By QB365 on 31 Dec, 2022
QB365 provides a detailed and simple solution for every Possible Questions in Class 12 Business Maths Subject - Revision Model Question Paper, English Medium. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.
12th Standard
Business Maths and Statistics
PART-I
Note : i) All Questions Are Compulsory.
ii) Choose The Most Suitable Answer From The Given
Four Correct Alternatives.
if T = \(_{ B }^{ A }\left( \begin{matrix} \overset { A }{ 0.4 } & \overset { B }{ 0.6 } \\ 0.2 & 0.8 \end{matrix} \right) \) is a transition probability matrix, then at equilibrium A is equal to ________.
\(\frac { 1 }{ 4 } \)
\(\frac { 1 }{ 5 } \)
\(\frac { 1 }{ 6 } \)
\(\frac { 1 }{ 8 } \)
If the number of variables in a non-homogeneous system AX = B is n, then the system possesses a unique solution only when _______.
\(\rho (A)=\rho (A,B)>n\)
\(\rho(A)=\rho(A, B)=n\)
\(\rho (A)=\rho (A,B) < n\)
none of these
ഽ\(\frac { { e }^{ x } }{ { e }^{ x }+1 } \) dx is _______.
log\(\left| \frac { { e }^{ x } }{ { e }^{ x }+1 } \right| +c\)
log\(\left| \frac { { e }^{ x }+1 }{ { e }^{ x } } \right| +c\)
log\(\left| { e }^{ x } \right| +c\)
log\(\left| { e }^{ x }+1 \right| +c\)
If MR and MC denotes the marginal revenue and marginal cost functions, then the profit functions is ________.
P = ഽ(MR − MC) dx + k
P = ഽ(MR + MC) dx + k
P = ഽ(MR)(MC)dx + k
P = ഽ(R −C)dx + k
The integrating factor of the differential equation \(\frac{dx}{dy}+Px=Q\) is ______.
eഽPdx
\(\int P d x\)
ഽPdy
eഽPdy
Which of the following is the homogeneous differential equation?
(3x−5)dx = (4y−1)dy
xy dx−(x3+y3)dy = 0
y2dx+(x2 − xy − y2)dy = 0
(x2+y)dx = (y2+x)dy
A formula or equation used to represent the probability distribution of a continuous random variable is called ________.
probability distribution
distribution function
probability density function
mathematical expectation
If Z is a standard normal variate, the proportion of items lying between Z = –0.5 and Z = –3.0 is ________.
0.4987
0.1915
0.3072
0.3098
An estimate of a population parameter given by two numbers between which the parameter would be expected to lie is called an………..interval estimate of the parameter.
point estimate
interval estimation
standard error
confidence
A typical control charts consists of ________.
CL, UCL
CL, LCL
CL, LCL, UCL
UCL, LCL
If number of sources is not equal to number of destinations, the assignment problem is called______.
balanced
unsymmetric
symmetric
unbalanced
\(\int { { 3 }^{ x+2 } } \) dx = ______________ +c
\(\frac { { 3 }^{ x } }{ log3 } \)
\(\frac { 9\left( { 3 }^{ x } \right) }{ log3 } \)
\(\frac { 3.{ 3 }^{ x } }{ log3 } \)
\(\frac { { 3 }^{ x } }{ 9log3 } \)
Choose the odd one out
\(\int_{0}^{\infty} e^{-t} d t\)
\(\int_{0}^{\infty} e^{-t} d t\)
\(\int_{0}^{\infty} x^{n-1} e^{-x} d x\)
\( \int_{0}^{\infty} e^{-4 x} x^{4} d x\)
The area bounded by the curve y = ex, the X-axis and the lines x = 0 and x = 2 is__________.
e2-1
e2+1
e2
e2-2
The solution of the equation of the type \(\frac{d x}{d y}+P x=\mathbf{Q}\) (P and Q are function of y) is _______________
\(\mathbf{y}=\int Q e^{\int p d x} d y+c\)
\(y e^{\int p d x}=\int Q e^{\int P d x} d x+c\)
\(x e^{\int_{P d} y}=\int Q e^{\int P d y} d y+c\)
\(x e^{\int p d y}=\int Q e^{\int P d x} d x+c\)
The degree of the differential equation \(\left(\frac{d^{2} y}{d x^{3}}\right)^{3}+\left(\frac{d y}{d x}\right)^{2}+\sin \left(\frac{d y}{d x}\right)+1=0 \text { is }\)________________
3
2
1
not defined
The nationality of the mathematician Joseph Louis Laguange is _________
German
Spain
Italian
French
If f(x) = kx(1-x), 0< x < 1 is a p.d.f. then the value of k is ________.
\(\frac{1}{5}\)
\(\frac{2}{5}\)
\(\frac{3}{5}\)
6
If a random variable X has the following probability distribution
\(\begin{array}{|c|c|c|c|c|} \hline x & -1 & -2 & 1 & 2 \\ \hline p(x) & 1 / 3 & 1 / 6 & 1 / 6 & 1 / 3 \\ \hline \end{array}\)Then the expected value of X-is _________
\(\frac{3}{2}\)
\(\frac{1}{6}\)
\(\frac{1}{2}\)
\(\frac{1}{3}\)
The point estimate variance of 6.33, 6.37, 6.36, 6.32, 6.37 is
0.0022
0.00055
0.0055
0.055
PART - II
Note : Answer Any Seven Questions and Question.No:30 is Compulsory.
Find the rank of the matrix A =\(\left( \begin{matrix} 1 & -3 \\ 9 & 1 \end{matrix}\begin{matrix} 4 & 7 \\ 2 & 0 \end{matrix} \right) \)
Integrate the following with respect to x.
\(\frac { { e }^{ 2x } }{ { e }^{ 2x }-2 } \)
Using Integration, find the area of the region bounded the line 2y + x = 8, the x axis and the lines x = 2, x = 4.
