OMTEX AD 2

10th Maths Part 1 Algebra March 2024 Board Question Paper Solution

Question Paper Page No. 1 Question Paper Page No. 2 Question Paper Page No. 3 Question Paper Page No. 4 Question Paper Page No. 5 Question Paper Page No. 6 Question Paper Page No. 7 Question Paper Page No. 8 Question Paper Page No. 9 Question Paper Page No. 10 Question Paper Page No. 11 For all your study Materials Visit : omtexclasses.com

Q. 1. (A) Choose the correct alternative from given: (4 Marks)

(i) If 3 is one of the root of the quadratic equation \(kx^2 - 7x + 12 = 0\), then \(k =\) ___
(A) 1
(B) -1
(C) 3
(D) -3
Solution:
Put \(x = 3\) in the equation \(kx^2 - 7x + 12 = 0\).
\(k(3)^2 - 7(3) + 12 = 0\)
\(9k - 21 + 12 = 0\)
\(9k - 9 = 0\)
\(9k = 9 \Rightarrow k = 1\)
Answer: (A) 1
(ii) To draw the graph of \(x + 2y = 4\), find \(x\) when \(y = 1\):
(A) 1
(B) 2
(C) -2
(D) 6
Solution:
Substitute \(y = 1\) in \(x + 2y = 4\).
\(x + 2(1) = 4\)
\(x + 2 = 4\)
\(x = 4 - 2 = 2\)
Answer: (B) 2
(iii) For an A.P., \(t_7 = 4\), \(d = -4\), then \(a =\) ___
(A) 6
(B) 7
(C) 20
(D) 28
Solution:
Formula: \(t_n = a + (n-1)d\)
\(4 = a + (7-1)(-4)\)
\(4 = a + 6(-4)\)
\(4 = a - 24\)
\(a = 4 + 24 = 28\)
Answer: (D) 28
(iv) In the format of GSTIN, there are ___ alpha-numerals.
(A) 9
(B) 10
(C) 15
(D) 16
Solution:
GSTIN (Goods and Services Tax Identification Number) is a 15-digit code.
Answer: (C) 15

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

(i) If \(17x + 15y = 11\) and \(15x + 17y = 21\), then find the value of \(x - y\).
Solution:
Let \(17x + 15y = 11\) ... (I)
Let \(15x + 17y = 21\) ... (II)
Subtracting equation (II) from (I):
\(2x - 2y = -10\)
Dividing both sides by 2:
\(\mathbf{x - y = -5}\)
(ii) Find first term of the sequence \(t_n = 3n - 2\).
Solution:
To find the first term, put \(n = 1\).
\(t_1 = 3(1) - 2\)
\(t_1 = 3 - 2\)
\(\mathbf{t_1 = 1}\)
(iii) If the face value of a share is ₹100 and market value is 150. If rate of brokerage is 2%, find brokerage paid on one share.
Solution:
Brokerage is calculated on Market Value (MV).
Brokerage = Rate of Brokerage \(\times\) MV
\(= \frac{2}{100} \times 150\)
\(= 2 \times 1.5\)
\(\mathbf{= \text{ ₹ }3}\)
(iv) Two digit numbers are formed using digits 2, 3 and 5 without repeating a digit. Write the sample space.
Solution:
Digits: 2, 3, 5.
Sample Space \(S = \{23, 25, 32, 35, 52, 53\}\)

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Q. 2. (A) Complete the following activities and rewrite it (any two): (4 Marks)

(i) If (0, 2) is the solution of \(2x + 3y = k\), then to find the value of k, complete the following activity:
(0, 2) is the solution of the equation \(2x + 3y = k\).
Put \(x = \Box\) and \(y = \Box\) in the given equation;
\(2 \times \Box + 3 \times 2 = k\)
\(0 + 6 = k\)
\(k = \Box\)
Answer:
Put \(x = \class{box}{0}\) and \(y = \class{box}{2}\)
\(2 \times \class{box}{0} + 3 \times 2 = k\)
\(0 + 6 = k\)
\(k = \class{box}{6}\)
(ii) If 2 and 5 are the roots of the quadratic equation, then complete the following activity to form quadratic equation:
Let \(\alpha = 2\) and \(\beta = 5\) are the roots of the quadratic equation.
Then quadratic equation is:
\(x^2 - (\alpha + \beta)x + \alpha\beta = 0\)
\(x^2 - (2 + \Box)x + \Box \times 5 = 0\)
\(x^2 - \Box x + \Box = 0\)
Answer:
\(x^2 - (2 + \class{box}{5})x + \class{box}{2} \times 5 = 0\)
\(x^2 - \class{box}{7}x + \class{box}{10} = 0\)
(iii) Two coins are tossed simultaneously. Complete the following activity to write the sample space and the given events A and B in the set form:
Sample space is \(S = \{ \text{HT, TH,} \Box, \Box \}\)
Event A: To get at least one head.
\(A = \{ \text{HT, TH,} \Box \}\)
Event B: To get no head.
\(B = \{ \Box \}\)
Answer:
Sample space \(S = \{ \text{HT, TH,} \class{box}{\text{HH}}, \class{box}{\text{TT}} \}\)
Event A: \(A = \{ \text{HT, TH,} \class{box}{\text{HH}} \}\)
Event B: \(B = \{ \class{box}{\text{TT}} \}\)

