Students can access the CBSE Sample Papers for Class 12 Applied Maths with Solutions and marking scheme Set 9 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 12 Applied Maths Set 9 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.
The least non-negative remainder, when 315 is divided by 7 is
(A) 1
(B) 5
(C) 6
(D) 7
Answer:
(C) 6
Explanation:
Since, 33 = 27 (mod 7)
33 = – 1(mod7)
(33)5 = (- 1) (mod 7)
315 = 6 (mod 7)
Question 2.
In a 2 km race, P can give Q a start of 200 m and R a start of 560 m. Then in the same race, Q can give R a start of
(A) 360 m
(B) 380 m
(C) 400 m
(D) 430 m
Answer:
(C) 400 m
Explanation:
Let P is complete the race 2000 m
and Q is at 1800 m
and R is at 1440 m
⇒ In a race of 1800 m
Q gives R a start at = 1800 – 1440
= 360
⇒ 1 m race = \(\frac{360}{1800}\)
⇒ 2000 m race = \(\frac{1}{2}\) × 2000
= 400 m
Question 3.
An observed set of the population that has been selected for analysis is called
(A) a sample
(B) a process
(C) a forecast
(D) a parameter
Answer:
(A) a sample
Explanation:
A selection of a group of individuals from a population in such a way that it represents the population is called as sample and the number of individuals in the sample is called the sample size.
Question 4.
A simple random sample consist of four observation 1, 3, 5, 7. What is the point estimate of population standard deviation?
(A) 2.3
(B) 2.52
(C) 0.36
(D) 0.4
Answer:
(B) 2.52
Explanation:
The point estimation of population standard deviation is sample deviation.
S = \(\sqrt{\frac{\left(x_i-\bar{x}\right)^2}{n-1}}\)
= \(\sqrt{\frac{(1-4)^2+(3-4)^2+(5-4)^2+(7-4)^2}{4-1}}\)
(∵ \(\bar{x}=\frac{\Sigma x}{n}\)
= \(\frac{16}{4}\) = 4)
= \(\sqrt{\frac{9+1+1+9}{3}}\)
= \(\sqrt{\frac{20}{3}}\)
= 2.52
Question 5.
A person can row in still water at the rate of 8 km/h. If it takes him thrice as long to row upstream as to row downstream then the speed of the stream is
(A) 2 km/h
(B) 3 km/h
(C) 4 km/h
(D) 6 km/h
Answer:
(C) 4 km/h
Explanation:
Let x be the speed of the stream
\(\frac{V_d}{t}=3 \frac{V_u}{t}\)
8 + x = 3 (8 – x)
∴ 4x = 16
x = 4 km/h
Question 6.
An automatic machine produces 20000 pins per day. On rare occasion it produces a perfect pin whose chance is \(\frac{1}{10000}\). Assuming Poisson distribution, the mean and variance of the number of perfect pins are — and —, respectively.
(A) √2, √2
(B) 2, 2
(C) 2, 4
(D) 4, 2
Answer:
(B) 2, 2
Explanation:
For Poisson distribution
Mean = variance
= np
= 20000 × \(\frac{1}{10000}\)
= 2
Question 7.
If a = b (mod n), then
(A) a ≡ a (mod n)
(B) b ≡ a (mod n)
(C) b ≡ b (mod n)
(D) None of these
Answer:
(B) b ≡ a (mod n)
Explanation:
If a ≡ b (mod n)
then n | (a – b)
Or, n|- (b – a)
Or, n| (b – a)
Or, b ≡ a (mod n)
Question 8.
The value of a machine purchased two years ago, depreciates at the annual rate of 10 %. If its present value is ₹ 97,200, then its value after 3 years is
(A) ₹ 70,859 approx
(B) ₹ 71,895 approx
(C) ₹ 80,859 approx
(D) ₹ 88,509 approx
Answer:
(A) ₹ 70,859 approx
Explanation:
Given, P = ₹ 97,200,
i = 10 % p.a.
i = \(\frac{10}{100}\)
So, value after 3 years
= 97,200x (1 – 0.1)3
= 97,200 × 0.729
= ₹ 70,858.80
= ₹ 70,859 (approx.)
