Students can access the CBSE Sample Papers for Class 12 Applied Maths with Solutions and marking scheme Set 3 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 12 Applied Maths Set 3 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.
If a ≡ b (mod n), then
(A) n|a and n|b
(B) n|b only
(C) n|(a – b)
(D) None of these
Answer:
(C) n|(a – b)
Explanation:
If a and b are integers and n > 0, then we write a ≡ b (mod n) to mean n |(a – b).
Question 2.
In a one-kilometre race, A, B and C are three participants. A can give B start of 50 m and C a start of 69 m. The start, which B can allow Cr is
(A) 17 m
(B) 18 m
(C) 19 m
(D) 20 m
Answer:
(D) 20 m
Explanation:
A : B : C = 1000 : (1000 – 50) : (1000 – 69)
= 1000 : 950 : 931
In a 950 m race, B can give C a start of (950 – 931) m = 19 m
In a 1000 m race, B can give C a start of (\(\frac{19}{950}\) × 1000) = 20 m
Question 3.
A machine makes car wheels and in a random sample of 26 wheels, the test statistic is found to be 3.07. As per the t-distribution test (of 5% level of significance), what can you say about the quality of wheels produced by the machine? (Use t25(0.05) = 2.06)
(A) Superior quality
(B) Inferior quality
(C) Same quality
(D) Cannot say
Answer:
(B) Inferior quality
Explanation:
n = 26
|t| = 3.07 > t25 (0.05)
= 3.07 > 2.06
Question 4.
Degree of freedom is calculated from the following formula
(A) N – 1
(B) N + 1
(C) \(\frac{N}{2}\)
(D) None of these
Answer:
(A) N – 1
Explanation:
Df = N —1
where: Df = degrees of freedom
and N = sample size
Question 5.
In what ratio water is mixed with milk to gain 16\(\frac{2}{3}\)% on selling the mixture at cost price?
(A) 1 : 6
(B) 6 : 1
(C) 2 : 3
(D) 3 : 2
Answer:
(A) 1 : 6
Explanation:
Let C.P of 1 litre milk be ₹ 1.
S.P of 1 litre of mixture = ₹ 1,
Gain = \(\frac{50}{3}\) %
∴ C.P. of 1 litre of mixture = (100 × \(\frac{3}{350}\) × 1) = \(\frac{6}{7}\)
By the rule of alligation, we have
∴ Ratio of water and milk = \(\frac{1}{7}\) : \(\frac{6}{7}\) = 1 : 6.
Question 6.
Mean of a constant ‘a’ is ________
(A) 0
(B) a
(C) all
(D) 1
Answer:
(B) a
Explanation:
Let f(x) be the pdf of the random variable X.
Now, E(a) = ∫ af (X) = a ∫ f(x) = a . 1 = a
Question 7.
Evaluate: (9 × 8) mod 6 = _________
(A) 2
(B) 1
(C) 0
(D) 6
Answer:
(C) 0
Explanation:
(9 × 8) mod 6 or 72 mod 6
On dividing 72 by 6, we get remainder = 0
Hence, (9 × 8) mod 6 =0
Question 8.
The reduction of the original price of fixed asset over a period of time due to use is called
(A) Appreciation
(B) Depreciation
(C) Amortized
(D) None of these
Answer:
(B) Depreciation
Explanation:
The monetary value of an asset decreases over time due to use, wear and tear or obsolescence. This decrease is measured as depreciation.
Question 9.
∫ \(\sqrt{1+x^2}\) dx is equal to
(A) \(\frac{x}{2} \sqrt{1+x^2}+\frac{1}{2} \log \left|\left(x+\sqrt{1+x^2}\right)\right|\) + C
(B) \(\frac{2}{3}\) (1 + x2)3/2 + C
(C) \(\frac{2}{3}\) x (1 + x2)3/2 + C
(D) \(\frac{x^2}{2} \sqrt{1+x^2}+\frac{1}{2} x^2 \log \left(x+\sqrt{1+x^2}\right)\) + C
Answer:
Question 10.
In the theory of time series, shortage of certain consumer goods before the annual budget is due to
(A) Seasonal Variation
(B) Secular Trend
(C) Irregular variations
(D) Cyclic Variation
Answer:
(A) Seasonal Variation
Explanation:
It is known that,
\(\int \sqrt{a^2+x^2} d x=\frac{x}{2} \sqrt{a^2+x^2}+\frac{a^2}{2} \log \left|x+\sqrt{x^2+a^2}\right|\) + C
∴ \(\int \sqrt{1+x^2} d x=\frac{x}{2} \sqrt{1+x^2}+\frac{1}{2} \log \left|x+\sqrt{1+x^2}\right|\) + C
Question 11.