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

(i) ABCD is a rectangle. Write two simultaneous equations using information given below in the diagram, in the form of \(ax+by=c\):
[Diagram Description: Rectangle ABCD. Side AB = \(2x + y + 8\), Side CD = \(4x - y\), Side AD = \(2y\), Side BC = \(x + 4\)]
Solution:
Opposite sides of a rectangle are equal.
Condition 1 (AB = CD):
\(2x + y + 8 = 4x - y\)
\(2x - 4x + y + y = -8\)
\(-2x + 2y = -8\)
Dividing by -2:
\(x - y = 4\) ... (I)

Condition 2 (AD = BC):
\(2y = x + 4\)
\(-x + 2y = 4\) ... (II) (or \(x - 2y = -4\))

The two simultaneous equations are:
1) \(x - y = 4\)
2) \(-x + 2y = 4\)
(ii) Solve the following quadratic equation using factorisation method: \(x^2 + x - 20 = 0\)
Solution:
\(x^2 + x - 20 = 0\)
Split the middle term (sum = +1, product = -20). Factors are +5 and -4.
\(x^2 + 5x - 4x - 20 = 0\)
\(x(x + 5) - 4(x + 5) = 0\)
\((x + 5)(x - 4) = 0\)
\(x + 5 = 0\) or \(x - 4 = 0\)
\(\mathbf{x = -5}\) or \(\mathbf{x = 4}\)
(iii) Find the 19th term of the following A.P.: 7, 13, 19, 25, ...
Solution:
Here, \(a = 7\), \(d = 13 - 7 = 6\), \(n = 19\).
Formula: \(t_n = a + (n-1)d\)
\(t_{19} = 7 + (19 - 1)6\)
\(t_{19} = 7 + 18(6)\)
\(t_{19} = 7 + 108\)
\(\mathbf{t_{19} = 115}\)
(iv) A card is drawn from well shuffled pack of 52 playing cards. Find the probability that the card drawn is a face card.
Solution:
Total sample space \(n(S) = 52\).
Let A be the event that the card is a face card.
Face cards are Jack, Queen, and King in 4 suits.
Total face cards \(n(A) = 3 \times 4 = 12\).
\(P(A) = \frac{n(A)}{n(S)}\)
\(P(A) = \frac{12}{52}\)
\(\mathbf{P(A) = \frac{3}{13}}\)
(v) The following table shows classification of number of workers and number of hours they work in software company. Prepare less than upper limit type cumulative frequency distribution table:
No. of hours dailyNo. of workers
8-10150
10-12500
12-14300
14-1650
Solution:
Class (Hours) Frequency (No. of Workers) Cumulative Frequency (Less than type)
8-10 150 150
10-12 500 150 + 500 = 650
12-14 300 650 + 300 = 950
14-16 50 950 + 50 = 1000

Q. 3. (A) Complete the following activity and rewrite it (any one): (3 Marks)

(i) The following frequency distribution table shows the classification of the number of vehicles and the volume of petrol filled in them. To find the mode of the volume of petrol filled, complete the following activity:
[Table shows Modal class 3.5-6.5 with freq 40. Previous freq 33, Next freq 27]

Activity:
From the given table, Modal class is 3.5-6.5.
\(Mode = \boxed{\Box} + \left[\frac{f_1 - f_0}{2f_1 - f_0 - \Box}\right] \times h\)
\(Mode = 3.5 + \left[\frac{40 - 33}{2(40) - 33 - 27}\right] \times \Box\)
\(Mode = 3.5 + \left[\frac{7}{80 - 60}\right] \times 3\)
\(Mode = \Box\)
The mode of the volume of petrol filled is \(\Box\).
Answer:
\(Mode = \class{box}{L} + \left[\frac{f_1 - f_0}{2f_1 - f_0 - \class{box}{f_2}}\right] \times h\)
\(Mode = 3.5 + \left[\frac{40 - 33}{2(40) - 33 - 27}\right] \times \class{box}{3}\)
\(Mode = 3.5 + \frac{7}{20} \times 3\)
\(Mode = 3.5 + 1.05\)
\(Mode = \class{box}{4.55}\)
The mode of the volume of petrol filled is 4.55 liters.
(ii) The total value (with GST) of remote controlled toy car is ₹2360. Rate of GST is 18% on toys. Complete the following activity to find the taxable value for the toy car:
\(2360 = \boxed{\Box} + \frac{\Box}{100} \times x\)
\(2360 = \frac{\Box}{100} \times x\)
\(2360 \times 100 = 118x\)
\(x = \frac{2360 \times 100}{\Box}\)
Taxable value for toy car is \(\Box\)
Answer:
\(2360 = \class{box}{x} + \frac{\class{box}{18}}{100} \times x\)
\(2360 = \frac{\class{box}{118}}{100} \times x\)
\(x = \frac{2360 \times 100}{\class{box}{118}}\)
\(x = 20 \times 100\)
Taxable value for toy car is \(\mathbf{\text{ ₹ } 2000}\)

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

(i) Solve the following quadratic equation by formula method: \(3m^2 - m - 10 = 0\)
Solution:
Compare \(3m^2 - m - 10 = 0\) with \(am^2 + bm + c = 0\).
\(a = 3, b = -1, c = -10\)
\(b^2 - 4ac = (-1)^2 - 4(3)(-10)\)
\(= 1 + 120 = 121\)
Formula: \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(m = \frac{-(-1) \pm \sqrt{121}}{2(3)}\)
\(m = \frac{1 \pm 11}{6}\)
Case 1: \(m = \frac{1 + 11}{6} = \frac{12}{6} = 2\)
Case 2: \(m = \frac{1 - 11}{6} = \frac{-10}{6} = -\frac{5}{3}\)
\(\mathbf{m = 2, m = -\frac{5}{3}}\)
(ii) Solve the following simultaneous equations using Cramer's rule: \(3x - 4y = 10\), \(4x + 3y = 5\)
Solution:
\(D = \begin{vmatrix} 3 & -4 \\ 4 & 3 \end{vmatrix} = (3 \times 3) - (-4 \times 4) = 9 - (-16) = 25\)
\(D_x = \begin{vmatrix} 10 & -4 \\ 5 & 3 \end{vmatrix} = (10 \times 3) - (-4 \times 5) = 30 - (-20) = 50\)
\(D_y = \begin{vmatrix} 3 & 10 \\ 4 & 5 \end{vmatrix} = (3 \times 5) - (10 \times 4) = 15 - 40 = -25\)
By Cramer's Rule:
\(x = \frac{D_x}{D} = \frac{50}{25} = \mathbf{2}\)
\(y = \frac{D_y}{D} = \frac{-25}{25} = \mathbf{-1}\)
Solution: \((x, y) = (2, -1)\)
(iii) 50 shares of face value 10 were purchased for market value of 25. Company declared 30% dividend on the shares, then find: (1) Sum invested (2) Dividend received (3) Rate of return.
Solution:
Given: No. of shares = 50, FV = ₹10, MV = ₹25, Dividend Rate = 30%.

(1) Sum Invested:
Sum Invested = Number of shares \(\times\) MV
\(= 50 \times 25 = \mathbf{\text{ ₹ }1250}\)

(2) Dividend Received:
Dividend per share = Rate \(\times\) FV = \(\frac{30}{100} \times 10 = \text{ ₹ }3\)
Total Dividend = Dividend per share \(\times\) No. of shares
\(= 3 \times 50 = \mathbf{\text{ ₹ }150}\)

(3) Rate of Return (RoR):
RoR = \(\frac{\text{Total Dividend}}{\text{Sum Invested}} \times 100\)
RoR = \(\frac{150}{1250} \times 100\)
RoR = \(\frac{15}{125} \times 100\)
RoR = \(\frac{3}{25} \times 100\)
RoR = \(3 \times 4 = \mathbf{12\%}\)
(iv) One coin and a die are thrown simultaneously. Find the probability of the following events:
Event A: To get a head and a prime number.
Event B: To get a tail and an odd number.
Solution:
Sample Space \(S = \{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6\}\)
\(n(S) = 12\)

Event A: Head and Prime number
Prime numbers on die: 2, 3, 5.
\(A = \{H2, H3, H5\}\)
\(n(A) = 3\)
\(P(A) = \frac{3}{12} = \mathbf{\frac{1}{4}}\)

Event B: Tail and Odd number
Odd numbers on die: 1, 3, 5.
\(B = \{T1, T3, T5\}\)
\(n(B) = 3\)
\(P(B) = \frac{3}{12} = \mathbf{\frac{1}{4}}\)

Q. 4. Solve the following subquestions (any two): (8 Marks)

(i) A tank can be filled up by two taps in 6 hours. The smaller tap alone takes 5 hours more than the bigger tap alone. Find the time required by each tap to fill the tank separately.
Solution:
Let the time taken by the bigger tap to fill the tank be \(x\) hours.
Then the time taken by the smaller tap is \((x + 5)\) hours.
In 1 hour, bigger tap fills \(\frac{1}{x}\) part.
In 1 hour, smaller tap fills \(\frac{1}{x+5}\) part.
Together they fill the tank in 6 hours, so in 1 hour they fill \(\frac{1}{6}\) part.

