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CHAPTER 4 — EXPRESSIONS USING LETTER-NUMBERS
Class VII Mathematics — NCERT Ganita Prakash 2024
Preeti Kushwah Classes — Unit Test — Set 4 🔥 (Challenge)
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. This is a CHALLENGE paper — partial marks are awarded for correct steps even if the final answer is wrong.
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. This is a CHALLENGE paper — partial marks are awarded for correct steps even if the final answer is wrong.
Section A — (1 Mark Each) [6 × 1 = 6]
Q1.1
If a + b = 10 and ab = 24, find a² + b². (Hint: Use the identity (a+b)² = a² + 2ab + b².)
Q2.1
A student writes: “3x + 4x = 7x²”. Is the student right? Correct the error and explain.
Q3.1
The sum of three consecutive integers is 63. Let the integers be n−1, n, n+1. Form an equation and find all three integers.
Q4.1
Write an algebraic expression for: “One-third of the sum of x and y, decreased by twice of z.”
Q5.1
Evaluate: 102² using the identity (a + b)² = a² + 2ab + b² (Take a=100, b=2).
Q6.1
If P = 2x + 3 and Q = 5 − x, for what value of x is P = Q? Show working.
Section B — (2 Marks Each) [5 × 2 = 10]
Q7.2
Two pipes A and B together can fill a tank. Pipe A alone fills the tank in x hours; Pipe B alone fills it in (x+4) hours.
(a) What fraction of the tank does Pipe A fill in 1 hour? And Pipe B? [1 mark]
(b) If together they fill the tank in 3 hours, form an equation. (You do NOT need to solve it.) [1 mark]
(a) What fraction of the tank does Pipe A fill in 1 hour? And Pipe B? [1 mark]
(b) If together they fill the tank in 3 hours, form an equation. (You do NOT need to solve it.) [1 mark]
Q8.2
Simplify the expression and then evaluate for x = −2:
5(x² − 2x + 1) − 2(3x² + x − 4)
5(x² − 2x + 1) − 2(3x² + x − 4)
Q9.2
The sum of two algebraic expressions is 5x² − 3x + 7. If one expression is 2x² + x − 5, find the other. Then evaluate both expressions at x = 3 and verify their sum equals the original expression at x = 3.
Q10.2
A sequence of square patterns is built:
• Pattern 1: A 1×1 square = 1 tile
• Pattern 2: A 2×2 square = 4 tiles
• Pattern 3: A 3×3 square = 9 tiles
(a) Write the expression for Pattern n. [1 mark]
(b) How many NEW tiles are added going from Pattern n to Pattern (n+1)? Simplify the expression. [1 mark]
• Pattern 1: A 1×1 square = 1 tile
• Pattern 2: A 2×2 square = 4 tiles
• Pattern 3: A 3×3 square = 9 tiles
(a) Write the expression for Pattern n. [1 mark]
(b) How many NEW tiles are added going from Pattern n to Pattern (n+1)? Simplify the expression. [1 mark]
Q11.2
In a class test, Riya's score was 5 more than twice Siya's score. Together their scores sum to 83. Form two equations using variables r (Riya) and s (Siya), and solve for both scores.
Section C — (3 Marks Each) [4 × 3 = 12]
Q12.3
Using 97 × 103 = (100 − 3)(100 + 3) = 100² − 9:
(a) Calculate 97 × 103 directly. [1 mark]
(b) Similarly find 98 × 102 using an identity. [1 mark]
(c) Find 95 × 105 using the same identity. [1 mark]
(a) Calculate 97 × 103 directly. [1 mark]
(b) Similarly find 98 × 102 using an identity. [1 mark]
(c) Find 95 × 105 using the same identity. [1 mark]
Q13.3
The perimeter of a rectangle is 2(l + b).