An annuity in which the periodic payment begins on fixed date and continue forever is called
(A) Sinking fund
(B) Perpetuity
(C) Bond
(D) None of these
Answer:
(B) Perpetuity
Explanation:
Seasonal variation is variation in a time series within one year that is repeated more or less regularly. So, shortage of consumer goods before the annual budget is due to seasonal variation.
Question 12.
The present value of a perpetuity of ₹ 5550 payable at the beginning of each year, if money is worth 7% effective is
(A) ₹ 84500
(B) ₹ 65500
(C) ₹ 84836
(D) ₹ 86542
Answer:
(C) ₹ 84836
Explanation:
Given, R = ₹ 5550
and i = \(\frac{7}{100}\) = 0.07
Present Value, P = R + \(\frac{R}{i}\)
= ₹ (5550 + \(\frac{5550}{0.07}\))
= ₹ 84836
Question 13.
Mr. Sharma purchased a home for ₹ 350,000 in 2018. 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 2022, 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 ₹ 450,000. His rate of return is
(A) 25.5%
(B) 28.5%
(C) 35.5%
(D) None of these
Answer:
(B) 28.5%
Explanation:
Given,
Current value = 350,000
Original value = 450,000
Rate of return = \(\frac{\text { current value }- \text { original value }}{\text { original value }}\) × 100
= \(\frac{45000-35000}{35000}\) × 100
= 28.5 %
Question 14.
The order and degree (if defined) of differential equation y”’ + y2 + y’ = 0 are
(A) 3, 1
(B) 3, not defined
(C) not defined, 3
(D) 1, 3
Answer:
(A) 3, 1
Explanation:
The highest order derivative present is y”’. so its order is 3.
Also, the power of highest order derivative is 1, so its degree is 1.
Question 15.
In the given figure (I), what is the LPP shaded region known as?
(A) Feasible region
(B) Feasible solution
(C) Optimal region
(D) Objective region
Answer:
(A) Feasible region
Question 16.
An insurance company has found that 50% of its claims are for damages resulting from accidents. The probability that a random sample of 10 claims will contain fewer than 2 for accidents is
(A) \(\frac{1}{1024}\)
(B) \(\frac{5}{512}\)
(C) \(\frac{11}{1024}\)
(D) \(\frac{15}{1024}\)
Answer:
(C) \(\frac{11}{1024}\)
Explanation:
P (r < 2) = P (0 or 1)
= \({ }^{10} C_0\left(\frac{1}{2}\right)^{10}+{ }^{10} C_1\left(\frac{1}{2}\right)^{10}\)
= \(\frac{1+10}{1024}\)
= \(\frac{11}{1024}\)
Question 17.
The best-fitted trend line is one for which sum of squares of residuals or errors is .
(A) Positive
(B) Minimum
(C) Negative
(D) Maximum
Answer:
(B) Minimum
Explanation:
The line is termed as the line of best fit from which the sum of squares of distances from the points is minimized.
Question 18.
A grain whole-seller visits the granary market. While going around to make a good purchase, he takes a handful of rice from random sacks of rice, in order to inspect the quality of farmers produce. The handful of rice taken from a sack of rice for quality inspection is a
(A) statistic
(B) population
(C) parameter
(D) sample
Answer:
(D) sample
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) : If y = x3 log x, then \(\frac{d^4 y}{d x^4}=\frac{6}{x}\)
Reason (R) : If y = x3 log x,
Answer:
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
Question 20.
Assertion (A) : The area under a standard normal curve is 1.
Reason (R) : The area under a standard normal curve is 2.
Answer:
(C) (A) is true but (R) is false.
Explanation:
For any probability distribution the sum of all probabilities is 1.
Area under normal curve refers to sum of all probabilities.
Section – B
(All Questions are compulsory. In case of internal Choice, attempt any one question only)
Question 21.
At what rate of interest will the present value of a perpetuity of ₹ 300 payable at the end of each quarter be ₹ 24000 ?
Answer:
Let the rate of investment be r% per annum, then
i = \(\frac{r}{4 \times 100}=\frac{r}{400}\)
Given, perpetuity, R = ₹ 300
and payment P = ₹ 24000
Using, P = \(\frac{R}{i}\)
⇒ i = \(\frac{R}{P}\)
⇒ \(\frac{r}{400}=\frac{300}{24000}\)
⇒ r = \(\frac{300 \times 400}{24000}\) = 5 %
Hence, rate of interest = 5 % per annum.
