Board Question Paper : JULY 2015
ALGEBRA
Solution:
Given the general term: \(t_n = n + 2\)
To find the first term (\(t_1\)), substitute \(n = 1\):
\(t_1 = 1 + 2 = 3\)
To find the second term (\(t_2\)), substitute \(n = 2\):
\(t_2 = 2 + 2 = 4\)
Ans: The first two terms are 3 and 4.
Solution:
Given equation: \(3y^2 = 10y + 7\)
To bring it to standard form, move all terms to the left side:
\(3y^2 - 10y - 7 = 0\)
Comparing with standard form \(ay^2 + by + c = 0\).
Ans: The equation in standard form is \(3y^2 - 10y - 7 = 0\).
Solution:
Value of determinant \(D = (4 \times 7) - (3 \times 2)\)
\(D = 28 - 6\)
\(D = 22\)
Ans: The value of the determinant is 22.
Solution:
When two coins are tossed simultaneously, the possible outcomes are Head (H) and Tail (T).
Sample Space (S) = \(\{HH, HT, TH, TT\}\)
Ans: \(S = \{HH, HT, TH, TT\}\)
1, 3, 6, 10, …
Solution:
Given sequence: 1, 3, 6, 10, …
Here, \(t_1 = 1, t_2 = 3, t_3 = 6, t_4 = 10\)
Find the common difference (d):
\(d_1 = t_2 - t_1 = 3 - 1 = 2\)
\(d_2 = t_3 - t_2 = 6 - 3 = 3\)
Since the difference between consecutive terms is not constant (\(2 \neq 3\)).
Ans: The given sequence is not an A.P.
Solution:
Let the length of the rectangle be \(x\) cm and breadth be \(y\) cm.
Perimeter of rectangle = \(2(length + breadth)\)
Given Perimeter = 36 cm.
\(\therefore 2(x + y) = 36\)
Dividing by 2:
\(x + y = 18\)
Ans: The equation is \(x + y = 18\).
Solution:
Since 4 is a root of the equation \(x^2 - 7x + k = 0\), substitute \(x = 4\) in the equation.
\((4)^2 - 7(4) + k = 0\)
\(16 - 28 + k = 0\)
\(-12 + k = 0\)
\(k = 12\)
Ans: The value of k is 12.
Solution:
Given A.P.: 7, 13, 19, 25, …
Here, first term \(a = 7\)
Common difference \(d = 13 - 7 = 6\)
We need to find \(t_{18}\).
Formula: \(t_n = a + (n - 1)d\)
\(t_{18} = 7 + (18 - 1)6\)
\(t_{18} = 7 + (17 \times 6)\)
\(t_{18} = 7 + 102\)
\(t_{18} = 109\)
Ans: The eighteenth term is 109.
Solution:
Sample Space (S) when a die is thrown:
\(S = \{1, 2, 3, 4, 5, 6\}\)
Event P: Getting an odd number.
Odd numbers in S are 1, 3, 5.
Ans: \(S = \{1, 2, 3, 4, 5, 6\}\), \(P = \{1, 3, 5\}\)
Solution:
Using Cramer's Rule:
\(x = \frac{D_x}{D} = \frac{18}{3} = 6\)
\(y = \frac{D_y}{D} = \frac{15}{3} = 5\)
Ans: \(x = 6\) and \(y = 5\).
Solution:
Let the roots be \(\alpha = 5\) and \(\beta = 7\).
Sum of roots (\(\alpha + \beta\)) = \(5 + 7 = 12\)
Product of roots (\(\alpha \cdot \beta\)) = \(5 \times 7 = 35\)
The required quadratic equation is given by formula:
\(x^2 - (\alpha + \beta)x + \alpha\beta = 0\)
\(x^2 - 12x + 35 = 0\)
Ans: The quadratic equation is \(x^2 - 12x + 35 = 0\).
Solution:
We use the empirical relationship between Mean, Median, and Mode:
Mode = 3(Median) - 2(Mean)
Given: Mode = 180, Median = 156
\(180 = 3(156) - 2(\text{Mean})\)
\(180 = 468 - 2(\text{Mean})\)
\(2(\text{Mean}) = 468 - 180\)
\(2(\text{Mean}) = 288\)
\(\text{Mean} = \frac{288}{2} = 144\)
Ans: The Mean is 144.
Solution:
Comparing \(2x^2 + 5x + 2 = 0\) with \(ax^2 + bx + c = 0\):
\(a = 2, b = 5, c = 2\)
Find discriminant (\(\Delta\)):
\(\Delta = b^2 - 4ac = (5)^2 - 4(2)(2) = 25 - 16 = 9\)
Formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(x = \frac{-5 \pm \sqrt{9}}{2(2)}\)
\(x = \frac{-5 \pm 3}{4}\)
Case 1: \(x = \frac{-5 + 3}{4} = \frac{-2}{4} = -\frac{1}{2}\)
Case 2: \(x = \frac{-5 - 3}{4} = \frac{-8}{4} = -2\)
Ans: The roots are \(-\frac{1}{2}\) and \(-2\).
