Students can access the CBSE Sample Papers for Class 12 Applied Maths with Solutions and marking scheme Set 2 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 12 Applied 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 is some internal ‘ choice in some questions.
- Section A has 18 MCQ’s and 02 Assertion Reason based questions of 1 mark each.
- Section B has 5 Very Short Answer (VSA) questions of 2 marks each.
- Section C has 6 Short Answer (SA) questions of 3 marks each.
- Section D has 4 Long Answer (LA) questions of 5 marks each.
- Section E has 3 source hased/case based/passage based/integrated units of assessment (04 marks each) with sub parts.
- Internal Choice is provided in 2 questions in Section-B, 2 questions in Section-C, 2 Questions in Section-D. You have to attempt only one alternatives in all such questions.
Section – A
(All Questions are compulsory. No internal choice is provided in this section)
Question 1.
If B > A, then which expression will have the highest value, given that A and B are positive integers.
(A) A – B
(B) A x B
(C) A + B
(D) Can’t say
Answer:
(D) Can’t say
Explanation:
As, B > A
⇒ A < B
⇒ A – B < 0 A + B > 0 and AB > 0
If A = 1, B = 4 then, AB < A + B If A = 2, B = 4 then, AB > A + B
Thus, we can’t say which one of A + B and AB has higher value.
Question 2.
Evaluate : 21 mod 2 = __________
(A) 2
(B) 19
(C) 10
(D) 1
Answer:
(D) 1
Explanation:
On dividing 21 by 2, we get remainder = 1
Hence, 21 mod 2 = 1.
Question 3.
The formula for standard error is
(A) s = \(\sqrt{\frac{\sum(x-\bar{x})^2+\sum(y-\bar{y})^2}{n_1+n_2-2}}\)
(B) s = \(\sqrt{\frac{\sum(x-\bar{x})^2+\sum(y-\bar{y})^2}{n_1-n_2-2}}\)
(C) s = \(\sqrt{\frac{\sum(x+\bar{x})^2+\sum(y+\bar{y})^2}{n_1+n_2-2}}\)
(D) s = \(\sqrt{\frac{\sum(x-\bar{x})^2+\sum(y-\bar{y})^2}{n_1+n_2}}\)
Answer:
(A) s = \(\sqrt{\frac{\sum(x-\bar{x})^2+\sum(y-\bar{y})^2}{n_1+n_2-2}}\)
Explanation:
The formula for standard error is
s = \(\sqrt{\frac{\sum(x-\bar{x})^2+\sum(y-\bar{y})^2}{n_1+n_2-2}}\)
Question 4.
Null hypothesis is represented by
(A) H0
(B) H1
(C) H2
(D) None of these
Answer:
(A) H0
Explanation:
Null hypothesis is represented by H whereas H1 represents the alternative hypothesis.
Question 5.
Speed of river is 6 km/hr. Speed of a motorboat in still water is 30 km/hr. How much distance can it cover downstream in 24 minutes?
(A) 9.8 km
(B) 12km
(C) 12.8 km
(D) 14.4 km
Answer:
(D) 14.4 km
Explanation:
Distance = Speed × Time
X + Y = 30 + 6 = 36 km/hr
24 minutes = \(\frac{24}{60}\) hours
Distance = 36 × \(\frac{24}{60}\) = 14.4 km
Question 6.
Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. Then the possible values of X are:
(A) 0, 1, 3, 5
(B) 0, 2, 4, 6
(C) 0, 2, 5, 6
(D) 1, 3, 4, 5
Answer:
(B) 0, 2, 4, 6
Explanation:
The coin is tossed six times and X represents the difference between the number of heads and the number of tails
∴ X(6H, 0T) = |6 – 0| = 6
X(5H, 1T) = |5 – 1| = 4
X(4H, 2T) = |4 – 2| = 2
X(3H, 3T) = |3 – 3| = 0
X(2H, 4T) = |2 – 4| = 2
X(1H, 5T) = |1 – 5| = 4
X(0H, 6T) = |0 – 6| = 6
thus, possible values are 0, 2, 4, 6.
Question 7.
For two distinct positive numbers x and y
(A) x + y > 2\(\sqrt{xy}\)
(B) \(\frac{x+y}{2}\) > xy
(C) \(\sqrt{xy}\) > \(\frac{x+y}{2}\)
(D) \(\frac{2 x y}{x+y}>\sqrt{x y}\)
Answer:
(A) x + y > 2\(\sqrt{xy}\)
Explanation:
For distinct x, y > 0;
A.M. > G.M.
⇒ \(\frac{x+y}{2}>\sqrt{x y}\)
⇒ x + y > 2 \(\sqrt{x y}\)
Question 8.
