Category: Class 12

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Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.4

Integrate the following with respect to x.

Question 1.
2 cos x – 3 sin x + 4 sec² x – 5 cosec² x
Solution:
∫2 cos x – 3 sin x + 4 sec² x – 5 cosec² x
= 2 ∫ cos x dx – 3 ∫ sin x dx + 4 ∫ sec² x – 5 ∫ cosec² x dx
= 2 sin x + 3 cos x + 4 tan x + 5 cot x + c

Question 2.
∫ sin³ x dx
Solution:
We know that, sin 3x = 3 sin x – 4 sin³ x

Question 3.
\(\frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x}\)
Solution:

Question 4.
\(\frac{1}{\sin ^{2} x \cos ^{2} x}\)
[Hint: sin² x + cos² x = 1]
Solution:

Question 5.
\(\sqrt{1-\sin 2} x\)
Solution:

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Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Miscellaneous Problems

Question 1.
Suppose that Q_d = 30 – 5P + 2 \(\frac{d p}{d t}+\frac{d^{2} p}{d t^{2}}\) and Q_s = 6 + 3P. Find the equilibrium price for market clearance.
Solution:
For equilibrium price Q_d = Q_s

Question 2.
Form the differential equation having for its general solution y = ax² + bx
Solution:
Given y = ax² + bx
Since there are 2 constants a, b we have to differentiate twice to eliminate them

Question 3.
Solve yx² dx + e^-x dy = 0
Solution:
The given equation can be written as

Question 4.
Solve: (x² + y²) dx + 2xy dy = 0
Solution:
The given equation can be written as \(\frac{d y}{d x}=-\frac{\left(x^{2}+y^{2}\right)}{2 x y}\)
It is a homogeneous differential equation

Question 5.
Solve: x \(\frac{d y}{d x}\) + 2y = x⁴
Solution:

Question 6.
A manufacturing company has found that the cost C of operating and maintaining the equipment is related to the length ‘m’ of intervals between overhauls by the equation \(m^{2} \frac{d C}{d m}\) + 2mC = 2 and c = 4 and when m = 2. Find the relationship between C and m.
Solution:

given that c = 4 when m = 2
4(4) = 2(2) + k
k = 12
So the relation ship between C and m is Cm² = 2m + 12 = 2(m + 6)

Question 7.
Solve (D² – 3D + 2)y = e^4x given y = 0 when x = 0 and x = 1.
Solution:
(D² – 3D + 2)y = e^4x
The auxiliary equations is m² – 3m + 2 = 0
(m – 2) (m – 1) = 0
m = 2, 1

Question 8.
Solve: \(\frac{d y}{d x}\) + y cos x = 2cos x
Solution:
The given equation can be written as \(\frac{d y}{d x}\) + (cos x)y = 2 cos x
It is of the form \(\frac{d y}{d x}\) + Py = Q
Where P = cos x, Q = 2 cos x
Now ∫Pdx = ∫cos x dx = sin x

Question 9.
Solve: x²y dx – (x³ + y³) dy = 0
Solution:

Question 10.
Solve: \(\frac{d y}{d x}\) = xy + x + y + 1
Solution:

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Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.6

Choose the correct answer.

Question 1.
The degree of the differential equation \(\frac{d^{4} y}{d x^{4}}-\left(\frac{d^{2} y}{d x^{2}}\right)^{4}+\frac{d y}{d x}=3\) is _________
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1
Hint:
Since the power of \(\frac{d^{4} y}{d x^{4}}\) is 1

Question 2.
The order and degree of the differential equation \(\sqrt{\frac{d^{2} y}{d x^{2}}}=\sqrt{\frac{d y}{d x}+5}\) are respectively
(a) 2 and 3
(b) 3 and 2
(c) 2 and 1
(d) 2 and 2
Answer:
(c) 2 and 1
Hint:
Squaring both sides, we get \(\frac{d^{2} y}{d x^{2}}=\frac{d y}{d x}+5\)
So order = 2, degree = 1

Question 3.
The order and degree of the differential equation \(\left(\frac{d^{2} y}{d x^{2}}\right)^{\frac{3}{2}}-\sqrt{\left(\frac{d y}{d x}\right)}-4=0\) are respectively.
(a) 2 and 6
(b) 3 and 6
(c) 1 and 4
(d) 2 and 4
Answer:
(a) 2 and 6
Hint:

Question 4.
The differential equation \(\left(\frac{d x}{d y}\right)^{3}+2 y^{\frac{1}{2}}=x\) is _________
(a) of order 2 and degree 1
(b) of order 1 and degree 3
(c) of order 1 and degree 6
(d) of order 1 and degree 2
Answer:
(b) of order 1 and degree 3

Question 5.
The differential equation formed by eliminating a and b from y = ae^x + be^-x is _______
(a) \(\frac{d^{2} y}{d x^{2}}-y=0\)
(b) \(\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}=0\)
(c) \(\frac{d^{2} y}{d x^{2}}=0\)
(d) \(\frac{d^{2} y}{d x^{2}}-x=0\)
Answer:
(a) \(\frac{d^{2} y}{d x^{2}}-y=0\)
Hint:

Question 6.
If y = cx + c – c³ then its differential equation is ______

Answer:
(a) \(y=x \frac{d y}{d x}+\frac{d y}{d x}-\left(\frac{d y}{d x}\right)^{3}\)

Question 7.
The integrating factor of the differential equation \(\frac{d x}{d y}\) + Px = Q is _____
(a) e^∫Pdx
(b) ∫Pdx
(c) ∫Pdy
(d) e^∫Pdy
Answer:
(d) e^∫Pdy

Question 8.
The complementary function of (D² + 4) y = e^2x is _______
(a) (Ax + B) e^2x
(b) (Ax + B) e^-2x
(c) A cos 2x + B sin 2x
(d) Ax^-2x + Be^2x
Answer:
(c) A cos 2x + B sin 2x
Hint:
A.E = m² + 4 = 0 ⇒ m = ±2i
C.F = e^0x (A cos 2x + B sin 2x)

Question 9.
The differential equation of y = mx + c is (m and c are arbitrary constants).
(a) \(\frac{d^{2} y}{d x^{2}}=0\)
(b) y = x \(\frac{d y}{d x}\)
(c) x dy + y dx = 0
(d) y dx – x dy = 0
Answer:
(a) \(\frac{d^{2} y}{d x^{2}}=0\)

Question 10.
The particular integral of the differential equation \(\frac{d^{2} y}{d x^{2}}-8 \frac{d y}{d x}+16 y=2 e^{4 x}\) is ________
(a) \(\frac{x^{2} e^{4 x}}{2 !}\)
(b) \(\frac{e^{4 x}}{2 !}\)
(c) x² e^4x
(d) xe^4x
Answer:
(c) x² e^4x
Hint:

Question 11.
Solution of \(\frac{d x}{d y}\) + px = 0
(a) x = ce^py
(b) x = ce^-py
(c) x = py + c
(d) x = cy
Answer:
(b) x = ce^-py
Hint:

Question 12.
If sec² x is an integrating factor of the differential equation \(\frac{d y}{d x}\) + Py = Q then P = _____
(a) 2 tan x
(b) sec x
(c) cos² x
(d) tan² x
Answer:
(a) 2 tan x
Hint:

Question 13.
The integrating factor of x\(\frac{d y}{d x}\) – y = x² is _____
(a) \(\frac{-1}{x}\)
(b) \(\frac{1}{x}\)
(c) log x
(d) x
Answer:
(b) \(\frac{1}{x}\)
Hint:

Question 14.
The solution of the differential equation \(\frac{d y}{d x}\) + Py = Q where P and Q are the function of x is ______

Answer:
(c) \(y e^{\int p d x}=\int \mathrm{Q} e^{\int p d x} d x+c\)

Question 15.
The differential equation formed by eliminating A and B from y = e^-2x (A cos x + B sin x) is _______
(a) y₂ – 4y₁ + 5y = 0
(b) y₂ + 4y₁ – 5y = 0
(c) y₂ – 4y₁ – 5y = 0
(d) y₂ + 4y₁ + 5y = 0
Answer:
(d) y₂ + 4y₁ + 5y = 0
Hint:

Question 16.
The particular integral of the differential equation f(D) y = e^ax where f(D) = (D – a)² ________
(a) \(\frac{x^{2}}{2} e^{2 x}\)
(b) xe^ax
(c) \(\frac{x}{2} e^{2 x}\)
(d) x² e^2x
Answer:
(a) \(\frac{x^{2}}{2} e^{2 x}\)

Question 17.
The differential equation of x² + y² = a² is _____
(a) x dy + y dx = 0
(b) y dx – x dy = 0
(c) x dx – y dx = 0
(d) x dx + y dy = 0
Answer:
(d) x dx + y dy = 0
Hint:
x² + y² = a²
⇒ 2x + 2y \(\frac{d y}{d x}\) = 0
⇒ x dx + y dy = 0

Question 18.
The complementary function of \(\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}=0\) is ______
(a) A + B e^x
(b) (A + B) e^x
(c) (Ax + B) e^x
(d) (Ae^x + B)
Answer:
(a) A + B e^x
Hint:
A.E is m² – m = 0
⇒ m(m – 1) = 0
⇒ m = 0, 1
CF is Ae^0x + Be^x = A + Be^x

Question 19.
The P.I of (3D² + D – 14)y = 13e^2x is _______
(a) \(\frac{x}{2}\) e^2x
(b) x e^2x
(c) \(\frac{x^{2}}{2}\) e^2x
(d) 13xe^2x
Answer:
(b) xe^2x
Hint:

Question 20.
The general solution of the differential equation \(\frac{d y}{d x}\) = cos x is _______
(a) y = sin x + 1
(b) y = sin x – 2
(c) y = cos x + c, c is an arbitrary constant
(d) y = sin x + c, c is an arbitrary constant
Answer:
(d) y = sin x + c, c is an arbitrary constant

Question 21.
A homogeneous differential equation of the form \(\frac{d y}{d x}=f\left(\frac{y}{x}\right)\) can be solved by making substitution
(a) y = vx
(b) v = yx
(c) x = vy
(d) x = v
Answer:
(a) y = vx

Question 22.
A homogeneous differential equation of the form \(\frac{d x}{d y}=f\left(\frac{x}{y}\right)\) can be solved by making substitution,
(a) x = vy
(b) y = vx
(c) y = v
(d) x = v
Answer:
(a) x = vy

Question 23.

Answer:
(d) \(\frac{1+v}{2 v^{2}} d v=-\frac{d x}{x}\)
Hint:

Question 24.
Which of the following is the homogeneous differential equation?
(a) (3x – 5) dx = (4y – 1) dy
(b) xy dx – (x³ + y³) dy = 0
(c) y² dx + (x² – xy – y²) dy = 0
(d) (x² + y) dx = (y² + x) dy
Answer:
(c) y² dx + (x² – xy – y²) dy = 0

Question 25.
The solution of the differential equation \(\frac{d y}{d x}=\frac{y}{x}+\frac{f\left(\frac{y}{x}\right)}{f^{\prime}\left(\frac{y}{x}\right)}\) is ______
(a) f(\(\frac{y}{x}\)) = kx
(b) x f(\(\frac{y}{x}\)) = k
(c) f(\(\frac{y}{x}\)) = ky
(d) y f(\(\frac{y}{x}\)) = k
Answer:
(a) f(\(\frac{y}{x}\)) = kx
Hint:

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Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.5

Solve the following differential equations:

Question 1.
\(\frac{d^{2} y}{d x^{2}}-6 \frac{d y}{d x}+8 y=0\)
Solution:
Given (D² – 6D + 8) y = 0, D = \(\frac{d}{d x}\)
The auxiliary equations is
m² – 6m + 8 = 0
(m – 4)(m – 2) = 0
m = 4, 2
Roots are real and different
The complementary function (C.F) is (Ae^4x + Be^2x)
The general solution is y = Ae^4x + Be^2x

Question 2.
\(\frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}+4 y=0\)
Solution:
The auxiliary equations A.E is m² – 4m + 4 = 0
(m – 2)² = 0
m = 2, 2
Roots are real and equal
The complementary function (C.F) is (Ax + B) e^2x
The general solution is y = (Ax + B) e^2x

Question 3.
(D² + 2D + 3) y = 0
Solution:
The auxiliary equations A.E is m² + 2m + 3 = 0
⇒ m² + 2m + 1 + 2 = 0
⇒ (m + 1)² = -2
⇒ m + 1 = ± √2i
⇒ m = – 1 ± √2i
It is of the form α ± iβ
The complementary function (C.F) = e^-x [A cos √2 x + B sin √2 x]
The general solution is y = e^-x [A cos √2 x + B sin √2 x]

Question 4.
\(\frac{d^{2} y}{d x^{2}}-2 k \frac{d y}{d x}+k^{2} y=0\)
Solution:
Given (D² – 2kD + k²)y = 0, D = \(\frac{d}{d x}\)
The auxiliary equations is m² – 2km + k = 0
⇒ (m – k)² = 0
⇒ m = k, k
Roots are real and equal
The complementary function (C.F) is (Ax + B) e^kx
The general solution is y = (Ax + B) e^kx

Question 5.
(D² – 2D – 15) y = 0 given that \(\frac{d y}{d x}\) = 0 and \(\frac{d^{2} y}{d x^{2}}\) = 2 when x = 0
Solution:
A.E is m² – 2m – 15 = 0
(m – 5)(m + 3) = 0
m = 5, -3
C.F = Ae^5x + Be^-3x
The general solution is y = Ae^5x + Be^-3x …….. (1)

Question 6.
(4D² + 4D – 3) y = e^2x
Solution:
The auxiliary equations is 4m² + 4m – 3 = 0

