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CHAPTER 4 — EXPRESSIONS USING LETTER-NUMBERS
Class VII Mathematics — NCERT Ganita Prakash 2024
Preeti Kushwah Classes — Unit Test — Set 3 (Advanced)
General Instructions:
1. All questions are compulsory.
2. Section A has 6 questions of 1 mark each.
3. Section B has 5 questions of 2 marks each.
4. Section C has 4 questions of 3 marks each.
5. Section D has 2 questions of 5 marks each.
6. Show all working clearly. Marks are awarded for steps.
1. All questions are compulsory.
2. Section A has 6 questions of 1 mark each.
3. Section B has 5 questions of 2 marks each.
4. Section C has 4 questions of 3 marks each.
5. Section D has 2 questions of 5 marks each.
6. Show all working clearly. Marks are awarded for steps.
Section A — (1 Mark Each) [6 × 1 = 6]
Q1.1
Write an algebraic expression with 3 terms using variable x, where one term has x², one has x, and the constant term is greater than 5.
Q2.1
If 3x + 7 = 22, find x. (Show one step of working.)
Q3.1
The expression (4a − 3b + 2c) has how many terms? Name each term and write its coefficient.
Q4.1
Is 2x − 3y the same as 3y − 2x? Justify with a numerical example by substituting values.
Q5.1
Write the expression for the nth term of the sequence: 5, 8, 11, 14, 17 …
Q6.1
Evaluate: (p + q)² − (p − q)² when p = 3, q = 2. (Expand both squares first, then substitute.)
Section B — (2 Marks Each) [5 × 2 = 10]
Q7.2
Simplify: 3(2x + y) − 2(x − 3y) + 5y
Q8.2
If x = 2, y = −1 and z = 3, evaluate: x² + y² + z² − xy + yz
Q9.2
The sum of two expressions is 7a − 3b + 2. One expression is 3a + b − 4. Find the other expression.
Q10.2
A number n is such that when 4 is added to twice of it, the result equals 1 less than three times of it. Form an equation and solve for n.
Q11.2
Write an expression for the total cost of x pens at ₹12 each, y notebooks at ₹25 each, and z erasers at ₹5 each. Find total cost when x = 3, y = 2, z = 5.
Section C — (3 Marks Each) [4 × 3 = 12]
Q12.3
Let A = 2x² + 3x − 1, B = x² − 2x + 3, C = x² + x − 2.
(a) Find A + B + C. [1 mark]
(b) Find A − B + C. [1 mark]
(c) Verify both answers when x = 1. [1 mark]
(a) Find A + B + C. [1 mark]
(b) Find A − B + C. [1 mark]
(c) Verify both answers when x = 1. [1 mark]
Q13.3
In a school competition, scores of three teams are:
Team A: (3n + 5) points | Team B: (2n − 1) points | Team C: (n + 8) points
(a) Find an expression for the total score of all three teams. [1 mark]
(b) For what values of n does Team B's score exceed Team A's? [1 mark]
(c) Find each team's score when n = 6. Which team won? [1 mark]
Team A: (3n + 5) points | Team B: (2n − 1) points | Team C: (n + 8) points
(a) Find an expression for the total score of all three teams. [1 mark]
(b) For what values of n does Team B's score exceed Team A's? [1 mark]
(c) Find each team's score when n = 6. Which team won? [1 mark]
Q14.3
Using the identity (a + b)² = a² + 2ab + b²:
(a) Expand (x + 3)². [1 mark]
(b) Expand (2y + 1)². [1 mark]
(c) Verify (x + 3)² for x = 2 by: (i) direct calculation and (ii) using your expansion from (a). [1 mark]
(a) Expand (x + 3)². [1 mark]
(b) Expand (2y + 1)². [1 mark]
(c) Verify (x + 3)² for x = 2 by: (i) direct calculation and (ii) using your expansion from (a). [1 mark]
Q15.3
L-shapes are made from unit squares: Shape 1 uses 3 squares, Shape 2 uses 5, Shape 3 uses 7.
(a) Find the algebraic expression for the number of squares in the nth shape. [1 mark]
(b) Which shape number uses exactly 21 squares? [1 mark]
(c) How many squares does Shape 50 use? [1 mark]
(a) Find the algebraic expression for the number of squares in the nth shape. [1 mark]
(b) Which shape number uses exactly 21 squares? [1 mark]
(c) How many squares does Shape 50 use? [1 mark]
Section D — (5 Marks Each) [2 × 5 = 10]
Q16.5
A factory produces x items on Day 1. On Day 2 it produces 20 more than Day 1. On Day 3 it produces twice of Day 2. On Day 4 production falls by 30 from Day 3.
