Watching TV is preferred by most of the studentsother than playing.

First of all, we construct \(\bar{AB}\) of length 12 cm.
Now, steps of construction are as follows:
Step I Draw a line segment \(\bar{AB}\) = 12 cm.
Step II Draw perpendicular bisector of AB, which meets \(\bar{AB}\) at O. (i.e. O is the mid-point of \(\bar{AB}\)) i.e. AO = OB.
              
Step III Now, draw perpendicular bisector of \(\bar{AO}\) , which meet \(\bar{AB}\) at P such that AP = PO.
Step IV Then, draw perpendicular bisector of BO, which meet \(\bar{AB}\) at Q such that BQ = OQ.
Step V The line segment AB is divided into 4 equal parts at P, O, and Q.
Step VI By actual measurement, we have \(\bar{AP}=\bar{PO}=\bar{OQ}=\bar{QB}\) = 3 cm.

On measuring, \(\angle\)COA = 45^o.

Let side of squared garden be x m.
Given that, perimeter of a squared garden = 48 m
∴ 4 x Side of a square = 48
⇒ 4x=48⇒x=\(\frac{48}{4}=12m\)
Now, area of the squared garden = (X)²
=(12)²=144m²
Also given, area of small flower bed cover inside the garden = 18 m²
∴ Area of the 9B.rden not covered by flower bed
= Area of squared garden - Area of flower bed
=144 m² -18 m² = 126 m²
The fractional part of the garden covered by flower bed=\(\frac { Area\quad covered\quad by\quad the\quad flower }{ Remaming\quad Area\quad of\quad the\quad squared\quad garden } \)
\(=\frac { 18 }{ 126 } =\frac { 2 }{ 14 } =\frac { 1 }{ 7 } \)
Hence, ratio of the area covered by the flower bed and the remaining area is 1 : 7.

(a) Here, length of the rectangle = 9 m
and breadth of the rectangle = 6 m
∴ Area of the rectangle = Length \(\times\) Breadth = 54 sq m
Hence, the area of the rectangle is 54 sq m.
(b) Here, length of the rectangle = 17 m
and breadth of the rectangle = 3 m
∴ Area of the rectangle = Length \(\times\) Breadth
= 17m \(\times\) 3m = 51 sq m
Hence, the area of the rectangle is 51 sq m.
(c) Here, length of the rectangle = 14 m
and breadth of the rectangle = 4 m
∴ Area of the rectangle = Length \(\times\) Breadth
= 14 m \(\times\) 4 m = 56 sq m
Hence, the area of the rectangle is 56 sq m.
Now, we have 56> 54> 51
Hence, the rectangle having sides 4 m and 14 m has the largest area and the rectangle having sides 17m and 3 m has the smallest area.

Given number line is a follows:

(i) From the figure, it is clear that A = 0.8 as the unit length between 0 and 1 has been divided into 10 equal parts and 8 parts from 0 have been taken.
(ii) From the figure, it is clear that B = 1.3 as the unit length between 1 and 2 has been divided into 10 equal parts and unit length between 0 to 1 and then 3 parts have been taken.
(iii) From the figure, it is clear that C = 2.2 as the unit length between 2 and 3 has been divided into 10 equal parts and unit length between 0 to 1, 1 to 2 and then 2 parts have been taken.
(iv) From the figure, it is clear that D = 2.9 as the unit length between 2 and 3 has been divided into 10 equal parts and unit length between 0 to 1, 1 to 2 and then 9 parts have been taken.

Given numbers can be written in decimals as follows:
(i) 5 hundreds + 3 tens + 8 ones + 1 tenth
 = 5 \(\times\) 100 + 3 \(\times\) 10 + 8 \(\times\) 1+ 1 \(\times\) \(\frac { 1 }{ 10 } \)
 = 500 + 30 + 8 + \(\frac { 1 }{ 10 } \)  = 538 + 0.1 = 538.1
(ii) 2 hundreds + 7 tens + 3 ones + 4 tenths
= 2 \(\times\) 100 + 7 \(\times\) 10 + 3 \(\times\) 1+ 4 \(\times\) \(\frac { 1 }{ 10 } \)
= 200 + 70 + 3 + \(\frac { 4 }{ 10 } \) = 273 + 0.4 = 273.4
(iii) 3 hundreds + 5 tens + 4 ones + 6 tenths
= 3 \(\times\) 100 + 5 \(\times\) 10 + 4 \(\times\) 1+ 6 \(\times\) \(\frac { 1 }{ 10 } \)
= 300 + 50 + 4 + \(\frac { 6 }{ 10 } \)  = 354 + 0.6 = 354.6