Solution:
Sample Space S contains numbers from 1 to 30.
\(S = \{1, 2, 3, \dots, 30\}\)
\(\therefore n(S) = 30\)
Event A: The number on the ticket is a perfect square.
Perfect squares between 1 and 30 are: \(1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25\) (Note: \(6^2=36 > 30\)).
\(A = \{1, 4, 9, 16, 25\}\)
\(\therefore n(A) = 5\)
Ans: \(S = \{1..30\}, n(S)=30, A=\{1, 4, 9, 16, 25\}, n(A)=5\).
Solution:
Given:
\(t_{18} = 52\)
\(t_{39} = 148\)
Using formula \(t_n = a + (n-1)d\):
1) \(a + 17d = 52\)
2) \(a + 38d = 148\)
We need to find sum of first 56 terms (\(S_{56}\)).
\(S_n = \frac{n}{2}[2a + (n-1)d]\)
\(S_{56} = \frac{56}{2}[2a + (56-1)d]\)
\(S_{56} = 28[2a + 55d]\)
Observe the equations (1) and (2). Let's add them:
\((a + 17d) + (a + 38d) = 52 + 148\)
\(2a + 55d = 200\)
Substitute this value back into the Sum equation:
\(S_{56} = 28 \times [200]\)
\(S_{56} = 5600\)
Ans: The sum of the first 56 terms is 5600.
Solution:
Equation: \(3x - y = -6\) which can be written as \(y = 3x + 6\).
To find intersection with X-axis (Put \(y=0\)):
\(3x - 0 = -6 \Rightarrow 3x = -6 \Rightarrow x = -2\)
Point: \((-2, 0)\)
To find intersection with Y-axis (Put \(x=0\)):
\(3(0) - y = -6 \Rightarrow -y = -6 \Rightarrow y = 6\)
Point: \((0, 6)\)
Graph: Plot points \((-2, 0)\) and \((0, 6)\) on a Cartesian plane and draw a line passing through them.
Ans: Intersection with X-axis: (-2, 0); Intersection with Y-axis: (0, 6).
Draw a pie diagram to represent this information.
Solution:
To draw a pie diagram, we must calculate the measure of the central angle for each part of the day.
Total Percentage = \(30 + 40 + 20 + 10 = 100\%\).
Formula for Central Angle \((\theta) = \frac{\text{Value}}{\text{Total}} \times 360^{\circ}\)
| Part of the Day | Percentage | Measure of Central Angle (\(\theta\)) |
|---|---|---|
| Morning | 30 | \(\frac{30}{100} \times 360^{\circ} = 108^{\circ}\) |
| Afternoon | 40 | \(\frac{40}{100} \times 360^{\circ} = 144^{\circ}\) |
| Evening | 20 | \(\frac{20}{100} \times 360^{\circ} = 72^{\circ}\) |
| Night | 10 | \(\frac{10}{100} \times 360^{\circ} = 36^{\circ}\) |
| Total | 100 | 360^{\circ} |
Note for Student: Draw a circle of convenient radius. Using a protractor, mark the sectors with angles \(108^{\circ}, 144^{\circ}, 72^{\circ}\), and \(36^{\circ}\) sequentially.
a. a king
b. a face card.
Solution:
Total cards in a pack, \(n(S) = 52\).
a. Event A: The card drawn is a king.
There are 4 kings in a deck (one of each suit).
\(n(A) = 4\)
\(P(A) = \frac{n(A)}{n(S)} = \frac{4}{52} = \frac{1}{13}\)
b. Event B: The card drawn is a face card.
Face cards are Jack, Queen, and King for each of the 4 suits.
\(n(B) = 3 \times 4 = 12\)
\(P(B) = \frac{n(B)}{n(S)} = \frac{12}{52} = \frac{3}{13}\)
Ans: a) \(\frac{1}{13}\), b) \(\frac{3}{13}\).
Solution:
Given: \(3x^4 - 13x^2 + 10 = 0\)
Let \(x^2 = m\). Then \(x^4 = m^2\).
Substituting \(m\) into the equation:
\(3m^2 - 13m + 10 = 0\)
Factorizing by splitting the middle term (\(3 \times 10 = 30\); factors adding to -13 are -10 and -3):
\(3m^2 - 3m - 10m + 10 = 0\)
\(3m(m - 1) - 10(m - 1) = 0\)
\((3m - 10)(m - 1) = 0\)
\(\therefore m = \frac{10}{3}\) or \(m = 1\)
Resubstitute \(m = x^2\):
Case 1: \(x^2 = 1 \Rightarrow x = \pm 1\)
Case 2: \(x^2 = \frac{10}{3} \Rightarrow x = \pm \sqrt{\frac{10}{3}}\)
Ans: The roots are \(1, -1, \sqrt{\frac{10}{3}}, -\sqrt{\frac{10}{3}}\).
