Students must start practicing the questions from CBSE Sample Papers for Class 12 Maths with Solutions Set 2 are designed as per the revised syllabus.
CBSE Sample Papers for Class 12 Maths Set 2 with Solutions
Time Allowed: 3 Hours
Maximum Marks: 80
General Instructions:
- This Question paper contains five sections A, B, C, D, and E. Each section is compulsory. However, there are internal choices in some questions.
- Section A has 18 MCQs and 2 Assertion-Reason-based questions of 1 mark each.
- Section B has 5 Very Short Answer (VSA) type questions of 2 marks each.
- Section C has 6 Short Answer (SA) type questions of 3 marks each.
- Section D has 4 Long Answer (LA) type questions of 5 marks each.
- Section E has 3 sources based/case based/passage based/integrated units of assessment (4 marks each) with sub-parts.
Section – A (20 Marks)
(Multiple Choice Questions Each question carries 1 mark)
Question 1.
Let A = {1, 2, 3}. A relation on the set A which contains (1, 2) and (2, 3) which are reflexive and transitive but not symmetric is: [1]
(a) (1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 1), (3, 3)
(b) (1, 1), (2, 2), (3, 3), (3, 1)
(c) (1, 1), (2, 1), (1, 2), (3, 1), (1, 3)
(d) (1, 2), (2, 1), (3, 1), (3, 2), (3, 3)
Solution:
(a) (1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 1), (3, 3)
Explanation: Total possible pairs using {1, 2, 3} is {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3}
For reflexive (1, 1), (2, 2), (3, 3) is a must.
For transitive having (1, 2) and (2, 3); (1, 3) is a must.
For not symmetric (2, 1) and (3, 2) should not be in relation to R.
Required R = {(1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 1), (3, 3)}
Question 2.
The range of sin-1 x is: [1]
(a) \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)
(b) (-π, -π)
(c) (-∞, 0)
(d) (0, ∞)
Solution:
(a) \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)
Explanation: The range of sin-1 x is \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)
Question 3.
The value of \(\sin \left[\cos ^{-1} \frac{1}{2}\right]\) is: [1]
(a) 1
(b) \(\frac{1}{\sqrt{2}}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{\sqrt{3}}{2}\)
Solution:
(d) \(\frac{\sqrt{3}}{2}\)
Explanation:
Question 4.
For the matrix A = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]\), a symmetric matrix is: [1]
(a) \(\left[\begin{array}{ll}
1 & 7 \\
5 & 6
\end{array}\right]\)
(b) \(\left[\begin{array}{ll}
1 & 5 \\
5 & 2
\end{array}\right]\)
(c) \(\left[\begin{array}{ll}
5 & 1 \\
6 & 7
\end{array}\right]\)
(d) \(\left[\begin{array}{ll}
5 & 1 \\
5 & 2
\end{array}\right]\)
Solution:
(b) \(\left[\begin{array}{ll}
1 & 5 \\
5 & 2
\end{array}\right]\)
Explanation:
Question 5.
The inverse of the matrix \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\) is: [1]
(a) \(\left[\begin{array}{cc}
\frac{1}{5} & \frac{-1}{5} \\
\frac{2}{5} & \frac{3}{5}
\end{array}\right]\)
(b) \(\left[\begin{array}{cc}
\frac{3}{5} & \frac{1}{5} \\
\frac{-2}{5} & \frac{1}{5}
\end{array}\right]\)
(c) \(\left[\begin{array}{cc}
3 & 2 \\
-1 & 1
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
-1 & 1 \\
3 & 2
\end{array}\right]\)
Solution:
(b) \(\left[\begin{array}{cc}
\frac{3}{5} & \frac{1}{5} \\
\frac{-2}{5} & \frac{1}{5}
\end{array}\right]\)
Explanation:
Question 6.
If A = \(\left[\begin{array}{ll}
x & 1 \\
1 & 0
\end{array}\right]\) and A2 is the identity matrix, then x is equal to: [1]
(a) 0
(b) -1
(c) 1
(d) 2
Solution:
(a) 0
Explanation:
Question 7.