The negative cash flows are classified as
(A) Present cash
(B) Future cash
(C) Cash outflows
(D) Cash inflows
Answer:
(C) Cash outflows
Explanation:
Negative cash flow describes a situation in which a firm spends more cash than it takes in.
Question 9.
\(\int e^x\left(\frac{1-x}{1+x^2}\right)^2\) is equal
(A) \(\frac{e^x}{1+x^2}\) + C
(B) \(\frac{-e^x}{1+x^2}\) + C
(C) \(\frac{e^x}{\left(1+x^2\right)^2}\) + C
(D) \(\frac{-e^x}{\left(1+x^2\right)^2}\) + C
Answer:
(A) \(\frac{e^x}{1+x^2}\) + C
Explanation:
Question 10.
The most commonly use mathematical method for measuring the trend is
(A) Moving average method
(B) Semi average method
(C) Method of least squares
(D) None of these
Answer:
(C) Method of least squares
Question 11.
The interest earned plus principal amount is considered as
(A) Semiannual Amount
(B) Compound Amount
(C) Discount Amount
(D) Annual Amount
Answer:
(B) Compound Amount
Explanation:
The sum of principal and its interest is known as Compound Amount.
Question 12.
Assume that Rohan holds a perpetual bond that generates an annual payment of ₹ 1500 each year. He believes that the borrower is creditworthy and that 7% interest rate will be suitable for this bond. The present value of this perpetuity is
(A) ₹ 21429
(B) ₹ 22520
(C) ₹ 23480
(D) ₹ 24192
Answer:
(A) ₹ 21429
Explanation:
PV of perpetuity = \(\frac{\text { Annual Payment }}{\text { Interest Rate }}\)
= \(\frac{\frac{1500}{7}}{\frac{7}{100}}\)
= \(\frac{150000}{7}\)
= ₹ 21429
Question 13.
A company buys a microscope at a cost of ₹ 25000. The company decides on a salvage value of ₹ 8000 and a useful life of 6 years, then annual depreciation cost is
(A) ₹ 2383
(B) ₹ 2523
(C) ₹ 2783
(D) ₹ 2833
Answer:
(D) ₹ 2833
Explanation:
We know that Annual Depreciation expense = \(\frac{\text { Asset Cost }- \text { Residual Value }}{\text { Useful life of asset }}\)
= \(\frac{25000-8000}{6}\)
= ₹ 2833
Question 14.
The order and degree (if defined) of differential equation x \(\frac{d y}{d x}\) + 2y = x2 x ≠ 0 are
(A) 1, 1
(B) 1, 0
(C) 0, 1
(D) Not defined
Answer:
(A) 1, 1
Explanation:
The highest order derivative present is \(\frac{d y}{d x}\) and its raised to power 1. So, its degree is 1 and order is also 1.
Question 15.
The position of points O (0, 0) and P (2, – 2) in the region of graph of inequation 2x – 3y < 5 will be
(A) O and P both inside
(B) O and P both outside
(C) O inside and P both outside
(D) O outside and P inside
Answer:
(C) O inside and P both outside
Explanation:
Question 16.
For a Poisson distribution with mean λ, \(\sum_{k=0}^{\infty} \frac{e^{-\lambda} \lambda^k}{k !}\)
(A) – 1
(B) 0
(C) \(\frac{1}{2}\)
(D) 1
Answer:
(D) 1
Explanation:
\(\sum_{k=0}^{\infty} \frac{e^{-\lambda} \lambda^k}{k !}\) = Total probability = 1
Question 17.
A set of observations recorded at an equal interval of time is called
(A) Array data
(B) Data
(C) Geometric Series
(D) lime series data
Answer:
(D) lime series data
Explanation:
A time series is a set of observations taken at specified times, usually at equal intervals.
Question 18.
A simple random sample consist of four observation 2,4,6,8. What is the point estimate of population standard deviation?
(A) 2.5
(B) 2.6
(C) 3.6
(D) 4.5
Answer:
(B) 2.6
Explanation:
The point estimation of population standard deviation is sample deviation.
S = \(\sqrt{\frac{\Sigma\left(x_i-\bar{x}\right)^2}{n-1}}\)
where \(\bar{x}=\frac{\Sigma x}{n}\)
Here,
\(\bar{x}=\frac{2+4+6+8}{4}\)
= \( \frac{20}{4}\) = 5
S = \(\sqrt{\frac{(2-5)^2+(4-5)^2+(6-5)^2+(8-5)^2}{4-1}}\)
= \(\sqrt{\frac{9+1+1+9}{3}}\)
= \( \sqrt{\frac{20}{3}}\) = 2.6
DIRECTION :
In the following questions, a statement of Assertion (A) is followed by a statement of 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) : Given y = \(\frac{x}{x^2+1}\), then the value of second derivative at x = 1 is – \(\frac{1}{2}\)
Reason (R) : Since, y” = \(\frac{\left(x^2+1\right)^2(-2 x)-\left(1-x^2\right)\left[2\left(x^2+1\right) 2 x\right]}{\left(x^2+1\right)^4}\)
Answer:
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation:
Question 20.
