Students can access the CBSE Sample Papers for Class 12 Applied Maths with Solutions and marking scheme Set 1 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 12 Applied Maths Set 1 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 based/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.
Let m ∈ Z+ consider the relation Rm defined as a Rm b iff a ≅ b (mod m), then Rm is
(A) reflexive but not symmetric
(B) symmetric but not transitive
(C) reflexive, symmetric but not transitive
(D) an equivalence relation
Answer:
(D) an equivalence relation
Explanation:
The relation Rm defined as a b (mod m) is reflexive, symmetric and transitive.
∴ Rm is an equivalent relation.
Question 2.
Pipe A and B can fill a tank in 5 hours and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the time taken to fill the tank is
(A) 2 hours
(B) 2 \(\frac{3}{4}\) hours
(C) 3 hours
(D) 3 \(\frac{9}{17}\) hours
Answer:
(D) 3 \(\frac{9}{17}\) hours
Explanation:
Here, LCM of 5,6 and 12 is 60.
i.e., LCM (5, 6, 12) = 60
Suppose capacity of the tank is 60 l.
Then, quantity filled by pipe A in 1 hr = \(\frac{60}{5}\) = 12 l
Quantity filled by pipe B in 1 hr = \(\frac{60}{6}\) = 10 l
Quantity emptied by pipe C in 1 hr = \(\frac{60}{12}\) = 5 l
So, quantity filled by all three pipes in 1 hr = (12 + 10 – 5) l = 17 l
Required time = \(\frac{60}{17}\) = 3 \(\frac{9}{17}\) hours
Question 3.
A statement made about a population parameter for testing purpose is called _________
(A) statistic
(B) parameter
(C) hypothesis
(D) level of significance
Answer:
(C) hypothesis
Explanation:
Statistical hypothesis is an assumption or statement, which may or may not be true, about a population.
Question 4.
The sample mean is given by
(A) Σ \(\frac{X_i}{n}\)
(B) Σ \(\frac{X_i}{n}-\bar{x}\)
(C) Σ \(\frac{X_i}{n}+\bar{x}\)
(D) Σ \(\frac{n}{X_i}\)
Answer:
(A) Σ \(\frac{X_i}{n}\)
Explanation:
Sample mean \((\bar{x})=\frac{\sum X_i}{n}\)
where n = sample size
Question 5.
The solution set of the inequation |x + 2| ≤ 5 is
(A) (- 7, 5)
(B) [- 7, 3]
(C) [- 5, 5]
(D) (- 7, 3)
Answer:
(B) [- 7, 3]
Explanation:
|x + 2| ≤ 5
⇒ – 5 ≤ x + 2 ≤ 5
[∵ |x| ≤ a,then – a ≤ x ≤ a]
⇒ – 5 – 2 ≤ x ≤ 5 – 2
⇒ – 7 ≤ x ≤ 3
⇒ x ∈ [- 7, 3]
Question 6.
A variable that can assume any value between two given points is called ___________
(A) Continuous Random Variable
(B) Discrete Random Variable
(C) Irregular Random Variable
(D) Uncertain Random Variable
Answer:
(A) Continuous Random Variable
Explanation:
This is the definition of a continuous random variable.
Question 7.
If x = – 4 (mod 3), then a solution for x is
(A) – 2
(B) 12
(C) 19
(D) 35
Answer:
(D) 35
Explanation:
Since, 3 |(x +4)| is true for x = 35
Question 8.
An investment of ₹ 10,000 becomes ₹ 60,000 in 4 years, then the CAGR (compound annual growth rate) is given by
(A) \(\frac{\sqrt[4]{6}-1}{100}\)
(B) \(\frac{\sqrt[4]{6}+1}{100}\)
(C) [\(\sqrt[4]{6}\) – 1] × 100
(D) [\(\sqrt[4]{6}\) + 1] × 100
Answer:
(C) [\(\sqrt[4]{6}\) – 1] × 100
Explanation:
CAGR = [\(\left(\frac{\text { Final investment }}{\text { Initial investment }}\right)^{\frac{1}{n}}\) – 1] × 100
= \(\left[\left(\frac{60000}{10000}\right)^{\frac{1}{4}}-1\right]\) × 100
= [\(\sqrt[4]{6}\) – 1] × 100
Question 9.
