SSC 10th Algebra (Maths-1) Board Paper Solution March 2026 | Maharashtra Board

SSC 10th Algebra (Mathematics Part - I) March 2026 Board Paper Solution

Subject: Mathematics (71) Algebra - Part I (E) | Paper Code: N 919 | Time: 2 Hours | Max. Marks: 40

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Q. 1. (A) Choose the correct answer and write the alphabet: [4 Marks]

(i) Rate of GST on essential commodities is ............

• (A) 5%
• (B) 12%
• (C) 0%
• (D) 18%

Answer: (C) 0%

(ii) Out of the following equations which one is not a quadratic equation?

• (A) x² + 4x = 11 + x²
• (B) x² = 4x
• (C) 5x² = 90
• (D) 2x - x² = x² + 5

Answer: (A) x² + 4x = 11 + x²

Explanation: Simplifying x² + 4x = 11 + x² gives 4x - 11 = 0, which is a linear equation as the highest index of the variable x is 1.

(iii) To draw graph of 4x + 5y = 19, find the value of y when x = 1.

• (A) 4
• (B) 3
• (C) 2
• (D) -3

Answer: (B) 3

Explanation: Substituting x = 1 in 4x + 5y = 19:
4(1) + 5y = 19
4 + 5y = 19
5y = 19 - 4
5y = 15
y = 15 / 5 = 3

(iv) Sum of first five multiples of 3 is ............

• (A) 45
• (B) 55
• (C) 15
• (D) 75

Answer: (A) 45

Explanation: The first five multiples of 3 are 3, 6, 9, 12, and 15.
Sum = 3 + 6 + 9 + 12 + 15 = 45.

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Q. 1. (B) Solve the following subquestions: [4 Marks]

(i) On certain article if rate of CGST is 9%, then what is the rate of GST?

Solution:

Given: Rate of CGST = 9%

We know that, Rate of GST = 2 × Rate of CGST

Rate of GST = 2 × 9% = 18%

Ans: The rate of GST is 18%.

(ii) For simultaneous equations in variables x and y, if D_x = -42 and D = -3, then find the value of x.

Solution:

Given: D_x = -42 and D = -3

By Cramer's Rule,

x = D_x / D

x = -42 / -3

x = 14

Ans: The value of x is 14.

(iii) When two coins are tossed simultaneously, write the sample space 'S'.

Solution:

When two coins are tossed simultaneously, the sample space 'S' is:

S = {HH, HT, TH, TT}

(iv) Write the following quadratic equation in the form ax² + bx + c = 0: 2y = 10 - y²

Solution:

Given quadratic equation: 2y = 10 - y²

Rearranging the terms in standard form ax² + bx + c = 0:

y² + 2y - 10 = 0

Q. 2. (A) Complete and write any two activities from the following: [4 Marks]

(i) If one die is rolled, then find the probability of the event, that the number on the upper face is prime.

Solution:

'S' is the sample space.

S = {1, 2, 3, 4, 5, 6}

∴ n(S) = [ 6 ]

Event A : Prime number on the upper face.

A = {2, 3, 5}

∴ n(A) = [ 3 ]

∴ P(A) = n(A) / [ n(S) ]

∴ P(A) = [ 3 ] / 6

∴ P(A) = 1 / 2

(ii) Write the correct number in the given boxes from the given A.P.; 70, 60, 50, 40, .............

Solution:

Here t₁ = [ 70 ], t₂ = 60, t₃ = [ 50 ]

∴ a = [ 70 ], d = [ -10 ]

(iii) Complete the following table to draw graph of the equation: x + y = 3

Solution:

x	3	[ -2 ]
y	[ 0 ]	5
(x, y)	[ (3, 0) ]	[ (-2, 5) ]

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Q. 2. (B) Solve any four subquestions from the following: [8 Marks]

(i) If two coins are tossed simultaneously, find the probability of the event, "getting at least one head."

Solution:

Let 'S' be the sample space when two coins are tossed simultaneously.

S = {HH, HT, TH, TT}

∴ n(S) = 4

Let Event A be the event of getting at least one head.

A = {HH, HT, TH}

∴ n(A) = 3

P(A) = n(A) / n(S)

P(A) = 3 / 4

Ans: The probability of getting at least one head is 3/4.

(ii) Solve the following simultaneous equations: x + y = 4 ; 2x - y = 2

Solution:

x + y = 4 — (Equation 1)

2x - y = 2 — (Equation 2)

Adding Equation 1 and Equation 2:

(x + y) + (2x - y) = 4 + 2

3x = 6

x = 6 / 3

x = 2

Substitute x = 2 in Equation 1:

2 + y = 4

y = 4 - 2

y = 2

Ans: (x, y) = (2, 2) is the solution of the given simultaneous equations.

