Andhra Pradesh SCERT Class 7 Maths Algebraic Expressions Question and Answers Solutions
Andhra Pradesh SCERT 7th Class Maths Solutions Chapter 10 Algebraic Expressions Question and answers. Students who are searching for Andhra Pradesh Class 7 Maths Chapter 10 can find here Solution of this chapter.
Board |
Andhra Pradesh (AP Board) |
Class |
7th |
Subject |
Maths |
Topic |
Solution |
EXERCISE 10.1
1.) Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.
(i) Subtraction of z from y.
ANSWER:
From the given statement, we have to make algebraic expression.
Algebraic expression is
Subtraction of z from y = y – z
(ii) One-half of the sum of numbers x and y.
ANSWER:
From the given statement, we have to make algebraic expression.
Algebraic expression is
One-half of the sum of numbers x and y = 1/2 (x + y)
(iii) The number z multiplied by itself.
ANSWER:
From the given statement, we have to make algebraic expression.
Algebraic expression is
The number z multiplied by itself. =z^{2}
(iv) One-fourth of the product of numbers p and q.
ANSWER:
From the given statement, we have to make algebraic expression.
Algebraic expression is
One-fourth of the product of numbers p and q = 1/4 pq
(v) Numbers x and y both squared and added.
ANSWER:
From the given statement, we have to make algebraic expression.
Algebraic expression is
Numbers x and y both squared and added = x ^{2} + y ^{2}
(vi) Number 5 added to three times the product of numbers m and n.
ANSWER:
From the given statement, we have to make algebraic expression.
Algebraic expression is
Number 5 added to three times the product of numbers m and n =5 + 3mn
(vii) Product of numbers y and z subtracted from 10.
ANSWER:
From the given statement, we have to make algebraic expression.
Algebraic expression is
Product of numbers y and z subtracted from 10 =10 – yz
(viii) Sum of numbers a and b subtracted from their product.
ANSWER:
From the given statement, we have to make algebraic expression.
Algebraic expression is
Sum of numbers a and b subtracted from their product =ab – (a + b)
2.) (i) Identify the terms and their factors in the following expressions Show the terms and factors by tree diagrams.
(a) x – 3
ANSWER:
Here we have to show the terms and factors by tree diagrams.
(b) 1 + x + x^{2}
ANSWER:
Here we have to show the terms and factors by tree diagrams.
(c) y – y^{3}
ANSWER:
Here we have to show the terms and factors by tree diagrams.
(d) 5xy^{2} + 7x^{2}y
ANSWER:
Here we have to show the terms and factors by tree diagrams.
(e) – ab + 2b ^{2} – 3a^{2}
ANSWER:
Here we have to show the terms and factors by tree diagrams.
(ii) Identify terms and factors in the expressions given below:
(a) – 4x + 5
ANSWER:
Here we have to show the terms and factors of given expression.
– 4x + 5
Term =– 4x, 5
Factor = – 4, x, 5
(b) – 4x + 5y
ANSWER:
Here we have to show the terms and factors of given expression.
– 4x + 5y
Term = – 4x, 5y
Factor = – 4, x,5, y
(c) 5y + 3y^{2}
ANSWER:
Here we have to show the terms and factors of given expression.
5y + 3y^{2}
Term = 5y,3y^{2}
Factor = 5, y, 3, y, y
(d) xy + 2x^{2}y^{2}
ANSWER:
Here we have to show the terms and factors of given expression.
xy + 2x^{2}y^{2}
Term = xy, 2x^{2}y^{2}
Factor = x, y, 2, x, x, y, y
(e) pq + q
ANSWER:
Here we have to show the terms and factors of given expression.
pq + q
Term = pq, q
Factor = p, q, q
(f) 1.2 ab – 2.4 b + 3.6 a
ANSWER:
Here we have to show the terms and factors of given expression.
1.2 ab – 2.4 b + 3.6 a
Term = 1.2 ab, -2.4 b, 3.6 a
Factor = 1.2, a, b, – 2.4, b, 3.6, a
(g) 3x /4 + 1/4
ANSWER:
Here we have to show the terms and factors of given expression.
