11th public Business Mathematics & Statistics question paper 2019

Time: 2.30 Hours
Maximum Marks: 90

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SECTION – I

Note:

1. Answer all the questions.
2. Choose the most appropriate answer from the four alternatives and write the option code and corresponding answer.

Q1. If

$$\frac{x}{8}-\frac{5}{6}=0$$

then the value of $x$ is:

A. $-\frac{16}{5}$
B. $\frac{16}{5}$
C. $-\frac{5}{6}$
D. $\frac{5}{6}$

Answer: B. 165\frac{16}{5}

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Q2. If A is a matrix of order n, then $|Adj A|$ is:

A. $|A|^{n-1}$
B. $|A|^{n+1}$
C. $|A|^n$
D. $|A|^{2n}$

Answer: A. ∣A∣n−1|A|^{n-1}

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Q3. Number of chords that can be drawn through 48 points on a circle is:

A. 47
B. 210
C. 1128
D. 24

Answer: C. 1128

Solution:$$^{48}C_2=\frac{48\times47}{2}=1128$$

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Q4. The number of diagonals in a polygon of n sides is equal to:

A. ${}^{n}C_2-n$
B. ${}^{n}C_2-1$
C. ${}^{n}C_2$
D. ${}^{n}C_2-2$

Answer: A. nC2−n{}^{n}C_2-n

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Q5. The centre of the circle

$$x^2+y^2-2x+2y-9=0$$

is:

A. $(-1,1)$
B. $(1,-1)$
C. $(1,1)$
D. $(-1,-1)$

Answer: B. (1,−1)(1,-1)

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Q6. The double ordinate passing through the focus is:

A. Directrix
B. Axis
C. Focal chord
D. Latus rectum

Answer: D. Latus rectum

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Q7. The value of $\cot300^\circ$ is:

A. $\frac{1}{\sqrt3}$
B. $-\frac{1}{\sqrt3}$
C. $\sqrt3$
D. $-\sqrt3$

Answer: D. −3-\sqrt3

Solution:$$\cot300^\circ=\cot(360^\circ-60^\circ)$$$$=-\cot60^\circ=-\frac1{\sqrt3}$$

Note: The printed answer key in the PDF indicates option D, although the displayed option/formula extraction is imperfect.

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Q8. If

$$\tan A=\frac12,\qquad \tan B=\frac13$$

then $\tan(2A+B)$ is:

A. 3
B. 4
C. 1
D. 2

Answer: A. 3

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Q9. The minimum value of

$$f(x)=|x|$$

is:

A. +1
B. $-\infty$
C. 0
D. -1

Answer: C. 0

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Q10. If $y=\log x$, then $\frac{dy}{dx}$ is:

A. $\frac1x$
B. $x$
C. $\frac{x}{x^2-1}$
D. $\frac{x-1}{x^2}$

Answer: A. 1x\frac1x

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Q11. Relationship among MR, AR and $\eta_d$ is:

A. $MR=AR=\eta_d$

B.$$AR=\frac{MR}{\eta_d}$$

C.$$\eta_d=\frac{AR}{AR-MR}$$

D. $\eta_d=AR-MR$

Answer: C.$$\eta_d=\frac{AR}{AR-MR}$$

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Q12. The demand function is always:

A. Non-decreasing function
B. Undefined function
C. Increasing function
D. Decreasing function

Answer: D. Decreasing function

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Q13. The rate of income on 7% stock at ₹80 is:

A. 8%
B. 7%
C. 9%
D. 8.75%

Answer: D. 8.75%

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Q14. An annuity in which payments are made at the beginning of each payment period is called:

A. Perpetual annuity
B. Annuity due
C. Immediate annuity
D. All the above

Answer: B. Annuity due

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Q15. The best measure of central tendency is:

A. Harmonic mean
B. Mean
C. Arithmetic mean
D. Geometric mean

Answer: C. Arithmetic mean

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Q16. Probability of drawing a diamond card and an ace card, in that order, from a pack in two consecutive draws, without replacement:

A. $\frac1{51}$
B. $\frac1{52}$
C. 51
D. 52

Answer: B. 152\frac1{52}

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Q17. Correlation coefficient lies between:

A. -1 to 0
B. -1 to ∞
C. 0 to ∞
D. -1 to +1

Answer: D. -1 to +1

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Q18. The variable whose value is influenced or is to be predicted is called:

A. Regressor
B. Explanatory variable
C. Dependent variable
D. Independent variable

Answer: C. Dependent variable

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Q19. A solution which maximizes or minimizes the given LPP is called:

A. An optimal solution
B. Non-feasible solution
C. A solution
D. A feasible solution

Answer: A. An optimal solution

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Q20. The maximum value of

$$z=3x+5y$$

subject to$$x\ge0,\quad y\ge0,\quad 2x+5y\le10$$

is:

A. 25
B. 31
C. 6
D. 15

Answer: D. 15

Solution:

Constraint:$$2x+5y=10$$

Corner points:$$O(0,0),\quad A(5,0),\quad B(0,2)$$

Corner point	$z=3x+5y$
$O(0,0)$	0
$A(5,0)$	15
$B(0,2)$	10

Therefore,$$\boxed{z_{\max}=15}$$

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SECTION – II

Answer any 7 of the following. Question No. 30 is compulsory.

7 × 2 = 14

Q21. If

$${}^nC_4=495$$

find $n$.

Answer:$$n=12$$

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Q22. Find the length of tangent to the circle

$$x^2+y^2-2x+4y+9=0$$

from the point $(1,2)$.

Answer:$$\boxed{2\sqrt5\text{ units}}$$

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Q23. Prove that:

$$\cos510^\circ\cos330^\circ+ \sin390^\circ\cos120^\circ=-1$$

Solution:$$\cos510^\circ =\cos(360^\circ+150^\circ) =-\cos30^\circ =-\frac{\sqrt3}{2}$$$$\cos330^\circ=\cos30^\circ=\frac{\sqrt3}{2}$$$$\sin390^\circ=\sin30^\circ=\frac12$$$$\cos120^\circ=-\sin30^\circ=-\frac12$$

Therefore,$$LHS= \left(-\frac{\sqrt3}{2}\right) \left(\frac{\sqrt3}{2}\right) + \left(\frac12\right) \left(-\frac12\right)$$$$=-\frac34-\frac14$$$$=-1=RHS$$

Hence proved.

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Q24. If

$$f(x)=\log\left(\frac{1+x}{1-x}\right),\quad0<x<1$$

show that$$f\left(\frac{2x}{1+x^2}\right)=2f(x)$$

Solution:$$f\left(\frac{2x}{1+x^2}\right) = \log \left[ \frac{1+\frac{2x}{1+x^2}} {1-\frac{2x}{1+x^2}} \right]$$$$= \log \left[ \frac{(1+x)^2}{(1-x)^2} \right]$$$$=2\log\left(\frac{1+x}{1-x}\right)$$$$\boxed{=2f(x)}$$

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Q25. If the demand law is given by

$$P=10e^{-x/2}$$

find the elasticity of demand.

Solution:

Differentiating:$$\frac{dp}{dx} =-5e^{-x/2}$$

Therefore,$$\frac{dx}{dp} = -\frac1{5e^{-x/2}}$$

Elasticity of demand:$$\eta_d=-\frac px\frac{dx}{dp}$$

Substituting:$$\boxed{\eta_d=\frac2x}$$

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Q26. If

$$u=x^2(y-x)+y^2(x-y)$$

show that$$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} =-2(x-y)^2$$

Solution:$$u=x^2y-x^3+xy^2-y^3$$

Therefore,$$\frac{\partial u}{\partial x} =2xy-3x^2+y^2$$

and$$\frac{\partial u}{\partial y} =x^2+2xy-3y^2$$

Adding,$$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} =-2x^2-2y^2+4xy$$$$=-2(x^2+y^2-2xy)$$$$\boxed{-2(x-y)^2}$$

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Q27. Find the market value of 62 shares available at ₹132 having par value ₹100.

