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Class XI Business Maths and Statistics Exam Answer Key 2024

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Samacheer Kalvi students, hope you wrote your Business Maths exam well. Here is the Class XI Business Maths and Statistics Exam Answer Key 2024 for your reference. If you need clarification for any answers, please let us know in the comments section.

Part I

1. The inventor of input – output analysis is

(a) Prof. Wassily W. Leontief (b) Sir Francis Galton(c) Arthur Caylay (d) Fisher  

 Answer: (a) Prof. Wassily W. Leontief   

2. The number of permutation of n different things taken r at a time, when the repetition is allowed is:

(a) \frac{n!}{(n-r)!}  (b) r^{n}(c) \frac{n!}{(n+r)!}  (d) n^{r}  

Answer: (d) n^{r} 

3. If y=e^{2x}, then \frac{d^{2}y}{dx^{2}} at x=0:

(a) 2 (b) 4(c) 0(d)9

Answer: (b) 4

4. The correlation coefficient:

(a) r=b_{xy \times b_{yx}} (b) r=\pm \sqrt{b_{xy}\times b_{yx}} (c) r=\pm \sqrt{\frac{1}{b_{xy}\times b_{yx}}}  (d) r=\frac{1}{b_{xy}\times b_{yx}}

Answer: (b) r=\pm \sqrt{b_{xy}\times b_{yx}} 

5. P(A) = \frac{3}{5} and P (B) = \frac{1}{5}. Find P (A\cap B) if A and B are independent events.

(a) \frac{3}{16} (b) \frac{3}{25} (c) \frac{4}{25}    (d) \frac{4}{10}

Answer: (b) \frac{3}{25}

6. If tan A =\frac{1}{2} and tan B=\frac{1}{3}, then tan (2A+B) is equal to

(a) 3  (b) 1(c) 4    (d) 2

Answer: (a) 3

7. For the function y=x^{3}+19 , find the values of x when the marginal value is equal to 27.

(a) \pm 3 (b) \pm 1 (c) \pm 4    (d) \pm 2

Answer: (a) \pm 3

8. If ‘a’ is the annual payment, ‘n’ is the number of periods and ‘i’ is compound interest for Rs. 1 then the future amount of the ordinary annuity is:

(a) P=\frac{a}{i}     (b) A=\frac{a}{i}(1+i[(1+i)^{n}-1])(c) P=\frac{a}{i}(1+i[1-(1+i)^{-n}])     (d) A=\frac{a}{i}[(1+i^{n})-1]

Answer: (d) A=\frac{a}{i}[(1+i^{n})-1]

9. If \begin{vmatrix} x &2 \\ 8 & 5 \end{vmatrix}=0, then the value of x is

(a) \frac{-16}{5} (b) \frac{-5}{6}(c) \frac{16}{5}  (d) \frac{5}{6}

Answer: (c) \frac{16}{5} 

10. If y= 4ae^{4x} then find \frac{dy}{dx}

(a) 16ae^{x}    (b) ae^{4x} (c) 16ae^{4x}     (d) 4ae^{4x}

Answer: (c) 16ae^{4x}   

11. The value of sin(-420^o) is:

(a) \frac{1}{2}   (b) \frac{\sqrt{3}}{2}(c) \frac{-1}{2}  (d) \frac{-\sqrt{3}}{2}

Answer: (d) \frac{-\sqrt{3}}{2}

12. If m_{1} and m_{2} are the slopes of the pair of lines given by ax^{2}+2hxy+by^{2}=0 then the value of m_{1} + m_{2} is:

(a) \frac{2h}{a}    (b) \frac{{2h}}{b}(c) \frac{-2h}{a}    (d) \frac{{-2h}}{b}

 Answer: (d) \frac{{-2h}}{b}

13. One of the conditions for the activity (i, j) to lie on the critical path is

(a) E_{j}-E_{i}=L_{i}-L_{j}=t_{ij}    (b) E_{j}-E_{i}=L_{j}-L_{i}=t_{ij}(c) E_{j}-E_{i}=L_{j}-L_{i}\neq t_{ij}  (d) E_{i}-E_{j}=L_{j}-L_{i}=t_{ij}

Answer: (b) E_{j}-E_{i}=L_{j}-L_{i}=t_{ij}

14. If q=1000+8p_{1}-p_{2} then \frac{\partial q}{\partial p_{1}} is:

(a) 1000   (b)-1q=1000-p_{2}   (d) 8

Answer: (d) 8

15. The dividend received on 200 shares of face value of Rs. 100 at 8% stock is:

(a) Rs. 1500     (b) Rs. 1600(c) Rs. 800   (d) Rs. 1000

Answer: (b) Rs. 1600

16. If clock wise and anti-clockwise circular permutations are considered to be same, the number of circular permutation of n objects taken all at a time is:

(a) \frac{n!}{2}     (b) \frac{n+1}{2!}   (c) \frac{(2n+1)!}{2}  (d) \frac{(n-1)!}{2} 

Answer: (b) \frac{n+1}{2!}   

17. If two events A and B are dependent then the conditional probability of P(B/A) is:

a) P \frac{(A\cap B)}{P(A)}      (b) P(A).P(B/A)  (c) P(A).P(A/B)  (d) P \frac{(A\cap B)}{P(B)} 

 Answer: a) P \frac{(A\cap B)}{P(A)} 

18. Co-relation coefficient lies between:

(a) -1 to 0    (b) 0 to \infty (c) -1 to \infty  (d) -1 to +1

 Answer: (d) -1 to +1

19. Find the equation of the circle with centre at (4,5) and radius 3 units

(a) x^{2}+y^{2}-8x-10y+32=0    (b) x^{2}+y^{2}-6x+2y-6=0(c) x^{2}+y^{2}-8x-6y=0 (d) x^{2}+y^{2}+8x-10y=0

Answer: (a) x^{2}+y^{2}-8x-10y+32=0 

20. Find the value of 100C_{99}.

(a) 1  (b) 100(c) 0 (d) -99

Answer: (b) 100

Part II

Answer any seven questions. Question 30 is compulsory.