(iii) Solve the following quadratic equation by factorisation method: x² + x - 20 = 0

Solution:

x² + x - 20 = 0

x² + 5x - 4x - 20 = 0

x(x + 5) - 4(x + 5) = 0

(x + 5)(x - 4) = 0

∴ x + 5 = 0 or x - 4 = 0

∴ x = -5 or x = 4

Ans: The roots of the given quadratic equation are -5 and 4.

(iv) Given A.P. is 12, 16, 20, 24, ................. Find the 24th term of this progression.

Solution:

Given A.P.: 12, 16, 20, 24, ...

Here, first term (a) = 12

Common difference (d) = 16 - 12 = 4

We know that,

t_n = a + (n - 1)d

Substitute n = 24, a = 12, and d = 4:

t₂₄ = 12 + (24 - 1) × 4

t₂₄ = 12 + 23 × 4

t₂₄ = 12 + 92

t₂₄ = 104

Ans: The 24th term of the given progression is 104.

(v) If ∑x_if_i = 1265 and N = 50, then find the mean of the given data.

Solution:

Given: ∑x_if_i = 1265 and N = ∑f_i = 50

Mean (x̄) = ∑x_if_i / N

Mean (x̄) = 1265 / 50

Mean (x̄) = 25.3

Ans: The mean of the given data is 25.3.

Q. 3. (A) Complete and write any one activity from the following: [3 Marks]

(i) Shreekar bought a Laptop with 10% discount on printed price. The printed price of that Laptop was ₹ 50,000. 18% GST was charged on discounted price. Find the amount of CGST and SGST. What amount did Shreekar pay?

Solution:

Discount = 10% of 50,000 = [ ₹ 5,000 ]

Taxable value of Laptop = 50,000 - [ 5,000 ] = ₹ 45,000

Rate of GST = [ 18% ]

∴ Rate of CGST = 9%

CGST = 9% of 45,000

= [ 9 / 100 ] × 45,000

= [ ₹ 4,050 ]

SGST = ₹ 4,050

Total amount of Laptop = 45,000 + 4,050 + 4,050 = ₹ [ 53,100 ]

(ii) One of the roots of the quadratic equation 5m² + 2m + k = 0 is -7/5. Complete the following activity to find the value of 'k'.

Solution:

-7/5 is a root of quadratic equation 5m² + 2m + k = 0.

∴ put m = -7/5 in the equation.

5 × [ -7/5 ]² + 2 × [ -7/5 ] + k = 0

[ 49/5 ] + [ -14/5 ] + k = 0

[ 7 ] + k = 0

k = [ -7 ]

Q. 3. (B) Solve any two subquestions of the following: [6 Marks]

(i) Two numbers differ by 3. The sum of twice the smaller number and thrice the greater number is 19. Find the numbers.

Solution:

Let the greater number be x and the smaller number be y.

According to the first condition, the two numbers differ by 3:

x - y = 3 — (Equation 1)

According to the second condition, the sum of twice the smaller number and thrice the greater number is 19:

3x + 2y = 19 — (Equation 2)

Multiply Equation 1 by 2:

2x - 2y = 6 — (Equation 3)

Add Equation 2 and Equation 3:

(3x + 2y) + (2x - 2y) = 19 + 6

5x = 25

x = 25 / 5

x = 5

Substitute x = 5 in Equation 1:

5 - y = 3

y = 5 - 3

y = 2

Ans: The required numbers are 5 and 2.

(ii) Sachin invested some amount in a National Saving Certificate Scheme. In the first year he invested ₹ 5,000, in the second year ₹ 7,000, in the third year ₹ 9,000 and so on. Find the total amount that he invested in 12 years.

Solution:

The amounts invested by Sachin each year are ₹ 5,000, ₹ 7,000, ₹ 9,000, ...

Since the common difference between consecutive investments is constant (₹ 2,000), the sequence forms an Arithmetic Progression (A.P.).

Here:

• First term (a) = 5000
• Common difference (d) = 7000 - 5000 = 2000
• Number of years (n) = 12

We know that the sum of the first n terms of an A.P. is:

S_n = (n / 2) × [2a + (n - 1)d]

Substitute n = 12, a = 5000, and d = 2000:

S₁₂ = (12 / 2) × [2(5000) + (12 - 1) × 2000]

S₁₂ = 6 × [10000 + 11 × 2000]

S₁₂ = 6 × [10000 + 22000]

S₁₂ = 6 × 32000

S₁₂ = 1,92,000

Ans: The total amount invested by Sachin in 12 years is ₹ 1,92,000.