The sum of cofactors of 7 and 12 in the determinant \(\left|\begin{array}{rrr}
1 & 2 & 4 \\
5 & 7 & 8 \\
9 & 10 & 12
\end{array}\right|\) is: [1]
(a) -27
(b) -24
(c) -18
(d) 0
Solution:
(a) -27
Explanation:
Question 8.
If y = \(\tan ^{-1} \sqrt{\frac{1-\sin x}{1+\sin x}}\), then value of \(\frac{d y}{d x}\) at x = \(\frac{\pi}{6}\) is: [1]
(a) \(\frac{1}{2}\)
(b) \(-\frac{1}{2}\)
(c) 1
(d) -1
Solution:
(b) \(-\frac{1}{2}\)
Explanation:
Question 9.
The value of \(\int_{-1}^1(x-[x]) d x\) is: [1]
(a) -1
(b) 0
(c) 1
(d) 2
Solution:
(c) 1
Explanation:
Question 10.
The value of \(\int \frac{1}{e^x-1} d x\) is: [1]
(a) log ex + c
(b) log|1 – e-x| + c
(c) \(\log \frac{1}{e^x}+c\)
(d) log|ex – 1| + c
Solution:
(b) log|1 – e-x| + c
Explanation:
Question 11.
The solution of \(\frac{d^2 y}{d x^2}\) + 9y – 6 cos 3x = 0 is: [1]
(a) y = x cos 3x
(b) y = x sin 3x
(c) y = x cos 2x
(d) y = x sin 2x
Solution:
(b) y = x sin 3x
Explanation:
y = x sin 3x
As \(\frac{d y}{d x}\) = sin 3x + 3x cos 3x
\(\frac{d^2 y}{d x^2}\) = 3 cos 3x + 3 cos 3x – 9x sin 3x
= 6 cos 3x – 9y
Thus, \(\frac{d^2 y}{d x^2}\) + 9y – 6 cos 3x = 0
Question 12.
For the vectors \(\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}\) and \(\vec{b}=-\hat{i}+\hat{j}-\hat{k}\), the unit vector in the direction 0f \(\vec{a}+\vec{b}\) is: [1]
(a) \((\hat{i}+\hat{j})\)
(b) \(\hat{i}+\hat{j}+\hat{k}\)
(c) \(\frac{1}{\sqrt{2}}(\hat{i}+\hat{k})\)
(d) \(\frac{\hat{i}-\hat{j} \times \hat{k}}{\sqrt{2}}\)
Solution:
(c) \(\frac{1}{\sqrt{2}}(\hat{i}+\hat{k})\)
Explanation:
Here \(\vec{a}+\vec{b}=\hat{i}+\hat{k}\)
So, \(|\vec{a}+\vec{b}|=\sqrt{1+1}=\sqrt{2}\)
So, required unit vector = \(\frac{1}{\sqrt{2}}(\hat{i}+\hat{k}) \text { i.e. } \frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}\)
= \(\frac{1}{\sqrt{2}}(\hat{i}+\hat{k})\)
Question 13.
The vector equation of the line joining the points (3, -2, -5)and (3, -2, 6) is: [1]
(a) \((3 \hat{i}-2 \hat{j}+5 \hat{k})+\lambda(11 \hat{k})\)
(b) \((4 \hat{i}-4 \hat{j}+5 \hat{k})+\lambda(12 \hat{k})\)
(c) \((6 \hat{i}-2 \hat{j}+2 \hat{k})+\lambda(5 \hat{k})\)
(d) \((9 \hat{i}-9 \hat{j}-2 \hat{k})+\lambda(2 \hat{k})\)
Solution:
(a) \((3 \hat{i}-2 \hat{j}+5 \hat{k})+\lambda(11 \hat{k})\)
Explanation:
The vector equation of a line joining the points (3, -2, -5) and (3, -2, 6) is
\(\vec{r}=(3 \hat{i}-2 \hat{j}-5 \hat{k})+\lambda[(3-3) \hat{i}+(-2+2) \hat{j}\) + \((6+5) \hat{k}\)]
i.e., \(\vec{r}=(3 \hat{i}-2 \hat{j}-5 \hat{k})+\lambda[11 \hat{k}]\)
Question 14.