Question 7.
\(\frac{d^{2} y}{d x^{2}}\) + 16y = 0
Solution:
Given (D² + 16) y =0
The auxiliary equation is m² + 16 = 0
⇒ m² = -16
⇒ m = ± 4i
It is of the form α ± iβ, α = 0, β = 4
The complementary function (C.F) is e^0x [A cos 4x + B sin 4x]
The general solution is y = [A cos 4x + B sin 4x]

Question 8.
(D² – 3D + 2)y = e^3x which shall vanish for x = 0 and for x = log 2
Solution:
A.E is m² – 3m + 2 = 0
⇒ (m – 2) (m – 1) = 0
⇒ m = 2, 1
CF = Ae^2x + Be^x

Question 9.
(D² + D – 6)y = e^3x + e^-3x
Solution:
A.E is m² + m – 6 = 0
(m + 3) (m – 2) = 0

Question 10.
(D² – 10D + 25)y = 4e^5x + 5
Solution:
A.E is m² – 10m + 25 = 0
⇒ (m – 5)² = 0
⇒ m = 5, 5
C.F = (Ax + B) e^5x

Question 11.
(4D² + 16D + 15) y = 4\(e^{\frac{-3}{2} x}\)
Solution:
A.E is 4m² + 16m + 15 = 0
(2m + 3) (2m + 5) = 0

Question 12.
(3D² + D – 14)y = 13e^2x
Solution:

Question 13.
Suppose that the quantity demanded Q_d = 13 – 6p + 2\(\frac{d p}{d t}+\frac{d_{2} p}{d t^{2}}\) and quantity supplied Q_s = -3 + 2p where p is the price. Find the equilibrium price for market clearance.
Solution:
For market clearance, the required condition is Q_d = Q_s

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Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.8

I. Using second fundamental theorem, evaluate the following:

Question 1.
\(\int_{0}^{1} e^{2 x} d x\)
Solution:

Question 2.
\(\int_{0}^{\frac{1}{4}} \sqrt{1-4 x} d x\)
Solution:

Question 3.
\(\int_{1}^{2} \frac{x d x}{x^{2}+1}\)
Solution:

Question 4.
\(\int_{0}^{3} \frac{e^{x} d x}{1+e^{x}}\)
Solution:

Question 5.
\(\int_{0}^{1} x e^{x^{2}} d x\)
Solution:

Question 6.
\(\int_{1}^{e} \frac{d x}{x(1+\log x)^{3}}\)
Solution:

Question 7.
\(\int_{-1}^{1} \frac{2 x+3}{x^{2}+3 x+7} d x\)
Solution:

Question 8.
\(\int_{0}^{\frac{\pi}{2}} \sqrt{1+\cos x} d x\)
Solution:

Question 9.
\(\int_{1}^{2} \frac{x-1}{x^{2}} d x\)
Solution:

II. Evaluate the following:

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

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Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Additional Problems

I. One Mark Questions

Choose the correct answer.

Question 1.
The differential equation of straight lines passing through the origin is ______
(a) \(\frac{x d y}{d x}=y\)
(b) \(\frac{d y}{d x}=\frac{x}{y}\)
(c) \(\frac{d y}{d x}=0\)
(d) \(\frac{x d y}{d x}=\frac{1}{y}\)
Answer:
(a) \(\frac{x d y}{d x}=y\)
Hint:

Question 2.
The solution of x dy + y dx = 0 is _______
(a) x + y = c
(b) x² + y² = c
(c) xy = c
(d) y = cx
Answer:
(c) xy = c
Hint:
x dy + y dx = 0
d(xy) = 0
xy = c

Question 3.
The solution of x dx + y dy = 0 is _____
(a) x² + y² = c
(b) \(\frac{x}{y}\) = c
(c) x² – y² = c
(d) xy = c
Answer:
(a) x² + y² = c
Hint:
x dx = -y dy
\(\frac{x^{2}}{2}=\frac{-y^{2}}{2}+c_{1}\)
x² + y² = c

Question 4.
The solution of \(\frac{d y}{d x}\) = e^x-y is ______
(a) e^y e^x = c
(b) y = log ce^x
(c) y = log(e^x + c)
(d) e^x+y = c
Answer:
(c) y = log(e^x + c)
Hint:

Question 5.
The solution of \(\frac{d p}{d t}\) = ke^-t (k is a constant) is ________
(a) \(c-\frac{k}{e^{t}}=p\)
(b) p = ke^t + c
(c) t = log\(\frac{c-p}{k}\)
(d) t = log_c p
Answer:
(a) \(c-\frac{k}{e^{t}}=p\)
Hint:

Question 6.
The integrating factor of (1 + x²) \(\frac{d y}{d x}\) + xy = (1 + x²)³ is _______
(a) \(\sqrt{1+x^{2}}\)
(b) log(1 + x²)
(c) \(e^{\tan ^{-1} x}\)
(d) \(\log ^{\left(\tan ^{-1} x\right)}\)
Answer:
(a) \(\sqrt{1+x^{2}}\)
Hint:

Question 7.
The complementary function of the differential equation (D² – D) y = e^x is _____
(a) A + B e^x
(b) (Ax + B) e^x
(c) A + B e^-x
(d) (A + Bx) e^-x
Answer:
(a) A + B e^x
Hint:
m² – m = 0
m(m – 1) = 0
CF = Ae^0x + Be^x = A + B e^x

Question 8.
Match the following
(a) \(\frac{d^{4} y}{d x^{4}}\) + sin y = 0 – (i) order 1, degree 1
(b) y’ + y = e^x – (ii) order 3, degree 2
(c) y'” + 2y” + y’ = 0 – (iii) order 4, degree 1
(d) (y”’)² + y’ + y⁵ = 0 – (iv) order 3, degree 1
Answer:
(a) – (iii)
(b) – (i)
(c) – (iv)
(d) – (ii)

Question 9.
Fill in the blanks
(a) The general solution of the equation \(\frac{d y}{d x}+\frac{y}{x}=1\) is ______
(b) Integrating factor of \(\frac{x d y}{d x}\) – y = sin x is ______
(c) The differential equation of y = A sin x + B cos x is _______
(d) The D.E \(\frac{d y}{d x}+\frac{y}{x \log x}=\frac{1}{x}\) is a ________ differential equation.
Answer:
(a) \(y=\frac{x}{2}+\frac{c}{x}\)
(b) \(\frac{1}{x}\)
(c) \(\frac{d^{2} y}{d x^{2}}+y=0\)
(d) linear

Question 10.
State true or false
(a) y = 3 sin x + 4 cos x is a particular solution of the differential equation \(\frac{d^{2} y}{d x^{2}}\) + y = 0
(b) The solution of \(\frac{d y}{d x}=\frac{x+2 y}{x}\) is x + y = kx²
(c) y = 13e^x + 4e^-x is a solution of \(\frac{d^{2} y}{d x^{2}}\) – y = 0
Answer:
(a) True
(b) True
(c) True

