\(2\frac{9}{35}\)Km

We know that, perpendicular bisector of a line segment is its axis of symmetry.
Step I Draw a line segment \(\bar{PQ}\) = 7 cm.
Step II With P and 0 as centres and radius more than half of \(\bar{PQ}\), draw two arcs which intersect each other at A and B.
                        
Step III Join A and B.
Thus, AB is the axis of symmetry of \(\bar{PQ}\).

For rectangle A,
Length = 12 cm and breadth = 6 cm
Perimeter of rectangle A = 2 x (Length + Breadth)
=2(12+ 6)=36cm
For rectangle B, Length = 11 cm and breadth = 9 cm
Perimeter of rectangle B = 2 (11 + 9) = 40 cm
\(\therefore\) Required ratio \(=\frac{36}{40}=\frac{9}{10}=9:10\)

For the LCM of 112, 168 and 266

112 = 2 x 2 x 2 x 2 x 7 = 24 x 7
168 = 2 x 2 x 2 x 3 x 7 = 23³ x 3 x 7
266 = 2 x 7 x19
\(\therefore\)LCM = 24 x 3 x 7 x 19 = 6384

Taking the mirror image of the letters A and B from the given line. These will look as shown below:

It is clear that, A look the same after reflection and B do not look the same because A has reflection symmetry but B does not have.
Now, the mirror image of other letters are shown below

From the above, we can say that A, O, M, H, T, V and X look the same after reflection because these letters are
symmetrical.
Also, we see that B, E, N, P, L and S do not look the same after reflection because these letters are not symmetrical.

In the given figures, we have,
\(\angle AOB\) is an obtuse angle.

(a) Cycling
(b) Reading books
(c) 160
(d) Music and shopping.

Given, length of piece ofland = 0.7 km
and breadth of piece of land = 0.5 km
∴ Perimeter of the rectangular piece of land
= 2 \(\times\) (Length + Breadth) = 2 \(\times\) (0.7 +05) km
= 2\(\times\)1.2 km = 2.4 km
Then, length of wire fencing for 1 row
= Perimeter of rectangular piece ofland = 2.4 km
∴ Length of wire fencing for 4 rows = 4 \(\times\) 2.4 km = 9.6 km
Hence, the wire of length 9.6 km is needed.

Given, the number of oranges in a smaller box = x
Since, one larger box is emptied to fill two smaller boxes. So, the number of oranges in two smaller boxes = 2 \(\times\) Th( number of oranges in one box = 2 \(\times\) x = 2x
Also, 10 oranges remain outside, when large box isemptied to fill two smaller boxes. So, number of oranges in the larger box = Numberr oranges in two smaller boxes + Oranges left over = 2x +1 Hence, the number of oranges in the larger box is(2x + 10)

Zero (0) is its own additive inverse.

We have, 968 x 73 + 968 x 27
       = 968 (73 + 27) [taking 968 as common term]
       = 968 x 100 = 96800.

We know that, a line is a line segment which can extend indefinitely in both directions. Name oflines in all possible ways are as follow:
(i) By taking A, all possible ways are \(\overleftrightarrow { AB } ,\overleftrightarrow { AC } and\overleftrightarrow { AD } \).
(ii) By taking B, all possible ways are \(\overleftrightarrow { BC } ,\overleftrightarrow { BD } and\overleftrightarrow {BA } \).
(iii) By taking C, all possible ways are \(\overleftrightarrow { CD } ,\overleftrightarrow { CA } and\overleftrightarrow {CB } \).
(iv) By taking D, all possible ways are \(\overleftrightarrow { DA } ,\overleftrightarrow { DB } and\overleftrightarrow {DC } \).

We have, 4536, 4892, 4370, 4452
Here, each of the given numbers is containing 4-digits and their digits at thousand places are also same. So, we compare the hundreds place digit.

∵ 8>5>4>3
∴ The greatest number is 4892 and the smallest number is 4370.

4.7 cm.

Ratio of 600 g to 1 kg \(=\frac{600}{1000}[\because\ 1\ kg=1000g]\)
\(=\frac{3}{5}=3:5\)

800 m²

We know that, 1 mm \(={1\over10}\) cm
\(\therefore\)     60 mm \(=60\times{1\over 10}\)cm = \({60\over10}\) cm = 6.0 cm

The given. figure is symmetric and Its line of symmetry is shown alongside.

(i) 2y+17, Here,' operation is multiplication and addition and to form an expression, firstly 2 is multiplied by y and then 17 is added to 2y.
(ii) 2y-17, Here, operation is multiplication and subtraction and to form an expression, firstly,y is 
multiplied by 2 and then 17 is subtracted from 2y.

