}$
$L.C.M$. of $9, 12, 18$ and $36 = (3 \times 3 \times 2 \times 3) = 36$
So, we convert each of the fractions whose denominator is not
equal to $36$ into an equivalent fraction with denominator $36.$
Thus, we have: $\frac{7}{9}=\frac{7\times4}{9\times4}=\frac{28}{36}$,
$\frac{5}{12}=\frac{5\times3}{12\times3}=\frac{15}{36}$,
$\frac{11}{18}=\frac{11\times2}{18\times2}=\frac{22}{36}$
Clearly, $\frac{28}{36}>\frac{22}{36}>\frac{17}{36}>\frac{15}{36}$
$\therefore\frac{7}{9}>\frac{11}{18}>\frac{17}{36}>\frac{5}{12}$
Hence, the given fractions can be arranged in
the descending order as follows: $\frac{7}{9},\frac{11}{18},\frac{17}{36},\frac{5}{12}$

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Question 55 Marks

Arrange the following fractions in ascending order:$\frac{3}{4},\frac{7}{8},\frac{11}{16}\ \text{and}\ \frac{23}{32}$

Answer

The given fractions are $\frac{3}{4},\frac{7}{8},\frac{11}{16}\ \text{and}\ \frac{23}{32}$
$\begin{array}{c|c}2&4,8,16,32\ \ \ \ \ \ \\\hline2&2,4,8,16\ \ \ \ \ \ \ \\\hline2&1,2,4,8\ \ \ \ \ \ \ \ \ \\\hline2&1,1,2,4\ \ \ \ \ \ \ \ \ \\\hline2&1,1,1,2\ \ \ \ \ \ \ \ \ \\\hline&1,1,1,1\ \ \ \ \ \ \ \ \ \end{array}$
$L.C.M$. of $4, 8, 16$ and $32 = (2 \times 2 \times 2 \times 2 \times 2) = 32$
So, we convert each of the fractions whose denominator is not
equal to $32$ into an equivalent fraction with denominator
Now, we have: $\frac{3}{4}=\frac{3\times8}{4\times8}=\frac{24}{32}$,
$\frac{7}{8}=\frac{7\times4}{8\times4}=\frac{28}{32}$,
$\frac{11}{16}=\frac{11\times2}{16\times2}=\frac{22}{32}$
Clearly, $\frac{22}{32}<\frac{23}{32}<\frac{24}{32}<\frac{28}{32}$
$\therefore\frac{11}{16}<\frac{23}{32}<\frac{3}{4}<\frac{7}{8}$
Hence, the given fractions can be arranged in
the ascending order as follows: $\frac{11}{16},\frac{23}{32},\frac{3}{4},\frac{7}{8}$

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Question 65 Marks

Find the sum $2\frac{4}{5}+1\frac{3}{10}+3\frac{1}{15}$.

Answer

We have, $2\frac{4}{5}+1\frac{3}{10}+3\frac{1}{15}$
$=\frac{14}{5}+\frac{13}{10}+\frac{46}{15}$
$\begin{array}{c|c}5&5,10,15\\\hline2&1,2,3\ \ \ \\\hline3&1,1,3\ \ \ \\\hline&1,1,1\ \ \ \end{array}$
$L.C.M$. of $5, 10$ and $15 = (5 × 2 × 3) = 30$
$(30\div5=6,6\times14=84)$
$(30\div10=3,3\times13=39)$ and $(30\div15=2,2\times46=92)$
$=\frac{84+39+92}{30}$
$=\frac{215}{30}$
$=\frac{43}{6}$
$=7\frac{1}{6}$

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Question 75 Marks

Arrange the following fractions in descending order:$\frac{2}{3},\frac{3}{15},\frac{7}{10}\ \text{and}\ \frac{8}{15}$

Answer

The given fractions are $\frac{2}{3},\frac{3}{15},\frac{7}{10}\ \text{and}\ \frac{8}{15}$
$\begin{array}{c|c}3&3,5,10,15 \ \ \ \ \ \ \\\hline5&1,5,10,5\ \ \ \ \ \ \ \\\hline2&1,1,2,1\ \ \ \ \ \ \ \ \ \\\hline&1,1,1,1\ \ \ \ \ \ \ \ \end{array}$
$L.C.M$. of $3, 5, 10$ and $15 = (2 × 3 × 5) = 30$
So, we convert each of the fractions into an
equivalent fraction with denominator $30$
Thus, we have: $\frac{2}{3}=\frac{2\times10}{3\times10}=\frac{20}{30}$,
$\frac{3}{5}=\frac{3\times6}{5\times6}=\frac{18}{30}$,
$\frac{7}{10}=\frac{7\times3}{10\times3}=\frac{21}{30}$,
$\frac{8}{15}=\frac{8\times2}{15\times2}=\frac{16}{30}$
Clearly, $\frac{21}{30}>\frac{20}{30}>\frac{18}{30}>\frac{16}{30}$
$\therefore\frac{7}{10}>\frac{2}{3}>\frac{3}{5}>\frac{8}{15}$
Hence, the given fractions can be arranged in the
descending order as follows: $\frac{7}{10},\frac{2}{3},\frac{3}{5},\frac{8}{15}$

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Question 85 Marks

The cost of a pen is $\text{Rs.}\ 16\frac{2}{3}$ and that of a pencil is $\text{Rs.}\ 4\frac{1}{6}$.
Which costs more and by how much?