21. Evaluate \begin{vmatrix}1& 3 & 4\\102&18 &36 \\17&3 & 6\\\end{vmatrix}

Answer:

= 6 \begin{vmatrix} 1& 3 & 4\\ 102&18 &36 \\ 17&3 & 6\\ \end{vmatrix}

\begin{vmatrix} 1& 3 & 4\\ 17&3 &6 \\ 17&3 & 6\\ \end{vmatrix}\\[.5 cm]

\because R_{2}\equiv R_{3}

=0

22. If nC_{4} = nC_{6} find 12C_{n}

Answer: nC_{x} = nC_{y}, then x + y = n

Here nC_{4} = nC_{6}

n = 4 + 6 = 10

= 12C_{n} = 12C_{10}

= nC_{4} = nC_{6}

= \frac{12\times 11}{1 \times 2}

=66

23. If f (x)=x^{3}-\frac{1}{x^{3}}, x\neq 0, then show that f (x) +\frac{1}{f(x)}=0,

Answer: f (x)=x^{3}-\frac{1}{x^{3}}

= f\left (\frac{1}{x} \right )=\left (\frac{1}{x} \right )^{3}-\frac{1}{\left(\frac{1}{x} \right ) ^{3}}

= \frac{1}{x^{3}}-x^{3}

=x^{3}-\frac{1}{x^{3}}+\frac{1}{x^{3}}-x^{3}

 = 0 (proved)

24. Find D2 and D6 for the following series 22, 4, 2, 12, 16, 6, 10, 18, 14, 20, 8

Answer: Arrange in ascending order: x= 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22

D_{2} = size of 2\left (\frac{n+1}{10} \right )^{th} value \\[.5 cm]

D_{6} = size of 6 \left (\frac{n+1}{10} \right )^{th} value\\ [.5 cm]

D_{2} = size of 2.4^{th} value = size of 2^{nd} value = 4\\[.5 cm]

D_{6} = size of 7.2^{th} value = size of 7^{nd} value = 14

25. Which is better investment? 20% stock at Rs. 140 (or) 10% stock at Rs. 70.

Let the investment in case be ₹ 140 × 70

Income from 20% stock at ₹ 140 = \frac{20}{140} \times 140 \times 70

= 20 x70

=Rs. 1400

For the same investment both stocks fetch the same income. Therefore they are both equally good investment options.

26. A manufacturing company has a contract to supply 4000 units of an item per year at uniform rate. The storage cost per unit per year amounts to ` 50 and the set up cost per production run is Rs. 160. If the production run can be started instantaneously and shortages are not permitted, determine the number of units which should be produced per run to minimize the total inventory cost.

Answer: Annual demand R= 4000

Storage cost C_{1} = Rs.50 \\

Setup cost per production: C_{3}= Rs. 160 \\

EOQ = \sqrt{\frac{2RC_{3}}{C_{1}}} \\ [.5 cm]

=\frac{\sqrt{2\times4000\times160}}{50}\\[.5 cm]

= 160

27. If the centre of the circle x^{2} + y^{2} + 2x - 6y + 1 lies on a straight line ax+2y+2=0, then find the value of ‘a’.

Answer. Centre C(–1, 3)

It lies on ax + 2y + 2 = 0

–a + 6 + 2 = 0

a =8

28. Find the 5^{th} term in the expansion (x-3y) ^{13}.

Answer: General term is t_{r+1} = nC_{r}x^{n-r}a^{r}

(x - 2y)^{13} = (x + (-2y))^{13}\\[.5cm]

Here x = x, a = (-2y) and n = 13

5^{th}= t_{5} = t_{4+1} = 13C_{4} x^{13-4} (-2y)^{4}\\[.5cm]

= 13C^{4} x^{9} 2^{4} y^{4}\\[.5cm]

\frac{13 \times 12 \times 11 \times 10} {4 \times 3 \times 2 \times 1} \times 2 \times 2 \times 2 \times 2 \times x^{9} y^{4}\\[.5cm]

= 13\times 11 \times 10 \times 8 x^{9}y^{4}\\[.5 cm]

= 13\times 880 x9y^{4} \\[.5 cm]

= 11440x^{9}y^{4}

29. If f(x)= 2^{x} show that f(x).f(y)= f (x+y)

Answer: f(x)= 2^{x}

f(x+y)= 2^{x+y}\\[.5cm]

= 2^{x}. 2^{y}\\[.5cm]

=f(x).f(y)

30. Find the rank of the word TABLE in English dictionary.

Number of words starting with A =4! =24 ways

Number of words starting with B =4! =24 ways

Number of words starting with E =4! =24 ways

Number of words starting with L =4! =24 ways

Number of words starting with TAB=2!=2 ways

Rank of the word TABLE= 24+24+24+24+2=98

Part III

Answer any seven questions. Question No 40 is compulsory.

