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Unit II: Algebra — Test 1

Concepts and applications · 11 questions · 28 practice marks · 60 minutes.
MCQs are scored automatically; written answers use step-wise self/teacher review.

Section A
Question 1 (1 Mark)
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Full question paper and step-wise model answers

Practice marks are not official examination weightage. Written answers require self- or teacher assessment.

Question 1 · 1 mark

Which is a polynomial in x?

  1. 1/x + 2
  2. √x + 1
  3. 3x² − 2x + 5
  4. x⁻²
Model answer and reasoning

(c) 3x² − 2x + 5

  1. A polynomial has non-negative integer exponents of its variable.
  2. 3x² − 2x + 5 has exponents 2, 1 and 0; the other options do not.

Question 2 · 1 mark

For y = −3x + 7, the slope and y-intercept are:

  1. 3 and 7
  2. −3 and 7
  3. 7 and −3
  4. −3 and −7
Model answer and reasoning

(b) −3 and 7

  1. Compare y = mx + c with y = −3x + 7.
  2. Each increase of 1 in x lowers y by 3; x = 0 gives y = 7.

Question 3 · 1 mark

For the sequence 4, 7, 10, …, which is an explicit rule for n ≥ 1?

  1. aₙ = 4n
  2. aₙ = 3n + 1
  3. aₙ = 3n − 1
  4. aₙ = 7n
Model answer and reasoning

(b) aₙ = 3n + 1

  1. The first term is 4 and the common difference is 3.
  2. aₙ = 4 + (n − 1)3 = 3n + 1.

Question 4 · 1 mark

The fifth term of the GP 3, 6, 12, … is:

  1. 24
  2. 48
  3. 96
  4. 30
Model answer and reasoning

(b) 48

  1. a = 3 and r = 2.
  2. a₅ = ar⁴ = 3 × 16 = 48.

Question 5 · 3 marks

A tank initially contains 120 litres and loses 5 litres per minute. Form a linear polynomial for the water remaining, state its degree and coefficients, and give the physically valid time interval.

Model answer and reasoning

W(t) = 120 − 5t; degree 1; coefficients −5 and 120; 0 ≤ t ≤ 24 minutes.

  1. Subtract 5t from the initial 120 litres.
  2. The non-zero highest power of t is 1; the coefficient of t is −5 and the constant term is 120.
  3. Require t ≥ 0 and W(t) ≥ 0; solve 120 − 5t ≥ 0.

Question 6 · 3 marks

A taxi charges ₹40 plus ₹12 per kilometre. Form an equation and a table for distances 0, 1, 2 and 3 km. Describe the graph, slope and intercept.

Model answer and reasoning

C = 12d + 40; costs ₹40, ₹52, ₹64, ₹76; slope 12 and intercept 40.

  1. The fixed charge is the constant term and the per-kilometre charge multiplies d.
  2. Substitute the four distances to form the table.
  3. Plot (0,40), (1,52), (2,64), (3,76). For continuously measured d ≥ 0 the graph is a ray on the line C = 12d + 40.

Question 7 · 4 marks

Expand (x + 3)² and factor x² + 6x + 9. Explain both using a square divided into an x-by-x square, two rectangles and a small square.

Model answer and reasoning

(x + 3)² = x² + 6x + 9; the factors are (x + 3)(x + 3).

  1. A square of side x + 3 has area (x + 3)².
  2. The pieces have areas x², 3x, 3x and 9.
  3. Their sum is x² + 6x + 9, establishing the identity for the geometric model x ≥ 0.
  4. Algebraic expansion extends the identity to all real x; reversing the expansion gives the factors.

Question 8 · 3 marks

An AP starts with 7 and has common difference −2. Give a recursive rule, an explicit rule, its tenth term, and explain how its points (n, aₙ) appear on a graph.

Model answer and reasoning

a₁ = 7, aₙ₊₁ = aₙ − 2; aₙ = 9 − 2n; a₁₀ = −11.

  1. The recursive rule subtracts 2 each time.
  2. aₙ = 7 + (n − 1)(−2) = 9 − 2n; n = 10 gives −11.
  3. Only positive integer indices are allowed. Plot separate collinear points with downward slope −2, not a continuous sequence curve.

Question 9 · 3 marks

Derive the sum of the first n natural numbers and use it to find 1 + 2 + … + 50.

Model answer and reasoning

n(n + 1)/2; the sum to 50 is 1275.

  1. Write S = 1 + 2 + … + n and the same sum in reverse.
  2. Adding corresponding terms gives 2S = n(n + 1).
  3. Divide by 2; for n = 50, S = 50 × 51/2 = 1275.

Question 10 · 4 marks

Solve x + y = 7 and y = 2x + 1 by substitution. State what the solution means on a graph.

Model answer and reasoning

x = 2, y = 5; the two lines meet at (2,5).

  1. Substitute y = 2x + 1 into x + y = 7.
  2. Then 3x + 1 = 7, giving x = 2.
  3. Use y = 2x + 1 to get y = 5; verify both equations.
  4. A common ordered-pair solution is the intersection of the two graphs.

Question 11 · 4 marks

A shop sells 2 notebooks and 3 pens for ₹90, and 3 notebooks and 2 pens for ₹110. Use elimination to find each price.

Model answer and reasoning

Notebook ₹30; pen ₹10.

  1. Let notebook price be n and pen price p. Then 2n + 3p = 90 and 3n + 2p = 110.
  2. Multiply the first equation by 3 and the second by 2: 6n + 9p = 270 and 6n + 4p = 220.
  3. Subtract to obtain 5p = 50, so p = 10.
  4. Substitute: 2n + 30 = 90, so n = 30. Check both purchase totals.