The price of plant reduced in a week = Rs. 6
\(\therefore\)The price reduced in 7 weeks = 7 x 6 = 4

∠COB = 50°
∠AOD = 50° (vertically opposite angles)
Now ∠AOC and ∠COB form a linear pair
Thus ∠AOC + ∠COB = 180°
⇒ ∠AOC + 50° = 180°.
∠AOC = 180° - 50° = 130°
Also ∠AOC and ∠BOD are vertically opposite angles
∠BOD = ∠AOC = 130°
Thus the three angles are
∠AOD = 50°
∠AOC = 130°
∠BOD = 130°

Length of the road = 600m
\(\cfrac { 2 }{ 3 } \) rd of the road = \(\cfrac { 2 }{ 3 } \times 600m=400m\)
In \(2\cfrac { 1 }{ 2 } \) hrs sumathi sweeps 600m
\(\therefore\) In 1 hr sumathi Sweeps \(\cfrac { 600 }{ 2\frac { 1 }{ 2 } } m=\cfrac { 600 }{ \frac { 5 }{ 2 } } m=120\times 2=240m\)
In \(1\frac { 1 }{ 2 } \) hrs Ramani sweeps 400 m.
\(\therefore\) In 1hr Ramani sweeps \(\cfrac { 400 }{ 1\frac { 1 }{ 2 } } m=\cfrac { 400 }{ 3 } m=400\times \cfrac { 2 }{ 3 } m=\cfrac { 800 }{ 3 } =266.66m\)
\(\therefore\) Ramanl sweeps more. in 1 hr
\(\therefore\) Ramani sweeps speedily.

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.

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 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

Let x be the required number of days. The decrease in number of workers lead to the increase in number of days. (Therefore, both are in inverse proportion)
For inverse proportion x₁y₁ = x₂ y₂

Hence 60 x 7 = 42 \(\times\) x
42 \(\times\) x = 60 x 7
x = \(\frac { 60\times 7 }{ 42 } \)
x = 10
In 10 days 42 workers can spin a bale of cotton.

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

In the given algebraic expression, x,y,z are the variables.
Let us group the like terms.
100x + 99y − 98z + 10x + 10y + 10z − x − y + z
= (100x + 10x − x) + (99y + 10y − y) + ( −98z + 10z + z)
= (100 + 10 − 1)x + (99 + 10 − 1)y + ( −98 + 10 + 1)z
= (110 − 1)x + (109 − 1)y + ( −98 + 11)z
= 109x + 108y + ( −87)z
= 109x + 108y − 87z.

Base = 17 cm, height = 8 cm
Area of the rhombus = b x h sq. units
= 17 x 8 = 136
Therefore, area of the rhombus = 136 sq. cm