Assertion (A) : If ‘p’, ‘q’ and ‘n’ are probability of success, failure and number of trials respectively in a Binomial Distribution, then its Standard deviation is \(\sqrt{npq}\).
Reason (R) : In a Binomial Distribution, the mean and variance are equal.
Answer:
(C) (A) is true but (R) is false.
Explanation:
Assertion (A) is true because the variance for a Binomial Distribution is given by npq.
So, Standard deviation = \(\sqrt{\text { Variance }}=\sqrt{n p q}\)
Reason (R) is false because
Mean of Binomial Distribution = np
and variance of Binomial Distribution = npq
Thus, mean is not equal to variance.
Section – B
(All Questions are compulsory. In case of internal Choice, attempt any one question only)
Question 21.
Anna owns a produce truck, invested ₹ 700 in purchasing the truck, some other initial admin related and insurance expenses of ₹ 1500 to get the business going and has now a day to day expense of ₹ 500/p.m. Consider hypothetically that her everyday profit is ₹ 550/p.m. (ideally, it will be based on sales). At the end of 6 months, Anna takes up her accounts and calculates her rate of return.
Answer:
Total Initial Investment: ₹ 700 + ₹ 1,500 = ₹ 2,200
Everyday Expenses: ₹ 500
Total Expenses for 6 months: ₹ 3,000
Everyday Returns: ₹ 550
Total Returns for 6 months: ₹ 3,300
So, Rate of Return = (Total Returns – Total Exenses
= \(\left(\frac{\text { Total Returns }- \text { Total Expenses }}{\text { Total Initial Investment }}\right)\) × 100
= \(\left(\frac{₹ 3,300-₹ 3,000}{₹ 2,200}\right)\) × 100
= 13.63%
Question 22.
Matrix A = \(\left[\begin{array}{ccc}
0 & 2 b & -2 \\
3 & 1 & 3 \\
3 a & 3 & -1
\end{array}\right]\) is given to be symmetric, find values of a and b.
Solution:
As A is a symmetric matrix,
Or, A = A’
∴ \(\left[\begin{array}{ccc}
0 & 3 & 3 a \\
2 b & 1 & 3 \\
-2 & 3 & -1
\end{array}\right]=\left[\begin{array}{ccc}
0 & 2 b & -2 \\
3 & 1 & 3 \\
3 a & 3 & -1
\end{array}\right]\)
∴ By equality of matrices, a = \(\frac{-2}{3}\) and b = \(\frac{3}{2}\).
OR
If A = \(\left[\begin{array}{ll}
p & 2 \\
2 & p
\end{array}\right]\) and |A| = 125, then find the value(s) of p.
Solution:
Given, A = \(\left[\begin{array}{ll}
p & 2 \\
2 & p
\end{array}\right]\)
and |A3| = 125
Since, |An| = |A|n
|A3| = |A|3
Now, |A| = \(\left[\begin{array}{ll}
p & 2 \\
2 & p
\end{array}\right]\)
= p2 – 4
According to given condition,
⇒ |A3| = 125
⇒ |A|3 = 125
⇒ (p2 – 4)3 = 125
⇒ (p2 – 4)3 = 53
⇒ p2 – 4 = 5
⇒ p2 = 9
Hence, values of p = ± 3.
Question 23.
The feasible solution for a LPP is shown in given figure. Let Z = 3x – 4y be the objective function. Find the minimum and maximum value of Z.
Answer:
Corner points | Corresponding value of Z = 3x – 4y |
(0, 0) | 0 |
(5, 0) | 15 ← Maximum |
(6, 5) | – 2 |
(6, 8) | – 14 |
(4, 10) | – 28 |
(0, 8) | – 32 ← Minimum |
Hence, the minimum of Z occurs at (0, 8) and its minimum value is (- 32)
And, the maximum of Z occurs at (5, 0) and its maximum value is 15.
Question 24.
Prove that 2n + 6 × 9n is always divisible by 7 for any positive integer n.
Answer:
2n + 6 × 9n = (mod 7)
2n + 6 × 9n = 7n for some integer 7
So, 2n + 6 × 9n is a multiple of 7.
When n = 0,
then 20 + 6 × 90 = 1 + 6 = 7 is a multiple of 7.
When n = 1,
then 2 + 6 × 91 = 2 + 54 = 56 is a multiple of 7.
When n = 2,
then 22 + 6 × 92 = 4 + 486 = 490 is a multiple of 7.
…………………..
Hence 2n + 6 × 9n is always divisible by 7 for any positive integer n.