Factors of 35 and 50 are as follows:
35 = 1 x 35 ; 35 = 5 x 7 and 50 = 1 x 50; 50 = 2 x 25 ; 50 = 5 x 10
Now, all factors of 35 =1, 5, 7, 35 and all factors of 50 = 1, 2, 5, 10, 25, 50
\(\therefore\) Common factors = 1, 5
Hence, common factors of 35 and 50 are 1 and 5.

The chords of the above circle are BE and CD.

We have, two set-squares in our box. In one of them, angles are of 30°, 60°, 90° and in the other, angles are of 45°, 45°, 90°.
(i)
Clearly, they have one angle measure in common, which is 90°.

On the number line, starting from 1 and moving 5 points towards left (each step being equal to 1 unit), we will reach at - 4.

7

7

To find 3 x 3
We have to multiply 3 by 3 i.e. 3 units x 3 (or 3 units 3 times). Let us start from 0, move 3 units to the right of 0.

After making 3 such moves, we reach at 9.
\(\therefore\)  3 x 3 = 9.

15 x 68 = 15 x (60 + 8)
= 15 x 60 + 15 x 8
(distributivity of multiplication over addition)
= 900 + 120 = 1020
17 x 23 = 17 x (20 + 3)
= 17 x 20 + 17 x 3
(distributivity of multiplication over addition)
=340+51=391
69 x 78 + 22 x 69
= 69 x 78 + 69 x 22
(commutativity of multiplication)
= 69 x (78 + 22)
(distributivity of multiplication over addition)
= 69 x 100 = 6900.

To construct an angle of 150^o, steps of construction are given below:
Step I Draw a line and mark point 0 and A on it such that A is in the right of O.
Step II With O as centre and with any convenient radius draw a semi-circle, cutting the line I at P and S.
Step III Now, take P as centre and radius same as in Step II, draw an arc which intersects the semi-circle at Q.
Step IV Now, take Qas centre and (same as step II) draw an arc which intersects the semi-circle at R.
Step V Now, bisect this angle. For this, take distance more than half of length RS as radius and with R and S as centre draw arcs such that both intersect each other at T.
Step VI Join OT and produce it up to point B.
                    
Thus, \(\angle\)AOB = 150^o.

First, we draw rough figure of each shape.
(ii) A rough figure of a square is given below:

In square, l₁,l₂,l₃ and l₄ are four lines of symmetry.
(iii) A rough figure of a rectangle is given below:

In rectangle,l₁ and l₂ are two lines of symmetry.
(iv)A rough figure of an isosceles triangle is given below:

In rhombus, l₁ and l₂ are two lines of symmetry.
(vi) A rough figure of a circle is given below:

In circle, there are infinite lines of symmetry.

\(\because\)   Total distance travelled by Namita
= 20 km 50 m = 20 km + 50 m
\(=20\quad km+50\times \frac { 1 }{ 1000 } km\)   \(\left[ \because 1\quad m=\frac { 1 }{ 1000 } km \right] \)
\(=20\quad km+\frac { 50 }{ 1000 } km\) = (20+0.050) km = 20.050 km
and distance travelled by Namita by bus
= 10 km 200 m = 10 km + 200 m
= 10 km + 200 \(\times \frac { 1 }{ 1000 } km\left[ \because \quad 1m=\frac { 1 }{ 1000 } km \right] \)
\(=10\quad km+\frac { 200 }{ 1000 } km=10\quad km+0.200\quad km\)
= ( 10 + 0.200) km = 10.200 km
\(\therefore\)  Distance travelled by auto
= 20.050 km - 10.200 km
= (20.050 - 10.200) km = 9.850 km
Hence, she travels 9.850 km by auto.

Given, length of floor of the room (l)= 9.68 m
and width of floor ofthe room (b)= 6.2 m
Area of the room = 9.68\(\times\) 6.2 sq m . ... (i)
Also, given that length of each tile = 22 cm
and width of each tile = 10 cm
Now, area of each tile = 22 \(\times\) 10 sq cm ... (ii)
Number of tiles required to cover the floor of the room
\(=\frac { 9.68\times 6.2\times 100\times 100 }{ 22\times 10 } \left[ \because 1m=100cm \right] \)
\(=\frac { 968\times 62\times 10 }{ 22\times 10 } =\frac { 968\times 62 }{ 22 } =2728\)
Total cost = Rs. 2728 \(\times\) 25 = Rs. 68200