Answer

Cost of a pen $=\text{Rs.}\ 16\frac{2}{3}=\text{Rs.}\ \frac{50}{3}$
$=\text{Rs.}\ \frac{50\times2}{3\times2}=\text{Rs.}\ \frac{100}{6}$
Cost of a pencil $=\text{Rs.}\ 4\frac{1}{6}=\text{Rs.}\ \frac{25}{6}$
$=\frac{100}{6}>\frac{25}{6}$
$\therefore\text{Rs.}\ 16\frac{2}{3}>\text{Rs.}\ 4\frac{1}{6}$
So, the cost of a pen is more than the cost of a pencil.
Difference between their costs: $=\text{Rs.}\ \Big(\frac{50}{3}-\frac{25}{6}\Big)$
$=\text{Rs.}\ \Big(\frac{100-25}{6}\Big)$
$=\text{Rs.}\ \Big(\frac{75}{6}\Big)$
$=\text{Rs.}\ \Big(\frac{25}{2}\Big)$
$=\text{Rs.}\ 12\frac{1}{2}$
Hence, the cost of a pen is $\text{Rs.}\ 12\frac{1}{2}$ more than the cost of a pencil.

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Question 95 Marks

Arrange the following fractions in ascending order:$\frac{2}{3},\frac{5}{6},\frac{7}{9}\ \text{and}\ \frac{11}{18}$

Answer

The given fractions are $\frac{2}{3},\frac{5}{6},\frac{7}{9}\ \text{and}\ \frac{11}{18}$
$\begin{array}{c|c}3&3,6,9,18\ \ \ \ \ \ \\\hline3&1,2,3,6\ \ \ \ \ \ \ \\\hline2&1,2,1,2\ \ \ \ \ \ \\\hline&1,1,1,1\ \ \ \ \ \ \end{array}$
$L.C.M$. of $3, 6, 9$ and $18 = (3 × 2 × 3) = 18$
So, we convert each of the fractions whose denominator is not
equal to $18$ into an equivalent fraction with denominator $18.$
Now, we have: $\frac{2}{3}=\frac{2\times6}{3\times6}=\frac{12}{18}$,
$\frac{5}{6}=\frac{5\times3}{6\times3}=\frac{15}{18}$,
$\frac{7}{9}=\frac{7\times2}{9\times2}=\frac{14}{18}$
Clearly, $\frac{11}{18}<\frac{12}{18}<\frac{14}{18}<\frac{15}{18}$
$\therefore\frac{11}{18}<\frac{2}{3}<\frac{7}{9}<\frac{5}{6}$
Hence, the given fractions can be arranged in
the ascending order as follows: $\frac{11}{18},\frac{2}{3},\frac{7}{9},\frac{5}{6}$

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Question 105 Marks

Arrange the following fractions in descending order:$\frac{5}{7},\frac{9}{14},\frac{17}{21}\ \text{and}\ \frac{31}{42}$

Answer

The given fractions are $\frac{5}{7},\frac{9}{14},\frac{17}{21}\ \text{and}\ \frac{31}{42}$
$\begin{array}{c|c}7&7,14,21,42\\\hline2&1,2,3,6 \ \ \ \ \ \\\hline3&1,1,3,3 \ \ \ \ \\\hline&1,1,1,1 \ \ \ \end{array}$
$L.C.M$. of $7, 14, 21$ and $42 = (2 \times 3 \times 7) = 42$
We convert each one of the fractions whose denominator is not
equal to $42$ into an equivalent fraction with denominator $42$ Thus,
we have: $\frac{5}{7}=\frac{5\times6}{7\times6}=\frac{30}{42}$,
$\frac{9}{14}=\frac{9\times3}{14\times3}=\frac{27}{42}$,
$\frac{17}{21}=\frac{17\times2}{21\times2}=\frac{34}{42}$,
Clearly, $\frac{34}{42}>\frac{31}{42}>\frac{30}{42}>\frac{27}{42}$
$\therefore\frac{17}{21}>\frac{31}{42}>\frac{5}{7}>\frac{9}{14}$
Hence, the given fractions can be arranged in the
descending order as follows: $\frac{17}{21},\frac{31}{42},\frac{5}{7},\frac{9}{14}$

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