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#### Question 1:

Classify the following algebraic expressions as monomials, binomials, trinomials or polynomials.
(i) 7x (ii) 5y $-$ 7z (iii) 3x3 $-$5x2 $-$11    (iv) 1 $-$8a$-$7a 2$-$7a3
(v) 5m$-$3 (vi) a (vii) 4
(viii) 3y2$-$7y + 5

It is known that, expressions with one term is called monomial, expressions with two terms are binomials, expressions with three terms are trinomials and expression with more than three terms are polynomials.
(i) 7x = Monomial
(ii) 5y$-$7z = Binomial
(iii) 3x3$-$5x2$-$11 = Trinomial
(iv) 1$-$8a$-$7a2$-$7a3 = Polynomial
(v) 5m$-$3 = Binomial
(vi) 4 = Monomial
(vii) 3y2$-$7y+5 = Trinomial

#### Question 1:

(i) 9p + 16q ; 13p + 2q       (ii) 2a + 6b + 8c; 16a + 13c+ 18
(iii) 13x2$-$12y2 ; 6x2 $-$ 8y2        (iv) 17a2 b2 + 16c ; 28c $-$ 28a2 b2
(v) 3y2 $-$ 10y + 16 ; 2y$-$7 ​       (vi) $-$3y2 + 10y  $-$16 ; 7y2 + 8

(i) (9p + 16q) + (13p + 2q)
= 9p + 16q + 13p + 2q
=(9p + 13p) + (16q + 2q)
=22p + 18q

(ii) (2a + 6b + 8c) + (16a + 13c + 18b)
= 2a + 6b + 8c + 16a + 13c + 18b
= (2a + 16a) + (6b + 18b) + (8c + 13c)
= 18a + 24b + 21c

(iii) (13x2 − 12y2) + (6x2 − 8y2)
= 13x2 − 12y2 + 6x2 − 8y2
= (13x2 + 6x2) + (−12y2 − 8y2)
=19x2 + (−20y2)
=19x− 20y2

(iv) (17a2b2 + 16c) + (28c − 28a2b2)
= 17a2b2 + 16c + 28c − 28a2b2
= (17a2b2 − 28a2b2) + (16c + 28c)
= −11a2b2 + 44c

(v) (3y2 − 10y + 16) + (2y − 7)
= 3y2 − 10y + 16 + 2y − 7
= 3y2 + (−10y + 2y) + (16 − 7)
=3y2 + (−8y) + 9
=3y2 − 8y + 9

(vi) (−3y2 + 10y − 16) + (7y2 + 8)
= −3y2 + 10y − 16 + 7y2 + 8
= (−3y2 + 7y2) + 10y  + (−16 + 8)
= 4y2 + 10y + (−8)
= 4y2 + 10y − 8

#### Question 1:

Subtract the second expression from the first.
(i) (4xy$-$9z) ; (3xy $-$ 16z) (ii) (5x + 4y + 7z) ; (x + 2y + 3z)
(iii) (14x2 + 8xy + 3y2 ) ; (26x2 $-$ 8xy$-$ 17y2) (iv) (6x2+ 7xy+ 16y2 ) ; (16x2 $-$ 17xy)
(v) (4x + 16z) ; (19y $-$14z + 16x)

(i) (4xy − 9z) − (3xy − 16z)
= 4xy − 9− 3xy + 16z
= (4xy − 3xy) + (16z − 9z)
xy + 7z

(ii) (5x + 4y + 7z) − (x + 2y + 3z)
= 5x + 4y + 7z − − 2− 3z
= (5x − x) + (4y − 2y) + ( 7z − 3z)
= 4x + 2y + 4z

(iii) (14x2 + 8xy + 3y2) − (26x2 − 8xy − 17y2)
= 14x2 + 8xy + 3y2 − 26x2 + 8xy + 17y2
= (14x2 − 26x2) + (8xy + 8xy) + (3y2 + 17y2)
= −12x2 + 16xy + 20y2

(iv) (6x2+ 7xy+ 16y2 ) − (16x2 $-$ 17xy)
= 6x2 + 7xy + 16y2 − 16x2 +17xy
= (6x2 − 16x2) + (7xy + 17xy) + 16y2
= −10x2 + 24xy + 16y2

(v) (4x + 16z) − (19y − 14z + 16x)
= 4x + 16z − 19y + 14z − 16x
= (4x − 16x) + (16z + 14z) − 19y
= −12x + 30z − 19y

