Maharashtra Board Class 7 Math Chapter 8 Algebraic Expressions and Operations on them Solution

Maharashtra Board Class 7 Math Solution Chapter 8 – Algebraic Expressions and Operations on them

Balbharati Maharashtra Board Class 7 Math Solution Chapter 8: Algebraic Expressions and Operations on them. Marathi or English Medium Students of Class 7 get here Algebraic Expressions and Operations on them full Exercise Solution.

Std

Maharashtra Class 7
Subject

Math Solution

Chapter

Algebraic Expressions and Operations on them


Practice Set 32

 ¤ Classify the following algebraic expressions as monomials, binomials, trinomials or polynomials.

(i) 7x

(ii) 5y – 7z

(iii) 3x3-5x2 – 11

(iv) 1 – 8a – 7a2 – 7a3

(v) 5m – 3

(vi) a

(vii) 4

(viii) 3y2 – 7y + 5

ANS:

Expressions containing only one term is called as Monomial.

Expressions containing two term is called as Binomial.

Expressions containing three term is called as Trinomial.

Expressions containing more than three term is called as polynomial.

Now, we are classify given algebraic expressions as monomials, binomials, trinomials or polynomials.

Monomial:

(i) 7x

(vi) a

(vii) 4

 

Binomial:

(ii) 5y – 7z

(v) 5m – 3

 

 

Trinomial:

(iii) 3x3-5x2 – 11

(viii) 3y2 – 7y + 5

 

Polynomial:

(iv) 1 – 8a – 7a2 – 7a3

Practice Set 33

 ¤ Add.

 (i) 9p + 16q; 13p + 2q

ANS:

Here we are adding 9p + 16q and 13p + 2q these two algebraic expressions.

We adding p variable terms with p variable and q variable terms with q variable.

9p + 16q

+     13p + 02q

________________________________

22p + 18q

Addition of 9p + 16q; 13p + 2q is 22p + 18q

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

ANS:

Here we are adding 2a + 6b + 8c and 16a + 13c + 18b these two algebraic expressions.

We adding a variable terms with a variable, b variable terms with b variable and c variable terms with c variable.

We rearranging 2nd expression

16a + 18b + 13c

2a + 6b  +   8c

+16a + 18b + 13c

——————————–

18a + 24b + 21c

Addition of 2a + 6b + 8c; 16a + 13c + 18b is 18a + 24b + 21c

(iii) 13x2-12y2; 6x2-8y2

ANS:

Here we are adding 13x2-12y2 and 6x2-8y2 these two algebraic expressions.

We adding x2 variable terms with x2 variable and y2 variable terms with y2 variable.

When there is two negative sign and addition operation is there we have to add both and give negative sign.

 

13x2-12y2

+       6x2– 8y2

———————————————————

19 x2 – 20 y2

Addition of 13x2-12y2; 6x2-8y2 is 19 x2 – 20 y2

(iv) 17a2 b2 + 16c; 28c-28a2 b2

ANS:

Here we are adding 17a2 b2 + 16c and 28c-28a2 b2 these two algebraic expressions.

We adding a2 b2 variable terms with a2 b2 variable and c variable terms with c variable.

We rearranging 2nd expression

-28a2 b2 + 28c

When there is one negative sign and one positive sign and addition operation is there we have to subtract smaller number from larger number and give sign of larger number.

 

17a2 b2 + 16c

+     -28a2 b2 + 28c

——————————-

-11 a2 b2 + 44c

Addition of 17a2 b2 + 16c; 28c-28a2 b2 is -11 a2 b2 + 44c

(v) 3y2-10y + 16; 2y-7

ANS:

Here we are adding 3y2-10y + 16 and 2y-7 these two algebraic expressions.

We adding y variable terms with y variable and constant term with constant term.

We writing 2nd expression as 0y2+ 2y  – 7

When there is one negative sign and one positive sign and addition operation is there we have to subtract smaller number from larger number and give sign of larger number.

3y2-10y + 16

+  0y2+ 2y  – 7

———————————-

3y2 – 8y + 9

Addition of 3y2-10y + 16; 2y-7 is 3y2 – 8y + 9.

