\(\frac{\mathrm{P}-\mathrm{Q}}{\mathrm{P}+\mathrm{Q}}=\frac{-15-5}{-15+5}=\frac{-20}{-10}=2\)

Let a, b be two integers.
The product of two integers = -135
a x b = -135
One number,a = -15
b = ?
Other number \(=\frac{135}{-15}=9\)
∴ The other number = 9

(5a - 3b + 2c) + Expression = a - 4b - 2c 
Expression = (a - 4b - 2c)- (5a- 3b + 2c)
= (a - 4b - 2c) + (-5a - 3b +2c)
= (a - 5a) + (-4b + 3b) + (-2c - 2c)
= (1 - 5)a + (-4 + 3)b + (-2 - 2)c
= -4a - b - 4c

From the figure
2x + 23 + 3x - 18 = 180^o [linear pair]
5x + 5 = 180^o
5x = 175^o
x = 35^o

Possible routes from Mandapam to Vivekandar Memorial are
route 1:
(a) . Mandapam \(\longrightarrow \) Pullivasal Island \(\longrightarrow \) Krusadai Island \(\longrightarrow \) Vivekanandar Memorial Hall.
Distance = 6 Km + 2 Km + 1.5 Km = 9.5 Km
route 2:
(b) Mandapam\(\longrightarrow \) Krusadai Island\(\longrightarrow \) Vivekanandar Memorial Hall.
Distance = 7 Km + 1.5 Km = 8.5 Km
8.5 km < 9.5 km
\(\therefore \) Shortest route: Mandapam \(\longrightarrow \) Krusadai Island\(\longrightarrow \) Vivekanandar Memorial

The level of the water in a pond decreases in a week  = -2 inches.
Level of water decreases in 6 weeks = 6 x -2 = -12 inches
The change in the level of the water over a period of 6 week is -12 inches

(i) 60°
Construction:
Step 1: Drawn a line and marked a point 'A' on it.
Step 2: With A as center drawn an arc of convenient radius to meet the line at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C.
Step 4: Joined AC. The ∠ABC is the required angle with the measure 60

(ii) 120°
Construction:
We know that there are two 60° angles in 120°.
∴ We can construct two 60° angles consecutively construct 120°
Step 1: Drawn a line and marked a point 'A' on it
Step 2: With 'A' as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center, drawn an arc to cut the previous arc at C
Step 4: With the same radius and C as center, drawn an arc to cut the arc drawn in step 2 at D
Step 5: Joined AD. Then ∠BAD is the required angle with measure 120°.

(iii) 30°
Constructions:
Since 30° is half of 60°, we can construct 30° by bisecting the angle. 60°.
Step 1: Drawn a line and marked a point A on it.
Step 2: With A as center drawn an arc of convenient radius to the line to meet at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C.
Step 4: Joined AC to get ∠BAC = 60°
Step 5: With B as center drawn an arc of convenient radius in the interior of ∠BAC
Step 6: With the same radius and C as center drawn an arc to cut the previous arc at D
Step 7: Joined AD. ∴ ∠BAD is the required angle of measure 30°.

(iv) 90°
Construction:
Step 1: Drawn a line and marked a point' A' on it.
Step 2: With 'A' as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at 'C'.
Step 4: With the same radius and C as center, drawn an arc to cut the arc drawn in step 2 at D.
Step 5: Joined AD. ∠BAD = 120°.
Step 6: With C as center, drawn an arc of convenient radius in the interior of ∠CAD.
Step 7: With the same radius and D as center, drawn an arc to cut the arc at E.
Step 8: Joined AF ∠BAE = 90°.