OR
A cistern has two taps attached to it. Tap B can empty the cistern in 45 minutes. But Tap A can fill the cistern in just 30 minutes. Rohit started both taps unknowingly but realized his mistake after 30 minutes. He immediately closed Tap B. Now, in how many minutes the cistern will be filled?
Answer:
In 1 minute, cistern filled by Tap A = \(\frac{1}{30}\)
In 1 minute, cistern emptied by Tap B = \(\frac{1}{45}\)
In 1 minute with Taps A and B, cistern filled = \(\frac{1}{30}-\frac{1}{45}=\frac{1}{90}\)
In 30 minutes, 30 × \(\frac{1}{90}\) = \(\frac{1}{3}\) amount of cistern is filled
Remaining cistern = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
This is filled by Tap A only
In 1 minute, \(\frac{1}{30}\) part of cistern is filled
In x minutes, \(\frac{2}{3}\) amount of cistern is filled
∴ x minutes = \(\frac{\frac{2}{3}}{\frac{1}{30}}\)
= 20 minuutes
20 minuutes = Time to fill after Tap B is closed
Question 25.
The mean weekly sales of a four-wheeler were 50 units per agency in 20 agencies. After an advertising campaign, the mean weekly sales increased to 55 units per agency with standard deviation of 10 units. Test whether the advertising campaign was successful. (Use t0.005 = 1.729 for 19 d.f.)
Answer:
Sample size, n = 20 agencies
Sample mean \(\bar{x}\) = 55 units
Sample SD, σ = 10 units
Population mean, µ = 50 units
Null Hypothesis:
The advertising compaign is not successful
i.e., H0 : µ = 50
(There is no significant difference between the mean weekly Sales of units in agencies before and after advertising campaign)
Alternate Hypothesis, H1 : µ > 50.
The advertising campaign was successful.
Test statistics,
z > t0.05 = 1.729
Thus the advertising campaign was successful.
Section – C
(All Questions are compulsory. In case of internal Choice, attempt any one question only)
Question 26.
\(\int_1^4\) {|x – 1| + |x – 2| + |x – 4|} dx
Answer:
I = \(\int_1^4\) {|x – 1| + |x – 2| + |x – 4|} dx
= \(\int_1^4\) (x – 1) dx – \(\int_1^2\) (x – 2) dx + \(\int_2^4\) (x – 2) dx – \(\int_1^4\) (x – 4) dx
= \(\left.\left.\left.\left.\frac{(x-1)^2}{2}\right]_1^4-\frac{(x-2)^2}{2}\right]_1^2+\frac{(x-2)^2}{2}\right]_2^4-\frac{(x-4)^2}{2}\right]_1^4\)
= \(\frac{9}{2}+\frac{1}{2}+2+\frac{9}{2}\)
= 11 \(\frac{1}{2}\) or \(\frac{33}{2}\)
OR
Find: ∫ \(\frac{\sqrt{x^2+1}\left\{\log \left(x^2+1\right)-2 \log x\right\}}{x^4}\) dx.
Answer:
Let I = ∫ \(\frac{\sqrt{x^2+1}\left\{\log \left(x^2+1\right)-2 \log x\right\}}{x^4}\) dx
= ∫ \(\sqrt{1+\frac{1}{x^2}}\left(\log \left(1+\frac{1}{x^2}\right)\right) \frac{1}{x^3}\) dx
Let, 1 + \(\frac{1}{x^2}\) = t2
Or, \(\frac{-2}{x^3}\) dx = 2t dt
Or, \(\frac{1}{x^3}\) dx = – t dt
= – ∫ t (2 log t) t dt
= – 2 ∫ log t . t2 dt
= – 2 log t . \(\frac{t^3}{3}+\int 2 \frac{1}{t} \cdot \frac{t^3}{3}\) dt
= – \(\frac{2}{3}\) log t . t3 + \(\frac{2}{9}\) t3 + C
= – \(\frac{2}{3}\) t3 (log t – \(\frac{1}{3}\)) + C
= \(-\frac{2}{3}\left(1+\frac{1}{x^2}\right)^{3 / 2}\left(\log \left(1+\frac{1}{x^2}\right)-\frac{1}{3}\right)\) + C
Question 27.
Using properties of determinants, solve for x:
\(\left|\begin{array}{lll}
a+x & a-x & a-x \\
a-x & a+x & a-x \\
a-x & a-x & a+x
\end{array}\right|\) = 0.
