Study MaterialsCBSE NotesAlgebraic Expressions Class 7 Extra Questions Maths Chapter 12

Algebraic Expressions Class 7 Extra Questions Maths Chapter 12

Algebraic Expressions Class 7 Extra Questions Maths Chapter 12

Extra Questions for Class 7 Maths Chapter 12 Algebraic Expressions

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    Algebraic Expressions Class 7 Extra Questions Very Short Answer Type

    Question 1.
    Identify in the given expressions, terms which are not constants. Give their numerical coefficients.
    (i) 5x – 3
    (ii) 11 – 2y2
    (iii) 2x – 1
    (iv) 4x2y + 3xy2 – 5
    Solution:
    Algebraic Expressions Class 7 Extra Questions Maths Chapter 12 Q1

    Question 2.
    Group the like terms together from the following expressions:
    -8x2y, 3x, 4y, \(\frac { -3 }{ 2 }\)x , 2x2y, -y
    Solution:
    Group of like terms are:
    (i) -8x2y, 2x2y
    (ii) 3x, \(\frac { -3 }{ 2 }\)x
    (iii) 4y, -y

    Question 3.
    Identify the pairs of like and unlike terms:
    (i) \(\frac { -3 }{ 2 }\)x, y
    (ii) -x, 3x
    (iii) \(\frac { -1 }{ 2 }\)y2x, \(\frac { 3 }{ 2 }\)xy2
    (iv) 1000, -2
    Solution:
    (i) \(\frac { -3 }{ 2 }\)x, y → Unlike Terms
    (ii) -x, 3x → Like Terms
    (iii) \(\frac { -1 }{ 2 }\)y2x, \(\frac { 3 }{ 2 }\)xy2 → Like Terms
    (iv) 1000, -2 → Like Terms

    Question 4.
    Classify the following into monomials, binomial and trinomials.
    (i) -6
    (ii) -5 + x
    (iii) \(\frac { 3 }{ 2 }\)x – y
    (iv) 6x2 + 5x – 3
    (v) z2 + 2
    Solution:
    (i) -6 is monomial
    (ii) -5 + x is binomial
    (iii) \(\frac { 3 }{ 2 }\)x – y is binomial
    (iv) 6x2 + 5x – 3 is trinomial
    (v) z2 + z is binomial

    Question 5.
    Draw the tree diagram for the given expressions:
    (i) -3xy + 10
    (ii) x2 + y2
    Solution:
    Algebraic Expressions Class 7 Extra Questions Maths Chapter 12 Q5

    Question 6.
    Identify the constant terms in the following expressions:
    (i) -3 + \(\frac { 3 }{ 2 }\)x
    (ii) \(\frac { 3 }{ 2 }\) – 5y + y2
    (iii) 3x2 + 2y – 1
    Solution:
    (i) Constant term = -3
    (ii) Constant term = \(\frac { 3 }{ 2 }\)
    (iii) Constant term = -1

    Question 7.
    Add:
    (i) 3x2y, -5x2y, -x2y
    (ii) a + b – 3, b + 2a – 1
    Solution:
    (i) 3x2y, -5x2y, -x2y
    = 3x2y + (-5x2y) + (-x2y)
    = 3x2y – 5x2y – x2y
    = (3 – 5 – 1 )x2y
    = -3x2y
    (ii) a + b – 3, b + 2a – 1
    = (a + b – 3) + (b + 2a – 1)
    = a + b – 3 + b + 2a – 1
    = a + 2a + b + b – 3 – 1
    = 3a + 2b – 4

    Question 8.
    Subtract 3x2 – x from 5x – x2.
    Solution:
    (5x – x2) – (3x2 – x)
    = 5x – x2 – 3x2 + x
    = 5x + x – x2 – 3x2
    = 6x – 4x2

    Question 9.
    Simplify combining the like terms:
    (i) a – (a – b) – b – (b – a)
    (ii) x2 – 3x + y2 – x – 2y2
    Solution:
    (i) a – (a – b) – b – (b – a)
    = a – a + b – b – b + a
    = (a – a + a) + (b – b – b)
    = a – b
    (ii) x2 – 3x + y2 – x – 2y2
    = x2 + y2 – 2y2 – 3x – x
    = x2 – y2 – 4x

    Algebraic Expressions Class 7 Extra Questions Short Answer Type

    Question 10.
    Subtract 24xy – 10y – 18x from 30xy + 12y – 14x.
    Solution:
    (30xy + 12y – 14x) – (24xy – 10y – 18x)
    = 30xy + 12y – 14x – 24xy + 10y + 18x
    = 30xy – 24xy + 12y + 10y – 14x + 18x
    = 6xy + 22y + 4x

    Question 11.
    From the sum of 2x2 + 3xy – 5 and 7 + 2xy – x2 subtract 3xy + x2 – 2.
    Solution:
    Sum of the given term is (2x2 + 3xy – 5) + (7 + 2xy – x2)
    = 2x2 + 3xy – 5 + 7 + 2xy – x2
    = 2x2 – x2 + 3xy + 2xy – 5 + 7
    = x2 + 5xy + 2
    Now (x2 + 5xy + 2) – (3xy + x2 – 2)
    = x2 + 5xy + 2 – 3xy – x2 + 2
    = x2 – x2 + 5xy – 3xy + 2 + 2
    = 0 + 2xy + 4
    = 2xy + 4