Find the modal bowling speed of players.
Solution:
To find the mode, we identify the class with the highest frequency.
| Speed (Class) | 85-100 | 100-115 | 115-130 | 130-145 |
|---|---|---|---|---|
| Frequency | 9 | 11 | 8 | 5 |
Maximum frequency is 11, so the Modal Class is 100-115.
Here:
Lower limit of modal class (\(L\)) = 100
Frequency of modal class (\(f_1\)) = 11
Frequency of class before (\(f_0\)) = 9
Frequency of class after (\(f_2\)) = 8
Class interval (\(h\)) = 15
Formula: \(\text{Mode} = L + \left[ \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right] \times h\)
\(\text{Mode} = 100 + \left[ \frac{11 - 9}{2(11) - 9 - 8} \right] \times 15\)
\(\text{Mode} = 100 + \left[ \frac{2}{22 - 17} \right] \times 15\)
\(\text{Mode} = 100 + \left[ \frac{2}{5} \right] \times 15\)
\(\text{Mode} = 100 + (2 \times 3)\)
\(\text{Mode} = 100 + 6 = 106\)
Ans: The modal bowling speed is 106 km/hr.
Solution:
Let the number of rows be \(x\) and number of students in each row be \(y\).
Total number of students = \(xy\).
Condition 1: 3 students less in each row, 10 more rows.
\((y - 3)(x + 10) = xy\)
\(xy + 10y - 3x - 30 = xy\)
\(10y - 3x = 30\) ... (Equation 1)
Condition 2: 5 students more in each row, 10 less rows.
\((y + 5)(x - 10) = xy\)
\(xy - 10y + 5x - 50 = xy\)
\(5x - 10y = 50\) ... (Equation 2)
Adding Equation (1) and (2):
\((10y - 3x) + (5x - 10y) = 30 + 50\)
\(2x = 80\)
\(x = 40\) (Number of rows)
Substitute \(x = 40\) in Equation (2):
\(5(40) - 10y = 50\)
\(200 - 10y = 50\)
\(150 = 10y\)
\(y = 15\) (Students per row)
Total students = \(xy = 40 \times 15 = 600\).
Ans: The number of students participating is 600.
Solution:
Let the temperatures from Monday to Friday be:
Mon: \(a - 2d\)
Tue: \(a - d\)
Wed: \(a\)
Thu: \(a + d\)
Fri: \(a + 2d\)
Condition 1: Sum of Mon, Tue, Wed is 0.
\((a - 2d) + (a - d) + a = 0\)
\(3a - 3d = 0\)
\(3a = 3d \Rightarrow a = d\) ... (Equation 1)
Condition 2: Sum of Thu, Fri is 15.
\((a + d) + (a + 2d) = 15\)
\(2a + 3d = 15\)
Substitute \(a = d\) from Eq 1 into this equation:
\(2(d) + 3d = 15\)
\(5d = 15\)
\(d = 3\)
Therefore, \(a = 3\).
Temperatures:
Mon: \(a - 2d = 3 - 6 = -3^\circ\)C
Tue: \(a - d = 3 - 3 = 0^\circ\)C
Wed: \(a = 3^\circ\)C
Thu: \(a + d = 3 + 3 = 6^\circ\)C
Fri: \(a + 2d = 3 + 6 = 9^\circ\)C
Ans: The temperatures are -3°C, 0°C, 3°C, 6°C, and 9°C.
Solution:
To draw the frequency polygon using a histogram, we first need the class midpoints (though the polygon connects the midpoints of the tops of the histogram bars).
| House Rent (₹) | Number of Families (Frequency) | Class Mark (Midpoint) |
|---|---|---|
| 400-600 | 200 | 500 |
| 600-800 | 240 | 700 |
| 800-1000 | 300 | 900 |
| 1000-1200 | 50 | 1100 |
Steps to Draw:
- Take "House Rent" on the X-axis (Scale: 1 cm = ₹200).
- Take "Number of Families" on the Y-axis (Scale: 1 cm = 50 families).
- Histogram: Draw adjacent rectangles for each class interval with heights corresponding to their frequencies (200, 240, 300, 50).
- Frequency Polygon: Mark the midpoints of the upper sides of these rectangles: (500, 200), (700, 240), (900, 300), (1100, 50).
- Join these midpoints with straight lines.
- To complete the polygon, join the first point to the midpoint of the previous imaginary class (300, 0) and the last point to the next imaginary class (1300, 0) on the X-axis.
