Students can access the CBSE Sample Papers for Class 12 Applied Maths with Solutions and marking scheme Set 6 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 12 Applied Maths Set 6 with Solutions
Time Allowed : 3 hours
Maximum Marks : 80
General Instructions:
(i) 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.
(ii) Section A has 18 MCQ’s and 02 Assertion Reason based questions of 1 mark each.
(iii) Section B has 5 Very Short Answer (VSA) questions of 2 marks each.
(iv) Section C has 6 Short Answer (SA) questions of 3 marks each.
(v) Section D has 4 Long Answer (LA) questions of 5 marks each.
(vi) Section E has 3 source based/case based/passage based/integrated units of assessment (04 marks each) with sub parts.
(vii) 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 value of – 70 mod 13 ¡s
(A) 5
(B) – 5
(C) 8
(D) – 8
Answer:
(C) 8
Explanation:
Here, X < 0 and Y > 0 , hence – 70 mod 13 is 8.
∴ 8 > 0.
Question 2.
If \(\frac{x+1}{x+2}\) ≥ 1 then
(A) x ∈ (- ∞, 2]
(B) x ∈ (- ∞, – 2)
(C) x ∈ (- ∞, 2]
(D) x ∈ (- ∞, 2)
Answer:
(B) x ∈ (- ∞, – 2)
Explanation:
x ∈ (- ∞, – 2)
Question 3.
Which of the following is a statistic?
(A) µ
(B) \(\bar{x}\)
(C) σ2
(D) None
Answer:
(B) \(\bar{x}\)
Explanation:
\(\bar{x}\) is a statistic.
Question 4.
In one sample tese the estimation for population mean is
(A) \(\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\)
(B) \(\frac{\bar{x}-\mu}{\frac{s}{n}}\)
(C) \(\frac{\bar{x}-\mu}{\frac{s^2}{n}}\)
(D) None
Answer:
(A) \(\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\)
Question 5.
A man can row 6 km/h in still water. It takes him twice as long to row up as to row down the river. Then the rate of the stream is
(A) 2 km/h
(B) 4 km/h
(C) 6 km/h
(D) 8 km/h
Answer:
(A) 2 km/h
Explanation:
Let man’s rate upstream = x km/h
Let man’s rate downstream = 2x km/h
Hence, man’s rate in still water = \(\frac{1}{2}\) (x + 2x)
= \(\frac{3 x}{2}\) km/h
Therefore, \(\frac{3 x}{2}\) = 6
⇒ x = 4 km/h
Man’s rate downstream = 8 km/h
Hence, rate of stream = \(\frac{1}{2}\) (8 – 4) = 2 km/h
Question 6.
If random variable X represents the number of heads when a coin is tossed twice then mathematical expectation of X is
(A) 0
(B) \(\frac{1}{4}\)
(C) \(\frac{1}{2}\)
(D) 1
Answer:
(D) 1
Explanation:
Mathematical Expectation E(X) = Σ pixi = 1.
Question 7.
The least non-negative remainder when 350 is divided by 7 is
(A) 4
(B) 3
(C) 2
(D) 1
Answer:
(C) 2
Explanation:
31 = 3 (mod 7)
⇒ 32 = 3 × 3 = 2 (mod 7)
⇒ 33 = 3 × 2 = 6 = – 1 (mod7)
⇒ (33)16= (- 1)16 (mod 7)
⇒ (33)16 = 1 (mod 7)
⇒ (33)16 32 = 1 x 31 (mod 7)
⇒ 350 = 2 (mod 7)
Question 8.
If the cash equivalent of a perpetuity of ₹ 300 payable at the end of each quarter is ₹ 24000 then rate of interest converted quarterly is
(A) 5 %
(B) 4 %
(C) 3 %
(D) 2 %
Answer:
(A) 5 %
Explanation:
i = \(\frac{r}{400}\)
P = \(\frac{R}{i}\)
⇒ 24000 = \(\frac{300 \times 400}{r}\)
⇒ r = \(\frac{120}{24}\) = 5 %
Question 9.
