Marks given for one correct answer = 4
Marks given for 10 correct answers = 10 x 4
= 40
Kalaivizhi’s final score = 20
Marks reduced for incorrect answers = 40 − 20
= 20
Therefore, number of questions answered incorrectly = 20 ÷ 2 = 10

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) (-11) - (-33)
= (-11) + (+33)
= 22
(ii) (-90)-(-50)
= -90 - (-50)
= -90 + 50
= -40

(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.

Given that, the price of one idly is ‘x’ rupees and the price of one dosa is ‘y’ rupees
So, Mani’s bill amount: \((2 \times x)+(2 \times y)=(2 x+2 y)\)
Mohamed’s bill amount: \((4 \times x)+(1 \times y)=4 x+y\)
Therefore, the total bill amount = (2x + 2y) + (4x + y)
= (2 + 4) x + (2 + 1)y = 6x + 3y.
Aliter:
In total, they had 2 + 4 = 6 idlies. i.e. \(6 \times x=6 x\)
and 2 + 1 = 3 dosas. i.e. \(3 \times y=3 y\)
Therefore, the total bill amount = 6x + 3y.

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.

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.

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.

Given, the length of one diagonal (d₁) = 8 cm
Let, the length of the other diagonal be d₂ cm
Area of the rhombus = 60 sq. cm (given)
\(\cfrac { 1 }{ 2 } \times ({ d }_{ 1 }\times { d }_{ 2 })=60\)
\(\cfrac { 1 }{ 2 } \times ({ 8 }\times { d }_{ 2 })=60\)
8 x d₂ = 60 x 2
\({ d }_{ 2 }=\cfrac { 120 }{ 8 } \)
= 15
Therefore, length of the other diagonal is 15 cm.

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

Opening balance = Rs. 5000
Debit amount = Rs. 1079 (−)
Balance amount
= Rs. 3921