II. 2 Marks Questions

Question 1.
Form the D.E of the family of curves y = ae^3x + be^x where a, b are parameters
Solution:
y = ae^3x + be^x
\(\frac{d y}{d x}\) = 3 ae^3x + be^x

Question 2.
Find the D.E of a family of curves y = a cos (mx + b), a and b are constants.
Solution:

Question 3.
Find the D.E by eliminating the constants a and b from y = a tan x + b sec x
Solution:

Question 4.
Solve: \(\frac{d y}{d x}=e^{7 x+y}\)
Solution:

Question 5.
Solve: (x² – ay) dx = (ax – y²) dy
Solution:
Writing the equation as
x² dx + y² dy = a (x dy + y dx)
x² dx + y² dy = a d(xy)
∫x² dx + ∫y² dy = a ∫d(xy) + c
\(\frac{x^{3}}{3}+\frac{y^{3}}{3}\) = axy + c
Hence the general solution is x³ + y³ = 3axy + c

Question 6.
Solve (sin x + cos x) dy + (cos x – sin x) dx = 0
Solution:
The given equation can be written as

III. 3 and 5 Marks Questions

Question 1.
Solve \(\frac{x d y}{d x}\) + cos y = 0, given y = \(\frac{\pi}{4}\) when x = √2
Solution:

Question 2.
The slope of a curve at any point is the reciprocal of twice the ordinate of the point. The curve also passes through the point (4, 3). Find the equation of the curve.
Solution:
Slope at any point (x, y) is the slope of the tangent at (x, y)
\(\frac{d y}{d x}=\frac{1}{2 y}\)
⇒ 2y dy = dx
⇒ ∫2y dy = ∫dx + c
⇒ y² = x + c
Since the curve passes through (4, 3)
we have 9 = 4 + c ⇒ c = 5
Equation of the curve is y² = x + 5

Question 3.
The net profit P and quantity x satisfy the differential equation \(\frac{d p}{d x}=\frac{2 p^{3}-x^{3}}{3 x p^{2}}\). Find the relationship between the net profit and demand given that p = 20 when x = 10
Solution:

Question 4.
Solve: \(\frac{d y}{d x}\) + ay = e^x (a ≠ -1)
Solution:
The given equation is of the form \(\frac{d y}{d x}\) + Py = Q
Here P = a, Q = e^x
The general solution is y (I.F) = ∫Q (I.F) dx + c

Question 5.
Solve: \(\frac{d y}{d x}\) + y cos x = \(\frac{1}{2}\) sin 2x
Solution:
Here P = cos x, Q = \(\frac{1}{2}\) sin 2x

Question 6.
Solve (D² – 6D + 25) y = 0
Solution:
The auxiliary equations is m² – 6m + 25 = 0

The Roots are complex and of the form,
α ± β with α = 3 and β = 4
The complementary function = e^3x (A cos 4x + B sin 4x)
The general solution is y = e^3x (A cos 4x + B sin 4x)

Question 7.
Solve (D² + 10D + 25) y = \(\frac{5}{2}\) + e^-5x
Solution:
The auxiliary equations is m² + 10m + 25 = 0
(m + 5)² = 0
m = -5, -5
The complementary function = (Ax + B) e^-5x

Question 8.
Suppose that the quantity demanded Q_d = 40 – 4p – 4\(\frac{d p}{d t}+\frac{d^{2} p}{d t^{2}}\) and quantity supplied Q_s = -6 + 8p where p is the price. Find the equilibrium price for market clearance.
Solution:
For market clearance, the required condition is Q_d = Q_s

Students can download 12th Business Maths Chapter 10 Operations Research Miscellaneous Problems and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems

Question 1.
The following table summarizes the supply, demand and cost information for four factors S₁, S₂, S₃, S₄ shipping goods to three warehouses D₁, D₂, D₃.

Find an initial solution by using the north-west corner rule. What is the total cost of this solution?
Solution:
Let ‘a_i‘ denote the supply and ‘b_j‘ denote the demand.
Then total supply = 5 + 8 + 7 + 14 = 34 and Total demand = 7 + 9 + 18 = 34
Σa_i = Σb_j. So the problem is a balanced transportation problem and we can find a basic feasible solution, by North-west comer rule.
First allocation:

Second allocation:

Third allocation:

Fourth allocation:

Fifth allocation:

We first allot 4 units to cell (S₃, D₃) and then the balance 14 units to cell (S₄, D₃).
Thus we get the following allocations:

The transportation schedule:
S₁ → D₁, S₂ → D₁, S₂ → D₂, S₃ → D₂, S₃ → D₃, S₄ → D₃
(i.e) x₁₁ = 5, x₂₁ = 2, x₂₂ = 6, x₃₂ = 3, x₃₃ = 4, x₄₃ = 14
Total cost = (5 × 2) + (2 × 3) + (6 × 3) + (3 × 4) + (4 × 7) + (14 × 2)
= 10 + 6+ 18 + 12 + 28 + 28
= 102
Thus the initial basic solution is got by NWC method and minimum cost is Rs. 102.

Question 2.
Consider the following transportation problem

Determine an initial basic feasible solution using
(a) Least cost method
(b) Vogel’s approximation method
Solution:
Let ‘a_i‘ denote the availability and ‘b_j‘ denote the requirement. Then,
Σa_i = 30 + 50 + 20 = 100 and Σb_j = 30 + 40 + 20 + 10 = 100
Since Σa_i = Σb_j. the given problem is a balanced transportation problem and we can get an initial basic feasible solution.
(a) Least Cost Method (LCM)
First allocation:

Second allocation:

Third allocation:

Fourth allocation:

Fifth allocation:

We first allocate 10 units to cell (O₂, D₂). Since it has a minimum cost. Then we allocate the balance 10 units to cell (O₁, D₂). Thus we get the final allocation table as given below.

Transportation schedule:
O₁ → D₂, O₁ → D₃, O₂ → D₁, O₂ → D₂, O₂ → D₄, O₃ → D₂
(i.e) x₁₂ = 10, x₁₃ = 20, x₂₁ = 30, x₂₂ = 10, x₂₄ = 10, x₃₂ = 20
The total cost is = (10 × 8) + (20 × 3) + (30 × 4) + (10 × 5) + (10 × 4) + (20 × 2)
= 80 + 60+ 120 + 50 + 40 + 40
= 390
Thus the LCM, we get the minimum cost for the transportation problem as Rs. 390.
(b) Vogel’s approximation method:
First allocation:

The largest penalty = 3. So allocate 20 units to the cell (O₃, D₂) which has the least cost in column D₂
Second allocation:

Largest penalty = 4. So allocate min (20, 30) units to the cell (O₁, D₃) which has the least cost in column D₃
Third allocation:

The largest penalty is 3. So allocate min (20, 50) to the cell (O₂, D₂) which has the least cost in column D₂.
Fourth allocation:

Largest penalty = 2. So allocate min (10, 30) to cell (O₂, D₄) which has the least cost in column D₄
Fifth allocation:

Largest penalty = 1. Allocate min (30, 20) to cell (O₂, D₁) which has the least cost in column D₁. Finally allot the balance 10 units to cell (O₁, D₁)
We get the final allocation table as follows.