(a) If the perimeter is (6x + 14) cm and the length is (2x + 5) cm, find the breadth as an expression in x. [1 mark]
(b) If x = 3 cm, find the actual dimensions and the area of the rectangle. [1 mark]
(c) Express the area as an algebraic expression in x and verify with x = 3. [1 mark]
(a) If the perimeter is (6x + 14) cm and the length is (2x + 5) cm, find the breadth as an expression in x. [1 mark]
(b) If x = 3 cm, find the actual dimensions and the area of the rectangle. [1 mark]
(c) Express the area as an algebraic expression in x and verify with x = 3. [1 mark]
Q14.3
Let E₁ = ax² + bx + c. Given that E₁ = 12 when x = 1, E₁ = 3 when x = 0, and E₁ = 22 when x = 2:
(a) Use x = 0 to find c. [1 mark]
(b) Use x = 1 to find a + b. [1 mark]
(c) Use x = 2 to find a and b individually. [1 mark]
(a) Use x = 0 to find c. [1 mark]
(b) Use x = 1 to find a + b. [1 mark]
(c) Use x = 2 to find a and b individually. [1 mark]
Q15.3
Triangle patterns use matchsticks: 1 triangle needs 3 sticks, 2 in-a-row need 5, 3 need 7.
(a) Write the expression for n triangles in a row. [1 mark]
(b) A chain of triangles uses exactly 99 matchsticks. How many triangles are there? [1 mark]
(c) A different pattern doubles: 3, 6, 12, 24 … Write the nth term and find the 6th term. [1 mark]
(a) Write the expression for n triangles in a row. [1 mark]
(b) A chain of triangles uses exactly 99 matchsticks. How many triangles are there? [1 mark]
(c) A different pattern doubles: 3, 6, 12, 24 … Write the nth term and find the 6th term. [1 mark]
Section D — (5 Marks Each) [2 × 5 = 10]
Q16.5
Rahul's age is 3 years more than twice Priya's age. Four years from now, the sum of their ages will be 43.
(a) Let Priya's current age = p. Write expressions for Rahul's current age and both ages 4 years from now. [2 marks]
(b) Form an equation using the condition about their future ages and solve for p. [2 marks]
(c) What will be the ratio of their ages 10 years from now? [1 mark]
(a) Let Priya's current age = p. Write expressions for Rahul's current age and both ages 4 years from now. [2 marks]
(b) Form an equation using the condition about their future ages and solve for p. [2 marks]
(c) What will be the ratio of their ages 10 years from now? [1 mark]
Q17.5
(a) Prove algebraically that the sum of any 5 consecutive integers is always 5 times the middle integer. Let the integers be (n−2), (n−1), n, (n+1), (n+2). [2 marks]
(b) Prove that the product of two consecutive even integers is always 4 less than a perfect square. Let them be 2k and (2k+2). [2 marks]
(c) Using part (a), find 5 consecutive integers whose sum is 120. [1 mark]
(b) Prove that the product of two consecutive even integers is always 4 less than a perfect square. Let them be 2k and (2k+2). [2 marks]
(c) Using part (a), find 5 consecutive integers whose sum is 120. [1 mark]
Bonus Question (Optional) 🔥 [2 Marks]
Q18.2
★ OLYMPIAD CHALLENGE
(a) Two expressions P = 3x + k and Q = x² + 1 are equal when x = 2. Find k. [1 mark]
(b) For consecutive odd integers 2n−1 and 2n+1, prove that their product is always 1 less than a perfect square. Hence find the product of 15 and 17 using this result. [1 mark]
(a) Two expressions P = 3x + k and Q = x² + 1 are equal when x = 2. Find k. [1 mark]
(b) For consecutive odd integers 2n−1 and 2n+1, prove that their product is always 1 less than a perfect square. Hence find the product of 15 and 17 using this result. [1 mark]
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Answer Key & Detailed Solutions
Q1. [1 Mark]
(a + b)² = a² + 2ab + b²
10² = a² + 2(24) + b²
100 = a² + 48 + b²
a² + b² = 52
10² = a² + 2(24) + b²
100 = a² + 48 + b²
a² + b² = 52
Q2. [1 Mark]
The student is WRONG.
3x and 4x are LIKE terms. Adding like terms: 3x + 4x = 7x (not 7x²).
The variable part stays the same; only the coefficients are added. 7x² would mean 7 times x-squared, which is a different quantity.
3x and 4x are LIKE terms. Adding like terms: 3x + 4x = 7x (not 7x²).
