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#### Page No 56:

#### Question 1:

*x*(ii) 5

*y*$-$ 7

*z*(iii) 3

*x*

^{3}$-$5x

^{2}$-$11 (iv) 1 $-$8

*a*$-$7

*a*

^{ 2}$-$7

*a*

^{3}

*m*$-$3 (vi)

*a*(vii) 4

*y*

^{2}$-$7

*y*+ 5

#### Answer:

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) 7*x* = Monomial

(ii) 5*y*$-$7*z* = Binomial

(iii) 3*x*^{3}$-$5*x*^{2}$-$11 = Trinomial

(iv) 1$-$8*a*$-$7*a*^{2}$-$7*a*^{3} = Polynomial

(v) 5*m*$-$3 = Binomial

(vi) 4 = Monomial

(vii) 3*y*^{2}$-$7*y*+5 = Trinomial

#### Page No 57:

#### Question 1:

*p*+ 16

*q*; 13

*p*+ 2

*q*(ii) 2

*a*+ 6

*b*+ 8

*c*; 16

*a*+ 13

*c*+ 18

*b*

(iii) 13

*x*

^{2}$-$12

*y*

^{2}; 6

*x*

^{2 }$-$ 8

*y*

^{2 }(iv) 17

*a*

^{2 }

*b*

^{2 }+ 16

*c*; 28

*c*$-$ 28

*a*

^{2 }

*b*

^{2}

(v) 3

*y*

^{2}$-$ 10

*y*+ 16 ; 2

*y*$-$7 (vi) $-$3

*y*

^{2}+ 10

*y*$-$16 ; 7

*y*

^{2 }+ 8

#### Answer:

(i) (9*p* + 16*q*) + (13*p* + 2*q*)

= 9*p* + 16*q* + 13*p* + 2*q*

=(9*p* + 13*p*) + (16*q* + 2*q*)

=22*p *+ 18*q*

(ii) (2*a* + 6*b* + 8*c*) + (16*a *+ 13*c* + 18*b*)

= 2*a* + 6*b* + 8*c* + 16*a* + 13*c* + 18*b*

= (2*a* + 16*a*) + (6*b* + 18*b*) + (8*c* + 13*c*)

= 18*a* + 24*b* + 21*c*

(iii) (13*x*^{2} − 12*y*^{2}) + (6*x*^{2} − 8*y*^{2})

= 13*x*^{2} − 12*y*^{2} + 6*x*^{2} − 8*y*^{2}

= (13*x*^{2} + 6*x*^{2}) + (−12*y*^{2} − 8*y*^{2})

=19*x*^{2} + (−20*y*^{2})

=19*x*^{2 }− 20*y*^{2}

(iv) (17*a*^{2}*b*^{2} + 16*c*) + (28*c* − 28*a*^{2}*b*^{2})

= 17*a*^{2}*b*^{2} + 16*c* + 28*c* − 28*a*^{2}*b*^{2}

= (17*a*^{2}*b*^{2} − 28*a*^{2}*b*^{2}) + (16*c* + 28*c*)

= −11*a*^{2}*b*^{2} + 44*c*

(v) (3*y*^{2} − 10*y *+ 16) + (2*y* − 7)

= 3*y*^{2} − 10*y* + 16 + 2*y* − 7

= 3*y*^{2} + (−10*y* + 2*y*) + (16 − 7)

=3*y*^{2} + (−8*y*) + 9

=3*y*^{2} − 8*y* + 9

(vi) (−3*y*^{2} + 10*y* − 16) + (7*y*^{2} + 8)

= −3*y*^{2} + 10*y* − 16 + 7*y*^{2} + 8

= (−3*y*^{2} + 7*y*^{2}) + 10*y* + (−16 + 8)

= 4*y*^{2} + 10*y* + (−8)

= 4*y*^{2} + 10*y* − 8

#### Page No 58:

#### Question 1:

*xy*$-$9

*z*) ; (3

*xy*$-$ 16

*z*) (ii) (5

*x*+ 4

*y*+ 7

*z*) ; (

*x*+ 2

*y*+ 3

*z*)

*x*

^{2 }+ 8

*xy*+ 3

*y*

^{2 }) ; (26

*x*

^{2}$-$ 8

*xy*$-$ 17

*y*

^{2}) (iv) (6

*x*

^{2}+ 7

*xy*+ 16

*y*

^{2}) ; (16

*x*

^{2}$-$ 17

*xy*)

*x*+ 16

*z*) ; (19

*y*$-$14

*z*+ 16

*x*)

#### Answer:

(i) (4*xy* − 9*z*) − (3*xy* − 16*z*)

= 4*xy* − 9*z *− 3*xy* + 16*z*

= (4*xy* − 3*xy*) + (16*z* − 9*z*)

= *xy* + 7*z*

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

= 5*x* + 4*y* + 7*z* − *x *− 2*y *− 3*z*

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

= 4*x* + 2*y* + 4*z*

(iii) (14*x*^{2} + 8*xy* + 3*y*^{2}) − (26*x*^{2} − 8*xy* − 17*y*^{2})

= 14*x*^{2} + 8*xy* + 3*y*^{2} − 26*x*^{2} + 8*xy* + 17*y*^{2}

= (14*x*^{2} − 26*x*^{2}) + (8*xy* + 8*xy*) + (3*y*^{2} + 17*y*^{2})

= −12*x*^{2} + 16*xy* + 20*y*^{2}

(iv) (6*x*^{2}+ 7*xy*+ 16*y*^{2} ) − (16*x*^{2} $-$ 17*xy*)

= 6*x*^{2} + 7*xy* + 16*y*^{2} − 16*x*^{2} +17*xy*

= (6*x*^{2} − 16*x*^{2}) + (7*xy* + 17*xy*) + 16*y*^{2}

= −10*x*^{2} + 24*xy* + 16*y*^{2}

(v) (4*x* + 16*z*) − (19*y* − 14*z* + 16*x*)

= 4*x* + 16*z* − 19*y* + 14*z* − 16*x*

= (4*x* − 16*x*) + (16*z* + 14*z*) − 19*y*

= −12*x* + 30*z* − 19*y*

#### Page No 59:

#### Question 1:

*xy*× 18

*xy*(ii) 23

*xy*

^{2 }× 4

*yz*

^{2 }

*a*+ 17

*b*) × 4

*c*(iv) (4

*x*+ 5

*y*) × (9

*x*+ 7

*y*)

#### Answer:

(i) 16*xy* × 18*xy*

= 16 × 18 ×* x* × *y* × *x* × *y*

= 288*x*^{2}*y*^{2}

(ii) 23*xy*^{2} × 4*yz*^{2}

= 23 × 4 × *x* × *y*^{2} × *y* × *z*^{2}

= 92*xy*^{3}*z*^{2}

(iii) (12*a* + 17*b*) × 4*c*

= 12*a* × 4*c* + 17*b* × 4*c*

= 12 × 4 × *a* ×* c* + 17 × 4 × *b* × *c*

= 48*ac* + 68*bc*

(iv) (4*x* + 5*y*) × (9*x* + 7*y*)

= 4*x*(9*x* + 7*y*) + 5*y*(9*x* + 7*y*)

= 4*x* × 9*x* + 4*x* × 7*y* + 5*y* × 9*x* + 5*y* × 7*y*

= 36*x*^{2} + 28*xy* + 45*xy* +^{ }35*y*^{2}

= 36*x*^{2} + 73*xy* +35y^{2}

#### Page No 59:

#### Question 2:

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

#### Answer:

Length of the rectangle = (8*x* + 5) cm

Breadth of the rectangle = (5*x* + 3) cm

Area of rectangle = length × breadth

= (8*x* + 5)(5*x* + 3)

= 8*x*(5*x* + 3) + 5(5*x* + 3)

= 8*x* × 5*x* + 8*x* × 3 + 5 × 5*x* + 5 × 3

= 40*x*^{2} + 24*x* + 25*x* + 15

= 40*x*^{2} + 49*x* + 15

So, area of rectangle is (40*x*^{2} + 49*x* + 15) cm^{2}.

#### Page No 60:

#### Question 1:

*x $-$*11

*y*)

*$-$*(17

*x*+ 13

*y*) and choose the right answer.

*x*

*$-$*12

*y*(ii)

*$-$*14

*x*

*$-$*54

*y*(iii)

*$-$*3 (5

*x*+ 4

*y*) (iv)

*$-$*2 (7

*x*+ 12

*y*)

#### Answer:

(3*x $-$* 11*y*) *$-$* (17*x * + 13*y*)

= 3*x* *$-$ *11*y* *$-$ *17*x* *$-$* 13*y*

= (3*x* *$-$ *17*x*) + (*$-$*11*y* *$-$* 13*y*)

= *$-$* 14*x* + (*$-$*24*y*)

= *$-$*14*x* *$-$ *24*y
= $-$*2(

*x*+ 12

*y*)

So, the correct option is (iv).

#### Page No 60:

#### Question 2:

*x*

^{2}

*y*

^{3 }

*z*) and ( $-$15

*x*

^{3}

*yz*

^{2}) is .................... .

*x*

^{5}

*y*

^{4}

*z*

^{3 }(ii) 345

*x*

^{2}

*y*

^{3 }

*z*

^{5}(iii) 145

*x*

^{3}

*y*

^{2}

*z*(iv) 170

*x*

^{3}

*y*

^{2}

*z*

^{3}

#### Answer:

(23*x*^{2}*y*^{3}*z*) × (−15*x*^{3}*yz*^{2})

= 23 × (−15) × *x*^{2} × *y*^{3} × *z* × *x*^{3} × *y* × *z*^{2}

= −345*x*^{5}*y*^{4}*z*^{3}

So, the correct option is (i).

#### Page No 60:

#### Question 3:

Solve the following equations .

(i) $4x+\frac{1}{2}=\frac{9}{2}$ (ii) 10 = 2*y* + 5 (iii) 5*m *$-$ 4 = 1

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

#### Answer:

$\left(\mathrm{i}\right)4x+\frac{1}{2}=\frac{9}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 4x=\frac{9}{2}-\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 4x=\frac{9-1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 4x=\frac{8}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 4x=4\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{4}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow x=1$

$\left(\mathrm{ii}\right)10=2y+5\phantom{\rule{0ex}{0ex}}\Rightarrow 2y+5=10\phantom{\rule{0ex}{0ex}}\Rightarrow 2y=10-5\phantom{\rule{0ex}{0ex}}\Rightarrow 2y=5\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{5}{2}$

$\left(\mathrm{iii}\right)5m-4=1\phantom{\rule{0ex}{0ex}}\Rightarrow 5m=1+4\phantom{\rule{0ex}{0ex}}\Rightarrow 5m=5\phantom{\rule{0ex}{0ex}}\Rightarrow m=\frac{5}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow m=1$

$\left(\mathrm{iv}\right)6x-1=3+8\phantom{\rule{0ex}{0ex}}\Rightarrow 6x-3x=8+1\phantom{\rule{0ex}{0ex}}\Rightarrow 3x=9\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{9}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow x=3$

$\left(\mathrm{v}\right)2\left(x-4\right)=4x+2\phantom{\rule{0ex}{0ex}}\Rightarrow 2\times x-2\times 4=4x+2\phantom{\rule{0ex}{0ex}}\Rightarrow 2x-8=4x+2\phantom{\rule{0ex}{0ex}}\Rightarrow 4x+2=2x-8\phantom{\rule{0ex}{0ex}}\Rightarrow 4x-2x=-8-2\phantom{\rule{0ex}{0ex}}\Rightarrow 2x=-10\phantom{\rule{0ex}{0ex}}\Rightarrow x=-\frac{10}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow x=-5$

$\left(\mathrm{vi}\right)5\left(x+1\right)=74\phantom{\rule{0ex}{0ex}}\Rightarrow 5\times x+5\times 1=74\phantom{\rule{0ex}{0ex}}\Rightarrow 5x+5=74\phantom{\rule{0ex}{0ex}}\Rightarrow 5x=74-5\phantom{\rule{0ex}{0ex}}\Rightarrow 5x=69\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{69}{5}$

#### Page No 60:

#### 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?

#### Answer:

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.

$\therefore x+\left(x-5\right)=27\phantom{\rule{0ex}{0ex}}\Rightarrow x+x-5=27\phantom{\rule{0ex}{0ex}}\Rightarrow 2x-5=27\phantom{\rule{0ex}{0ex}}\Rightarrow 2x=27+5\phantom{\rule{0ex}{0ex}}\Rightarrow 2x=32\phantom{\rule{0ex}{0ex}}\Rightarrow x=16$

So, age of Sania = *x* years = 16 years

age of Rakesh = (*x* − 5) years = (16 − 5) years = 11 years

#### Page No 60:

#### 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 ?

#### Answer:

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.

$\Rightarrow x+\left(x+60\right)=200\phantom{\rule{0ex}{0ex}}\Rightarrow x+x+60=200\phantom{\rule{0ex}{0ex}}\Rightarrow 2x+60=200\phantom{\rule{0ex}{0ex}}\Rightarrow 2x=200-60\phantom{\rule{0ex}{0ex}}\Rightarrow 2x=140\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{140}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow x=70$

Number of ashoka trees planted in the forest = *x* = 70

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

#### Page No 60:

#### 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 ?

#### Answer:

Let the number of 50-rupee notes = *x*

Then, the number of 20-rupee notes = 2 × number of 50-rupee notes = 2*x*

Amount of 50-rupee notes = Rs(50 × *x*) = Rs 50*x*

Amount of 20-rupee notes = Rs(20 × 2*x*) = Rs 40*x*

It is given that total amount is Rs 2700.

$\therefore 50x+40x=2700\phantom{\rule{0ex}{0ex}}\Rightarrow 90x=2700\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{2700}{90}\phantom{\rule{0ex}{0ex}}\Rightarrow x=30$

Number of 50-rupee notes = *x* = 30

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

#### Page No 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 ?

#### Answer:

Let the runs scored by Rohit = *x*

The, the runs scored by Virat = 2 × runs scored by Rohit = 2*x*

According to question, total of their scores is 2 less than a double century.

$\therefore x+2x=200-2\phantom{\rule{0ex}{0ex}}\Rightarrow 3x=198\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{198}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow x=66$

So, runs scored by Rohit = *x* = 66

Runs scored by Virat = 2*x* = 2 × 66 = 132

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