Question 22.
If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\), find k so that A2 = 5A + kI.
Answer:
A2 = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\)
A2 = \(\left[\begin{array}{cc}
8 & 5 \\
-5 & 3
\end{array}\right]\)
5A = \(\left[\begin{array}{cc}
15 & 5 \\
-5 & 10
\end{array}\right]\)
kI = \(\left[\begin{array}{ll}
k & 0 \\
0 & k
\end{array}\right]\)
∵ A2 – 5A = kI
\(\left[\begin{array}{cc}
-7 & 0 \\
0 & -7
\end{array}\right]=\left[\begin{array}{ll}
k & 0 \\
0 & k
\end{array}\right]\)
⇒ k = – 7.
OR
Without expanding at any stage, find the value of determinant:
∆ = \(\left|\begin{array}{lll}
2 & x & y+z \\
2 & y & z+x \\
2 & z & x+y
\end{array}\right|\)
Answer:
Given, ∆ = \(\left|\begin{array}{lll}
2 & x & y+z \\
2 & y & z+x \\
2 & z & x+y
\end{array}\right|\)
By applying, C3 ⇒ C2 + C3
∆ = \(\left|\begin{array}{lll}
2 & x & x+y+z \\
2 & y & x+y+z \\
2 & z & x+y+z
\end{array}\right|\)
= 2 (x + y + z) \(\left|\begin{array}{lll}
1 & x & 1 \\
1 & y & 1 \\
1 & z & 1
\end{array}\right|\)
= 2 × 0 = 0
[Since, C1 = C2
∴ ∆ = 0]
Question 23.
Define the following:
(a) Decision variables
(b) Feasible Region
Answer:
Decision variables: In the objective function Z = ax + by, the variables x, y are said to be decision variables.
Feasible region:
The common region determined by all the constraints including non-negative constraints x, y > 0 of linear programming problem is known as the feasible region.
Question 24.
Solve for x: \(\frac{x+3}{x-2}\) ≤ 2
Answer:
We have,
\(\frac{x+3}{x-2}\) ≤ 2
\(\frac{x+3}{x-2}\) – 2 ≤ 0
\(\frac{x+3-2 x+4}{x-2}\) ≤ 0
\(\frac{-x+7}{x-2}\) ≤ 0
(x – 2) (- x + 7) ≤ 0
Multiplying both sides by (x – 2)2
∴ – x + 7 = 0 and x – 2 = 0
x = 7 and x = 2.
Thus, solution set is (- ∞, 2) ∪ [7, ∞).
OR
A pipe A can fill a tank in 3 hours. There are two outlet pipes B and C from the tank which can empty it in 7 and 10 hours respectively. If all the three pipes are opened simultaneously, how long will it take to fill the tank?
Answer:
Pipe A can fill the part of tank in 1 hour = \(\frac{1}{3}\) part
Pipe B can empty the part of tank in 1 hour = \(\frac{1}{7}\) part
Pipe C can empty the part of tank in 1 hour = \(\frac{1}{10}\) part
Net part filled in 1 hour = \(\frac{1}{3}-\left(\frac{1}{7}+\frac{1}{10}\right)\)
= \(\frac{1}{3}-\frac{17}{70}\)
= \(\frac{19}{210}\)
Thus, tank to will be filled in \(\frac{210}{19}\) hour i.e., 11 \(\frac{1}{9}\) hours or 11 hours 6 mins 10 sec.
Question 25.
Find the student’s – t for the following variable values in a sample of eight: – 4, – 2, – 2, 0, 2, 2, 3, 3 taking the mean of the universe to be zero.
Answer:
x | x – \(\bar{x}\) | (x – \(\bar{x}\))2 |
– 4 | – 4.25 | 18.0625 |
– 2 | – 2.25 | 5.0625 |
– 2 | – 2.25 | 5.0625 |
0 | – 0.25 | 0.0625 |
2 | 1.75 | 3.0625 |
3 | 2.75 | 7.5625 |
3 | 2.75 | 7.5625 |
Σ x = 2 | Σ (x – \(\bar{x}\))2 = 49.5000 |
\(\bar{x}\) = mean
= \(\frac{\sum x}{n}\)
= \(\frac{2}{8}\)
= 0.25
Now, compute the standard deviation using formula as,
Section – C
(All Questions are compulsory. In case of internal Choice, attempt any one question only)
Question 26.
