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Maharashtra SSC March 2026 Mathematics Part I (Algebra) Question Paper with Step-by-Step Solutions

AI-written · numbers calculator-checked on 23 Oct 2026

Step-by-step working for 27 questions of the board's own exam paper “MATHEMATICS (71) ALGEBRA – PART I (E), REVISED COURSE, paper N 919” (40 marks, 2 hours), written by Shishya. Each question is in our own shorter words; open the board's paper for the exact wording, figures and instructions.

The board's paper (PDF) ↗Where the board lists it (mahahsscboard.in) ↗

How to answer this paper: All five questions are compulsory, with choices inside them: Q.2 (A) any two activities, Q.2 (B) any four, Q.3 (A) any one, Q.3 (B) any two, Q.4 any two and Q.5 any one. A calculator is not allowed.

How these were checked: the solutions were drafted with AI from the board's English-medium paper (read on 23 Oct 2026). On 23 Oct 2026, every final number was recomputed from the question's own numbers by a calculator check that runs before each release. They were not compared with a board answer key: none was listed beside this paper on the board's page when Shishya read it. Graphs, constructions and diagrams are described step by step, not drawn. If something looks wrong, your teacher and the board's own answer key, where one exists, are the authority.

Maths in the Maharashtra SSC is two separate 40-mark papers: Part I (Algebra) and Part II (Geometry). This page solves Part I; Part II (Geometry) is solved on its own page.

Free to read: the first 5 solutions are open to everyone; the other 22 open with a free Shishya account (for students 13 and above).

Q.1 (A)

  1. Q1(A)(i)

    1 markFinancial planning (GST)

    The rate of GST on essential commodities.

    1. Essential commodities are nil-rated under GST: no GST is charged on them.

    Answer: 0% — option (C).

  2. Q1(A)(ii)

    1 markQuadratic equations

    Which equation is not a quadratic equation: x² + 4x = 11 + x², x² = 4x, 5x² = 90 or 2x − x² = x² + 5?

    1. Simplify each one and look at the highest power of x left.
    2. x² + 4x = 11 + x²: the x² terms cancel, leaving 4x = 11 — degree 1, a linear equation.
    3. The other three keep an x² term: x² − 4x = 0, 5x² − 90 = 0 and 2x² − 2x + 5 = 0.

    Answer: x² + 4x = 11 + x² — option (A).

  3. Q1(A)(iii)

    1 markLinear equations in two variables

    For the graph of 4x + 5y = 19, find y when x = 1.

    1. Put x = 1: 4 + 5y = 19.
    2. 5y = 15, so y = 3.

    Answer: y = 3 — option (B).

  4. Q1(A)(iv)

    1 markArithmetic progression

    The sum of the first five multiples of 3.

    1. The first five multiples of 3 are 3, 6, 9, 12, 15.
    2. Sum = 3 + 6 + 9 + 12 + 15 = 45 (or 5/2 × (3 + 15) = 45).

    Answer: 45 — option (A).

Q.1 (B)

  1. Q1(B)(i)

    1 markFinancial planning (GST)

    If the CGST rate on an article is 9%, what is the GST rate?

    1. GST = CGST + SGST, and the two are equal.
    2. GST = 9% + 9% = 18%.

    Answer: 18%

  2. Q1(B)(ii)

    1 markLinear equations in two variables

    For simultaneous equations in x and y, Dx = −42 and D = −3. Find x.

    1. By Cramer's rule, x = Dx / D.
    2. x = (−42) / (−3) = 14.

    Answer: x = 14

    Steps and answer: open with a free account

  3. Q1(B)(iii)

    1 markProbability

    Two coins are tossed together. Write the sample space S.

    1. Each coin shows H or T, so there are 2 × 2 = 4 outcomes.

    Answer: S = {HH, HT, TH, TT}

    Steps and answer: open with a free account

  4. Q1(B)(iv)

    1 markQuadratic equations

    Write 2y = 10 − y² in the form ay² + by + c = 0.

    1. Bring every term to the left side: y² + 2y − 10 = 0.

    Answer: y² + 2y − 10 = 0 (a = 1, b = 2, c = −10)

    Steps and answer: open with a free account

Q.2 (A)

  1. Q2(A)(i)

    2 marksProbability

    Activity: one die is rolled. Find the probability that the number on the upper face is prime.

    1. S = {1, 2, 3, 4, 5, 6}, so n(S) = 6.
    2. Event A (prime number) = {2, 3, 5}, so n(A) = 3.
    3. P(A) = n(A) / n(S) = 3/6 = 1/2.

    Answer: Boxes: n(S) = 6, n(A) = 3, P(A) = 3/6 = 1/2

    Steps and answer: open with a free account

  2. Q2(A)(ii)

    2 marksArithmetic progression

    Activity: fill the boxes for the A.P. 70, 60, 50, 40, …: t₁, t₃, a and d.

    1. t₁ is the first term: 70. t₃ is the third term: 50.
    2. a = t₁ = 70.
    3. d = t₂ − t₁ = 60 − 70 = −10.

