Q. 1 (A) Choose the correct alternative from given: (4 Marks)
(i) \( \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} \). Write the degree of the given determinant.
- (A) 1
- (B) 2
- (C) 3
- (D) 4
Solution:
(B) 2
(B) 2
Note: The determinant is of order \(2 \times 2\).
(ii) From the following equations which one is the quadratic equation?
- (A) \( \frac{5}{x} - 3 = x^2 \)
- (B) \( x(x+5) = 2 \)
- (C) \( n - 1 = 2n \)
- (D) \( \frac{1}{x^2}(x+2) = x \)
Solution:
(B) \( x(x+5) = 2 \)
Explanation: \( x(x+5) = 2 \Rightarrow x^2 + 5x - 2 = 0 \). This is in the form \(ax^2+bx+c=0\) with \(a \neq 0\).
(B) \( x(x+5) = 2 \)
Explanation: \( x(x+5) = 2 \Rightarrow x^2 + 5x - 2 = 0 \). This is in the form \(ax^2+bx+c=0\) with \(a \neq 0\).
(iii) Find the common difference of the following A.P.: 4, 4, 4, ...
- (A) 1
- (B) 8
- (C) 4
- (D) 0
Solution:
(D) 0
Explanation: \( d = t_{2} - t_{1} = 4 - 4 = 0 \).
(D) 0
Explanation: \( d = t_{2} - t_{1} = 4 - 4 = 0 \).
(iv) Which number cannot represent a probability?
- (A) \( \frac{2}{3} \)
- (B) \( \frac{15}{10} \)
- (C) 15%
- (D) 0.7
Solution:
(B) \( \frac{15}{10} \)
Explanation: Probability cannot be greater than 1. \( \frac{15}{10} = 1.5 > 1 \).
(B) \( \frac{15}{10} \)
Explanation: Probability cannot be greater than 1. \( \frac{15}{10} = 1.5 > 1 \).
Q. 1 (B) Solve the following subquestions: (4 Marks)
(i) If \( 2x + y = 7 \) and \( x + 2y = 11 \), then find the value of \( x + y \).
Solution:
$$ 2x + y = 7 \quad \text{...(I)} $$
$$ x + 2y = 11 \quad \text{...(II)} $$
Adding equations (I) and (II):
$$ (2x + x) + (y + 2y) = 7 + 11 $$
$$ 3x + 3y = 18 $$
Dividing both sides by 3:
$$ x + y = 6 $$
(ii) Find the first term of the given sequence: \( t_n = 3n - 4 \).
Solution:
Substitute \( n = 1 \) in the given formula:
$$ t_1 = 3(1) - 4 $$
$$ t_1 = 3 - 4 $$
$$ t_1 = -1 $$
The first term is -1.
(iii) How many alpha numerals are there in the format of GSTIN?
Solution:
There are 15 alpha numerals in a GSTIN.
(iv) Two coins are tossed simultaneously. Write the sample space S.
Solution:
Sample space \( S = \{ HH, HT, TH, TT \} \).
Q. 2 (A) Complete and write any two activities from the following: (4 Marks)
(i) Complete the following table to draw the graph of \( x + 2y = 4 \).
Solution:
Reasoning:
When \( x = -2 \): \( -2 + 2y = 4 \Rightarrow 2y = 6 \Rightarrow y = 3 \).
When \( y = 1 \): \( x + 2(1) = 4 \Rightarrow x + 2 = 4 \Rightarrow x = 2 \).
| x | -2 | 2 |
|---|---|---|
| y | 3 | 1 |
| (x, y) | (-2, 3) | (2, 1) |
When \( x = -2 \): \( -2 + 2y = 4 \Rightarrow 2y = 6 \Rightarrow y = 3 \).
When \( y = 1 \): \( x + 2(1) = 4 \Rightarrow x + 2 = 4 \Rightarrow x = 2 \).
(ii) Complete the following activity to form a quadratic equation.
Solution:
Activity:
I am a quadratic equation.
My standard form is \( ax^2 + bx + c = 0 \).
My roots are 5 and 12.
Sum of my roots \( 5 + 12 = 17 \).
Product of my roots \( 5 \times 12 = 60 \).
My quadratic equation is \( x^2 - 17x + 60 = 0 \).
(iii) Pushpmala has invested ₹ 24,000 and purchased share of FV ₹ 20 at a premium of ₹ 4. Complete the following activity to find the number of shares she purchased.
Solution:
FV = ₹ 20
Premium = ₹ 4
MV = FV + Premium
= 20 + 4
= ₹ 24
Number of shares = \( \frac{\text{Total investment}}{\text{MV}} \)
= \( \frac{24,000}{\fbox{24}} \)
= 1000 shares.