(a) In first figure, fraction for shaded portion = \(\frac { 1 }{ 5 } \)
In second figure, fraction for shaded portion = \(\frac { 2 }{ 5 } \)
and in third figure, fraction for shaded portion = \(\frac { 3 }{ 5 } \)
Here, third figure represents more shaded portion than first and second figures.
So, \(\frac { 1 }{ 5 } +\frac { 2 }{ 5 } =\frac { 1+2 }{ 5 } =\frac { 3 }{ 5 } \)
∴ The given figure will be as follows:

(b) In first figure, fraction for shaded portion = 1or \(\frac { 5 }{ 5 } \)
In second figure, fraction for shaded portion = \(\frac { 3 }{ 5 } \)
and in third figure, fraction for shaded portion =\(\frac { 2 }{ 5 } \)
Here, third figure represents less shaded portion than first and second figure
So, \(1-\frac { 3 }{ 5 } =\frac { 5-3 }{ 5 } =\frac { 2 }{ 5 } \)
∴ The given figure is represented as follows:

(c) ∴ The given figure is represented as:

(a) Here, 25 cm : 1 m = 25 cm : 1 x 100 cm [\(\because\)1 m=100em]
= 25 cm: 100 cm
\(=25:100=\frac{25}{100}=\frac{25\div25}{100\div25}=\frac{1}{4}=1:4\)
[dividing numerator and denominator both by 25]
and Rs.40:Rs.160=40:160=\(\frac{40}{160}=\frac{4}{16}=\frac{1}{4}=1:4\)
[dividing numerator and denominator both by 10]
Here, 1: 4 = 1: 4 i.e. 25 cm : 1 m = Rs.40 : Rs.160
So, the ratios of 25 cm : 1 m and Rs.40 : Rs.160 are in proportion.
i.e. 25 cm : 1 m: : Rs.40 : Rs.160
Now, middle terms are 1 m and Rs.40 and extreme terms are 25 cm and Rs.160.
(b) Here,39L:65L=39:65=\(\frac{39}{65}\)
\(=\frac{39\div13}{65\div13}\) [\(\because\) HCF of 39 and 65 = 13]
\(=\frac{3}{5}=3:5\)
and 6 bottles: 10 bottles = 6 :10 \(=\frac{6}{10}\)
\(=\frac{6\div2}{10\div2}=\frac{3}{5}=3:5\)
[dividing numerator and denominator both by 2]
Here, 3 : 5 = 3 :5 i.e. 39 L : 65 L = 6 bottles: 10 bottles.
So, the ratio of 39 L : 65 Land 6 bottles: 10 bottles are in proportion.
i.e. 39 L : 65 L : : 6 bottles: 10 bottles
Now, middle terms of ratios are 65 Land 6 bottles and extreme terms of ratios are 39 Land 10 bottles.
(c) Here, 2kg:80kg=2:80=\(=\frac{2}{80}=\frac{2\div2}{80\div2}=\frac{1}{40}=1:40\)
[dividing numerator and denominator both by 2]
and 25 g : 625 g = 25 :625 \(=\frac{25}{625}=\frac{25\div25}{625\div25}=\frac{1}{25}=1:25\)
[\(\because\)HCF of 25 and 625 = 5 X 5 = 25]
Since, both ratios are not equal.
\(\therefore\) 2 kg : 80 kg \(\ne\) 25 g : 625 g
Hence, the given ratios are not in proportion.
(d) Here, 200mL:2.5L=200 \(\times\frac{1}{1000}L:2.5\ L\)
\([\because\ 1\ mL=\frac{1}{1000}L]\)
\(=\frac{200}{1000}L:2.5L=0.200L:2.5L\)
\(=0.200:2.5=\frac{0.200}{2.5}=\frac{2}{25}=2:25\)
[multiplying numerator and denominator both by 10]
and Rs.4:Rs.50=4:50=\(\frac{4}{50}=\frac{4\div2}{50\div2}=\frac{2}{25}=2:25\)
[dividing numerator and denominator both by 2]
Here, 2 : 25 = 2 : 25 i.e. 200 mL : 2.5 L = Rs. 4 : Rs.50
So, the ratios of 200 mL : 2.5 L and Rs. 4 : Rs. 50 are in proportion.
i.e. 200 mL : 2.5 L: : Rs.4 : Rs. 50
Now, middle terms of ratios are 2.5 L and Rs.4 and extreme terms of ratios are 200 mL and Rs.50.

(a) Given, Anil's present age = x yr
Anil's age after 15 yr = (x + 15) yr
(b) 5 yr ago Anil's age = (x - 5) yr
5 times Anil's age, 5 yr ago = 5(x - 5) yr
(c) Given, Anil's age = x yr
Father's age = (2x + 15) yr
(d) 5 yr ago Anil's age = (x - 5) yr
Now, cousin's age = \([\frac{1}{3}(x-5)-3]\)yr
(e) Given, Anil's age = x yr
Anil's father age = (x + 20) yr
Anil's grandfather age = 2(x + 20) yr

(a) Class X has minimum number of scouts i.e. 10.
(b) Class VIII has maximum number of scouts i.e. 60.
(c) Number of scouts in class VI = 40
(d) Number of scouts in class X =10
Hence. class VI has exactly four times the scouts as that of class X.
(e) Total number of scouts in classes VI to X
                = 40 + 20 + 60 + 30 + 10 = 160.

(i) AB and CD, BC and AD,
(ii) \(\angle A\ and\ \angle B,\angle B\ and\ \angle C,\angle C\ and\ \angle D,\angle D\ and\ \angle A\)

According to the rules of addition and subtraction of integers,
(a) 30 + (-50) + (-20) = 30 -(50 + 20) = 30 -70 = -40
(b) 40 + (-10) + (-15) = 40 -(10 + 15) = 40 - 25 = 15
(c) 45 - (-5) + 10 = 45 + 5 + 10 = 60
(d) 60 - 10 + 5 - (-5 ) = 60 + 5 + 5 - 10 = 70 -10 = 60
(e) 16 + (-13) + (-10) =16 -(13 + 10) =16 - 23 = -7
(f) 0 + 5 - (-6) + (-7) = 0 + 5 + 6 - 7 ⇒ 11 - 7 = 4

Let the required unknown digit be x.
Then, the number becomes
\(\begin{matrix} 8 \\ \downarrow \\ E \end{matrix}\begin{matrix} x \\ \downarrow \\ O \end{matrix}\begin{matrix} 9 \\ \downarrow \\ E \end{matrix}\begin{matrix} 4 \\ \downarrow \\ O \end{matrix}\begin{matrix} 8 \\ \downarrow \\ E \end{matrix}\begin{matrix} 4 \\ \downarrow \\ O \end{matrix}\)
Sum of digits at odd places from right = 4 + 4 + x = 8+x
Sum of digits at even places from right = 8 + 9 + 8= 25
\(\because\) Number is divisible by 11.
\(\therefore\) Difference of digits will be 0 or 11.
\(\Rightarrow\)25 - (8 + x) = 0 or 11
\(\Rightarrow\)25 - 8 - x = 0 or 11 \(\Rightarrow\) 17 - x = 0 or 11
Taking difference 0, 17-x=0 \(\Rightarrow\) x=17+0 \(\Rightarrow\) x=17 [but 17 is not a single digit number, so it is not possible]
Taking difference 11, 17 - x = 11 \(\Rightarrow\) x = 17 - 11= 6
So, required digit to write in the blank space is 6.

We know that,
An angle which is smaller than a right angle is called an acute angle.
An angle with measure 90° is called a right angle.
An angle which is greater than a right angle but smaller than a straight angle is called an obtuse angle.
An angle with measure 180° is called a straight angle.
An angle which is greater than a straight angle but smaller than a complete angle is called a reflex angle.
Now, from these results
(a) Given angle is smaller than a right angle, so it is an acute angle.
(b) Given angle is more than a right angle, so it is an obtuse angle.
(c) Given angle is a right angle.
(d) Given angle is more than a straight angle, so it is a reflex angle.
(e) Given angle is a straight angle.
(f) Both angles are smaller than a right angle, so both these are acute angles.

(a) Weight of apples sold during the last year = 2457 kg
Weight of oranges sold during the last year = 3004 kg
∴ Total weight of apples and oranges Raman sold, during the last year = 2457 + 3004= 5461 kg
(b) Total money Raman got by selling apples
=Total number of apples \(\times\) Cost per kg
= 2457 \(\times\) 40 = Rs. 98280
(c) Total money Raman got by selling oranges
=Total number of oranges \(\times\) Cost per kg
=3004 \(\times\) 30 = Rs. 90120
∴ Total money Raman got by selling apples and oranges together =98280 + 90120 = Rs. 188400
(d) Following table shows the money received from selling items:

The entries of amount of money received in descending order are as follows:
Rs. 253670 >Rs. 240012 >Rs. 160040 > Rs. 98280 > Rs. 90120 >Rs. 68280 > Rs. 38350
The item, which brought him the highest amount is tooth brushes. This amount is Rs. 253670.

Length of side = \(\frac{9}{3}\)=3 m

Given, integer is -20
Precedessor of -20 = -20 - 1 = -21
Successor of -21 = -21 + 1 = -20