The acute angle between two lines whose direction ratios are 2, 3, 6 and 1, 2, 2 is: [1]
(a) \(\cos ^{-1} \frac{20}{21}\)
(b) \(\cos ^{-1} \frac{19}{21}\)
(c) \(\cos ^{-1} \frac{1}{3}\)
(d) \(\cos ^{-1} \frac{1}{7}\)
Solution:
(a) \(\cos ^{-1} \frac{20}{21}\)
Explanation:
Question 15.
A dice is tossed thrice. The probability of getting an odd number at least once is: [1]
(a) \(\frac{7}{8}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{3}{8}\)
(d) \(\frac{1}{8}\)
Solution:
(a) \(\frac{7}{8}\)
Explanation:
Required probability = 1 – Probability of getting no odd number
= 1 – \(\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right)\)
= 1 – \(\frac{1}{8}\)
= \(\frac{7}{8}\)
Question 16.
If the following table represents a probability distribution for a random variable X: [1]
The value of k is:
(a) 0.01
(b) 0.1
(c) \(\frac{1}{1000}\)
(d) \(\frac{2}{5}\)
Solution:
(b) 0.1
Explanation:
In the probability distribution of X,
ΣP(X) = 1
⇒ (0.1) + 2k + k + (0.2) + 3k + (0.1) = 1
⇒ 6k = 0.6
⇒ k = 0.1
Question 17.
If direction cosines of a line are \(\left(\frac{1}{c}, \frac{1}{c}, \frac{1}{c}\right)\), then: [1]
(a) 0 < c < 1
(b) c > 2
(c) c = ±√2
(d) c = ±√3
Solution:
(d) c = ±√3
Explanation:
since D C’s of a line are \(\left(\frac{1}{c}, \frac{1}{c}, \frac{1}{c}\right)\)
∴ \(\left(\frac{1}{c}\right)^2+\left(\frac{1}{c}\right)^2+\left(\frac{1}{c}\right)^2=1\)
⇒ c2 = 3
⇒ c = ±√3
Question 18.
The mean of the number obtained on throwing a dice having written 1 on three faces, 2 on two faces, and 5 on the face are: [1]
(a) 0
(b) 1
(c) 2
(d) 4
Solution:
(c) 2
Explanation:
Assertion-Reason Based Questions
In the following questions, a statement of assertion (A) is followed by a statement of the reason (R).
Choose the correct answer out of the following choices.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Question 19.
Assertion (A): A dice marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let ‘A’ be the event ‘number is even’ and ‘B’ be the event ‘the number is marked’. Events A and B are independent.
Reason (R): Two events A and B are independent, if P(A ∩ B) = P(A) . P(B) [1]
Solution:
(d) A is false but R is true.
Explanation:
A: number is blue
B: number is red
A = {2, 4, 6}, B = {1, 2, 3}
A ∩ B = {2}, n(A) = 3, n(B) = 3, n(A ∩ B) = 1
P(A) = \(\frac{3}{6}=\frac{1}{2}\),
P(B) = \(\frac{3}{6}=\frac{1}{2}\),
P(A ∩ B) = \(\frac{1}{6}\)
P(A) × P(B) = \(\frac{1}{4}\) ≠ P(A ∩ B)
So, A and B are not independent
Question 20.
Assertion (A): The area of a parallelogram with diagonals \(\vec{a}\) and \(\vec{b}\) is \(\frac{1}{2}[\vec{a} \times \vec{b}]\)
Reason (R): If \(\vec{a}\) and \(\vec{b}\) represents the adjacent side of a triangle, then the area of a triangle can be obtained by evaluating \(|\vec{a} \times \vec{b}|\). [1]
Solution:
(c) A is true but R is false
Explanation:
The area of a triangle, with adjacent sides represented by \(\vec{a}\) and \(\vec{a}\) is \(\frac{1}{2}[\vec{a} \times \vec{b}]\)
Section – B (10 Marks)
This section comprises very short answer type questions (VSA) of 2 marks.
Question 21.