Question 9.
If f (a + b – x) = f(x), then \(\int_a^b\) x f(x) dx is equal to
(A) \(\frac{a+b}{2} \int_a^b\) f (b – x) dx
(B) \(\frac{a+b}{2} \int_a^b\) f (b + x) dx
(C) \(\frac{-b-a}{2} \int_a^b\) f(x) dx
(D) \(\frac{a+b}{2} \int_a^b\) f(x) dx
Answer:
(D) \(\frac{a+b}{2} \int_a^b\) f(x) dx
Explanation:
Let I = \(\int_a^b\) x f(x) dx
I = \(\int_a^b\) (a + b – x) (a + b – x) dx
[∵ \(\int_a^b\) f(x) dx = \(\int_a^b\) f (a + b – x) dx]
⇒ I = \(\int_a^b\) (a + b – x) f(x) dx
⇒ I = (a + b) \(\int_a^b\) f(x) dx – I [Using Eq. (i)]
⇒ I + I = (a + b) \(\int_a^b\) f(x) dx
∵ 2I = (a + b) \(\int_a^b\) f(x) dx
⇒ I = \(\left(\frac{a+b}{2}\right)\) \(\int_a^b\) f(x) dx
Question 10.
Moving average method is used for measurement of trend when
(A) Trend is linear
(B) Trend is non-linear
(C) Trend is curvilinear
(D) None of these
Answer:
(A) Trend is linear
Explanation:
When trend is linear, then only we use moving average method for measurement.
Question 11.
The present value of perpetuity of ₹ 600 payable at end of every 6 month be ₹ 10,000. Find rate of interest.
(A) 10 %
(B) 12 %
(C) 6 %
(D) 12.5 %
Answer:
(B) 12 %
Explanation:
Let rate of interest be r % per annum.
then i = \(\frac{r}{200}\)
Given, R = ₹ 600
and P = ₹ 10000
Using P = \(\frac{R}{i}\)
⇒ i = \(\frac{R}{P}=\frac{600}{10,000}\)
⇒ \(\frac{r}{200}=\frac{600}{10,000}\)
⇒ r = \(\frac{120,000}{10,000}\)
= 12 %
Question 12.
A company buys a machine at a cost of ₹ 5000. The company decides on a salvage value of ? 1000 and a useful life of 5 years, then annual depreciation cost is
(A) ₹ 500
(B) ₹ 600
(C) ₹ 700
(D) ₹ 800
Answer:
(D) ₹ 800
Explanation:
We know that
Annual Depredation expense = \(\frac{\text { (Asset cost – Residual Value) }}{\text { Useful life of the asset }}\)
= \(\left(\frac{₹ 5000-₹ 1000}{5}\right)\)
= ₹ 800
Question 13.
EMI depends on the following factors
(A) The amount of loan
(B) The loan tenure
(C) The interest rate
(D) All of these
Answer:
(D) All of these
Question 14.
The solution of the differential equation \(\frac{d x}{x}+\frac{d y}{y}\) = 0
(A) \(\frac{1}{x}+\frac{1}{y}\) = C
(B) xy = C
(C) log x log y = C
(D) x + y = C
Answer:
(B) xy = C
Explanation:
Given differential equation is
\(\frac{d x}{x}\) + \(\frac{d y}{y}\) = 0
Integrating both sides, we get
\(\int \frac{d x}{x}=-\int \frac{d y}{y}\)
log x = – log y + log C
log x + log y = log C
log (xy) = log C
xy = C
Question 15.
Feasible region in the set of points which satisfy
(A) The objective functions
(B) Some of the given constraints
(C) All of the given constraints
(D) None of these
Answer:
(C) All of the given constraints
Question 16.
A table which contains possible value of a random variable and its corresponding probabilities is called
(A) Probability Mass Function
(B) Probability Density Function
(C) Cumulative Distribution Function
(D) Probability Distribution
Answer:
(D) Probability Distribution
Explanation:
The given statement is the definition of a probability distribution.
Question 17.
