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DAV Class 7 Maths Chapter 6: all worksheets All Class 7 Maths chapters →
DAV Class 7 Secondary Mathematics Chapter 6: Key Concepts Overview
Before attempting the Brain Teasers and HOTS problems, keep these fundamental algebraic rules in mind:
| Concept | Definition & Rule | Example |
|---|---|---|
| Constant & Variable | A constant has a fixed numerical value. A variable takes various numerical values. | In 5x + 3, 3 is constant and x is variable. |
| Like & Unlike Terms | Terms having the same algebraic factors (variables with identical powers) are like terms. Only like terms can be added or subtracted. | 7x2y and -3yx2 are like terms. |
| Numerical Coefficient | The constant factor multiplying the variable part in a term. | In -5xy2, numerical coefficient is -5. |
| Degree of Polynomial | The highest sum of exponents of variables in any single term of the polynomial. | 4x3y2 has degree 3 + 2 = 5. |
Chapter 6 Brain Teasers Solutions
Q.1 (A). Tick the correct option:
(a) The numerical coefficient of -5xy2 is:
(i) 5
(ii) -5
(iii) 1
(iv) -1
Answer: (ii) -5
Explanation: The numerical factor multiplying the variables xy2 is -5.
(b) The sum of 3x, -5x and 7x is:
(i) 5x
(ii) -5x
(iii) 15x
(iv) 9x
Answer: (i) 5x
Explanation: 3x + (-5x) + 7x = (3 – 5 + 7)x = 5x.
(c) The degree of the polynomial 4x3y2 – 3x2y3 + 5xy4 is:
(i) 3
(ii) 4
(iii) 5
(iv) 6
Answer: (iii) 5
Explanation: Sum of exponents of variables in each term:
Term 1: 4x3y2 → degree = 3 + 2 = 5.
Term 2: -3x2y3 → degree = 2 + 3 = 5.
Term 3: 5xy4 → degree = 1 + 4 = 5.
The highest power sum is 5.
(d) The value of 2x2 – 3x + 5 when x = -2 is:
(i) 7
(ii) 19
(iii) 15
(iv) 3
Answer: (ii) 19
Explanation: Substitute x = -2:
2(-2)2 – 3(-2) + 5 = 2(4) + 6 + 5 = 8 + 6 + 5 = 19.
(e) Which of the following is a pair of like terms?
(i) 7x, 7y
(ii) 3x2y, 3xy2
(iii) -5a2b, 2ba2
(iv) 4mn, 4m2n
Answer: (iii) -5a2b, 2ba2
Explanation: In -5a2b and 2ba2, the algebraic factors are identical (a2b = ba2 by commutative property of multiplication). Hence, they are like terms.
Q.1 (B). Answer the following questions:
(a) Write an algebraic expression for: “7 subtracted from one-third of the product of x and y“.
Solution:
Product of x and y = xy.
One-third of the product = (1/3)xy.
7 subtracted from it gives: (1/3)xy – 7 or (xy / 3) – 7.
(b) Identify the terms and their numerical factors in the expression: -3p2q + 5pq – 7.
Solution:
• Term 1: -3p2q → Numerical factor (coefficient) = -3.
• Term 2: 5pq → Numerical factor (coefficient) = 5.
• Term 3: -7 → Numerical factor (constant term) = -7.
(c) Classify the following into monomials, binomials and trinomials:
(i) 4x2y
(ii) 3a – 2b
(iii) x2 – 5x + 6
Solution:
(i) 4x2y has 1 term → Monomial.
(ii) 3a – 2b has 2 terms → Binomial.
(iii) x2 – 5x + 6 has 3 terms → Trinomial.
(d) What must be added to 3x2 – 5x + 2 to obtain 5x2 + 2x – 3?
Solution:
Let the required expression to be added be E.
(3x2 – 5x + 2) + E = 5x2 + 2x – 3
E = (5x2 + 2x – 3) – (3x2 – 5x + 2)
= 5x2 + 2x – 3 – 3x2 + 5x – 2
= (5 – 3)x2 + (2 + 5)x + (-3 – 2)
= 2x2 + 7x – 5.
(e) Subtract (4a – 3b + 2c) from (7a + 2b – 5c).