∫ (x – 1) e-x dx is equal to
(A) (x – 2) e-x + C
(B) x-x + C
(C) – xe-x + C
(D) (x + 1 )e-x + C
Answer:
(C) – xe-x + C
Explanation:
Let I = ∫ (x – 1)e-x dx
Let – x = t
⇒ – dx = dt
∴ I = – ∫ et (- t – 1) dt
= et (t + 1) dt
[∵ ∫ ex [f(x) +f'(x)] dx = ex f(x) + c]
= tet + c
I = – x e-x + C
Question 10.
The graph of time series is called
(A) Histogram
(B) Straight line
(C) Historigram
(D) Ogive
Answer:
(B) Straight line
Question 11.
If the nominal rate of interest is 12.5 % and the inflation is 2 %, then the effective rate of interest is
(A) 10.5 %
(B) 10 %
(C) 9.5 %
(D) 9 %
Answer:
(A) 10.5 %
Question 12.
Which of the following is not the value of depreciation asset at the end of useful life ? 1
(A) Scrap value
(B) Salvage value
(C) Depreciate value
(D) Useful value
Answer:
(D) Useful value
Explanation:
Salvage value is sometimes referred to as disposal value, residual value, terminal value, or scrap value. The estimated salvage value is deducted from the cost of the assets in order to determine the total amount of depreciation expenses that will be reported during the assets useful life.
Question 13.
Mr. Roy purchased a home for ₹ 2,00,000 in 2012. In the next few years, homes in neighbourhood have been selling well due to the new shopping plaza a couple of miles away, which increased the market value of his home. So in 2017, he decided to downsize and sell his home. Based on the current market value during this time, he was able to sell his home for ₹ 2,50,000. The rate of return is
(A) 15 %
(B) 20 %
(C) 22.5 %
(D) 25 %
Answer:
(D) 25 %
Explanation:
Given,Current value = 2,50,000
Original value = 2,00,000
Rate of return = \(\frac{\text { Current Value – Original Value }}{\text { Original Value }}\) × 100
Original Value = \(\frac{2,50,000-2,00,000}{2,00,000}\) × 100
= 25 %
Question 14.
The order and degree (if defined) of differential equation \(\frac{d y}{d x}\) = ky, where k is a scalar are 1
(A) 1, 1
(B) 0, 1
(C) 1, 0
(D) Not defined
Answer:
(A) 1, 1
Explanation:
The highest order derivative present is \(\frac{d y}{d x}\) and it is raised to power 1. So, its order is 1 and degree is also 1.
Question 15.
Objective function of an LPP is
(A) a constant
(B) a function to be optimized
(C) a relation between the variables
(D) None of these
Answer:
(B) a function to be optimized
Question 16.
E(X = λ) is for which distribution?
(A) Bernoulli’s
(B) Binomial
(C) Poisson’s
(D) None of these
Answer:
(C) Poisson’s
Explanation:
In Poisson’s distribution, there is a positive constant A which is the mean of the distribution and variance of the distribution.
Question 17.
A set of observations recorded at an equal interval of time is called
(A) Array data
(B) Data
(C) Geometric Series
(D) Time series data
Answer:
(D) Time series data
Explanation:
A time series is a set of observations taken at specified times, usually at equal intervals.
Question 18.
The process by which one makes inferences about a population, based on the information obtained from a sample is known as
(A) sample mean
(B) estimation
(C) hypothesis
(D) none of these
Answer:
(B) estimation
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) : The function f(x) = (x + 2)e-x is increasing in the interval (- 1, ∞).
Reason (R) : A function fix) is increasing, if f(x) > 0.
Answer:
(D) (A) is false but (R) is true.
Explanation:
We have assertion,
f(x) = (x + 2)e-x
f’(x) = e-x – (x + 2)e-x
= e-x(1 – x – 2)
= e-x (x + 1)
Here, e-x will be always positive, so the function will increasing when – (x + 1) > 0
Or, x < – 1
So,f(x) > 0 when x ∈(- ∞ , – 1).