Increase in the number of patients in the hospital due to heat stroke is
(A) Secular trend
(B) Irregular variation
(C) Seasonal variation
(D) Cyclical variations
Answer:
(C) Seasonal variation
Explanation:
In seasonal variation, tendency movements are due to the nature which repeat themselves periodically in every season.
Question 18.
For testing the significance of difference between the means of two independent samples, the degree of freedom (v) is taken as
(A) n1 – n2 + 2
(B) n1 – n2 – 2
(C) n1 + n2 – 2
(D) n1 + n2 – 1
Answer:
(C) n1 + n2 – 2
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 g(x) = log x does not has maxima or minima.
Reason (R) : The function g(x) = log x has maxima.
Answer:
(C) (A) is true but (R) is false.
Explanation:
g(x) = log x
∴ g’(x) = \(\frac{1}{x}\)
Since, log x is defined for a positive number x, g’(x) > 0 for any x.
Therefore, there does not exist c E R such that g'(c) = 0.
Hence, function g does not have maxima or minima.
Question 20.
A random variable X has the following distribution:
Assertion (A) : Value of C is \(\frac{1}{10}\)
Reason (R) : Using formula ΣPi = 1, we get value of C.
Answer:
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation:
We know that,
ΣPi = 1
∴ C + 2C + 2C + 3C + C2 + 2C2 + 7C2 + C = 1
⇒ 10C2 + 9C – 1 = 0
⇒ (10C – 1) (C + 1) = 0
⇒ C = \(\frac{1}{10}\) or C = – 1
(but C cannot be negative)
⇒ C = \(\frac{1}{10}\).
Section – B
(All Questions are compulsory. In case of internal Choice, aftempt any one question only)
Question 21.
Define useful life and scrap value.
Answer:
Useful Life:
The time period in which an asset is expected to be functional and fit for purpose is called useful life. It is generally measured in years.
Scrap Value:
The value of a depreciable asset at the end of its useful life is called scrap value or salvage value or depreciated value.
Question 22.
If the matrix A = \(\left[\begin{array}{ccc}
0 & a & -3 \\
2 & 0 & -1 \\
b & 1 & 0
\end{array}\right]\) is skew symmetric. Find the values of ’a’ and ‘b’.
Answer:
A is a skew symmetric matrix.
⇒ A’ = – A
A = \(\left[\begin{array}{ccc}
0 & a & -3 \\
2 & 0 & -1 \\
b & 1 & 0
\end{array}\right]\)
A’ = \(\left[\begin{array}{ccc}
0 & 2 & b \\
a & 0 & 1 \\
-3 & -1 & 0
\end{array}\right]\)
A’ = – A
Now,
\(\left[\begin{array}{ccc}
0 & 2 & b \\
a & 0 & 1 \\
-3 & -1 & 0
\end{array}\right]=-\left[\begin{array}{ccc}
0 & a & -3 \\
2 & 0 & -1 \\
b & 1 & 0
\end{array}\right]\)
\(\left[\begin{array}{ccc}
0 & 2 & b \\
a & 0 & 1 \\
-3 & -1 & 0
\end{array}\right]=\left[\begin{array}{ccc}
0 & -a & 3 \\
-2 & 0 & 1 \\
-b & -1 & 0
\end{array}\right]\)
On comparing, we get
a = – 2
b = 3.
OR
If A and B are square matrices of the same order 3, such that | A | = 2 and AB = 2I, write the value of | B |.
Answer:
Given, AB = 2I
Or, AB = 2 \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{array}\right]\)
∴ |AB| = 2 (4 – 0) – 0 + 0 = 8
Since, |AB| = |A| |B|
∴ |B| = \(\frac{|A B|}{A}\)
= \(\frac{8}{2}\) = 4
[given, |A| = 2]
Question 23.
Draw the feasible region for given inequation system y ≤ 6, x + y ≤ 3, x ≥ 0, y ≥ 0.
Answer:
Thus, feasible region is OAB.
Question 24.