Find the values of a and b if the equation (a - 2) x^{2} + by^{2} +](b - 8)xy + 4x + 4y - 1 = 0

Answer: Conditions of the equation are: (i) coefficient of xy = 0

b-8 =0

b=8

coefficient of x^{2} = Coefficient of y^{2}\\[ .5 cm]

= a-1=8

a=9

\therefore The equation of the circle is 9x^{2} + 8y^{2} + 4x + 4y - 1 = 0

32. Evaluate \lim_{x\rightarrow 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}

Answer: \lim_{x\rightarrow 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x} \times [latex]\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}[/latex]

\lim_{x\rightarrow 0} \frac{{(1+x)}-(1-x)}{x {\sqrt{1+x}+\sqrt{1-x}} }\\[.5cm]

\lim_{x\rightarrow 0} \frac{(1+x-1+x)}{x {\sqrt{1+x}+\sqrt{1-x}} }\\[.5cm]

\lim_{x\rightarrow 0} \frac{2x}{x\sqrt{1+x}+\sqrt{1-x}}\\[.5cm]

\lim_{x\rightarrow 0} \frac{2}{ \sqrt{1+x}+\sqrt{1-x}}\\[.5cm]

\frac{2}{ \sqrt{1+0}+\sqrt{1-0}}\\ [.5cm]

=\frac{2}{2} =1

33. Draw the event oriented network for the following data.

Events1234567
Immediate Predecessors–112,334,55,6
Q 33

34. Solve the following linear programming problem by graphical method.

Maximize: z=40x_{1}+50x_{2} subject to constraints 3x_{1}+x_{2}\leq 9; x_{1}+2x_{2}\leq 8; and x_{1}, x_{2}\geq 0

Let 3x_{1}+x_{2}= 9;

x_{1}03
x_{2}90

Also, x_{1}+2x_{2}=8;

x_{1}08
x_{2}40

3x_{1} + x_{2} = 9 \cdots (1)\\[.5cm]

x_{1} + 2 \times 2 = 8 \cdots (2)\\[.5cm]

\times 2 = 6 \times 1 + 2 \times 2 = 18 \cdots (3)\\[.5cm]

+ (3) = -5 \times 1 = -10\\[.5cm]

x_{1} = 2\\[.5cm]

Substitute x_{1} = 2 in (1) \\[.5cm]

3(2) + x_{2} = 9\\[.5cm]

x_{2} = 3\\[.5cm]

graph

The co-ordinates of the corner points are (0, 0), (3, 0), (2, 3), (0, 4)

Corner Points z=40x_{1}+50x_{2}
(0,0)=40(0)+50(0) =0
(3,0)=40(3)+50 (0) = 120
(2,3)=40(2)+50 (3)= 230
(0,4)=40(0)+50 (4) =200

The maximum value of Z occurs at (2, 3).

Therefore, the optimal solution is at x_{1}=2 and x_{2}=3

35. Find the interval in which the function f(x)=x^{2}–4x+6 is strictly increasing and strictly decreasing.

Given: f(x) = x^{2}-4x+6 \\[.5cm]

Differentiate with respect to x, f{}' (x)= 2x-4 \\[.5cm]

When f{}'(x) = 0; x-4=0; x = 2 \\[.5cm]

Then the real line is divided into two intervals namely (-\infty,2) and (2,\infty)

36. Find the value of tan 75\degree.

tan 75\degree = tan (45\degree+30\degree)\\[.5cm]

=\frac{tan 45\degree +tan 30\degree}{1-tan 45\degree tan 30\degree}\\[.5cm]

=\frac{1 +\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}\\[.5cm]

=\frac{\sqrt{3}+1}{\sqrt{3}-1}\\[.5cm]

=2+\sqrt{3}

Q. 37. A person purchases tomatoes from each of the 4 places at the rate of 1kg., 2kg., 3kg., and 4kg. per rupee respectively .On the average, how many kilograms has he purchased per rupee?

Using harmonic mean:

H.M = \Large\frac{n}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}\\[.5cm]

\Large\frac{4}{\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}\\[.5cm]

\frac{4 \times 12}{25}\\[.5cm]

=1.92 kg per rupee.

38. The technology matrix of an economic system of two industries is
\begin{bmatrix} .8&.2 \\[.5cm] .9& .7 \end{bmatrix}

Test whether the system is viable as per Hawkins – Simon conditions.

B =\begin{bmatrix} .8&.2 \\ .9& .7 \end{bmatrix}\\[.5cm]

I-B =\begin{bmatrix} 1&0 \\ 0& 1 \end{bmatrix}-\begin{bmatrix} .8&.2 \\ .9& .7 \end{bmatrix}\\[.5cm]

=\begin{bmatrix} .2&-.2 \\ -.9& .3 \end{bmatrix} \\[.5cm]

\left | I-B \right | =\begin{bmatrix} .2&-.2 \\ -.9& .3 \end{bmatrix}\\[.5cm]

= (0.2)(0.3)-(-0.2)(-0.9)\\[.5cm]

= 0.06 – 0.18

=-0.12 < 0\\[.5cm]

Since \left |I - B \right | is negative, Hawkins – Simon conditions are not satisfied.

