Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Mathematics with Solutions Set 11 are designed as per the revised syllabus.
CBSE Sample Papers for Class 12 Applied Mathematics Set 11 with Solutions
Maximum Marks: 80 Marks
Time Allowed : 3 Hours
General Instructions:
- This question paper contains five sections A, B, C, D and E. Each section is compulsory.
- Section – A carries 20 marks weightage, Section – B carries 10 marks weightage, Section – C carries 18 marks weightage, Section – D carries 20 marks weightage and Section – E carries 3 case-based with total weightage of 12 marks.
- Section A: It comprises of 20 MCQs of 1 mark each.
- Section B: It comprises of 5 VSA type questions of 2 marks each.
- Section C: It comprises of 6 SA type of questions of 3 marks each.
- Section D: It comprises of 4 LA type of questions of 5 marks each.
- Section E: It has 3 case studies. Each case study comprises of 3 case-based questions, where 2 VSA type questions are of 1 mark each and 1 SA type question is of 2 marks. Internal choice is provided in 2 marks question in each case-study.
- 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 of the alternatives in all such questions.
Section – A [20 Marks]
(All questions are compulsory. No internal choice is provided in this section)
Question 1.
In 1 km race, A beats B by 18 m or 9 sec. What is the A’s time over the course?
(a) 481 sec
(b) 491 sec
(c) 502 sec
(d) 512 sec [1]
Answer:
(b) 491 sec
Explanation: B covers 18 m in 9 sec. So, B’s time over the course = \(\left(\frac{9}{18}\right)\) × 100 = 500 sec.
Whereas A’s time over the course = 500 – 9
= 491 sec.
Question 2.
The speed of a boat in still water is 8 km/h and the speed of the stream is 3 km/h. what are the upstream and downstream speeds of the boat?
(a) 11 km/h, 5km/h
(b) 6 km/h, 11 km/h
(c) 5 km/h, 11 km/h
(d) 4 km/h, 10 km/h [1]
Answer:
(c) 5 km/h, 11 km/h
Question 3.
If two pipes function together, the tank is filled in 12 hours. One pipe atone fills the tank in 10 hours faster than the other. How many hours does the faster pipe take to fill up the tank?
(a) 16 hours
(b) 18 hours
(c) 20 hours
(d) 22 hours
Answer:
(c) 20 hours
Explanation: Let the faster pipe takes x hr to fill the tank one hour = \(\frac{1}{x}\)
Given,
The slower pipe takes 10 hr more that the faster pipe = (x + 10) hr
filled by slower pipe = \(\frac{1}{x+10}\)
Now, some tank can be filled = 12 hr
1 hr = \(\frac{1}{12}\) hr
A.T.Q., \(\frac{1}{x}\) + \(\frac{1}{x+10}\) = \(\frac{1}{12}\)
\(\frac{x+x+10}{x(x+10)}\) = \(\frac{1}{12}\)
\(\frac{2 x+10}{x(x+10)}\) = \(\frac{1}{12}\)
⇒ 12(2x + 10) = x2 + 10x
x2 – 14x – 120 = 0
x2 – 14x + 6x – 120 = 0
x(x – 20) + 6 (x – 20) = 0
(x – 20) (x + 6) = 0
x = 20
Question 4.
The solution set of the inequality \(\frac{x}{2}\) + \(\frac{x}{3}\) > -10 is:
(a) x > -12
(b) -12 < x < 12 (c) x ≥ 12 (d) x ≤ -12 [1] Answer: (a) x > -12
Explanation:The given inequality \(\frac{x}{2}\) + \(\frac{x}{3}\) > -10 simplifies to \(\frac{5 x}{6}\) > -10 , simplifies to 5x > -60, or x > -12.
Question 5.
The values of x which satisfy both the linear equations simultaneously, are:
4x – 1 < 0, 3 – 4x < 0 (a) \(\left(\frac{1}{4}, \frac{3}{4}\right)\) (b) \(\left(-\frac{1}{4}, \frac{3}{4}\right)\) (c) \(\left(\frac{3}{4}, \frac{1}{4}\right)\) (d) no value [1] Answer: (d) no value Explanation: The given inequalities gives values of x as x ≤ \(\frac{1}{4}\) and x > \(\frac{3}{4}\). Since no value of x satisfies both x ≤ \(\frac{1}{4}\) anc| x > \(\frac{3}{4}\) there is no value of x.