Evaluate: ∫ \(\frac{x^3}{x^4+3 x^2+2}\) dx
Answer:
Put x2 = t
⇒ 2x dx = dt
⇒ x dx = \(\frac{1}{2}\) dt
\(\int \frac{x^3}{x^4+3 x^2+2} d x=\frac{1}{2} \int \frac{t d t}{t^2+3 t+2}\)
Now, \(\frac{t}{t^2+3 t+2}=\frac{t}{(t+2)(t+1)}\)
= \(\frac{A}{(t+2)}+\frac{B}{(t+1)}\)
t = A (t + 1) + B (t + 2)
Equating the coefficients of t and constant terms, we get
A + B = 1
and A + 2B = 0
⇒ A = 2 and B = – 1
∴ \(\int \frac{x^3}{x^4+3 x^2+2} d x=\frac{1}{2} \int\left(\frac{2}{t+2}-\frac{1}{t+1}\right) d t\)
= \(\frac{1}{2}\) 2 log |t + 2| – \(\frac{1}{2}\) log |t + 1| + C
= log |x2 + 2| – \(\frac{1}{2}\) log |x2 + 2| + C
= 1
OR
Evaluate: \(\int_1^3\) |x2 – 2x| dx
Answer:
Consider I = \(\int_1^3\) |x2 – 2x| dx
|x2 – 2x| = \(\left\{\begin{array}{ccc}
-\left(x^2-2 x\right) & \text { where } & 1 \leq x<2 \\
\left(x^2-2 x\right) & \text { where } & 2 \leq x \leq 3
\end{array}\right.\)
I = \(\int_1^2\) |x2 – 2x| dx + \(\int_2^3\) |x2 – 2x| dx
I = \(\int_1^2\) |x2 – 2x| dx + \(\int_2^3\) |x2 – 2x| dx
I = \(-\left[\frac{x^3}{3}-x^2\right]_1^2+\left[\frac{x^3}{3}-x^2\right]_2^3\)
I = \(-\left[\frac{8}{3}-4-\frac{1}{3}+1\right]+\left[9-9-\frac{8}{3}+4\right]\)
= – \(\frac{7}{3}\) + 3 – \(\frac{8}{3}\) + 4
I = 7 – 5 = 2
Question 27.
Show that the matrix BT AB is symmetric or skew-symmetric accordingly when A is symmetric or skew-symmetric.
Answer:
Case I :
Let A be a symmetric matrix.
Then AT = A.
Now, (BTAB)T = BTAT(BT)T [By reversal law]
= BTATB [∵ (BT)T = B]
Or, (BTAB)T = BTAB [∵ AT = A]
∴ BTAB is a symmetric matrix.
Case II:
Let A be a skew-symmetric matrix. Then,
AT =- A.
Now, (BTAB)T = BTAT(BT)T [By reversal lawl
Or, (BTAB)T = BTATB [∵ (BT)T = B]
Or, (BTAB)T = BT(- A)B [∵ AT = – A]
Or, (BTAB)T = – BTAB
∴ BTAB is a skew-symmetric matrix.
Question 28.
Find the intervals in which the function f(x) = \(\frac{x^4}{4}\) – x3 – 5x2 + 24x + 12 is
(a) strictly increasing,
(b) strictly decreasing.
Answer:
\(\frac{x^4}{4}\) – x3 – 5x2 + 24x + 12
f'(x) = x3 – 3x2 – 10x + 24
= (x – 2) (x – 4) (x + 3)
f'(x) = 0
⇒ x = – 3, 2, 4
sign of f'(x):
∴ f(x) is strictly increasing on (- 3, 2) ∪ (4, ∞)
and f(x) is strictly decreasing on (- ∞, – 3) ∪ (2, 4)
Question 29.
If the supply function for a commodity is p = 4 + x, and 12 units of goods are sold, then find the producer’s surplus.
Answer:
Given, the supply function is
p = 4 + x …………….(i)
and the market demand
x0 = 12.
At equilibrium, P0 = 4 + x0
Substituting this value of x0 in (i), we get
p0 = 4 + 12
⇒ p0 = 16
So, Producers Surplus = x0 × p0 – \(\int_0^{x_0}\) p dx
= 12 × 16 – \(\int_0^{12}\) (4+x) dx
= 192 – \(\left[4 x+\frac{x^2}{2}\right]_0^{12}\)
= 192 – (48 + 72) + 0
= 192 – 120
= 72
OR
Find the area of the region bounded above by y = x2 + 1, bounded below by y = x and bounded on the sides by x = 0 and x = 1.