(v) 45°
Construction:
Step 1: Drawn a line and marked a point A on it
Step 2: With A as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C.
Step 4: With the same radius and C as center, drawn an arc to cut the arc drawn in step 2 at D.
Step 5: Joined AD. ∠BAD = 120°.
Step 6: With G as center and any convenient radius drawn an arc in the interior of ∠GAB
Step 7: With the same radius and B as center drawn an arc to cut the arc at F.
Step 8: Joined AF. ∠BAF = 45°

(vi) 150°
Construction:
Since 50° = 60° + 60° + 30°; we construct as follows
Step 1: Drawn a line and marked a point A on it.
Step 2: With' A' as center, drawn a full arc of convenient radius to the line at a point B and at E the other end.
Step 3: With the same radius and B as center, drawn an arc to cut the previous arc at C.
Step 4: With the same radius and C as center drawn an arc to cut the already drawn arc at D.
Step 5: With D as center, drawn an arc of convenient radius in the interior of ∠DAE
Step 6: With E as center and with the same radius drawn an arc to cut the previous arc at F.
Step 7: Joined AF, ∠FAB = 150°.

(vii) 135°
Construction:
Step 1: Drawn a line and marked a point A on it.
Step 2: With 'A' as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C.
Step 4: With the same radius and C as center, drawn an arc to cut the arc at D.
Step 5: With C and D as centers drawn arcs of convenient (same) radius in the interior of ∠CAD. Marked the point of intersection as E.
Step 6: Joined AE, through G. ∠BAE = 90°.
Step 7: Drawn angle bisector to ∠GAH through F
Now ∠BAF = 135°.

base = 4 times of the height
b = 4h
Area of parallelogram = 576
b x h = 576
4h x h = 576
4h² = 576
h² = I44 cm
h = 12 cm
Ans : Height of the parallelogram is 12 cm.

Given: b : h = 7 : 3 and h = 45 cm
\(\Rightarrow \ \frac{\mathrm{b}}{\mathrm{h}}=\frac{7}{3} \)
\(\Rightarrow \ \frac{\mathrm{b}}{45}=\frac{7}{3} \)
\(\Rightarrow \ b=\frac{7}{3} \times 45 \)
b = 105 cm
Area of the parallelogram = b x h sq.units
= 105 x 45 = 4725 sq.cm

Given: Height of parallelogram \(=\frac{1}{4} of\ its\ base \)
h \(=\frac{1}{4} b \)
Area of the parallelogram = 676
b x h = 676
\(b \times \frac{1}{4} b=676\)
b x b = 676 x 4
b x b = 26 x 26 x 2 x 2
b = 26 x 2
Base (b) = 52 cm
\(h=\frac{1}{4} \times b \)
\(b=\frac{1}{4} \times 52 \)
h = 13

Given: A = 1586 sq.cm, h = 26 cm, a = 84 cm, b = ?
Area of trapezium = 1586
\(\frac{1}{2} \times \mathrm{h}(\mathrm{a}+\mathrm{b}) =1586 \)
\(\frac{1}{2} \times 26(84+\mathrm{b}) =1586 \)
\(84+\mathrm{b} =\frac{1586 \times 2}{26} \)
b = 122 - 84
b = 38 cm

Let x be the amount of diesel to cover 1650 km.

This is a direct proportion.
\(\frac{x_{1}}{y_{1}} =\frac{x_{2}}{y_{2}} \)
\(\frac{108}{594} =\frac{x}{1650} \)
\(x =\frac{108 \times 1650}{594} \)
x = 300
300 litres diesel required to cover 1650 km.

-47 -(-12) = -47 + 12 = -35
A student got -35.
No, he is not correct

From the figure the pair ofadjacent angles are
i) \(\angle\)ABG and, \(\angle\)CBG
ii) \(\angle\) DCE and \(\angle\)ECF
iii) \(\angle\)ECE and \(\angle\)ACF
iv) \(\angle\)ACF and \(\angle\)ECD.
v) \(\angle\)BCF and \(\angle\)FCE

This is an inverse proportion,
x₁ y₁ = x₂ y₂
6 x 120 = x \(\times\) 30
\(x=\frac{7 \times 180}{84} \Rightarrow x=24\)
24 gardeners are needed. Already 6 gardeners are available. Therefore, 18 more gardeners are needed.

The weight of 72 books = 9 kg
The weight of 1 book = \(\frac{9}{72} \mathrm{~kg}\)
The weight of 40 books \(=\frac{9}{72} \times 40=5 \mathrm{~kg}\)

Rs. 840 paid for 21 students.
Rs. 1680 (double of its amount) paid for 42 students (double the students).