Express \(\tan ^{-1}\left(\frac{\sqrt{x^2+1}-1}{x}\right)\) in the simplest form. [2]
OR
Write in the simplest form of \(\tan ^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)\)
Solution:
Put x = tan θ so that tan-1x = θ
Now,
Question 22.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]\), then show that |2A| = 4|A| . [2]
Solution:
A = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]\)
|A| = 2 – 8 = -6
2A = \(\left[\begin{array}{ll}
2 & 4 \\
8 & 4
\end{array}\right]\)
|2A| = 8 – 32 = -24
4|A| = 4 × (-6) = -24
|2A| = 4|A|
Question 23.
Find \(\int \frac{x+3}{x^2-2 x-5} d x\) [2]
Solution:
Let \(\frac{x+3}{x^2-2 x-5}=\mathrm{A} \frac{d}{d x}\left(x^2-2 x-5\right)+\mathrm{B}\)
⇒ x + 3 = A(2x – 2) + B
Comparing the coefficient of x and constant terms,
we get, A = \(\frac{1}{2}\) and B = 4
So,
Question 24.
Solve \(x \frac{d y}{d x}-y=x^2\) [2]
Solution:
Question 25.
Find the angle between the vectors \(\vec{a}=\hat{i}+\hat{j}-\hat{k}\) and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\). [2]
OR
Find the value of λ and µ, if \(\vec{a} \times \vec{b}=0\), where \(\vec{a}=2 \hat{i}+6 \hat{j}+27 \hat{k}\) and \(\vec{b}=\hat{i}+\lambda \hat{j}+\mu \hat{k}\).
Solution:
Let θ be the angle between the given two vectors. then,
Section – C (18 Marks)
This section comprises short answer type questions (SA) of 3 marks
Question 26.
Let R be a relation on the set A of ordered pairs of positive integers defined as (x, y) R(u, v) if and only if xv = yu. Show that, R is an equivalence relation. [3]
Solution:
Clearly, (x, y) S(u, v) for all (x, y) ∈ A,
since xy = yx for all positive integers x and y.
This show that S is reflexive.
Further, (x, y) S(u, v)
⇒ xv = yu
⇒ uy = vx
⇒ (u, v) S(x, y)
and hence (u, v) S(x, y)
This shows that S is symmetric.
Suppose, (x, y) S(u, v) and (u, v) S(a, b)
So, xv = yu and ub = va
⇒ \(x v \frac{a}{u}=y u \frac{a}{u}\)
⇒ \(x v \cdot \frac{b}{v}=y u \frac{a}{u}\) [∵ ub = va ⇒ \(\frac{a}{u}=\frac{b}{v}\)]
⇒ xb = ya
⇒ (x, y) S(a, b)
This shows that S is transitive.
Hence, S is an equivalence relation.
Question 27.
Find the derivative of the function given by f(x) = (1 + x)(1 + x2)(1 + x4)(1 + x8) and hence find f'(1). [3]
Solution:
Question 28.
Find \(\int \sqrt{1-4 x-x^2} d x\). [3]
OR
Evaluate \(\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x\)
Solution:
Question 29.
If y = (log x)x + xlog x, then find \(\frac{d y}{d x}\) [3]
Solution:
Question 30.
Using integration, find the area of the region bounded by the line y = x + 1, the x-axis, and the abscissae x = -2 and x = 3. [3]
OR
Solve \((x+1) \frac{d y}{d x}=2 e^{-y}-1\); given y = 0 when x = 0.
Solution:
Since, y = x + 1, is a straight line that cuts the x-axis at (-1, 0).
Intersection point of y = x + 1 and x = -2 is (-2, -1).
The intersection point of y = x + 1 and x = 3 is (3, 4).
From figure,
Question 31.
Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, -1, 4), and C(4, 5, -1). [3]
OR
If \(\vec{a}=2 \hat{i}-3 \hat{j}-\hat{k}\) and \(\vec{b}=\hat{i}+4 \hat{j}-2 \hat{k}\), then find the value of \(\vec{a} \times \vec{b}\).
Solution:
Given: Coordinates of vertices of ∆ABC are A(1, 2, 3), B(2, -1, 4) and C(4, 5, -1)
∴ \(\overrightarrow{\mathrm{AB}}=\hat{i}-3 \hat{j}+\hat{k}\) and \(\overrightarrow{\mathrm{AC}}=3 \hat{i}+3 \hat{j}-4 \hat{k}\)
We know that,
Section – D (20 Marks)
This section comprises long answer type questions (LA) of 5 marks each
Question 32.