Thus, Assertion is incorrect.
Here, Reason is the base correct condition for a function be an increasing function.
Question 20.
Given:
If f(X) = \(\begin{cases}X-\frac{5}{2}, & 0<X<1 \\ 2 X, & 1<X<2 \\ 0, & \text { otherwise }\end{cases}\)
Assertion (A) : The mean of the random variable (X) is 3.75.
Reason (R) : The mean of the random variable (X) is 3.
Answer:
(C) (A) is true but (R) is false.
Explanation:
Given, f(x) = \(\begin{cases}X-\frac{5}{2}, & 0<X<1 \\ 2 X, & 1<X<2 \\ 0, & \text { otherwise }\end{cases}\)
Then, E(X) = \(\int_0^1\) X (X – \(\frac{5}{2}\)) dX + \(\int_1^2\) X (2X) dx
= \(\left[\frac{X^3}{3}-\frac{5 X^2}{4}\right]_0^1+\left[\frac{2 X^3}{3}\right]_1^2\)
= \(\frac{1}{3}-\frac{5}{4}+\frac{16}{3}-\frac{2}{3}\)
= 3.75
Section – B
(All Questions are compulsory. In case of internal Choice, attempt any one question only)
Question 21.
₹ 5000 is invested in a Term Deposit Scheme that fetches interest 6 % per annum compounded quarterly. What will be interest after one year? [Given (1.015)4 = 1.0613]
Answer:
We know that,
Compound interest, I = P [(1 + i)n – 1]
Here, P = ₹ 5000,
i = 6 % p.a.
= 0.06 p.a.
= 0.06 × (1/4) per quarter
= 0.015 per quarter
∴ I = 5000 [(1 + 0.015)4 – 1]
= 5000 {(1.015)4 – 1}
= 5000 (1.0613 – 1)
= 5000 × 0.0613
= 306.82
Question 22.
Determine the integral value(s) of x for which the matrix A is singular.
A = \(\left[\begin{array}{ccc}
x+1 & -3 & 4 \\
-5 & x+2 & 2 \\
4 & 1 & x-6
\end{array}\right]\)
Answer:
We have,
A = \(\left[\begin{array}{ccc}
x+1 & -3 & 4 \\
-5 & x+2 & 2 \\
4 & 1 & x-6
\end{array}\right]\) is singular
∴ |A| = 0
⇒ \(\left|\begin{array}{ccc}
x+1 & -3 & 4 \\
-5 & x+2 & 2 \\
4 & 1 & x-6
\end{array}\right|\) = 0
So, [(x + 1) {(x + 2) (x – 6) – 2) + 3{- 5 (x – 6) – 8) + 4 {- 5 – 4 (x + 2)}] = 0
[(x + 1) {(x + 2) (x – 6) – 2} + 3 {- 5 (x – 6) – 8} + 4 {- 5 – 4 (x + 2)}] = 0
((x + 1) (x2 – 4x – 14) + 3 (- 5x + 22) + 4(- 13 – 4x)] = 0
x3 – 4x2 – 14x + x2 – 4x – 14 – 15x + 66 – 52 – 16x = 0
x3 – 3x2 – 49x = 0
x (x2 – 3x – 49) = 0
∴ x = 0 or x2 – 3x – 49 = 0
x = 0 or x = \(\frac{3 \pm \sqrt{9+4(49)}}{2}\)
x = 0 or x = \(\frac{1}{2}(3 \pm \sqrt{205})\)
OR
If A = \(\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]\), then find (A2 – 5A).
Answer:
Given A = \(\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]\)
Question 23.
The sum of two positive integers is atmost 5. The difference between two times of second number and first number is atmost 4. If first number is x and second number is y, then for maximizing the product of these two numbers, formulate LPP.
Answer:
The required LPP is
Max. Z = x . y
Subject to:
x + y ≤ 5
2y – x ≤ 4
x ≥ 0
y ≥ 0
Question 24.
In a one-kilometer race, A beats B by 30 seconds and B beats C by 15 seconds. 1f A beats C by 180 metres, then find the time taken by A to run 1 kilometre.