Answer:
\(\left|\begin{array}{lll}
a+x & a-x & a-x \\
a-x & a+x & a-x \\
a-x & a-x & a+x
\end{array}\right|\) = 0
Applying R1 → R1 + R2 + R3
\(\left|\begin{array}{ccc}
3 a-x & 3 a-x & 3 a-x \\
a-x & a+x & a-x \\
a-x & a-x & a+x
\end{array}\right|\) = 0
Applying C2 → C2 – C1
\(\left|\begin{array}{ccc}
3 a-x & 0 & 3 a-x \\
a-x & 2 x & a-x \\
a-x & 0 & a+x
\end{array}\right|\) = 0
Applying C3 → C3 – C1
\(\left|\begin{array}{ccc}
3 a-x & 0 & 0 \\
a-x & 2 x & 0 \\
a-x & 0 & 2 x
\end{array}\right|\) = 0
Expanding along C3
4x2 (3a – x) = 0
∴ x = 0, 3a
Question 28.
Find the intervals in which the function f(x) = 3x4 – 4x3 – 12x2 + 5 is (a) strictly increasing (b) strictly decreasing.
Answer:
f'(x) = 12x3 – 12x2 – 24x
= 12x (x + 1) (x – 2)
f’(x) > 0,∀ x ∈ (- 1, 0) ∪ (2, ∞)
f’(x) < 0,∀ x ∈ (- ∞, – 1) ∪ (0, 2)
∴ f(x) is strictly increasing in (- 1, 0) ∪ (2, ∞) and strictly decreasing in (- ∞, – 1) ∪ (0, 2).
Question 29.
Under the pure market competition scenario, the demand function pd and the supply function ps for a certain commodity are given as pd = \(\frac{8}{x+1}\) – 2 and ps = \(\frac{x+3}{2}\) respectively, where p is the price and x is the quantity of the commodity. Using integrals, find the producer’s surplus.
Answer:
Under pure competition, pd = ps
⇒ \(\frac{8}{x+1}\) – 2 = \(\frac{x+3}{2}\)
⇒ x2 – 8x – 9 = 0
⇒ x = – 9, 1
∴ x = 1
when x0 = 1
⇒ p0 = 2
Producer’s surplus = p0 – \(\int_0^{x_0}\) ps dx
= 2 – \(\int_0^1 \frac{x+3}{2}\) dx
= 2 – \(\left[\frac{x^2}{4}+\frac{3 x}{2}\right]_0^1\)
= \(\frac{1}{4}\)
OR
The demand function p for maximising a profit monopolist is given by p = 274 – x2 while the marginal cost is 4 + 3x, for x units of the commodity. Using integrals, find the consumer surplus.
Answer:
p = 274 – x2
R = px = 274x – x3
\(\frac{d R}{d x}\) = 274 – 3x2
Given MR = 4 + 3x
In profit monopolist market,
MR = \(\frac{d R}{d x}\)
⇒ 4 + 3x = 274 – 3x2
⇒ x2 + x – 90 = 0
⇒ x = – 10, 9
∴ x = 9.
When x0 = 9
⇒ p0 = 193
Consume?s Surplus = \(\int_0^{x_0}\) p dx – p0 × x0
= \(\int_0^9\) (274 – x2) dx – 193 × 9
= \(\left[274 x-\frac{x^3}{3}\right]_0^9\)
= 2223 – 1737
= 486
Question 30.
Riya invested ₹ 20,000 in a mutual fund in year 2016. The value of mutual fund increased to ₹ 32,000 in year 2021. Calculate the compound annual growth rate of her investment. [Given, log (1.6) = 0.2041, antilog (0.04082) = 1.098]
Answer:
Given, beginning value of investment = ₹ 20,000
Final value of the investment = ₹ 32,000
No. of years = 5
So, CAGR = \(\left(\frac{\text { End Value }}{\text { Beginning Value }}\right)^{\frac{1}{n}}\) – 1
= \(\left(\frac{32000}{20000}\right)^{\frac{1}{5}}\) – 1
= \((1.6)^{\frac{1}{5}}\) – 1
Let, x = \((1.6)^{\frac{1}{5}}\)
Taking log both sides, we get
log x = \(\frac{1}{5}\) log (1.6)
⇒ log x = \(\frac{1}{5}\) × 0.2041
⇒ log x = 0.04802
⇒ x = antilog (0.04802) = 1.098
CAGR = 1.098 – 1
= 0.098
= 9.8%
Question 31.
A person amortizes a loan of ₹ 150000 for renovation of his house by 8 years mortgage at the rate of 12% p.a. compounded monthly. Find the equated monthly instalment. [Given (1.01)% = 2.5993]
Answer:
Given, P = ₹ 1,50,000,
i = \(\frac{12}{12 \times 100}\)
= \(\frac{1}{100}\) = 0.01
and n = 8 × 12 = 96
EMI = \(\frac{P \times i \times(1+i)^n}{(1+i)^n-1}\)
= \(\frac{15,00,000 \times 0.01 \times(1.01)^{96}}{(1.01)^{96}-1}\)
= \(\frac{15,00,000 \times 0.01 \times 2.5993}{2.5993-1}\)
= \(\frac{15,00,000 \times 0.01 \times 2.5993}{1.5993}\)
= ₹ 24,379.10
Section – D
(This section comprises of long answer type questions (LA) of 5 marks each)
Question 32.