The value of ∫ \(\frac{\log x}{x}\) dx is
(A) \(\frac{\log x}{2}\) + C
(B) \(\frac{(\log x)^2}{2}\) + C
(C) log x + C
(D) None
Answer:
(B) \(\frac{(\log x)^2}{2}\) + C
Explanation:
∫ \(\frac{log x}{x}\) dx
put log x = t
Differentiating \(\frac{1}{x}\) dx = dt
Hence, ∫ \(\frac{log x}{x}\) dx = ∫ t dt
= \(\frac{t^2}{2}\) + C
= \(\frac{(\log x)^2}{2}\) + C
Question 10.
The supply of finished good was delayed for a month due to landslide in hilly terrain. Under which trend oscillation does this situation fall
(A) Seasonal
(B) Cyclical
(C) Secular
(D) Irregular
Answer:
(D) Irregular
Question 11.
A machine costing ₹ 30,000 is expected to have a useful life of 4 years and a final scrap value of ₹ 4000. The annual depreciation is
(A) ₹ 5500
(B) ₹ 6500
(C) ₹ 7500
(D) ₹ 8500
Answer:
(B) ₹ 6500
Explanation:
D = \(\frac{C-S}{n}\)
⇒ D = \(\frac{30,000-4,000}{4}\)
= \(\frac{26,000}{4}\)
= 6,500
Question 12.
The effective rate of interest equivalent to the nominal rate 6% compounded semi-annually is
(A) 6.05%
(B) 6.07%
(C) 6.09%
(D) None
Answer:
(C) 6.09%
Explanation:
reff = [(1 + \(\frac{r}{m}\))m – 1] × 100
reff = [(1.03)2 – 1] × 100
= (1.0609 – 1) 100
6.09 %
Question 13.
If the investment of ₹ 20000 in the mutual fund in 2015 increased to ₹ 32000 in year 2020, then CAGR (Compound Annual Growth rate is) is [Given(1.6)1/3 = 1.098]
(A) 9.08%
(B) 9.8%
(C) 0.098%
(D) 0.09%
Answer:
(B) 9.8%
Explanation:
CAGR = [\(\left(\frac{E V}{S V}\right)^{\frac{1}{5}}\) – 1] × 100
= [\(\left(\frac{32000}{20000}\right)^{\frac{1}{5}}\) – 1] × 100
= [\((1.6)^{\frac{1}{5}}\) – 1] × 100
= [1.098 – 1] × 100
= 0.98 × 100
= 9.8 %
Question 14.
The integrating factor of the differential equation x \(\frac{d y}{d x}\) + 2y = x3 (x ≠ 0) is
(A) x
(B) log x
(C) x2
(D) \(\frac{1}{x^2}\)
Answer:
(C) x2
Explanation:
\(\frac{xd y}{d x}\) + 2y = x3
\(\frac{d y}{d x}+\frac{2 y}{x}=\frac{x^3}{x}\)
\(\frac{d y}{d x}+\frac{2 y}{x}\) = x2
I.F. = \(e^{\int \frac{2}{x} d x}\)
= e2 log x
= elog x2
= x2
Question 15.
Besides non negativity constraint the figure given below is subject to which of the following constraints
(A) x + 2y ≤ 5; x + y ≤ 4
(B) x + 24 ≥ 5; x + y ≤ 4
(C) x + 2y ≥ 5; x + y ≥ 4
(D) x + 2y ≤ 5; x + y ≥ 4
Answer:
(A) x + 2y ≤ 5; x + y ≤ 4
Explanation:
Line CD :
\(\frac{x}{4}+\frac{y}{4}\) ≤ 1 or x + y ≤ 4
Line AE:
\(\frac{x}{5}+\frac{y}{2.5}\) ≤ 1 or x + 2y ≤ 5
Question 16.
If X is a Poisson variate such that 3P(X = 2) = 2P(X = 1) then the mean of the distribution is equal to
(A) \(\frac{4}{3}\)
(B) \(\frac{3}{4}\)
(C) \(-\frac{4}{3}\)
(D) \(-\frac{3}{4}\)
Answer:
(A) \(\frac{4}{3}\)
Explanation:
3 P(X = 2) = 2P (X = 1)
⇒ \(3 \frac{m^2 e^{-m}}{2 !}=2 \frac{m e^{-m}}{1 !}\)
⇒m = \(\frac{4}{3}\)
Question 17.
