3x - 5 - x + 9 = 3(3) - 5 - 3 + 9
= 9 - 5 - 3 + 9 =18 - 8 =10

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.

Initial position of submarine = 300 feet below
= -300 feet
Distance ascended by submarine = 175 feet
= +175 feet
New position of submarine = (−300) + (+175)
= −125
That is, the submarine is 125 feet below the sea level.

(i) –3 – (–2)
To subtract –2 from –3 using number line,

Therefore, –3 – (–2) = –3 + 2 = –1
(ii) +6 – (–5)
To subtract –5 from 6 using number line,

Therefore, +6 – (–5) = +6 + 5 = 11.
Now, let us see how to subtract negative integers using additive inverse.

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.

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.

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