(a) In a Poisson distribution probability of zero success is 10%. Find the mean, given that loge 10 = 2.3026.
Answer:
Let m be the mean. Then
P(X = x) = e-m \(\frac{m^x}{x !}\)
⇒ P(0) = e-m
e-m = 10% = \(\frac{10}{100}\) = 0.1
Or, e-m = 10
m = loge 10 = 2.3026
(b) A car-hire-firm has two cars, which it hires, out by the day by day. The number of demands for a car on each day is distributed as Poisson distribution with mean 1.5. Calculate the proportion of days on which neither car is used and proportion of days on which some demand is refused, [e– 1.5 = 0.2231]
Answer:
The proportion of days when no car will be required is
e-m \(\frac{m^0}{0 !}\) = 0.2231
The probability when no car, one car, two cars will be required is
= \(e^{-m} \frac{m^0}{0 !}+e^{-m} \frac{m^1}{1 !}+e^{-m} \frac{m^2}{2 !}\)
= e-m (1 + m + \(\frac{1}{2}\) m2)
= 0.2231 (1 + 1.5 + 1.125)
= 0.80807375
∴ The proportion of days on which some demand is refused = 1 – 80807375 = 0.1912625.
OR
A box has 20 pens of which 2 are defective. Calculate the probability that out of 5 pens drawn one by one with replacement, at most 2 are defective.
Answer:
Let X be a random variable that denotes the number of defective pens is a draw of 5 pens.
Then, X can take values 0, 1, 2, 3, 4 and 5.
Here, p = probability of getting a defective pen in a single draw = \(\frac{2}{20}=\frac{1}{10}\)
and q = probability of not getting a defective pen in a single draw = 1 – p
= 1 – \(\frac{1}{10}=\frac{9}{10}\)
Clearly, X follows binomial distribution with parameters n = 5, p = \(\frac{1}{10}\) and q = \(\frac{9}{10}\)
∴ P(X = r) = \({ }^5 C_r\left(\frac{1}{10}\right)^r\left(\frac{9}{10}\right)^{5-r}\)
r = 0, 1, 2, 3, 4 and 5
Now, P(getting at most 2 defective pen)
= P (x ≤ 2)
= P(X = 0) + P(X = 1) + (X = 2)
= \({ }^5 C_0\left(\frac{1}{10}\right)^0\left(\frac{9}{10}\right)^5+{ }^5 C_1\left(\frac{1}{10}\right)\left(\frac{9}{10}\right)^4\) + \({ }^5 C_2\left(\frac{1}{10}\right)^2\left(\frac{9}{10}\right)^3\)
= \(\frac{9^3}{10^5}\) [92 + 5 × 9 + 10]
= \(\frac{729}{100000}\) [81 + 45 + 10]
= \(\frac{729 \times 136}{100000}\)
= \(\frac{99144}{100000}\)
= 0.99144
Question 33.
A company manufactures two types of toys A and B. A toy of type A requires 5 minutes for cutting and 10 minutes for assembling. A toy of type B requires 8 minutes for cutting and 8 minutes for assembling. There are 3 hours available for cutting and 4 hours available for assembling the toys in a day. The profit is ₹ 50 each on a toy of type A and ₹ 60 each on a toy of type B. How many toys of each type should the company manufacture in a day to maximize the profit? Use linear programming to find the solution.
Answer:
Let x be the no. of toys of Type A
and y be the no. of toys of Type B.
Toy A | Toy B | Time in a day | |
Cutting time | 5 min | 8 min | 180 min |
Assembling time | 10 min | 8 min | 240 min |
Objective function
Minimize: Z = 50x + 60y
Subject to: 10x + 8y ≤ 240
Or, 5x + 4y ≤ 120 ……………..(i)
x ≥ 0, y ≥ 0 ……………(ii)
5x + 8y ≤ 180 ……………(iii)
For, 5x + 8y = 180
A | B | E | |
x | 0 | 36 | 12 |
y | 22.5 | 0 | 15 |
For, 5x + 4y = 120
C | D | E | |
x | 0 | 24 | 12 |
y | 30 | 0 | 15 |
Corner points | Z = 50x + 60y |
At O (0,0) | 0 |
At D(24, 0) | 1200 |
At E(12, 15) | 1500 (Max.) |
At A(0, 22.5) | 1350 |
Hence, the maximum profit is ₹ 1500 at E (12, 15).
Therefore 12 units of type A and 15 units of type B is produced.
Question 34.