Answer:
The upper boundary curve is y = x2 + 1
and the lower boundary curve is y = x.
Using the formula for area bounded by curves
A = \(\int_0^1\) [(x2 + 1) – x] dx
= \(\int_0^1\) (x2 – x + 1) dx
= \(\left.\frac{x^3}{3}-\frac{x^2}{2}+x\right]_0^1\)
= \(\frac{1}{3}-\frac{1}{2}\) + 1
= \(\frac{5}{6}\) unit2
Question 30.
Rohan has completed his MBA and now he wants to start a new business. So, he approaches to many banks. One bank is agreed to give loan to Rohan. So, Rohan has borrowed ₹ 5 lakhs from a bank on the interest rate of 12 per cent for 10 years.
Calculate monthly instalment using (1.01)120 = 3.300
Answer:
Given, i = \(\left[\frac{\left(\frac{\text { annual rate }}{12}\right)}{100}\right]\)
= \(\left[\frac{\left(\frac{12}{12}\right)}{100}\right]\)
= \(\left[\frac{\left(\frac{12}{12}\right)}{100}\right]\)
= 0.01
n = 10 × 12
= 120
P = ₹ 5, 00,000
Instalment Amount = \(\frac{(1+i)^n}{(1+i)^n-1} \times(P \times i)\)
Instalment Amount = \(\frac{(1+0.01)^{120}}{(1+0.01)^{120}-1}\) × (5,00,000 × 0.01)
= \(\frac{3.300}{3.300-1}\) × 5,000
= \(\frac{16,500}{2.300}\)
= ₹ 7173.91 – ₹ 7174
So, EMI that Rohan has to pay ₹ 7174.
Question 31.
A textile company has raised funds in the form of 5,000 zero-coupon bonds worth ₹ 1,100 each. The company wants to set up a sinking fund for repayment of the bonds, which will be after 7 years. Determine the amount of the periodic contribution if the annualized rate of interest is 6% and the contribution will be done quarterly. [Given (1.015)28 = 1.51721]
Answer:
Given, Sinking fund,
A = Par value of bond × No. of bonds
= ₹ 1,100 × 5,000
= ₹ 5,500,000
No. of years, n = 7
No. of payments per year, m = 4
Annualized rate of interest, r = 6%
∴ i = \(\frac{6}{400}\) = 0.015
Therefore, the amount of the periodic contribution can be calculated using the above formula as,
Section – D
(This section comprises of long answer type questions (LA) of 5 marks each)
Question 32.
A bank manager has observed that the length of time the customers have to wait for being attended by the teller is normally distribution with mean time of 5 minutes and standard deviation of 0.7 minutes. Find the probability that a customer has to wait
(i) for less than 6 minutes
Answer:
Let X be the waiting time of a customer in the queue
and it is normally distributed with mean 5 and SD 0.7.
for less than 6 minutes
Z = \(\frac{X-\mu}{\sigma}\)
= \(\frac{6-5}{0.7}\)
= 1.4285
z = 0 z = 1.43
P(X < 6) = P(Z < 1.43)
= 0.5 + 0.4236
= 0.9236
(ii) between 3.5 and 6.5 minutes
Answer:
between 3.5 and 6.5 minutes
When X = 3.5
Z = \(\frac{X-\mu}{\sigma}\)
= \(\frac{3.5-5}{0.7}\)
= – 2.1429
When X = 6.5
= \(\frac{X-\mu}{\sigma}\)
= \(\frac{6.5-5}{0.7}\)
= 2.1429
P (3.5 < X < 6.5) = P (- 2.1429 < Z < 2.1429)
= P (0 < Z < 2.1429) + P (0 < Z < 2.1429)
= 2P (0 < Z < 2.1429)
= 2 × .4838
= 0.%76
OR
(i) If P(X = 0) = P(X = 1) = a in a Poisson distribution, show that a = \(\frac{1}{e}\).
Answer:
Let, P (X = x) = e-m \(\frac{m^x}{x !}\)
x = 0, 1, 2, …………., ∞
Then, P (X = 0) = e-m
and P (X = 1) = e-m
Since, given P(X = 0) = P(X = 1), therefore
e-m = e-m m
⇒ m = 1
Also, P (X = 0) = a
⇒ e-a = a
⇒ a = \(\frac{1}{e}\)
(ii) There are 50 telephone lines in an exchange. The probability that any one of them will be busy is 0.1. Find the probability that all the lines are busy.