(vi) -3y2 + 10y-16; 7y2 + 8

ANS:

Here we are adding -3y2 + 10y-16 and 7y2 + 8 these two algebraic expressions.

We adding y2 variable terms with y2 variable and constant term with constant term.

We writing 2nd expression as 7y2 + 0y + 8

When there is one negative sign and one positive sign and addition operation is there we have to subtract smaller number from larger number and give sign of larger number.

-3y2 + 10y-16

+  7y2 + 0y + 8

————————————

4y2+ 10y – 8

Addition of -3y2 + 10y-16; 7y2 + 8 is 4y2+ 10y – 8

Practice Set 34

¤ Subtract the second expression from the first.

 (i) (4xy – 9z) ; (3xy – 16z)

ANS:

Here we are subtracting (3xy – 16z) from (4xy – 9z)

We subtract xy term from xy term and z term from z term.

(4xy – 9z)

– (3xy – 16z)

———————

When negative sign is in front of any bracket sign changes of the terms inside the bracket.

– (3xy – 16z) becomes – 3xy + 16z

When there is one negative sign and one positive sign and subtraction operation is there we have to subtract smaller number from larger number and give sign of larger number.

4xy – 9z

 –  3xy + 16z

——————————

xy + 7z.

Subtraction of (4xy – 9z) ; (3xy – 16z) is xy  + 7z.

 (ii) (5x + 4y + 7z) ; (x + 2y + 3z)

ANS:

Here we are subtracting (x + 2y + 3z) from (5x + 4y + 7z).

We subtract x term from x term, y term from y term and z term from z term.

(5x + 4y + 7z)

-(x + 2y + 3z)

———————————-

When negative sign is in front of any bracket sign changes of the terms inside the bracket.

-(x + 2y + 3z) becomes – x – 2y – 3z.

5x + 4y + 7z

– x – 2y –  3z

—————————————-

4x + 2y + 4z

Subtraction of (5x + 4y + 7z) ; (x + 2y + 3z) is 4x + 2y + 4z.

 (iii) (14x2 + 8xy + 3y2 ) ; (26x2 – 8xy – 17y2 )

ANS:

Here we are subtracting (26x2 – 8xy – 17y2 ) from (14x2 + 8xy + 3y2 )

We subtract x2 term from x2 term, xy term from xy term and y2  term from y2 term.

(14x2 + 8xy + 3y2 )

-(26x2 – 8xy – 17y2 )

—————————————

When negative sign is in front of any bracket sign changes of the terms inside the bracket

-(26x2 – 8xy – 17y2 ) becomes -26x2 + 8xy + 17y2

 

14x2 + 8xy + 3y2

-26x2 + 8xy + 17y2

————————————————————————

-12 x2 +16xy + 20 y2

 Subtraction of (14x2 + 8xy + 3y2) ; (26x2 – 8xy – 17y2 ) is -12 x2 +16xy + 20 y2

(iv) (6x2 + 7xy + 16y2) ;( 16x2 – 17xy)

ANS:

Here we are subtracting ( 16x2 – 17xy) from (6x2 + 7xy + 16y2)

We subtract x2 term from x2 term, xy term from xy term and y2  term from y2 term.

We rearranging 2nd expression ( 16x2 – 17xy + 0y2)

(6x2 + 7xy + 16y2)

-( 16x2 – 17xy + 0y2)

———————————-

When negative sign is in front of any bracket sign changes of the terms inside the bracket

-( 16x2 – 17xy + 0y2) becomes -16x2 + 17xy – 0y2

6x2 + 7xy + 16y2

-16x2 + 17xy – 0y2

—————————————————————

-10x2 + 24xy + 16y2

Subtraction of (6x2 + 7xy + 16y2) ;( 16x2 – 17xy) is -10x2 + 24xy + 16y2

(v) (4x + 16z) ; (19y – 14z + 16x)

ANS:

Here we are subtracting (19y – 14z + 16x) from (4x + 16z).

We subtract x term from x term, y term from y term and z term from z term.