Solve the following differential equations
\(\frac{d^2 y}{dx^2}-2k\frac{dy}{dx}+k^2y = 0\)
Evaluate Δ (log ax).
Prove that, V(X+b) = V(X)
Define critical value.
Construct a forward difference table for \(\mathrm{y}=\mathrm{f}(x)=x^{2}+2 x+2 \text { for } x=1,2,3,4\)
Find the probability distribution of X, the number of heads in 2 tosses of a coin
The following data shows the value of sample mean (\(\bar{X}\)) and the range R for 10 samples of size 5 each. Calculate the control limits for : mean chart and range chart.
Sample No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Mean \(\bar{X}\) | 11.2 | 11.8 | 10.8 | 11.6 | 11.0 | 9.6 | 10.4 | 9.6 | 10.6 | 10.0 |
Range | 7 | 4 | 8 | 5 | 7 | 4 | 8 | 4 | 7 | 9 |
(Given for n = 5, A2 = .577, D3 = 0, D4 = 2.115)
PART - III
Note : Answer Any Seven Questions and Question.No:40 is Compulsory.
Find the rank of the matrix \(\left( \begin{matrix} 0 & -1 & 5 \\ 2 & 4 & -6 \\ 1 & 1 & 5 \end{matrix} \right) \)
Akash bats according to the following traits. If he makes a hit (S), there is a 25% chance that he will make a hit his next time at bat. If he fails to hit (F), there is a 35% chance that he will make a hit his next time at bat. Find the transition probability matrix for the data and determine Akash’s long- range batting average.
Integrate the following with respect to x.
x log x
Evaluate the following integrals:
ഽ\(\frac { dx }{ { { 2-3x-2x }^{ 2 } } } \)
The marginal cost function MC = 2 + 5ex Find AC.
Given y3 = 2, y4 = −6, y5 = 8, y6 = 9 and y7 = 17 Calculate Δ4y3
Two unbiased dice are thrown simultaneously and sum of the upturned faces considered as random variable. Construct a probability mass function.
What is the probability of guessing correctly atleast six of the ten answers in a TRUE/FALSE objective test?
Solve: \(\left(x^{2}-y x^{2}\right) d y+\left(y^{2}+x^{2} y^{2}\right) d x=0\)
A random variable X has the folowing probability distribution
x | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
p(x) | a | 3a | 5a | 7a | 9a | 11a | 13a | 15a | 17a |
(i) Find a
\(\text {(ii) } \mathrm{P}(x<3), \mathrm{P}(x>3) \text { and } \mathrm{P}(0<x<5)\)
PART-IV
Note : i ) Write all the following questions.
The total number of units produced (P) is a linear function of amount of over times in labour (in hours) (l), amount of additional machine time (m) and fixed finishing time (a)
i.e, P = a + bl + cm
From the data given below, find the values of constants a, b and c
Day | Production (in Units P) |
Labour (in Hrs l) |
Additional Machine Time (in Hrs m) |
Monday Tuesday Wednesday |
6,950 6,725 7,100 |
40 35 40 |
10 9 12 |
Estimate the production when overtime in labour is 50 hrs and additional machine time is 15 hrs.
Integrate the following with respect to x.
\(e^{x}\left[\frac{x-1}{(x+1)^{3}}\right]\)
The marginal cost of production of a firm is given by C'(x) = 20 + \(\frac { x }{ 20 } \) the marginal revenue is given by R'(x) = 30 and the fixed cost is Rs. 100. Find the profit function
Solve x2ydx-(x3+y3)dy = 0
Find the missing entries from the following
x | 0 | 1 | 2 | 3 | 4 | 5 |
y = f(x) | 0 | - | 8 | 15 | - | 35 |
The marks obtained in a certain exam follow normal distribution with mean 45 and SD 10. If 1,300 students appeared at the examination, calculate the number of students scoring
(i) less than 35 marks and
(ii) more than 65 marks.
Calculate the seasonal index for the quarterly production of a product using the method of simple averages.
Year | I Quarter | II Quarter | III Quarter | IV Quarter |
2005 | 255 | 351 | 425 | 400 |
2006 | 269 | 310 | 396 | 410 |
2007 | 291 | 332 | 358 | 395 |
2008 | 198 | 289 | 310 | 357 |
2009 | 200 | 290 | 331 | 359 |
2010 | 250 | 300 | 350 | 400 |
Obtain an initial basic feasible solution to the following transportation problem using Vogel’s approximation method.
A new transit system has just gone into operation in a city. Of those who use the transit system this year, 10% will switch over to using their own car next year and 90% will continue to use the transit system. Of those who use their cars this year, 80% will continue to use their cars next year and 20% will switch over to the transit system. Suppose the population of the city remains constant and that 50% of the commuters use the transit system and 50% of the commuters use their own car this year,
(i) What percent of commuters will be using the transit system after one year?
(ii) What percent of commuters will be using the transit system in the long run?
The total cost of production y and the level of output x are related to the marginal cost of production by the equation (6x2 + 2y2) dx - (x2+ 4xy) dy = 0. What is the relation between total cost and output if y = 2, when x = 1
A continuous random variable has the following p.d.f, \(f(x)= \begin{cases}k x^{2}, & 0 \leq x \leq 10 \\ 0, & \text { otherwise }\end{cases}\) find k and evaluate
\((i)\ \mathrm{P}(0.2 \leq x \leq 0.5) \)
(ii) \( \mathrm{P}(x \leq 3)\)
A sample of 400 students is found to have a mean height of 171.38 cms. Can it reasonably be regarded as a sample from a large population with mean height of 171.17 cms and standard deviation of 3.3 cms (Test at 5% level)
The followingdata relateto the life(inhours) of 10 samples of 6 electricbulbs each drawn at an intervalof one hour from a production process.Draw the controlchart for \(\overline { X } \) and \(\overline { R } \) and comment.