Answer:
Given, 50 telephone line in an exchange probability that any of them will be busy is 0.1.
L.et X be the Poisson variable.
∴ P(X = R) = \(\frac{e^{-5}(5)^r}{r !}\), r = 0, 1, 2, ………….
Here, n = 50, p = 0.1
∴ m = np
= 50 × 0.1 = 5
∴ P(X = r) = \(\frac{e^{-5}(5)^r}{r !}\), r = 0, 1, 2, ……….., 50
Thus, P (all lines are busy) = P (X = 50)
∴ = \(\frac{e^{-5}(5)^r}{r !}\)
For r = 50,
P = \(\frac{e^{-5} 5^{50}}{50 !}\)
Question 33.
A dietician wishes to mix two kinds of food X and Y in such a way that the mixture contains atleast 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:
Food | Vitamin A | Vitamin B | Vitamin C |
X | 1 unit | 2 units | 3 units |
Y | 2 units | 2 units | 1 unit |
One kg of food X costs ₹ 24 and one kg of food Y costs ₹ 36. Using Linear Programming, find the least cost of the total mixture which will contain the required vitamins.
Answer:
Let x be the number of units of food X and y be the number of units of food Y be mixed to obtain the desired diet.
Then LPP of the given problem is:
Minimise, Z = 24x + 36y
Subject to the constraints,
x + 2y ≥ 10, 2x + 2y ≥ 12
i.e., x + y ≥ 6 and 3x + y ≥ 8
x ≥ 0, y ≥ 0
We draw the lines x + 2y = 10, x + y = 6, 3x + y = 8 and obtain the feasible region (unbounded and convex) as shown in the figure.
Thus, corner points are A(0, 8), B(1, 5), C(2, 4) and D(10,0).
The values of Z (in ₹) at these points are given in the following table:
Corner Point | Objective Function |
A(0, 8) | Z = 24 × 0 + 36 × 8 = 288 |
B(1, 5) | Z = 24 × 1 + 36 × 5 = 204 |
C(2, 4) | Z = 24 × 2 + 36 × 4 = 192 (Min.) |
D(10, 0) | Z = 24 × 10 + 36 × 0 = 240 |
As the feasible region is unbounded, we draw the graph of the half plane 24x + 36y < 192 i.e., 2x + 3y < 16 and note that there is no point common with the feasible region. The minimum value of Z is ₹ 192. It occurs at C(2, 4).
i.e., when 2 kg of food X and 4 kg of food Y are mixed to get the desired diet.
Question 34.
Show that the surface area of cloed cuboid with square base and given volume is minimum when it is a cube.
Answer:
Let the length and breadth be x units (square base),
height of cuboid be h unit.
Volume of cuboid (v) = x2h
∴ h = \(\frac{v}{x^2}\)
Surface area of cuboid (s) = 2 (x2 + xh + xh)
∴ s = 2x2 + 4xh
⇒ s = 2x2 + 4x × \(\frac{v}{x^2}\)
⇒ s = 2x2 + 4v \(\frac{1}{x}\)
∴ \(\frac{d s}{d x}\) = 4x – 4v \(\frac{1}{x^2}\)
For maximum and minimum surface, \(\frac{d s}{d x}\) = 0
∴ 4x – 4v \(\frac{1}{x^2}\) = 0
⇒ 4x = \(\frac{4 v}{x^2}\)
⇒ x3 = v
⇒ x3 = x2 h
⇒ x = h
Now, \(\frac{d^2 s}{d x^2}=4+\frac{8 v}{x^3}\)
At x = h,
\(\frac{d^2 s}{d x^2}=4+\frac{8 v}{h^3}\) > 0
∴ Surface area is minimum, when x = h.
i.e., when x = h, it is a cube, surface area will be minimum. x = h
Hence Proved.
OR
The cost function for the manufacture of x number of goods by a company is C(x) = x3 – 9x2 + 24x. Find the level of output at which the marginal cost is minimum. Further, if the selling price of a unit is 2x3 + 9x2, find the average profit.