Question 6.
If x > y and z < 0, Then, (a) xz > yz
(b) \(\frac{x}{z}\) > \(\frac{x}{z}\)
(c) xz ≥ yz
(d) no value [1]
Answer:
(d) \(\frac{x}{z}\) < \(\frac{y}{z}\) Explanation: x > y and z < 0 implies \(\frac{x}{z}\) < \(\frac{y}{z}\), because dividing an inequality by a negative real number, change the sign of the inequality.
Question 7.
The number of all possible matrices of order 2 × 3 with entry 0 or 1 is:
(a) 10
(b) 12
(c) 64
(d) 512 [1]
Answer:
(c) 64
Explanation: Total number of elements in a matrix of order 2 × 3 = 6
Number of choices for each element = 2
So, number of all possible matrices of order 2 × 3 = 26, i.e., 64.
Question 8.
If A is a 3 × 4 matrix and B is a matrix such that ATB and BAT are both defined, then the order of matrix B is:
(a) 3 × 4
(b) 3 × 3
(c) 4 × 4
(d) 4 × 3 [1]
Answer:
(a) 3 × 4
Explanation: Let the order of matrix B be m × n.
Here the order of AT is 4 × 3
Since ATB and BAT are both defined, m = 3 and n = 4
Question 9.
If AB = A and BA = B, then (B2 + B) equals:
(a) 2A
(b) O
(c) 21
(d) 2B [1]
Answer:
(d) 2B
Explanation: B2 = BB = (BA)B = B(AB) = BA = B. So, B2 + B = 2B.
Question 10.
If A = [2 -3 4], B = \(\left[\begin{array}{l}
3 \\
2 \\
2
\end{array}\right]\), X = [1 2 3] and Y = \(\left[\begin{array}{l}
2 \\
3 \\
4
\end{array}\right]\), then AB + XY is
(a) [20]
(b) [22]
(c) [24]
(d) [28]
Answer:
(d) [28]
Explanation:
Question 11.
If y = e-2x, then its third derivative y3 \(\left(\text { or } \frac{d^3 y}{d x^3}\right)\) is
(a) 2e-2x
(b) e-4x
(c) 4e-4x
(d) -8e-2x [1]
Answer:
(d) -8e-2x
Explanation Given y = e-2x, we have
\(\frac{d y}{d x}\) = -2e-2x; \(\frac{d^2 y}{d x^2}\) = (-2)(-2)e-2x
\(\frac{d^3 y}{d x^3}\) = (-2)(-2)(-2)e-2x = -8e-2x
Thus, (d) is the correct option.
Question 12.
\(\int \frac{x^2+1}{x^2-1}\) = log k, then the value of k is
(a) 3
(b) 9
(c) \(\frac{9}{2}\)
(d) \(\frac{3}{2}\) [1]
Answer:
(b) 9
Explanation: Here,
\(\int_0^{40} \frac{d x}{2 x+1}\) = log k
⇒ \(\int_0^{40} \frac{d x}{2 x+1}\) = log k
⇒ \(\left[\frac{1}{2} \log (2 x+1)\right]_0^{40}\) = log k
⇒ \(\frac{1}{2}\)log 81 = log k
⇒ k = 9
Question 13.
An observed set of a population that has been selected for analysis is catted
(A) a process
(b) a forecast
(c) a parameter
(d) a sample [1]
Answer:
(d) a sample
Question 14.
If we reject the null hypothesis when it is true, we might be making
(a) Type – I error
(b) Type — III error
(c) a correct decision
(d) a wrong decision [1]
Answer:
(a) Type – I error
Question 15.
Laspeyre’s index formula uses the weights of the:
(a) ease year
(b) current year
(c) average of the weights of number of years
(d) none of these [1]
Answer:
(a) base year
Question 16.
At what rate of Interest will the present value of a perpetuity of ₹ 500 payable at the end of every 6 months be ₹ 10,000?
(a) 6%
(b) 8%
(c) 10%
(d) 12% [1]
Answer:
(c) 10%
Explanation: Here,
i = \(\frac{r}{200}\) : R = ₹ 500 and P = ₹ 10,000
Using P = \(\frac{\mathrm{R}}{i}\), we get r = 10
so, (c) is the correct option
Question 17.
