(i) -275°C
T =  -275°C
T^oC = (T + 273)K
= (-275 + 273)K = -2K
-275°C =  -2 Kelvin
(ii) 45°C
T = 45°C
T^oC = (T + 273)K
(45 + 273)K = 318K
45°C = 318 Kelvin
(iii) -400°C
T = -400°C
T^oC = (T + 273)K
= (-400 + 273) K = -127 K
-400°C = -127 Kelvin
(iv) -273°C
T = -273°C
T^oC = (T + 273)K
= (-273 + 273) K = 0K.
-273°C = 0 Kelvin

(i) (−85) ÷ 5 =−17 
(ii) 250 ÷ (-25) = +10
(iii) 120 ÷ (–6) = −20  
(iv) 182 ÷ (–2) = −91

(a) 60° c
Construction:
Step 1 : Drawn the given angle ∠ABC with the measure 60° using protractor.
Step 2: With B as centre and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as E on BA and F on BC.
Step 3: With the same radius and E as centre drawn an arc in the interior of ∠ABC and another arc of same measure with centre at F to cut the previous arc.
Step 4: Marked the point of intersection as G. Drawn a ray BX through G. BG is the required bisector of the given ∠ABC
Now ∠ABG = ∠CBG = 30°

(b) 100°
Construction:
Step 1: Drawn the given angle ∠ABC with the measure 100° using protractor.
Step 2: With B as centre and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as E on BA and F on BC.
Step 3: With the same radius and E as centre drawn an arc in the interior of ∠ABC and another arc of the same measure with centre at F to cut the previous arc.
Step 4: Marked the point of intersection at G. Drawn a ray BX through G. BG is the required bisector of angle ∠ABC
∠ABG = ∠GBC = 50°

(c) 90°
Construction :
Step 1: Drawn the given angle ∠ABC with the measure 90° using protractor.
Step 2: With B as center and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as E on BA and F on BC
Step 3: With the same radius and E as center drawn an arc in the interior of ∠ABC and another arc of same measure with center at F to cut the previous arc
Step 4: Mark the point of interaction as G. Drawn a ray BX through G BG is the required bisector of the given angle ∠ABC
∠ABG = ∠GBC = 45°

(d) 48°
Construction:
Step 1: Drawn the given angle ∠ABC with the measure 48° using protractor.
Step 2: With B as center and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as E on BA and F on BC.
Step 3: With the same radius and E as center drawn an arc in the interior of ∠ABC and another arc of the same measure with center at F to cut the previous arc.
Step 4: Marked the point of intersection as G. Drawn a ray BX through G. BG is the required bisector of the given angle ∠ABC
Now ∠ABC = ∠GBC = 24°

(e) 110°
Construction:
Step 1: Drawn the given angle ∠ABC with the measure 110° using protractor
Step 2: With B as center and convenient radius, drawn an arc to cut BA and BC. Marked points of intersection as E on BA and F BC.
Step 3: With the same radius and E as center, drawn an arc in the interior of ∠ABC and another arc of same measure with center at F to cut the previous arc.
Step 4: Mark the point of intersection as G. Drawn a ray BX through G. BG is the required bisector of the given angle ∠ABC
∠ABG = ∠GBC = 55°.

Construction:
Step 1:

Drawn a line. Marked two points A and B on it so that AB = 8 cm
Step2:

Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length one above AB and one below AB
Step3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2. Marked the points of intersection of the arcs as C and D.
Step4:

Joined C and D, CD intersect AB. Marked the point of intersection as 'O'. CD is the required perpendicular bisector of AB.
(b) 7 cm
Step 1:

Drawn a line and marked points A and B on it so that AB = 7cm.
Step 2:

Using compass with A as, centre and radius more than half of the length of AB drawn two arcs of same length one above AB and one below AB.
Step 3: With the same radius and B as centre drawn two arcs to cut the already drawn arcs in step 2. Marked the intersection of the arcs as C and D.
Step 4:

Joined C and D, CD is the required perpendicular bisector of AB.
(c) 5.6 cm.
Construction:

Step 1: Drawn a line and marked two points A and B on it so that AB = 5.6 cm
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length, one above AB and one below AB
Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2 and marked the points of intersection of the arcs as C and D
Step 4: Joined C and D. CD intersects AB. Marked the point of intersection as 'O' CD is the required perpendicular bisector of AB.
(d) 10.4 cm
Construction :