Transportation schedule:
O₁ → D₁, O₁ → D₃, O₂ → D₁, O₂ → D₂, O₂ → D₄, O₃ → D₂
(i.e) x₁₁ = 10, x₁₃ = 20, x₂₁ = 20, x₂₂ = 20, x₂₄ = 10, x₃₂ = 20
Total cost is given by = (10 × 5) + (20 × 3) + (20 × 4) + (20 × 5) + (10 × 4) + (20 × 2)
= 50 + 60 + 80+ 100 + 40 + 40
= 370
Hence the minimum cost by YAM is Rs. 370.

Question 3.
Determine an initial basic feasible solution to the following transportation problem by using (i) North-West Corner rule (ii) least-cost method.

Solution:
Total supply = 25 + 35 + 40 = 100 = Σa_i
Total requirement = 30 + 25 + 45 = 100 = Σb_j
Since Σa_i = Σb_j the given transportation problem is balanced and we can find an initial basic feasible solution.
(i) North West Corner Rule
First allocation:

Second allocation:

Third allocation:

Fourth allocation:

We first allot 5 units to cell (S₂, D₃). Then balance 40 units we allot to cell (S₃, D₃).
The final allotment is given as follows

Transportation schedule:
S₁ → D₁, S₂ → D₁, S₂ → D₂, S₂ → D₃, S₃ → D₃
(i.e) x₁₁ = 25, x₂₁ = 5, x₂₂ = 25, x₂₃ = 5, x₃₃ = 40
Total cost is = (25 × 9) + (5 × 6) + (25 × 8) + (5 × 4) + (40 × 9)
= 225 + 30 + 200 + 20 + 360
= 835
Hence the minimum cost is Rs. 835 by NWC method.

(ii) Least cost method (LCM)
First allocation:

Second allocation:

Third allocation:

Fourth allocation:

We first allot 15 units to cell (S₃, D₁) since it has the least cost. Then we allot the balance 15 units to cell (S₁, D₁).
The final allotment is given as follows

Transportation schedule:
S₁ → D₁, S₁ → D₃, S₂ → D₃, S₃ → D₁, S₃ → D₂
(i.e) x₁₁ = 15, x₁₃ = 10, x₂₃ = 35, x₃₁ = 15, x₃₂ = 25
Total cost is = (15 × 9) + (10 × 5) + (35 × 4) + (15 × 7) + (25 × 6)
= 136 + 50 + 140 + 105 + 150
= 580
The optimal cost by LCM is Rs. 580.

Question 4.
Explain Vogel’s approximation method by obtaining an initial basic feasible solution to the following transportation problem.

Solution:
Total supply (a_i) = 6+1 + 10=17
Total demand (b_j) = 7 + 5 + 3 + 2 = 17
Σa_i = Σb_j. So given transportation problem is a balanced problem and hence we can find a basic feasible solution by VAM.
First allocation:

The largest penalty is 6. So allot min (2, 1) to the cell (O₂, D₄) which has the least cost in column D₄.
Second allocation:

The largest penalty is 5. So allot min (5, 6) to the cell (O₁, D₂) which has the least cost in column D₂.
Third allocation:

The largest penalty is 5. So allot min (7, 1) to the cell (O₁, D₁) which has the least cost in row O₁.
Fourth allocation:

Largest penalty = 4. So allocate min (6, 10) to the cell (O₃, D₁) which has the least cost in row O₃
Fifth allocation:

The largest penalty is 6. So allot min (1, 4) to the cell (O₃, D₄) which has the least cost in row O₃. The balance 3 units is allotted to the cell (O₃, D₃). We get the final allocation as given below.

Transportation schedule:
O₁ → D₁, O₁ → D₂, O₂ → D₄, O₃ → D₁, O₃ → D₃, O₃ → D₄
(i.e) x₁₁ = 1, x₁₂ = 5, x₂₄ = 1, x₃₁ = 6, x₃₃ = 3, x₃₄ = 1
Total cost is given by = (1 × 2) + (5 × 3) + (1 × 1) + (6 × 5) + (3 × 15) + (1 × 9)
= 2 + 15 + 1 + 30 + 45 + 9
= 102
Thus the optimal (minimal) cost of the given transportation problem by Vogel’s approximation method is Rs. 102.

Question 5.
A car hire company has one car at each of five depots a, b, c, d and e. A customer in each of the fine towers A, B, C, D and E requires a car. The distance (in miles) between the depots (origins) and the towers (destinations) where the customers are given in the following distance matrix.

How should the cars be assigned to the customers so as to minimize the distance travelled?
Solution:
In the given assignment problem, the number of rows equals the number of columns. So the problem is balanced and we can get an optimal solution.
Step 1:

We have subtracted the minimum distance of each row from all elements of that row.
Step 2:

We have subtracted the minimum distance of each column from all elements of that column.
Step 3: (Assignment)

We are not able to assign depots for C and E. So we proceed as below.

We have drawn a minimum number of lines to cover all zeros. Now 15 is the smallest element from all uncovered elements. Subtract this from all uncovered elements and add to the elements which lie at the intersection of two lines. Thus we obtain the following reduced matrix for fresh assignment.

Assignment Schedule

Thus the minimum distance is 570 miles.

Question 6.
A natural truck-rental service has a surplus of one truck in each of the cities 1, 2, 3, 4, 5 and 6 and a deficit of one truck in each of the cities 7, 8, 9, 10, 11 and 12. The distance (in kilometres) between the cities with a surplus and the cities with a deficit are displayed below:

How should the truck be dispersed so as to minimize the total distance travelled?
Solution:
The number of rows equals the number of columns. So the problem is a balanced assignment problem and we can get an optimal solution.
Step 1: Subtract the minimum element of each row from all elements of that row.

Step 2: Subtract the least element of each column from all elements of that column.

Step 3: (Assignment)

We are not able to assign city for starting place 4. We proceed as follows.
Draw a minimum number of lines to cover all the zeros of the matrix. Subtract the smallest element from all uncovered elements and add to the elements which lie at the intersection of two lines.

Then we obtain a new reduced matrix for fresh assignment.

Final dispersal

Thus the minimum total distance the trucks should travel is 125 km.

Question 7.
A person wants to invest in one of three alternative investment plans: Stock, Bonds and Debentures. It is assumed that the person wishes to invest all of the funds in a plan. The pay-off matrix based on three potential economic conditions is given in the following table:

Determine the best investment plan using each of the following criteria (i) Maximin (ii) Minimax
Solution:
(i) Maximin rule

Max (3000, 1000, 6000) = 6000. Since the maximum value is 6000, he must invest in debentures by the maximin criteria.
(ii) Minimax rule

Min (10000, 8000, 6000) = 6000. Since the minimum value is 6000, he must invest in debentures by the minimax criteria.

