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CHAPTER 4 — EXPRESSIONS USING LETTER-NUMBERS
Class VII Mathematics — NCERT Ganita Prakash 2024
Preeti Kushwah Classes — Unit Test — Set 2 (Standard)
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 the number of terms in: 4x² − 3xy + 7y − 2
Q2.1
Are 5xy and 5yx the same term? Justify your answer.
Q3.1
Write an expression for: “the sum of twice a and three times b.”
Q4.1
Identify the coefficient of ab in: 8ab − 3a + 2b
Q5.1
Find the value of x² − x + 1 when x = 2.
Q6.1
Write the expression if a number n is first multiplied by 4 and then 9 is subtracted from it.
Section B — (2 Marks Each) [5 × 2 = 10]
Q7.2
Classify the following as monomials, binomials, or trinomials:
(a) 3x² + 2x − 1 (b) 7ab (c) 4p − 3q (d) 5
(a) 3x² + 2x − 1 (b) 7ab (c) 4p − 3q (d) 5
Q8.2
Add (2x² + 3x − 4) and (x² − 5x + 7). Show your working.
Q9.2
If A = 3m + 2n and B = m − n, find:
(a) A + B (b) A − B
(a) A + B (b) A − B
Q10.2
Evaluate 2a² + 3ab − b² when a = 2 and b = 1.
Q11.2
The sides of a triangle are (2x+1) cm, (3x−2) cm and (x+4) cm. Find the perimeter in simplified form.
Section C — (3 Marks Each) [4 × 3 = 12]
Q12.3
Simplify: (3x² + 5x − 2) + (x² − 3x + 4) − (2x² + x − 1)
Q13.3
Priya is p years old. Her mother is 3 times her age. Her father is 4 years older than her mother.
(a) Write expressions for mother’s and father’s ages. [1]
(b) Write and simplify an expression for the total age of all three. [1]
(c) Find the total age when p = 12. [1]
(a) Write expressions for mother’s and father’s ages. [1]
(b) Write and simplify an expression for the total age of all three. [1]
(c) Find the total age when p = 12. [1]
Q14.3
In a class there are b boys and g girls. The number of boys increases by 5 and the number of girls decreases by 3.
(a) Write new expressions for boys and girls. [1]
(b) Write and simplify an expression for the new total strength. [1]
(c) If originally b = 28 and g = 22, find the new total. [1]
(a) Write new expressions for boys and girls. [1]
(b) Write and simplify an expression for the new total strength. [1]
(c) If originally b = 28 and g = 22, find the new total. [1]
Q15.3
A rectangle has length (3x + 2) cm and breadth (2x − 1) cm.
(a) Find the expression for perimeter. Simplify. [1]
(b) Find the area expression. [1]
(c) Find both when x = 3. [1]
(a) Find the expression for perimeter. Simplify. [1]
(b) Find the area expression. [1]
(c) Find both when x = 3. [1]
Section D — (5 Marks Each) [2 × 5 = 10]
Q16.5
Ravi scores m marks in Maths, (m − 10) in Science and (m + 15) in English.
(a) Find an expression for total marks scored. [1]
(b) Simplify the expression. [1]
(c) Find the total if m = 72. [1]
(d) Each subject has maximum 100 marks. Write the expression for total maximum marks. [1]
(e) What percentage of total maximum marks does Ravi score when m = 72? [1]
(a) Find an expression for total marks scored. [1]
(b) Simplify the expression. [1]
(c) Find the total if m = 72. [1]
(d) Each subject has maximum 100 marks. Write the expression for total maximum marks. [1]
(e) What percentage of total maximum marks does Ravi score when m = 72? [1]
Q17.5
Matchstick pattern — a row of equilateral triangles sharing one side:
1 triangle: 3 sticks; 2 triangles: 5 sticks; 3 triangles: 7 sticks.