Solution:
(7a + 2b – 5c) – (4a – 3b + 2c)
= 7a + 2b – 5c – 4a + 3b – 2c
= (7 – 4)a + (2 + 3)b + (-5 – 2)c
= 3a + 5b – 7c.
Q.2. Add the following algebraic expressions:
(i) 3x2 – 4xy + 2y2, -2x2 + 7xy – 5y2, and 5x2 – 2xy + 3y2
Solution:
Sum = (3x2 – 4xy + 2y2) + (-2x2 + 7xy – 5y2) + (5x2 – 2xy + 3y2)
Group like terms together:
= (3 – 2 + 5)x2 + (-4 + 7 – 2)xy + (2 – 5 + 3)y2
= 6x2 + (1)xy + (0)y2
= 6x2 + xy.
(ii) 4a2b – 3ab2 + 5, -2a2b + 7ab2 – 8, and 3a2b – 2ab2 + 6
Solution:
Sum = (4a2b – 3ab2 + 5) + (-2a2b + 7ab2 – 8) + (3a2b – 2ab2 + 6)
Group like terms:
= (4 – 2 + 3)a2b + (-3 + 7 – 2)ab2 + (5 – 8 + 6)
= 5a2b + 2ab2 + 3.
Result: 5a2b + 2ab2 + 3.
Q.3. Subtraction & Compound Polynomial Operations:
(i) From the sum of (4x2 – 3x + 7) and (-2x2 + 5x – 4), subtract (3x2 – 2x + 5).
Solution:
Step 1: Find the sum of first two expressions:
(4x2 – 3x + 7) + (-2x2 + 5x – 4)
= (4 – 2)x2 + (-3 + 5)x + (7 – 4)
= 2x2 + 2x + 3.
Step 2: Subtract (3x2 – 2x + 5) from this sum:
(2x2 + 2x + 3) – (3x2 – 2x + 5)
= 2x2 + 2x + 3 – 3x2 + 2x – 5
= (2 – 3)x2 + (2 + 2)x + (3 – 5)
= –x2 + 4x – 2.
(ii) Subtract 2a2 – 3ab + 5b2 from the sum of (4a2 + 5ab – 2b2) and (-a2 – 2ab + 3b2).
Solution:
Step 1: Calculate the sum:
(4a2 + 5ab – 2b2) + (-a2 – 2ab + 3b2)
= (4 – 1)a2 + (5 – 2)ab + (-2 + 3)b2
= 3a2 + 3ab + b2.
Step 2: Subtract (2a2 – 3ab + 5b2):
(3a2 + 3ab + b2) – (2a2 – 3ab + 5b2)
= 3a2 + 3ab + b2 – 2a2 + 3ab – 5b2
= (3 – 2)a2 + (3 + 3)ab + (1 – 5)b2
= a2 + 6ab – 4b2.
Q.4. Simplify the following expressions removing brackets:
(i) 3(2x – 3y) – 2(3x – 4y) + 5(x – 2y)
Solution:
Distribute coefficients:
= 6x – 9y – 6x + 8y + 5x – 10y
Group like terms:
= (6 – 6 + 5)x + (-9 + 8 – 10)y
= 5x – 11y.
Result: 5x – 11y.
(ii) x2 – [2x – {3x2 – (5x – 4)}]
Solution:
Work from the innermost parentheses outward:
Innermost round brackets ( ):
3x2 – (5x – 4) = 3x2 – 5x + 4.
Curly braces { }:
2x – {3x2 – 5x + 4} = 2x – 3x2 + 5x – 4 = -3x2 + 7x – 4.
Square brackets [ ]:
x2 – [-3x2 + 7x – 4] = x2 + 3x2 – 7x + 4
= 4x2 – 7x + 4.
Q.5. Evaluate the following algebraic expressions:
(i) If a = 2, b = -3, find the numerical value of: 3a2 – 2ab + 4b2.
Solution:
Substitute a = 2 and b = -3:
= 3(2)2 – 2(2)(-3) + 4(-3)2
= 3(4) – 2(-6) + 4(9)
= 12 + 12 + 36 = 60.
(ii) If x = -1, y = 2, z = -2, find the value of: x3 + y3 + z3 – 3xyz.
Solution:
Substitute values:
x3 = (-1)3 = -1
y3 = (2)3 = 8
z3 = (-2)3 = -8
-3xyz = -3(-1)(2)(-2) = -3(4) = -12.