Students can download 12th Business Maths Chapter 10 Operations Research Ex 10.4 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Ex 10.4

Choose the correct answer.

Question 1.
The transportation problem is said to be unbalanced if _______
(a) Total supply ≠ Total demand
(b) Total supply = Total demand
(c) m = n
(d) m + n – 1
Answer:
(a) Total supply ≠ Total demand

Question 2.
In a non-degenerate solution number of allocation is ________
(a) Equal to m + n – 1
(b) Equal to m + n + 1
(c) Not equal to m + n – 1
(d) Not equal to m + n + 1
Answer:
(a) Equal to m + n – 1

Question 3.
In a degenerate solution number of allocations is ________
(a) equal to m + n – 1
(b) not equal to m + n – 1
(c) less than m + n – 1
(d) greater than m + n – 1
Answer:
(c) less than m + n – 1

Question 4.
The penalty in VAM represents the difference between the first ________
(a) Two largest costs
(b) Largest and Smallest costs
(c) Smallest two costs
(d) None of these
Answer:
(c) Smallest two costs

Question 5.
Number of basic allocation in any row or column in an assignment problem can be ________
(a) exactly one
(b) at least one
(c) at most one
(d) none of these
Answer:
(a) exactly one

Question 6.
North-West Comer refers to _________
(a) top left corner
(b) top right comer
(c) bottom right comer
(d) bottom left comer
Answer:
(a) top left corner

Question 7.
Solution for transportation problem using _________ method is nearer to an optimal solution.
(a) NWCM
(b) LCM
(c) VAM
(d) Row Minima
Answer:
(c) VAM

Question 8.
In an assignment problem the value of decision variable xij is _______
(a) 1
(b) 0
(c) 1 or 0
(d) none of them
Answer:
(c) 1 or 0

Question 9.
If the number of sources is not equal to the number of destinations, the assignment problem is called _______
(a) balanced
(b) unsymmetric
(c) symmetric
(d) unbalanced
Answer:
(d) unbalanced

Question 10.
The purpose of a dummy row or column in an assignment problem is to _________
(a) prevent a solution from becoming degenerate
(b) the balance between total activities and total resources
(c) provide a means of representing a dummy problem
(d) none of the above
Answer:
(b) the balance between total activities and total resources

Question 11.
The solution for an assignment problem is optimal if _________
(a) each row and each column has no assignment
(b) each row and each column has at least one assignment
(c) each row and each column has at most one assignment
(d) each row and each column has exactly one assignment
Answer:
(d) each row and each column has exactly one assignment

Question 12.
In an assignment problem involving four workers and three jobs, total numbers of assignments possible are ______
(a) 4
(b) 3
(c) 7
(d) 12
Answer:
(b) 3

Question 13.
Decision theory is concerned with ________
(a) analysis of information that is available
(b) decision making under certainty
(c) selecting optimal decisions in sequential problem
(d) All of the above
Answer:
(d) All of the above

Question 14.
A type of decision-making environment is _______
(a) certainty
(b) uncertainty
(c) risk
(d) all of the above
Answer:
(d) all of the above

Students can download 12th Business Maths Chapter 10 Operations Research Additional Problems and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Additional Problems

I. One Mark Questions

Choose the correct Answer

Question 1.
Which of the following methods is used to verify the optimality of the current solution of the transportation problem?
(a) Least cost method
(b) Vogel’s method
(c) North-west comer rule
(d) None of these
Answer:
(a) Least cost method

Question 2.
The degeneracy’in the transportation problem indicates that _________
(a) Dummy allocations need to be added
(b) The problem has no feasible solution
(c) Multiple optimal solutions exist
(d) All of the above
Answer:
(c) Multiple optimal solutions exist

Question 3.
The Hungarian method can also be used to solve ______
(a) Transportation problem
(b) Travelling salesman problem
(c) A linear programming problem
(d) All the above
Answer:
(b) Travelling salesman problem

Question 4.
An optimal solution of an assignment problem can be obtained only if, _________
(a) each row and column has only one zero element
(b) each row and column has at least one zero element
(c) The data is arranged in a square matrix
(d) None of the above
Answer:
(d) None of the above

Question 5.
Say True or False.

1. In a transportation problem, a single source may supply something to all destinations.
2. A transportation model must have the same number of rows and columns.
3. It is usually possible to find an optimal solution to a transportation problem that is degenerate.
4. In a transportation problem, a dummy source is given a zero cost, while in an assignment problem, a dummy source is given a very high cost.
5. The Hungarian method operates on the principle of matrix reduction, whereby the cost table is reduced to a set of opportunity costs.

Answer:

1. True
2. False
3. True
4. False
5. True

Question 6.
Fill in the blanks.

1. In a transportation problem, we must make the number of ________ and _______ equal.
2. ______ or ______ are used to balance an assignment problem.
3. The method of finding an initial solution based on opportunity costs is called _______
4. ________ occurs when the number of occupied squares is less than the number of rows plus the number of columns minus one.
5. Both transportation and assignment problems are members of a category of LP problems called ________
6. In the case of an unbalanced problem, shipping cost coefficients of ______ are assigned to each dummy factory or warehouse.

Answer:

1. units supplied, units demanded
2. Dummy rows, dummy columns
3. Vogel’s approximation method
4. Degeneracy
5. Network flow problems
6. zero

Question 7.
Match the following.

(a) Dummy column	(i) Finding initial solution
(b) Northwest comer rule	(ii) Rows ≠ Columns
(c) Hungarian method	(iii) Supply ≠ Demand
(d) Feasible solution	(iv) Assignment problem
(e) Unbalanced problem	(v) All demand and supply constraints are met

Answer:
(a) – (iii)
(b) – (i)
(c) – (iv)
(d) – (v)
(e) – (ii)

Question 8.
The objective function of transportation problem is to ________
(a) Maximise total cost
(b) Minimise the total cost
(c) Total cost should be zero
(d) All the above
Answer:
(b) Minimise the total cost

Question 9.
In transportation problem, optimal solution can be verified by using _______
(a) NWC
(b) LCM
(c) MODI method
(d) Matrix method
Answer:
(c) MODI method

Question 10.
The cells in the transportation problem can be classified as _______
(a) assigned cells and empty cells
(b) allocated cells and unallocated cells
(c) occupied and unoccupied cells
(d) assigned and unoccupied cells
Answer:
(c) occupied and unoccupied cells

Question 11.
In transportation problem if total supply > total demand we add _________
(a) dummy row with cost 0
(b) dummy column with cost 0
(c) dummy row with cost 1
(d) dummy column with cost 1
Answer:
(b) dummy column with cost 0

Question 12.
In an LPP the objective function is to be ________
(a) Minimised
(b) Maximised
(c) (a) or (b)
(d) only (b)
Answer:
(c) (a) or (b)

Question 13.
The method used for solving an assignment problem is called ________
(a) Reduced matrix method
(b) MODI method
(c) Hungarian method
(d) Graphical method
Answer:
(c) Hungarian method

II. 2 Mark Questions

Question 1.
Consider 3 jobs to be assigned to 3 machines. The cost for each combination is shown in the table below. Find the minimal job machine combinations.