3x /4 + 1/4
Term = 3x /4, 1/4
Factor = 3/4, x,1/4
(h) 0.1 p^{2} + 0.2q^{2}
ANSWER:
Here we have to show the terms and factors of given expression.
0.1 p^{2} + 0.2q^{2}
Term = 0.1 p^{2}, 0.2q^{2}
Factor = 0.1, p, p, 0.2, q, q
3.) Identify the numerical coefficients of terms (other than constants) in the following expressions:
(i) 5 – 3t^{2}
ANSWER:
We have to find the numerical coefficients of terms (other than constants) in the given expression.
5 – 3t^{2}
Terms = – 3t^{2}
Coefficients = – 3
(ii) 1 + t + t^{2} + t^{3}
ANSWER:
We have to find the numerical coefficients of terms (other than constants) in the given expression.
1 + t + t^{2} + t^{3}
Terms = t, t^{2}, t^{3}
Coefficients =1, 1, 1
(iii) x + 2xy + 3y
ANSWER:
We have to find the numerical coefficients of terms (other than constants) in the given expression.
x + 2xy + 3y
Terms = x, 2xy, 3y
Coefficients =1, 2, 3
(iv) 100m + 1000n
ANSWER:
We have to find the numerical coefficients of terms (other than constants) in the given expression.
100m + 1000n
Terms = 100m, 1000n
Coefficients =100, 1000
(v) – p^{2}q^{2} + 7pq
ANSWER:
We have to find the numerical coefficients of terms (other than constants) in the given expression.
– p^{2}q^{2} + 7pq
Terms = – p^{2}q^{2}, 7pq
Coefficients = -1,7
(vi) 1.2 a + 0.8 b
ANSWER:
We have to find the numerical coefficients of terms (other than constants) in the given expression.
1.2 a + 0.8 b
Terms = 1.2 a, 0.8 b
Coefficients = 1.2, 0.8
(vii) 3.14r^{2}
ANSWER:
We have to find the numerical coefficients of terms (other than constants) in the given expression.
3.14r^{2}
Terms = 3.14r^{2}
Coefficients = 3.14
(viii) 2(l + b)
ANSWER:
We have to find the numerical coefficients of terms (other than constants) in the given expression.
2(l + b)
Terms = 2l, 2b
Coefficients = 2, 2
(ix) 0.1 y + 0.01y^{2}
ANSWER:
We have to find the numerical coefficients of terms (other than constants) in the given expression.
0.1 y + 0.01y^{2}
Terms = 0.1 y, 0.01y^{2}
Coefficients = 0.1, 0.01
^{ }
4.) (a) Identify terms which contain x and give the coefficient of x.
(i) y^{2}x + y
ANSWER:
We have to find terms which contain x and the coefficient of x.
y^{2}x + y
Terms which contain x = y^{2}x
Coefficient of x = y^{2}
(ii) 13y^{2} – 8yx
ANSWER:
We have to find terms which contain x and the coefficient of x.
13y^{2} – 8yx
Terms which contain x = – 8yx
Coefficient of x = – 8y
(iii) x + y + 2
ANSWER:
We have to find terms which contain x and the coefficient of x.
x + y + 2
Terms which contain x = x
Coefficient of x = 1
(iv) 5 + z + zx
ANSWER:
We have to find terms which contain x and the coefficient of x.
5 + z + zx
Terms which contain x = zx
Coefficient of x = z
(v) 1 + x + xy
ANSWER:
We have to find terms which contain x and the coefficient of x.
1 + x + xy
Terms which contain x = x, xy
Coefficient of x = 1, y
(vi) 12xy^{2} + 25
ANSWER:
We have to find terms which contain x and the coefficient of x.
12xy^{2} + 25
Terms which contain x = 12xy^{2}
Coefficient of x = 12y^{2}
(vii) 7x + xy^{2}
ANSWER:
We have to find terms which contain x and the coefficient of x.