We rearranging 1st expression (4x + oy + 16z)

Also we rearranging 2nd expression (19y – 14z + 16x) as 16x + 19y – 14z

 

(4x + oy + 16z)

-(16x + 19y – 14z)

——————————-

When negative sign is in front of any bracket sign changes of the terms inside the bracket

-(16x + 19y – 14z) becomes  -16x – 19y + 14z

4x + oy + 16z

-16x – 19y + 14z

——————————-

-12x – 19y + 30z

Subtraction of (4x + 16z) ; (19y – 14z + 16x) is -12x – 19y + 30z

Practice Set 35

1.) Multiply.

 (i) 16xy × 18xy

ANS:

Here we have to multiply 16xy × 18xy.

When there is multiplication of monomial with monomial we have to first do multiplication of coefficients of expressions and then we have to do multiplication of variables.

Coefficient of 1st monomial is 16.

Coefficient of 2st monomial is 18.

Variable of 1st monomial is xy.

Variable of 2nd monomial is xy.

16xy × 18xy

16 x 18 x (xy) x (xy)

= 288x2y2

Multiplication of 16xy × 18xy is 288x2y2

 (ii) 23xy2 × 4yz2

ANS:

Here we have to multiply 23xy2 × 4yz2

When there is multiplication of monomial with monomial we have to first do multiplication of coefficients of expressions and then we have to do multiplication of variables.

Coefficient of 1st monomial is 23.

Coefficient of 2st monomial is 4.

Variable of 1st monomial is xy2.

Variable of 2nd monomial is yz2.

23xy2 × 4yz2

= 23 x 4 x (xy2) x (yz2)

= 92xy3 z2

Multiplication of 23xy2 × 4yz2 is 92xy3 z2

(iii) (12a + 17b) × 4c

ANS:

Here we have to multiply (12a + 17b) × 4c

When there is multiplication of binomial with monomial we have to do multiplication of monomial and each term of binomial expressions.

(12a + 17b) × 4c

= (4c x 12a) + ( 4c x 17b)

  = (48ac + 68bc)

Multiplication of (12a + 17b) × 4c is (48ac + 68bc).

 (iv) (4x + 5y) × (9x + 7y)

ANS:

Here we have to multiply (4x + 5y) × (9x + 7y)

When there is multiplication of binomial with binomial we have to do multiplication of each term of first binomial with the 2nd binomial expressions.

(4x + 5y) × (9x + 7y)

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

= 36x2 + 28xy + 45 xy + 35y2

= 36x2 + 73 x2y2 + 35y2

Multiplication of (4x + 5y) × (9x + 7y) is 36x2 + 73 x2y2 + 35y2.

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

ANS:

Given that

Length of rectangle is (8x + 5) cm

Breadth of rectangle is (5x + 3) cm

We know,

 The area of rectangle = Length x Breadth

The area of rectangle = (8x + 5) cm x (5x + 3) cm

When there is multiplication of binomial with binomial we have to do multiplication of each term of first binomial with the 2nd binomial expressions.

(8x + 5)  x (5x + 3)

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

= 40x2+ 24x + 25x + 15

= 40x2+ 49x + 15

The area of rectangle is 40x2+ 49x + 15

Practice Set 36

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

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

ANS:

Here we are subtracting (17x + 13y) from (3x-11y)

We subtract x term from x term and y term from y term.

(3x-11y)

– (17x + 13y)

———————

When negative sign is in front of any bracket sign changes of the terms inside the bracket.

– (17x + 13y) becomes -17x – 13y

 

3x-11y

-17x – 13y

——————————

-14x – 24y

We take -2 common we get,

-2 (7x + 12y)

Simplification of (3x – 11y) – (17x + 13y)  is -2 (7x + 12y)

(iv) -2 (7x + 12y)

 2.) The product of (23 x2 y3 z) and (-15x3yz2) is……………….. .

(i) -345 x5 y4 z3 (ii) 345 x2 y3 z5 (iii) 145 x3 y2 z (iv) 170 x3 y2 z3

ANS:

Here we have to multiply (23 x2 y3 z) and (-15x3yz2).

When there is multiplication of monomial with monomial we have to first do multiplication of coefficients of expressions and then we have to do multiplication of variables.

Coefficient of 1st monomial is 23.

Coefficient of 2st monomial is – 15.

Variable of 1st monomial is x2 y3 z.

Variable of 2nd monomial is x3 y z2.