For the given five values 35, 70, 36, 59, 64, the three years moving averages are given by
(A) 47, 53, 55
(B) 53, 47,45
(C) 47, 55 , 53
(D) 45, 55, 57
Answer:
(C) 47, 55 , 53
Explanation:
Values | 3 – years moving total | 3 – years moving average |
35 | – | – |
70 | 35 + 70 + 36 = 141 | \(\frac{141}{3}\) = 47 |
36 | 70 + 36 + 59 = 165 | \(\frac{165}{3}\) = 55 |
59 | 36 + 59 + 64 = 159 | \(\frac{159}{3}\) = 53 |
64 | – | – |
Question 18.
The data point of a normal variate with mean 12, standard deviation 4 and Z – score 5 is
(A) 28
(B) 304
(C) 34
(D) 32
Answer:
(D) 32
Explanation:
Z = \(\frac{x-\mu}{\sigma}\)
⇒ 5 = \([\frac{x-12}{4}/latex]
⇒ x = 32
Assertion Reason Based Questions
In the following questions, a statement of Assertion(A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false and R is true.
Question 19.
Assertion (A) : The maximum profit that a company makes if profit function is given by P(x) = 41 + 24x – 8x2; where ‘x’ is the number of units and P is the profit in rupees is 59.
Reason : The profit is maximum at x = a if P'(a) = 0 and P”(a) > 0
Answer:
(C) A is true but R is false.
Explanation:
P(x) = 41 + 24x – 8x2
P’(x) = 24 – 16x
P’(x) = 0
⇒ 24 – 16x = 0
⇒ x = [latex]\frac{24}{16}=\frac{3}{2}\)
P”(x) = – 16 < 0
⇒ x = \(\frac{3}{2}\) is a point of maxima
Max Profit = P
= 41 + 24 × \(\frac{3}{2}\) – 8 × \(\frac{9}{4}\)
= 41 + 36 – 18 = 59
Assertion is true but Reason is false, for Maximum P’(x) = 0 and P”(x) < 0.
Question 20.
Assertion (A) : The probability of getting 6 heads when a unbiased coin is tossed 10 Limes is C (10, 6) \(\left(\frac{1}{2}\right)^{10}\).
Reason (R) : In a Binomial distribution the probability is given by P (X = r) = C (n, r) (p)r (q)n-r
Answer:
(A) Both A and R are true and R is the correct explanation of A.
Explanation:
Given, n = 10
Probability of getting head = p = 1/2
Probability of getting tail = q = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
The probability of getting 6 heads when a unbiased coin is tossed 10 times is
P(X = 6) = C (10, 6) \(\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^{10-6}\)
= C (10, 6) \(\left(\frac{1}{2}\right)^{10}\)
[Using P(X = r) = C(n, r)(p)r(q)n-r]
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 perpetuity of ₹ 1500 payable at the end of every 6 months be ₹ 20,000 ?
Answer:
Let rate of interest be r% per annum, then i = \(\frac{r}{200}\)
Given R = ₹ 1500 and P = ₹ 20,000
P = \(\frac{R}{i}\)
⇒ i = \(\frac{R}{P}\)
= \(\frac{1500}{20000}\)
⇒ \(\frac{r}{200}=\frac{1500}{20000}\)
⇒ r = 15%
Question 22.
If A is a square matrix \(\left[\begin{array}{cc}
2 & -2 \\
-2 & 2
\end{array}\right]\) such that A2 = pA, then find the value of p.
Answer:
OR
If \(\left[\begin{array}{ccc}
0 & a & 3 \\
2 & b & -1 \\
c & 1 & 0
\end{array}\right]\) is skew-symmetric matrix, then find value of a + b + c.
Answer:
\(\left[\begin{array}{ccc}
0 & a & 3 \\
2 & b & -1 \\
c & 1 & 0
\end{array}\right]=-\left[\begin{array}{ccc}
0 & 2 & c \\
a & b & 1 \\
3 & -1 & 0
\end{array}\right]\)
Comparing a = – 2;
b = – b
⇒ 2b = b
⇒ b = 0 and c= – 3
Hence, a + b + c = – 2 + 0 – 3 = – 5.
Question 23.
A Cooperative Society of farmers has 10 hectares of land to grow two crops A and B. To control weeds, pesticide has to be used for crops A and B at the rate of 30 grams and 15 grams per hectare. Further, not more than 750 grams of pesticide should be used. The profit from crops A and B per hectare are estimated as ₹ 8000 and ₹ 9500. Formulate the above problem as LPP, in order to allocate land to each crop for maximum use.