Q. 2 (B) Solve any four subquestions from the following: (8 Marks)
(i) Solve the following simultaneous equations: \( x + y = 3 \); \( 3x - 2y = 4 \)
Solution:
$$ x + y = 3 \quad \text{...(I)} $$
$$ 3x - 2y = 4 \quad \text{...(II)} $$
Multiply equation (I) by 2:
$$ 2x + 2y = 6 \quad \text{...(III)} $$
Add equations (II) and (III):
$$ 3x - 2y + 2x + 2y = 4 + 6 $$
$$ 5x = 10 \Rightarrow x = 2 $$
Substitute \( x = 2 \) in equation (I):
$$ 2 + y = 3 \Rightarrow y = 1 $$
Solution: \( (x, y) = (2, 1) \)
(ii) Solve the following quadratic equation by factorisation method: \( m^2 + 14m + 13 = 0 \)
Solution:
$$ m^2 + 14m + 13 = 0 $$
Split the middle term (14m) such that product is 13 and sum is 14:
$$ m^2 + 13m + 1m + 13 = 0 $$
$$ m(m + 13) + 1(m + 13) = 0 $$
$$ (m + 13)(m + 1) = 0 $$
$$ m + 13 = 0 \text{ or } m + 1 = 0 $$
\( m = -13 \text{ or } m = -1 \)
(iii) Find the 19th term of the following A.P.: 7, 13, 19, 25, ...
Solution:
Here, \( a = 7 \), \( d = 13 - 7 = 6 \).
Formula: \( t_n = a + (n-1)d \)
For \( n = 19 \):
$$ t_{19} = 7 + (19 - 1)6 $$
$$ t_{19} = 7 + 18 \times 6 $$
$$ t_{19} = 7 + 108 $$
$$ t_{19} = 115 $$
The 19th term is 115.
(iv) A share is sold for the market value of ₹ 2,000. Brokerage is paid at the rate of 0.5%. What is the amount received after the sale?
Solution:
MV = ₹ 2000, Brokerage Rate = 0.5%
$$ \text{Brokerage} = \text{MV} \times \frac{0.5}{100} $$
$$ = 2000 \times \frac{5}{1000} = 2 \times 5 = \text{₹} 10 $$
$$ \text{Amount Received} = \text{MV} - \text{Brokerage} $$
$$ = 2000 - 10 = \text{₹} 1990 $$
Amount received is ₹ 1,990.
(v) Find the mean time spent by the students for their studies.
Solution:
Mean \( \bar{X} = \frac{\sum f_i x_i}{\sum f_i} = \frac{300}{60} = 5 \).
| Class (Time in hours) | Class Mark (\(x_i\)) | No. of Students (\(f_i\)) | \(f_i x_i\) |
|---|---|---|---|
| 0-2 | 1 | 8 | 8 |
| 2-4 | 3 | 14 | 42 |
| 4-6 | 5 | 18 | 90 |
| 6-8 | 7 | 10 | 70 |
| 8-10 | 9 | 10 | 90 |
| Total | \(\sum f_i = 60\) | \(\sum f_i x_i = 300\) |
Mean time spent is 5 hours.
Q. 3 (A) Complete and write any one activity from the following: (3 Marks)
(i) Rate of return on investment (Shri Maniklal).
Solution:
Activity:
(a) Sum invested = M.V. × No. of shares
= 120 × 300
= ₹ 36,000
(b) Dividend per share = F.V. × rate of dividend
= 100 × \( \frac{7}{100} \)
= ₹ 7
Total dividend received = 300 × 7 = ₹ 2,100
(c) Rate of return = \( \frac{\text{Dividend income}}{\text{Sum invested}} \times 100 \)
= \( \frac{2,100}{36,000} \times 100 \)
= 5.83 %.
(ii) Two digit number probability activity (digits 2, 3, 5).
Solution:
Activity:
Let S be the sample space.
\( S = \{23, 25, 32, \fbox{35}, 52, 53\} \)
\( n(S) = \fbox{6} \)
Event A: The number so formed is an odd number.