(a) Write the pattern as an expression for n triangles. [1]
(b) Verify your expression for n = 4 and n = 5 by drawing/counting. [1]
(c) How many sticks for 20 triangles? [1]
(d) How many complete triangles can be made from 50 sticks? [1]
(e) For what value of n does the number of sticks equal 51? [1]
1 triangle: 3 sticks; 2 triangles: 5 sticks; 3 triangles: 7 sticks.
(a) Write the pattern as an expression for n triangles. [1]
(b) Verify your expression for n = 4 and n = 5 by drawing/counting. [1]
(c) How many sticks for 20 triangles? [1]
(d) How many complete triangles can be made from 50 sticks? [1]
(e) For what value of n does the number of sticks equal 51? [1]
Bonus Question (Optional) [2 Marks]
Q18.2
★ The sum of three consecutive even numbers starting from 2n is S.
(a) Write an expression for S in terms of n. [1]
(b) Find n if S = 66. [1]
(a) Write an expression for S in terms of n. [1]
(b) Find n if S = 66. [1]
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Answer Key & Detailed Solutions
Q1. [1 Mark]
4x², −3xy, 7y, −2 — that is 4 terms.
Q2. [1 Mark]
Yes, 5xy = 5yx. Multiplication is commutative, so xy = yx. Both terms have the same variables x and y with the same coefficient 5.
Q3. [1 Mark]
Twice a = 2a; three times b = 3b.
Expression: 2a + 3b
Expression: 2a + 3b
Q4. [1 Mark]
In 8ab − 3a + 2b, the coefficient of ab is 8.
Q5. [1 Mark]
x² − x + 1 when x = 2:
= 2² − 2 + 1 = 4 − 2 + 1 = 3
= 2² − 2 + 1 = 4 − 2 + 1 = 3
Q6. [1 Mark]
Multiply n by 4: 4n. Then subtract 9: 4n − 9
Q7. [2 Marks]
(a) 3x² + 2x − 1 — 3 terms → Trinomial
(b) 7ab — 1 term → Monomial
(c) 4p − 3q — 2 terms → Binomial
(d) 5 — 1 term (constant) → Monomial
(b) 7ab — 1 term → Monomial
(c) 4p − 3q — 2 terms → Binomial
(d) 5 — 1 term (constant) → Monomial
Q8. [2 Marks]
(2x² + 3x − 4) + (x² − 5x + 7)
x²-terms: 2x² + x² = 3x²
x-terms: 3x − 5x = −2x
constants: −4 + 7 = 3
Answer: 3x² − 2x + 3
x²-terms: 2x² + x² = 3x²
x-terms: 3x − 5x = −2x
constants: −4 + 7 = 3
Answer: 3x² − 2x + 3
Q9. [2 Marks]
A = 3m + 2n, B = m − n
(a) A + B = (3m + 2n) + (m − n) = 4m + n
(b) A − B = (3m + 2n) − (m − n) = 3m + 2n − m + n = 2m + 3n
(a) A + B = (3m + 2n) + (m − n) = 4m + n
(b) A − B = (3m + 2n) − (m − n) = 3m + 2n − m + n = 2m + 3n
Q10. [2 Marks]
2a² + 3ab − b² when a = 2, b = 1:
= 2(4) + 3(2)(1) − (1)
= 8 + 6 − 1 = 13
= 2(4) + 3(2)(1) − (1)
= 8 + 6 − 1 = 13
Q11. [2 Marks]
P = (2x+1) + (3x−2) + (x+4)
= (2x + 3x + x) + (1 − 2 + 4)
= 6x + 3 cm
= (2x + 3x + x) + (1 − 2 + 4)
= 6x + 3 cm
Q12. [3 Marks]