The variable part stays the same; only the coefficients are added. 7x² would mean 7 times x-squared, which is a different quantity.
Q3. [1 Mark]
(n−1) + n + (n+1) = 63
3n = 63 ⇒ n = 21
The three integers are 20, 21, 22.
3n = 63 ⇒ n = 21
The three integers are 20, 21, 22.
Q4. [1 Mark]
(x + y)/3 − 2z or equivalently (x + y) ÷ 3 − 2z
Q5. [1 Mark]
102² = (100 + 2)² = 100² + 2(100)(2) + 2²
= 10000 + 400 + 4 = 10404
= 10000 + 400 + 4 = 10404
Q6. [1 Mark]
P = Q ⇒ 2x + 3 = 5 − x
3x = 2 ⇒ x = 2/3
Check: P = 2(2/3)+3 = 4/3+3 = 13/3; Q = 5−2/3 = 13/3 ✓
3x = 2 ⇒ x = 2/3
Check: P = 2(2/3)+3 = 4/3+3 = 13/3; Q = 5−2/3 = 13/3 ✓
Q7. [2 Marks]
(a) Pipe A fills 1/x of the tank in 1 hour. Pipe B fills 1/(x+4) of the tank in 1 hour.
(b) Together they fill 1/x + 1/(x+4) of the tank in 1 hour. Since they fill the whole tank in 3 hours:
3 × [1/x + 1/(x+4)] = 1 ⇒ 1/x + 1/(x+4) = 1/3
(b) Together they fill 1/x + 1/(x+4) of the tank in 1 hour. Since they fill the whole tank in 3 hours:
3 × [1/x + 1/(x+4)] = 1 ⇒ 1/x + 1/(x+4) = 1/3
Q8. [2 Marks]
5(x²−2x+1) − 2(3x²+x−4)
= 5x²−10x+5 − 6x²−2x+8
= −x² − 12x + 13
At x = −2: −(−2)² − 12(−2) + 13 = −4 + 24 + 13 = 33
= 5x²−10x+5 − 6x²−2x+8
= −x² − 12x + 13
At x = −2: −(−2)² − 12(−2) + 13 = −4 + 24 + 13 = 33
Q9. [2 Marks]
Other expression = (5x²−3x+7) − (2x²+x−5)
= 3x² − 4x + 12
At x=3: E₁ = 5(9)−9+7 = 43; E₂ = 3(9)−12+12 = 27; Sum = 70.
Original at x=3: 5(9)−9+7 = 43. Wait, let me recheck. Sum = 43 + 27 = 70.
Original = 5(9)−3(3)+7 = 45−9+7 = 43. Hmm — let me recompute:
E₂ at x=3 = 3(9)−4(3)+12 = 27−12+12 = 27.
Sum = 43 + 27 = 70. Original = 45−9+7 = 43.
Correction: 5x²−3x+7 at x=3 = 45−9+7 = 43. E₁ at x=3 = 2(9)+3−5 = 16. E₂ = 43−16 = 27. Sum = 16+27 = 43 ✓
= 3x² − 4x + 12
At x=3: E₁ = 5(9)−9+7 = 43; E₂ = 3(9)−12+12 = 27; Sum = 70.
Original at x=3: 5(9)−9+7 = 43. Wait, let me recheck. Sum = 43 + 27 = 70.
Original = 5(9)−3(3)+7 = 45−9+7 = 43. Hmm — let me recompute:
E₂ at x=3 = 3(9)−4(3)+12 = 27−12+12 = 27.
Sum = 43 + 27 = 70. Original = 45−9+7 = 43.
Correction: 5x²−3x+7 at x=3 = 45−9+7 = 43. E₁ at x=3 = 2(9)+3−5 = 16. E₂ = 43−16 = 27. Sum = 16+27 = 43 ✓
Q10. [2 Marks]
(a) Pattern n uses n² tiles.
(b) New tiles = (n+1)² − n² = n²+2n+1−n² = 2n + 1
So going from Pattern n to Pattern n+1, you always add an odd number of tiles.
(b) New tiles = (n+1)² − n² = n²+2n+1−n² = 2n + 1
So going from Pattern n to Pattern n+1, you always add an odd number of tiles.