Solution:

Market value of one share:$$₹132$$

Therefore,$$62\times132=₹8184$$

Answer:$$\boxed{₹8184}$$

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Q28. A man travelled by car for 3 days. He covered 480 km each day. On the first day he drove for 10 hours at 48 km/hour. On the second day he drove for 12 hours at 40 km/hour and on the third day he drove for 15 hours at 32 km/hour. What is the average speed?

Solution:

Speed	Time
48 km/hr	10 hr
40 km/hr	12 hr
32 km/hr	15 hr

Using harmonic mean:$$H=\frac{N}{\sum\frac fx}$$$$H=\frac{37}{0.9771}$$$$\boxed{H=37.86\text{ km/hr}}$$

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Q29. Draw the network diagram for the following activities.

Activity	A	B	C	D	E	F	G
Predecessor	–	–	A	A	B	C	D,E

Answer: Network diagram is constructed according to the above predecessor relationships.

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Q30. Differentiate the following with respect to $x$:

$$y=x^{\log x}$$

Solution:

Taking logarithm:$$\log y=\log(x^{\log x})$$$$=(\log x)(\log x)$$

Differentiate:$$\frac1y\frac{dy}{dx} = \frac{2\log x}{x}$$

Therefore,$$\boxed{\frac{dy}{dx} =\frac{2y\log x}{x}}$$

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SECTION – III

Answer any 7 of the following. Question No. 40 is compulsory.

7 × 3 = 21

Q31. Find the number of arrangements that can be made out of the letters of the word “MATHEMATICS”.

There are 11 letters.

M occurs twice.
T occurs twice.
A occurs twice.

Therefore,$$\text{Number of arrangements} = \frac{11!}{2!2!2!}$$$$\boxed{4,989,600}$$

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Q32. A question paper has two parts, Part A and Part B. Each part contains 10 questions. If the student has to choose 8 from Part A and 5 from Part B, in how many ways can he choose the questions?

$${}^{10}C_8\times{}^{10}C_5$$$$={}^{10}C_2\times{}^{10}C_5$$$$=45\times252$$$$\boxed{11340}$$

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Q33. Prove that:

$$\tan^{-1}\frac34+\tan^{-1}\frac17=\frac{\pi}{4}$$

Solution:

Using$$\tan^{-1}a+\tan^{-1}b = \tan^{-1} \left(\frac{a+b}{1-ab}\right)$$

we get$$= \tan^{-1} \left( \frac{\frac34+\frac17} {1-\frac34\cdot\frac17} \right)$$$$= \tan^{-1} \left( \frac{\frac{25}{28}} {\frac{25}{28}} \right)$$$$=\tan^{-1}(1)$$$$\boxed{\frac{\pi}{4}}$$

Hence proved.

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Q34. Evaluate:

$$\lim_{x\to\infty} \frac{6-5x^2}{4+15x^2}$$

Divide numerator and denominator by $x^2$:$$= \lim_{x\to\infty} \frac{\frac6{x^2}-5} {\frac4{x^2}+15}$$$$=\frac{-5}{15}$$$$\boxed{-\frac13}$$

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Q35. For

$$A=(a_{ij})_{2\times2}$$

defined by$$a_{ij}=2i-j$$

prove that$$AA^{-1}=I$$

Solution:$$A= \begin{pmatrix} 1&0\\ 3&2 \end{pmatrix}$$$$|A|=2$$

Therefore,$$A^{-1} = \frac12 \begin{pmatrix} 2&0\\ -3&1 \end{pmatrix}$$

Hence,$$AA^{-1} = \begin{pmatrix} 1&0\\ 3&2 \end{pmatrix} \frac12 \begin{pmatrix} 2&0\\ -3&1 \end{pmatrix}$$$$= \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}$$$$\boxed{AA^{-1}=I}$$

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Q36. Find the stationary points and stationary values for

$$f(x)=2x^3+9x^2+12x+1$$

Solution:$$f'(x)=6x^2+18x+12$$$$=6(x^2+3x+2)$$$$=6(x+2)(x+1)$$

For stationary points:$$f'(x)=0$$$$x=-2,\,-1$$

When $x=-2$:$$f(-2)=-3$$

When $x=-1$:$$f(-1)=-4$$

Therefore stationary points are:$$\boxed{(-2,-3),\;(-1,-4)}$$

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Q37. Find the present value of ₹2,000 p.a. for 14 years at an interest rate of 10% per annum.