A tank with rectangular base and rectangular sides open at the top is to be constructed so that its depth is 3 m and volume is 75 m3. If building of tank costs ₹ 100 per square metre for the base and 50 per square metre for the sides, find the cost of least expensive tank.
Answer:
Let l, b, h be the length, breadth and depth.
∴ l × b × 3 = 75
⇒ l × b = 25
Let C be the cost, then
C = 100(l × b) + 50 × 2[h (b + l)]
= 100 (1 × \(\frac{25}{l}\)) + 300 (\(\frac{25}{l}\) + 1)
= 2500 + 300 (\(\frac{25}{l}\) + l)
Differentiating w.r.t l,
\(\frac{d C}{d t}\) = 0 + 300 \(\left(\frac{-15}{l^2}+1\right)\)
For maximum and minimum cost,
\(\frac{d C}{d t}\) = 0
⇒ 300 \(\left(-\frac{25}{l^2}+1\right)\) = 0
⇒ l2 = 25 or l = ± 5
Getting \(\left(\frac{d^2 C}{d l^2}\right)=300\left(\frac{50}{l^3}\right)\)
⇒ \(\left(\frac{d^2 C}{d l^2}\right)_{\text {at } l=5}=\frac{15000}{125}\) > 0
i.e., C is minimum, when l = 5
⇒ b = 5
∴ C = 100 (25) + 300 (10)
= 2,500 + 3,000
= 5,500
Hence the minimum cost is ₹ 5,500.
OR
AB is a diameter of a circle and C is any point on the circle. Show that the area of ∆ABC is maximum, when it is isosceles.
Answer:
Let the sides of rt. ∆ABC be x and y.
∴ x2 + y2 = 4r2
and A = Area of ∆ = \(\frac{1}{2}\) xy
Let, S = A2
= \(\frac{1}{4}\) x2y2
= \(\frac{1}{4}\) x2 (4r2 – x2)
= \(\frac{1}{4}\) (4r2x2 – x4)
∴ \(\frac{d S}{d x}\) = \(\frac{1}{4}\) [8r2x – 4x3]
Or, \(\frac{d S}{d x}\) = 0
or x2 = 2r2
or x = √2r
and y2 = 4r2 – 2r2
= 2r2
or y = √2r
i.e., x = y
and \(\frac{d^2 S}{d x^2}\) = (2r2 – 3x2)
= 2r2 – 6r2 < 0
or Area is maximum, when ∆ is isosceles.
Hence Proved.
Question 35.
The equilibrium conditions for three competitive markets are described as given below, where p1, p2 and p3 are the equilibrium price for each market respectively.
P1 + 2p2 + 3p3 = 85
3p1 + 2p2 + 2p3 = 105
2p1 + 3p2 + 2p3 = 110
Using matrix method, find the values of respective equilibrium prices.
Answer:
A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 2 & 2 \\
2 & 3 & 2
\end{array}\right]\)
|A| = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
3 & 2 & 2 \\
2 & 3 & 2
\end{array}\right|\)
= \(1\left|\begin{array}{ll}
2 & 2 \\
3 & 2
\end{array}\right|-2\left|\begin{array}{ll}
3 & 2 \\
2 & 2
\end{array}\right|+3\left|\begin{array}{ll}
3 & 2 \\
2 & 3
\end{array}\right|\)
= – 2 – 4 + 15 = 9
∴ A-1 exists.
∴ p1 = 15,
p2 = 20,
p3 = 10.
Section – E
(This section comprises of 3 source based questions (Case Studies) of 4 marks each)
CASE STUDY-I
Question 36.
Read the following text and answer the following questions on the basis of the same:
Akshay and Ritesh are the neighbours. They are studying in the same school. Both are taking tuitions from ( same teacher. One day they have combined tuition class in which teacher explain about the topic Alligation and Mixtures.
Teacher said, according to alligation rule, if two ingredients at a given price are mixed to produce a mixture at given price, the ratio of quantity of cheaper ingredient and quantity of dearer ingredient is given by
\(\frac{\text { Quantity of cheaper }}{\text { Quantity of dearer }}=\frac{\text { Cost price of dearer }- \text { Mean price }}{\text { Mean price }- \text { Cost price of cheaper }}\)(i) In what ratio must a grocer mix two varieties of pulses costing ₹ 15 and ₹ 20 per kg respectively so as to get a mixture worth ₹ 16.50 kg?
Answer:
By the rule of alligation:
(ii) In what ratio must rice at ₹ 9.30 per kg be mixed with rice at ₹ 10.80 per kg so that the mixture be worth ₹ 10 per kg?
Answer:
By the rule of alligation, we have
∴ Required ratio = 80 : 70 = 8 : 7
(iii) Five litres of water is added to a certain quantity of pure milk costing ₹ 60 per litre. If by selling the mixture at same price as before, a profit of 20% is made, what is the amount of pure milk in the mixture?