16xy × 18xy

23 x – 15 x (x2 y3 z) x (x3 y z2)

= -345 x5 y4 z3

The product of (23 x2 y3 z) and (-15x3yz2) is-345 x5 y4 z3

(i) -345 x5 y4 z3

3.) Solve the following equations.

(i) 4x + 1/ 2 = 9/ 2

ANS:

We have to solve 4x + 1/ 2 = 9/ 2 this equation.

4x + 1/ 2 = 9/ 2

4x = 9/ 2 – 1/ 2

When denominator is same we doing subtraction of numerator only.

4x = 8 / 2

4x = 4

X = 4/ 4

X = 1

Value of x is 1.

 

(ii) 10 = 2y + 5

ANS:

We have to solve 10 = 2y + 5this equation.

10 = 2y + 5

2y = 10 – 5

2y = 5

y = 5/2

Value of y is 5/2.

 

(iii) 5m – 4 = 1

ANS:

We have to solve 5m – 4 = 1this equation.

5m – 4 = 1

5m = 1 + 4

5m = 5

m = 5 / 5

m = 1

Value of m is 1.

 

(iv) 6x-1 = 3x + 8

ANS:

We have to solve6x-1 = 3x + 8 this equation.

6x-1 = 3x + 8

We are taking variable term one side and constant term on other side.

When we move from one side to another and in between = sign is there then its sign changes.

6x- 3x = 8 + 1

3x = 9

X = 9/3

X= 3

Value of x is 3.

 

(v) 2 (x-4) = 4x + 2

ANS:

We have to solve 2 (x-4) = 4x + 2 this equation.

2 (x-4) = 4x + 2

Firstly we are multiplying monomial with binomial.

2 (x-4) = 2x – 8

Now,

2x – 8 = 4x + 2

We are taking variable term one side and constant term on other side.

When we move from one side to another and in between = sign is there then its sign changes.

2x – 4x = 2 + 8

-2x = 10

X = 10 / -2

X = -5

Value of x is -5.

 

(vi) 5 (x + 1) = 74

ANS:

We have to solve 5 (x + 1) = 74 this equation.

5 (x + 1) = 74

Firstly we are multiplying monomial with binomial.

5 (x + 1) = 5x + 5

Now,

5x + 5 = 74

5x = 74 – 5

5x = 69

X = 69 /5

Value of x is 69 /5.

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?

ANS:

Given that,

Rakesh’s age is less than Sania’s age by 5 years.

Let,

We are taking Sania’s age is x years.

Therefore

Rakesh’s age is less than Sania’s age by 5 years.

Rakesh’s age = (x – 5) years.

Now,

The sum of their ages is 27 years

X + (x – 5) = 27

2x – 5 = 27

2x = 27 + 5

2x = 32

X = 32/2

X = 16 years.

Bur x is Sania’s age

Sania’s age is 16 years.

Rakesh’s age is less than Sania’s age by 5 years.

Rakesh’s age = 16 – 5 = 11 years.

Rakesh’s age =11 years.

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?

ANS:

The number of jambhul trees planted was greater than the number of ashoka trees by 60.

Let,

The number of ashoka trees planted is x.

Therefore

The number of jambhul trees planted = x + 60

There are altogether 200 trees

We are adding both trees.

X + (x + 60) = 200

2x + 60 = 200

2x = 200 – 60

2x = 140

X = 140 / 2

X = 70

But,

X is the number of ashoka trees planted.

The number of ashoka trees planted is 70.

The number of jambhul trees planted was greater than the number of ashoka trees by 60.

= 70 + 60

= 130.

The number of jambhul trees planted= 130.

6 .) 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?

ANS:

Given that,

Virat made twice as many runs as Rohit.

Let,

Rohit score x runs.

Therefore

Viratscore 2x runs.

The total of their scores is 2 less than a double century.

Addition of their scores is 198.

X + 2x = 198

3x = 198

X = 198 /3

X = 66

But,

Rohit score x runs.

Rohit score 66 runs.

Virat made twice as many runs as Rohit.

Virat score 2x runs.

Virat score 2x 66 runs.

Virat score 132 runs.

Updated: July 29, 2021 — 10:43 pm

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