Step 1: Drawn a line and marked two points A and B on it so that AB = 10A cm.
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of same length one above AB and one below AB
Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2 and marked the points of intersection of the arcs as C and D.
Step 4: Joined C and D. CD intersects AB. Marked the points of intersection as 0 I CD is the required perpendicular bisector
(e) 58 mm
Construction :

Step 1: Drawn a line. Marked two points A and B on it so that AB = 5.8 cm = 58 mm.
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length one above AB and one below AB.
Step 3: With the same radius and B as centre drawn two arcs to cut the arcs of drawn in step 2. Marked the points of intersection of the arcs as C and D
Step 4: Joined C and D. CD intersects AB. Marked the point of intersection as O. CD is the required perpendicular bisector.

(i) x + 5 = 8
x + 5 - 5 = 8 - 5
[Subtract 5 on both sides]
x = 3
(ii) p -3 = 7
P - 3 + 3 = 7 + 3
[ Add 3 on both sides]
P = 10
(iii) 2x = 30
\(\frac{2 x}{2}=\frac{30}{2}\)
[ Divide by 2 on both sides]
x = 15
\((iv) \frac{m}{6}=5 \)
\(\frac{m}{6} \times 6=5 \times 6 \)
[ Multiply by 6 on both sides]
m = 5 x 6
m = 30
(v) 7x + 10 = 80
7x + 10 - 10 = 80 - 10
[Subtract 10 on both sides]
7x = 70
\(\frac{7 x}{7}=\frac{70}{7}\)
[Divide by 7 onboth sides]
x = 10

i) 7p + 6q, 5p - q, q + 16p
= (7p + 6q) + (5p - q) + (q + 16p)
=(7p+5p+ 16p) + (6q - q + q)
= (7 +5 + 16)p + (6- 1 + 1)q
= 28p + 6q
(ii) a + 5b + 7c, 2a + 10b + 9c
= (a + 2a) + (5b + 10b) + (7c + 9c)
= (1 + 2)a+ (5 + 10)b + (7 +9)c
=3a + 15b  +16c
(iii) mn + (,2mn - 2t, - 3t + 3mn
(mn + t) + (2mn -2t) + (-3t + 3mn)
- (mn + 2mn + 3mn) + (t - 2t- 3t)
=(1 + 2 + 3)mn + (1 -2 - 3)t
= 6mn + (-4)t = 6mn - 4t
(iv) u + v, u - v, 2u+ 5v, 2u - 5v
(u + v) + (u -v) + (2u + 5v) + (2u- 5v)
- (u + u + 2u + 2u) + (v - v + 5v - 5v)
= (1 + 1 +2 + 2)a+ (1 - 1 + 5 - 5)v
= 6u + 0v = 6u
(v) 5xyz - 3xy,3zxy - 5yx
(5xyz - 3xy) + (3zxy - 5yx)
= (5xyz + 3zxy) + (-3xy - 5yx)
= 8xyz + (-8xy) = 8xyz - 8xy

i) (pq − 1) + (3pq + 2) = (pq + 3pq) + ( − 1 + 2)
= (1 + 3)pq + 1
= 4pq + 1
ii) (8x + 3) + (1 − 7x) = 8x + 3 + 1 − 7x
= (8x − 7x) + (3 + 1)
= (8 − 7) x + 4
= x + 4.

(i) 

∴. 8 + (-12) = -4
(ii) 

∴ (-3) + (-5) = -8
(iii) (-100) + (-10) = -100-10 = -110
(iv) 20 + (-72) = 20 - 72 = -52
(v) 82 + (-75) = 82 - 75 = 7
(vi) -48 + (-15) = -48 - 15 =- 63
(vii) -225 + (-63) = 225 - 63 = -288

The vertically opposite angles are
\((i) \angle P T R \ and\ \angle S T Q \)
\((ii) \angle P T S \ and\ \angle R T Q\)

i) The other angle of the linear pair
= 180^o - 84^o = 96^o
ii) The other angle of the linear pair
= 180^o - 86^o = 94^o
iii) The other angle of the linear pair
= 180^o - 159^o = 21^o

i) From this figure \(\angle B A C \text { and } \angle D A C\) are adjacent angles
ii) From this figures \(\angle \mathrm{ZWY} \text { and } \angle \mathrm{XWY}\) are adjacent angles

i) In this figure, AC is the common arm and A is the common vertex.
ii) In the figure, OQ is the common arm and O is the common vertex.

(i) -8
(ii) -1
(iii) 2
(iv) -20
(v) -8 

(i) 900
(ii) 72
(iii) 1200
(iv) -1000

\(\angle \mathrm{AOC}+\angle \mathrm{BOC} =180^{\circ} \)
\(43^{\circ}+\angle \mathrm{BOC} =180^{\circ} \)
\(\therefore \angle \mathrm{BOC} =180^{\circ}-43^{\circ}=137^{\circ} \)

(i) 7xy, 3yx, 12xy,15xy are like terms containing same algebraic variables x and y.
(ii) 19 x, x, 6x, -9x, 23x are like terms containing same algebraic variable x.
(iii) 5y, -13y, 16y, 45y, -8y, -y are the like terms  containing same algebraic variable y.
(iv) 1, 15, -5, 23,11 are like terms containing numerical constants

(i) Rs. 690 + Rs. 485 = Rs. 1175
(ii) Rs. 1175 - Rs. 500 = Rs. 675
(iii)  Rs. 675 - Rs. 350 = Rs. 325
(iv)  Rs. 325 + Rs. 89 = Rs.414
(v)  Rs. 1214 - Rs. 300 = Rs. 914

(I) \(\frac{-75}{5}{=-15}\)
(II) \(\frac{-100}{-20}=5\)
(III) \(\frac{45}{-9}=-5\)
(IV) \(\frac{-82}{82}=-1\)

From the figure
2x + 3x = 90^o
5x = 99^o
x = 18^o

Possible pairs are
(1, -50), (2, -25), (5, -10), (50, -1), (25, -2), (10, -5)

(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°.

(i)  positive integer.
(ii)  negative integer.

(i) any two pairs of adjacent angles are
(a) ∠PQT and ∠TOS 
(b) ∠PQU and ∠RQU 
(ii) two pairs of vertically opposite angles are
(a) ∠PQT and ∠RQU 
(b) ∠RQT and ∠PQU 

From the figure
∴ ∠JlL = ∠JIK + ∠KIL
= 38° + 27°
= 65°
∴ ∠JlL = 65°

(i) 

∴ (-3) - (-4) = +1
(ii) 

7 - (-10) = 7 + 10 = 17
(iii) 35 - (-64) = 35 + 64 = 99
(iv) -200 - (+100) = -200 -100 = -300

Since the given data are in direct proportion.
\(\frac{738}{6}=\frac{x}{9}\)

6x = 738 x 9
\(x=\frac{738 \times 9}{6}=1107\)
1107 letters can be sorted in 9 hours.

Since the given data are in direct proportion
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
\(\frac{8}{120}=\frac{12}{x}\)
8x = 120 x 12
\(\therefore x=\frac{120 \times 12}{8}=180\)
In 12 trips, 180 people would have attended the party.

(i) Given: d₁ = 8 cm, d₂ = 16 cm
Area of the rhombus
\(=\frac{1}{2} \times\left(\mathrm{d}_{1} \times \mathrm{d}_{2}\right) \text { sq.units } \)
\(=\frac{1}{2} \times(8 \times 16)=\frac{128}{2} \)
Area of the rhombus = 64 sq.cm
(ii) b = 15 cm, h = 11 cm
Area of the rhombus = b x h sq.units
= 15 x 11
Area of the rhombus = 165 sq.cm

Given x = 2; y = 3.
(i) 2x - 3y = 2 (2) - 3 (3) = 4 - 9
= 4 + (Additive inverse of 9)
= 4 + (-9) = - 5
(ii) x + y =2 + 3 = 5 
(iii) 4y - x = 4 (3) - 2 = 12 - 2 = 10
(iv) x + 1- y = 2 + 1 - 3 = 3 - 3 = 0

(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'.

(11 + 7) + 10 = 18 + 10 = 28
11 + (7 + 10) = 11 + 17 = 28
Yes, (11 + 7) + 10 and 11 + (7 + 10) are equal.
This is Associative property. a+ (b + c) = (a + b) + c.