Tap P alone fills a cistern in 2 hours; while tap Q alone fills the same cistern in 3 hours. A new tap R is attached to the bottom of the cistern which can empty the completely filled cistern in 6 hours. Sunny started all three taps together at 9 a.m. When will the tank be full?
Answer:
Cistern fiiled or work done by Tap P in 1 hour = \(\frac{1}{2}\)
Cistern filled or work done by Tap Q in 1 hour = \(\frac{1}{3}\)
Tank emptied or work done by Tap R in 1 hour = \(\frac{1}{6}\)
Tank filled or work done by all three pipes in 1 hour = \(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3}\)
So, Tank gets completely filled in \(\frac{3}{2}\) hours.
= 1.5 hours = 1 h 30 min.
So, time will be 9.00 a.m. + 1 hour 30 min = 10.30 a.m.
OR
There is a road beside a river. Two friends started from a place A, moved to a temple situated at another place B and then returned to A again. One of them moves on a cycle at a speed of 12 km/h, while the other sails on a boat at a speed of 10 km/h. If the river flows at the speed of 4 km/h, which of the two friends will return to place A first?
Answer:
Here, cyclist moves both ways at a speed of 12 km/h.
So, average speed of the cyclist = 12 km/h
The boat sailor moves at rate of (10 ± 4) i.e., 6 km/h
So, average speed of the boat sailor = \(\left(\frac{2 \times 14 \times 6}{14+6}\right)\) km/h
= \(\frac{42}{5}\) km/h = 8.4 km/h
Since, the average speed of the cyclist is greater, he will return to A first.
Question 25.
A machine which produces mica insulating washers of use in the electric devices is set to turn out washers having a thicknes of 10 mils (1 mil = 0.001 inch). A sample of 10 washers has an average thickness of 9.52 mils with a standard deviation of 0.60 mil. Find out t.
Answer:
Here, \(\bar{X}\) = 9.52,
Mean μ = 10,
σ = 0.60
and n = 10
Section – C
(All Questions are compulsory. In case of internal Choice, attempt any one question only)
Question 26.
Evaluate: ∫ \(\frac{d x}{x\left(x^5+3\right)}\)
Answer:
Let I = ∫ \(\frac{d x}{x\left(x^5+3\right)}\)
= ∫ \(\frac{x^4}{x^5\left(x^5+3\right)}\) dx
Put x5 = t
5x4 dx = dt
x4 dx = \(\frac{1}{5}\) dt
OR
Evaluate: \(\int_0^1 \log \left(\frac{1}{x}-1\right)\) dx
Answer:
Let I = \(\int_0^1 \log \left(\frac{1}{x}-1\right)\) dx
= \(\int_0^1 \log \left(\frac{1-x}{x}\right)\) dx
Using property,
I = – I
2I = 0
∴ I = 0
∴ \(\int_0^1 \log \left(\frac{1}{x}-1\right)\) dx = 0
Question 27.
Using properties of determinants, prove that: \(\left|\begin{array}{ccc}
1+a & 1 & 1 \\
1 & 1+b & 1 \\
1 & 1 & 1+c
\end{array}\right|\) = abc + bc + ca + ab.
Answer:
R1 → \(\frac{1}{a}\) R1,
R2 → \(\frac{1}{b}\) R2,
R3 → \(\frac{1}{c}\) R3
Expanding along R1
= abc (1 + \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\))
= abc + bc + ca + ab
= RHS
Hence proved.
Question 28.
Find all the points of local maxima and local minima for the function f(x) = x3 – 6x2 + 9x – 8.
Answer:
Given, y = f(x) = x3 – 6x2 + 9x – 8
∴ y’ = 3x2 – 12x + 9
Now, put y’ = 0
∴ 3x2 – 12x + 9 = 0
⇒ x = \(\frac{12 \pm \sqrt{144-4 \times 3 \times 9}}{6}\)
⇒ x = \(\frac{12 \pm \sqrt{36}}{6}\)
⇒ x = \(\frac{12 \pm 6}{6}\)
⇒ x = \(\frac{12+6}{6}\) and x = \(\frac{12-6}{6}\)
⇒ x = 3 and x = 1
Now, y” = 6x – 12
At x = 3, y” = 6.3 – 12 = 6 > 0
At x = 1, y” = 6.1 – 12 = – 6 < 0
Thus, at x = 3, the function is at its local minimum
and at x = 1, the function is at its local maximum.