(a) AB (b) b (c) 45

Diameter

We have, 15 + 0.632 + 13.8
\(\quad Tens\quad Ones \quad Tenths \quad Hundreaths \quad Thousandhs\\ \quad1\quad\quad\quad\quad5\quad\quad\quad0\quad\quad\quad\quad\quad0\quad\quad\quad\quad\quad\quad0\\ \quad0\quad\quad\quad\quad0\quad\quad\quad6\quad\quad\quad\quad\quad3\quad\quad\quad\quad\quad\quad2\\+1\quad\quad\quad\quad3\quad\quad\quad8\quad\quad\quad\quad\quad0\quad\quad \quad \quad \quad\quad 0\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\\quad 2\quad\quad\quad\quad9\quad\quad\quad4\quad\quad\quad\quad\quad3\quad\quad\quad\quad\quad\quad2\\\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\)
\(\therefore\)     15+0.632+13.8=29.432

Given, figure is covered by 7 full squares, 8 half squares, 3 more than half squares and 3 less than half squares.
∴ Area = \(\left( 7\times 1+8\times \frac { 1 }{ 2 } +3\times 1+3\times 0 \right) sq\quad units\)
=(7 +4 +3 +0)sq unit
= 14 sq units

140 cm

When y is multiplied by -5, then result is -5 y and then this result is added to 16, then expression will be -5y+16.

\(\because\) Distance covered by Rohan in 1 h = 12 km and distance covered by Rakesh in 1 h = 8 km
\(The\ required\ ratio=\frac{Distance\ covered\ by\ Rohan}{Distance\ covered\ by\ Rakesh}=\frac{12}{8}\)
Now, divide numerator and denominator both by the HCF of 12 and 8 i.e. 4, \(\frac{12\div4}{8\div4}\ \ [\because HCF\ of\ 12\ and\ 8\ is\ 4]\)
\(=\frac{3}{2}=3:2\)

From the figure, it can be observed that it will require five matchsticks.
∴ The number of matchsticks required is 5n.

Perimeter of a squared park = 4 \(\times\) Length of a side = 4 \(\times\) 60 = 240 m
Distance covered in one round = 240 m
Distance covered in three rounds = 3 \(\times\) 240 m = 720 m

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

The required circles, are given below.

No, they are parallel to each other.

(a) 6 : 5
(b) 12 : 11

Converting the given decimals into like decimals
we get
53.5 - 34.68 + 64.75 - 28.9
= (53.50 + 64.75)-(34.68 + 28.90)
= 118.25 - 63.58
= 54.67

For Raju, cost of 10 pens = Rs.150
\(\therefore\) Cost of 1 pen \(=Rs.\frac{150}{10}=Rs.15\)
For Manish, cost of 7 pens = Rs.84
\(\therefore\) Cost of 1 pen \(=Rs.\frac{84}{7}=Rs.12\)
Here, Rs.12 < Rs.15
Hence, Manish got the pens cheaper.

50

Given, length of a rectangular plot = 500 m
and breadth of a rectangular plot = 200 m
∴ Area of the rectangular plot = Length \(\times\) Breadth
= 500 m\(\times\) 200 m
= 100000 sq m
∴ Cost of tiling per hundred square metres = Rs. 8
∴ Cost of 1 sq m = Rs.\(\frac { 8 }{ 100 } \)
Now, cost of 100000 sq metre =Rs.100000\(\times\frac { 8 }{ 100 } \)
=Rs. 8000
Hence, the cost of tiling rectangular plot is Rs. 8000

(a) E
(b) B
(c) 5 yr.

8

\(\frac { 100 }{ 5 } \times 2\) = 40

\(\frac { 25 }{ 25+25 } =\frac { 25 }{ 30 } =\frac { 1 }{ 2 } \)=1:2

Third side = 12 - (5 + 4) = 3 cm

Length of side =\(\frac{8}{4}\)=2 m