    Answer: t₁ = 70, t₃ = 50, a = 70, d = −10

    Steps and answer: open with a free account

  3. Q2(A)(iii)

    2 marksLinear equations in two variables

    Activity: complete the table for x + y = 3 — x = 3 in one column and y = 5 in the other.

    1. Column 1: x = 3, so y = 3 − 3 = 0, giving the point (3, 0).
    2. Column 2: y = 5, so x = 3 − 5 = −2, giving the point (−2, 5).

    Answer: y = 0 and (3, 0); x = −2 and (−2, 5)

    Steps and answer: open with a free account

Q.2 (B)

  1. Q2(B)(i)

    2 marksProbability

    Two coins are tossed together. Find the probability of getting at least one head.

    1. S = {HH, HT, TH, TT}, n(S) = 4.
    2. A (at least one head) = {HH, HT, TH}, n(A) = 3.
    3. P(A) = 3/4.

    Answer: P(A) = 3/4

    Steps and answer: open with a free account

  2. Q2(B)(ii)

    2 marksLinear equations in two variables

    Solve x + y = 4 and 2x − y = 2.

    1. Add the two equations: 3x = 6, so x = 2.
    2. Put x = 2 in x + y = 4: y = 2.
    3. Check: 2(2) − 2 = 2 ✓.

    Answer: x = 2, y = 2

    Steps and answer: open with a free account

  3. Q2(B)(iii)

    2 marksQuadratic equations

    Solve x² + x − 20 = 0 by factorisation.

    1. Find two numbers with product −20 and sum 1: 5 and −4.
    2. x² + 5x − 4x − 20 = 0 → x(x + 5) − 4(x + 5) = 0 → (x + 5)(x − 4) = 0.
    3. x + 5 = 0 or x − 4 = 0.

    Answer: x = −5 or x = 4

    Steps and answer: open with a free account

  4. Q2(B)(iv)

    2 marksArithmetic progression

    For the A.P. 12, 16, 20, 24, …, find the 24th term.

    1. a = 12, d = 16 − 12 = 4.
    2. tₙ = a + (n − 1)d, so t₂₄ = 12 + 23 × 4 = 12 + 92.

    Answer: t₂₄ = 104

    Steps and answer: open with a free account

  5. Q2(B)(v)

    2 marksStatistics

    If Σxᵢfᵢ = 1265 and N = 50, find the mean.

    1. Mean x̄ = Σxᵢfᵢ / N.
    2. x̄ = 1265 / 50 = 25.3.

    Answer: Mean = 25.3

    Steps and answer: open with a free account

Q.3 (A)

  1. Q3(A)(i)

    3 marksFinancial planning (GST)

    Activity: a laptop printed at ₹50,000 is sold at 10% discount; 18% GST is charged on the discounted price. Find CGST, SGST and the amount paid.

    1. Discount = 10% of 50,000 = ₹5,000.
    2. Taxable value = 50,000 − 5,000 = ₹45,000.
    3. GST rate = 18%, so CGST = SGST = 9%.
    4. CGST = 9/100 × 45,000 = ₹4,050, and SGST = ₹4,050.
    5. Total = 45,000 + 4,050 + 4,050 = ₹53,100.

    Answer: CGST = ₹4,050, SGST = ₹4,050; Shreekar paid ₹53,100

    Steps and answer: open with a free account

  2. Q3(A)(ii)

    3 marksQuadratic equations

    Activity: −7/5 is a root of 5m² + 2m + k = 0. Find k.

    1. Put m = −7/5: 5 × (−7/5)² + 2 × (−7/5) + k = 0.
    2. 5 × 49/25 = 49/5 and 2 × (−7/5) = −14/5, so 49/5 − 14/5 + k = 0.
    3. 35/5 + k = 0 → 7 + k = 0.

    Answer: k = −7

    Steps and answer: open with a free account

Q.3 (B)

  1. Q3(B)(i)

    3 marksLinear equations in two variables

    Two numbers differ by 3. Twice the smaller plus thrice the greater is 19. Find the numbers.

    1. Let the greater number be x and the smaller y: x − y = 3 … (1) and 2y + 3x = 19 … (2).
    2. From (1), x = y + 3. Put it in (2): 2y + 3(y + 3) = 19 → 5y + 9 = 19 → y = 2.
    3. x = 2 + 3 = 5. Check: 2(2) + 3(5) = 4 + 15 = 19 ✓.

    Answer: The numbers are 5 and 2

    Steps and answer: open with a free account

  2. Q3(B)(ii)

    3 marksArithmetic progression

    Sachin invests ₹5,000 in year 1, ₹7,000 in year 2, ₹9,000 in year 3 and so on. Find the total invested in 12 years.

    1. The yearly amounts form an A.P.: a = 5,000, d = 2,000, n = 12.
    2. Sₙ = n/2 [2a + (n − 1)d] = 12/2 [10,000 + 11 × 2,000].
    3. = 6 × [10,000 + 22,000] = 6 × 32,000 = 1,92,000.