Under a settlement of property, Mr. Mahesh is entitled to receive ₹ 2400 per annum forever, the first being due at the end of first year. The present value of Mr. Mahesh’s right at 6% compounded annually is
(a) ₹ 24,000
(b) ₹ 34,000
(c) ₹ 40,000
(d) ₹ 48,000
Answer:
(c) ₹ 40,000
Explanation: Here, i = 0.06; R = ₹ 2400
Using P = \(\frac{\mathrm{R}}{i}\), we get P = 40,000
(c) is the correct option
Question 18.
The solution set of the inequality 3x + 5y < 4 is
(a) open half plane not containing the origin
(b) whole xy – plane except the points lying on the line 3x + 5y = 4
(c) open half plane containing the origin
(d) none of these [1]
Answer:
(c) open half plane containing the origin
Assertion-Reason Questions
Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
(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 degree of the given differential equation
\(\left(\frac{d^2 y}{d x^2}\right)^2\) + \(\left(\frac{d y}{d x}\right)^2\) + sin \(\left(\frac{d y}{d x}\right)\) + 1 = 0 is 3
Reason (R) : The highest order derivative involved in a differential equation, when it is written a polynomial in derivatives, is called its degree. [1]
Answer:
(d) (A) is false but (R) is true.
Explanation: Given differentiate equations is not a polynomial equation i.e., its derivatives. So, its degree is not defined.
Question 20.
Assertion (A): If the demand function for a commodity is p = 20 – 2x – x2 and the market demand is 3 units, then the consumer’s surplus is 27.
Reason (R): Consumer’s surplus is the gain made by the consumer when they purchase some commodity from the market at a price lower than what they would have been willing to pay for it.
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation: Here,
x0 = 3
This given p0 = 3
So, C.S. = \(\int_0^3\left(20-2 x-x^2\right) d x-p_0\)
x0 = \(\left[20 x-x^2-\frac{x^3}{3}\right]_0^3\) – 5 × 3
= 27
Section – B [10 Marks]
(All questions are compulsory. In case of internal choice, attempt any one question only)
Question 21.
If the cash equivalent to a perpetuity ₹ x payable at the end of every six months is ₹ 20,000, if money is 5% p.a. compounded half-yearly. Find the value of x.
OR
Find the EMI of a loan of ₹10,00,000 for 15 years at 11% per annum
[Given (1.0092)180 = 5.19876] [2]
Answer:
We know that the present value of perpetuity of ₹ R payable at the end of every period forever at a rate of i per period is given by \(\frac{\mathrm{R}}{i}\),
ie., P∞ = \(\frac{R}{i}\)
Here, R = x, i per period = \(\frac{5}{100}\) = 0.025 P∞ = 20,000
So, P∞ = \(\frac{R}{i}\)
⇒ 20,000 = \(\frac{x}{0.025}\)
⇒ x = 500
So, the value of x is ₹500
OR
When P denotes the Loan amount r the interest rate (monthly) and n the number of payments, then EMI = \(\frac{\mathrm{P} \times r \times(1+r)^n}{(1+r)^n-1}\)
Here, P = ₹10,00,000; r = 0.0092 and n = 15 × 12 = 180 months
So, EMI = ₹ 10,00,000 × 0.0092 × \(\frac{5.19876}{4.19876}\)
= 11391.12
Thus, the EMI is ₹ 11391.12
Question 22.
If X is a random variable and a, b are real numbers, then prove that
E(ax + b) = aE(X) + b [2]
OR
A salesman wants to know the average number of units he sells per sales call. He checks his past sales records and comes up with the following probabilities:
| Sales(in Units) | Probability |
| 0 | 0.15 |
| 1 | 0.20 |
| 2 | 0.10 |
| 3 | 0.05 |
| 4 | 0.30 |
| 5 | 0.20 |
Answer:
E(X) = Σxipi gives
E(aX + b) = Σ(axi + b)pi = Σxipi gives
E(aX + b) = Σxipi + b {∵ Σpi = 1}
= E(X) + b
OR
E(X) = Σxipi
= 0 × 0.15) + (1 × 0.20) + (2 × 0.10) + (3 × 0.05) + (4 × 0.30) + (5 × 0.20)
= 0 + 0.20 + 0.20 + 0.15 + 1.2 + 1 = 2.75
Thus, the expected value of the number of units he would sell per sale call is 2.75.
Question 23.
Using integration, find the area of the region bounded by the curve x = y2 and the line y = 16. [2]
Answer:
y = x2 gives x = \(\sqrt{y}\)
Question 24.
Using Cramer’s rule, solve the system of equations :
{3x – y = 2
2x + y = 11} [2]
Answer:
Here, D = \(\left|\begin{array}{cc}
3 & -1 \\
2 & 1
\end{array}\right|\) = 5 ;
Dx = \(\left|\begin{array}{cc}
2 & -1 \\
11 & 1
\end{array}\right|\) = 13;
Dy = \(\left|\begin{array}{cc}
3 & 2 \\
2 & 11
\end{array}\right|\) = 29
Since D = 5 ≠ 0, the system of equations has a unique solution given by
x = \(\frac{D_x}{D}\) = \(\frac{13}{5}\),
y = \(\frac{D_y}{D}\) = \(\frac{29}{5}\)
Hence,
x = \(\frac{13}{5}\), y = \(\frac{29}{5}\)
Question 25.
A boat goes 48 km downstream in 20 hours. It takes 4 hours more to cover the same distance against the stream. Find the speed of the boat in the still water. [2]
Answer:
Let the speed of the boat in stilt water be x km/h; and the speed of the stream be y km/h. Here.
Speed of the downstream = x + y = \(\frac{48}{5}\) = 2.4
Speed of the upstream = x – y = \(\frac{48}{20}\) = 2
x = 2.2
Thus, the speed of the boat in still water be 2.2 km/h
Section – C [18 marks]
(All questions are compulsory. In case of internal choice, attempt any one question only)
Question 26.
If b > a > 0 and c > 0 then show that \(\frac{a+c}{b+c}\) > \(\frac{a}{b}\) [3]
Answer:
Consider \(\frac{a+c}{b+c}\) – \(\frac{a}{b}\)
= \(\frac{b(a+c)-a(b+c)}{b(b+c)}\)
= \(\frac{b a+b c-a b-a c}{b(b+c)}\)
= \(\frac{b c-a c}{b(b+c)}\) = \(\frac{(b-a) c}{b(b+c)}\)
= \(\left[\frac{c}{b(b+c)}\right]\)(b – a)
Since b – a > 0 and \(\frac{c}{b(b+c)}\) > 0, \(\left[\frac{c}{b(b+c)}\right]\)
(b – a) > 0
\(\frac{a+c}{b+c}\) – \(\frac{a}{b}\) > 0
or \(\frac{a+c}{b+c}\) > \(\frac{a}{b}\)
Question 27.
A random variable X follows Poisson distribution with parameter 4. Find the probabilities that the variable assumes the values
(A) 0, 1, 2, 3, 4;
(B) less than 4;
(C) at least 3 {Given e-4 = 0.0183}
OR
A business firm receives on an average 1.8 telephone calls per day during the period 10.00 a.m. – 10.05 a.m. Find the probability that on a particular day, the firm receives
(A) no call;
(B) exactly 3 calls during the same period
{Given e-1.8 = 0.1653} [3]
Answer:
Here, λ = 4. So,
(A) P(X = 0) = \(\frac{4^0}{0 !} \cdot e^{-4}\) = e-4 = 0.0183
P(X = 1) = \(\frac{4}{1}\) P(X = 0)
= 4 × 0.0183 = 0.0732
P(X = 2) = \(\frac{4}{2}\)P(X = 1)
= 2 × 0.0732 = 0.1464
P(X = 3) = \(\frac{4}{3}\)P(X = 2)
= \(\frac{4}{3}\) × 0.1464 = 0.1952
P(X = 4) = \(\frac{4}{4}\)P(X=3)
= 1 × 0.1952 = 0.1952
(B) P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.0183 + 0.0732 + 0.1465 + 0.1952
= 4.331.
(C) P(X ≥ 3) = 1 – {P(X = 0) + P(X = 1) + P(X = 2)}
= 1 – {0.0183 + 0.0732 + 0.1464}
= 1 – 0.2379. i.e., 0.7621
OR
Here, λ = 1.8. So,
(A) P(X = 0) = \(\frac{(1.8)^0}{0 !}\) .e-(1.8)
= e-(1.8) = 0.1653
(B) P(X = 3) = \(\frac{(1.8)^3}{3 !}\) .e-(1.8)
= 0.1653 × e-(1.8)
0.972 × 0.1653 = 0.160672.
Question 28.
In a certain factory turning out razor blades, there is a small chance of 0.002 for any blade to be defective. The blades are supplied in packets of 10. Calculate the number of packets containing no defective, one defective and two defective blades in a consignment of 10,000 packets.
{Given e-0.02 = 0.9802} [3]
Answer:
Here, p’ = 0.002. n = 10. So, np = λ = 0.02
Probability of having no defective blade
P(X = 0) = \(\frac{(0.2)^0}{0 !}\) .e-(0.2)
= e-(0.2) = 0.9802
⇒ Approximate number of packets having no defective blade in the consignment of 10,000 packets = 0.9802 × 10,000 = 9802
Probability of having one defective blade
P(X = 1) = \(\frac{0.02}{1}\).P(X = 0)
= 0.02 × 0.9802 = 0.019604
⇒ Approximate number of packets having one defective blade in the consignment of 10,000 packets = 0.019604 × 10,000 = 196
Probability of having two defective blade
P(X = 2) = \(\frac{0.02}{2}\). P(X = 1)
= 0.01 × 0.019604 = 0.000196
Approximate number of packets having two defective blades in the consignment of 10.000 pockets = 0.000196 × 10.000 = 1.96 or 2.
Question 29.
A plans to buy a new flat after 5 years, which will cost him 780,00,000. How much money he deposit annually to accumulate this much amount, if he gets interest of 5% compounded continuously? [3]
Answer:
Here, S = 80.00,000. A = ?, r = 0.05 and n = 5
So, A = \(\frac{80,00,000 \times 0.05}{e^{0.25}-1}\) gives
= \(\frac{4,00,000}{1.2840254153-1}\)
= \(\frac{4,00,000}{0.2840254153}\)
= 1408324.70
Thus, he should deposit ₹ 14,08,324.70 annually.
Question 30.
A machine costs the company 71,00,000 and its effective life is estimated to be 12 years. If the scrap realises ₹ 5000 only, what amount should be retained out of profits at the end of each year to accumulate at a compound interest of 5% per annum? [Given (1.05)12 = 1.797) [3]
Answer:
If ₹A be the sum of money set aside periodically, S is the amount accumulated after n periods and r is the rate of interest per period, then S is given by
S = \(\frac{\mathrm{A}}{r}\)[(1 + r)n – 1]
Thus periodic payment is given by A = \(\frac{\mathrm{Sr}}{(1+r)^n-1}\)
Here, S = 1,00,000 – 5000 = 95,000; r = 0.05
and n = 12
So, A = \(\frac{95,000 \times 0.05}{(1.05)^{12}-1}\) = \(\frac{47500}{1.797-1}\) = 5959.85
Thus, an amount of ₹5959.85 should be retained out of profits at the end of each year.
Question 31.
Solve the following LPP graphically using corner point method:
Minimise Z = 2x + 3y subject to the constraints
x + 2y ≤ 10; x + 2y ≥ 1
x ≥ 0, y ≥ 0.
OR
Solve the following LPP graphically using corner point method:
Maximise Z = 6x + 5y subject to the constraints
3x + 5y ≤ 15; 5x + 2y ≤ 10
x ≥ 0, y ≥ 0. [3]
Answer:
Here, the graph of the given cônstraints and the feasible region, so obtained, is shaded in the graph given below.
| Corner point | Corresponding value of Z |
| A(1, 0) | 2(1) + 3(0) = 2 |
| B(10, 0) | 2(10) + 3(0) = 20 |
| C(0, 5) | 2(0) + 3(5) = 15 |
| D(0, \( \frac{1}{2} \)) | 2(0) + 3(\( \frac{1}{2} \)) = \( \frac{3}{2} \) |
Thus, minimum value of Z is \(\frac{3}{2}\) which occurs at D.
(0, \(\frac{1}{2}\))
OR
Here, the graph of the given constraints and the feasible region, so obtained, is shaded in the graph given below
| Corner point | Corresponding value of Z |
| O(0, 0) | 6(0) + 5(0) = 0 |
| A(2, 0) | 6(2) + 5(0) = 0 |
| B(\( \frac{20}{19} \), \( \frac{45}{19} \)) | 6(\( \frac{20}{19} \)) + 5(\( \frac{45}{19} \)) = \( \frac{345}{19} \)) = 18\( \frac{3}{19} \)) |
| C(0, 3) | 6(0) + 5(3) = 15 |
Section – D
(All questions are compulsory. In case of internal choice, attempt any one question only)
Question 32.
A steel plant is capable of producing V tonnes per day of a low grade steel and ‘y’ tonnes per day of a high grade steel, where y = \(\frac{42-5 x}{10-x}\). If the fixed market price of low grade is half of high grade steel, show that 6 tonnes of low grade steel are produced per day for the maximum total revenue.
OR
In a college hostel accommodating 1000 students, one of them came in carrying Covid-19 virus, then the hostel was isolated. If the rate at which the virus spreads is assumed to be proportional not only to the number N of infected students, but also to the number of non- infected students, and if the number of infected students is 50 after 4 days, then show that more than 95% of the students will be infected after 10 days. [5]
Answer:
Let the fixed market price of low grade steel be ₹ p per tonne. Then, the fixed price of high grade steel is ₹ 2p per tonne.
Let R be the total revenue. Then, R = px + 2py
⇒ R = px + 2p\(\left(\frac{42-5 x}{10-x}\right)\)
∴ \(\frac{d R}{d x}\) = p + 2p\(\left[\frac{-5(10-x)-(42-5 x)(-1)}{(10-x)^2}\right]\)
= p – \(\frac{16 p}{(10-x)^2}\) For R to be maximum, \(\frac{d R}{d x}\) = 0
⇒ (10 – x)2 = 16
⇒ 10 – x = ±4
⇒ x = 6 or 14
Now, \(\frac{d^2 R}{d x^2}\) = –\(\frac{32 p}{(10-x)^3}\)
At x = 6, \(\frac{d^2 R}{d x^2}\) = –\(\frac{32 p}{(10-x)^3}\) < 0 So, R is maximum when x = 6 At x = 14, \(\frac{d^2 R}{d x^2}\) = –\(\frac{32 p}{(10-14)^3}\) > 0. So, R is minimum when x = 14
Hence, revenue is maximum when 6 tonnes of low grade steel is produced per day.
OR
Let N denote the number of infected students at any time t. Then,
\(\frac{d N}{d t}\) = kN(1000 – N).
where k is constant of proportionality.
⇒ \(\frac{d N}{N(1000 \quad N)}\) = k dt
Integrating both sides,
we have \(\int \frac{d N}{N(1000 N)}\) = \(\int k d t\)
⇒ 20046 (approx.)
Solving, we get N = 952 (approx..)
Hence, more than 95% of the students will be infected after 10 days.
Question 33.
An unbiased die is thrown again and again until three sixes are obtained. Find the probability of obtaining third six in the sixth throw of a die.
OR
An aptitude test for selecting officers in a bank is conducted on 1000 candidates. The mean score is 42 and the standard deviation of score is 24. Assuming normal distribution for the scores, find :
(A) The number of candidates whose scores exceed 60,
(B) The number of candidates whose scores lie between 30 and 60. [5]
Answer:
Let A be the event of obtaining two sixes in the first five throws of a die and B be the event of obtaining a six in the sixth throw of a die.
Then, the required probability = P(AB) = P(A)P(B)
Now, P(B) = \(\frac{1}{6}\)
Now we shall find P(A) using binomial distribution.
Let p denote the probability of success, i. e.. getting a six in a single throw of a die.
Then, p = \(\frac{1}{6}\) and q = \(\frac{5}{6}\)
Let X denote the number of successes in 5 throws of a die. Then, X is a binomial variate with parameters n = 5 and p = \(\frac{1}{6}\).
Probability of getting r successes is given by P(X = r) = nCrprqn-r
So, probability of getting 2 successes is given by P(X = 2)
Now P(X = 2) = 5C2\(\left(\frac{1}{6}\right)^2\left(\frac{1}{6}\right)^3\) = \(\frac{625}{3888}\)
Hence, P(A) = \(\frac{625}{3888}\)
Therefore, the required probabiLity = P(A) P(B) = \(\frac{625}{3888}\) × \(\frac{1}{6}\) = \(\frac{625}{3888}\)
OR
Here, Z = \(\frac{X-42}{24}\)
X = 35 gives
Z = \(\frac{35-\mu}{\sigma}\) = z2 (say);
X = 63 gives
Z = \(\frac{63-\mu}{\sigma}\) = z2 (say);
Here. P(X < 35) = P(Z < z1)
= 0.07
P( X < 63) = P(Z < z2) = 0.89
P(Z < 0) = P(z1 < Z < 0)
= 0.070.5 + P(0 < Z < z2)
= 0.89
{P( Z < z,sub>1) is less than 0.5, z1 < 0}
{P( Z < z2) is greater than 0.5, z2 > 0]
0.5 – P(0 < Z < |z1| = 0.07
P(o < z < z2) = 0.39
P( 0 < Z < |z1|) = 0.43
z2 = 1.23
Question 34.
Compute the trend for the following data using method of least squares. Find out an estimate of the year 2014.
|
Year |
2005 | 2006 | 2007 | 2008 | 2009 | 2010 |
2011 |
| Value | 80 | 90 | 92 | 83 | 94 | 99 |
106 |
Answer:
Here, n = 7(odd).
Let the equation of the straight line of best fit, with the origin at the middle year 2008 and units of x as 1 year, be y = a + bx
By the method of least squares, the values of ‘a and ‘b are given by
a = \(\frac{\Sigma y}{n}\) and b = \(\frac{\Sigma x y}{\Sigma x^2}\) … (A)
Calculation for fitting the line of best fit
Hence, the required equation of the best fitted straight line is y = 92 + 3.5x
| Year | x | Trend values (y = 92 + 3.5x) |
| 2005 | -3 | 92 + 3.5(-3) = 81.5 |
| 2006 | -2 | 92 + 3.5(-2) = 85 |
| 2007 | -1 | 92 + 3.5(-1) = 88.5 |
| 2008 | 0 | 92 + 3.5(0) = 92 |
| 2009 | 1 | 92 + 3.5(1) = 95.5 |
| 2010 | 2 | 92 + 3.5(2) = 99 |
| 2011 | 3 | 92 + 3.5(3) = 102.5 |
Thus, the trend values are 81.5. 85, 88.5, 92, 95.5. 99 and 102.
Estimate fór the year 2014 =92 + 3.5 (6) = 113
Question 35.
Mrs. Sunita is considering ₹ 1000 par value bond bearing a coupon rate of 11% that matures after 5 years. She wants a minimum rate of return of 15%. The bond is currently sold at ₹ 870. Should she buy the bond? (Given (1.15)-5 = 0.497). [5]
Answer:
Here,
F = Face value of the bond = ₹ 1000
n = number of periodic dividend payments = 5
i = Annual yield rate = 0.15
R = F × id = ₹ (1000 × 0.11) = ₹ 110
Since the bond is to be redeemed at par,
C = Redemption price or Maturity value = Face value = ₹ 1000
Let V be the purchase value of the bond. Then,
V = \(R\left[\frac{1-(1+i)^{-n}}{i}\right]\) + F(1 + i)-n
= ₹ {110\(\left[\frac{1-(1.15)^{-5}}{0.15}\right]\)1000(1.15)-5}
= ₹{733[1 – (1.15)-5 + 1000(1.15)-5)}
= ₹{733 [1 – 0.497) + 1000(0.497)}
= ₹{368.70 + 497} or ₹865.70
Since the present value of the bond is ₹ 865.70, it is not advisable to buy it at 870.
Section – E [12 marks]
(All questions are compulsory. In case of internal choice, attempt any one question only)
Question 36.
On her birthday, Radha decided to donate some money to children of orphanage home.
If there are 8 children less, everyone will get ₹ 10 more. However, if there are 16 children more, everyone will get ₹ 10 less. Let the number of children be x and amount distributed by Radha to each child be ‘y’.
Based on the given information, answer the following question:
(A) What is the system of equate one is formed for the given condition? [1]
(B) Write the system of equations in matrix form? [1]
(C) What are the values of x and y ?
OR
What is the inverse of the matrix \(\left[\begin{array}{cc}
-5 & 8 \\
5 & -4
\end{array}\right]\) ? [2]
Answer:
(A) Total amount distributed = xy
A.T.Q.
(x – 8)(y + 10) = xy
⇒ xy – 8y + 10x – 80 = xy
⇒ 5x – 4y = 40
and (x + 16)(y – 10) = xy
⇒ xy + 16y – 10x – 160 = xy
⇒ 16y – 10x = 160
⇒ 8y – 5x = 80 …..(ii)
Equation (i) and (ii) are the required equations.
(B) The system of equations in matrix form is
\(\left[\begin{array}{cc}
5 & -4 \\
-5 & 8
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
40 \\
80
\end{array}\right]\)
[As the equations obtained in Part (A)]
(C) The equations firmed in part (A) is
5x – 4y = 40
and -5x + 8y = 80
The equation in matrix form is:
∴ x = 32, y = 30.
OR
Question 37.
One very useful application of linear programming is using a graphical method for solving problems in two variable. Mrs. Meena wanted to use that concept to help students figure out how the area of the 3D-model is composed of 3 staight lines.
On the below diagram (cross section of 3D model), 0 is the origin. The shaded region R is defined by three inequalities one of the three inequalities is x + y ≤ 6.
Based on the given information, answer the following question:
(A) Given that the point (x, y) is in the region R, then what is the maximum value of x + 2y? [1]
(B) Does the (3, 2.5) lies inside the region R? [1]
(C) What is the area of the region R?
OR
What are the two other inequalities on the graph? [2]
Answer:
(A) Corner points of region R are – (4, 2), (2, 1) and (2, 4)
Z = x + 2y
Z(4, 2) = 4 + 2 × 2 = 4 + 4 = 8
Z (2, 1) = 2 + 2 × 1 = 4
Z(2, 4) = 2 + 2 × 4 = 10
Hence, maximum value of x + 2y is 10.
(B) Yes, b the given graph it could be clearly seem that point (3, 2.5) lies inside the region R.
(C) Corner points of R are (4, 2), (2, 1) and (2, 4).
Area of triangle with the coordinates (4, 2), (2, 1) and (2, 4) is given as
A = \(\frac{1}{2}\) |[4(1 – 4) + 2(4 – 2) + 2(2 – 1)]|
= \(\frac{1}{2}\) |[-12 + 4 + 2]|
= \(\frac{1}{2}\) × 6 = 3 sq. units
OR
The line which is parallel to y-axis and touches the x-axis at point 2.
∴ Equation of this line is x = 2.
∴ Required inequality is x ≥ 2.
For the second line, the x-coordinate is double the y-coordinate
∴ Equation of this line is x = 2y or x – 2y = o
By the feasible region the inequality is x – 2y ≤ 0 or x ≤ 2y.
Question 38.
A wire of length 34 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a rectangle whose length is twice its breadth.
Another pieces of wire of the same length is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. Assume that the length of the piece which is made into a square is ‘x’ m.
Based on the given information, answer the following question:
(A) What is the combined area of the rectangle and the square? [1]
(B) What is the combined area of the circle and the square ? [1]
(C) For what value of x, the combined area of square and the rectangle is minimum?
OR
For what values of x, the combined area of the square and the circle to be minimum? [2]
Answer:
(A) Perimeter of square = x
Side of square = \(\frac{x}{4}\)
Area of square = \(\frac{x^2}{16}\)
Perimeter of rectangle = (34 – x) and length is twice its breadth
Let, breadth = 6
Then, 2(b + 2b) = 34 – x
b = \(\left(\frac{34-x}{6}\right)\)
Area of rectangle = b × 2b = b2
= 2\(\left(\frac{34-x}{6}\right)^2\)
∴ combined area = \(\left(\frac{x}{4}\right)^2\) + 2\(\left(\frac{34-x}{6}\right)^2\)
(B) Area of square = \(\left(\frac{x}{4}\right)^2\)
Circumference of a circle = (34 – x)
Let radius = r
Then, 2πr = 34 – x
r = \(\frac{34-x}{2 \pi}\)
Area of circle = πr²
(C)
⇒ f'(x) = \(\frac{17}{72}\) > 0
‘So at x = 16 area is minimum.
OR
Combined area is:
