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10th Algebra Question Paper Solution July 2024 Maharashtra Board

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SSC Class 10 Algebra Question Paper Solution - July 2024

Q. 1. (A) Choose the correct answer and write the alphabet of it in front of the subquestion number: (4 Marks)
(i) For simultaneous equations in variables x and y, $D_{x}=49$, $D_{y}=-63$, $D=7$, then what is the value of x?
Options: (A) 7, (B) -7, (C) 1/7, (D) -1/7
Solution:
By Cramer's Rule, $x = \frac{D_x}{D}$
$x = \frac{49}{7} = 7$
Answer: (A)
(ii) Which equation from the following is a quadratic equation?
Options:
(A) $\frac{5}{x}-3=x^{2}$ (Becomes cubic: $5-3x=x^3$)
(B) $x(x+5)=2$ (Becomes $x^2+5x-2=0$, degree is 2)
(C) $n-1=2n$ (Linear)
(D) $\frac{1}{x^{2}}(x+2)=x$ (Becomes $x+2=x^3$)
Answer: (B)
(iii) The sequence -10, -6, -2, 2, ...
Options:
(A) is an A.P., Reason $d=-16$
(B) is an A.P., Reason $d=4$
(C) is an A.P., Reason $d=-4$
(D) is not an A.P.
Solution: $d = t_2 - t_1 = -6 - (-10) = -6 + 10 = 4$.
Answer: (B)
(iv) Which number from the following cannot represent probability?
Options: (A) 2/3, (B) 1.5, (C) 15%, (D) 0.7
Solution: Probability ranges from 0 to 1. 1.5 is greater than 1.
Answer: (B)
Q. 1. (B) Solve the following subquestions: (4 Marks)
(i) Find the value of the following determinant: $\begin{vmatrix}5 & -2 \\ -3 & 1\end{vmatrix}$
Value = $(5 \times 1) - (-2 \times -3)$
$= 5 - (6)$
$= 5 - 6 = -1$
Value = -1
(ii) Find the first term and common difference for the following A.P.: 5, 1, -3, -7...
First term ($a$) = 5
Common difference ($d$) = $t_2 - t_1 = 1 - 5 = -4$
a = 5, d = -4
(iii) Face value of a share is Rs. 100 and premium is Rs. 65, then what is market value of that share?
Market Value (MV) = Face Value + Premium
MV = $100 + 65 = 165$
Market Value = Rs. 165
(iv) Write sample space if one die is thrown.
Sample Space $S = \{1, 2, 3, 4, 5, 6\}$
S = {1, 2, 3, 4, 5, 6}

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Q. 2. (A) Complete any two activities and rewrite it: (4 Marks)
(i) Complete the following table for drawing the graph of the equation: $x+2y=4$
x -2 0 2
y 3 2 1
(x, y) (-2, 3) (0, 2) (2, 1)
Calculation check: If x=-2, -2+2y=4 → 2y=6 → y=3. If y=1, x+2(1)=4 → x=2.
(ii) Determine the nature of roots of the following quadratic equation: $m^{2}+2m+9=0$
Comparing with $ax^2+bx+c=0$:
$a = 1, b=2, c=9$
$b^{2}-4ac = 2^{2} - 4 \times 1 \times$ 9
$= 4 - 36$
$b^{2}-4ac =$ -32
$b^{2}-4ac < 0$
Roots of quadratic equation are not real.
(iii) Smita has invested Rs. 12,000 and purchased share of FV Rs. 10 at a premium of Rs. 2. Find the number of shares she purchased.
$FV=10$, Premium = Rs. 2
$MV = FV + \text{Premium} =$ 10 $+ 2 = \text{Rs. } 12$
Number of shares = $\frac{\text{Total investment}}{MV}$
$= \frac{12,000}{\boxed{12}}$
$= \boxed{1000} \text{ shares.}$
Q. 2. (B) Solve the following subquestions (any four): (8 Marks)
(i) Solve the following simultaneous equations: $x+y=5$, $x-y=3$
$x+y=5$ ... (I)
$x-y=3$ ... (II)
Adding (I) and (II):
$2x = 8 \Rightarrow x = 4$
Substituting $x=4$ in (I):
$4 + y = 5 \Rightarrow y = 1$
Solution: (x, y) = (4, 1)
(ii) Find $k$ if $x=3$ is a root of quadratic equation $kx^{2}-10x+3=0$.
Substitute $x=3$ in the equation:
$k(3)^2 - 10(3) + 3 = 0$
$9k - 30 + 3 = 0$
$9k - 27 = 0$
$9k = 27 \Rightarrow k = 3$
k = 3
(iii) Find the 19th term of the following A.P.: 7, 13, 19, 25...
$a = 7, d = 13 - 7 = 6, n = 19$
Formula: $t_n = a + (n-1)d$
$t_{19} = 7 + (19-1)6$
$t_{19} = 7 + (18 \times 6)$
$t_{19} = 7 + 108 = 115$
The 19th term is 115.
(iv) The taxable value of a wrist watch belt is Rs. 586. Rate of GST is 18%, then what is the price of the belt for the customer?
Taxable Value = Rs. 586
GST Amount = $586 \times \frac{18}{100} = 586 \times 0.18 = 105.48$
Total Price = Taxable Value + GST
Total Price = $586 + 105.48 = 691.48$
Price for customer = Rs. 691.48
(v) The annual investments of a family are shown in a pie diagram. Shares: $60^\circ$, Immovable Property: $120^\circ$, Mutual Fund: $60^\circ$, Bank Deposit: $90^\circ$, Post: $30^\circ$. If investment in shares is Rs. 2,000, find total investment and amount invested in Post.
(a) Measure of arc for Shares = $\frac{\text{Investment in Shares}}{\text{Total Investment}} \times 360^\circ$
$60^\circ = \frac{2000}{\text{Total}} \times 360^\circ$
$\text{Total Investment} = \frac{2000 \times 360}{60} = 2000 \times 6 = 12,000$

(b) Investment in Post:
Angle for Post = $30^\circ$
$\text{Amount} = \frac{30}{360} \times 12,000 = \frac{1}{12} \times 12,000 = 1,000$
Total Investment = Rs. 12,000; Post Investment = Rs. 1,000
Q. 3. (A) Complete any one activity: (3 Marks)
(i) A share is sold for the market value of Rs. 1,000. Brokerage is paid at 0.1%. Find amount received.
Brokerage = MV $\times$ rate of brokerage
$= 1000 \times \frac{0.1}{\boxed{100}}$
$= 10 \times 0.1 =$ 1
Amount received = MV - Brokerage
$= 1000 -$ 1
$= \text{Rs. }$ 999
(ii) Numbers 2, 4, 6, 8, 10, 12 are on a die. Probability of getting a perfect square.
$S = \{2, 4, 6, 8, 10, 12\}$
$n(S) =$ 6
Event B: Perfect square. Only 4 is a perfect square ($2^2$).
$B = \{ \boxed{4} \}$
$n(B) =$ 1
$P(B) = \frac{n(B)}{n(S)} = \frac{\boxed{1}}{6}$
Q. 3. (B) Solve the following subquestions (any two): (6 Marks)
(i) Solve by Cramer's method: $4m-2n=-4; 4m+3n=16$.
$D = \begin{vmatrix}4 & -2 \\ 4 & 3\end{vmatrix} = (12) - (-8) = 20$
$D_m = \begin{vmatrix}-4 & -2 \\ 16 & 3\end{vmatrix} = (-12) - (-32) = 20$
$D_n = \begin{vmatrix}4 & -4 \\ 4 & 16\end{vmatrix} = (64) - (-16) = 80$
$m = \frac{D_m}{D} = \frac{20}{20} = 1$
$n = \frac{D_n}{D} = \frac{80}{20} = 4$
Solution: (m, n) = (1, 4)
(ii) Solve by formula method: $y^{2}+\frac{1}{3}y=2$
Multiply by 3: $3y^2 + y = 6 \Rightarrow 3y^2 + y - 6 = 0$
$a=3, b=1, c=-6$
$b^2 - 4ac = (1)^2 - 4(3)(-6) = 1 + 72 = 73$
$y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$y = \frac{-1 \pm \sqrt{73}}{6}$
Roots: $\frac{-1+\sqrt{73}}{6}, \frac{-1-\sqrt{73}}{6}$
(iii) Two dice are rolled. Write S, n(S), and events A and B.
$S = \{(1,1)...(6,6)\}$, $n(S) = 36$.
Event A: Sum is multiple of 5.
Sums can be 5 or 10.
$A = \{(1,4), (2,3), (3,2), (4,1), (4,6), (5,5), (6,4)\}$
$n(A) = 7$