Question 29.
Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the X-axis in the first quadrant.
Answer:
The area of the region bounded by the curve, y2 =
9x, x = 2 and x = 4, and the X-axis is the area ABCD.
Area of ABCD = \(\int_2^4\) y dx
= \(\int_2^4\) 3√x dx
= 3 \(\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_2^4\)
= 2 \(\left[x^{\frac{3}{2}}\right]_2^4\)
= 2 \(\left[(4)^{\frac{3}{2}}-(2)^{\frac{3}{2}}\right]\)
= 2 [8 – 2√2]
= (16 – 4√2) units2
OR
If the demand curve is given by D(x) = 50 – 0.06x2. Find the surplus or profit of the consumers if the level of sale amounts to twenty units.
Answer:
As the number of units is 20,
its price rises up = D (20)
= 50 – 0.06 × 202 = 26
The profit of consumer’s:
CS = \(\int_0^{Q_e}\) D(x) dx – Qe . Pe
= \(\int_0^{20}\) 20 (50 – 0.06 x2) dx – 20.26
= \(50[x]_0^{20}-0.06\left[\frac{x^3}{3}\right]_0^{20}\) – 520
= 50 (20 – 0) – 0.06[\(\frac{20^3}{3}\) – 0] – 520
= 50 (20) – 0.06 [latex]\frac{20^3}{3}[/latex] – 520
= 1000 – 160 – 520
= 1000 – 680
= 320
The consumer gain is ₹ 320, if the level of sales is twenty units.
Question 30.
An asset costs ₹ 4,50,000 with an estimated useful life of 5 years and a scrap value of ₹ 1,00,000. Using linear depreciation method, find the annual depreciation of the asset and construct a yearly depreciation schedule.
Answer:
We know that,
Amount of annual depreciation = \(\frac{\text { Cost of asset }- \text { scrap value of asset }}{\text { Useful life in years }}\)
= \(\frac{4,50,000-1,00,000}{5}\)
= \(\frac{3,50,000}{5}\)
= ₹ 70,000
Depreciation Schedule
Year | Book Value (Beginning of the Year) | Depreciation | Book Value (End of the Year) |
1. | ₹ 4,50,000 | ₹ 70,000 | ₹ 3,80,000 |
2. | ₹ 3,80,000 | ₹ 70,000 | ₹ 3,10,000 |
3. | ₹ 3,10,000 | ₹ 70,000 | ₹ 2,40,000 |
4. | ₹ 2,40,000 | ₹ 70,000 | ₹ 1,70,000 |
5. | ₹ 1,70,000 | ₹ 70,000 | ₹ 1,00,000 |
Question 31.
Amrita bought a car worth ₹ 12,50,000 and make a down payment of ₹ 3,00,000. The balance amount is to be paid in 4 years by equal monthly instalments at an interest rate of 15% p.a. Find the EMI that Amrita has to pay for the car. {Given (1.0125)-48 = 0.5508565}
Answer:
We have,
Cost of car purchased by Amrita = ₹ 12,50,000
Amount of down payment made by Arnrita = ₹ 3,00,000
So, balanced amount = ₹ 12,50,000 – ₹ 3,00,000
= ₹ 9,50,000
Here, P = ₹ 9,50,000
i = \(\frac{15}{1200}\) = 0.0125
and n = 4 years = 4 × 12 months = 48 months.
Let E be the monthly installment Amrita has to pay.
Then,
Section – D
(This section comprises of long answer type questions (LA) of 5 marks each)
Question 32.
If X is a normal variate with mean 30 and SD 5. Find the probabilities that
(i) 26 < X < 40 (ii) X > 45.