Answer:
We calculate the marginal cost
\(\frac{d c(x)}{d x}\) = \(\frac{d}{d x}\) (x3 – 9x2 + 24x)
[Given c(x) = x3 – 9x2 + 24x]
In order to be a minimum at x = x0 (say), its derivative must vanish at x0
Thus, \(\left[\frac{d c(x)}{d x}\right]_{x=x_0}\) = 0
Or, (3x2 – 18x + 24) = 0
Thus, x0 = 2, 4
Now, \(\frac{d^2}{d x^2}\) (x3 – 9x2 + 24x) = 6x – 18
\(\left[\frac{d^2 c(x)}{d x^2}\right]_{x_0=2}\) = – 6
and \(\left[\frac{d^2 c(x)}{d x^2}\right]_{x_0=4}\) = 6
By the second derivative test, we can conclude that at x = 4, the function asumes a minima.
Thus, for an output = 4 finished goods, the marginal cost would be minimum.
Now, the average profit as
\(\frac{P(x)}{x}=\frac{S P(x)-C(x)}{x}\)
⇒ \(\frac{\left(2 x^3+9 x^2\right)-\left(x^3-9 x^2+24 x\right)}{x}\)
⇒ \(\frac{x^3+18 x^2-24 x}{x}\)
⇒ x2 + 18x – 24
Question 35.
If A = \(\left[\begin{array}{ccc}
3 & 1 & 2 \\
3 & 2 & -3 \\
2 & 0 & -1
\end{array}\right]\), find A-1.
Hence, solve the system of equations: 3x + 3y + 2z = 1 x + 2y = 4 2x – 3y – z = 5.
Answer:
A = \(\left[\begin{array}{ccc}
3 & 1 & 2 \\
3 & 2 & -3 \\
2 & 0 & -1
\end{array}\right]\)
|A| = 3 (- 2) – 1 (3) + 2 (- 4)
= – 6 – 3 – 8
= – 17 ≠ 0
∴ A-1 exists.
Cofactor matrix of A = \(\left[\begin{array}{ccc}
-2 & -3 & -4 \\
1 & -7 & 2 \\
-7 & 15 & 3
\end{array}\right]\)
A-1 = \(\frac{1}{|A|}\) Adj A
= \(\frac{1}{-17}\left[\begin{array}{ccc}
-2 & 1 & -7 \\
-3 & -7 & 15 \\
-4 & 2 & 3
\end{array}\right]\)
Now for given system of equations.
\(\left[\begin{array}{ccc}
3 & 3 & 2 \\
1 & 2 & 0 \\
2 & -3 & -1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
1 \\
4 \\
5
\end{array}\right]\)
(AT)X = B
⇒ X = (AT)-1 B
⇒ X = (A-1)TB
[∵ X = (AT)-1 = (A-1)T]
X = \(\frac{1}{-17}\left[\begin{array}{ccc}
-2 & -3 & -4 \\
1 & -7 & 2 \\
-7 & 15 & 3
\end{array}\right]\left[\begin{array}{l}
1 \\
4 \\
5
\end{array}\right]\)
X = \(\frac{1}{-17}\left[\begin{array}{c}
-34 \\
-17 \\
68
\end{array}\right]=\left[\begin{array}{c}
2 \\
1 \\
-4
\end{array}\right]\)
x = 2, y = 1, z = – 4.
Section – E
(This section comprises of 3 source based questions (Case Studies) of 4 marks each)
CASE STUDY – 1
Question 36.
Read the following text and answer the following questions on the basis of the same:
You may never have heard of modular arithmetic, but you use it every day without the slightest difficulty. In this system, numbers wrap around when they reach a certain size called the modulus; it is the arithmetic of ‘ remainders. When reckoning hours, we count up to 12 and start again from one. Thus, four hours after 9 o’clock it is 1 o’clock. Numbers that differ by a multiple of the modulus 12 are said to be congruent modulo 12.
A similar situation arises for the days of the week, which are computed modulo seven. Suppose today is Thursday. What weekday will it be 1,000 days from today? We don’t have to count through the thousand days, just calculate the remainder when 1,000 is divided by seven, which is six. It is much the same for other time measurements. With 52 weeks in a year, the week number is reset to 1 at the beginning of each year, so it is confined to the range 1 to 52. Likewise, for the months, we use modulo 12 arithmetic.
(i) Assume it is 18 : 00 hours. What time will it be 59 hours from now?
Answer:
Here, we use the 24 hour clock.
i.e., (18 + 59) (mod 24) = 77 (mod 24)
= 5
So, time will 5:00 hours.
OR
What time is it in 79 hours if it is 4 o’clock now?
Answer:
We have (4 + 79) mod 24 = 83 mod 24 = 11.
Thus 79 hours from now, it is 11 o’clock.
(ii) It’s 7:00 (am/pm doesn’t matter). Where will the hour hand be in 7 hours?