A tank with a rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 cu m. If the building of the tank costs ₹ 70 per sq meter for the base and ₹ 45 per sq meter for the sides, what is the cost of the least expensive tank? [5]
Solution:
Let x and y be the length and breadth of the rectangular base of the open tank.
Then, the volume (V) and the surface area (S) of the tank are given by,
V = 2xy; S = xy + 2(2x + 2y)
⇒ 2xy = 8
⇒ xy = 4 (∵ V = 8 cu. m)
and S = \(4+4\left(x+\frac{4}{x}\right)\) ……(i)
Now \(\frac{d S}{d x}=4\left(1-\frac{4}{x^2}\right)\)
and \(\frac{d^2 S}{d x^2}=\frac{32}{x^3}\)
Equating \(\frac{d S}{d x}\) to 0, we have,
\(4\left(1-\frac{4}{x^2}\right)=0\)
⇒ x2 = 4
⇒ x = 2
At x = 2,
\(\frac{d^2 S}{d x^2}=\frac{32}{8}\) = 4 > 0
So, S is the least when x = 2
Now, when x = 2, y = 2 (∵ xy = 4)
Thus, area of the base = xy = 4 m2
⇒ cost of the base = ₹ 280
and Area of four walls = 4(x + y) = 16
⇒ cost of walls = ₹ 720
⇒ Total cost of the tank = (₹ 280 + ₹ 720) = ₹ 1000
Question 33.
If y = \(x \log \frac{x}{(a+b x)}\), then show that \(x^3 \frac{d^2 y}{d x^2}=\left(x \frac{d y}{d x}-y\right)^2\). [5]
OR
Find whether the following function f is differentiable at x = 1 and x = 2 or not:
f(x) = \(\left\{\begin{array}{cl}
x, & \text { if } x<1 \\ 2-x, & \text { if } 1 \leq x \leq 2 \\ -2+3 x-x^2, & \text { if } x>2
\end{array}\right\}\)
Solution:
Question 34.
Solve the following linear programming problem graphically. [5]
Maximise: Z = 5x + 3y subject to constraints 2x + 5y ≤ 15; 5x + 2y ≤ 10; x ≥ 0; y ≥ 0
OR
Solve the following linear programming problem graphically.
Minimise: Z = 5x + 10y subject to constraints x + 2y ≤ 120; x + y ≥ 10; x – 2y ≥ 0, x ≥ 0, y ≥ 0
Solution:
We have, subject to the constraints
2x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0
Changing inequations to equations
For 2x + 5y = 15
For 5x + 2y ≤ 10
Plotting the equation on a graph paper, we get the bounded feasible region (Shaded region) with corners A(0, 3), B(\(\frac{20}{21}, \frac{55}{21}\)), C(2, 0)
Z = 5x + 3y
ZA = 5 × 0 + 3 × 3 = 9
ZB = \(5 \times \frac{20}{21}+3 \times \frac{55}{21}=\frac{265}{21}\)
ZC = 5 × 2 + 3 × 0 = 10
Z is maximum at \(\left(\frac{20}{21}, \frac{55}{21}\right)\) and maximum value is \(\frac{265}{21}\).
OR
We have, subject to the constraints
x + 2y ≤ 120, x + y ≥ 10, x – 2y ≥ 10, x – 2y ≥ 0, x ≥ 0, y ≥ 0
Changing inequations to equations
For x + 2y = 120
For x + y = 10
For x – 2y = 0
Plotting the equations on a graph paper, we get the bounded feasible region (Shaded region) with corners A(10, 0), B\(\left(\frac{20}{3}, \frac{10}{3}\right)\), C = (60, 30), D(120, 0)
Z = 5x + 10y
ZA = 5 × 10 + 10 × 0 = 50
ZB = \(5 \times \frac{20}{3}+10 \times \frac{10}{3}=\frac{200}{3}\)
ZC = 5 × 60 + 10 × 30 = 600
ZD = 5 × 120 + 10 × 0 = 600
Z is minimum at (10, 0) and the minimum value is 50.
Question 35.