Answer:
Here, mean µ = 30 and
standard deviation σ = 5
(i) When X = 26
Z = \(\frac{(X-\mu)}{\sigma}\)
= \(\frac{(26-30)}{5}\)
= – 0.8
And when X = 40,
Z = \(\frac{40-30}{5}\) = 2
Therefore,
P(26 < X < 40) = P(- 0.8 ≤ Z < 2)
=P(- 0.8 ≤ Z ≤ 0) + P(0 ≤ Z ≤ 2)
= P(0 ≤ Z ≤ 0.08) + P(0 ≤ Z ≤ 2)
= 0.2881 + 0.4772 (By tables)
= 0.7653
(ii) The probability that X ≥ 45
When X = 45
X = \(\frac{X-\mu}{\sigma}\)
= \(\frac{45-30}{5}\)
= 3
P(X ≥ 45) = P (Z ≥ 3)
= 0.5 – 0.49865
= 0.00135
OR
Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X).
Answer:
Let X denotes the number of aces obtained.
Therefore, X can take any of the values of 0, 1, or 2.
In a deck of 52 cards, 4 cards are aces.
Therefore, there are 48 non-ace cards.
∴ P(X = 0) = P(0 ace and 2 non-ace cards)
= \(\frac{{ }^4 C_0 \times{ }^{48} C_2}{{ }^{52} C_2}\)
= \(\frac{1128}{1326}\)
P(X = 1) = P (1 ace and I non-acc cards)
= \(\frac{{ }^4 C_1 \times{ }^{48} C_1}{{ }^{52} C_2}\)
= \(\frac{192}{1326}\)
P(X = 2) = P(2 ace and 0 non-ace cards)
= \(\frac{{ }^4 C_2 \times{ }^{48} C_0}{{ }^{52} C_2}\)
= \(\frac{6}{1326}\)
Thus, the probability distribution is as follows:
X | 0 | 1 | 2 |
P(X) | 1128/1326 | 192/1326 | 6/1326 |
Then,
E(X) = Σ pixi
= \(0 \times \frac{1128}{1326}+1 \times \frac{192}{1326}+2 \times \frac{6}{1326}\)
= \(\frac{204}{1326}\)
= \(\frac{2}{13}\)
Question 33.
A manufacturer manufactures two types of tea-cups, A and B. Three machines are needed for manufacturing the tea cups. The time in minutes required for manufacturing each cup on the machines is given below:
Each machine is available for a maximum of six hours per day. If the profit on each cup of type A is ₹ 1.50 and that on each cup of type B is ₹ 1.00, find the number of cups of each type that should be manufacturing in a day to get maximum profit.
Answer:
Let x be the number of A type tea cups
and y be the number of B type tea cups.
The problem can be formulated as
Max.Z = 1.50 x + 1.00 y
i.e., Max. Z = 1.50 x + y
Subject to constraints:
12x + 6y ≤ 360, i.e., 2x + y ≤ 60
18x + 0.y ≤ 360, i.e., x ≤ 20
6x + 9y ≤ 360, i.e., 2x + 3y ≤ 120
and x ≥0, y ≥ 0 (Non-negative constraints)
We draw the straight lines:
2x + y = 60, x = 20, 2x + 3y = 120
For 2x + y = 60
x | 30 | 0 | 40 |
y | 0 | 60 | 20 |
For 2x + 3y = 120
x | 60 | 0 | 30 |
y | 0 | 40 | 20 |
The shaded portion shows the feasible region.
The points B and C are points of intersection of lines 2x + y = 60,with 2x + 3y = 120 and x = 20 with 2x + y = 60 respectively.
Thus, the 5 corner points of the feasible region are
O (0, 0), A (0, 40), B (15, 30), C (20, 20) and D (20, 0).
Z = 1.5x + y
Corner Point | Value of Z = 1.5x + y |
O (0, 0) | 0 |
A (0, 40) | 40 |
B(15, 30) | 52.5 (Max.) |
C(20, 20) | 50 |
D(20, 0) | 30 |
Clearly, the maximum profit is at x = 15 and y = 30.
Thus, the manufacturer should manufacture 15 cups of type A and 30 cups of type B to get maximum profit in a day.
Question 34.
Show that y = log (1 + x) – \(\frac{2 x}{2+x}\), x > – 1 an increasing function of x throughout its domain.