(i) 7x, -8x, -12x are like terms containing same algebraic variable 'x'.
(ii) 5y, 12 -9y are like terms containing same algebraic variable 'y'.
(ili) 6z, z and 11 z are like terms containing same algebraic variables 'z'.

h = 14 m,b = 14 + 8 = 22 m
Area of ground (parallelogram) = b x h sq.units
= 22 x 14 = 308 m²
Cost of levelling the ground = Rs. 15  per sq.m
Cost of levelling the ground = 308 x 15 = Rs. 4620

(i) Area of a parallelogram
= b x h sq.units
= 11 x 3 = 33 cm²
Perimeter of the parallelogram
= sum of the length of the four sides.
= 11+ 4 + 11 + 4 = 30 cm.
(ii) Area of a parallelogram
= b x h sq.units
= 7 x 10 = 70 cm²
Perimeter of the parallelogram
= sum of the length of the four sides.
= 13 + 7 + 13 + 7 = 40 cm

(i) Both Team A and B scored 10 marks. Quiz is draw.
(ii) Yes, we can add integers in any orders since addition is commutative

let the two consecutive odd numbers be 'x' and
x + 2
Given that their sum is 200
x + (x + 2) = 200
x + x + 2 = 200
(1 + 1) x + 2 = 200
2x + 2 = 20A
2x + 2 - 2 = 200 - 2
[Subtract 2 on both sides]
2x = 198
\(\frac{2 x}{2}=\frac{198}{2}\)
[Divide by 2 on both sides]
x = 99
x + 2 = 99 + 2
x + 2 = 101
The two consecutive odd numbers are 99 and 101.

(i) From the figure
∠AOC + ∠BOC = 180°
72° + x^o = 180°
72° + x^o - 72° = 180° - 72°
x^o = 108°
(ii) From the figure
∠AOC + ∠BOC = 180°
3x + 42° - 180°
3x^o + 42° - 42°= 180^o - 42°
3x^o = 138°
x^o = \(\frac{{138}^o}{3}\) = 46^o
x^o = 46°
(iii) From the figure
∠AOC + ∠BOC = 180°
4x^o + 2x^o = 180°
6x^o = 180°
x^o = \(\frac{{138}^o}{6}\)
x^o = 30°

(i) (x +y - z) + (3x .: 5y + 7z) - (14x + 7y - 6z)
= (x +y -z) + (3x - 5y + 7z) + (-14x -7y + 6z)
= (x + 3x - 14x) + (y - 5y -7y) + (-z + 7z + 6z)
= (1 + 3 - 14) x + (1-5 -7) y + (-1 + 7 + 6) z
= -10x -11y + 12z
(ii) p + p + 2 + p + 3 - p - 4 - P - 5 + p + 10
= (p + p + p - p - p + p) + (2 + 3 -4-5 + 10)
=(1 + 1 + 1 - 1 - 1 + 1)p  +(15 -9)
= 2p + 6
(iii) n + (m + 1) + (n + 2) + (m + 3) + (n + 4) + (m + 5)
n + (m + 1) + (n + 2) + (m + 3) + (n + 4) + (m + 5)
=(n + n + n) + (m + m + m) + (1 + 2 + 3 + 4 + 5)
= (1 + 1 + 1) n + (1 + 1 + l)m + 15
= 3m + 3n + 15

Statement           Expression
Me                            x
Add 2 to me        x + 2
Multiply by 5       ( x + 2) x 5
Subtract 10       ( x + 2) x 5 -10
Divide by 4        \(\frac{(x+2) 5-10}{4}\)
Resulting 15       \(\frac{(x+2) 5-10}{4}=15\)
\(\frac{(x+2) 5-10}{4}=15\)
(x + 2)5 - 10 = 15 x 4
(x + 2)5 - 10 = 60
5x + 10 - 10 = 60
5x = 60
\(x=\frac{60}{5}\)
x = 12
I am 12

Lift in the ground floor denotes 0.
It goes 5 floors down = -5
It moves up to 10 floors = +10
Position of the lift = 10 - 5 = 5
Lift will be in the 5m floor