\( A = \{23, 25, \fbox{35}, 53\} \)
\( n(A) = 4 \)
\( P(A) = \fbox{\( \frac{n(A)}{n(S)} \)} \) ... (Formula)
\( P(A) = \frac{\fbox{4}}{6} \)
\( P(A) = \frac{\fbox{2}}{3} \)
Q. 3 (B) Solve any two subquestions from the following: (6 Marks)
(i) Solve by Cramer's rule: \( 4x + 3y = 18 \); \( 3x - 2y = 5 \)
Solution:
$$ D = \begin{vmatrix} 4 & 3 \\ 3 & -2 \end{vmatrix} = (4)(-2) - (3)(3) = -8 - 9 = -17 $$
$$ D_x = \begin{vmatrix} 18 & 3 \\ 5 & -2 \end{vmatrix} = (18)(-2) - (3)(5) = -36 - 15 = -51 $$
$$ D_y = \begin{vmatrix} 4 & 18 \\ 3 & 5 \end{vmatrix} = (4)(5) - (18)(3) = 20 - 54 = -34 $$
$$ x = \frac{D_x}{D} = \frac{-51}{-17} = 3 $$
$$ y = \frac{D_y}{D} = \frac{-34}{-17} = 2 $$
Solution: \( x = 3, y = 2 \)
(ii) Solve using formula method: \( x^2 - 2x - 3 = 0 \)
Solution:
Comparing with \( ax^2 + bx + c = 0 \): \( a=1, b=-2, c=-3 \).
$$ b^2 - 4ac = (-2)^2 - 4(1)(-3) = 4 + 12 = 16 $$
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
$$ x = \frac{-(-2) \pm \sqrt{16}}{2(1)} $$
$$ x = \frac{2 \pm 4}{2} $$
$$ x = \frac{2+4}{2} = \frac{6}{2} = 3 \quad \text{or} \quad x = \frac{2-4}{2} = \frac{-2}{2} = -1 $$
Roots are 3 and -1.
(iii) A committee of two members is to be formed from three boys and two girls. Find probability.
Solution:
Let boys be \( B_1, B_2, B_3 \) and girls be \( G_1, G_2 \).
Sample space \( S = \{ B_1B_2, B_1B_3, B_2B_3, B_1G_1, B_1G_2, B_2G_1, B_2G_2, B_3G_1, B_3G_2, G_1G_2 \} \).
\( n(S) = 10 \).
Event A: At least one girl. \( A = \{ B_1G_1, B_1G_2, B_2G_1, B_2G_2, B_3G_1, B_3G_2, G_1G_2 \} \). \( n(A) = 7 \). \( P(A) = \frac{7}{10} \).
Event B: One boy and one girl. \( B = \{ B_1G_1, B_1G_2, B_2G_1, B_2G_2, B_3G_1, B_3G_2 \} \). \( n(B) = 6 \). \( P(B) = \frac{6}{10} = \frac{3}{5} \).
Event A: At least one girl. \( A = \{ B_1G_1, B_1G_2, B_2G_1, B_2G_2, B_3G_1, B_3G_2, G_1G_2 \} \). \( n(A) = 7 \). \( P(A) = \frac{7}{10} \).
Event B: One boy and one girl. \( B = \{ B_1G_1, B_1G_2, B_2G_1, B_2G_2, B_3G_1, B_3G_2 \} \). \( n(B) = 6 \). \( P(B) = \frac{6}{10} = \frac{3}{5} \).
(iv) Find the Median of the prices from the frequency distribution table.
Solution:
\( N = 400 \Rightarrow N/2 = 200 \).
This lies in the class 20-40.
\( L = 20, f = 100, cf = 140, h = 20 \).
$$ \text{Median} = L + \left( \frac{N/2 - cf}{f} \right) \times h $$
$$ = 20 + \left( \frac{200 - 140}{100} \right) \times 20 $$
$$ = 20 + \left( \frac{60}{100} \right) \times 20 $$
$$ = 20 + (0.6 \times 20) = 20 + 12 = 32 $$
| Class | Frequency (f) | Cumulative Freq (cf) |
|---|---|---|
| 0 - 20 | 140 | 140 |
| 20 - 40 | 100 | 240 |
| 40 - 60 | 80 | 320 |
| 60 - 80 | 60 | 380 |
| 80 - 100 | 20 | 400 |
Median price is ₹ 32.
Q. 4 Solve any two subquestions from the following: (8 Marks)
(i) Find the value of 'm' if the quadratic equation \( (m - 12) x^2 + 2(m - 12) x + 2 = 0 \) has real and equal roots.
Solution:
Here \( a = m-12 \), \( b = 2(m-12) \), \( c = 2 \).