Answer:
Let ‘x’ hectares and ‘y’ hectares of land be allocated to crop A and Crop B.
Max Z = 8000 x + 9500 y
Subject to x + y ≤ 10; 2x + y ≤ 50 ; x ≥ 0 and y ≥ 0
Detailed Solution:
Let ‘x’ hectares and ‘y’ hectares of land be allocated to crop A and Crop B.
Crop A |
Crop B | |
₹ 8000 | ₹ 9500 |
Profit |
x |
y | 10 hectares |
30 gram/hectares | 15 grams/hectares |
750 grams atmost |
Max Z = 8000 x + 9500 y
Subject to x + y ≤ 10
2x + y ≤ 50
x ≥ 0 and y ≥ 0
Question 24.
A boatman takes twice as long as to go upstream to a point as to return downstream to the starting point. If the speed of a boat in still water is 15 km/h, what is the speed of the stream.
Answer:
\(\frac{\text { Time taken upstream }}{\text { Time taken downstream }}=\frac{2}{1}\)
Let speed of boat = 15 km/h
and speed of stream = y km/h
Hence, \(\frac{15+y}{15-y}=\frac{2}{1}\)
⇒ 15 + y = 30 – 2y
⇒ 3y = 15
⇒ y = 5 km/h
OR
‘A’ can run 40 meters while ‘B’ runs 50 meters in the same time. In a 1000 m race, find by how much distance ‘B’ beats ‘A’.
Answer:
When B runs 50 m, A runs 40 m
When B runs 1 m, A runs = \(\frac{40}{50}=\frac{4}{5}\)
When B runs 1000 m,
A runs = \(\frac{4}{5}\) × 1000 = 800 m
Hence, B beats A by 200 m.
Question 25.
A machine produces washers of thickness 0.50 mm. To determine whether the machine is in proper working order, a sample of 10 washers is chosen for which the mean thickness is 0.53 mm and the standard deviation is 0.03 mm. Test the hypothesis at 5% level of significance that the machine is working in proper order. [Given t0.025 = 2.262 at 9 degree of freedom]
Answer:
Define Null hypothesis H0,
Alternate hypothesis H1
as follows:
H0 : µ = 0.50 mm
H1 : µ ≠ 0.50 mm
Thus, a two-tailed test is applied under hypothesis H0, we have
t = \(\frac{\bar{x}-\mu}{s} \sqrt{n-1}\)
= \(\frac{0.53-0.50}{0.03}\) × 3 = 3
Since, the calculated value of t i.e., tcal (= 3) > ttab (= 2.262), the null hypothesis H0 can be rejected.
Hence, we conclude that machine is not working properly.
Section – C
(All Questions are compulsory. In case of internal Choice, attempt any one question only)
Question 26.
Evaluate: ∫ \(\frac{x^3}{(x+2)}\) dx
Answer:
∫ \(\frac{x^3}{(x+2)}\) dx = (∫ x2 – 2x + 4 – \(\frac{8}{x+2}\)) dx
[By Performing Division]
= \(\frac{x^3}{3}\) – x2 + 4x – 8 ln (x + 2) + C
(where C is an arbitrary constant of integration)
OR
Evaluate: ∫ (x2 + 1) loge x dx
Answer:
∫ (x2 + 1) ln x dx
Integrating by parts
\(\ln x\left(\frac{x^3}{3}+x\right)-\int \frac{1}{x}\left(\frac{x^3}{3}+x\right)\) dx
\(\ln x\left(\frac{x^3}{3}+x\right)-\int\left(\frac{x^2}{3}+1\right)\) dx
\(\ln x\left(\frac{x^3}{3}+x\right)-\left(\frac{x^3}{9}+x\right)\) + C
Question 27.
Cost of two toys A and B are ₹ 50 and ₹ 75. On a particular Sunday shopkeeper P sells 7 toys of type A and 10 toys of type B whereas shopkeeper Q sells 8 toys of type A and 6 toys of type B. Find income of both shopkeepers using matrix Algebra.