Q. 39 If the payment of Rs.2,000 is made at the end of every quarter for 10 years at the rate of 8% per year, then find the amount of annuity. [(1.02)^{40} =2.2080 ].

a=2000 n= 10 years \frac{i}{k}=\frac{\frac{8}{100}}{4} = \frac{2}{100} =0.02\\[.5cm]

A= \frac{a}{i}[(i+\frac{i}{k})^{nk}-1]\\[.5cm]

A= \frac{2000}{0.02}[(1+ 0.02)^{10 \times4}-1]\\[.5cm]

A= \frac{200000}{2}[(1+ 0.02)^{40}-1]\\[.5cm]

= {100000}[(2.2080-1)] \because (1.02)^{40}=2.2080\\[.5cm]

= {100000}(1.2080) \\[.5cm]

= Rs.120800

Q 40. Prove that tan^{-1}\frac{2}{11} +tan^{-1}\frac{7}{24} =tan^{-1}\frac{1}{2}

tan^{-1}\frac{2}{11} +tan^{-1}\frac{7}{24} \\[.5cm]

tan ^{-1} x + tan ^{-1} y = tan ^{-1} \frac{x+y}{1-xy}\\[.5cm]

= tan ^{-1} \Large \frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}\times \frac{7}{24}}\\[.5cm]

= tan ^{-1 }\Large \left( \frac{\frac{24\times2+7\times11}{24\times11}{}}{\frac{\frac{11\times12 -7}{}}{11\times 12}}\right )\\[.5cm]

= tan ^{-1 } \Large \left( \frac{\frac{48+7}{24\times11}{}}{\frac{\frac{132 -7}{}}{11\times 12}}\right )\\[.5cm]

= tan ^{-1 }\Large \left ( \frac{\frac{125}{24\times11}{}}{\frac{\frac{125}{}}{11\times 12}}\right )\\[.5cm]

= tan ^{-1 }\left ( \frac{125}{24\times11}\times \frac{11\times12}{125}\right )\\[.5cm]

= tan ^{-1 } \left (\frac{1}{2} \right )\\[.5cm]

=RHS. Proved.

Part IV

Answer all the questions.

41.(a) An economy produces only coal and steel. These two commodities serve as intermediate inputs in each other’s production. 0.4 tonne of steel and 0.7 tonne of coal are needed to produce a tonne of steel. Similarly 0.1 tonne of steel and 0.6 tonne of coal are required to produce a tonne of coal. No capital inputs are needed. Do you think that the system is viable? 2 and 5 labour days are required to produce a tonnes of coal and steel respectively. If economy needs 100 tonnes of coal and 50 tonnes of steel, calculate the gross output of the two commodities and the total labour days required.

The technology matrix B = \begin{bmatrix} .4 &.1 \\ .7 & .6 \end{bmatrix}\\[.5cm]

I-B = \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix} -\begin{bmatrix} .4 &.1 \\ .7 & .6 \end{bmatrix}\\[.5cm]

=\begin{bmatrix} .6 &-.1 \\ -.7 & .4 \end{bmatrix}\\[.5cm]

\left | I-B \right |=\begin{bmatrix} .6 &-.1 \\ -.7 & .4 \end{bmatrix}\\[.5cm]

= (0.6)(0.4)-(-0.7)(-0.1)\\[.5cm]

= 0.24 - 0.07 = 0.17 \\[.5cm]

Since the diagonal elements of \left |I - B \right | are positive and value of I – B is positive, the system is viable. 

*** QuickLaTeX cannot compile formula:
\\[.5cm] 

*** Error message:
There's no line here to end.
leading text: \\[.5cm]

adj \left | (I-B) \right |=\begin{bmatrix} .4 &.1 \\ .7 & .6 \end{bmatrix}\\[.5cm]

(I-B)^{-1} = \frac{1}{\left |I-B \right |}adj (I-B) \\[.5cm]

=\frac{1}{0.17}\begin{bmatrix} .4&.1 \\ .7&.6 \end{bmatrix} \\[.5cm]

X =(I-B)^{-1} D, where D=\begin{bmatrix} 50\\ 100 \end{bmatrix} \\[.5cm]

= \frac{1}{0.17}\begin{bmatrix} .4&.1 \\ .7&.6 \end{bmatrix}\begin{bmatrix} 50\\ 100 \end{bmatrix} \\[.5cm]

= \frac{1}{0.17}\begin{bmatrix} 30\\ 95 \end{bmatrix} \\[.5cm]

=\begin{bmatrix} 176.5\\ 558.8 \end{bmatrix} \\[.5cm]

Steel output = 176.5 tonnes

Coal output = 558.8 tonnes

Total labour days required = 5(steel output) + 2(coal output)

= 5(176.5) + 2(558.8)

882.5 + 1117.6 = 2000.1

= 2000 labour days (approx)

41.(b) Verify Euler’s theorem for the function u=x^{3}+y^{3}+3xy^{2}\\[.5 cm]

u (x,y)=x^{3}+y^{3}+3xy^{2}\\[.5 cm]

u (tx,ty)=tx^{3}+ty^{3}+3(tx)(ty)^{2}\\[.5 cm]

=t^{3}x^{3}+t^{3}y^{3}+3(tx) (t^{2}y^{2})\\[.5 cm]

=t^{3}(x^{3}+y^{3}+3xy^{2})\\[.5 cm]

=t^{3}(u)\\[.5 cm]

Therefore, u is a homogeneous function in x and y of degree 3.