#### Question 1:

Multiply.
(i) 16xy × 18xy (ii) 23xy2 × 4yz
(iii) (12a + 17b) × 4 (iv) (4x + 5y) × (9x + 7y)

(i) 16xy × 18xy
= 16 × 18 × x × y × x × y
= 288x2y2

(ii) 23xy2 × 4yz2
= 23 × 4 × x × y2 × y × z2
= 92xy3z2

(iii) (12a + 17b) × 4c
= 12a × 4c + 17b × 4c
= 12 × 4 × a × c + 17 × 4 × b × c
= 48ac + 68bc

(iv) (4x + 5y) × (9x + 7y)
= 4x(9x + 7y) + 5y(9x + 7y)
= 4x × 9x + 4x × 7y + 5y × 9x + 5y × 7y
= 36x2 + 28xy + 45xy + 35y2
= 36x2 + 73xy +35y2

#### Question 2:

A rectangle is (8x + 5) cm long and (5x + 3) cm broad. Find its area.

Length of the rectangle = (8x + 5) cm
Breadth of the rectangle = (5x + 3) cm
Area of rectangle = length × breadth
= (8x + 5)(5x + 3)
= 8x(5x + 3) + 5(5x + 3)
= 8x × 5x + 8x × 3 + 5 × 5x + 5 × 3
= 40x2 + 24x + 25x + 15
= 40x2 + 49x + 15
So, area of rectangle is (40x2 + 49x + 15) cm2.

#### Question 1:

Simplify (3x $-$ 11y)$-$ (17 + 13y) and choose the right answer.

(i) 7$-$12y  (ii) $-$ 14$-$54y (iii) $-$3 (5 + 4y )    (iv) $-$2 (7+ 12y )

(3$-$ 11y$-$ (17 + 13y)
= 3x $-$ 11y $-$ 17x $-$ 13y
= (3x $-$ 17x) + ($-$11y $-$ 13y)
$-$ 14x + ($-$24y)
$-$14x $-$ 24y
$-$
2(x + 12y)
So, the correct option is (iv).

#### Question 2:

The product of (23 x2 y3 z) and ( $-$15x3 yz2 ) is .................... .

(i) $-$345  x5 y4 z3 (ii) 345  x2 y3 z5 (iii) 145 x3y2 z    (iv) 170 x3y2 z3

(23x2y3z) × (−15x3yz2)
= 23 × (−15) × x2 × y3 × z × x3 × y × z2
= −345x5y4z3
So, the correct option is (i).

#### Question 3:

Solve the following equations .

(i)     (ii) 10 = 2y + 5   (iii)  5​m $-$ 4 = 1

(iv) 6x  $-$ 1 = 3x + 8   (v) 2 ( x $-$ 4) = 4x  + 2   (vi)  5( x + 1 ) = 74

#### Question 4:

Rakesh’s age is less than Sania’s age by 5 years. The sum of their ages is 27 years. How old are they?

Let the age of Sania = x years
Then, the age of Rakesh = (x − 5) years
According to question, sum of ages of Sania and Rakesh is 27 years.

So, age of Sania = x years = 16 years
age of Rakesh = (x − 5) years = (16 − 5) years = 11 years

#### Question 5:

When planting a forest, the number of jambhul trees planted was greater than the number of ashoka trees by 60. If there are altogether 200 trees of these two types, how many jambhul trees were planted ?

Let the number of ashoka trees planted in the forest = x
Then, the number of jambhul trees planted in the forest = x + 60
According to question, total number of ashoka and jambhul trees planted in the forest is 200.

Number of ashoka trees planted in the forest = x = 70
Number of jambhul trees planted in the forest = x + 60 = 70 + 60 = 130

#### Question 6:

Shubhangi has twice as many 20-rupee notes as she has 50-rupee notes. Altogether,she has 2700 rupees. How many 50-rupee notes does she have ?

Let the number of 50-rupee notes = x
Then, the number of 20-rupee notes = 2 × number of 50-rupee notes = 2x
Amount of 50-rupee notes = Rs(50 × x) = Rs 50x
Amount of 20-rupee notes = Rs(20 × 2x) = Rs 40x
It is given that total amount is Rs 2700.

Number of 50-rupee notes = x = 30
Number of 20-rupee notes = 2x = 2 × 30 = 60

#### Question 7:

Virat made twice as many runs as Rohit. The total of their scores is 2 less than a double century. How many runs did each of them make ?