Answer:
Given,
A beats B by 30 seconds
B beats C by 15 seconds
So, A beats C by (30 + 15) seconds 45 seconds
Now, time taken by C to travel 180 m = 45 seconds
∴ Time taken by C to cover the distance of 1 km (1000 m) = \(\frac{45}{180}\) × 1000 = 250 seconds
Thus, time taken by A to cover the distance of 1 km = (250 – 45) = 205 seconds.
OR
Find the remainder when 561 is divided by 7.
Answer:
561 = (53)20 51
= (126 – 1)20 51
= [12620 – \({ }^{20} C_1\) 12619 + \({ }^{20} C_2\) 12618 + …………. + \({ }^{20} C_19\) 1261 + \({ }^{20} C_20\) 1260 120] × 5
= 126 [12619 – \({ }^{20} C_1\) 12618 + \({ }^{20} C_2\) 12617 + …………… + \({ }^{20} C_19\)] × 5 + 1 × 5
= 126 × 5 [12619 – \({ }^{20} C_1\) 12618 + \({ }^{20} C_2\) 12617 + ……………. + \({ }^{20} C_19\)] + 5
Here, 126 × 5 =630, which is divisible by 7, thus remainder is 5.
Question 25.
What is statistic ? Define parameter in Statistics.
Answer:
A statistic is used to estimate the value of a population parameter. For instance, we selected a random sample of 100 students from a school with 1000 students. The average height of the sampled students would be an example of a statistic.
A parameter is any numerical quantity that characterizes a given population or some aspect of it. This means the parameter tells us something about the whole population. For example, the population means µ, variance σ2, population proportion P, population correlation ρ.
Section – C
(All Questions are compulsory. In case of internal Choice, attempt any one question only)
Question 26.
If [x] denotes the greatest integer function, then find \(\int_0^{3 / 2}\) [x2] dx.
Answer:
\(\int_0^{3 / 2}\) [x2] dx
For 0 ≤ x < 1, 0 ≤ x2 < 1,
hence [x2] = 0
For 1 ≤ x < √2, 1 ≤ x2 < 2,
hence [x2] = 1
For √2 ≤ x < \(\frac{3}{2}\), 2 ≤ x2 < \(\frac{9}{4}\),
hence [x2] = 2
\(\int_0^{3 / 2}\) [x2] dx = \(\int_0^1 0 d x+\int_1^{\sqrt{2}} 1 d x+\int_{\sqrt{2}}^{3 / 2} 2 d x\)
= \(0[x]_0^1+1[x]_1^{\sqrt{2}}+2[x]_{\sqrt{2}}^{3 / 2}\)
= 0 + √2 – 1 + 3 – 2√2
= 2 – √2
Find : ∫ \(\frac{\left(x^4-x\right)^{\frac{1}{4}}}{x^5}\) dx.
Answer:
Let I = ∫ \(\frac{\left(x^4-x\right)^{\frac{1}{4}}}{x^5}\) dx
Question 27.
Using properties of determinants, prove the following:
\(\left|\begin{array}{ccc}
1+a^2-b^2 & 2 a b & -2 b \\
2 a b & 1-a^2+b^2 & 2 a \\
2 b & -2 a & 1-a^2-b^2
\end{array}\right|\) = (1 + a2 + b2)3.
Answer:
∆ = \(\left|\begin{array}{ccc}
1+a^2-b^2 & 2 a b & -2 b \\
2 a b & 1-a^2+b^2 & 2 a \\
2 b & -2 a & 1-a^2-b^2
\end{array}\right|\)
Using C1 → C1 – bC3
C2 → C2 + aC3
= \(\left|\begin{array}{ccc}
1+a^2+b^2 & 0 & -2 b \\
0 & 1+a^2+b^2 & 2 a \\
b\left(1+a^2+b^2\right) & -a\left(1+a^2+b^2\right) & 1-a^2-b^2
\end{array}\right|\)
Taking (1 + a2 + b2) common from C1 and C2
= (1 + a2 + b2)2 \(\left|\begin{array}{ccc}
1 & 0 & -2 b \\
0 & 1 & 2 a \\
b & -a & 1-a^2-b^2
\end{array}\right|\)
Using R3 → R3 + aR2 – bR1
= (1 + a2 + b2)2 \(\left|\begin{array}{ccc}
1 & 0 & -2 b \\
0 & 1 & 2 a \\
0 & 0 & 1+a^2+b^2
\end{array}\right|\)
Taking (1 + a2 + b2)2 common from R
= (1 + a2 + b2)3 \(\left|\begin{array}{ccc}
1 & 0 & -2 b \\
0 & 1 & 2 a \\
0 & 0 & 1
\end{array}\right|\)
Expanding with respect to R3
∆ = (1 + a2 + b2)3 (1)
Or, ∆ = (1 + a2 + b2)3
Question 28.
