(a) No. of students like to study English
= \(\frac { 1 }{ 5 } \times 40\)
= 8
(b) No. of students like to study Maths
= \(\frac { 2 }{ 5 } \times 40\)
= 2\(\times\) 8
= 16
(c) No. of students who like to study Science
=40 - (8 + 16)
= 40-24
= 16
In part (ii), we know
\(\frac { 2 }{ 5 } \times 40\) = 16
So,\(\frac { 2 }{ 5 } \)of the total number of students like to study Science

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.

(i) Number of circles= 12

∴ \(\frac{1}{2}\)of the circles in box (a) =\(\frac{1}{2}\times 12=\frac{12}{2}=6\)
Hence, the shaded figure is shown below:
(ii) Number of triangles = 9

∴ \(\frac{2}{3}\)of the triangles in box (b) =\(\frac{2}{3}\times 9\)
\(=\frac{2\times 9}{3}=\frac{18}{3}=6\)
Hence, the shaded figure is shown below:
(iii)Number of squares = 15

∴ \(\frac{3}{5}\)of the squares in box (c)
=\(\frac{3}{5}\times 15=\frac{45}{5}=9\)
Hence, the shaded figure is shown alongside.

Time taken by Michael to finished colouring the picture \(\frac{7}{12}\)hr
Time taken by Vaibhav to finished colouring the picture = \(\frac{3}{4}\)hr

LCM of 12 and 4 = 3 X 4 = 12
On converting these fractions into like fractions, we get
\(\frac{3}{4}=\frac{3\times 3}{4\times 3}=\frac{9}{12} and \frac{7}{12}=\frac{7}{12}\)
It is clear, 9>7 [numerators of fractions]
So, \(\frac{9}{12}>\frac{7}{12} i.e., \frac{3}{4}>\frac{7}{12}\)
So, Vaibhav takes longer time to finish colouring the picture. Difference between time taken by Vaibhav and Michael
\(=\frac{3}{4}-\frac{7}{12}\)

LCM of 4 and 12 = 2 x 2 x3 = 12
\(=\frac{3}{4}-\frac{7}{12}=\frac{3\times 3-7\times 1}{12}\)
=\(\frac{9-7}{12}=\frac{2}{12}=\frac{1}{6}\) [dividing numerator and denominator by 2]
Hence, Vaibhav takes \(\frac{1}{6}\) h more than Michael.

- Multiply 27 and 135. We get 3645.
- In 2.7, there is 1 digit to the right of the decimal point and in 1.35, there are 2 digits to the right of the decimal point.
- So, count 1 + 2 = 3 digits starting from the right most digit (i.e., 5) in 3645 and move towards left.
- We get 3.645.

We have,
Product of two numbers = 20\(\frac { 5 }{ 7 } \)=\(\frac { 145 }{ 7 } \)
One of the number =6\(\frac { 2 }{ 3 } \)=\(\frac { 20 }{ 3 } \)
The other number = (Product of the numbers ÷ One of the numbers)
= \(\frac { 145 }{ 7 } \div \frac { 20 }{ 3 } \)
=\(\frac { 145 }{ 7 } \times \frac { 3 }{ 20 } =\frac { 145\times 3 }{ 7\times 20 } =\frac { 29\times 3 }{ 7\times 4 } =\frac { 87 }{ 28 } =3\frac { 3 }{ 28 } \)
Hence, the other number is 3 \(\frac { 3 }{ 28 } \)

\(2\frac{2}{3}+12\frac{2}{4}+5\frac{1}{2}\)
ஃ \(\frac{(2\times 3)+2}{3}+\frac{(12\times 4)+2}{4}+\frac{(5\times 2)+1}{2}\)
=\(\frac{6+2}{3}+\frac{48+2}{3}+\frac{10+1}{3}=\frac{8}{3}+\frac{50}{4}+\frac{11}{2}\)
=\(\frac{(8\times 4)+(50 \times 3)+(11\times 6)}{12}\)     [∵ LCM of 3,4 and 2 is 12]
=\(\frac{32+150+66}{12}=\frac{248}{12}=\frac{124}{6}=\frac{62}{3}=20\frac{2}{3}\)