For real and equal roots, discriminant \( \Delta = 0 \).
$$ b^2 - 4ac = 0 $$
$$ [2(m-12)]^2 - 4(m-12)(2) = 0 $$
$$ 4(m-12)^2 - 8(m-12) = 0 $$
Divide by 4:
$$ (m-12)^2 - 2(m-12) = 0 $$
Take \( (m-12) \) common:
$$ (m-12)[ (m-12) - 2 ] = 0 $$
$$ (m-12)(m-14) = 0 $$
$$ m = 12 \text{ or } m = 14 $$
Since it is a quadratic equation, \( a \neq 0 \), so \( m-12 \neq 0 \Rightarrow m \neq 12 \).
Therefore, \( m = 14 \).
(ii) A farmer borrows ₹ 1,000 and agrees to repay with a total interest of ₹ 140, in 12 instalments. Each instalment being less than the preceding instalment by ₹ 10. Find first and last instalment.
Solution:
Total amount to repay \( S_n = 1000 + 140 = 1140 \).
Number of instalments \( n = 12 \).
The instalments form an A.P. with difference \( d = -10 \).
Formula: \( S_n = \frac{n}{2}[2a + (n-1)d] \)
$$ 1140 = \frac{12}{2}[2a + (11)(-10)] $$
$$ 1140 = 6[2a - 110] $$
$$ \frac{1140}{6} = 2a - 110 $$
$$ 190 = 2a - 110 $$
$$ 2a = 300 \Rightarrow a = 150 $$
First instalment \( a = 150 \).
Last instalment \( t_{12} = a + 11d = 150 + 11(-10) = 150 - 110 = 40 \).
First instalment: ₹ 150, Last instalment: ₹ 40.
(iii) Find the value of 'x' and draw histogram. Total students = 180.
Solution:
Sum of frequencies = 180.
$$ 25 + x + 30 + 2x + 65 = 180 $$
$$ 3x + 120 = 180 $$
$$ 3x = 60 \Rightarrow x = 20 $$
Frequencies are: 25, 20, 30, 40 (since 2x=40), 65.
Histogram Data:
- 0-10: 25
- 10-20: 20
- 20-30: 30
- 30-40: 40
- 40-50: 65
(Draw a histogram with Class Intervals on X-axis and Number of Students on Y-axis using the data above.)
Q. 5 Solve any one of the following subquestions: (3 Marks)
(i) Draw graphs of \( 2x = y + 2 \) and \( 4x + 3y = 24 \). Find area of triangle formed by these lines and X-axis.
Solution:
Equations:
1. \( 2x - y = 2 \) (Intercepts: \( (1,0), (0,-2) \))
2. \( 4x + 3y = 24 \) (Intercepts: \( (6,0), (0,8) \))
Intersection Point:
Solving the system, we get \( x = 3, y = 4 \). Vertex \( C(3,4) \).
Vertices of Triangle on X-axis:
Line 1 crosses X-axis at \( y=0 \Rightarrow 2x=2 \Rightarrow x=1 \). Point \( A(1,0) \).
Line 2 crosses X-axis at \( y=0 \Rightarrow 4x=24 \Rightarrow x=6 \). Point \( B(6,0) \).
Area Calculation:
Base \( AB = 6 - 1 = 5 \) units.
Height \( h = y\)-coordinate of intersection \( = 4 \) units.
$$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 4 = 10 \text{ sq. units}. $$
Equations:
1. \( 2x - y = 2 \) (Intercepts: \( (1,0), (0,-2) \))
2. \( 4x + 3y = 24 \) (Intercepts: \( (6,0), (0,8) \))
Intersection Point:
Solving the system, we get \( x = 3, y = 4 \). Vertex \( C(3,4) \).
Vertices of Triangle on X-axis:
Line 1 crosses X-axis at \( y=0 \Rightarrow 2x=2 \Rightarrow x=1 \). Point \( A(1,0) \).
Line 2 crosses X-axis at \( y=0 \Rightarrow 4x=24 \Rightarrow x=6 \). Point \( B(6,0) \).
Area Calculation:
Base \( AB = 6 - 1 = 5 \) units.
Height \( h = y\)-coordinate of intersection \( = 4 \) units.
$$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 4 = 10 \text{ sq. units}. $$
(ii) Pie-diagram questions on Blood Group.
Solution:
From the diagram: A = 20%, AB = 5%, B = 30%.
Remaining is Group O: \( 100 - (20+5+30) = 45\% \).
(a) Central Angles:
$$ \theta = \frac{\text{Percentage}}{100} \times 360^\circ $$
- Group A: \( 0.20 \times 360 = 72^\circ \)
- Group B: \( 0.30 \times 360 = 108^\circ \)
- Group AB: \( 0.05 \times 360 = 18^\circ \)
- Group O: \( 0.45 \times 360 = 162^\circ \)
Total persons = 2000.