Determine for what values of x, the function f(x) = x3 + \(\frac{1}{x^3}\) (x ≠ 0) is strictly increasing or strictly decreasing.
Answer:
Here, f'(x) = 3x2 – 3x-4
= \(\frac{3\left(x^6-1\right)}{x^4}\)
= \(\frac{3\left(x^4+x^2+1\right)}{x^4}\) (x + 1) (x – 1)
Critical points are – 1 and 1.
Or, f'(x) > 0 if x > 1 or x < – 1, and f'(x) < 0 if – 1 < x < 1 (∵ \(\frac{3\left(x^4+x^2+1\right)}{x^4}\) always +ve)
Hence, f(x) is strictly increasing for x > 1.
Or, x < – 1 and strictly decreasing in (- 1, 0) ∪ (0, 1).
Question 29.
Find the area of the region bounded by x = 4y, y = 2, x = 4 and the Y-axis in the first quadrant.
Answer:
The area of the region bounded by the curve, x2 = 4y, y = 2 and y = 4 and the y-axis is the area ABCD.
Area of ABCD = \(\int_2^4\) x dy
= \(\int_2^4\) 2√y dy
= 2 \(\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_2^4\)
= \(\frac{4}{3}\left[(4)^{\frac{3}{2}}-(2)^{\frac{3}{2}}\right]\)
= \(\frac{4}{3}\) [8 – 2√2]
= \(\left(\frac{32-8 \sqrt{2}}{3}\right)\) units2
OR
If the demand function for a commodity is p = 35 – 2x – x2. Then find the consumer’s surplus at equilibrium price p0 = 20.
Answer:
Given, the demand function is
p = 25 – 2x – x2
and the equilibrium price, p0 = 20.
Note:
At equilibrium demand function is
p0 = 35 – 2x0 – x02
Substituting this value of p0 in (i), we get
20 = 35 – 2x0 – x02
⇒ x02 + 2x0 – 15 = 0
⇒ (x0 + 5) (x0 – 3) = 0
⇒ x0 = – 5, or x0 = 3
∵ x0 ≠ – 5, so x0 = 3
We know that, Consumer’s Surplus
CS = \(\int_0^{Q_e}\) D(x) dx – Qe Pe
Where, D(x) = Demand Function
Qe = Quantity Sold
Pe = Unit Market Price
Here, D(x) = p, Pe = p0 = 20 and Qe = x0 = 3
So, Consumers Surplus
= \(\int_0^3\) (35 – 2x – x2) dx – 3 × 20
= \(\left[35 x-x^2-\frac{x^3}{3}\right]_0^3\) – 60
= (105 – 9 – 9) – 60 = 27
Question 30.
Madhu exchanged her old car valued at ₹ 1,50,000 with a new one priced at ₹ 6,50,000. She paid ₹ x as down payment and the balance in 20 monthly equal instalments of ₹ 21,000 each. The rate of interest offered to her is 9% p.a. Find the value of x.
Answer:
Cash down price = ₹ (6,50,000 – 1,50,000) = ₹ 5,00,000
Now, Cash down price = down payment (x) + present value (PV) of 21,000 for 20 months at 9% compounded monthly
= ₹ 388,668.42
Down payment (x) = Cash down price – PV
= 5,00,000 – 388,668.42
= ₹ 111,331.58
Question 31.