(i) \(\frac{2}{3}\times \Box=\frac{10}{30}\)
Here, 2 X5 = 10 and 3 X10 = 30
Hence, the required number in the box is \(\frac{5}{10}\).
(ii) Simplest form of \(\frac{5}{10}=\frac{1}{2}\)

Given, width of picture = 7\(\frac{3}{5}\)cm
\(=\frac{7\times5+3}{5}=​​\frac{35+3}{5}=\frac{38}{5}\)cm
and width of the frame = 7\(\frac{3}{10}\)cm=\(\frac{7\times 10+3}{10}\)
=\(\frac{70+3}{10}=\frac{73}{10}\)cm
ஃ To fit the picture in frame, picture should be trimmed
= Width of the picture - Width of the frame
\(=\frac{38}{5}-\frac{73}{10}=\frac{38\times 2-73\times 1}{10}\)   [∵ LCM of 5 and 10=10]
=\(\frac{76-73}{10}=\frac{3}{10}\)cm
Hence,\(\frac{3}{10}\)ern picture should be trimmed.

\(\frac{1}{7}\ of\ \frac{6}{5}=\frac{1}{7}\times\frac{6}{5}=\frac{1\times6}{7\times5}=\frac{6}{35}\)
\(\frac{1}{7}\ of\ \frac{3}{10}=\frac{1}{7}\times\frac{3}{10}=\frac{1\times3}{7\times10}=\frac{3}{70}\)

\(\frac{8}{3}\times\)\(\frac{4}{7}\)​​​=\(\frac{8\times4}{3\times7}\)=\(\frac{32}{21}\)
\(\frac{3}{4}\times\)\(\frac{2}{3}\)=\(\frac{3\times2}{4\times3}\)=\(\frac{1\times1}{2\times1}\)=\(\frac{1}{2}\)

Place value of 5 in 103.2523 - Place value of 2 in 115.552
= 0.0500 - 0.002
= 0.050 - 0.002 
= \(\frac { 50 }{ 1000 } -\frac { 2 }{ 1000 } \)
= \(\frac { 50-2 }{ 1000 } =\frac { 48 }{ 1000 } \)
= 0.048

Given, \(\frac{1}{9}\times 2\frac{3}{7}=\frac{1}{9}\times \frac{(2\times 7)+3}{7}​=\frac{1}{9}​\times \frac{(14+3)}{7}\)
=\(\frac{1}{9}\times \frac{17}{7}=\frac{1\times 17}{9\times 7}=\frac{17}{63}\)

20.1\(\times\)4 =\(\frac { 201 }{ 10 } \times 4=\frac { 804 }{ 10 } \)= 80.4

Given numbers, 70.01 or 70.1

Here, digits to right side of decimal at tenth place are 0 and 1.
1> 0 i.e. 70.01 < 70.1

We have, \(12 \div \frac{3}{4}=12\times \frac{4}{3}= \frac{12\times 4}{3}= \frac{48}{3}=16\) [∵ reciprocal of \( \frac{3}{4} is \frac{4}{3}\)]

We have, \(\frac{8}{3}\times \frac{4}{7}=\frac{8\times 4}{3 \times 7} =\frac{32}{21}\)

We have, \(4\times \frac{1}{3}\)=\( \frac{4\times1}{3}=\frac{4}{3}\)

This is the required lowest form,
Now, dividing 4 (numerator) by 3
(denominator), we get, quotient = 1and remainder = 1
∴ Mixed fraction =Quotient \(\frac{Reminder}{Divisor}=1 \frac{1}{3}\)

8

\(1.44\div1.2=\frac{144}{100}\div\frac{12}{10}=\frac{144}{100}\times\frac{10}{12}=\frac{12}{10}=1.2\)
or
\(\frac{1.44}{1.2}=\frac{144}{120}=\frac{12}{10}=1.2\)

Meenu drank water\(=\frac{1}{8}\ of 2=\frac{1}{8}\times2=\frac{1\times2}{8}=\frac{2}{8}=\frac{1}{4}\ litre\)

73>10