2 Mark Question

Question 12 Marks

Verify the following comparisons using a number line.
i. $2<3$ but $-2>-3$ ii. $\frac{5}{4}<\frac{7}{4}$ but $\frac{-5}{4}<\frac{-7}{4}$

Answer

We know that, on a number line the number to the left is smaller than the other.
∴ 2 < 3 and -3 < -2
i.e. 2 < 3 and -2 > -3
$\frac{5}{4}<\frac{7}{4}$ and $\frac{-7}{4}<\frac{-5}{4}$
i.e. $\frac{5}{4}<\frac{7}{4}$ and $\frac{-5}{4}>\frac{-7}{4}$

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Question 22 Marks

$-\frac{7}{11},-\frac{3}{4}$

Answer

$
\frac{7}{11},-\frac{3}{4}
$
Here, the denominators of the given numbers are not the same.
LCM of 11 and $4=44$
$
\begin{array}
& -\frac{7}{11}=-\frac{7 \times 4}{11 \times 4}=-\frac{28}{44}, \\
-\frac{3}{4}=-\frac{3 \times 11}{4 \times 11}=-\frac{33}{44}
\end{array}
$
Since, $28<33$
$
\begin{array}{ll}
\therefore & \frac{28}{44}<\frac{33}{44} \\
\therefore & -\frac{28}{44}>-\frac{33}{44} \\
\therefore & -\frac{7}{11}>-\frac{3}{4}
\end{array}
$

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Question 32 Marks

$\frac{12}{15}, \frac{3}{5}$

Answer

$\frac{12}{15}, \frac{3}{5}$
Here, the denominators of the given numbers are not the same.
LCM of 15 and $5=15$
$
\frac{3}{5}=\frac{3 \times 3}{5 \times 3}=\frac{9}{15}
$
Since, $12>9$
$
\begin{array}{ll}
\therefore & \frac{12}{15}>\frac{9}{15} \\
\therefore \quad \frac{12}{15} & >\frac{3}{5}
\end{array}
$

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Question 42 Marks

$-\frac{25}{8},-\frac{9}{4}$

Answer

$-\frac{25}{8},-\frac{9}{4}$
Here, the denominators of the given numbers are not the same.
LCM of 8 and $4=8$
$
-\frac{9}{4}=-\frac{9 \times 2}{4 \times 2}=-\frac{18}{8}
$
Since, $25>18$
$
\begin{array}{ll}
\therefore & \frac{25}{8}>\frac{18}{8} \\
\therefore & -\frac{25}{8}<-\frac{18}{8} \\
\therefore & -\frac{25}{8}<-\frac{9}{4}
\end{array}
$

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Question 52 Marks

$\frac{15}{12}, \frac{7}{16}$

Answer

$\frac{15}{12}, \frac{7}{16}$
Here, the denominators of the given numbers are not the same.
LCM of 12 and $16=48$
$
\begin{array}
\frac{15}{12}=\frac{15 \times 4}{12 \times 4}=\frac{60}{48}, \\
\frac{7}{16}=\frac{7 \times 3}{16 \times 3}=\frac{21}{48} \\
\text { Since, } 60>21 \\
\therefore \quad \frac{60}{48}>\frac{21}{48} \\
\therefore \quad \frac{15}{12}>\frac{7}{16}
\end{array}
$
Alternate method:
$
\begin{array}
15 \times 16=240 \\
12 \times 7=84 \\
\text { Since, } 240>84 \\
\therefore 15 \times 16>12 \times 7 \\
\therefore \quad \frac{\mathbf{1 5}}{\mathbf{1 2}}>\frac{7}{\mathbf{1 6}} \quad \ldots\left[\text { If } a \times d>b \times c \text {, then } \frac{\mathrm{a}}{\mathrm{b}}>\frac{\mathrm{c}}{\mathrm{d}}\right]
\end{array}
$

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Question 62 Marks

$-\frac{17}{20},-\frac{13}{20}$

Answer

$-\frac{17}{20},-\frac{13}{20}$
Here, the denominators of the given numbers are the same.
Since, $17<13$
$
\begin{array}
& \therefore-17<-13 \\
\therefore-\frac{17}{20}<-\frac{13}{20}
\end{array}
$

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Question 72 Marks

$\frac{40}{29}, \frac{141}{29}$

Answer

$\frac{40}{29}, \frac{141}{29}$
Here, the denominators of the given numbers are the
same.
Since, $40<141$
$
\therefore \frac{40}{29}<\frac{141}{29}
$

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Question 82 Marks

$-\frac{5}{4}, \frac{1}{4}$

Answer

$-\frac{5}{4}, \frac{1}{4}$
We know that, a negative number is always less than a positive number.
$
\therefore-\frac{5}{4}<\frac{1}{4}
$

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Question 92 Marks

$\frac{8}{7}, 0$

Answer

$\frac{8}{7}, 0$
On a number line, zero is to the left of $\frac{8}{7}$.
$
\therefore \frac{8}{7}>0
$

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Question 102 Marks

$0, \frac{-9}{5}$

Answer

$0, \frac{-9}{5}$
On a number line, $\frac{-9}{5}$ is to the left of zero.
$\therefore 0>\frac{-9}{5}$

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Question 112 Marks

$7,-2$

Answer

7, -2
If a and b are positive numbers such that a < b, then -a > -b.
Since, 2 < 7 ∴ -2 > -7

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