A company XYZ borrowed ₹ 1.5 lakhs for reformation. The company plans to set up a sinking fund that will pay . back the loan at the end of 3 years. Assuming a rate of 9% compounded quarterly and the sinking fund of the ordinary annuity. Given that (1.0225)12 = 1.3060.
Answer:
Given, P = 1,50,000, r = 9 % or 0.09,
No. of years, n = 3 years
and No. of payments per year, m = 4 (quarterly)
Sinking Fund, A = \(\frac{\left[\left(1+\left(\frac{r}{m}\right)\right)^{n \times m}\right]-1}{\left(\frac{r}{m}\right)}\) × P
= \(\frac{\left[\left(1+\left(\frac{0.09}{4}\right)\right)^{12}\right]-1}{\left(\frac{0.09}{4}\right)}\) × 1,50,000
= \(\frac{\left[(1+0.0225)^{12}\right]-1}{0.0225}\) × 1,50,000
= \(\frac{\left[(1.0225)^{12}\right]-1}{0.0225}\) × 1,50,000
= \(\frac{1.3060-1}{0.0225}\) × 1,50,000
= \(\frac{0.3060}{0.0225}\) × 1,50,000
= ₹ 2,040,000
Section – D
(This section comprises of long answer type questions (LA) of 5 marks each)
Question 32.
In the proportion of defective in a bulk is 4%. Find the probability of not more than 2 defective in a sample of 10. It is known that e-4 = 0.6703.
Answer:
Proportion of defective units
p = \(\frac{4}{100}\) = 0.4
m = np = 0.04 × 10 = 0.4
Probability of not more than 2 defective means the sum of probabilities of 0, 1 and 2 defectives:
Hence, the probability of not more than 2 defectives = 0.9920 or 99.2%
OR
Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of diamond cards drawn. Also, find the mean and the variance of the distribution.
Answer:
Let X = denote the random variable
X = 0, 1, 2,
n = 2,
p = 1/4,
q = 3/4
Mean = Σ xipi = \(\frac{1}{2}\)
Variance = Σ xi2pi – (Σ xipi)2
= \(\frac{5}{8}-\frac{1}{4}=\frac{3}{8}\)
Question 33.
A farmer has a supply of chemical fertilizer of type A which contains 10 % nitrogen and 6 % phosphoric acid and of type B which contains 5 % nitrogen and 10 % phosphoric acid. After soil test, it is found that atleast 7 kg of nitrogen and same quantity of phosphoric acid is required for a good crop. The fertilizer of type A costs ₹ 5.00 per kg and the type B costs ₹ 8.00 per kg. Using Linear Programming, find how many kilograms of each type of the fertilizer should be bought to meet the requirement and for the cost to be minimum. Find the feasible region in the graph.
Answer:
Let the fertilizer of type A be x kg and type B be y kg.
According to Question:
Nitrogen | Phosphoric Acid |
Quantity of Nitrogen in A = 10 %
Quantity of Nitrogen in B = 5 % Minimum availability = 7 kg |
Quantity of Phosphoric Acid in A = 6 %
Quantity of Phosphoric Acid in B = 10% Minimum availability = 7 kg |
∴ 10 % of x + 5 % of y ≥ 7
\(\frac{10}{100}\) x + \(\frac{5}{100}\) y ≥ 7 2x + y ≥ 140 and x ≥ 0, y ≥ 0 |
∴ 6 % of x + 10 % of y ≥ 7
\(\frac{6}{100}\) x + \(\frac{10}{100}\) y ≥ 7 3x + 5y ≥ 350 and x ≥ 0, y ≥ 0 |
∴ Min.Z = 5x + 8y
Subject to constraints
2x + y ≥ 140
3x + 5y ≥ 350
x ≥ 0, y ≥ 0
2x + y = 140 …………….(i)
x | 70 | 0 | 50 |
y | 0 | 140 | 40 |
3x + 5y = 350 ……………(ii)
x | 0 | \(\frac{350}{3}\) |
y | 70 | 0 |
We draw the lines 2x + y = 140 and 6x + y = 70 and obtain the feasible region (unbounded and convex) as shown in figure.