Answer:
Given, y = log (1 + x) – \(\frac{2 x}{2+x}\), x > – 1
Differentiating , w.r.t. ‘x’
For increasing function,
\(\frac{d y}{d x}\) ≥ 0
\(\frac{x^2}{(2+x)^2(x+1)}\) ≥ 0
⇒ \(\left(\frac{x}{(2+x)}\right)^2 \cdot\left(\frac{1}{x+1}\right)\) > 0
⇒ \(\left(\frac{1}{x+1}\right)\) > 0
This is possible only when 1 + x > 0 i.e., x > – 1
So, \(\frac{d y}{d x}\) > 0 for x > – 1
When x > – 1
\(\frac{d y}{d x}\) is always greater than zero
∴ y = log(1 + x) – \(\frac{2 x}{2+x}\) is always increasing throughout its domain.
OR
Gymnast Clothing manufactures expensive hockey jerseys for sale to college bookstores in runs of up to 150. Its cost (in rupees) for a run of x hockey jerseys is
C(x) = 1500 + 10x + 0.2 x2, (0 ≤ x ≤ 150)
(a) Gymnast Clothing sells the jerseys at ₹ 90 each. Find the revenue function.
Answer:
Since, the manufacturer sells the jerseys for ₹ 90 each, the revenue function is
R(x) = 90x
(b) Find the profit function.
Answer:
Profit is defined to be revenue minus cost, so the profit function is
P(x) = R(x) – C(x)
= 90x – (1500 + 10x + 0.2x2)
= – 1500 + 80x – 0.2x2
(c) How many should Gymnast Clothing manufacture to make a profit? (Round your answer up to the nearest whole number.)
Answer:
In order to make a profit, P(x) must be greater than zero.
So, find how many jerseys we need to make in order to make a profit, we should find the breakeven point.
So, we put P(x) = 0, i.e.,
– 0.2x2 + 80x – 1500 = 0
x = \(\frac{-80 \pm \sqrt{80^2-4(-0.2)(-1500)}}{2(-0.2)}\)
This simplifies to x = 19.7 or x = 380.2,
but since the problem specifies that the domain is between x = 0 and x = 150, we can reject the larger answer.
Rounding the other answer up, we get x = 20.
So, if we make 20 jerseys, we will make a profit.
Question 35.
Express the matrix A = \(\left[\begin{array}{ccc}
2 & 4 & -6 \\
7 & 3 & 5 \\
1 & -2 & 4
\end{array}\right]\) as the sum of a symmetric and skew symmetric matrix.
Answer:
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:
While making the notes on linear inequalities, Riya note down the following points in her notebook.
Inequality: Two real numbers or two algebraic expressions related by the symbol ‘<‘, ‘>’, ‘≤’, form an inequality.
Linear Inequality: An inequality is said to be linear, if each variable occurs in first degree only and there is no term involving the product of the variables.
e.g., ax + b < 0, ax + by + c > 0, ax < 4.
An inequality in one variable in which degree of variable is 2, is called quadratic inequality in one variable, e.g., ax2 + bx + c > 0, 3x2 + 2x + 4 < 0.
Linear equality in one Variable:
A linear inequality which has only one variable, is called linear inequality in one variable.
e.g., ax + b < 0, where a ≠ 0, 4c + 7 ≥ 0
(i) Rules of solving inequalities:
- If a ≥ b then a ± k ≥ b ± k where k is any number.
- If a ≥ b then ka is not always ≥ kb If k > 0 (i.e., positive) then a ≥ b ⇒ ka ≥ kb
- If k > 0 (i.e., negative) thena ≥ b ⇒ ka ≤ kb
Thus, always reverse the sign of inequality while multiplying or dividing both sides of an inequality by a negative number.
(ii) Procedure to solve a linear inequality in one variable:
- Simplify both sides by collecting like terms.
- Remove fractions (or decimals) by multiplying both sides by appropriate factor (L.C.M. of denominator or a power of 10 in case of decimals.)
- Isolate the variable on one side and all constants on the other side. Collect like terms whenever possible.
