By QB365 on 31 Dec, 2022
QB365 provides a detailed and simple solution for every Possible Questions in Class 12 Business Maths Subject - Important 2 Mark 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
Answer all the following Questions.
Find the rank of each of the following matrices.
\(\left( \begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix} \right) \)
Find the rank of each of the following matrices.
\(\left( \begin{matrix} 1 & -1 \\ 3 & -6 \end{matrix} \right) \)
Evaluate \(\int { \left( { x }^{ 3 }+7 \right) \left( x-4 \right) dx } \)
Integrate the following with respect to x.
\(\sqrt { 1-\sin2x } \)
Using Integration, find the area of the region bounded the line 2y + x = 8, the x axis and the lines x = 2, x = 4.
he cost of over haul of an engine is ₹10,000 The operating cost per hour is at the rate of 2x − 240 where the engine has run x km. Find out the total cost if the engine run for 300 hours after overhaul.
Solve \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -4\frac { dy }{ dx } +5y\) = 0
Find the order and degree of the following differential equation
\(\frac { { d }^{ 2 }y }{ { dx }^{ 3 } } -3{ \left( \frac { dy }{ dx } \right) }^{ 6 }+2y={ x }^{ 2 }\)
Evaluate \({ \Delta }^{ 2 }\left( \frac { 1 }{ x } \right) \) by taking ‘1’ as the interval of differencing.
Prove that
E ∇ = Δ = ∇E
The discrete random variable X has the probability function
X | 1 | 2 | 3 | 4 |
P(X=x) | k | 2k | 3k | 4k |
Show that k = 0.1.
Describe what is meant by a random variable.
Write the conditions for which the poisson distribution is a limiting case of binomial distribution.
Write down the conditions in which the Normal distribution is a limiting case of binomial distribution.
Define parameter.
Define alternative hypothesis.
Explain cyclic variations.
The following table gives the number of small-scale units registered with the Directorate of Industries between 1985 and 1991. Show the growth on a trend line by the free hand method.
Years | 1985 | 1986 | 1987 | 1988 | 1989 | 1990 | 1991 | 1992 |
No. of units (in‘000) | 10 | 22 | 36 | 62 | 55 | 40 | 34 | 50 |
What is the Assignment problem?
What is the difference between Assignment Problem and Transportation Problem?
Find the rank of the matrix \(\left( \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right) \)
\(\int \frac{e^{5 \log _{e} x}-e^{+\log _{1} x}}{e^{3 \log _{e} x}-e^{2 \log _{1} x}} d x\)
Find the area under the curve y = \(4 x^{2}-8 x+6\) bounded by the y-axis, x-axis and the ordinate x = 2
Write down the order and degree of the following differential equations.
\(\left( \frac { dy }{ dx } \right) ^{ 2 }-7\frac { d^{ 3 }y }{ { dx }^{ 3 } } +y\frac { { d }^{ 2 }y }{ dx^{ 2 } } +4\frac { dy }{ dx } \)- log x = 0
Calculate a forward difference table for the following data
x | 20 | 30 | 40 | 50 |
y | 51 | 43 | 34 | 24 |
If \(\mathrm{F}(x)=\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} x\right)-\infty<x<\infty\) distribution function of a continuous variable X, find \(\mathrm{P}(0 \leq x \leq 1)\)
It is given that 3% of the electric bulbs manufactured by a company are defective find the probability that a simple of 100 bulbs will contain
(i) no defective
(ii) exactly one defective (e-3 = 0.0498)
State any 2 demerits of simple random sampling.
Calculate the 3-yearlymoving averages of the production figures (in tonnes) for the following data.
Year | 1973 | 1974 | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 | 1984 | 1985 | 1986 | 1987 |
Production | 15 | 21 | 30 | 36 | 42 | 46 | 50 | 56 | 63 | 70 | 74 | 82 | 90 | 95 | 102 |
Consider the following pay-off (profit) matrix action, states
Action | States | |
B1 | B2 | |
A1 | 8 | 6 |
A2 | 9 | 2 |
A3 | 6 | 4 |
Determine the best action using maximin principle.
Answers
Let \(A=\left( \begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix} \right) \)
Order of A is 2 \(\times\) 2
\(\therefore \rho (A)\le 2\) [Since minimum of (2, 2) is 2]
Consider the second order minor
\(\left| \begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix} \right| =40-42\)
= \(-2\neq 0\)
There is a minor of order 2, which is not zero
\(\therefore \rho (A)=2\)
Let \(A=\left( \begin{matrix} i & -1 \\ 3 & -6 \end{matrix} \right) \)
Order of A is 2 \(\times\) 2
\(\therefore \rho (A)\le 2\) [Since minimum of (2, 2) is 2]
Consider the second order minor
\(\left| \begin{matrix} 1 & -1 \\ 3 & -6 \end{matrix} \right| =-6-(-3)\)
= -6 + 3 = -3
\(\neq 0\)
There is a minor of order 2, which is not zero
\(\therefore \rho (A)=2\)
\(\int { \left( { x }^{ 3 }+7 \right) \left( x-4 \right) dx } \)
= \(\int { \left( { x }^{ 4 }-{ 4x }^{ 3 }+7x-28 \right) dx } \)
\(= \frac { { x }^{ 5 } }{ 5 } -{ x }^{ 4 }+\frac { { 7x }^{ 2 } }{ 2 } -28x+c\)
\(\int { \sqrt { 1-\sin2x } dx } \)
\(=\int { \sqrt { { \sin }^{ 2 }x+{ \cos }^{ 2 }x-2\sin x \cos x } dx } \) [∵1 = sin2x + cos2x sin2x = 2sin x cos x]
\(=\int { \sqrt { { \left( \sin x-\cos x \right) }^{ 2 } } } dx\) [∵(a−b)2=a2−2ab+b2]
=∫(sinx − cosx)dx
=−cosx − sinx+ c
=−(cosx + sinx)+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.
₹28,000
Given \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -4\frac { dy }{ dx } +5y\) = 0
(D2−4D+5)y = 0
The auxiliary equation is m2−4m + 5 = 0
⇒ (m−2)2−4 + 5 = 0
(m− 2)2 = –1
m - 2 = 土\(\sqrt{-1}\)
m = 2 土 i , it is if the form α 土 iβ
∴ C.F = e2x[A cos x + B sin x]
The general solution is y = e2x[A cos x + B sin x]
\(\frac { { d }^{ 2 }y }{ { dx }^{ 3 } } -3{ \left( \frac { dy }{ dx } \right) }^{ 6 }+2y={ x }^{ 2 }\)
∴ order = 3,
∴ Degree = 1
\({ \Delta }^{ 2 }\left( \frac { 1 }{ x } \right) =\Delta \left( \Delta \left( \frac { 1 }{ x } \right) \right) \)
Now \(\Delta \left[ \frac { 1 }{ x } \right] =\frac { 1 }{ 1+x } -\frac { 1 }{ x } \)
\({ \Delta }^{ 2 }\left( \frac { 1 }{ x } \right) ={ \Delta }\left( \frac { 1 }{ 1+x } -\frac { 1 }{ x } \right) \)
\(=\Delta \left( \frac { 1 }{ 1+x } \right) -\Delta \left( \frac { 1 }{ x } \right) \)
Similarly \({ \Delta }^{ 2 }\left( \frac { 1 }{ x } \right) =\frac { 2 }{ x(x+1)(x+2) } \)
LHS = E∇
[∵ Δ = \(\frac { E-1 }{ E } \)] |
[∵ Δ = E - 1]
= E - 1 = Δ
RHS
= E - 1 = Δ = RHS
∴ E∇ = Δ = ∇ E
Hence proved.
The given probability function is
X | 1 | 2 | 3 | 4 |
P(X = x) | k | 2k | 3k | 4k |
Since the given function is a probability function, each ρi>0 and Σρi = 1
⇒ k + 2k + 3k + 4k = 1
⇒10k = 1 ⇒ k = \(\frac{1}{10}\)
⇒ k = 0.1
When we perform any experiment, we expect an outcome. We associate a real numbers with each outcome of an experiment. In other words, we considering a function whose domain is the set of possible outcomes and whose range is subset of the set of real numbers such a function is called random variable.
Poisson distribution is a limiting case of binomial distribution under the following conditions.
(i) n, the number of trials is indefinitely large ie., n⟶∞.
(ii) p, the constant probability of success in each trial is very small ie., p ⟶0.
(iii) np = λ is finite.Thus p = λ/n and q = 1 -(λ/n) where λ is a positive real number.
Normal distribution is a limiting case of Binomial distribution under the following conditions.
(i) n, the number of trials is infinitely large i.e. n ⟶ ∞
(ii) Neither p(or q) is very small.
The statistical constants of the population like mean (μ), variance (σ2) are referred as population parameters.
Any hypothesis which is complementary to the null hypothesis is called as the alternative hypothesis and it is usually denoted by H1.
Cyclic uniformly periodic in nature. They may or may not follow exactly similar patterns after equal intervals of time. Generally one cyclic period ranges from 7 to.9 years and there is no hard and fast rule in the fixation of years for a cyclic period. For example, every business cycle has a Start-Boom-Depression- Recover maintenance during booms and depressions, changes in government monetary policies, changes in interestrates.
To assign the different jobs to the different machines (one job per machine) to minimize the overall cost is known as assignment problem.
The assignment problem is a special case of transportation problem where the number of sources and destinations are equal. Here, jobs represent sources and machines represent destinations.
Let A = \(\left( \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right) \)
The order of A is 2 \(\times\) 2
\(\rho (A)\le min(2,2)\)
\(\Rightarrow \rho (A)\le 2\)
\(\left| \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right| =4-4=0\)
Since the second order minor vanishes \(\rho (A)\neq 2\)
We have to try for atleast one non-zero first order minor.
ie. atleast one non-zero element of A.
This is possible because A has non-zero element
\(\therefore \rho (A)-1\)
\(\int \frac{e^{5 \log _{e} x}-e^{4 \log _{e} x}}{e^{3 \log _{e} x}-e^{2 \log _{e} x}} d x=\int \frac{e^{\log _{e} x^{3}}-e^{\log _{e} x^{4}}}{e^{\log _{e} x^{3}}-e^{\log _{e} x^{2}}} d x\)
\(
=\int \frac{x^{5}-x^{4}}{x^{3}-x^{2}} d x=\int \frac{x^{2}\left(x^{3}-x^{2}\right)}{x^{3}-x^{2}} d x \\
\)
\(=\int x^{2} d x=\frac{x^{3}}{3}+c
\)
\( \mathrm{A} =\int_{a}^{b} y d x \quad \frac{x=0 \mid}{(0,0)} \mid(2,0) \)
\(=\int_{0}^{2}\left(4 x^{2}-8 x+6\right) d x \)
\( =\left[\frac{4 x^{3}}{3}-\frac{8 x^{2}}{2}+6 x\right]_{0}^{2}=\frac{32}{3}-16+12 \)
\( =\frac{32}{3}-4=\frac{20}{3} \text { sq. units } \)
The highest derivative if of order 3 and its power is 1
∴ order is 3 and degree is 1.
x | y | \(\Delta y\) | \(\Delta^2 y\) | \(\Delta^3 y\) |
20 | 51 | |||
-8 | ||||
30 | 43 | -1 | ||
-9 | 0 | |||
40 | 34 | -1 | ||
-10 | ||||
50 | 42 |
\(\mathrm{F}(x)=\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} x\right)\)
\(
P(0 \leq x \leq 1) =\mathrm{F}(1)-\mathrm{F}(0)
\)
\( =\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} 1\right)-\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} 0\right)
\)
\( =\frac{1}{2}+\frac{1}{\pi}\left(\frac{\pi}{4}\right)-\frac{1}{2}-\frac{1}{\pi}(0)=\frac{1}{4}
\)
Let X be the random variable denoting the number of defective items
\(
\mathrm{p}=\frac{3}{100}, \mathrm{n}=100, \lambda=\mathrm{np}=3
\)
\(\mathrm{P}(\mathrm{X}=\mathrm{x})=\frac{e^{-\dot{\lambda}} \lambda^{x}}{x !}
\)
\(\mathrm{P}(\mathrm{X}=0)=\frac{e^{-3} 3^{0}}{0 !}=0.0498
\)
\(\mathrm{P}(\mathrm{X}=1)=\frac{e^{-3} 3^{1}}{1 !}=3(0.0498)=0.1494
\)
(i) This requires a complete list of the population but such upto date lists are not available in many enquiries.
(ii) If the size of the sample is small, then it will not be a representative of the population.
Year | Production | 3-yearly moving total | 3-yearly moving average |
1973 | 15 | - | - |
1974 | 21 | 22.00 | |
1975 | 30 | 66 | 29.00 |
1976 | 36 | 87 | 36.00 |
1977 | 42 | 108 | 41.33 |
1978 | 46 | 124 | 46.00 |
1979 | 50 | 138 | 50.67 |
1980 | 56 | 152 | 56.33 |
1981 | 63 | 169 | 63.00 |
1982 | 70 | 189 | 69.00 |
1983 | 74 | 207 | 75.33 |
1984 | 82 | 226 | 82.00 |
1985 | 90 | 246 | 89.00 |
1986 | 95 | 267 | 95.67 |
1987 | 102 | 287 | - |
Action | States | Minimum | |
B1 | B2 | ||
A1 | 8 | 6 | 6 |
A2 | 9 | 2 | 2 |
A3 | 6 | 4 | 4 |
Max (6, 2, 4) = 6
∴ Action Al is the best according to maximin principle