(a) Write expressions for Day 2, Day 3, and Day 4 production. [2 marks]
(b) Write an expression for total 4-day production and simplify it. [1 mark]
(c) Find total production when x = 100. [1 mark]
(d) If total production must exceed 600, what is the minimum integer value of x? [1 mark]
(a) Write expressions for Day 2, Day 3, and Day 4 production. [2 marks]
(b) Write an expression for total 4-day production and simplify it. [1 mark]
(c) Find total production when x = 100. [1 mark]
(d) If total production must exceed 600, what is the minimum integer value of x? [1 mark]
Q17.5
(a) Simplify: 4(3x − 2y) − 3(2x + y) + 2(x − 4y). Show every step. [2 marks]
(b) If a + b = 8 and a − b = 2, find the values of a and b using the method:
a = [(a + b) + (a − b)] ÷ 2 and b = [(a + b) − (a − b)] ÷ 2. [2 marks]
(c) Using the values of a and b from part (b), evaluate: 3a² − 2ab + b². [1 mark]
(b) If a + b = 8 and a − b = 2, find the values of a and b using the method:
a = [(a + b) + (a − b)] ÷ 2 and b = [(a + b) − (a − b)] ÷ 2. [2 marks]
(c) Using the values of a and b from part (b), evaluate: 3a² − 2ab + b². [1 mark]
Bonus Question (Optional) [2 Marks]
Q18.2
★ CHALLENGE
(a) Two numbers are in the ratio 3 : 5. If each number is decreased by 4, the ratio becomes 1 : 2. Let the numbers be 3k and 5k. Form an equation and find k. What are the original numbers? [1 mark]
(b) Using the identity (a − b)(a + b) = a² − b², find the value of 97 × 103 without direct multiplication. [1 mark]
(a) Two numbers are in the ratio 3 : 5. If each number is decreased by 4, the ratio becomes 1 : 2. Let the numbers be 3k and 5k. Form an equation and find k. What are the original numbers? [1 mark]
(b) Using the identity (a − b)(a + b) = a² − b², find the value of 97 × 103 without direct multiplication. [1 mark]
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Answer Key & Detailed Solutions
Q1. [1 Mark]
Any valid expression with 3 terms — e.g. x² + 3x + 7 (constant 7 > 5). Other examples: x² + 5x + 8, 2x² + x + 6. Accept any correct form.
Q2. [1 Mark]
3x + 7 = 22
3x = 22 − 7 = 15
x = 5
3x = 22 − 7 = 15
x = 5
Q3. [1 Mark]
The expression has 3 terms:
• 4a — coefficient is 4
• −3b — coefficient is −3
• 2c — coefficient is 2
• 4a — coefficient is 4
• −3b — coefficient is −3
• 2c — coefficient is 2
Q4. [1 Mark]
No, they are NOT the same.
Let x = 2, y = 1:
2x − 3y = 4 − 3 = 1
3y − 2x = 3 − 4 = −1
1 ≠ −1, so 2x − 3y ≠ 3y − 2x. (In fact 2x−3y = −(3y−2x).)
Let x = 2, y = 1:
2x − 3y = 4 − 3 = 1
3y − 2x = 3 − 4 = −1
1 ≠ −1, so 2x − 3y ≠ 3y − 2x. (In fact 2x−3y = −(3y−2x).)
Q5. [1 Mark]
Differences = 3 each time (arithmetic sequence).
nth term = first term + (n−1) × 3 = 5 + 3n − 3 = 3n + 2
Verify: n=1: 5 ✓, n=2: 8 ✓, n=3: 11 ✓
nth term = first term + (n−1) × 3 = 5 + 3n − 3 = 3n + 2
Verify: n=1: 5 ✓, n=2: 8 ✓, n=3: 11 ✓
Q6. [1 Mark]
(p + q)² = p² + 2pq + q²
(p − q)² = p² − 2pq + q²
Difference = 4pq = 4 × 3 × 2 = 24
(p − q)² = p² − 2pq + q²
Difference = 4pq = 4 × 3 × 2 = 24
Q7. [2 Marks]
3(2x + y) − 2(x − 3y) + 5y
= 6x + 3y − 2x + 6y + 5y
= (6x − 2x) + (3y + 6y + 5y)
= 4x + 14y
= 6x + 3y − 2x + 6y + 5y
= (6x − 2x) + (3y + 6y + 5y)
= 4x + 14y
Q8. [2 Marks]
x=2, y=−1, z=3
x² + y² + z² − xy + yz
= 4 + 1 + 9 − (2)(−1) + (−1)(3)
= 14 + 2 − 3
= 13
x² + y² + z² − xy + yz
= 4 + 1 + 9 − (2)(−1) + (−1)(3)
= 14 + 2 − 3
= 13
Q9. [2 Marks]
Other expression = (7a − 3b + 2) − (3a + b − 4)
= 7a − 3b + 2 − 3a − b + 4
= 4a − 4b + 6
= 7a − 3b + 2 − 3a − b + 4
= 4a − 4b + 6
Q10. [2 Marks]
Equation: 2n + 4 = 3n − 1
4 + 1 = 3n − 2n
n = 5
Verify: LHS = 2(5)+4 = 14; RHS = 3(5)−1 = 14 ✓
4 + 1 = 3n − 2n
n = 5
Verify: LHS = 2(5)+4 = 14; RHS = 3(5)−1 = 14 ✓
Q11. [2 Marks]
Expression: 12x + 25y + 5z
When x=3, y=2, z=5:
= 12(3) + 25(2) + 5(5)
= 36 + 50 + 25 = ₹111
When x=3, y=2, z=5:
= 12(3) + 25(2) + 5(5)
= 36 + 50 + 25 = ₹111
Q12. [3 Marks]
A = 2x²+3x−1, B = x²−2x+3, C = x²+x−2
(a) A+B+C = (2+1+1)x² + (3−2+1)x + (−1+3−2) = 4x² + 2x + 0 = 4x² + 2x
(b) A−B+C = (2−1+1)x² + (3+2+1)x + (−1−3−2) = 2x² + 6x − 6
(c) x=1: A=4, B=2, C=0
A+B+C = 6; 4(1)+2(1) = 6 ✓
A−B+C = 2; 2(1)+6(1)−6 = 2 ✓
(a) A+B+C = (2+1+1)x² + (3−2+1)x + (−1+3−2) = 4x² + 2x + 0 = 4x² + 2x
(b) A−B+C = (2−1+1)x² + (3+2+1)x + (−1−3−2) = 2x² + 6x − 6
(c) x=1: A=4, B=2, C=0
A+B+C = 6; 4(1)+2(1) = 6 ✓
A−B+C = 2; 2(1)+6(1)−6 = 2 ✓
Q13. [3 Marks]
(a) Total = (3n+5)+(2n−1)+(n+8) = 6n + 12
(b) Team B > Team A: 2n−1 > 3n+5 ⇒ −6 > n ⇒ n < −6.