Answer:
Here Row 1 and Row 2 indicate shopkeeper 1 and shopkeeper 2
Cost Matrix = \(\left[\begin{array}{l}
50 \\
75
\end{array}\right]\)
Amount = \(\left[\begin{array}{cc}
7 & 10 \\
8 & 6
\end{array}\right]\left[\begin{array}{l}
50 \\
75
\end{array}\right]\)
= \(\left[\begin{array}{l}
350+750 \\
400+450
\end{array}\right]=\left[\begin{array}{c}
1100 \\
850
\end{array}\right]\)
Income of shopkeeper P is ₹ 1100 and shopkeeper Q is ₹ 850.
Question 28.
Find the intervals in which the function f(x) = 2x3 – 9x2 + 12x – 5 is increasing or decreasing.
Answer:
f(x) = 2x3 – 9x2 + 12x – 5
f ‘(x) = 6x2 – 18x + 12
= 6(x2 – 3x + 2)
f ‘(x) = 6(x – 1 )(x – 2)
f ‘(x) = 0
⇒ x = 1
and x = 2 are the critical points.
The intervals are (- ∞, 1); (1, 2); (2, ∞)
Increasing in (- ∞, 1) ∪ (2, ∞) Decreasing in (1, 2)
Question 29.
The demand and supply functions under the pure market competition are pd = 16 – x2 and ps = 2x2 + 4 respectively, where p is the price and x is the quantity of the commodity. Using integrals find Consumer’s surplus.
Answer:
Under pure competition
pd = ps
16 – x2 = 2x2 + 4
3x2 = 12
x = 2,- 2;
since, x can’t be – ve, so x = 2
When x0 = 2; p0 = 12
Hence, Consumer’s surplus = \(\int_0^2\) pd dx – p0x0
= \(\int_0^2\) (16 – x2) dx – 12 × 2
= \(\left[16 x-\frac{x^3}{3}\right]_0^2\) – 24
= \(\frac{16}{3}\) units
OR
The demand and supply functions under the pure market competition are pd = 56 – x2 and ps = 8 + \(\frac{x^2}{3}\) respectively, where p is the price and x is the quantity of the commodity. Using integrals find Producer’s surplus.
Answer:
pd = ps
⇒ 56 – x2 = 8 + \(\frac{x^2}{3}\)
⇒ \(\frac{4}{3}\) x2 = 48
⇒ x2 = 36
⇒ x = 6, – 6
since, x can’t be – ve, so x = 6
When x0 = 6; p0 = 20
Hence, Producer’s surplus = p0x0 – \(\int_0^6\) ps dx
= 6 20 – \(\int_0^6\left(8+\frac{x^2}{3}\right)\) dx
= 120 – \(\left[8 x+\frac{x^3}{9}\right]_0^6\)
= 120 – [48 + 24]
= 48 units.
Question 30.
Mr Surya borrows a sum of ₹ 5,00,000 with total interest paid ₹ 2,00,000 (flat) and he is paying an EMI of ₹ 12,500. Calculate loan tenure.
Answer:
Here P = 5,00,000;
I = 2,00,000;
EMI = 12,500
EMI = \(\frac{P+I}{n}\)
12,500 = \(\frac{5,00,000+2,00,000}{n}\)
⇒ n = \(\frac{7,00,000}{12,500}\)
= 56 months.
Question 31.
Mr Sharma wants to send his daughter abroad for higher studies after 10 years. He sets up a sinking fund in order to have ₹ 5,00,000 after 10 years. How much should he set aside bi-annually into an account paying 5% per annum compounded annually. [Use (1.025)20 = 1.6386]
Answer:
Let ₹ R be set aside biannually for 10 years in order to have ₹ 5,00,000 after 10 years.
Here, S = ₹ 5,00,000; n = 10 × 2 = 20
i = \(\frac{5}{2 \times 100}\) = 0.025
R = \(\frac{i S}{(1+i)^n-1}\)
= \(\frac{0.025 \times 5,00,000}{(1.025)^{20}-1}\)
= \(\frac{12.500}{1.6386-1}\)
= ₹ 19,574.07
Section – D
(This section comprises of long answer type questions (LA) of 5 marks each)
Question 32.
On doing the proof reading of a book on an average 4 errors in 10 pages were detected. Using Poisson’s . distribution find the probability of (i) No error and (ii) one error, in 1000 pages of first printed edition of the book. (Given e-0.4 = 0.6703)
Answer:
Here m = 0.4
OR
How many time must Sunil toss a fair coin so that the probability of getting at least one head is more than 90 %?