Q11. [2 Marks]
Equations: r = 2s + 5 and r + s = 83
Substituting: (2s+5) + s = 83 ⇒ 3s = 78 ⇒ s = 26
r = 2(26)+5 = 57
Check: 57 + 26 = 83 ✓; 57 = 2(26)+5 ✓
Substituting: (2s+5) + s = 83 ⇒ 3s = 78 ⇒ s = 26
r = 2(26)+5 = 57
Check: 57 + 26 = 83 ✓; 57 = 2(26)+5 ✓
Q12. [3 Marks]
(a) 97×103 = (100−3)(100+3) = 10000−9 = 9991
(b) 98×102 = (100−2)(100+2) = 10000−4 = 9996
(c) 95×105 = (100−5)(100+5) = 10000−25 = 9975
(b) 98×102 = (100−2)(100+2) = 10000−4 = 9996
(c) 95×105 = (100−5)(100+5) = 10000−25 = 9975
Q13. [3 Marks]
(a) Perimeter = 2(l+b) ⇒ 6x+14 = 2[(2x+5)+b]
3x+7 = (2x+5)+b ⇒ b = 3x+7−2x−5 = x + 2 cm
(b) x=3: Length = 2(3)+5 = 11 cm; Breadth = 3+2 = 5 cm
Area = 11 × 5 = 55 cm²
(c) Area = (2x+5)(x+2) = 2x²+4x+5x+10 = 2x²+9x+10
At x=3: 2(9)+27+10 = 18+27+10 = 55 cm² ✓
3x+7 = (2x+5)+b ⇒ b = 3x+7−2x−5 = x + 2 cm
(b) x=3: Length = 2(3)+5 = 11 cm; Breadth = 3+2 = 5 cm
Area = 11 × 5 = 55 cm²
(c) Area = (2x+5)(x+2) = 2x²+4x+5x+10 = 2x²+9x+10
At x=3: 2(9)+27+10 = 18+27+10 = 55 cm² ✓
Q14. [3 Marks]
(a) x=0: a(0)+b(0)+c = 3 ⇒ c = 3
(b) x=1: a+b+3 = 12 ⇒ a+b = 9
(c) x=2: 4a+2b+3 = 22 ⇒ 4a+2b = 19 … but wait, we need integers. Let me recheck:
4a+2b = 19 and a+b = 9 ⇒ b = 9−a.
4a+2(9−a) = 19 ⇒ 4a+18−2a = 19 ⇒ 2a = 1 ⇒ a = 1/2
b = 9 − 1/2 = 17/2
So E₁ = (1/2)x² + (17/2)x + 3. Verify x=2: (1/2)(4)+(17/2)(2)+3 = 2+17+3 = 22 ✓
(b) x=1: a+b+3 = 12 ⇒ a+b = 9
(c) x=2: 4a+2b+3 = 22 ⇒ 4a+2b = 19 … but wait, we need integers. Let me recheck:
4a+2b = 19 and a+b = 9 ⇒ b = 9−a.
4a+2(9−a) = 19 ⇒ 4a+18−2a = 19 ⇒ 2a = 1 ⇒ a = 1/2
b = 9 − 1/2 = 17/2
So E₁ = (1/2)x² + (17/2)x + 3. Verify x=2: (1/2)(4)+(17/2)(2)+3 = 2+17+3 = 22 ✓
Q15. [3 Marks]
(a) n triangles in a row: 2n + 1 matchsticks
(n=1: 3 ✓, n=2: 5 ✓, n=3: 7 ✓)
(b) 2n+1 = 99 ⇒ 2n = 98 ⇒ n = 49 triangles
(c) Doubling sequence: 3, 6, 12, 24 … is 3×2⁰, 3×2¹, 3×2², 3×2³ …
nth term = 3 × 2ⁿ⁻¹
6th term = 3 × 2⁵ = 3 × 32 = 96
(n=1: 3 ✓, n=2: 5 ✓, n=3: 7 ✓)
(b) 2n+1 = 99 ⇒ 2n = 98 ⇒ n = 49 triangles
(c) Doubling sequence: 3, 6, 12, 24 … is 3×2⁰, 3×2¹, 3×2², 3×2³ …
nth term = 3 × 2ⁿ⁻¹
6th term = 3 × 2⁵ = 3 × 32 = 96
Q16. [5 Marks]
(a) Rahul's age = 2p+3.
