Short and Long Question: Algebraic Expressions | Short & Long Answer Questions for Class 7 PDF Download

Q1: Group the like terms together from the following expressions:
-8x²y, 3x, 4y,  -3/2x , 2x²y, -y
Ans:
Group of like terms are:
(i) -8x²y, 2x²y
(ii) 3x, -3/2x
(iii) 4y, -y 

Q2: Identify the pairs of like and unlike terms:
(i) −3/2x, y
(ii) -x, 3x
(iii) −1/2y²x, 3/2xy²
(iv) 1000, -2
Ans:
(i) −3/2x, y → Unlike Terms
(ii) -x, 3x → Like Terms
(iii) −1/2y²x, 3/2xy² → Like Terms
(iv) 1000, -2 → Like Terms

Q3: Classify the following into monomials, binomial and trinomials.
(i) -6
(ii) -5 + x
(iii) 3/2x – y
(iv) 6x² + 5x – 3
(v) z² + 2
Ans:
(i) -6 is monomial
(ii) -5 + x is binomial
(iii) 3/2x – y is binomial
(iv) 6x² + 5x – 3 is trinomial
(v) z² + z is binomial

Q4: Identify the constant terms in the following expressions:
(i) -3 + 3/2x
(ii) 3/2 – 5y + y²
(iii) 3x² + 2y – 1
Ans:
(i) Constant term = -3
(ii) Constant term = 3/2
(iii) Constant term = -1

Q5: Add:
(i) 3x²y, -5x²y, -x²y
(ii) a + b – 3, b + 2a – 1
Ans:
(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

Q6: Subtract 3x² – x from 5x – x².
Ans: 
(5x – x²) – (3x2 – x)
= 5x – x² – 3x² + x
= 5x + x – x² – 3x²
= 6x – 4x²

Q7: Simplify combining the like terms:
(i) a – (a – b) – b – (b – a)
(ii) x² – 3x + y² – x – 2y²
Ans:
(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

Q8: Subtract 24xy – 10y – 18x from 30xy + 12y – 14x.
Ans:
(30xy + 12y – 14x) – (24xy – 10y – 18x)
= 30xy + 12y – 14x – 24xy + 10y + 18x

= 30xy – 24xy + 12y + 10y – 14x + 18x
= 6xy + 22y + 4x

Q9: From the sum of 2x² + 3xy – 5 and 7 + 2xy – x² subtract 3xy + x2 – 2.
Ans:
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

Q10: Subtract 3x² – 5y – 2 from 5y – 3x² + xy and find the value of the result if x = 2, y = -1.
Ans:
(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

Q11: 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)
Ans:
(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

Q12: Find the value of t if the value of 3x² + 5x – 2t equals to 8, when x = -1.
Ans:
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.

Q13: To what expression must 99x³ – 33x² – 13x – 41 be added to make the sum zero?
Ans:
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