Given:$$(1.1)^{-14}=0.2632$$

Formula:$$P=\frac{a}{i}[1-(1+i)^{-n}]$$$$=\frac{2000}{0.1}[1-0.2632]$$$$=20000(0.7368)$$$$\boxed{₹14,736}$$

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Q38. Solve the following LPP:

Maximize$$z=3x_1+4x_2$$

subject to:$$2x_1+x_2\le40$$$$2x_1+5x_2\le180$$$$x_1,x_2\ge0$$

Answer:

The optimum point is:$$\boxed{x_1=2.5,\quad x_2=35}$$

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Q39. If

$$u=\frac{x^2+y^2}{x+y}$$

prove that$$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \frac32u$$

by using Euler’s theorem.

Solution:

Since $u$ is homogeneous of degree:$$n=1$$

the Euler theorem gives:$$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} =nu$$

Thus,$$\boxed{x\frac{\partial u}{\partial x} +y\frac{\partial u}{\partial y}=u}$$

Note: The PDF’s extracted formula around Q39 is visually garbled; the original page should be referred to for the exact intended expression.

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Q40. Find $y_2$, if

$$x=a\sec\theta,\qquad y=a\tan\theta$$

Solution:$$\frac{dx}{d\theta}=a\sec\theta\tan\theta$$$$\frac{dy}{d\theta}=a\sec^2\theta$$

Therefore,$$y_1=\frac{dy}{dx} = \frac{a\sec^2\theta} {a\sec\theta\tan\theta}$$$$=\frac{\sec\theta}{\tan\theta}$$$$=\frac1{\sin\theta}$$

Then,$$\boxed{y_2=-\frac{\cos\theta}{a\sin^3\theta}}$$

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SECTION – IV

Answer the following.

7 × 5 = 35

Q41. (a)

In an economy, there are two industries $P_1$ and $P_2$. The following table gives the supply and demand position in crores of rupees.

Production Sector	$P_1$	$P_2$	Final Demand	Total Output
$P_1$	10	25	15	50
$P_2$	20	30	10	60

Determine the outputs when final demand is 35 for $P_1$ and 42 for $P_2$.

Answer:$$\boxed{P_1=₹150\text{ crores}}$$$$\boxed{P_2=₹204\text{ crores}}$$

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OR

Q41. (b)

Find the term independent of $x$ in the expansion of$$\left(2x^2+\frac1x\right)^{12}$$

General term:$$T_{r+1} = {}^{12}C_r(2x^2)^{12-r} \left(\frac1x\right)^r$$$$={}^{12}C_r2^{12-r}x^{24-3r}$$

For the term independent of $x$:$$24-3r=0$$$$r=8$$

Therefore,$$T_9 = {}^{12}C_8 2^4$$$$={}^{12}C_4\times16$$$$\boxed{7920}$$

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Q42. (a)

If$$\tan A-\tan B=x$$

and$$\cot B-\cot A=y$$

prove that$$\cot(A-B)=\frac1x+\frac1y$$

Solution:$$\cot(A-B) = \frac{1+\tan A\tan B} {\tan A-\tan B}$$

Since$$\tan A-\tan B=x$$

and$$\cot B-\cot A=y$$

we obtain:$$\tan A\tan B= \frac{x}{y}$$

Hence,$$\cot(A-B) = \frac{1+\frac xy}{x}$$$$=\frac1x+\frac1y$$

Hence proved.

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OR

Q42. (b)

Show that:$$\cos20^\circ\cos40^\circ\cos60^\circ\cos80^\circ =\frac1{16}$$

Solution:

Let$$x=\cos20^\circ\cos40^\circ\cos80^\circ$$

Multiplying by $2\sin20^\circ$:$$2x\sin20^\circ = 2\sin20^\circ\cos20^\circ \cos40^\circ\cos80^\circ$$$$=\sin40^\circ\cos40^\circ\cos80^\circ$$$$=\frac12\sin80^\circ\cos80^\circ$$$$=\frac14\sin160^\circ$$$$=\frac14\sin20^\circ$$

Therefore,$$2x=\frac14$$$$x=\frac18$$

Since$$\cos60^\circ=\frac12$$

we get:$$\cos20^\circ\cos40^\circ\cos60^\circ\cos80^\circ = \frac18\times\frac12$$$$\boxed{\frac1{16}}$$

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Q43. (a)

Find the axis, vertex, focus, equation of directrix and length of latus rectum for the parabola:$$y^2-4x-4y+8=0$$

Solution:$$4x=y^2-4y+8$$$$4x=(y-2)^2+4$$$$4x-4=(y-2)^2$$

Therefore,$$(y-2)^2=4(x-1)$$

Comparing with:$$Y^2=4aX$$

we get:$$a=1$$

Therefore:

Axis:$$\boxed{x=1}$$

Vertex:$$\boxed{(1,2)}$$

Focus:$$\boxed{(1,3)}$$

Directrix:$$\boxed{x=0}$$

Length of latus rectum:$$4a=4$$$$\boxed{4}$$

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OR

Q43. (b)

Find the equation of the circle passing through:$$(1,1),\quad(2,-1),\quad(2,3)$$

Let the equation be:$$x^2+y^2+2gx+2fy+c=0$$

Using $(1,1)$:$$2+2g+2f+c=0$$$$2g+2f+c=-2$$

Using $(2,-1)$:$$5+4g-2f+c=0$$$$4g-2f+c=-5$$

Using $(2,3)$:$$13+4g+6f+c=0$$$$4g+6f+c=-13$$

Solving:$$f=-1$$$$g=-\frac72$$$$c=7$$

Therefore:$$\boxed{x^2+y^2-7x-2y+7=0}$$

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Q44. (a)

A person sells a 20% stock of face value ₹5,000 at a premium of 62%. With the money obtained he buys a 15% stock at a discount of 22%. What is the change in his income if brokerage paid is 2%?

Solution:

Face value:$$₹5000$$

Dividend rate:$$20\%$$

Income from 20% stock:$$\frac{5000}{100}\times20 =₹1000$$

Selling price of one share after brokerage:$$100+62-2=160$$

Sale proceeds:$$\frac{5000}{100}\times160 =₹8000$$

Investment:$$₹8000$$

Market price of second stock:$$100-22+2=80$$

Income from 15% stock:$$\frac{8000}{80}\times15$$$$=₹1500$$

Change in income:$$1500-1000$$$$\boxed{₹500}$$

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OR

Q44. (b)

Calculate coefficient of correlation from the following data:

X	46	54	56	56	58	60	62
Y	36	40	44	54	42	58	54

Taking assumed means:$$\bar X=56,\qquad \bar Y=42$$

The calculated values are:$$\sum dx^2=160$$$$\sum dy^2=588$$$$\sum dxdy=200$$

Using:$$r= \frac{N\sum dxdy-(\sum dx)(\sum dy)} {\sqrt{[N\sum dx^2-(\sum dx)^2] [N\sum dy^2-(\sum dy)^2]}}$$

we get:$$r\approx0.77$$

Answer:$$\boxed{r=0.77}$$

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Q45. (a)

Bag I contains 3 red and 4 black balls while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag I.

Solution:

Let:$$E_1=\text{Bag I is chosen}$$$$E_2=\text{Bag II is chosen}$$$$A=\text{red ball is drawn}$$

Since either bag is selected randomly:$$P(E_1)=P(E_2)=\frac12$$

Also,$$P(A/E_1)=\frac37$$

and$$P(A/E_2)=\frac5{11}$$

By Bayes’ theorem:$$P(E_1/A) = \frac{P(E_1)P(A/E_1)} {P(E_1)P(A/E_1)+P(E_2)P(A/E_2)}$$$$= \frac{\frac12\times\frac37} {\frac12\times\frac37+\frac12\times\frac5{11}}$$$$= \frac{33}{68}$$

Answer:$$\boxed{\frac{33}{68}}$$

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OR

Q45. (b)

The demand for commodity A is:$$q=80-p_1^2+5p_2-p_1p_2$$

Find the partial elasticities$$E_{p_1} \quad\text{and}\quad E_{p_2}$$

when$$p_1=2,\qquad p_2=1$$

For $p_1$:

$$\frac{\partial q}{\partial p_1} =-2p_1-p_2$$

Partial elasticity:$$E_{p_1} = -\frac{p_1}{q} \frac{\partial q}{\partial p_1}$$

At $p_1=2,\ p_2=1$:$$E_{p_1}=\frac{10}{79}$$

Therefore:$$\boxed{E_{p_1}=\frac{10}{79}}$$

For $p_2$:

$$\frac{\partial q}{\partial p_2} =5-p_1$$

Therefore:$$E_{p_2} = -\frac{p_2}{q} \frac{\partial q}{\partial p_2}$$

At $p_1=2,\ p_2=1$:$$E_{p_2}=-\frac3{79}$$

Therefore:$$\boxed{E_{p_2}=-\frac3{79}}$$

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Q46. (a)

Compute the mean deviation about mean for the following data:

Class Interval	Frequency
0–5	3
5–10	5
10–15	12
15–20	6
20–25	4

Mid-values:$$2.5,\ 7.5,\ 12.5,\ 17.5,\ 22.5$$$$N=30$$$$\sum fx=390$$

Therefore,$$\bar X=\frac{390}{30}=13$$

Now:$$\sum f|D|=130$$

Mean deviation about mean:$$MD=\frac{\sum f|D|}{N}$$$$=\frac{130}{30}$$$$\boxed{4.34}$$

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OR

Q46. (b)

The following table gives activities in a construction project:

Activity	1–2	1–3	2–3	2–4	3–4	4–5
Duration (days)	22	27	12	14	6	12

Calculate earliest start time, earliest finish time, latest start time, latest finish time and critical path.

The calculated values are:

Activity	Duration	EST	EFT	LST	LFT
1–2	22	0	22	0	22
1–3	27	0	27	7	34
2–3	12	22	34	22	34
2–4	14	22	36	26	40
3–4	6	34	40	34	40
4–5	12	40	52	40	52

Therefore, the critical path is:$$\boxed{1-2-3-4-5}$$

Duration:$$\boxed{52\text{ days}}$$

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Q47. (a)

The total revenue function for a commodity is:$$R=15x+\frac{x^2}{3}-\frac{x^4}{36}$$

Show that at the highest point, average revenue is equal to marginal revenue.

Solution:

Average revenue:$$AR=\frac Rx$$$$AR=15+\frac{x}{3}-\frac{x^3}{36}$$

For maximum AR:$$\frac{d(AR)}{dx}=0$$

This gives:$$x=2$$

At $x=2$:$$AR=15+\frac23-\frac{8}{36}$$$$AR\approx15.45$$

Marginal revenue:$$MR=\frac{dR}{dx}$$$$MR=15+\frac{2x}{3}-\frac{x^3}{9}$$

At $x=2$:$$MR=15+\frac43-\frac89$$$$MR\approx15.45$$

Therefore:$$\boxed{AR=MR}$$

Hence proved.

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OR

Q47. (b)

Show that the function$$f(x)=|x|$$

is not differentiable at $x=0$.

We have:$$f(x)= \begin{cases} x,&x\ge0\\ -x,&x<0 \end{cases}$$

Left-hand derivative at $x=0$:$$L[f'(0)]=-1$$

Right-hand derivative at $x=0$:$$R[f'(0)]=1$$

Since:$$L[f'(0])\ne R[f'(0)]$$

therefore,$$\boxed{f(x)=|x|\text{ is not differentiable at }x=0}$$

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ANSWER KEY – QUICK LIST

1. B
2. A
3. C
4. A
5. B
6. D
7. D*
8. A
9. C
10. A
11. C
12. D
13. D
14. B
15. C
16. B
17. D
18. C
19. A
20. D