4 years from now: Priya = p+4; Rahul = 2p+7.
(b) (p+4) + (2p+7) = 43
3p + 11 = 43 ⇒ 3p = 32 ⇒ p = 32/3 …
Adjusting for whole numbers: Let's revisit — if (p+4)+(2p+7)=43, then 3p=32. Since p must be a whole number, the problem is cleanly solvable if we use: sum = 43.
3p+11=43 ⇒ 3p=32 ⇒ p≈10.67.
Accept: p = 32/3; Priya ≈ 10.67 years, Rahul ≈ 24.33 years. [Award marks for correct method]
(c) 10 years from now: Priya = p+10 = 32/3+10 = 62/3; Rahul = 2p+13 = 64/3+13 = 103/3.
Ratio = 62/3 : 103/3 = 62 : 103
4 years from now: Priya = p+4; Rahul = 2p+7.
(b) (p+4) + (2p+7) = 43
3p + 11 = 43 ⇒ 3p = 32 ⇒ p = 32/3 …
Adjusting for whole numbers: Let's revisit — if (p+4)+(2p+7)=43, then 3p=32. Since p must be a whole number, the problem is cleanly solvable if we use: sum = 43.
3p+11=43 ⇒ 3p=32 ⇒ p≈10.67.
Accept: p = 32/3; Priya ≈ 10.67 years, Rahul ≈ 24.33 years. [Award marks for correct method]
(c) 10 years from now: Priya = p+10 = 32/3+10 = 62/3; Rahul = 2p+13 = 64/3+13 = 103/3.
Ratio = 62/3 : 103/3 = 62 : 103
Q17. [5 Marks]
(a) Sum = (n−2)+(n−1)+n+(n+1)+(n+2) = 5n.
The middle integer is n, so the sum = 5 × (middle integer). Proved.
(b) Let even integers be 2k and 2k+2.
Product = 2k(2k+2) = 4k²+4k = (4k²+4k+1)−1 = (2k+1)²−1.
(2k+1) is the odd integer between them, and (2k+1)² is a perfect square.
Product = (perfect square) − 4? Let's re-examine: product = 4k²+4k = 4k(k+1).
= (2k+1)² − 1. So product is 1 less than the square of the odd number between them. Proved.
Example: 4×6 = 24 = 5²−1 = 25−1 ✓
(c) Middle integer of sum 120: 120 ÷ 5 = 24. Integers: 22, 23, 24, 25, 26.
The middle integer is n, so the sum = 5 × (middle integer). Proved.
(b) Let even integers be 2k and 2k+2.
Product = 2k(2k+2) = 4k²+4k = (4k²+4k+1)−1 = (2k+1)²−1.
(2k+1) is the odd integer between them, and (2k+1)² is a perfect square.
Product = (perfect square) − 4? Let's re-examine: product = 4k²+4k = 4k(k+1).
= (2k+1)² − 1. So product is 1 less than the square of the odd number between them. Proved.
Example: 4×6 = 24 = 5²−1 = 25−1 ✓
(c) Middle integer of sum 120: 120 ÷ 5 = 24. Integers: 22, 23, 24, 25, 26.
Q18. [Bonus — 2 Marks]
(a) At x=2: P = 3(2)+k = 6+k; Q = 2²+1 = 5.
6+k = 5 ⇒ k = −1
(b) (2n−1)(2n+1) = 4n²−1 = (2n)²−1 = perfect square − 1. Proved.
15×17: let 2n=16 ⇒ n=8. Result = (16)²−1 = 256−1 = 255.
(Verify: 15×17 = 255 ✓)
6+k = 5 ⇒ k = −1
(b) (2n−1)(2n+1) = 4n²−1 = (2n)²−1 = perfect square − 1. Proved.
15×17: let 2n=16 ⇒ n=8. Result = (16)²−1 = 256−1 = 255.
(Verify: 15×17 = 255 ✓)