- Make the coefficient of the variable equal to 1.
- Choose the solution set from the replacement set.
Solution set:
A solution to an inequality is a number which when substituted for the variable, makes the inequality true. The set of all solutions of an inequality is called the solution set of the inequality.
(i) Write the solution set for the following figure.
Answer:
The given figure represents all value of x greater than 5 excluding 5 on the real number line.
So x ∈ (5, ∞)
(ii) Write the solution set for the following figure.
Answer:
The given figure represent all real values of x less than and equal to – 2.
So x ∈ (- ∞, – 2]
(iii) Find the solution set for the given inequality is:
4x + 3 ≥ 2x + 17, 3x – 5 < – 2
Answer:
OR
Find the solution set for the given inequality is |x + 2| < 9.
Answer:
We have,
4x + 3 ≥ 2x + 17
⇒ 4x – 2x ≥ 17 – 3
⇒ 2x ≥ 14
⇒ x ≥ \(\frac{14}{2}\)
⇒ x ≥ 7
Also, we have 3x – 5 < – 2
⇒ 3x < – 2 + 5
⇒ 3x < 3
⇒ x < 1
On combining Eqs. (i) and (ii), we see that solution is not possible because nothing is common between these two solutions,
(i.e.,x < 1, x ≥ 7).
Given, |x + 2| ≤ 9
⇒ – 9 ≤ x + 2 ≤ 9
⇒ – 9 – 2 ≤ x ≤ 9 – 2
⇒ – 11 ≤ x ≤ 7
⇒ x ∈ [- 11, 7].
CASE STUDY – II
Question 37.
Read the following text and answer the following questions on the basis of the same:
Today in mathematics class Mr. Lai teaches the topic of Binomial distribution. Four friends Rohan, Rohit, Roshan and Raman have few doubts in this topic, so they decided for group studt at Rohan’s place. They are trying to solve a question in which a pair of dice is thrown 7 times. If getting a total 7 is considered a success, then answer the following questions:
(i) If X denote the number of success in 7 throws of a pair of dice, then find X Binomial variate with parameters.
Answer:
Let p denote the probability of getting a total of 7 in a single throw of a pair of dice.
Then
p = \(\frac{6}{36}=\frac{1}{6}\)
(∵ The sum can be 7 in any one of the ways: (1, 6), (6, 1), (2, 5), (5, 2), (3, 4) and (4, 3)]
So, the parameters are:
n = 7 and p = \(\frac{1}{6}\)
(ii) What is the probability of no success?
Answer:
The X is a binomial variate with parameters n = 7
and p = \(\frac{1}{6}\)
P(X = r) = \({ }^7 C_r\left(\frac{1}{6}\right)^r\left(\frac{5}{6}\right)^{7-r}\)
r = 0, 1, 2, ……., …………..(i)
[∵ q = 1 – p
= 1 – \(\frac{1}{6}\)
= \(\frac{5}{6}\)]
Probability of no success = P(X = 0)
= \({ }^7 C_0\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^{7-0}\)
= \(\left(\frac{5}{6}\right)^7\)
OR
What is the probability of 6 success?
Answer:
From eq. (1),
Probability of 6 success = P(X = 6)
= \({ }^7 C_6\left(\frac{1}{6}\right)^6\left(\frac{5}{6}\right)^{7-6}\)
= 35 \(\left(\frac{1}{6}\right)^7\)
(iii) What is the probability of atleast 6 success?
Answer:
Probability of atleast 6 success = P(X ≥ 6)
= P (X = 6) + P (X = 7)
CASE STUDY – III
Question 38.
Fit a straight line trend by the method of least squares to the data given below:
Answer:
Now, a = \(\frac{\sum y_i}{n}\)
= \(\frac{91}{7}\) = 13
b = \(\frac{\sum x_i y_i}{\sum x_i^2}\)
= \(\frac{33}{28}\) = 1.179
The equation of the straight line trend is
y = ax + b
y = 13x + 1.179
OR
Calculate trend values from the following data assuming 5-yearly and 7-yearly moving average.
Answer:
Calculation of Trend values by moving average method.