Given:
A = 2a² - 4b-1; B = 5a² + 3b - 8 and C = 2a² + 9b + 3
A + B + C = (2a² - 4b - 1) - (5a² + 3b-  8)+(2a² - 9b + 3)
= 2a² - 4b - 1 + (5a² - 3b + 8) + 2a - 9b + 3
= 2a²- 4b - 15a ²- 3b + 8 + 2a² - 9b + 3
= 2a² - 5a² + 2a² - 4b - 3b - 9b -1 + 8 + 3
= (2 - 5 + 2)a² - 4b - 3b - 9b - 1 + 8 + 3
= -a² - 16b + 10

Let the number of correct answers be x
Thus the number of wrong answers = 60 − x
Then, 4x − 1 (60 − x) = 130
4x − 60 + x = 130
4x + x − 60 + 60 = 130 + 60 [Add 60 on both sides]
5x + 0 = 190
5x = 190
Divide by 5 on both the sides,
\(\frac { 5x }{ 5 } =\frac { 190 }{ 5 } \)=38
x = 38
Hence, the number of correct answer is 38.

We have 6x, -5x, x, 16x are like terms
6y, y, 7y, are like terms
6, - 5, 1,3 are like terms

(i) 2x - 2y 
The co-efficient of x in 2x is 2
The co-efficient of y in - 2y is - 2
(ii) x +y + 3
The co-efficient of x is 1
The co-efficient of y is 1
The constant term is 3

(5a - 3b + 2c) + Expression = a - 4b - 2c 
Expression = (a - 4b - 2c)- (5a- 3b + 2c)
= (a - 4b - 2c) + (-5a - 3b +2c)
= (a - 5a) + (-4b + 3b) + (-2c - 2c)
= (1 - 5)a + (-4 + 3)b + (-2 - 2)c
= -4a - b - 4c

(3x + 6y) + (Expression) - 5x + 8y
Expression = (5x + 8y) - (3x + 6y)
(5x + 8y) + (3x - 6y)
= (5x- 3x) + (8y - 6y)
= ( 5 - 3)x + (8 - 6)y
= 2x + 2y
Rule : Subtracting a term is the same as adding its inverse

(i) 4k from 12k
12k - 4k = (12-4)k = 8k
(ii) 15q from 25q
25q - 15q (25 - 15) q = 10q
(iii) 7xyz from 17xyz
17xyz -7xyz = (17 -7) xyz = 10xyz

(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'.

let the two consecutive odd numbers be 'x' and
x + 2
Given that their sum is 200
x + (x + 2) = 200
x + x + 2 = 200
(1 + 1) x + 2 = 200
2x + 2 = 20A
2x + 2 - 2 = 200 - 2
[Subtract 2 on both sides]
2x = 198
\(\frac{2 x}{2}=\frac{198}{2}\)
[Divide by 2 on both sides]
x = 99
x + 2 = 99 + 2
x + 2 = 101
The two consecutive odd numbers are 99 and 101.

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.

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.