Sample No | Lifetime (inhour) | |||||
1 | 2 | 3 | 4 | 5 | 6 | |
1 | 620 | 687 | 666 | 689 | 738 | 686 |
2 | 501 | 585 | 524 | 585 | 653 | 668 |
3 | 673 | 701 | 686 | 567 | 619 | 660 |
4 | 646 | 626 | 572 | 628 | 631 | 743 |
5 | 494 | 984 | 659 | 643 | 660 | 640 |
6 | 634 | 755 | 625 | 582 | 683 | 555 |
7 | 619 | 710 | 664 | 693 | 770 | 534 |
8 | 630 | 723 | 614 | 535 | 550 | 570 |
9 | 482 | 791 | 533 | 612 | 497 | 499 |
10 | 706 | 524 | 626 | 503 | 661 | 754 |
(For n = 6,A2= 0.483,D3 = 0,D4 = 2.004)
Solve the transportation problem using
(i) North west corner Rule
(ii) Least cost method
(iii) Vogel Approximation method
Answers
\(\frac { 1 }{ 4 } \)
\(\rho(A)=\rho(A, B)=n\)
log\(\left| { e }^{ x }+1 \right| +c\)
P = ഽ(MR − MC) dx + k
eഽPdy
y2dx+(x2 − xy − y2)dy = 0
probability density function
0.3072
interval estimation
CL, LCL, UCL
unbalanced
\(\frac { 9\left( { 3 }^{ x } \right) }{ log3 } \)
\(\int_{0}^{\infty} e^{-t} d t\)
e2-1
\(x e^{\int_{P d} y}=\int Q e^{\int P d y} d y+c\)
not defined
Italian
\(\frac{1}{5}\)
\(\frac{1}{6}\)
0.00055
Given A =\(\left( \begin{matrix} 1 & -3 \\ 9 & 1 \end{matrix}\begin{matrix} 4 & 7 \\ 2 & 0 \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & -3 \\ 0 & 28 \end{matrix}\begin{matrix} 4 & 0 \\ -34 & -63 \end{matrix} \right) { R }_{ 2 }\rightarrow { R }_{ 2 }-9{ R }_{ 1 }\)
\(-\left( \begin{matrix} 1 & -3 \\ 0 & 0 \end{matrix}\begin{matrix} 4 & 0 \\ \frac { 10 }{ 3 } & -63 \end{matrix} \right) { R }_{ 2 }\rightarrow { R }_{ 2 }+\frac { 28 }{ 3 } .{ R }_{ 1 }\)
The last equivalent matrix is in echelon form and there are 2 non-zero rows
\(\therefore \rho (A)=2\)
\(Let\ I=\int { \frac { { e }^{ 2x } }{ { e }^{ 2x }-2 } } \)
\(put\ t={ e }^{ 2x }-2\)
\(\Rightarrow dt=2{ e }^{ 2x }dx\)
\(\Rightarrow \frac { dt }{ 2 } ={ e }^{ 2x }dx\)
\(\therefore I=\int { \frac { \frac { dt }{ 2 } }{ t } } =\frac { \log { \left| t \right| } }{ 2 } +c\)
\(=\log { \frac { \left| { e }^{ 2x }-2 \right| }{ 2 } } +c\)
2y + x = 8
x | 0 | 8 |
y | 4 | 0 |
Given 2y + x = 8
2y = 8-x
y = \(\frac{1}{2}\) (8-x)
Given limits are x = 2 and x = 4
Area of the shaded region between the given limits
\(A=\int _{ a }^{ b }{ y\quad dx } =\int _{ 2 }^{ 4 }{ \frac { 1 }{ 2 } (8-x)dx } \)
\(\frac { 1 }{ 2 } \int _{ 2 }^{ 4 }{ (8-x)dx } =\frac { 1 }{ 2 } { \left[ 8x-\frac { { x }^{ 2 } }{ x } \right] }_{ 2 }^{ 4 }\)
\(=\frac { 1 }{ 2 } \left[ \left( 8(4)-\frac { { 4 }^{ 2 } }{ 2 } \right) \left( 8(2)-\frac { { 2 }^{ 2 } }{ 2 } \right) \right] \)
\(=\frac { 1 }{ 2 } [(32-8)-(16-21)]\)
\(=\frac { 1 }{ 2 } [24-14]\)
A = 5 sq. units.
The auxiliary equation is m2-2km+k2 = 0
⇒ (m - k)2 = 0
⇒ m = k.k
The roots are real and equal
∴ Complimentary function CF is (Ax + B)ekx
The general solution is y= (Ax + B)ekx
log(1+h/ax)
Prove that V(X+b) = V(X)
LHS = V(X+b) = E[(X+b)-E(X+b)]2
= E[(X+b)-E(X+b)]2
[∵E(b) = b]
= E[X-E(X)]2
= E[X-\(\bar { X } \)]2 [∵\(\bar { X } \) = E(X)]
= Var(X) = RHS
Hence proved.
The value of test statistic which separates the critical (or rejection) region and the acceptance region is called the critical value or significant value.
\(\mathrm{y}=\mathrm{f}(x)=x^{2}+2 x+2\)
x | y | \(\Delta y\) | \(\Delta^2 y\) | \(\Delta^3 y\) |
1 | 5 | |||
5 | ||||
2 | 10 | 2 | ||
7 | ||||
3 | 17 | 2 | ||
9 | 0 | |||
4 | 26 | 2 | ||
11 | ||||
5 | 37 |
Let X be the random variable denoting the number of heads.