(iii) There are six cards in a box, each bearing a number from 0 to 5. Find the probability of each of the following events, that: (1) The number on the card drawn is a natural number. (2) The number on the card drawn is a whole number.

Solution:

Sample space 'S' for six cards bearing numbers from 0 to 5:

S = {0, 1, 2, 3, 4, 5}

∴ n(S) = 6

(1) Let Event A be the event that the number on the card drawn is a natural number.

Natural numbers in the sample space are 1, 2, 3, 4, 5.

A = {1, 2, 3, 4, 5}

∴ n(A) = 5

P(A) = n(A) / n(S)

P(A) = 5 / 6

(2) Let Event B be the event that the number on the card drawn is a whole number.

Whole numbers in the sample space are 0, 1, 2, 3, 4, 5.

B = {0, 1, 2, 3, 4, 5}

∴ n(B) = 6

P(B) = n(B) / n(S) = 6 / 6

P(B) = 1

Ans: (1) Probability that the card bears a natural number is 5/6.
(2) Probability that the card bears a whole number is 1.

(iv) Draw a histogram of the following data:

Height of Students (cm)	No. of Students
135—140	4
140—145	12
145—150	16
150—155	8

Solution (Steps to construct the Histogram on Graph Paper):

1. Since the given classes (135—140, 140—145, 145—150, 150—155) are already continuous, no class adjustment is required.
2. Draw two perpendicular axes: the horizontal axis (X-axis) and the vertical axis (Y-axis) intersecting at the origin O.
3. On the X-axis, represent the Height of Students (in cm). Use a krink mark (zig-zag line) between the origin O and 135 because the class intervals start from 135 instead of 0.
4. On the Y-axis, represent the No. of Students (Frequency).
5. Choose a suitable scale:
   • X-axis: 1 cm = 5 cm (Height)
   • Y-axis: 1 cm = 2 students
6. Draw adjacent rectangles over each class interval with heights corresponding to their respective frequencies:
   • For class 135—140: draw a rectangle of height 4 units.
   • For class 140—145: draw a rectangle of height 12 units.
   • For class 145—150: draw a rectangle of height 16 units.
   • For class 150—155: draw a rectangle of height 8 units.

Q. 4. Solve any two subquestions of the following: [8 Marks]

(i) The sum of two roots of a quadratic equation is 7 and sum of their cubes is 91, find the equation.

Solution:

Let α and β be the two roots of the required quadratic equation.

According to the given conditions:

α + β = 7 — (1)

α³ + β³ = 91 — (2)

We know the algebraic identity:

α³ + β³ = (α + β)³ - 3αβ(α + β)

Substitute the values from (1) and (2) into the identity:

91 = (7)³ - 3αβ(7)

91 = 343 - 21αβ

21αβ = 343 - 91

21αβ = 252

αβ = 252 / 21

αβ = 12

A quadratic equation having roots α and β is given by:

x² - (α + β)x + αβ = 0

Substitute α + β = 7 and αβ = 12:

x² - 7x + 12 = 0

Ans: The required quadratic equation is x² - 7x + 12 = 0.

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(ii) Smt. Tejaswini purchased 100 shares of FV ₹ 80 when the MV is ₹ 900. She paid brokerage at the rate of 0.5% and 18% GST on brokerage. Find the following:
(1) Net amount paid for 100 shares.
(2) Brokerage paid on sum invested.
(3) GST paid on brokerage.
(4) Total amount paid for 100 shares.

Solution:

Given:

• Number of shares = 100
• Face Value (FV) = ₹ 80
• Market Value (MV) = ₹ 900
• Rate of Brokerage = 0.5%
• Rate of GST on brokerage = 18%

(1) Net amount paid for 100 shares (Sum Invested):

Sum Invested = Number of shares × Market Value (MV)

Sum Invested = 100 × 900

Net amount paid for 100 shares = ₹ 90,000

(2) Brokerage paid on sum invested:

Brokerage = 0.5% of Sum Invested

Brokerage = (0.5 / 100) × 90,000

Brokerage = 0.5 × 900

Brokerage paid = ₹ 450

(3) GST paid on brokerage:

GST = 18% of Brokerage

GST = (18 / 100) × 450

GST = 18 × 4.5

GST paid on brokerage = ₹ 81

(4) Total amount paid for 100 shares:

Total amount paid = Sum Invested + Brokerage + GST

Total amount paid = 90,000 + 450 + 81

Total amount paid for 100 shares = ₹ 90,531

Ans:

1. Net amount paid for 100 shares = ₹ 90,000
2. Brokerage paid on sum invested = ₹ 450
3. GST paid on brokerage = ₹ 81
4. Total amount paid for 100 shares = ₹ 90,531

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(iii) The following table shows the funds collected by 105 students for flood affected people. Find the median of the data.