Answer:
Let quantity of pure milk be x litres. It is given that the mixture is sold at a profit of 20% and the selling price of the mixtures is 60 per litre.
∴ S.P = C.P. (1 + \(\frac{\text { Profit }}{100}\))
⇒ 60 = C.P. (1 + \(\frac{20}{100}\))
⇒ C.P. = 60 × \(\frac{5}{6}\) = 50
So, the mixture is sold at ₹ 50 per litre using alligation rule, we have
Thus, \(\frac{\text { Quantity of pure milk }}{\text { Quantity of water }}=\frac{50}{10}\)
⇒ \(\frac{x}{5}=\frac{5}{1}\)
⇒ x = 25
Hence, the amount of pure milk in the mixture is 25 litres.
OR
Two liquids one mixed in the proportion 3 : 2 and the mixture is sold at ₹ 110 per litre at 10% profit. If first liquid costs ₹ 20 more per litre than the second, what does it cost per litre?
Answer:
We have, S.P of mixture = ₹ 110,
profit = 10%
∴ S.P. = C.P. (1 + \(\frac{\text { Profit }}{100}\))
⇒ 110 = C.P. (1 + \(\frac{10}{100}\))
⇒ 110 = C.P. \(\frac{110}{100}\)
⇒ C.P. = ₹ 100
Let the C.P. of second liquid be ₹ x, then, C.P. of first liquid = ₹ (x + 20)
Thus, we have
C.P of first liquid = ₹ (x + 20)
C.P of mixture = ₹ 100
The alligation grid is as given below:
Using alligation rule, we get
\(\frac{\text { Quantity of first liquid }}{\text { Quantity of second liquid }}=\frac{100-x}{x-80}\)
But it is given that two liquids are mixed in the proportion 3 : 2.
∴ \(\frac{100-x}{x-80}=\frac{3}{2}\)
⇒ 200 – 2x = 3x – 240
⇒ 5x = 440
⇒ x = 88
hence, cost price of second liquid is ₹ 88.
CASE STUDY-II
Question 37.
Read the following text and answer the following questions on the basis of the same:
Three friends Rohan, Ajay and Sameer were playing cards. They were trying to learn new games with cards. While playing the different games with cards, Sameer was thinking about the lesson of probability which was taught in yesterday’s class where the teacher told about the probability of drawing different cards in different situations. Soon, he came up with the problem of probability in which two cards were drawn simultaneously (or successively without replacement) from a well shuffled pack of 52 cards.
Based on the above information answer the following questions:
(i) Find the probability of drawing one king and one non-king card.
Answer:
P(one king and one non-king) = P(one king and one non-king)
= \(\frac{4}{52} \times \frac{48}{51} \times 2\)
= \(\frac{32}{221}\)
(ii) Find the probability of drawing two king cards.
Answer:
P(two kings) = \(\frac{48}{52} \times \frac{47}{51}\)
= \(\frac{188}{221}\)
OR
Find the probability of drawing no king card.
Answer:
P (no king card) = \(\frac{48}{52} \times \frac{47}{51}=\frac{188}{221}\)
(iii) Find the mean of the probability distribution.
Answer:
Here, X = no. of kings = 0, 1, 2
P(no king card) = \(\frac{48}{52} \times \frac{47}{51}\)
= \(\frac{188}{221}\)
P(one king card) = \(\frac{4}{52} \times \frac{48}{51} \times 2\)
= \(\frac{32}{221}\)
P(two kings) = \(\frac{4}{52} \times \frac{3}{51}\)
= \(\frac{1}{221}\)
Probability distribution is given by
Now, Mean = Σ X . P(X)
= \(\frac{34}{221}\) or \(\frac{2}{13}\)
CASE STUDY-III
Question 38.
The following data shows the percentage of rural, urban and suburban Indians who have a high speed internet connection at home:
Fit a straight line trend by the method of least square for the Rural Indians.
Answer:
Here, n = 5,
a = \(\frac{\Sigma y}{n}\)
and b = \(\frac{\Sigma x y}{\Sigma x^2}\)
So, a = \(\frac{58}{5}\) = 11.6
and b = \(\frac{52}{10}\) = 5.2
Thus, trend equation is given by yt = a + bx
i.e. yt = 11.6 + 5.2x
OR
Fit a straight line trend by the method of least square for the Urban Indians.
Answer:
Here, n = 5,
a = \(\frac{\Sigma y}{n}\)
and b = \(\frac{\Sigma x y}{\Sigma x^2}\)
So, a = \(\frac{115}{5}\) = 23
and b = \(\frac{69}{10}\) = 6.9
Thus, trend equation is given by yt = a + bx
i.e. yt = 23 + 6.9 x