X takes values 0, 1, 2
P(X = 0) = P(TT)
\(=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)
P(X = 1) = 2P(HT)
\(=2 \times \frac{1}{2} \times \frac{1}{2}=\frac{2}{4}\)
P(X = 2) = P(HH)
\(=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)
The required probability distribution X is
X | 0 | 1 | 2 |
P(X) | \(\frac{1}{4}\) | \(\frac{1}{2}\) | \(\frac{1}{4}\) |
\(\bar{\bar{X}}\) = \(\frac{11.2 + 11.8 + 10.8 + 11.6 + 11.0+ 9.6 + 10.4 + 9.6 + 10.6 + 10.0}{10}\)
\(=\frac{106.6}{10}=10.66\)
\(\bar{R}=\frac{7+4+8+5+7+4+8+4+7+9}{10}\)
\(=\frac{63}{10}=6.3\)
Control limits for mean chart
UCL = \(\bar{\bar{X}}\) + A2\(\bar{R}\)
= 10.66 + .577(6.3) = 14.295
CL = \(\bar{\bar{X}}\) = 10.66
Control limits for R-chart
UCL = D2\(\bar{R}\) = 2.115 \(\times\) 6.3
= 13.324
CL = \(\bar{R}\) = 6.3
LCL = D3\(\bar{R}\) = 0
Let A =\(\left( \begin{matrix} 0 & -1 & 5 \\ 2 & 4 & -6 \\ 1 & 1 & 5 \end{matrix} \right) \)
Order of A is 3 \(\times\) 3
∴ \(\rho \)(A)\(\le \)3
Consider the third order minor \(\left| \begin{matrix} 0 & -1 & 5 \\ 2 & 4 & -6 \\ 1 & 1 & 5 \end{matrix} \right| =6\neq 0\)
There is a minor of order 3, which is not zero
\(\therefore \rho (A)=3\)
The Transition probability matrix is T = \(\left( \begin{matrix} 0.25 & 0.75 \\ 0.35 & 0.65 \end{matrix} \right) \)
At equilibrium, (S F) \(\left( \begin{matrix} 0.25 & 0.75 \\ 0.35 & 0.65 \end{matrix} \right) \) = (S F)
where S + F = 1
0.25 S + 0.35 F = S
0.25 S + 0.35 (1 – S) = S
On solving this, we get S = \(\frac { 0.35 }{ 1.10 } \)
⇒ S = 0.318 and F = 0.682
\(\therefore \) Akash’s batting average is 31.8%
Let I = ∫ x logx dx
Let u = log x; dv = x dx
\(du=\frac { 1 }{ x } dx;v=\frac { { x }^{ 2 } }{ 2 } \)
Using integration by parts,
∫ udv = uv - ∫ vdu
\(\Rightarrow \int { x\ \log\ x\ dx=\log x\left( \frac { { x }^{ 2 } }{ 2 } \right) } -\int { \frac { { x }^{ 2 } }{ 2 } } \frac { 1 }{ x } dx\)
\(=\frac { { x }^{ 2 } }{ 2 } \log x\frac { 1 }{ 2 } \int { x } dx\)
\(=\frac { { x }^{ 2 } }{ 2 } \log x-\frac { 1 }{ 2 } \frac { { x }^{ 2 } }{ 2 } +c\)
\(=\frac { { x }^{ 2 } }{ 2 } \log x-\frac { { x }^{ 2 } }{ 4 } +c\)
\(=\frac { { x }^{ 2 } }{ 2 } \left( \log\ x-\frac { 1 }{ 2 } \right) +c\)
\(I=\int { \cfrac { dx }{ 2-3x-{ 2x }^{ 2 } } } \)= \(-\cfrac { 1 }{ 2 } \int { \cfrac { dx }{ { x }^{ 2 }+\frac { 3 }{ 2 } x-1 } } \)
\(=-\cfrac { 1 }{ 2 } \int { \cfrac { dx }{ \left( x+\frac { 3 }{ 4 } \right) ^{ 2 }-1-\frac { 9 }{ 16 } } } \)
= \(-\cfrac { 1 }{ 2 } \int { \cfrac { dx }{ \left( x+\frac { 3 }{ 4 } \right) ^{ 2 }-\frac { 25 }{ 16 } } } \)
= \(\cfrac { 1 }{ 2 } \int { \cfrac { dx }{ \left( \frac { 5 }{ 4 } \right) ^{ 2 }-\left( x+\frac { 3 }{ 4 } \right) ^{ 2 } } } \)
= \(\left[ \because \int { \cfrac { dx }{ { a }^{ 2 }-{ x }^{ 2 } } =\cfrac { 1 }{ 2a } log\left( \cfrac { a+x }{ a-x } \right) } +c \right] \)
= \(\cfrac { 1 }{ 2 } \left[ \cfrac { 1 }{ 2\times \frac { 5 }{ 4 } } log\left[ \cfrac { \frac { 5 }{ 4 } +x+\frac { 3 }{ 4 } }{ \frac { 5 }{ 4 } -x\frac { 3 }{ 4 } } - \right] \right] +c\)
= \(\cfrac { 1 }{ 5 } log\left[ \cfrac { 2+x }{ 1-2x } \right] +c\)
Given MC = 2 + 5ex
\(C=\int { MC } dx+k\)
\(=\int { (2+5{ e }^{ x })dx } +k\)
= 2x + 5ex + k
Average cost = \(\frac { C }{ x } =\frac { 2x+5{ e }^{ x }+95 }{ x } \)
\(AC=2+\frac { 5{ e }^{ x } }{ x } +\frac { 95 }{ x } \)
Given y3 = 2, y4 = −6, y5 = 8, y6 = 9 and y7 = 17
Δ4y3 = (E−1)4y3
= (E4 − 4E3 + 6E2 − 4E+1)y3
= E4y3 − 4E3y3 + 6E2y3− 4Ey3 + y3
= y7 − 4y6 + 6y5 −4y4+ y3
= 17 – 4(9) + 6(8) –4(–6) + 2
= 17 – 36 + 48 + 24 + 2 = 55