Answer:
(7 + 7) mod 12 = (14) mod 12 = 2 mod 12
[2 is the remainder when 14 is divided by 12]
So, its 2 o’clock and hour hand is at 2.
(iii) Which month is it 721 months from now if this month is November?
Also, find which day of the week is it in 110 days from today if today is Friday ?
Answer:
Months wrap around after 12 months.
We have 721 mod 12 = 1.
Since, December is one month after November, months from now, it will be December.
The days of the week wrap around after seven days.
So, 110 mod 7 = 5 days
Thus, 110 days after Friday is the same day of the week as 5 days after Friday, i.e., Wednesday.
CASE STUDY – II
Question 37.
Read the following text and answer the following questions on the basis of the same:
According to an educational board survey, it was observed that class XII students apply at least one to four weeks ahead of colleges application deadline. Let X represent the week when an average student applies ahead of a college’s application deadline and the probability of student to get admission in the college P(X = x) is given as follows:
P(X = x) = \(\begin{cases}\frac{k x}{6} & \text { when } x=0,1 \text { or } 2 \\ \frac{(1-k) x}{6} & \text { when } x=3 \\ \frac{k x}{2} & \text { when } x=4 \\ 0 & \text { when } x>4\end{cases}\)
Based on the above information, answer the following questions. Show steps to support your answers.
(i) Find the value of k.
Answer:
\(\frac{k}{6}+\frac{2 k}{6}+\frac{3(1-k)}{6}+\frac{4 k}{2}\) = 1
k = \(\frac{1}{4}\)
(ii) What is the probability that Sonali will get admission in the college, given that she applied at least 2 weeks ahead of application deadline?
Answer:
P (getting admission on applying at least 2 weeks ahead of application deadline)
= P(X = 2, 3, 4)
= \(\frac{1}{12}+\frac{3}{8}+\frac{1}{2}=\frac{23}{24}\)
[alternate method : 1 – P(X = 1) = 1 – \(\frac{1}{24}\) = \(\frac{23}{24}\)]
(iii) Calculate the mathematical expectation of number of weeks taken by a student to apply ahead of a college’s application deadline.
Answer:
X = week applied ahead of application deadline
E(X) = \(\frac{80}{24}\)
= 3 \(\frac{1}{3}\) weeks
OR
To promote early admissions, the college is offering scholarships to the students for applying ahead of deadline as follows:
₹ 50000 for applying 4 weeks early, ₹ 20000 for applying 3 weeks early, ₹ 12000 for applying 2 weeks early and ₹ 9600 for applying 1 week early What is the expected scholarship offered by the college?
Answer:
X = Scholarship money awarded for the week applied in, before the deadline
CASE STUDY – III
Question 38.
Read the following text and answer the following questions on the basis of the same:
Strikes is a very powerful weapon used .by trade unions and other labour associations to get their demand accepted generally involves quitting of work by a group of workers for the purpose of bringing the pressure qn their employer so that their demands get accepted. When workers collectively cease to work in a particular industry/they are said to be on strike.
The table give below shows the average number in lakhs, of working days lost in strikes during each year of period 2009-2018 was
Answer:
Calculate three yearly moving average. Draw the graphs for moving averages.
Answer:
Year | Working days lost in Strikes (in lakhs) | Three yearly moving totals | Three yearly moving averages |
2009 | 1.5 | – | – |
2010 | 1.8 | 5.2 | 1.73 |
2011 | 1.9 | 5.9 | 1.96 |
2012 | 2.2 | 6.7 | 2.23 |
2013 | 2.6 | 8.5 | 2.83 |
2014 | 3.7 | 8.5 | 2.83 |
2015 | 2.2 | 12.3 | 4.1 |
2016 | 6.4 | 12.2 | 4.06 |
2017 | 3.6 | 15.4 | 5.13 |
2018 | 5.4 | – | – |
OR
Below are given the figures of production (in m. tonnes) of a rice factory:
Fit the straight line trend to these figures and estimate the likely sales of the company during 2019.
Answer:
Now,
a = \(\frac{\Sigma y}{n}\)
= \(\frac{630}{7}\) = 90
b = \(\frac{\Sigma x y}{\Sigma x^2}\)
= \(\frac{56}{28}\) = 2
Thus, trend equation is:
yt = 90 + 2x
Now,
For 2019, x would be 4.
Putting x = 4 in trend equation, we get
y2019 = 90 + 2 (4) = 98 m. tonnes.
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