(a) AB (b) b (c) 45

\(\frac { 5 }{ 6 } -\frac { 1 }{ 3 } =\frac { 1 }{ 2 } \)

We know that, 1 g \(={1\over 1000}\) kg
\(\therefore\)     5 kg 8 g = 5 kg + 8 g = 5 kg \(+8\times{1\over 1000}\) kg
= 5 kg \(+{8\over1000}\) kg = (5+0.008) kg = 5.008 kg.

We have\(\frac { 1 }{ 7 } \Box { } \frac { 1 }{ 4 } \)
Here,numerators are same of both fractions. So, fraction having smaller denominator will be greater.
∴  \(\frac { 1 }{ 7 } \boxed { < } \frac { 1 }{ 4 } \)

0.5 or 0.05
Whole part for both is 0.
Tenth part for 0.5 is 5 and for 0.05 is 0.
\(\therefore \) 5 >0
\(\therefore \) 0.5> 0.05

\(200+60+5+\frac { 1 }{ 10 } =265.1\)

We know that, one day = 24 h
\(\therefore\) The required fraction = \(\frac { 15\ h }{ 24\ h } =\frac { 15 }{ 14 } =\frac { 5 }{ 8 } \)

\(12\frac{4}{5}\)

Rs 313.86

First of all, we draw a horizontal and a vertical line. On the horizontal line, mark points at equal distance. Write the subject on these points whose percentage is to be plotted. Now, choose a suitable scale, such as
1 unit = 10 small squares = 10%.
For the given table, bar graph is shown alongside:

Given, numerator of an equivalent fraction = 9
 ∴  \(\frac { 9 }{ D } =\frac { 3 }{ 5 } \) ⇒ D\(\times\)3 = 9\(\times\)5
So,  D\(\times\)3 = 3\(\times\)3\(\times\)5     [∵ 9=3\(\times\)3]
⇒   D\(\times\)3 =15\(\times\)3
On comparing, we get D = 15
∴  Required equivalent fraction of \(\frac { 3 }{ 5 } =\frac { N }{ D } =\frac { 9 }{ 15 } \)

\(2\frac{3}{20}\)

Weight of rice purchased by Ravi
= 5 kg 400 g = 5 kg + 400 g
\(= 5 kg+ 400\times{1\over 1000}kg\)   \(\begin{bmatrix} \because 1\ g={1\over 1000} kg\end{bmatrix}\)
\(=5\ kg+{400\over1000}\)kg = (5+ 0.400) kg = 5.400 kg
Weight of sugar purchased by Ravi
= 2 kg 20 g = 2 kg+ 20 g
\(= 2 kg+ 20\times{1\over1000}kg\)    \(\begin{bmatrix} \because 1\ g={1\over 1000} kg\end{bmatrix}\)
\(=2\ kg+{20\over1000}kg\) = (2+ 0.020) kg = 2.020 kg
Weight of flour purchased by Ravi
= 10 kg 850 g = 10 kg + 850 g
\(=10\ kg+850\times{1\over 1000}kg\)   \(\begin{bmatrix} \because 1\ g={1\over 1000} kg\end{bmatrix}\)
\(=10\ kg+{850\over1000}kg\) (10 + 0.850) kg = 10.850 kg
\(\therefore\) Total weight of his purchases
\(\quad5.400\\\quad2.020\\+10.850\\\_\_\_\_\_\_\_\_\\\ \ 18.270\\\_\_\_\_\_\_\_\_\)

Hence, the total weight of all his purchases is 18.270 kg.