7x + xy^{2}
Terms which contain x = 7x, xy^{2}
Coefficient of x = 7, y^{2}
(b) Identify terms which contain y^{2} and give the coefficient of y^{2}.
(i) 8 – xy^{2}
ANSWER:
We have to find terms which contain y^{2} and the coefficient of y^{2}.
8 – xy^{2}
Terms which contain y^{2} = – xy^{2}
Coefficient of y^{2 }= -x
(ii) 5y^{2} + 7x
ANSWER:
We have to find terms which contain y^{2} and the coefficient of y^{2}.
5y^{2} + 7x
Terms which contain y^{2} = 5y^{2}
Coefficient of y^{2 }= 5
(iii) 2x^{2}y – 15xy^{2} + 7y^{2}
ANSWER:
We have to find terms which contain y^{2} and the coefficient of y^{2}.
2x^{2}y – 15xy^{2} + 7y^{2}
Terms which contain y^{2} = – 15xy^{2}, 7y^{2}
Coefficient of y^{2 }= – 15x, 7
5.) Classify into monomials, binomials and trinomials.
(i) 4y – 7z
(ii) y^{2}
(iii) x + y – xy
(iv) 100
(v) ab – a – b
(vi) 5 – 3t
(vii) 4p 2q – 4pq^{2}
(viii) 7mn
(ix) z^{2} – 3z + 8
(x) a^{2} + b^{2}
(xi) z^{2} + z
(xii) 1 + x + x^{2}
ANSWER:
We have to classify given algebraic expressions as monomials, binomials and trinomials.
We know,
An expression with only one term is called a monomial.
Monomials are
(ii) y^{2}
(iv) 100
(viii) 7mn
Now,
An expression which contains two unlike terms is called a binomial.
Binomials are,
(i) 4y – 7z
(vi) 5 – 3t
(vii) 4p 2q – 4pq^{2}
(x) a^{2} + b^{2}
(xi) z^{2} + z
Now,
An expression which contains three terms is called a trinomial.
Trinomials are
(iii) x + y – xy
(v) ab – a – b
(ix) z^{2} – 3z + 8
(xii) 1 + x + x^{2}
6.) State whether a given pair of terms is of like or unlike terms.
(i) 1, 100
ANSWER:
We have to state that given pair of terms is of like or unlike terms.
We know,
When terms have the same algebraic factors, they are like terms.
When terms have different algebraic factors, they are unlike terms.
1, 100 is pair of like term.
(ii) –7x, 5x /2
ANSWER:
We have to state that given pair of terms is of like or unlike terms.
We know,
When terms have the same algebraic factors, they are like terms.
When terms have different algebraic factors, they are unlike terms.
–7x, 5x /2 is pair oflike term.
(iii) – 29x, – 29y
ANSWER:
We have to state that given pair of terms is of like or unlike terms.
We know,
When terms have the same algebraic factors, they are like terms.
When terms have different algebraic factors, they are unlike terms.
– 29x, – 29yis pair of unlike terms.
(iv) 14xy, 42yx
ANSWER:
We have to state that given pair of terms is of like or unlike terms.
We know,
When terms have the same algebraic factors, they are like terms.
When terms have different algebraic factors, they are unlike terms.
14xy, 42yx is pair oflike term.
(v) 4m^{2}p, 4mp^{2}
ANSWER:
We have to state that given pair of terms is of like or unlike terms.
We know,
When terms have the same algebraic factors, they are like terms.
When terms have different algebraic factors, they are unlike terms.
4m^{2}p, 4mp^{2}is pair of unlike terms.
(vi) 12xz, 12x^{2}z
ANSWER:
We have to state that given pair of terms is of like or unlike terms.
We know,
When terms have the same algebraic factors, they are like terms.
When terms have different algebraic factors, they are unlike terms.
12xz, 12x^{2}z is pair of unlike terms.
7.) Identify like terms in the following:
(a) – xy^{2} , – 4yx^{2} , 8x^{2 }, 2xy^{2} , 7y, – 11x^{2} , – 100x, – 11yx, 20x^{2}y, – 6x^{2} , y, 2xy, 3x
ANSWER:
From above expressions we have to Identify like terms.
We know,
When terms have the same algebraic factors, they are like terms.
Like terms are,
– xy^{2}, 2xy^{2}
– 4yx^{2}, 20x^{2}y
8x^{2}, –11x^{2}, – 6x^{2}
7y, y
– 100x, 3x
– 11yx,2xy.
(b) 10pq, 7p, 8q, – p^{2}q^{2}, – 7qp, – 100q, – 23, 12q^{2}p^{2}, – 5p^{2}, 41, 2405p, 78qp, 13p^{2}q, qp^{2}, 701p^{2}
ANSWER:
From above expressions we have to Identify like terms.
We know,
When terms have the same algebraic factors, they are like terms.
Like terms are,
10pq, –7qp, 78qp
7p, 2405p
8q, – 100q
–p^{2}q^{2}, 12q^{2}p^{2}
–23, 41
–5p^{2} , 701p^{2}
13p^{2}q, qp^{2}
EXERCISE 10.2
1.) If m = 2, find the value of:
(i) m – 2
ANSWER:
We have to find value of m – 2 when m = 2
We put m = 2
m – 2 = 2 – 2 = 0
(ii) 3m – 5
ANSWER:
We have to find value of 3m – 5 when m = 2
We put m = 2
3m – 5 = 3x 2 – 5 = 1
(iii) 9 – 5m
ANSWER:
We have to find value of 9 – 5m when m = 2
We put m = 2
9 – 5m = 9 – 5 x 2 = -1
(iv) 3m^{2} – 2m – 7
ANSWER:
We have to find value of 3m^{2} – 2m – 7 when m = 2
We put m = 2
3m^{2} – 2m – 7 = 3x (2)^{2} – 2x 2 – 7
3m^{2} – 2m – 7 = 12 -4 -7
3m^{2} – 2m – 7 = 1
(v) 5 m /2 4
ANSWER:
We have to find value of 5 m /2 – 4 when m = 2
We put m = 2
5 m /2 – 4 = 5 x 2 /2 – 4
5 m /2 – 4 = 1
2.) If p = – 2, find the value of:
(i) 4p + 7
ANSWER:
We have to find value of 4p + 7 when p = -2
We put p = -2
4p + 7 = 4 x -2 + 7
4p + 7 = -1
(ii) – 3p^{2} + 4p + 7
ANSWER:
We have to find value of – 3p^{2} + 4p + 7 when p = -2
We put p = -2
– 3p^{2} + 4p + 7 = – 3(-2) ^{2} + 4(-2) + 7
– 3p^{2} + 4p + 7 = -12 -8 + 7
– 3p2 + 4p + 7 = -13
(iii) – 2p^{3} – 3p^{2} + 4p + 7
ANSWER:
We have to find value of – 2p^{3} – 3p^{2} + 4p + 7 when p = -2
We put p = -2
– 2p^{3} – 3p^{2} + 4p + 7 = – 2(-2)^{3} – 3(-2)^{2} + 4(-2) + 7
– 2p^{3} – 3p^{2} + 4p + 7 = 16 – 12 – 8 + 7
– 2p^{3} – 3p^{2} + 4p + 7 = 3
3.) Find the value of the following expressions, when x = –1:
(i) 2x – 7
ANSWER:
We have to find value of 2x – 7 when x = -1
We put x = -1
2x – 7 = 2(-1) – 7
2x – 7 = -2 -7 = -9
(ii) – x + 2
ANSWER:
We have to find value of – x + 2 when x = -1
We put x = -1
– x + 2 = – (-1) + 2
– x + 2 = 1 + 2 = 3
(iii) x^{2} + 2x + 1
ANSWER:
We have to find value of x^{2} + 2x + 1 when x = -1
We put x = -1
x^{2} + 2x + 1 = (-1)^{2} + 2(-1) + 1
x^{2} + 2x + 1 = 1 -2 + 1
x^{2} + 2x + 1 = 0
(iv) 2x^{2} – x – 2
ANSWER:
We have to find value of x^{2} + 2x + 1 when x = -1
We put x = -1
2x^{2} – x – 2 = 2(-1)^{2} – (-1) – 2
2x^{2} – x – 2 = 2 + 1 – 2
2x^{2} – x – 2 = 1
4.) If a = 2, b = – 2, find the value of:
(i) a^{2} + b^{2}
ANSWER:
We have to find value of a^{2} + b^{2} when a = 2, b = – 2
We put a = 2, b = – 2
a^{2} + b^{2 } = 2^{2} + (-2)^{2}
a^{2} + b^{2 } = 4 + 4
a^{2} + b^{2 } = 8
(ii) a^{2} + ab + b^{2}
ANSWER:
We have to find value of a^{2} + ab + b^{2 }when a = 2, b = – 2
We put a = 2, b = – 2
a^{2} + ab + b^{2 }= 2^{2} + 2 x -2 + (-2)^{2}
a^{2} + ab + b^{2} = 4 – 4 + 4
a^{2} + ab + b^{2} = 4
(iii) a^{2} – b^{2}
ANSWER:
We have to find value of a^{2} – b^{2} when a = 2, b = – 2
We put a = 2, b = – 2
a^{2} – b^{2} = 2^{2} – (-2)^{2}
a^{2} – b^{2 }= 4 – 4
a^{2} – b^{2} = 0
5.) When a = 0, b = – 1, find the value of the given expressions:
(i) 2a + 2b
ANSWER:
We have to find value of 2a + 2b when a = 0, b = – 1
We put a = 0, b = – 1
2a + 2b = 2(0) + 2(-1)
2a + 2b = -2
(ii) 2a^{2} + b^{2} + 1
ANSWER:
We have to find value of2a^{2} + b^{2} + 1 when a = 0, b = – 1
We put a = 0, b = – 1
2a^{2} + b^{2} + 1 = 2(0)^{2} + (-1)^{2} + 1
2a^{2} + b^{2} + 1 = 1 + 1
2a^{2} + b^{2} + 1 = 2
(iii) 2a^{2}b + 2ab^{2} + ab
ANSWER:
We have to find value of 2a^{2}b + 2ab^{2} + ab when a = 0, b = – 1
We put a = 0, b = – 1
2a^{2}b + 2ab^{2} + ab = 2(0)^{2}(-1) + 2x 0 x -1 ^{2} + 0 x -1
2a^{2}b + 2ab^{2} + ab = 0
(iv) a^{2} + ab + 2
ANSWER:
We have to find value of a^{2} + ab + 2 when a = 0, b = – 1
We put a = 0, b = – 1
a^{2} + ab + 2 = 0^{2} + 0 x -1 + 2
a^{2} + ab + 2 = 2
6.) Simplify the expressions and find the value if x is equal to 2
(i) x + 7 + 4 (x – 5)
ANSWER:
x + 7 + 4 (x – 5)
x + 7 + 4x -20
5x – 13 simplified expression.
We have to find value of x + 7 + 4 (x – 5) when x = 2
We put x = 2
x + 7 + 4 (x – 5) = 2 + 7 + 4 (2 – 5)
x + 7 + 4 (x – 5) = 9 + 8 – 20
x + 7 + 4 (x – 5) = -3
(ii) 3 (x + 2) + 5x – 7
ANSWER:
3 (x + 2) + 5x – 7
3x + 6 + 5x – 7
8x – 1simplified expression
We have to find value of 8x – 1 when x = 2
We put x = 2
8x – 1 = 8 x 2 – 1
8x – 1 = 15
(iii) 6x + 5 (x – 2)
ANSWER:
6x + 5 (x – 2)
6x + 5x – 10
11x – 10 simplified expression
We have to find value of 11x – 10when x = 2
We put x = 2
11x – 10 = 11 x 2 – 10
11x – 10 = 22 – 10
11x – 10 = 12
(iv) 4(2x – 1) + 3x + 11
ANSWER:
4(2x – 1) + 3x + 11
8x – 4 + 3x + 11
11x + 7 simplified expression
We have to find value of 11x + 7 when x = 2
We put x = 2
11x + 7 = 11 x 2 + 7
11x + 7 = 22 + 7
11x + 7 = 29
7.) Simplify these expressions and find their values if x = 3, a = – 1, b = – 2.
(i) 3x – 5 – x + 9
ANSWER:
3x – 5 – x + 9
3x – x -5 + 9
2x + 4 simplified expression.
We have to find value of 2x + 4 when x = 3
We put x = 3
2x + 4 = 2 x 3 + 4
2x + 4 = 10
(ii) 2 – 8x + 4x + 4
ANSWER:
2 – 8x + 4x + 4
– 8x + 4x + 2 + 4
-4x + 6 simplified expression.
We have to find value of -4x + 6 when x = 3
We put x = 3
-4x + 6 = -4 x 3 + 6
-4x + 6 = -12 + 6
-4x + 6 = -6
(iii) 3a + 5 – 8a + 1
ANSWER:
3a + 5 – 8a + 1
– 8a + 3a + 5 + 1
-5a + 6 simplified expression
We have to find value of -5a + 6 when a = -1
We put a = -1
-5a + 6 = -5 x (-1) + 6
-5a + 6 = 5 + 6
-5a + 6 = 11
(iv) 10 – 3b – 4 – 5b
ANSWER:
10 – 3b – 4 – 5b
10 – 4 -3b – 5b
6 – 8b simplified expression
We have to find value of 6 – 8b when b = -2
We put b = -2
6 – 8b = 6 – 8 x (-2)
6 – 8b = 6 + 16
6 – 8b = 22
(v) 2a – 2b – 4 – 5 + a
ANSWER:
2a – 2b – 4 – 5 + a
3a – 2b – 9 simplified expression
We have to find value of 3a – 2b – 9 when a= -1, b = -2
We put a= -1, b = -2
3a – 2b – 9 = 3(-1) – 2(-2) – 9
3a – 2b – 9 = -3 + 4 – 9
3a – 2b – 9 = 1 – 9
3a – 2b – 9 = -8
- (i) If z = 10, find the value of z^{3} – 3(z – 10).
ANSWER:
We have to find value of z^{3} – 3(z – 10) when z = 10
We put z = 10
z^{3} – 3(z – 10) = 10^{3} – 3(10 – 10)
z^{3} – 3(z – 10) = 1000 – 0
z^{3} – 3(z – 10) = 1000
(ii) If p = – 10, find the value of p^{2} – 2p – 100
ANSWER:
We have to find value ofp^{2} – 2p – 100whenp = – 10
We put p = – 10
p^{2} – 2p – 100 = (-10)^{2} – 2(-10) – 100
p^{2} – 2p – 100 = 100 + 20 – 100
p^{2} – 2p – 100 = 20
9.) What should be the value of a if the value of 2x^{2} + x – a equals to 5, when x = 0?
ANSWER:
We have to find value of a when 2x^{2} + x – a = 5 when x = 0
We put x = 0 in 2x^{2} + x – a = 5
2x^{2} + x – a = 5
2 x 0^{2}+ 0 – a = 5
a = -5
10.) Simplify the expression and find its value when a = 5 and b = – 3. 2(a^{2} + ab) + 3 – ab
ANSWER:
2(a^{2} + ab) + 3 – ab
2a^{2} + 2ab – ab + 3
2a^{2} + ab + 3 is simplified expression.
We have to find value of2a^{2} + ab + 3 when a = 5 and b = – 3
We put when a = 5 and b = – 3
2a^{2} + ab + 3 = 2 x 5^{2} + 5 x -3 + 3
2a^{2} + ab + 3 = 50 – 15 + 3
2a^{2} + ab + 3 = 38
Also See: Previous Chapter No. 9 Question Answer