By \:Euler's\: theorem,\: x \frac{\partial u}{\partial x} +y \frac{\partial u}{\partial y}=3u \\[.5 cm]

Verification:

=x^{3}+y^{3}+3xy^{2}\\[.5 cm]

\frac{\partial u}{\partial x} =3x^{2}+0+3y^{2} \frac{\partial }{\partial x}(x) \\[.5 cm]

=3x^{2}+3y^{2} (1) \\[.5 cm]

=3x^{2}+3y^{2}\cdots(1)\\[.5 cm]

\therefore x \frac{\partial u}{\partial x} =3x^{3}+3xy^{2} \\[.5 cm]

\frac{\partial u}{\partial y} =0+3y^{2}+3x (2y )=3y^{2}+6xy\\[.5 cm]

y. \frac{\partial u}{\partial y} = 3y^{3}+6xy^{2}\cdots (2)\\[.5 cm]

Adding (1) and (2)

x \frac{\partial u}{\partial x} +y \frac{\partial u}{\partial y}=3x^{3}+3y^{3}+9xy^{2} \\[.5 cm]

=3 (x^{3}+y^{3}+3xy^{2}) \\[.5 cm]

=3u

Hence Euler’s theorem is verified.

42. (a). Show that the equation 2x^{2}+7xy+3y^{2}+5x+5y+2=0\\[.5 cm] represent two straight lines and find their separate equations.

Comparing 2x^{2}+7xy+3y^{2}+5x+5y+2=0 with ax^{2}+2hxy+by^{2}+2gx+2fy+c=0\\[.5 cm]

we\: get\: a=2; b=3; h=\frac{7}{2}; g=\frac{5}{2}; f=\frac{5}{2}; c=2 \\[.5 cm]

Hence the given equation represents a pair of parallel straight lines.

Now\: 4x^{2}-12xy+9y^{2}= (2x-3y)^{2}\\[.5 cm]

Consider\: 4x^{2}-12xy+9y^{2}+18x -27y+8= 0\\[.5 cm]

(2x-3y)^{2} +9 (2x-3y)+8 = 0\\[.5 cm]

Put 2x–3y = z

z^{2} +9 z+8 = 0\\[.5 cm]

(z+1 )(z+8) = 0\\[.5 cm]

(z+1 )= 0; (z+8) = 0\\[.5 cm]

2x–3y+1 = 0; 2x–3y+8 =0

Therefore, the two separate equations are: 2x–3y+1 = 0 and 2x–3y+8 =0

42 (b) Solve tan ^{-1}(x+1 )+ tan^{-1}(x-1 )=tan^{-1}(\frac{4}{7}) .

We\: know\: that\:tan ^{-1}(x)+ tan^{-1}(y)= \frac{x+y}{1- xy}\\[.5 cm]

=tan ^{-1} \frac{(x+1)+(x-1)}{1- (x+1)(x-1)}=tan^{-1}(\frac{4}{7})\\[.5 cm]

=tan ^{-1} \left (\frac{2x}{1- (x^{2}-1)} \right )=tan^{-1}(\frac{4}{7})\\[.5 cm]

=tan ^{-1} \left (\frac{2x}{1-x^{2}+1} \right )=tan^{-1}(\frac{4}{7})\\[.5 cm]

=tan ^{-1} \left (\frac{2x}{2-x^{2}} \right )=tan^{-1}(\frac{4}{7})\\[.5 cm]

= \frac{2x}{2-x^{2}} =\frac{4}{7}\\[.5 cm]

= \frac{x}{2-x^{2}} =\frac{2}{7}\\[.5 cm]

 7x = 2(2 – x2)

7x = 4 – 2x2

2x2 + 7x – 4 = 0

(x + 4) (2x – 1) = 0

x + 4 = 0 ; 2x – 1 = 0

= x=-4; x=\frac{1}{2}\\[.5 cm]

Ignore x=-4 as it does not satisfy the equation. Therefore, x=\frac{1}{2}

Q 43. (a) Construct the network and calculate the earliest start time, earliest finish time, latest start time and latest finish time of each activity and determine the Critical path of the project and duration to complete the project.

Activity1-21-32-43-43-54-95-65-76-87-88-109-10
Time411165481257
Q 43a

E1 = 0

E2 = 0 + 4 = 4

E3 = 0 + 1 = 1

E4 = (E2 + 1) (or) (E3 + 1) (whichever is maximum)

= (4 + 1) (or) (1 + 1) (whichever is maximum)

= 5 (or) 2 (whichever is maximum)

E4 = 5

E5 = 1 + 6 = 7

E6 = 7 + 4 = 11

E7 = 8 + 7 = 15

E8 = (E7 + 2) (or) (E6 + 1)

= (15 + 2) or (11 + 1) (whichever is maximum)

= 17 or 12 (whichever is maximum)

E8 = 17

E9 = 5 + 5 = 10

E10 = (E9 + 7) (or) (E8 + 5) (whichever is maximum)

= (10 + 7) (or) (17 + 5) (whichever is maximum)

= 17 (or) 22 (whichever is maximum)