    Question 12.
    Subtract 3x2 – 5y – 2 from 5y – 3x2 + xy and find the value of the result if x = 2, y = -1.
    Solution:
    (5y – 3x2 + xy) – (3x2 – 5y – 2)
    = 5y – 3x2 + xy – 3x2 + 5y + 2
    = -3x2 – 3x2 + 5y + 5y + xy + 2
    = -6x2 + 10y + xy + 2
    Putting x = 2 and y = -1, we get
    -6(2)2 + 10(-1) + (2)(-1) + 2
    = -6 × 4 – 10 – 2 + 2
    = -24 – 10 – 2 + 2
    = -34

    Question 13.
    Simplify the following expressions and then find the numerical values for x = -2.
    (i) 3(2x – 4) + x2 + 5
    (ii) -2(-3x + 5) – 2(x + 4)
    Solution:
    (i) 3(2x – 4) + x2 + 5
    = 6x – 12 + x2 + 5
    = x2 + 6x – 7
    Putting x = -2, we get
    = (-2)2 + 6(-2) – 7
    = 4 – 12 – 7
    = 4 – 19
    = -15
    (ii) -2(-3x + 5) – 2(x + 4)
    = 6x – 10 – 2x – 8
    = 6x – 2x – 10 – 8
    = 4x – 18
    Putting x = -2, we get
    = 4(-2) – 18
    = -8 – 18
    = -26

    Question 14.
    Find the value of t if the value of 3x2 + 5x – 2t equals to 8, when x = -1.
    Solution:
    3x2 + 5x – 2t = 8 at x = -1
    ⇒ 3(-1)2 + 5(-1) – 2t = 8
    ⇒ 3(1) – 5 – 2t = 8
    ⇒ 3 – 5 – 2t = 8
    ⇒ -2 – 2t = 8
    ⇒ 2t = 8 + 2
    ⇒ -2t = 10
    ⇒ t = -5
    Hence, the required value of t = -5.

    Question 15.
    Subtract the sum of -3x3y2 + 2x2y3 and -3x2y3 – 5y4 from x4 + x3y2 + x2y3 + y4.
    Solution:
    Sum of the given terms:
    Algebraic Expressions Class 7 Extra Questions Maths Chapter 12 Q15
    Required expression

    Question 16.
    What should be subtracted from 2x3 – 3x2y + 2xy2 + 3y2 to get x3 – 2x2y + 3xy2 + 4y2? [NCERT Exemplar] Solution:
    We have
    Algebraic Expressions Class 7 Extra Questions Maths Chapter 12 Q16
    Required expression

    Question 17.
    To what expression must 99x3 – 33x2 – 13x – 41 be added to make the sum zero? [NCERT Exemplar] Solution:
    Given expression:
    99x3 – 33x2 – 13x – 41
    Negative of the above expression is
    -99x3 + 33x2 + 13x + 41
    (99x3 – 33x2 – 13x – 41) + (-99x3 + 33x2 + 13x + 41)
    = 99x3 – 33x2 – 13x – 41 – 99x3 + 33x2 + 13x + 41
    = 0
    Hence, the required expression is -99x3 + 33x2 + 13x + 41

    Algebraic Expressions Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

    Question 18.
    If P = 2x2 – 5x + 2, Q = 5x2 + 6x – 3 and R = 3x2 – x – 1. Find the value of 2P – Q + 3R.
    Solution:
    2P – Q + 3R = 2(2x2 – 5x + 2) – (5x2 + 6x – 3) + 3(3x2 – x – 1)
    = 4x2 – 10x + 4 – 5x2 – 6x + 3 + 9x2 – 3x – 3
    = 4x2 – 5x2 + 9x2 – 10x – 6x – 3x + 4 + 3 – 3
    = 8x2 – 19x + 4
    Required expression.

    Question 19.
    If A = -(2x + 3), B = -3(x – 2) and C = -2x + 7. Find the value of k if (A + B + C) = kx.
    Solution:
    A + B + C = -(2x + 13) – 3(x – 2) + (-2x + 7)
    = -2x – 13 – 3x + 6 – 2x + 7
    = -2x – 3x – 2x – 13 + 6 + 7
    = -7x
    Since A + B + C = kx
    -7x = kx
    Thus, k = -7

    Question 20.
    Find the perimeter of the given figure ABCDEF.
    Algebraic Expressions Class 7 Extra Questions Maths Chapter 12 Q20
    Solution:
    Required perimeter of the figure
    ABCDEF = AB + BC + CD + DE + EF + FA
    = (3x – 2y) + (x + 2y) + (x + 2y) + (3x – 2y) + (x + 2y) + (x + 2y)
    = 2(3x – 2y) + 4(x + 2y)
    = 6x – 4y + 4x + 8y
    = 6x + 4x-4y + 8y
    = 10x + 4y
    Required expression.

    Question 21.
    Rohan’s mother gave him ₹ 3xy2 and his father gave him ₹ 5(xy2 + 2). Out of this total money he spent ₹ (10 – 3xy2) on his birthday party. How much money is left with him? [NCERT Exemplar] Solution:
    Money give by Rohan’s mother = ₹ 3xy2
    Money given by his father = ₹ 5(xy2 + 2)
    Total money given to him = ₹ 3xy2 + ₹ 5 (xy2 + 2)
    = ₹ [3xy2 + 5(xy2 + 2)] = ₹ (3xy2 + 5xy2 + 10)
    = ₹ (8xy2 + 10).
    Money spent by him = ₹ (10 – 3xy)2
    Money left with him = ₹ (8xy2 + 10) – ₹ (10 – 3xy2)
    = ₹ (8xy2 + 10 – 10 + 3x2y)
    = ₹ (11xy2)
    Hence, the required money = ₹ 11xy2

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