For positive n, Team B's score is always less than Team A's.
(c) n=6: Team A = 23, Team B = 11, Team C = 14.
Team A won with 23 points.
(b) Team B > Team A: 2n−1 > 3n+5 ⇒ −6 > n ⇒ n < −6.
For positive n, Team B's score is always less than Team A's.
(c) n=6: Team A = 23, Team B = 11, Team C = 14.
Team A won with 23 points.
Q14. [3 Marks]
(a) (x+3)² = x² + 2(x)(3) + 3² = x² + 6x + 9
(b) (2y+1)² = (2y)² + 2(2y)(1) + 1² = 4y² + 4y + 1
(c) x=2: Direct: (2+3)² = 5² = 25
Expansion: (2)² + 6(2) + 9 = 4+12+9 = 25 ✓
(b) (2y+1)² = (2y)² + 2(2y)(1) + 1² = 4y² + 4y + 1
(c) x=2: Direct: (2+3)² = 5² = 25
Expansion: (2)² + 6(2) + 9 = 4+12+9 = 25 ✓
Q15. [3 Marks]
(a) Differences = 2. nth term = 3 + (n−1)×2 = 2n + 1
(b) 2n+1 = 21 ⇒ 2n = 20 ⇒ n = 10 (Shape 10)
(c) Shape 50: 2(50)+1 = 101 squares
(b) 2n+1 = 21 ⇒ 2n = 20 ⇒ n = 10 (Shape 10)
(c) Shape 50: 2(50)+1 = 101 squares
Q16. [5 Marks]
(a) Day 2 = x + 20; Day 3 = 2(x+20) = 2x+40; Day 4 = 2x+40−30 = 2x+10
(b) Total = x + (x+20) + (2x+40) + (2x+10) = 6x + 70
(c) x=100: 6(100)+70 = 670 items
(d) 6x+70 > 600 ⇒ 6x > 530 ⇒ x > 88.33
Minimum integer value of x = 89
(b) Total = x + (x+20) + (2x+40) + (2x+10) = 6x + 70
(c) x=100: 6(100)+70 = 670 items
(d) 6x+70 > 600 ⇒ 6x > 530 ⇒ x > 88.33
Minimum integer value of x = 89
Q17. [5 Marks]
(a) 4(3x−2y) − 3(2x+y) + 2(x−4y)
= 12x−8y − 6x−3y + 2x−8y
= (12−6+2)x + (−8−3−8)y
= 8x − 19y
(b) a = (8+2)÷2 = 5; b = (8−2)÷2 = 3
(c) 3a²−2ab+b² = 3(25)−2(5)(3)+9 = 75−30+9 = 54
= 12x−8y − 6x−3y + 2x−8y
= (12−6+2)x + (−8−3−8)y
= 8x − 19y
(b) a = (8+2)÷2 = 5; b = (8−2)÷2 = 3
(c) 3a²−2ab+b² = 3(25)−2(5)(3)+9 = 75−30+9 = 54
Q18. [Bonus — 2 Marks]
(a) Numbers = 3k and 5k.
(3k−4)/(5k−4) = 1/2 ⇒ 2(3k−4) = 5k−4
6k−8 = 5k−4 ⇒ k = 4
Original numbers: 12 and 20
(b) 97 × 103 = (100−3)(100+3) = 100²−3² = 10000−9 = 9991
(3k−4)/(5k−4) = 1/2 ⇒ 2(3k−4) = 5k−4
6k−8 = 5k−4 ⇒ k = 4
Original numbers: 12 and 20
(b) 97 × 103 = (100−3)(100+3) = 100²−3² = 10000−9 = 9991