Total = -1 + 8 + (-8) – 12 = -1 + 0 – 12 = -13.
HOTS (Higher Order Thinking Skills) Questions
HOTS 1: What should be subtracted from 1 + 2x – 3x2 to get 2x2 – 3x + 1?
Solution:
Let P be the expression to be subtracted.
(1 + 2x – 3x2) – P = 2x2 – 3x + 1
P = (1 + 2x – 3x2) – (2x2 – 3x + 1)
= 1 + 2x – 3x2 – 2x2 + 3x – 1
= (-3 – 2)x2 + (2 + 3)x + (1 – 1)
= -5x2 + 5x or 5x(1 – x).
HOTS 2: If P = 3x2 – 4x + 1, Q = -2x2 + 5x – 3, and R = 5x2 – x + 4, find:
(a) (P + Q) – R
(b) P – (Q + R)
Solution:
(a) Finding (P + Q) – R:
P + Q = (3x2 – 4x + 1) + (-2x2 + 5x – 3) = x2 + x – 2.
Now subtract R:
(x2 + x – 2) – (5x2 – x + 4)
= x2 + x – 2 – 5x2 + x – 4
= -4x2 + 2x – 6.
(b) Finding P – (Q + R):
Q + R = (-2x2 + 5x – 3) + (5x2 – x + 4) = 3x2 + 4x + 1.
Now subtract from P:
(3x2 – 4x + 1) – (3x2 + 4x + 1)
= 3x2 – 4x + 1 – 3x2 – 4x – 1
= (3 – 3)x2 + (-4 – 4)x + (1 – 1)
= -8x.
HOTS 3: A piece of wire of length (12x – 8) cm is bent to form an equilateral triangle. What is the length of each side of the triangle? If the same wire is rebent into a square, what will be the length of each side of the square?
Solution:
Total perimeter of wire = (12x – 8) cm.
Case 1: Equilateral Triangle:
An equilateral triangle has 3 equal sides.
Side of triangle = Perimeter / 3 = (12x – 8) / 3 = (4x – 8/3) cm.
Case 2: Square:
A square has 4 equal sides.
Side of square = Perimeter / 4 = (12x – 8) / 4 = (12x/4) – (8/4) = (3x – 2) cm.
Value Based Questions (VBQs)
Question: Rohan’s parents gave him Rs (5x2 – 3x + 10) on his birthday. He decided to donate Rs (2x2 + 4x – 5) to buy books and notebooks for underprivileged students in a nearby community centre. He also spent Rs (x2 – 2x + 7) on school stationery for himself.
(a) Find the algebraic expression representing the amount of money left with Rohan.
Solution:
Total money given = 5x2 – 3x + 10.
Total expenditure = Money donated + Money spent on stationery
= (2x2 + 4x – 5) + (x2 – 2x + 7)
= (2 + 1)x2 + (4 – 2)x + (-5 + 7)
= 3x2 + 2x + 2.
Money remaining with Rohan = Total money – Total expenditure
= (5x2 – 3x + 10) – (3x2 + 2x + 2)
= 5x2 – 3x + 10 – 3x2 – 2x – 2
= (5 – 3)x2 + (-3 – 2)x + (10 – 2)
= Rs (2x2 – 5x + 8).
(b) If x = 5, calculate how much money he donated for underprivileged students.
Solution:
Amount donated = 2x2 + 4x – 5.
Substituting x = 5:
= 2(5)2 + 4(5) – 5
= 2(25) + 20 – 5
= 50 + 20 – 5 = Rs 65.
(c) What moral value is highlighted by Rohan’s conduct?
Solution:
Rohan exemplifies empathy, compassion, social responsibility, and generosity by sharing his resources to support the education of underprivileged children.
Summary & Study Tips for Algebraic Expressions
Here are crucial tips to avoid common algebraic mistakes in exams:
- Watch the Signs when Subtracting: Remember that subtracting an expression reverses the sign of every term inside the bracket (e.g., -(3x – 5) = -3x + 5).
- Never Combine Unlike Terms: 5x + 3y cannot be written as 8xy. They remain separate terms.
- Follow BODMAS with Brackets: Always clear innermost round brackets ( ) first, then curly braces { }, and finally square brackets [ ].
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