Event B: Sum is 25.
Max sum is 12 (6+6). Sum 25 is impossible.
$B = \{ \}$ (Empty Set)
$n(B) = 0$
Q. 4. Solve the following subquestions (any two): (8 Marks)
(i) Find the mean of the toll by 'assumed mean' method.
Let Assumed Mean $A = 550$ (Class Mark of 500-600).
Class (Toll) Class Mark ($x_i$) $d_i = x_i - 550$ Freq ($f_i$) $f_i d_i$
300-400350-20080-16000
400-500450-100110-11000
500-60055001200
600-700650100707000
700-800750200408000
Total $\sum f_i = 420$ $\sum f_i d_i = -12000$
$\bar{d} = \frac{\sum f_i d_i}{\sum f_i} = \frac{-12000}{420} \approx -28.57$
Mean $\bar{X} = A + \bar{d} = 550 + (-28.57) = 521.43$
Mean Toll = Rs. 521.43
(ii) Manisha distributes 540 bananas. If 30 students more, each gets 3 less. Find number of students.
Let original students = $x$. Bananas per student = $540/x$.
New students = $x+30$. New bananas per student = $540/(x+30)$.
Condition: Old share - New share = 3
$\frac{540}{x} - \frac{540}{x+30} = 3$
Dividing by 3: $\frac{180}{x} - \frac{180}{x+30} = 1$
$180(x+30) - 180x = x(x+30)$
$180x + 5400 - 180x = x^2 + 30x$
$x^2 + 30x - 5400 = 0$
Factors of 5400 diff 30: 90 and 60.
$(x+90)(x-60) = 0$
$x = 60$ (Since students cannot be negative).
Number of students = 60
(iii) Simple Interest check for AP and calculation.
Principal = 2000, Rate = 10%.
Interest Year 1: $(2000 \times 10 \times 1)/100 = 200$
Interest Year 2: $(2000 \times 10 \times 2)/100 = 400$
Interest Year 3: 600...
Sequence: 200, 400, 600...
Difference $d = 200$. Constant. It is an A.P.
Amount after 10 years (Interest only asked? "Interest amount after 10 years"):
$t_{10} = a + 9d = 200 + 9(200) = 200 + 1800 = 2000$.
Interest after 10 years = Rs. 2,000
(iv) [Statistics - Mode] Prepare frequency table and find Mode.
Given: 0-20 (10%), 20-40 (20%), 40-60 (35%), 60-80 (20%), 80-100 (Remaining 30 students).
Total % used = $10+20+35+20 = 85\%$. Remaining = $15\%$.
If $15\% = 30$ students, then Total Students $N = (30/15) \times 100 = 200$.
Frequency Table:
0-20: $10\% \text{ of } 200 = 20$
20-40: $20\% \text{ of } 200 = 40$ ($f_0$)
40-60: $35\% \text{ of } 200 = 70$ ($f_1$ - Modal Class)
60-80: $20\% \text{ of } 200 = 40$ ($f_2$)
80-100: 30

Mode Calculation:
Modal Class: 40-60. $L=40, h=20, f_1=70, f_0=40, f_2=40$.
Mode $= L + [\frac{f_1-f_0}{2f_1-f_0-f_2}] \times h$
$= 40 + [\frac{70-40}{140-40-40}] \times 20$
$= 40 + [\frac{30}{60}] \times 20$
$= 40 + 10 = 50$
Mode = 50 marks
Q. 5. Solve any one subquestion: (3 Marks)
(i) Draw histogram of the following data:
Data given: 60-80 (4), 80-100 (12), 100-120 (16), 120-140 (8).
Draw histogram of the following data:Data given: 60-80 (4), 80-100 (12), 100-120 (16), 120-140 (8). Note: This requires a drawing. Plot 'Student IQ' on X-axis and 'Number of Students' on Y-axis. Bars should be adjacent with heights corresponding to frequencies 4, 12, 16, and 8 respectively.
(ii) Find length and breadth of rectangle using given figure.
Find length and breadth of rectangle using given figure Opposite sides of a rectangle are equal.
1) $2m + 3n + 5 = 5m + n$
$5m - 2m + n - 3n = 5$
$3m - 2n = 5$ ... (I)

2) $3m - 2n + 7 = m + n - 3$
$3m - m - 2n - n = -3 - 7$
$2m - 3n = -10$ ... (II)

Multiply (I) by 3 and (II) by 2:
$9m - 6n = 15$
$4m - 6n = -20$
Subtracting: $5m = 35 \Rightarrow m = 7$.

Substitute $m=7$ in (I):
$3(7) - 2n = 5$
$21 - 5 = 2n$
$16 = 2n \Rightarrow n = 8$.

Dimensions:
Length $= 5m + n = 5(7) + 8 = 35 + 8 = 43$.
Breadth $= m + n - 3 = 7 + 8 - 3 = 12$.
Length = 43 units, Breadth = 12 units
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