Solution:
Step 1:

Step 2:

Step 3: (Assignment)

Optimal assignment:

Question 2.
Find an initial basic feasible solution by LCM.

Solution:

Total cost = (1 × 2) + (6 × 1) + (4 × 4) + (4 × 6)
= 2 + 6 + 16 + 24
= 48

Question 3.
Find an initial basic feasible solution by the North West Corner Rule (NWC).

Solution:
Total demand = Total supply = 60

Total cost = (10 × 9) + (11 × 6) + (12 × 8) + (2 × 3) + (25 × 11)
= 90 + 66 + 96 + 6 + 275
= 533

Question 4.
Find an initial basic feasible solution using Least cost method.

Solution:
Total Demand = 5 + 8 + 7 + 14 = 34
Total Supply = 7 + 9 + 18 = 34
Since they are equal, problem is balanced.

The minimum total transportation cost is = (7 × 10) + (2 × 70) + (7 × 40) + (3 × 40) + (8 × 8) + (7 × 20)
= 70 + 140 + 280 + 120 + 64 + 140
= Rs. 814

Question 5.
Find the investment option using Maximin rule for the following:

Solution:

Max (5, -13, -5) = 5. Since the maximum payoff is 5, by maximin criteria, the decision is to invest in bonds.

III. 3 and 5 Marks Questions

Question 1.
Find an optimal solution to the following transportation problem by North West Corner Method.

Solution:
Total supply = 65 = Total demand. So the given problem is balanced.
First allocation:

Second allocation:

Third allocation:

Fourth allocation:

Total Transportation cost

Question 2.
Find an initial basic solution for the following transportation problem by Vogel’s Approximation method.

Solution:
Total demand = 72 + 102 + 41 = 215 and
Total supply = 76 + 82 + 77 = 235.
Total supply > Total demand. So we add a dummy constraint with 0 unit cost and with allocation 20 (235 – 215). The modified table is

First allocation:

The maximum penalty is 16. Allot 20 units to cell (S₂, D_dummy)
Second allocation:

Third allocation:

Fourth allocation:

The final allocation table is given below.

The minimum total cost = (76 × 8) + (21 × 24) + (41 × 16) + (20 × 0) + (72 × 8) + (5 × 16) = 2424

Question 3.
A company has 4 men available for 4 separate jobs. Only one man can work on anyone job. The cost of assigning each man to each job is given below. Find the optimal solution by the Hungarian method.

Solution:
The number of rows and columns are equal. So the given problem is a balanced assignment problem and we can get an optimal solution.
Step 1:

Step 2:

Step 3: (Assignment)

We are not able to assign job for person B. Proceed as follows. Draw a minimum number of vertical and horizontal lines to cover all the zeros.

Subtract the smallest element 1 from all the uncovered elements and add it to the elements which lie at the intersection of two lines. Thus we obtain another reduced matrix for fresh assignment.

Total cost is

Question 4.
There are five machines and five jobs are to be assigned and the cost matrix is given below. Find the proper assignment.

Solution:
Step 1: (Row-reduction)

Step 2: (Column – reduction)

Step 3: (Assignment)

We are not able to assign a machine to job D. We proceed as follows.

The smallest uncovered element is 2. Subtract 2 from all those elements which are not covered. Add 2 all elements which are at the intersection of two lines. Then proceed with the new matrix.

The assignment is as follows

Question 5.
The cost of transportation from 3 sources to four destinations are given in the follow¬ing table. Obtain an initial basic feasible solution using
(i) North West Corner Rule (NWC)
(ii) Least Cost Method (LCM) and
(iii) Vogel’s Approximation Method (VAM)

Solution:
(i) North West Corner Rule
We start by allotting the units to the North -West Comer cell. We show all the allocations in a single table.

Total transportation cost is (200 × 4) + (50 × 2) + (350 × 7) + (100 × 5) + (200 × 3) + (1 × 300)
= 800 + 100 + 2450 + 500 + 600 + 300
= Rs. 4750

(ii) Least cost method (LCM)

Transportation cost is = (250 × 2) + (200 × 3) + (150 × 7) + (100 × 5) + (200 × 3) + (300 × 1)
= 500 + 600+ 1050 + 500 + 600 + 300
= Rs. 3550

(iii) Vogel Approximation Method (VAM)

There are five penalties which have the maximum value 2. The cell with the least cost is row 3 and hence select cell (3, D) for allocation.

There are four penalties which have maximum value 2. Select cell (1, B) which has the least cost for allocation.

The largest penalty is 6. Allot units to cell (2, A)

The largest penalty is 3. Allot units to cell (3, B)

We first allot 50 units to cell (3, C) which has less cost. Then the balance units we allot to cell (2, C). We get the final allocation table as follows.

Transportation cost is = (250 × 2) + (200 × 3) + (250 × 5) + (150 × 4) + (50 × 3) + (300 × 1)
= 500 + 600 + 1250 + 600 + 150 + 300
= Rs. 3400

Students can download 12th Business Maths Chapter 8 Sampling Techniques and Statistical Inference Additional Problems and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Additional Problems

Choose the correct answer:

Question 1.
Non sampling error is reduced by ________
(a) Increasing sample size
(b) Decreasing sample size
(c) Reducing amount of data
(d) None of these
Answer:
(d) None of these

Question 2.
Any numerical value calculated from sample data is called ______
(a) Error
(b) Statistic
(c) Bias
(d) Mean
Answer:
(b) Statistic

Question 3.
In sampling with replacement a sampling unit can be selected _______
(a) Only once
(b) More than one time
(c) Less than one time
(d) None of above
Answer:
(b) More than one time

Question 4.
Standard deviation of sampling distribution of any statistic is called ________
(a) Sampling error
(b) Type-I error
(c) Standard error
(d) Non-sampling error
Answer:
(c) Standard error

Question 5.
The difference between statistic and parameter is called ________
(a) Random error
(b) Sampling error
(c) Standard error
(d) Bias
(e) Error
Answer:
(e) Error

Question 6.
In random sampling, the probability of selecting an item from the population is _______
(a) unknown
(b) known
(c) undecided
(d) zero
Answer:
(b) known

Question 7.
Match the following:

(i) Type I error	(a) determine whether a statistical result is significant
(ii) Type II error	(b) Left-tailed test
(iii) Hypothesis testing	(c) reject a true null hypothesis
(iv) H1 : µ > µ0	(d) do not reject a false null hypothesis
(v) H1 : µ < µ0	(e) Right-tailed test

Answer:
(i) – (c)
(ii) – (d)
(iii) – (a)
(iv) – (e)
(v) – (b)

Question 8.
Fill in the blanks:

1. Any statement whose validity is tested on the basis of a sample is called ________
2. The alternative hypothesis is also called _______
3. The probability of rejecting the null hypothesis when it is true is called ________
4. The hypothesis µ ≤ 10 is a _______
5. If a hypothesis specifies the population distribution it is called ______