Sample space \((s)=\left\{ \begin{matrix} (1,1) & (1,2) & (1,3) \\ (2,1) & (2,2) & (2,3) \\ \begin{matrix} (3,1) \\ (4,1) \\ \begin{matrix} (5,1) \\ (6,1) \end{matrix} \end{matrix} & \begin{matrix} (3,2) \\ (4,2) \\ \begin{matrix} (5,2) \\ (6,2) \end{matrix} \end{matrix} & \begin{matrix} (3,3) \\ (4,3) \\ \begin{matrix} (5,3) \\ (6,3) \end{matrix} \end{matrix} \end{matrix}\begin{matrix} (1,4) & (1,5) & (1,6) \\ (2,4) & (2,5) & (2,6) \\ \begin{matrix} (3,4) \\ (4,4) \\ \begin{matrix} (5,4) \\ (6,4) \end{matrix} \end{matrix} & \begin{matrix} (3,5) \\ (4,5) \\ \begin{matrix} (5,5) \\ (6,5) \end{matrix} \end{matrix} & \begin{matrix} (3,6) \\ (4,6) \\ \begin{matrix} (5,6) \\ (6,6) \end{matrix} \end{matrix} \end{matrix} \right\} \)
Total outcomes : n(S) = 36
Probability p of guessing an answer correctly is p = \(\frac{1}{2}\)
⇒ q = \(\frac{1}{2}\)
Probability of guessing correctly x answers in 10 questions
\(P(X=x)=p(x)^{ n }{ C }_{ x }{ q }^{ n-x }=10Cx\left( \frac { 1 }{ 2 } \right) \left( \frac { 1 }{ 2 } \right) \)
The required probability P(X ≥ 6) = P(6) + P(7) + P(8) + P(9) + P(10)
\(={ \left( \frac { 1 }{ 2 } \right) }^{ 10 }\left[ { 10C }_{ 6 }+{ 10C }_{ 7 }+{ 10C }_{ 8 }+{ 10C }_{ 9 }+{ 10C }_{ 10 } \right] \)
\(=\left[ \frac { 1 }{ 1024 } \right] [210+120+45+10+1]\)
\(=\frac { 193 }{ 512 } \)
\(
\left(x^{2}-\mathrm{y} x^{2}\right) \mathrm{dy} =-\left(\mathrm{y}^{2}+x^{2} \mathrm{y}^{2}\right) \mathrm{d} x
\)
\(x^{2}(1-\mathrm{y}) \mathrm{dy} =-\mathrm{y}^{2}\left(1+x^{2}\right) \mathrm{d} x
\)
\(\int \frac{1-y}{y^{2}} d y =-\int \frac{\left(1+x^{2}\right)}{x^{2}} d x
\)
\(\int \frac{d y}{y^{2}}-\int \frac{y d y}{y^{2}} =-\int x^{-2} d x-\int d x
\)
\(\int y^{-2} d y-\int \frac{d y}{y} =-\int x^{-2} d x-\int d x
\)
\(\frac{y^{-1}}{-1}-\log |y| =\frac{x^{-1}}{1}-x+c
\)
\(\log |y|+\frac{1}{y}+\frac{1}{x}-x =\mathrm{c}
\)
\((i)\ \sum p_{i}=1
a+3 a+5 a+7 a+9 a \div 11 a+13 a+15 a + 17 a=1
\)
\(81 a=1
\)
\( a=\frac{1}{81}\)
\((ii)\ \mathrm{P}(x<3)=\mathrm{P}(X=0)+\mathrm{P}(X=1)+\mathrm{P}(X=2)
\)
\(=a+3 a+5 a
\)
\(=9 \mathrm{a}=\frac{9}{81}=\frac{1}{9}
\)
\(\mathrm{P}(x>3)=\mathrm{P}(X=4)+\mathrm{p}(X=5)+\mathrm{p}(X=7)
+\mathrm{p}(X=8)
\)
\(=9 a+11 a+13 a+15 a+17 a
\)
\(=65 \mathrm{a}=\frac{65}{81}
\)
\(\mathrm{P}(0<x<5)=\mathrm{P}(X=1)+\mathrm{P}(X=2)+
\mathrm{P}(X=3)+\mathrm{P}(X=4)
\)
\(=3 a+5 a+7 a+9 a
\)
\(=24 \mathrm{a}=\frac{24}{81}=\frac{8}{27}\)
We have, P = a + bl + cm
Putting above values we have
6,950 = a + 40b + 10c
6,725 = a + 35b + 9c
7,100 = a + 40b + 12c
The Matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 40 & 10 \\ 1 & 35 & 9 \\ 1 & 40 & 12 \end{matrix} \right) \left( \begin{matrix} a \\ b \\ c \end{matrix} \right) =\left( \begin{matrix} 6950 \\ 6725 \\ 7100 \end{matrix} \right) \)
Augmented matrix [A,B] | Elementary Transformation |
\(\left( \begin{matrix} 1 & 40 & 10 \\ 1 & 35 & 9 \\ 1 & 40 & 12 \end{matrix}\begin{matrix} 6950 \\ 6725 \\ 7100 \end{matrix} \right) \) \(\left( \begin{matrix} 1 & 40 & 10 \\ 0 & -5 & -1 \\ 0 & 0 & 2 \end{matrix}\begin{matrix} 6950 \\ -225 \\ 150 \end{matrix} \right) \) |
\({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }\) |
\(\rho (A)=3,\rho ([A,B])=3\) |
\(\therefore \) The given system is equivalent to the matrix equation
\(\left( \begin{matrix} 1 & 40 & 10 \\ 0 & -5 & -1 \\ 0 & 0 & 2 \end{matrix} \right) \left( \begin{matrix} a \\ b \\ c \end{matrix} \right) =\left( \begin{matrix} 6950 \\ -225 \\ 150 \end{matrix} \right) \)
a + 40b + 10c = 6950 (1)
-5b - c = -225 (2)
2c = 150 (3)
c- = 75
Now, (2) \(\Rightarrow \) -5b - 75 = -225
b = 30
and (1) \(\Rightarrow \) a + 1200 + 750 = 6950
a = 5000
a = 5000, b = 30, c = 75
\(\therefore \) The production equation is P = 5000 + 301 + 75m
\(\therefore \) Pat l = 50, m = 15 = 5000 + 30(50) + 75(15)
= 7625 units.
\(\therefore \) The production = 7,625 units.
\(Let\ I=\int { { e }^{ x } } \left[ \frac { x-1 }{ { \left( x+1 \right) }^{ 3 } } \right] \)
\(=\int { { e }^{ x } } \left[ \frac { x-1+1-1 }{ { \left( x+1 \right) }^{ 3 } } \right] dx\)
[Adding and subtracting 1 in the numerator]
\(=\int { { e }^{ x } } \left[ \frac { x+1-2 }{ { \left( x+1 \right) }^{ 3 } } \right] dx\)
\(=\int { { e }^{ x } } \left( \frac { x+1 }{ { \left( x+1 \right) }^{ 3 } } -\frac { 2 }{ { \left( x+1 \right) }^{ 3 } } \right) dx\)
\(=\int { { e }^{ x } } \left( \frac { 1 }{ { \left( x+1 \right) }^{ 3 } } -\frac { 2 }{ { { \left( x+1 \right) } }^{ 3 } } \right) dx\)
\(Let\ f\left( x \right) =\frac { 1 }{ { \left( x+1 \right) }^{ 2 } } ={ \left( x+1 \right) }^{ -2 }\)
\(\Rightarrow f^{ ' }\left( x \right) =-2{ \left( x+1 \right) }^{ -2-1 }\)
\(=-2{ \left( x+1 \right) }^{ -3' }\)
\(=\frac { -2 }{ { \left( x+1 \right) }^{ 3 } } \)
\(\therefore I=\int { { e }^{ x }\left[ f\left( x \right) +f^{ ' }\left( x \right) \right] } dx\)
\(={ e }^{ x }\quad f\left( x \right) +c\)
\(={ e }^{ x }\frac { 1 }{ { \left( x+1 \right) }^{ 2 } } +c\)
\(=\frac { { e }^{ x } }{ { \left( x+1 \right) }^{ 2 } } +c\)
Given C'(x) = 20 + \(\frac{x}{20}\)
R'(x) = 30
C'(x) = 20 + \(\frac{x}{20}\)
\(\Rightarrow \int { C'(x) } =\int { \left( 20+\frac { x }{ 20 } \right) dx } \)
\(=20x+\frac { { x }^{ 2 } }{ 40 } +{ k }_{ 1 }\)
Since fixed cost is Rs. 100
When x = 0, C = 100 ⇒ k1 = 100
\(\therefore \ C(x)=20x+\frac { { x }^{ 2 } }{ 40 } +100\) ...(1)
Also, R'(x) = 30
\(\Rightarrow \int { R'(x) } =\int { 30 } dx\)
⇒ R(x) = 30x + k2
When x = 0, R = 0 ⇒ k2 = 0
∴ R(x) = 30x ...(2)
Profit function = R(x) - C(x)
\(=30x-\left( 20x+\frac { { x }^{ 2 } }{ 40 } +100 \right) \)
\(=30x-20x-\frac { { x }^{ 2 } }{ 40 } -100\)
\(P=10x-\frac { { x }^{ 2 } }{ 40 } -100\)
x2y dx = (x3+y3)dy
⇒ \(\frac { dy }{ dx } =\frac { { x }^{ 2 }y }{ { x }^{ 3 }+{ y }^{ 3 } } \)
The numerator and denominator homogeneous functions of degree 3,
So put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
v+\(x\frac { dv }{ dx } =\frac { { x }^{ 2 }vx }{ { x }^{ 3 }+{ v }^{ 3 }{ x }^{ 3 } } =\frac { { x }^{ 3 }-v }{ { x }^{ 3 }(1+{ v }^{ 3 }) } =\frac { v }{ 1+{ v }^{ 3 } } \)
⇒ \(x\frac { dv }{ dx } =\frac { v }{ 1+{ v }^{ 3 } } =\frac { v-v(1+{ v }^{ 3 }) }{ 1+{ v }^{ 3 } } =\frac { v-v-{ v }^{ 4 } }{ 1+{ v }^{ 3 } } \)
= \(\frac { -{ v }^{ 4 } }{ 1+{ v }^{ 3 } } \)
⇒ X\(\frac { dv }{ dx } =\frac { -{ v }^{ 4 } }{ 1+{ v }^{ 3 } } \)
Separating the variable we get,
\(\frac { (1+{ v }^{ 3 })dv }{ { v }^{ 4 } } =-\frac { dx }{ x } \)
⇒ \(\frac { 1 }{ { v }^{ 4 } } dv+\frac { { v }^{ 3 } }{ { v }^{ 4 } } dv=-\frac { dx }{ x } \)
⇒ \({ v }^{ -4 }dv+\frac { 1 }{ v } dv=-\frac { dx }{ x } \)
Integrating both sides we get,
\(\int { { v }^{ -4 } } dv+\int { \frac { 1 }{ v } } dv=-\int { \frac { dx }{ x } } \)
⇒ \(\frac { { v }^{ -3 } }{ -3 } \)+ logv = -log x + C
⇒ \(\frac { -1 }{ 3v^{ 3 } } \) = -log x = -log v + C
⇒ \(\frac { 1 }{ 3v^{ 3 } } \) = log vx + K
Replacing v by\(\frac { y }{ x } \) we get
\(\frac { { x }^{ 3 } }{ 3{ y }^{ 3 } } =log\frac { y }{ x } .x+K\)
⇒ \(\frac { { x }^{ 3 } }{ 3{ y }^{ 3 } } \) = log y + C
Let the missing entries by y1 and y4
Since only four values of f(x) are given, the polynomial which fits the data is of degree 3.
Hence fourth differences are zero.
∴ Δ4yk = 0 ⇒ (E - 1)4yk = 0
(E4 - 4E3 + 6E2- 4E + 1) yk = 0
Put k = 0 in (1) we get,
(E4 - 4E3 + 6E2 - 4E + 1)y0 = 0
⇒ y4 - 4y3 + 5y2 - 4y1 + y0 = 0
⇒ y4 - 4(15) + 6(8) - 4(y1) + 0 = 0
⇒ y4 - 60 + 48 - 4y1 = 0
⇒ y4 - 4y1 - 12 = 0
⇒ y4 - 4y1 = 12 (2)
put k = 1 in (1) we get
(E4- 4E3 + 6E2 - 4E + 1) y1 = 0
⇒ y5 - 4y4 + 6y3 - 4y2 + y1 = 0
⇒ 35 - 4y4 + 6 (15) - 4 (8) +y1 = 0
⇒ 35 - 4y4 + 90 - 32 +y1 = 0
⇒ 4y4 +y1 = -93 (3)
(2) x 4 ➝ | 4y4 - 16y1 | = 48 |
(3) ➝ | -4y4 + y1 | = -93 |
Adding | -15y1 | = -45 |
⇒ y1 = \(\frac{-45}{-15}\) = 3
⇒ y1 = 3.
Substituting y1 = 3 in (2) we get,
y4 - 4(3) = 12
y4 - 12 = 12
⇒ y4 = 12 + 12
⇒ y4 = 24
Hence the missing entries are 3 and 24.
Let X be the normal variate showing the score of the candidate with mean 45 and standard deviation 10.
(i) less than 35 marks
When X = 35
\(Z=\frac { X-\mu }{ \sigma } =\frac { 35-45 }{ 10 } =-1\)
P(X < 35) = P(Z < –1)
P(Z > 1) = 0.5 – P(0 < Z < 1)
= 0.5 – 0.3413
= 0.1587
Expected number of students scoring less than 35 marks are 0.1587 × 1300
= 206
(ii) more than 65 marks
When X = 65
\(Z=\frac { X-\mu }{ \sigma } =\frac { 65-45 }{ 10 } =2.0\)
P(X > 65) = P(Z > 2.0)
0.5 – P(0 < Z < 2.0)
0.5 – 0.4772
= 0.0228
Expected number of students scoring more than 65 marks are 0.0228 x 1300 = 30
Computation of Seasonal Index by the method of simple averages.
Year | I Quarter | II Quarter | III Quarter | IV Quarter |
2005 | 255 | 351 | 425 | 400 |
2006 | 269 | 310 | 396 | 410 |
2007 | 291 | 332 | 358 | 395 |
2008 | 198 | 289 | 310 | 357 |
2009 | 200 | 290 | 331 | 359 |
2010 | 250 | 300 | 350 | 400 |
Quarterly Total |
1463 | 1872 | 2170 | 2321 |
Quarterly Averages |
243.83 | 312 | 361.67 | 386.83 |
S.I for I Quarter = \(\frac{Average\ of \ I\ quarter }{Grand\ average} \times100\)
Grand Average = \(\frac{1304.333}{4}=326.0833\)
S.I for I Q = \(\frac{243.8333}{326.0833} \times100= 74.77;\)
S.I for II Q = \(\frac{312}{326.0833} \times100=95.68;\)
S.I for III Q = \(\frac{361.6667}{326.0833} \times 100=110.91; \)
S.I for IV Q = \(\frac{386.833}{326.0833} \times 100=118.63\)
Here \(\sum { { a }_{ i }=\sum { { b }_{ j } } =80 } \)
(i.e) Total Availability = Total Requirement
\(\therefore\) The given problem is balanced transportation problem.
Hence there exists a feasible solution to the given problem.
First Allocation:
Second Allocation:
Third Allocation:
Fourth Allocation:
Fifth Allocation:
Sixth Allocation:
Thus we have the following allocations:
Transportation schedule :
A⟶I, A⟶II, A⟶III, A⟶IV, B⟶I, C⟶IV, D⟶II
Total transportation cost:
= (6×5)+(6+1)+(17×3)+(5×3)+(15×3)+(12×3)+(19×1)
= 30+6+51+15+45+36+19
= Rs.202
Let A represents the percent of commuters who use the transit system and B represents the percent of commuters who use their own car. Transition probability matrix
Given 50% of commuters use the transit system and 50% of the commuters use their own car this year.
(i) Percentage of commuters after one year
\(\left( \cdot 5\cdot 5 \right) \left( \begin{matrix} \cdot 9 & \cdot 1 \\ \cdot 2 & \cdot 8 \end{matrix} \right) \)
= (-5\(\times\)·9+·5\(\times\).2 ·5\(\times\)·1+·5\(\times\)·8)
= (-45 + ·10 ·05 +.40)
= (-55 - 45)
A = 55% and B = 45%
(ii) Equilibrium will be reached in the long run at equilibrium, we must have
(A B)T = (A B) wher A+B = 1
\(\Rightarrow \left( \begin{matrix} A & B \end{matrix} \right) \left( \begin{matrix} \cdot 9 & \cdot 1 \\ \cdot 2 & \cdot 8 \end{matrix} \right) =\left( \begin{matrix} A & B \end{matrix} \right) \)
\(\left( \begin{matrix} \cdot 9A+\cdot 2B & \cdot 1A+8B \end{matrix} \right) =\left( \begin{matrix} A & B \end{matrix} \right) \)
Equating the corresponding entries on both sides we get,
\(\cdot 9A+\cdot 2B=A\Rightarrow \cdot 9A+\cdot 2(1-A)=A\)
[Since A + B = 1, B = 1 -A]
\(\Rightarrow \cdot 9A+\cdot 2-\cdot 2A=A\)
\(\Rightarrow \cdot 2=A-\cdot 9A+\cdot 2A\)
\(\Rightarrow \cdot 2=A(1-\cdot 9+\cdot 2)\)
\(\Rightarrow \cdot 2=A=(\cdot 3)\)
\(\therefore\) 67% of the commuters will be using the transit system in the long run.
Given \(\left(6 x^{2}+2 y^{2}\right) \mathrm{d} x-\left(x^{2}+4 x y\right) \mathrm{dy}=0\)
\(\frac{d y}{d x}=\frac{6 x^{2}+2 y^{2}}{x^{2}+4 x y}\) is a homogeneous equation in x and y
y = vx
\( \frac{d y}{d x} =v+x \frac{d v}{d x} v+x \frac{d v}{d x}=\frac{6 x^{2}+2 v^{2} x^{2}}{x^{2}+4 x(v x)} =\frac{6+2 v^{2}}{1+4 v} \)
\(x \frac{d v}{d x} =\frac{6+2 v^{2}}{1+4 v}-v \ =\frac{6+2 v^{2}-v-4 v^{2}}{1+4 v} \)
\(x \frac{d v}{d x} =\frac{6-v-2 v^{2}}{1+4 v} \)
\(-\int \frac{(1+4 v)}{6-v-2 v^{2}} d v =-\int \frac{d x}{x} \)
\(\log \left(6-v-2 v^{2}\right) =-\log x+\log \mathrm{k} \)
\( \log \left(6-\frac{y}{x}-\frac{2 y^{2}}{x^{2}}\right)+\log x =\log \mathrm{k} \)
\(\log \left(\frac{6 x^{2}-x y-2 y^{2}}{x^{2}}\right) x =\log \mathrm{k} \)
f is a p.d.f
\( \int_{-\infty}^{\infty} f(x) d x =1 \)
\(k \int_{0}^{10} x^{2} d x =1 \Rightarrow k\left[\frac{x^{3}}{3}\right]_{0}^{10}=1 \)
\(k\left(\frac{1000}{3}\right) \Rightarrow 1 \mathrm{k}=\frac{3}{1000}=0.003 \)
\(\text {(i) } \mathrm{P}(0.2
\( =0.003\left[\frac{x^{3}}{3}\right]_{0.2}^{0.5} \)
\(\text {(ii) } \mathrm{P}(x \leq 3) =\int_{0}^{3} k x^{2} d x \)
= 0.003\(\left[\frac{x^{3}}{3}\right]_{0}^{3} \)
= 0.001(27-0)
= 0.027
\(\mathrm{n}=400,\ \bar{x}=171.38, \ \sigma=3.3\)
Null Hypothesis :
\(\mathrm{H}_{0}: \mu=171.17\)
Alternative Hypothesis :
\(\mathrm{H}_{1}: \mu \neq 171.17\)
Test statistic :
\( \mathrm{Z} =\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \sim N(0,1) \)
\(=\frac{171.38-171.17}{\frac{3.3}{\sqrt{400}}}=1.273 \)
Since |z| = 1.273 < 1.96 we accept null hypothesis at 5% level of significance. Thus the sample of 400 has come from the population with height of 171.17 cms
Sample No | Total | \(\overline { X } \) | R = Xmax - Xmin |
1 | 4086 | 681 | 118 |
2 | 3516 | 586 | 167 |
3 | 3906 | 651 | 134 |
4 | 3846 | 641 | 171 |
5 | 4080 | 680 | 490 |
6 | 3834 | 639 | 200 |
7 | 3990 | 665 | 236 |
8 | 3622 | 604 | 188 |
9 | 3414 | 569 | 309 |
10 | 3774 | 629 | 251 |
6345 | 2264 |
\(\bar{\bar{X}}\) = 634.5, \(\bar{R}\) = 226.4
Control limits for \(\overline { X } \) - chart are
UCL = \(\bar{\bar{X}}\) + A2\(\bar{R}\)
= 634.5+ 0.483x 226.4
= 743.85
CL = 634.5
LCL = \(\bar{\bar{X}}\)- A2\(\bar{R}\) = 525.15
Control limits of R-chart are
UCL = D4\(\bar{R}\) = 2.004 x226.4
= 453.7056
CL = 226.4
LCL = D3\(\bar{R}\) = 0
\(\overline { X } \)-chart
R - chart
Conclusion: Since one point in R-chart lie outside the control limits, the given system is not in control
Total demand = Total supply Penalty
It is a balanced transportation problem
Hence it has a feasible solution
(i) North west corner :
Total cost
\(
=(35 \times 8)+(10 \times 9)+(20 \times 12)+
(20 \times 13)+(10 \times 16)+(30 \times 5)
\)
\(= 280+90+240+260+160+150
\)
= 1180
(ii) Least cost method:
Total optimum cost
\(
= (30 \times 9)+(20 \times 6)+(15 \times 8)+
(20 \times 13)+(10 \times 16)+(30 \times 5)
\)
= 270+120+120+260+160+150
= 1080
(iii) Vogel's approximation method:
Penalty 1 3 3 2
Penalty 1 3 3 -
Penalty 1 6 3 -
Penalty 1 3
Penalty - 3
Penalty - 3
The final allotment is
Total transportation cost
\(
= (10 \times 6)+(25 \times 10)+(45 \times 9)+
(5 \times 13)+(10 \times 9)+(30 \times 5)
\)
60 + 250 + 405 + 65 + 90 + 150 = 1020