E10 = 22

L10 = 22

L9 = 22 – 7 = 15

L8 = 22 – 5 = 17

L7 = 17 – 2 = 15

L6 = 17 – 1 = 16

L5 = (16 – 4) (or) (15 – 8) (whichever is minimum)

L5 = 7

L4 = 15 – 5 = 10

L3 = (10 – 1) (or) (7 – 6) (whichever is minimum)

L3 = 1

L2 = 10 – 1 = 9

L1 = 0

ActivityDurationESTEFTLSTLFT
1-24049-4=59
1-31011-1=01
2-414510-1=910
3-411210-1=910
3-56177-6=17
4-9551015-5=1015
5-6471116-4=1216
5-7871515-8=715
6-81111217-1=1617
7-82151717-2=1517
8-105172222-5=1722
9-107161722-7=1522
PathTime
1-2-4-9-104+1+5+7=17
1-3-4-9-101+1+5+7=14
1-3-5-6-8-101+6+4+1+5=17
1-3-5-7-8-101+6+8+2+5=22

The critical path is 1-3-5-7-8-10. It takes 22 units to complete the project.

Q. 43 (b). Compute Quartile deviation from the following data.

CI10-2020-3030-4040-5050-6060-7070-80
f1219510966
CIfcf
10-201212
20-301931
30-40536
40-501046
50-60955
60-70661
70-80667

Q_{1}=size\: of\: \left (\frac{N}{4} \right )^{th}\: value\\[.5 cm]

= \left (\frac{67}{4} \right )^{th}\: value\\[.5 cm]

= 16.75^{th} \:value\\[.5 cm]

Thus, Q_{1} lies in the class 20 – 30; and the corresponding values are:

L=20; \frac{N}{4}=16.75; pcf=12; f=19; c=10 \\[.5 cm]

Q_{1}=L+\Large \left (\frac{\frac{N}{4}-pcf}{f} \right )\times c \\[.5 cm]

Q_{1}=20+\left (\frac{16.75-12}{19} \right )\times 10 \\[.5 cm]

=20+2.5=22.5

= \left (\frac{3N}{4} \right )^{th}\: value\\[.5 cm]

= 50.25^{th} value\\[.5 cm]

Thus, Q_{3} lies in the class 50 – 60; and the corresponding values are:

L=50; \frac{3N}{4}=50.25; pcf=46; f=9; c=10 \\[.5 cm]

Q_{3}=L+\left (\frac{\frac{3N}{4}-pcf}{f} \right )\times c \\[.5 cm]

Q_{3}=50+\Large \left (\frac{50.25-46}{9} \right )\times 10 \\[.5 cm]

=54.72

QD =\frac{1}{2}(Q_{3}-Q_{1}) \\[.5 cm]

QD =\frac{54.72-22.5}{2} \\[.5 cm]

=16.11

Q.44 (a) By Mathematical Induction, prove that 1^{2}+2^{2}+3^{2}+\cdots +n^{2}=\frac{n (n+1)(2n+1)}{6} \:for\: all\: n \in N.

Let P(n)= 1^{2}+2^{2}+3^{2}+\cdots +n^{2}=\frac{n (n+1)(2n+1)}{6} \:for\: all\: n \in N\\[.5 cm]

Put n = 1

LHS=1^{2}=1\\[.5 cm]

RHS = \frac{1(1+1)(2+1)}{6}=1 \\[.5 cm]

LHS=RHS=1

Therefore, P(1) is true.

Assume that P(k) is true.

Let P(K)= 1^{2}+2^{2}+3^{2}+\cdots +k^{2}=\frac{k (k+1)(kn+1)}{6} \\[.5 cm]

=1^{2}+2^{2}+3^{2}+\cdots +k^{2}+(k+1)^{2}=\ p(k) + (k+1)^{2} \\[.5 cm]

=\frac{k(k+1)(2k+1)}{6}+ (k+1)^{2} \\[.5 cm]

=\large \frac{k(k+1)(2k+1)+6(k+1)^{2}}{6} \\[.5 cm]

=(k+1)[k(2k+1)+6(k+1)] \\[.5 cm]

=\large \frac{(k+1)(2k^{2}+7k+6)}{6}) \\[.5 cm]

=\large \frac{(k+1)(k+2)(2k+3) }{6}\\[.5 cm]

P(k + 1) is true whenever P(k) is true.

\therefore P(n)\: is\: true\: for \:all \:n\in N

Q. 44 (b) If sin y= x sin (a+y) then prove that \frac{dy}{dx}= \frac{sin^{2}(a+y)}{sin a} .

sin y = x sin (a+y)

x=\Large \frac{sin y}{sin(a+y)}\\[.5 cm]

Differentiating with respect to y,

\Large \frac{dx}{dy}=\frac{sin(a+y)cos y-sin y\:cos(a+y)}{sin^{2}(a+y)}\\[.5 cm]

=\Large \frac{sin(a+y-y)}{sin^{2}(a+y)}\\[.5 cm]

=\Large \frac{sina}{sin^{2}(a+y)}\\[.5 cm]

\Large \frac{dy}{dx}= \frac{sin^{2}(a+y)}{sin a} .

Q.45 (a) The following are the ranks obtained by 10 students in Commerce and Accountancy are given below:

Commerce64312798105
Accountancy41675810932

To what extent is the knowledge of students in the two subjects related?