Answer:

1. Statistical hypothesis
2. Research hypothesis
3. Level of significance
4. Composite hypothesis
5. Simple hypothesis

Question 9.
Null and alternative hypothesis are statements about ________
(a) population parameters
(b) sample parameters
(c) sample statistics
(d) none of the above
Answer:
(a) population parameters

2 and 3 Mark Questions

Question 1.
A random sample of size 50 with mean 67.9 is drawn from a normal population. If the S.E of the sample mean is √0.7, find a 95% confidence interval for the population mean.
Solution:
n = 50, \(\bar{x}\) = 67.9
95% confidence limits for the population mean µ are

Thus the 95% confidence intervals for estimating µ is given by (66.26, 69.54)

Question 2.
A random sample of 500 apples was taken from large consignment and 45 of them were found to be bad. Find the limits at which the bad apples lie at 99% confidence level.
Solution:
Sample size n = 500
Proportion of bad apples P = \(\frac{45}{500}\) = 0.09
So proportion of good apples Q = 1 – 0.09 = 0.91
The confidence limits for population proportion are given by,

= (0.09 – (2.58) (0.013), 0.09 + (2.58) (0.013)) .
= (0.09 – 0.034, 0.09 + 0.034)
= (0.056, 0.124)
Thus the bad apples lie between 5.6% and 12.4%

Question 3.
A sample of 400 students is found to have a mean height of 171.38 cms can it be regarded as a sample from a large population with mean height 171.17 cms and S.D 3.30 cms?
Solution:
Given
Sample size n = 400
Sample mean \(\bar{x}\) = 171.38 cm
Population mean µ = 171.17 cm
Population SD σ = 3.30 cm
H₀ : µ = 171.17 cm
H₁ : µ ≠ 171.17 cm
Test statistic:

The table value of \(z_{\alpha / 2}\) at 5% level is 1.96. Since Z < \(z_{\alpha / 2}\), H₀ is accepted. Therefore the given sample can be regarded as one from the population with mean 171.17 cm.

Question 4.
An automatic machine fills tea in sealed tins with a mean weight of tea as 1 kg and S.D 1 gram. A random sample of 50 tins was examined and it was found that their mean weight was 999.5 grams. Is the machine working properly?
Solution:
Given
Sample size n = 50
Sample mean \(\bar{x}\) = 999.5 grams
Population mean µ = 1000 grams
Population SD σ = 1 gram
H₀ : µ = 1 kg
H₁ : µ ≠ 1 kg
Test statistic

The table value of \(z_{\alpha / 2}\) at 1% level = 2.58. Since |Z| > \(z_{\alpha / 2}\), H₀ is rejected. Therefore the machine is not working properly.

5 Marks Questions

Question 1.
A simple random sample of size 100 has mean
(а) 15, the population variance being 25. Find an interval estimate of the population mean with a confidence level of 95% and 99%
(b) If the population variance is not given, what should be done to find out the required estimates?
Solution:
(a) Given
Sample size n = 100
Sample mean \(\bar{x}\) = 15
Population SD σ = 5
The 95% confidence interval for the population mean is \(\bar{x} \pm \mathrm{Z}_{\alpha / 2} \frac{\sigma}{\sqrt{n}}\)
Here \(z_{\alpha / 2}\) = 1.96. So we get
= 15 ± (1.96) (\(\frac{5}{\sqrt{100}}\))
= 15 ± (1.96) (0.5)
= 15 ± 0.98
= 14.02 and 15.98
Therefore 95% confidence interval for population mean µ is (14.02, 15.98)
The 99% confidence interval is \(\bar{x} \pm 2.58 \frac{\sigma}{\sqrt{n}}\)
= 15 ± 2.58 (\(\frac{5}{\sqrt{100}}\))
= 15 ± 2.58 (0.5)
= 13.71 and 16.29
Therefore 99% confidence interval for the population mean µ is (13.71, 16.29)
(b) If population S.D a is not known, then the sample S.D can be used in the place of o in estimating the confidence interval.

Question 2.
A factory is producing 50,000 pairs of shoes daily. From a sample of 500 pairs, 2% were found to be of sub-standard quality. Estimate the number of pairs that can be reasonably expected to be spoiled at 95% level of confidence.
Solution:
N = 50,000, n = 500, P = \(\frac{2}{100}\), Q = \(\frac{98}{100}\)
The estimated percentage of spoiled pairs in daily production = \(\frac{2}{100}\) × 50,000 = 1000
The limits for the number of spoiled pans at 95% level of confidence

Question 3.
A company that packages peanuts states that at a maximum 6% of the peanut shells contain no nuts. At random, 300 peanuts were selected and 21 of them were empty. With a significance level of 1% can the statement made by the company be accepted?
Solution:
The population proportion P = 6% = 0.06
The null hypothesis: H₀ : P ≤ 0.06
Alternative hypothesis: H₁ : P > 0.06
For α = 1% = 0.01. Z_α = 2.33
The test statistic is P + 2.33 (\(\sqrt{\frac{(0.06)(0.94)}{300}}\)) = 0.092
(where n = 300, P = 0.06, Q = 0.94)
Since the calculated value is less than the table value, 0.092 < 2.33,we accept the null hypothesis H₀.
Hence the statement of the company can be accepted.

Question 4.
A school principal claims that the students in his school are above average intelligence. A random sample of 30 students IQ scores has a mean score of 112.5. The mean population IQ is 100 with an SD of 15. Is there sufficient evidence to support the principal’s claim?
Solution:
Given
Population mean µ = 100
Population SD σ = 15
Sample size n = 30
Sample mean \(\bar{x}\) = 112.5
Null hypothesis H₀ : µ = 100
(the students have average I.Q)
Alternative hypothesis H₁ : µ > 100
(the students have above average I.Q scores)

Let the significance level α = 0.05. The table value Z_α = 1.645, since this is one-tailed test.
Test Statistic:

Since 4.56 > 1.645 (ie) Z > Z_α at 5% level, we reject the null hypothesis. Hence we conclude that the students have above average IQ scores. So the principal’s claim is right.

Question 5.
Boys of a certain age are known to have a mean weight of 85 pounds. A complaint is made that the boys living in hostels are underfed. So a sample of 25 boys are weighed and ‘ found to have a mean weight of 80.94 pounds. The population S.D is 11.6. What should be concluded about the complaint?
Solution:
Given n = 25, µ = 85, \(\bar{x}\) = 80.94, and σ = 11.6
Null hypothesis H₀ : µ = 85 (the boys are not underfed)
Alternative hypothesis H₁ : µ < 85 (the boys are underfed)
Test statistic:

Let us take the significance level α = 0.05. The table value is Z_α = -1.645

Now -1.75 < -1.645 (i.e) Z < Z_α. Therefore we reject the null hypothesis. Since the calculated value falls in the rejection region. Hence we conclude that the boys are underfed. So the complaint should be addressed immediately.