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(a) 9. (b) 4

2/5

Let the total number of pages be x.
Reemu read part of book = \(\frac{1}{5}\)of x=\(\frac{1}{5}\times x=\frac{x}{5}\)
She further reads 40 pages and covers \(\frac{7}{10}\)th part of book.
Accor diIng to the question, \(\frac{x}{5}+40=​​\frac{7x}{10}\)
\(⇒ \frac{x}{5}-\frac{7x}{10}=-40 ⇒ \frac{(2\times x)-(1\times 7x)}{10}=-40\)
\(\Rightarrow -\frac{5x}{10}=-\frac{40}{1}\)
⇒ -5x=-40 x10 [by cross-multiplication]
⇒ \(x=\frac{-40\times 10}{-5}=80\)
Total pages = 80
She read the pages = \(\frac{x}{5}+40=\frac{80}{5}+40\)
=16 + 40 = 56 pages 
Left pages to be read = (80 - 56) pages = 24 pages Hence, 24 pages are left to be read.

Given, distance between A and B = 7.5 km
Distance between Band C = 12.7 km
∴ Distance travelled by Dinesh = AB + BC
= 7.5+ 12.7
= 20.2km
Distance between A and D = 9.3 km
Distance between D and C = 11.8 km
∴ Distance travelled by Ayub = AD + DC
= 9.3+ 11.8
= 21.1 km
Hence, it is clear that 21.1>20.2.
∴ Ayub travelled more distance i.e. 21.1 km.
Now, difference between distances, they both travelled
= (21.1- 20.2) km
=0.9km
Hence, Ayub travelled 0.9 km more than Dinesh.

\(\frac{22}{25}\)m

We have
Cost of 1 kg of sugar = Rs.17\(\frac { 3 }{ 4 } \)=Rs\(\frac { 71 }{ 4 } \)k
∴ Cost of 8 kg of sugar = Rs \(\left( \frac { 71 }{ 4 } \times 8\frac { 1 }{ 2 } \right) \)
=Rs \(\left( \frac { 71 }{ 4 } \times \frac { 17 }{ 2 } \right) =Rs.\left( \frac { 71\times 17 }{ 4\times 2 } \right) \)
=Rs\(\left( \frac { 1207 }{ 8 } \right) \) =Rs.150\(\frac { 7 }{ 8 } \)
Hence, the cost of 8\(\frac { 1 }{ 2 } \) kg of sugar is Rs.150\(\frac { 7 }{ 8 } \)

(I) We have, \(\frac{2}{7}\)of  \(\frac{3}{4} =\frac{2}{7}\times \frac{3}{4}=\frac{1}{7}\times \frac{3}{2}=\frac{3}{14}\) [dividing numerator and denominator by 2]
and \(\frac{3}{5}\)of \( \frac{5}{8}=\frac{3}{5}\times \frac{5}{8}\) [dividing numerator and denominator by 5]
\(=\frac{3\times 1}{1\times 8}=\frac{3}{8}\)

LCM of 14 and 8 = 2 X 2 X 2 X 7 = 56
On converting both the fractions into like fractions,
we get \(\frac{3}{14}=\frac{3\times 4}{14\times 4}=\frac{12}{56} \) and \( \frac{3}{8}=\frac{3\times 7}{8 \times 7}=\frac{21}{56}\)
∴ 21> 21, So, \(\frac{21}{56}>\frac{12}{56 }\) ie., \(\frac{3}{8}>\frac{3}{14}\)
Hence, \(\frac{3}{5}\) of \( \frac{5}{8}\) is greater than \(\frac{2}{7} \)of\( \frac{3}{4}\).
Alternate Method
If two fractions have the same numerator but different denominators, then fraction with smaller
denominators will be greater.
We have,\(\frac{2}{7}\)of \(\frac{3}{4}=\frac{2}{7}\times \frac{3}{4}=\frac{6}{28}=\frac{3}{14}\)  [dividing numerator and denominator by 2]
and \(\frac{3}{5} \)of \( \frac{5}{8}=\frac{3}{5}\times \frac{5}{8}=\frac{3\times 5}{5\times 8}=\frac{15}{40}=\frac{3}{8}\) [dividing numerator and denominator by 5]
Here, both fractions are like fractions.
Since, 8<14, So, \(\frac{2}{7} \)of \( \frac{3}{4}\).

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