R_{x} R_{y} d=R_{x}-R_{y} d^{2}
6424
4139
36-39
17-636
25-39
78-11
910-11
89-11
103749
5239
\sum d^{2}=128

N=11\: \large \sum d^{2}=128 \\[.5 cm]

N=11 \rho = \large 1-\frac{6\sum d^{2}}{N(N^{2}-1)} \\[.5 cm]

= 1-\large \frac{6\times 128}{10(10^{2}-1)} \\[.5 cm]

= 1-\large \frac{6\times 128}{990} \\[.5 cm]

=0.224

Q.45 (b) A person sells a 20% stocks of face value Rs. 10,000 at a premium of 42%. With the
money obtained he buys a 15% stock at a discount of 22%. What is the change in his income if the brokerage paid is 2%.

For 20% stock:

Face Value FV= Rs.100

Income = \frac{20}{100}\times 10000= Rs. 2000 \\[.5 cm]

Investment= Rs.10000

Face Value = Rs. 100

Market value = Rs.100 +42 -2 =140

Number of shares = = \large \frac{investment}{FV} \\[.5 cm]

= \frac{10000}{100} =100\\[.5 cm]

Sales proceeds = 100 × 140 = 14,000

For 15% stock:

Market value = Rs.100 -22 +2 =80

Number of shares = = \large \frac{investment}{FV} \\[.5 cm]

= \frac{14000}{80}\\[.5 cm]

=175\\[.5 cm]

Income = = 175\times \frac{15}{100}\times 100\\[.5 cm]

= Rs. 2625\\[.5 cm]

Change of income = Rs. 2625 – Rs. 2000

=Rs.625

Q. 46 (a) The sum of three numbers is 20. If we multiply the first by 2 and add the second number and subtract the third we get 23. If we multiply the first by 3 and add second and third to it, we get 46. By using matrix inversion method find the numbers.

Let the three numbers be x, y, and z.
x + y + z = 20
2x + y – z = 23
3x + y + z = 46
The system can be written as:

\begin{bmatrix} 1& 1 & 1\\ 2 & 1 & -1\\ 3& 1 & 1 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 20\\ 23\\ 46 \end{bmatrix}\\[.5 cm]

We know that AX=B where A= \begin{bmatrix} 1& 1 & 1\\ 2 & 1 & -1\\ 3& 1 & 1 \end{bmatrix} X= \begin{bmatrix} x\\ y\\ z \end{bmatrix} B =\begin{bmatrix} 20\\ 23\\ 46 \end{bmatrix}\\[.5 cm]

To find \left | A \right | \\[.5 cm]

1 \begin{vmatrix} 1& -1\\ 1 & 1 \end{vmatrix}-1 \begin{vmatrix} 2& -1\\ 3 & 1 \end{vmatrix}+1 \begin{vmatrix} 2& 1\\ 3 & 1 \end{vmatrix}\\[.5 cm]

=1(1+1)-1(2+3)+1(2-3)

= 1(2)-1(5)+1(-1)

=2-5-1

=4

A_{11} = \begin{bmatrix} 1 & -1 \\ 1 & 1 \\ \end{bmatrix}=2 \\[.5 cm]

A_{12} = \begin{bmatrix} 2 & -1 \\ 3 & 1 \\ \end{bmatrix}=-5 \\[.5 cm]

A_{13} = \begin{bmatrix} 2 & 1 \\ 3 & 1 \\ \end{bmatrix}=-1 \\[.5 cm]

A_{21} = \begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}=0 \\[.5 cm]

A_{22} = \begin{bmatrix} 1 & 1 \\ 3 & 1 \\ \end{bmatrix}=-2\\[.5 cm]

A_{23} = \begin{bmatrix} 1 & 1 \\ 3 & 1 \\ \end{bmatrix}=2\\[.5 cm]

A_{31} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \\ \end{bmatrix}=-2 \\[.5 cm]

A_{32} = \begin{bmatrix} 1 & 1 \\ 2 & -1 \\ \end{bmatrix}=3 \\[.5 cm]

A_{33} = = \begin{bmatrix} 1 & 1 \\ 2 & 1 \\ \end{bmatrix}=-1 \\[.5 cm]

A_{ij} = \begin{bmatrix} 2 & -5 &-1 \\ 0 & -2 & 2\\ -2& 3 & -1 \end{bmatrix}\\[.5 cm]

(A_{ij})^{T} = \begin{bmatrix} 2&0 &-2 \\ -5& -2 &3 \\ -1 & 2 & -1 \end{bmatrix} \\[.5 cm]

A^{-1} =\frac{1}{\left |A \right |} adj A \\[.5 cm]

=\frac{-1}{4} \begin{bmatrix} 2&0 &-2 \\ -5& -2 &3 \\ -1 & 2 & -1 \end{bmatrix} \\[.5 cm]

X=A^{-1}B\\[.5 cm]

\begin{bmatrix} x\\ y\\ z \end{bmatrix} =\frac{-1}{4} \begin{bmatrix} 2&0 &-2 \\ -5& -2 &3 \\ -1 & 2 & -1 \end{bmatrix} \begin{bmatrix} 20\\ 23\\ 46 \end{bmatrix}\\[.5 cm]

=\frac{-1}{4} \begin{bmatrix} 40+0-92 \\ -100-46+138 \\ -20+46-46 \end{bmatrix}\\[.5 cm]

=\frac{-1}{4} \begin{bmatrix} -52 \\ -8 \\ -20 \end{bmatrix}\\[.5 cm]

=\frac{-1}{4} \begin{bmatrix} 13\\ 2 \\ 5 \end{bmatrix}\\[.5 cm]

Therefore the numbers are 13,2 and 5.

