OMTEX AD 2

10th Algebra Question Paper July 2025 with Solutions - SSC Maharashtra Board

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Q.1 (A) Four alternative answers are given for every subquestion. Choose the correct alternative and write its alphabet with subquestion number.
(i) The value of determinant \(\begin{vmatrix} 5 & 3 \\ -7 & -4 \end{vmatrix}\) is:
  • (A) -1
  • (B) -41
  • (C) 41
  • (D) 1

Solution:
$$ D = (5 \times -4) - (3 \times -7) $$ $$ D = -20 - (-21) $$ $$ D = -20 + 21 = 1 $$ Answer: (D) 1

(ii) Out of the following equations which one is not a quadratic equation?
  • (A) \(x^2+4x=11+x^2\)
  • (B) \(x^2=4x\)
  • (C) \(5x^2=90\)
  • (D) \(2x-x^2=x^2+5\)

Solution:
Option (A): \(x^2+4x=11+x^2 \Rightarrow 4x=11\). This is a linear equation (degree 1), not quadratic. Answer: (A)

(iii) For given A.P. \(a=3.5\), \(d=0\), then \(t_2 =\) ?
  • (A) 0
  • (B) 3.5
  • (C) 7
  • (D) 10.5

Solution:
\(t_2 = a + d = 3.5 + 0 = 3.5\) Answer: (B) 3.5

(iv) From the following numbers which number cannot represent a probability?
  • (A) 0.6
  • (B) 2.0
  • (C) 0.15
  • (D) 0.75

Solution:
Probability range is \(0 \le P(A) \le 1\). The value 2.0 exceeds 1. Answer: (B) 2.0

Q.1 (B) Solve the following subquestions:
(i) For simultaneous equations in variables x and y, \(D_x=49, D_y=-63, D=7\), then find the value of x.

By Cramer's Rule: $$ x = \frac{D_x}{D} $$ $$ x = \frac{49}{7} = 7 $$ Ans: \(x = 7\)

(ii) Write the following quadratic equation in standard form: \(2y=10-y^2\)

Given: \(2y = 10 - y^2\)
Rearranging terms to \(ax^2 + bx + c = 0\):
Ans: \(y^2 + 2y - 10 = 0\)

(iii) Face value of one share is 100 and premium is 10. Find the market value of the share.

Market Value (MV) = Face Value (FV) + Premium
\(MV = 100 + 10 = 110\)
Ans: Market Value is ₹ 110.

(iv) Find the class mark of the class 6-10.

Class Mark = \(\frac{\text{Lower Limit} + \text{Upper Limit}}{2}\)
Class Mark = \(\frac{6 + 10}{2} = \frac{16}{2} = 8\)
Ans: 8

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Q.2 (A) Complete the following activities and rewrite it (any two):
(i) Complete the following activity to solve the quadratic equation \(x^2+8x-48=0\) by completing square method.

\(x^2 + 8x - 48 = 0\)
\(x^2 + 8x + 16 - \)16\( - 48 = 0\)
\((x+4)^2 - \)64\( = 0\)
\((x+4)^2 = 64\)
\(x+4 = \)8 or \(x+4 = -8\)
\(x = 4\) or \(x = \)-12

(ii) Courier service agent charged total 590 to courier a parcel... taxable value is 500 on which CGST is 45 and SGST is 45. Complete the activity to find rate of GST.

Total GST = CGST + SGST
= 45 + 45
= 90
Rate of GST = \(\frac{90}{500} \times \) 100
\(\therefore\) Rate of GST charged by agent is 18%

(iii) Complete the following activity to calculate the related central angles.
Items Percentage Measure of central angle
Food 40 \(\frac{40}{100}\times360^{\circ}=\) 144\(^{\circ}\)
Clothing 20 \(\frac{20}{100}\times360^{\circ}=\) 72\(^{\circ}\)
Education 30 \(\frac{30}{100}\times360^{\circ}=\) 108\(^{\circ}\)
Other 10 \(\frac{10}{100}\times360^{\circ}=\) 36\(^{\circ}\)
Total 100 \(360^{\circ}\)
Q.2 (B) Solve the following subquestions (any four):
(i) Find the values of \((x+y)\) and \((x-y)\): \(101x+99y=501\); \(99x+101y=499\)

Adding both equations:
\(200x + 200y = 1000 \Rightarrow x+y = 5\)
Subtracting eq(2) from eq(1):
\(2x - 2y = 2 \Rightarrow x-y = 1\)
Ans: \(x+y=5, x-y=1\)

(ii) Solve by factorisation method: \(x^2-15x+54=0\)

\(x^2 - 9x - 6x + 54 = 0\)
\(x(x-9) - 6(x-9) = 0\)
\((x-9)(x-6) = 0\)
\(x=9\) or \(x=6\)
Ans: \(x=9, 6\)

(iii) Which term of the following A.P. is 560? 2, 11, 20, 29...

\(a=2, d=9, t_n=560\)
\(t_n = a + (n-1)d\)
\(560 = 2 + (n-1)9\)
\(558 = 9(n-1)\)
\(62 = n-1 \Rightarrow n = 63\)
Ans: The 63rd term is 560.

(iv) Market value of a share is 200. If brokerage rate is 0.3%, find purchase value.

Brokerage = \(0.3\%\) of \(200 = 0.60\)
Purchase Value = MV + Brokerage
= \(200 + 0.60 = 200.60\)
Ans: ₹ 200.60

(v) If \(\sum f_id_i=10,000, \sum f_i=100\) and \(A=2000\), then find mean (\(\bar{X}\)).

\(\bar{d} = \frac{\sum f_id_i}{\sum f_i} = \frac{10000}{100} = 100\)
Mean \(\bar{X} = A + \bar{d}\)
\(\bar{X} = 2000 + 100 = 2100\)
Ans: Mean = 2100

Q.3 (A) Complete the following activities and rewrite it (any one):
(i) In an A.P. sum of three consecutive terms is 27 and product is 504. Find terms.

Let terms be \(a-d, a, a+d\)
\(a-d+a+a+d = \) 27
\(3a = 27 \Rightarrow a = \) 9
\((a-d) \times a \times (a+d) = 504\)
\((9^2 - d^2) \times 9 = 504\)
\(d^2 = 81 - 56 = 25 \Rightarrow d = \pm 5\)
For \(a=9, d=5\), terms = 4, 9, 14
For \(a=9, d=-5\), terms = 14, 9, 4

(ii) Probability activity (Red Card, Face Card).

\(n(S) = \) 52
Event A: Red Card
\(n(A) = \) 26
\(P(A) = \frac{26}{52} = \) \(\frac{1}{2}\)
Event B: Face Card
\(n(B) = \) 12
\(P(B) = \frac{12}{52} = \) \(\frac{3}{13}\)

Q.3 (B) Solve the following subquestions (any two):
(i) Draw histogram for investment made by 210 families.

[Graph Required]
Steps:
X-axis: Investment (Class Intervals: 10-20, 20-30, etc.)
Y-axis: Number of Families (Scale: 1cm = 10 families)
Bars drawn with heights: 30, 50, 60, 55, 15 respectively.

(ii) Shri Shivajirao purchased 150 shares of F.V. 100, for M.V. 120. Dividend 7%. Find Rate of Return (RoR).

Total Investment = \(150 \times 120 = 18,000\)
Dividend per share = \(7\%\) of \(100 = 7\)
Total Dividend = \(150 \times 7 = 1,050\)
Rate of Return = \(\frac{\text{Total Dividend}}{\text{Investment}} \times 100\)
\(RoR = \frac{1050}{18000} \times 100 = 5.83\%\)
Ans: 5.83%

(iii) Product of Shraddha's age 2 years ago and 3 years hence is 84. Find present age.

Let present age be \(x\).
\((x-2)(x+3) = 84\)
\(x^2 + x - 6 = 84\)
\(x^2 + x - 90 = 0\)
\((x+10)(x-9) = 0\)
\(x=9\) (Age cannot be negative)
Ans: 9 years.

(iv) Solve graphically: \(x+y=6, x-y=4\).

Line 1: \((0,6), (6,0)\)
Line 2: \((0,-4), (4,0)\)
Intersection Point: \((5,1)\)
Ans: \(x=5, y=1\)

Q.4 Solve the following subquestions (any two):
(i) Ratio of salary of skilled and unskilled workers is 5:4. Total salary of one day for both is 900. Find daily wages.

Let skilled wage = \(5x\), unskilled wage = \(4x\).
\(5x + 4x = 900\)
\(9x = 900 \Rightarrow x = 100\)
Skilled: \(5(100) = 500\)
Unskilled: \(4(100) = 400\)
Ans: Skilled: ₹500, Unskilled: ₹400.

(ii) Two dice thrown. Find probability of:

\(n(S) = 36\)
(a) Sum \(\ge\) 9: \(\{(3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)\}\). Count=10. \(P(A)=10/36 = 5/18\).
(b) Sum divisible by 5: \(\{(1,4),(2,3),(3,2),(4,1),(4,6),(5,5),(6,4)\}\). Count=7. \(P(B)=7/36\).
(c) First > Second: 15 outcomes. \(P(C)=15/36=5/12\).

(iii) Find the median age of patients.

Total \(N=300\), \(N/2 = 150\).
Median Class (cf > 150): 30-40 (f=55, cf=102).
\(L=30, h=10, f=55, cf=102\)
Median \(= L + \left[ \frac{N/2 - cf}{f} \right] \times h\)
\(= 30 + \left[ \frac{150 - 102}{55} \right] \times 10\)
\(= 30 + \frac{480}{55} = 30 + 8.73 = 38.73\) years.

Q.5 Solve the following subquestions (any one):
(i) Write the quadratic equation satisfying the condition:

(a) \(\Delta = 0\) (Real & Equal): e.g., \(x^2 - 4x + 4 = 0\)
(b) \(\Delta > 0\) (Real & Unequal): e.g., \(x^2 - 5x + 6 = 0\)
(c) \(\Delta < 0\) (Not Real): e.g., \(x^2 + x + 1 = 0\)

(ii) If first term of A.P. is 'p', second is 'q', last is 'r', show sum is \(\frac{(p+r)(q+r-2p)}{2(q-p)}\).

Here, \(a = p\), \(d = q - p\), \(t_n = r\).
Using \(t_n = a + (n-1)d\):
\(r = p + (n-1)(q-p)\)
\(r - p = (n-1)(q-p)\)
\(n-1 = \frac{r-p}{q-p}\)
\(n = \frac{r-p}{q-p} + 1 = \frac{r-p+q-p}{q-p} = \frac{q+r-2p}{q-p}\)
Sum \(S_n = \frac{n}{2}(t_1 + t_n)\)
\(S_n = \frac{q+r-2p}{2(q-p)} \times (p+r)\)
\(S_n = \frac{(p+r)(q+r-2p)}{2(q-p)}\)
Hence Proved.

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