Fund (Rupees)	No. of Students
10—20	10
20—30	37
30—40	15
40—50	20
50—60	7
60—70	8
70—80	8

Solution:

Let us prepare the cumulative frequency table (less than type):

Class Interval (Fund in ₹)	Frequency (No. of Students) (f)	Cumulative Frequency (Less than type) (c.f.)
10—20	10	10
20—30	37	10 + 37 = 47 (c.f.)
30—40 (Median Class)	15 (f)	47 + 15 = 62
40—50	20	62 + 20 = 82
50—60	7	82 + 7 = 89
60—70	8	89 + 8 = 97
70—80	8	97 + 8 = 105
Total	N = ∑f = 105

Here, N = 105

∴ N / 2 = 105 / 2 = 52.5

The cumulative frequency just greater than or equal to 52.5 is 62, which corresponds to the class interval 30—40.

∴ Median Class = 30—40

• Lower boundary of the median class (L) = 30
• Frequency of the median class (f) = 15
• Cumulative frequency of the class preceding the median class (c.f.) = 47
• Class width (h) = 40 - 30 = 10

Formula for Median:

Median = L + [ (N/2 - c.f.) / f ] × h

Substitute the values into the formula:

Median = 30 + [ (52.5 - 47) / 15 ] × 10

Median = 30 + (5.5 / 15) × 10

Median = 30 + (55 / 15)

Median = 30 + (11 / 3)

Median = 30 + 3.67

Median = 33.67

Ans: The median of the funds collected is ₹ 33.67.

Q. 5. Solve any one subquestion of the following: [3 Marks]

(i) A survey was conducted in the school. Student using different modes of transport is given below:

Mode of Transport	Percentage of Students
Bicycle	35
Bus	a
Railway	2a
Rickshaw	20

Find the value of a and draw the Pie diagram.

Solution:

Step 1: Find the value of 'a'

The sum of all percentages is always 100.

35 + a + 2a + 20 = 100

55 + 3a = 100

3a = 100 - 55

3a = 45

a = 45 / 3

a = 15

• Percentage of students using Bus = a = 15%
• Percentage of students using Railway = 2a = 2 × 15 = 30%

Step 2: Calculate the measure of central angle (θ) for each mode of transport

Formula: Measure of Central Angle (θ) = (Percentage of Students / 100) × 360°

Mode of Transport	Percentage of Students	Measure of Central Angle (θ)
Bicycle	35%	(35 / 100) × 360° = 126°
Bus	15%	(15 / 100) × 360° = 54°
Railway	30%	(30 / 100) × 360° = 108°
Rickshaw	20%	(20 / 100) × 360° = 72°
Total	100%	360°

Step 3: Pie Diagram

Ans: The value of a = 15.

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(ii) There are some students in group A and group B. If one student goes from group A to group B, then the number of students in both groups is same. But one student goes from group B to group A, then the number of students in group A becomes twice of group B. Find the number of students in each group.

Solution:

Let the number of students in Group A be x and the number of students in Group B be y.

Case 1: If one student goes from Group A to Group B:

• Remaining students in Group A = x - 1
• New number of students in Group B = y + 1

According to the first condition, the number of students in both groups becomes the same:

x - 1 = y + 1

x - y = 1 + 1

x - y = 2 — (Equation 1)

Case 2: If one student goes from Group B to Group A:

• New number of students in Group A = x + 1
• Remaining students in Group B = y - 1

According to the second condition, the number of students in Group A becomes twice that of Group B:

x + 1 = 2(y - 1)

x + 1 = 2y - 2

x - 2y = -2 - 1

x - 2y = -3 — (Equation 2)

Subtracting Equation 2 from Equation 1:

(x - y) - (x - 2y) = 2 - (-3)

x - y - x + 2y = 2 + 3

y = 5

Substitute y = 5 in Equation 1:

x - 5 = 2

x = 2 + 5

x = 7

Ans: The number of students in Group A is 7 and the number of students in Group B is 5.