Thus, corner points are A (0, 140), B (50, 40), C (\(\frac{350}{3}\), 0)
The values of Z at these points are given in following table:
Corner Points | Objective function, Z = 5x + 8y |
A(0, 140) | Z = 5 × 0 + 8 × 140 = 1120 |
B(50, 40) | Z = 5 × 50 + 8 × 40 = 570 → Minimum |
C (\(\frac{350}{3}\), 0) | Z = 5 × \(\frac{350}{3}\) + 8 × 0 = 583 \(\frac{1}{3}\) |
As the feasible area is unbounded. Hence, 570 may or may not be the minimum value of Z.
For this, we need to graph inequality.
5x + 8y < 570
x | 114 | 0 |
y | 0 | 71.25 |
Since, there is no common point between the feasible region and 5x + 8y < 570
Hence, the cost will be minimum, if
Fertilizer of type A used = 50 kg
Fertilizer of type B used = 40 kg
Minimum cost = ₹ 570
Question 34.
Find a point on the curve y2 = 2x which is nearest to the point (1, 4).
Answer:
Given the equation of curve (parabola) is y2 = 2x.
Iet (x, y) be the nearest to the point (1, 4).
We have,
x = \(\frac{y^2}{2}\) ………………(i)
Using distance formula, we get
d = \(\sqrt{(x-1)^2+(y-4)^2}\)
z = d2 = (x – 1)2 + (y – 4)2
[from (i)]
Or, z = \(\left(\frac{y^2}{2}-1\right)^2\) + (y – 4)2
z = \(\frac{y^4}{4}\) – 8y + 17
∴ \(\frac{d z}{d y}\) = y3 – 8
Put \(\frac{d z}{d y}\) = 0
⇒ y3 – 8 = 0
⇒ y = 2
Now, from eq (1), we get
x = \(\frac{(2)^2}{2}\) = 2
Again differentiating w.r.t. y, we get
\(\frac{d^2 z}{d y^2}\) = 3y2
\(\left.\frac{d^2 z}{d y^2}\right|_{\text {at } y=2}\) = 3 (2)2 = 12 > 0 Minima
So, the nearest point is (2, 2).
OR
Suppose the total cost C(x) (in millions) for manufacturing x air-planes per year is given by the function C(x) = 6 + \(\sqrt{4 x+4}\) 0 ≤ x ≤ 30
(a) Find the marginal cost at a production level of x air-planes per year.
Answer:
‘The marginal cost at a production level of x air planes is
C(x) = \(\frac{d}{d x}\) (6 + \(\sqrt{4 x+4}\)),
= \(\frac{2}{\sqrt{4 x+4}}\)
(b) Find the marginal cost at a production level of 15 and 24 air-planes per year; and interpret the results.
Answer:
The marginal cost at a production level of 15 air planes is
C'(15) = \(\frac{2}{\sqrt{4(15)+4}}\) = 0.23
At a production level of 15 air-planes per year, the total cost is increasing at the rate of ₹ 250,000 per one air-plane.
The marginal cost at a production level of 24 air planes is
C'(24) = \(\frac{2}{\sqrt{4(24)+4}}\) = 0.2
At a production levelof 24 air-planes per year, the total cost is increasing at the rate of ₹ 200,000 per one air-plane.
Question 35.
Find the inverse of the matrix:
A = \(\left[\begin{array}{ccc}
-1 & 1 & 2 \\
3 & -1 & 1 \\
-1 & 3 & 4
\end{array}\right]\)
and hence show that AA-1 = 1.
Answer:
Given matrix is
A = \(\left[\begin{array}{ccc}
-1 & 1 & 2 \\
3 & -1 & 1 \\
-1 & 3 & 4
\end{array}\right]\)
A-1 exists if |A| ≠ 0
Now, |A| = \(\left[\begin{array}{ccc}
-1 & 1 & 2 \\
3 & -1 & 1 \\
-1 & 3 & 4
\end{array}\right]\)
= – 1 (- 4 – 3) – 1 (12 + 1) + 2 (9 – 1)
= – 1 (- 7) – 1 (13) + 2 (8)
= 7 – 13 + 16
= 10 ≠ 0
Now, A-1 = \(\frac{\ {adj} A}{|A|}\)
adj A = \(\left[\begin{array}{lll}
c_{11} & c_{12} & c_{13} \\
c_{21} & c_{22} & c_{23} \\
c_{31} & c_{32} & c_{33}
\end{array}\right]^{\prime}\)
where, cofactor cij = (- 1)i + j Mij
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:
The 12 hour clock is most commonly represented on an analogue clock with number 1 – 12, when shown on a digital clock, it usually accompanied by a.m. or p.m.