Answer:
Here, p = \(\frac{1}{2}\)
and q = \(\frac{1}{2}\)
P(X = r) = C (n, r) prqn-r
1 – P (r = 0) > \(\frac{90}{100}\)
1 – C(n, 0)\(\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^n>\frac{9}{10}\)
⇒ \(\frac{n !}{0 !(n-0) !}\left(\frac{1}{2}\right)^n<\frac{1}{10}\)
⇒ \(\left(\frac{1}{2}\right)^n<\frac{1}{10}\)
⇒ 2n > 10
⇒ n is 4 or more times
Question 33.
A manufacturer has three machines I, II and III installed in his factory. Machines I and II are capable of being operated for at most 12 hours whereas machine III must be operated for at least 5 hours a day. He produces only two items M and N, each requiring the use of all the three machines. The number of hours required for producing 1 unit of M and N on three machines are given in the following table:
He makes a profit of ₹ 600 and ₹ 400 on one unit of items M and N respectively. How many units of each item be produced so as to maximize the profit. What is the maximum Profit ?
Answer:
Let ‘x’ and ‘y’ be the number of units of items M and
N respectively. We have : x ≥ 0, y ≥ 0
x + 2y ≤ 12;
2x + y ≤ 12; x + \(\frac{5}{4}\) y ≥ 5.
Max Z = 600 x + 400 y
Corner Point | Z = 600x + 400y |
E : (5, 0) | 3000 |
C : (6, 0) | 3600 |
G : (4, 4) | 4000 (Maximum) |
B : (0, 6) | 2400 |
F : (0, 4) | 1600 |
Hence, maximum profit is 4000 when 4 units or each of the items M and N are produced.
Question 34.
A company produces a certain commodity with ₹ 24000 fixed cost. The variable cost is estimated to be 25% of the total revenue received on selling the product at a rate of ₹ 8 per unit. Find the following
(i) Cost Function.
(ii) Revenue Function
(iii) Breakeven Point
(iv) Profit Function
Answer:
Let ‘x’ units of product be produced and sold.
As selling price of one unit is ₹ 8 total revenue on ‘x’ units = ₹ 8x.
(i) Cost Function C(x) = Fixed Cost + 25% of 8x
= 24000 + \(\frac{25}{100}\) × 8x
= 24000 + 2x
(ii) Revenue Function = 8x
(üi) Breakeven Point 8x = 24000 + 2x
x = 4000
(iv) Profit function = R(x) – C(x) = 6x – 24000
OR
The production manager of a company plans to include 180 sq. cms. of actual printed matter in each page of a book under production. Each page should have a 2.5 cm wide margin along the top and bottom and 2 cm wide margin along the sides. What are the most economical dimensions of each printed page?
Answer:
Let x and y be the dimension of the printed pages,
then x.y = 180
A = Area of the page = (x + 4) (y + 5)
= xy + 5x + 4y + 20
= 180 + 5x + 4 × (\(\frac{180}{x}\)) + 20
= 200 + 5x + \(\frac{720}{x}\)
For most economical dimension \(\frac{d A}{dx}\) = 0
⇒ 5 – \(\frac{720}{x^2}\) = 0
⇒ x2 = 144
⇒ x = 12
⇒ \(\frac{d^2 A}{d x^2}=\frac{1440}{x^3}\)
\(\left(\frac{d^2 A}{d x^2}\right)_{x=12}=\frac{1440}{12^3}\) > 0
∴ A is minimum
Hence, the most economical dimensions are 16 cm and 20 cm.
Question 35.
The management committee of a Welfare Club decided to award some of its members (say x) for sincerity, some (say y) for helping others selflessly and some others (say z) for effective management. The sum of all the awardees is 12. Three times the sum of all awardees for helping others selflessly and effective management added to two times the number of awardees for sincerity is 33. If the sum of the nuniber of awardees for sincerity and effective management is twice the number of awardees for helping others, use matrix method to find the number of awardees of each category.
Answer:
x + y + z = 12
2x + 3y + 3z = 33
x – 2y + z = 0
\(\left[\begin{array}{ccc}
1 & 1 & 1 \\
2 & 3 & 3 \\
1 & -2 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
12 \\
33 \\
0
\end{array}\right]\)
|A| ≠ 3
Hence, x = 3, y = 4, z = 5.
Section – E
(This section comprises of 3 source based questions (Case Studies) of 4 marks each)
Question 36.
Case Study 1:
Pipes and Cisterns (Mark 2 + 1 + 1) (Internal choice is in the iii part)
A, B and C are three pipes connected to a tank. A and B together fill the tank in 6 hours B and C together fill the tank in 10 hours. A and C together fill the tank in 7 hours. Based on above information answer the following questions.
(i) In how much time will A, B and C fill the tank?
Answer:
Case Study-I
(i) A + B fill the tank in 6 hours
B + C fill the tank in 10 hours
A + C fill the tank in hours
2 (A + B + C) = \(\frac{6 \times 10 \times \frac{15}{2}}{6 \times 10+6 \times \frac{15}{2}+10 \times \frac{15}{2}}\)
= \(\frac{450}{60+45+75}\)
= \(\frac{450}{180}\)
= \(\frac{5}{2}\) hours
Hence A, B and C together will fill the tank in 5 hours.
(ii) In how much time will A separately fill the tank?
Answer:
A will fill in [(A + B + C) – (B + C)]
= \(\frac{10 \times 5}{10-5}\)
= 10 hours
(iii) In how much time will B separately fill the tank?
Answer:
B will fill in \(\frac{\frac{15}{2} \times 5}{\frac{15}{2}-5}\) = 15 hours
OR
In how much time will C separately fill the tank?
Answer:
C will fill in \(\frac{5 \times 6}{6-5}\) = 15 hours
Question 37.
Case Study 2:
Read the following passage and answer the questions below (Internal Choice is in option iii.) (Mark 1 + 1 + 2)
Let X denote the number of hours a person watches television during cl randomly selected day. The probability that X can take the values x., has the following form, where ‘k’ is some unknown constant.
P(X = xi) = \(\begin{cases}0.2, & \text { if } x_i=0 \\ k x_i, & \text { if } x_i=1 \text { or } 2 \\ k\left(5-x_i\right), & \text { if } x_i=3 \\ 0, & \text { otherwise }\end{cases}\)
(i) Find the value of k.
Answer:
Since, ΣP = 1
⇒ 0.2 + k + 2k + 2k = 1
⇒ 0.2 + 5k = 1
⇒ 5k = 1 – 0.2
⇒ k = \(\frac{4}{25}\)
(ii) What is the probability that a person watches two hours of television on a selected day?
Answer:
P(X = 2) = 2k = \(\frac{8}{25}\)
(iii) What is the probability that the person watches at least two hours of television on a selected day ?
Answer:
P(X ≥ 2) = 4k = \(\frac{16}{25}\)
OR
What is the probability that the person watches at most two hours of television on a selected day ?
Answer:
P(X ≤ 2) = 0.2 + 3k = \(\frac{17}{25}\)
Question 38.
Case Study 3:
When observed over a long period of time, a time series data can predict trend that can forecast increase or decrease or stagnation of a variable under consideration. Such analytical studies can benefit a business for forecasting or prediction of future estimated sales or production.
The table below shows the welfare expenses(in lakh ₹) of Steel Industry during 2001-2005. Fit a straight line trend by the method of least squares and estimate the trend for the year 2008.
Answer:
a = \(\frac{\Sigma Y}{n}\)
= \(\frac{1375}{5}\) = 275
b = \(\frac{\Sigma X Y}{\Sigma x^2}\)
= \(\frac{815}{10}\) = 81.5
Ye = a + bX
Ye = 275 + 81.5X
The estimated value for 2008 will be 275 + 151.5 × 5 = 275 + 757.5 = 1032.5
OR
The annual rainfall (in mm) was recorded in Cherrapunji, Meghalaya.
Determine the trend of rainfall by three years moving average and draw the moving averages graph.
Answer:
Year | Rainfall (in cm) | 3 years moving total | 3 years moving average |
2001 | 1.2 | – | – |
2002 | 1.9 | 5.1 | 1.70 |
2003 | 2 | 5.3 | 1.77 |
2004 | 1.4 | 5.5 | 1.83 |
2005 | 2.1 | 4.8 | 1.60 |
2006 | 1.3 | 5.2 | 1.73 |
2007 | 1.8 | 4.2 | 1.40 |
2008 | 1.1 | 4.2 | 1.40 |
2009 | 1.3 | – | – |