Show that the differential equation \(\left[x \sin ^2\left(\frac{y}{x}\right)-y\right] d x+x \cdot d y=0\) is homogeneous. Find the particular solution of this differential equation, given that y = \(\frac{\pi}{4}\), when x = 1. [5]
Solution:
Section – E (12 Marks)
This section comprises 3 case-study/passage-based questions of 4 marks each with two sub-parts. First, two case-study questions have three sub-parts (A), (B), (C) of marks 1, 1, 2 respectively. The third case-study question has two subparts of 2 marks.
Question 36.
A trust having a fund of ₹ 30000 invests in two different types of bonds. The first bond pays 5% interest per annum which will be given to an orphanage and the second bond pays 7% interest per annum which will be given to ‘Cancer Aid Society’ an NGO. The trust wishes to divide ₹ 30000 among two types of bonds in such a way that they earn an annual total interest of ₹ 1800. [4]
Based on the above information, answer the following questions:
(A) If the amount invested in the first bond by ₹ x and in the second bond is ₹ y, then what is the system of equations formed?
(B) Write the system of equations in matrix form.
(C) Find the values of ‘x’ and ‘y’.
OR
What is the inverse of the matrix \(\left[\begin{array}{ll}
1 & 1 \\
5 & 7
\end{array}\right]\)?
Solution:
(A) Let x be invested in the first bond and ₹ y is invested in the second bond, and the total amount invested is ₹ 30,000
⇒ x + y = 30000
5% of x + 7% of y = 1800
⇒ 5x + 7y = 18000
(B) We have equations,
x + y = 30000 and 5x + 7y = 1800000
Then their Matrix form is,
\(\left[\begin{array}{ll}
1 & 1 \\
5 & 7
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{r}
30000 \\
180000
\end{array}\right]\)
(C) We have,
x + y = 30000 …….(i)
and 5x + 7y = 180000 ……(ii)
Multiplying equation (i) by 5 and subtracting equations (ii) from (i),
(5x + 5y) – (5x + 7y) = 150000 – 180000
⇒ -2y = -30000
⇒ y = 15000
From (i), x = 30000 – 15000 = 15000
Thus, x = 15000, y = 15000
OR
Question 37.
In a town, it’s rainy one-third of the day. Given that it is rainy, there will be heavy traffic with probability \(\frac{1}{2}\). Given that it is not rainy, there will be heavy traffic with probability \(\frac{1}{4}\). If it’s rainy and there is heavy traffic, I arrive late for work with probability \(\frac{1}{2}\). On the other hand, the probability of being late is reduced to \(\frac{1}{8}\) if it, is not rainy and there is no heavy traffic. In other situations (rainy and no heavy traffic, net rainy and heavy traffic), the probability of being late is \(\frac{1}{4}\). You pick a random day. [4]
Based on the above information, answer the following questions:
(A) What is the probability that it’s not raining and there is heavy traffic and I am not late?
(B) What is the probability that I am late?
(C) Given that I arrived late at work, what is the probability that it rained that day?
OR
If P(not A) = 0.7, P(B) = 0.7 and P(\(\frac{B}{A}\)) = 0.5, then find the (\(\frac{A}{B}\))
Solution:
From the given passage we can form a tree as below:
(A) (from the data given)
(B) P(I am late) = Sum of probabilities corresponds to “I am late”
P(RTL) + (RTCL) + (RCTL) + (RCTCL)
= \(\frac{1}{12}+\frac{1}{24}+\frac{1}{24}+\frac{1}{16}\)
= \(\frac{11}{48}\)
(C) \(P\left(\frac{R}{L}\right)=\frac{P(R \cap L)}{P(L)}\)
P(R ∩ L) = Sum of probabilities in which R and L are common
Question 38.
A student going to appear for the class XII exam had to attempt a few questions based on vector algebra. Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) by any 3 non-zero vectors.
Based on the above information, answer the following questions:
(A) If \(\vec{a}\) and \(\vec{b}\) are such that \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\). Then how is \(\vec{a}\) and \(\vec{b}\) are related to each other?
(B) Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) be unit vectors such that \(\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}=0\) and angle between \(\vec{a} \cdot \vec{c}=\frac{\pi}{6}\), then what is the value \(\vec{a}\). [4]
Solution:
