Step 1: Draw the given angle ∠ABC with the measure 80° using protractor.

Step 2: With B as center and convenient radius, draw an arc to cut BA and BC. Mark the points of intersection as E on BA and F on BC.

Step 3: With the same radius and E as center, draw 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. Draw a ray BX through G. BG is the required bisector of the given angle ∠ABC

Since neither profit nor loss
Profit + Loss = 0
(ie) profit = -Loss
Profit earned from 1 notebook = Rs. 5
Profit earned from 20 notebooks = 20 x Rs. 5
= Rs. 100
Loss incurred by selling pens = Rs. 100 Loss
= −100
Loss for 1 pen = Rs. 2
loss = −2
Total number of pens sold = (−100) ÷ (−2)
= 50 pens.

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

Given l is parallel to m and n is transversal to l and m.
We get, y = 2x [Vertically opposite angles are equal]
y + 4x = 180° [sum of interior angles that lie on the same side of the transversal]
2x + 4x = 180° [since y = 2x]
6x = 180°
Dividing by 6 on both sides
\(\frac{x}{6}=\frac{180^o}{6}\)gives, x = 30°
Now, y = 2(30°) = 60°.

(i) The corresponding angles to b° are d° and 68°.
(ii) The measure of b° is 68° (Because b° corresponds to 68°)
(iii) The angles having measure 68° are b°, d°, g°, i° and k°.
(iv) The angles having measure 112° are a°, c°, e°, f°, h° and j°.

(i) The angle that corresponds to \(\angle\)1 is \(\angle\)5
(ii) The angle that is alternate interior to \(\angle\)3 is \(\angle\)5
(iii) The angle that is alternate exterior to \(\angle\)8 is \(\angle\)2
(iv) The angle that corresponds to \(\angle\)8 is \(\angle\)4
(v) The angle that is alternate exterior to \(\angle\)7 is \(\angle\)1
(vi) The angle that is alternate interior to \(\angle\)6 is \(\angle\)4

The cost of one box = Rs. 200
Increased cost of one box = Rs. 260
Let x be the number of boxes bought by Vendan.
As cost of boxes increases the number of boxes decreases.
This is in inverse proportion. Therefore, x₁y₁ = x₂ y₂

13 x 200 = x \(\times\) 260
x x 260 = 13 x 200
x = \(\frac { 13\times 200 }{ 260 } \)
x = 10
Therefore, he can buy 10 boxes for the same amount.

(i) 70 - (-40)
= 70 + (additive inverse of−40)
= 70 + 40
= 110
(ii) (-20 )- (-12)
= (-20) + (additive inverse of (-12))
= (-20) + 12
= -8

The amount earned on the first day is Rs.200.
Let the amount spent on the first day be Rs. x. Amount with her at the end of the first day is (200 − x).
Amount earned on the second day is Rs.300 and the amount spent on the second day is Rs.2x. The amount left on the second day is 300 − 2x.
Similarly, the net amount that she would have on the third day is 400 − 4x. Therefore, the total amount that she would have at the end of three days is (200 − x) + (300 − 2x) + (400 − 4x).
That is, 200 + 300 + 400 + ( − 1 − 2 − 4)x
= 900 + ( −7)x
Thus the required algebraic expression is 900 − 7x.

i) (3x − 4y + z) + (2x − z + 3y)
= (3x + 2x) + ( −4y + 3y) + (z − z)
= (3 + 2)x + ( −4 + 3)y + (1 − 1)z
= 5x − 1y + 0z
= 5x − y.

ii) (4x + 3y) − (2x − 5y)
= (4x + 3y) + ( −2x + 5y)
= (4x − 2x) + (3y + 5y)
= (4 − 2)x + (3 + 5)y
= 2x + 8y.

i) 11pq – 7pq. Additive inverse of 7pq is − 7pq
11pq + ( − 7pq) = 11pq − 7pq = (11 − 7)pq = 4pq
ii) a – (–a). Additive inverse of −a is a.
So, a + a = 2a
iii) 21x + 9 – (5x + 7). Additive inverse of 5x + 7 is – (5x + 7).
(21x + 9) + [– (5x + 7)] = (21x + 9) – (5x + 7)
= 21x + 9 − 5x − 7
= (21 − 5)x + (9 − 7)
= 16x + 2.

Given, the length of the diagonals of a rhombus shaped tile are 40 cm and 25 cm
The area of one tile = \(\cfrac { 1 }{ 2 } \times \left( { d }_{ 1 }\times { d }_{ 2 } \right) sq.units\)
= \(\cfrac { 1 }{ 2 } \times 40\times 25\)
= 500 sq.cm
Therefore, the area of 200 such tiles = 200 x 500
= 100000 sq. cm
= \(\cfrac { 100000 }{ 10000 } \)(1sq.m = 1000 sq.cm)
= 10 sq.m
Therefore, the cost of polishing 200 such tiles at the rate of Rs 45 per sq. cm
= 10 x 45 = Rs. 450.

Given: d₁ = 6 m, d₂ = 8 m
Area of the rhombus = \(\cfrac { 1 }{ 2 } \times ({ d }_{ 1 }\times { d }_{ 2 })sq.units\)
= \(\cfrac { 1 }{ 2 } \times (6\times 8)\)
= \(\cfrac { 48 }{ 2 } \)
= 24 sq.m
Hence, area of the rhombus is 24 sq.m.

The temperature at Srinagar was −3°C on Friday. If the temperature decreases by 1°C
then, temperature on the next day = −3°C – 1°C = −4°C

Amount with Chitra = Rs. 150
Cost of bag = Rs. 225
Amount to be borrowed = Rs. 225 − Rs. 150
= Rs. 75