    Answer: ₹1,92,000

    Steps and answer: open with a free account

  3. Q3(B)(iii)

    3 marksProbability

    Six cards numbered 0 to 5; one is drawn. Find the probability that the number is (1) a natural number, (2) a whole number.

    1. S = {0, 1, 2, 3, 4, 5}, n(S) = 6.
    2. (1) Natural numbers on the cards: {1, 2, 3, 4, 5}, 5 outcomes, so P = 5/6.
    3. (2) Every card is a whole number (0 is a whole number): 6 outcomes, so P = 6/6 = 1.

    Answer: (1) 5/6 (2) 1

    Steps and answer: open with a free account

  4. Q3(B)(iv)

    3 marksStatistics

    Draw a histogram: heights 135–140: 4, 140–145: 12, 145–150: 16, 150–155: 8 students.

    1. Take height (cm) on the X-axis and number of students on the Y-axis; mark a break (kink) near the origin on the X-axis, since the classes start at 135.
    2. Choose a scale, e.g. X-axis: 1 cm = 5 cm of height; Y-axis: 1 cm = 2 students.
    3. Draw touching rectangles on the class limits with heights 4, 12, 16 and 8.
    4. Write the scale in the top-right corner of the graph.

    Answer: Four touching bars of heights 4, 12, 16 and 8 on 135–140, 140–145, 145–150 and 150–155

    Steps and answer: open with a free account

Q.4

  1. Q4(i)

    4 marksQuadratic equations

    The sum of the two roots of a quadratic equation is 7 and the sum of their cubes is 91. Find the equation.

    1. Let the roots be α and β: α + β = 7 and α³ + β³ = 91.
    2. α³ + β³ = (α + β)³ − 3αβ(α + β), so 91 = 343 − 21αβ.
    3. 21αβ = 252, so αβ = 12.
    4. The equation is x² − (α + β)x + αβ = 0.

    Answer: x² − 7x + 12 = 0

    Steps and answer: open with a free account

  2. Q4(ii)

    4 marksFinancial planning (shares)

    100 shares of FV ₹80 bought at MV ₹900; brokerage 0.5% and 18% GST on brokerage. Find (1) the amount for 100 shares, (2) the brokerage, (3) the GST, (4) the total paid.

    1. (1) Shares are bought at market value: 100 × 900 = ₹90,000 (the sum invested, before brokerage).
    2. (2) Brokerage = 0.5% of 90,000 = 0.5/100 × 90,000 = ₹450.
    3. (3) GST = 18% of 450 = ₹81.
    4. (4) Total = 90,000 + 450 + 81 = ₹90,531.

    Answer: (1) ₹90,000 (2) ₹450 (3) ₹81 (4) ₹90,531

    Steps and answer: open with a free account

  3. Q4(iii)

    4 marksStatistics

    Find the median: funds 10–20: 10, 20–30: 37, 30–40: 15, 40–50: 20, 50–60: 7, 60–70: 8, 70–80: 8 students (105 in all).

    1. Cumulative frequencies: 10, 47, 62, 82, 89, 97, 105. N = 105, so N/2 = 52.5.
    2. The first cumulative frequency above 52.5 is 62, so the median class is 30–40.
    3. L = 30, cf (before the class) = 47, f = 15, h = 10.
    4. Median = L + [(N/2 − cf) / f] × h = 30 + (52.5 − 47)/15 × 10 = 30 + 55/15 = 30 + 3.67.

    Answer: Median ≈ ₹33.67 (exactly 101/3)

    Steps and answer: open with a free account

Q.5

  1. Q5(i)

    3 marksStatistics

    Transport: bicycle 35%, bus a%, railway 2a%, rickshaw 20%. Find a and draw the pie diagram.

    1. The percentages add to 100: 35 + a + 2a + 20 = 100 → 3a = 45 → a = 15.
    2. So bus = 15%, railway = 30%.
    3. Central angle = percentage × 360/100 = percentage × 3.6: bicycle 126°, bus 54°, railway 108°, rickshaw 72° (total 360°).
    4. Draw a circle, mark the angles one after another from a radius with a protractor, and name each sector.

    Answer: a = 15; angles: bicycle 126°, bus 54°, railway 108°, rickshaw 72°

    Steps and answer: open with a free account

  2. Q5(ii)

    3 marksLinear equations in two variables

    If one student moves from group A to B, both groups are equal; if one moves from B to A, A becomes twice B. Find the number in each group.

    1. Let A have x students and B have y.
    2. x − 1 = y + 1 → x − y = 2 … (1).
    3. x + 1 = 2(y − 1) → x − 2y = −3 … (2).
    4. (1) − (2): y = 5, then x = 7. Check: 7 + 1 = 8 = 2 × (5 − 1) ✓.

    Answer: Group A: 7 students, group B: 5 students

    Steps and answer: open with a free account

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