(3x² + 5x − 2) + (x² − 3x + 4) − (2x² + x − 1)
Step 1 — Remove brackets (change signs of last group):
= 3x² + 5x − 2 + x² − 3x + 4 − 2x² − x + 1
Step 2 — Group like terms:
x²: (3 + 1 − 2)x² = 2x²
x: (5 − 3 − 1)x = x
constant: −2 + 4 + 1 = 3
Answer: 2x² + x + 3
Step 1 — Remove brackets (change signs of last group):
= 3x² + 5x − 2 + x² − 3x + 4 − 2x² − x + 1
Step 2 — Group like terms:
x²: (3 + 1 − 2)x² = 2x²
x: (5 − 3 − 1)x = x
constant: −2 + 4 + 1 = 3
Answer: 2x² + x + 3
Q13. [3 Marks]
(a) Mother = 3p; Father = 3p + 4
(b) Total = p + 3p + (3p + 4) = 7p + 4
(c) When p = 12: 7(12) + 4 = 84 + 4 = 88 years
(b) Total = p + 3p + (3p + 4) = 7p + 4
(c) When p = 12: 7(12) + 4 = 84 + 4 = 88 years
Q14. [3 Marks]
(a) New boys = b + 5; New girls = g − 3
(b) New total = (b + 5) + (g − 3) = b + g + 2
(c) b = 28, g = 22: 28 + 22 + 2 = 52 students
(b) New total = (b + 5) + (g − 3) = b + g + 2
(c) b = 28, g = 22: 28 + 22 + 2 = 52 students
Q15. [3 Marks]
l = (3x + 2) cm, b = (2x − 1) cm
(a) P = 2[(3x+2) + (2x−1)] = 2[5x + 1] = 10x + 2 cm
(b) Area = (3x+2)(2x−1) = 6x² − 3x + 4x − 2 = 6x² + x − 2 cm²
(c) When x = 3: P = 10(3)+2 = 32 cm; Area = 6(9)+3−2 = 55 cm²
(a) P = 2[(3x+2) + (2x−1)] = 2[5x + 1] = 10x + 2 cm
(b) Area = (3x+2)(2x−1) = 6x² − 3x + 4x − 2 = 6x² + x − 2 cm²
(c) When x = 3: P = 10(3)+2 = 32 cm; Area = 6(9)+3−2 = 55 cm²
Q16. [5 Marks]
(a) Total = m + (m−10) + (m+15)
(b) = 3m + 5
(c) When m = 72: 3(72) + 5 = 216 + 5 = 221 marks
(d) Total maximum = 100 + 100 + 100 = 300 marks
(e) Percentage = (221/300) × 100 ≈ 73.67%
(b) = 3m + 5
(c) When m = 72: 3(72) + 5 = 216 + 5 = 221 marks
(d) Total maximum = 100 + 100 + 100 = 300 marks
(e) Percentage = (221/300) × 100 ≈ 73.67%
Q17. [5 Marks]
Pattern: 3, 5, 7, 9... (increases by 2 each time)
(a) Expression for n triangles: 2n + 1
(b) n=4: 2(4)+1 = 9 sticks ✓ n=5: 2(5)+1 = 11 sticks ✓
(c) For 20 triangles: 2(20)+1 = 41 sticks
(d) 2n+1 ≤ 50 → 2n ≤ 49 → n ≤ 24.5 → 24 complete triangles
(e) 2n+1 = 51 → 2n = 50 → n = 25
(a) Expression for n triangles: 2n + 1
(b) n=4: 2(4)+1 = 9 sticks ✓ n=5: 2(5)+1 = 11 sticks ✓
(c) For 20 triangles: 2(20)+1 = 41 sticks
(d) 2n+1 ≤ 50 → 2n ≤ 49 → n ≤ 24.5 → 24 complete triangles
(e) 2n+1 = 51 → 2n = 50 → n = 25
Q18. [Bonus — 2 Marks]
Three consecutive even numbers starting from 2n: 2n, 2n+2, 2n+4
(a) S = 2n + (2n+2) + (2n+4) = 6n + 6
(b) 6n + 6 = 66 → 6n = 60 → n = 10
(Numbers are 20, 22, 24 and 20+22+24 = 66 ✓)
(a) S = 2n + (2n+2) + (2n+4) = 6n + 6
(b) 6n + 6 = 66 → 6n = 60 → n = 10
(Numbers are 20, 22, 24 and 20+22+24 = 66 ✓)