The 24 hour clock is more often shown on digit clocks and is written in a 4 digit form, with the first two digits representing the hours and the last two digits representing the minutes.
There is no need for a.m. or p.m. as each time represents each hour in a 24 hour day.
(i) It is currently 7:00 a.m. in 12 hour clock. What will the time in next 493 hours ?
Answer:
We know that time repeat after every 24 hours.
So, we find 493 (mod 24)
So, 493 (mod 24) = 13
∴ 493 hours is equivalent to 13 hours.
Now, 7 : 00 a.m. + 13 hours = 8 p.m.
OR
If the time after 640 hours from now will be 9 a.m., then what is the current time ?
Answer:
Here, we find 640 (mod 24)
640 (mod 24) = 16
∴ 640 hours is equivalent to 16 hours
Now,
9: 00 a.m. – 16 hours = 5 p.m.
(ii) Shamita has a flight arriving at 3 pm. It’s getting delayed 14 hours. What time will it land?
Answer:
We have, 14 mod 12 = 2
So, the flight will land at “3 + 2 = 5 am”.
(iii) What time in 24 hour dock is equivalent to 7 a.m. in 12 hour clock?
Answer:
7 : 00 hour in 24 hour clock is equivalent to 07 : 00 a.m. in 12 hour clock.
CASE STUDY-II
Question 37.
Read the following text and answer the following questions on the basis of the same:
To fit a straight line by the method of least squares, Rohan constructed the following table:
The trend equation can be considered as yt = a + bx.
(i) Find the values of V and ‘V in the trend equation.
Answer:
We know that,
a = \(\frac{\Sigma y}{n}\)
and b = \(\frac{\Sigma x y}{\Sigma x^2}\)
Here,
Σy = 972
Σxy = 214
n = 7
So, a = \(\frac{972}{7}\)
= 138.86
and b = \(\frac{214}{28}\) = 7.64
(ii) Write the trend equation.
Answer:
Since, a = 138.86
and b = 7.64
Thus, trend equation is given by
yt = 138.86 + 7.64x
(iii) Find the trend value for year 2015.
Answer:
For year 2015, x = – 3
yt = 138.86 + 7.64 (- 3)
= 115.94
OR
Find the trend value for year 2018.
Answer:
For year 2018, x = 0
yt = 138.86 + 7.64(0)
= 138.86
CASE STUDY-III
Question 38.
The amount of bread (in hundreds of kilograms) X that a certain bakery is able to sell in a day is found to be a numerical valued random phenomenon with a probability function specified by the probability density function f(x), given by
f(x) = \(\left\{\begin{array}{lll}
A x ; & \text { for } & 0 \leq x<5 \\
A(10-x), & \text { for } & 5 \leq x<10 \\
0 & \text { otherwise } &
\end{array}\right.\)
(i) Find the value of A such that f(x) is the probability density function.
What is the probability that number of kilogram of bread that will be sold tomorrow is more than 500 kgs?
Answer:
For f(x) to be the probability density function, we must have
OR
(ii) Find the value of A such that f(x) is the probability density function.
What is the probability that number of kilograms of bread that will be sold tomorrow is between 250 kgs and 750 kgs?
Answer:
From part (a), value of A = \(\frac{1}{25}\)
Since,
f(x) = \(\left\{\begin{array}{lll}
\frac{x}{25}, & \text { for } & 0 \leq x<5 \\
\frac{1}{25}(10-x) & \text { for } & 5 \leq x<10 \\
0 & \text { otherwise }
\end{array}\right.\)
Required Probability = P (2.5 ≤ X ≤ 7.5)