Q.46 (b) A factory has 3 machines A1, A2, A3 producing 1000, 2000, 3000 screws per day respectively. A1 produces 1% defectives, A2 produces 1.5% and A3 produces 2% defectives. A screw is chosen at random at the end of a day and found defective. What is the probability that it comes from machines A1?

P (A_{1}) \:= P(that\: the\: machine\: A_{1} \: produces\: screws)\: = \large \frac{1000}{6000}=\frac{1}{6}\\[.5 cm]

P (A_{2}) \:= P(that\: the\: machine\: A_{2} \: produces\: screws)\: = \large \frac{2000}{6000}=\frac{1}{3}\\[.5 cm]

P (A_{3}) \:= P(that\: the\: machine\: A_{3} \: produces\: screws)\: =\large \frac{3000}{6000}=\frac{1}{2}\\[.5 cm]

Let B be the event that the chosen screw is defective.

P(B/A_{1}) = P(that\: the\: defective \: screw\: is\: from \: machine\: A_{1}= 0.01\\[.5 cm]

P(B/A_{2}) = P(that\: the\: defective \: screw\: is\: from \: machine\: A_{2}= 0.015 \\[.5 cm]

P(B/A_{3}) = P(that\: the\: defective \: screw\: is\: from \: machine\: A_{3}= 0.02 \\[.5 cm]

P(A_{1}/B) = \Large \frac{P(A_{1}) P(B/A_{1})}{P(A_{1}) P(B/A_{1})+P(A_{2}) P(B/A_{2})+P(A_{3}) P(B/A_{3})}\\[.5 cm]

\Large \frac{\frac{1}{6}(0.01)}{\frac{1}{6}(0.01)+\frac{1}{3}(0.015)+\frac{1}{2}(0.02)} \\[.5 cm]

= \Large \frac{.001}{.01+.03+.06}\\[.5 cm]

= \Large \frac{.01}{.1} = \frac{1}{10}\\[.5 cm]

Q. 47 (a) Find the equation of the circle passing through the points (1,0), (-1,0) and (0,1).

Let the equation of the circle be x^{2} + y^{2} + 2gx + 2fy + c = 0\\[.5 cm]

The circle passes through the point (1,0)

1+2g+c=0\cdots (1) \[.5 cm]

1-2g+c=0\cdots (2) \[.5 cm]

1+2f+c=0\cdots (3) \\[.5 cm]

Solving (1) (2) and (3) we get g=0; f=0; and c=-1

Substituting these values in the equation of the circle we get the equation of the circle x^{2}+y^{2}=1 \\[.5 cm] .

Q. 47 (b). Resolve into partial fraction:\large \frac{x-2}{(x+2)(x-1)^{2}} \\[.5 cm]

Let \large \large \frac{x-2}{(x+2)(x-1)^{2}}= \frac{A}{x+2}+\frac{B}{x-1}+\frac{c}{(x-1)^{2}}\cdots (1) \\[.5 cm]

Let \large \frac{x-2}{(x+2)(x-1)^{2}}= \frac{A(x-1)^{2}+{B(x-1)(x+2)}+C(x+2)}{(x+2)(x-1)^{2}} \\[.5 cm]

\large (x-2) = A(x-1)^{2}+{B(x-1)(x+2)+C(x+2)}\cdots (2) \\[.5 cm]

Put \: x=1\: in \cdots (2) \\[.5 cm]

1-2=A(1-1)^{2}+B(1-1)(1+2)+C(1+2) \\[.5 cm]

-1=0+0+3C \\[.5 cm]

C=\frac{-1}{3} \\[.5 cm]

Put \: x=-2\: in \cdots (2) \\[.5 cm]

-2-2=A(-2-1)^{2}+B(-2-1)(-2+2)+C(-2+2)) \\[.5 cm]

-4=A(-3)^{2}+0+0 \\[.5 cm]

-4=9A \\[.5 cm]

A=\frac{-4}{9} \\[.5 cm]

Put \: x=0\: in \cdots (2) \\[.5 cm]

(0-2) = A(0-1)^{2}+{B(0-1)(0+2)+C(0+2)} \\[.5 cm]

We get -2 = A -2B + 2C \\[.5 cm]

Substituting A=\frac{-4}{9} and \:C=\frac{-1}{3}, we\: get\: \\[.5 cm]

We get -2 = \frac{-4}{9} -2B + 2(\frac{-1}{3}) \\[.5 cm]

We get -2+ \frac{4}{9}+\frac{2}{3}= -2B \\[.5 cm]

We get -2+ \frac{4}{9}+\frac{2}{3}= -2B \\[.5 cm]

\frac{-18+4+6}{9}= -2B \\[.5 cm]

\frac{-8}{9}= -2B \\[.5 cm]

B =\frac{4}{9} \\[.5 cm]

Substituting the values of A, B and C, we get the partial equation:

\large \large \frac{x-2}{(x+2)(x-1)^{2}}= \frac{4}{9(x-1)}-\frac{4}{9(x+2)}-\frac{1}{3(x-1)^{2}} \\[.5 cm]

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