\(\frac{1}{x} + \frac{1}{x+5} = \frac{1}{6}\)
\(\frac{x + 5 + x}{x(x+5)} = \frac{1}{6}\)
\(\frac{2x + 5}{x^2 + 5x} = \frac{1}{6}\)
\(6(2x + 5) = x^2 + 5x\)
\(12x + 30 = x^2 + 5x\)
\(x^2 - 7x - 30 = 0\)
Factors of -30 summing to -7 are -10 and +3.
\((x - 10)(x + 3) = 0\)
\(x = 10\) or \(x = -3\)
Time cannot be negative, so \(x = 10\).

Time for Bigger tap = \(\mathbf{10 \text{ hours}}\).
Time for Smaller tap = \(10 + 5 = \mathbf{15 \text{ hours}}\).
(ii) The following table shows the classification of percentage of marks of students and the number of students. Draw frequency polygon from the table without drawing histogram:
Result (%)No. of Students
20-4025
40-6065
60-8080
80-10015
Solution:
To draw a frequency polygon, we need class marks (midpoints). We also add a class before and after with zero frequency.
ClassClass Mark (x)Frequency (y)Coordinates (x, y)
0-20100(10, 0)
20-403025(30, 25)
40-605065(50, 65)
60-807080(70, 80)
80-1009015(90, 15)
100-1201100(110, 0)
(Note: Plot these coordinates on a graph paper and join them in order with straight lines to form the frequency polygon.)
(iii) In a 'Mahila Bachat Gat' Kavita invested from the first day of month 20 on first day, 40 on second day and 60 on third day. If she saves like this, then what would be her total saving in the month of February 2020?
Solution:
Savings: 20, 40, 60, ...
This is an Arithmetic Progression (A.P.) where \(a = 20\), \(d = 20\).
February 2020 is a leap year, so it has 29 days. Thus, \(n = 29\).
We need to find total saving \(S_n\).
\(S_n = \frac{n}{2}[2a + (n-1)d]\)
\(S_{29} = \frac{29}{2}[2(20) + (29-1)20]\)
\(S_{29} = \frac{29}{2}[40 + 28(20)]\)
\(S_{29} = \frac{29}{2}[40 + 560]\)
\(S_{29} = \frac{29}{2}[600]\)
\(S_{29} = 29 \times 300\)
\(S_{29} = 8700\)
\(\mathbf{\text{Total saving is ₹8700.}}\)

Q. 5. Solve the following subquestions (any one): (3 Marks)

(i) In the given figure (Pie Diagram), Cricket \(160^\circ\), Kabaddi \(55^\circ\), Football \(45^\circ\), Hockey \(100^\circ\). If the money spent on football is 9,000, answer the following questions:
(a) What is the total amount spent on sports?
(b) What is the amount spent on cricket?
Solution:
(a) Total amount spent on sports:
Let total amount be \(x\).
Central angle for Football = \(45^\circ\). Amount = ₹9000.
Formula: \(\text{Measure of arc} = \frac{\text{Item Value}}{\text{Total Value}} \times 360^\circ\)
\(45^\circ = \frac{9000}{x} \times 360^\circ\)
\(x = \frac{9000 \times 360}{45}\)
\(x = 9000 \times 8\)
\(x = 72,000\)
\(\mathbf{\text{Total Amount = ₹72,000}}\)

(b) Amount spent on Cricket:
Angle for Cricket = \(160^\circ\).
Amount = \(\frac{160}{360} \times 72000\)
\(= \frac{4}{9} \times 72000\)
\(= 4 \times 8000\)
\(\mathbf{= \text{ ₹ }32,000}\)
(ii) Draw the graph of the equation \(x + y = 4\) and answer the following questions:
(a) Which type of triangle is formed by the line with X and Y-axes based on its sides.
(b) Find the area of that triangle.
Solution:
Equation: \(x + y = 4\)
Intercepts: When \(x=0, y=4\) (Point A: 0, 4). When \(y=0, x=4\) (Point B: 4, 0). Origin O is (0,0).

(a) Type of Triangle:
The triangle formed is \(\Delta OAB\).
Length OA = 4 units. Length OB = 4 units.
Since two sides are equal and the angle between axes is \(90^\circ\), it is an Isosceles Right-Angled Triangle.

(b) Area of Triangle:
Area = \(\frac{1}{2} \times \text{base} \times \text{height}\)
Area = \(\frac{1}{2} \times 4 \times 4\)
Area = \(\mathbf{8 \text{ sq. units}}\)
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