Since the required value is at the end of the table, apply backward interpolation formula

x	y	\(\Delta y\)	\(\Delta ^{ 2 }y\)	\(\Delta ^{ 2 }y\)	\(\Delta ^{ 4 }y\)
1	1
		7
2	8		12
		19		6
3	27		18		0
		37		6
4	64		24		0
		61		6
5	125		30		0
		91		6
6	216		36		0
		127		6
7	343		42
		169
8	512

.\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+...\)
To find y at x = 7.5
\(\therefore\) xn + nh = 7.5, xn = 8, h = 1 \(\Rightarrow\) n = –0.5
\({ y }_{ (x=7.5) }=512+\frac { -0.5 }{ 1! } 169+\frac { 0.5(-0.5+10 }{ 2! } 42+\frac { -0.5(-0.5+1)(-0.5+2) }{ 3! } 6\)
= 421.88

Here, the number of rows and columns are equal
∴ The given assignment problem is balanced.
Step I : Select a smallest element in each row and subtract this from all the elements in its row.
∴ The given assignment problem is

Here column 2 has no zero. Go to Step 2.
Step 2 : Select the smallest element (10) and subtract it from all the elements in its column.

Step 3 : Examine the rows with only one zero mark that zero by Ԡ. Mark other zeros in its column by X.
Row 1 and Row 3 contains only one zero. Mark the other zeros by X
Column 2 contains exactly one zero. Mark it by Ԡ
Thus, all the 3 assignments have been made.
Hence, the optimal assignment schedule and total cost is

Programmers	Programmes	Cost
1	R	80
2	Q	90
3	P	110

Total cost = Rs. 280
Thus, the optimal assignment (minimum) cost  =  Rs. 280

(D² − 2D + 1)y = e^2x + e^x
The auxiliary equation is
m² − 2m + 1 = 0
⇒ (m−1)(m−1) = 0
m = 1, 1
C.F = (Ax +B)e^x
\(PI=\frac { 1 }{ \phi (D) } f(x)\) = \(\frac { 1 }{ { D }^{ 2 }-2D+1 } \)(e^2x + e^x)
P.I₁ = \(\frac { 1 }{ { D }^{ 2 }-2D+1 } \)e^2x
= \(\frac { 1 }{ 4-4+1 } \)e^2x (replace D by 2)
= e^2x
and P.I₂ = \(\frac { 1 }{ { D }^{ 2 }-2D+1 } \) e^x
= \(\frac { 1 }{ { (D-1) }^{ 2 } } { e }^{ x }\)
Replace D by 1. (D −1)² = 0 when D = 1
∴ \(P.{ I }_{ 2 }={ x }_{ 2 }.\frac { 1 }{ { 2(D-1) }^{ 2 } } { e }^{ x }\)
Replace D by 1. (D −1) = 0 when D = 1
∴ P.I₂ = x²\(\frac { 1 }{ 2 } { e }^{ x }\)
The general solution is
y = C.F + P.I₁ + P.I₂
y = (Ax +B)e^x + e^2x + \(\frac { { x }^{ 2 } }{ 2 } { e }^{ x }\)

Commodities Price Weights
(V) \(P=\frac { { P }_{ 1 } }{ { P }_{ 0 } } \times 100\) PV 2007
(P₀) 2011
(P₁) A 350 400 40 114.285 4571.4 B 175 250 35 142.857 4999.99 C 100 115 15 115 1725 D 75 105 20 114 2280 E 60 80 25 133.333 3333.33     Total 135   17429 76 Value of Cost of Living Index Number\(\frac { \sum { PV } }{ \sum { V } } =\frac { 17429 76 }{ 135 } =129.1093\)
Hence, the Cost of Living Index Number for a particular class of people for the year 2011 is increased by 129.1093 % as compared to the year 2007.

Sample size n = 1600
Sample mean \(\bar { X } \) = 99
Population mean μ = 100
Population standard deviation σ = 15
Null Hypotheses H_o : μ = 100
Alternative Hypotheses H₁ : μ ≠ 100
Level of significance α = 0.05
Applying the test statistic,
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 99-100 }{ \frac { 15 }{ \sqrt { 1600 } } } =\frac { -1 }{ \frac { 15 }{ 40 } } =-\frac { 40 }{ 15 } =-2.667\)
∴ |Z| = 2.667
Critical value at 5% level of significance is \({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
Here |Z|>\({ Z }_{ \frac { \alpha }{ 2 } }\) as 2.667>1.96
Inference: since |Z|>\({ Z }_{ \frac { \alpha }{ 2 } }\) at 5% level of sinificance, the null hypothesis is rejected. 
Hence, we can conclude that the sample has not been taken from the population with mean μ = 100.

\(\int _{ a }^{ b }{ f\left( x \right) } dx=\lim _{ n\rightarrow \infty \\ h\rightarrow 0 }{ \sum _{ r=1 }^{ n }{ h.f(a+rh) } } \)
Here a = 1, b = 3
\(h=\frac { b-a }{ n } =\frac { 3-1 }{ n } =\frac { 2 }{ n } \)
and f (x) = x
Now \(f(a+rh)=f\left( 1+r\cfrac { 2 }{ n } \right) =f\left( 1+\cfrac { 2r }{ n } \right) \)
\(\therefore \int _{ 1 }^{ 3 }{ xdx } =\underset { n\rightarrow \infty }{ lim } .\cfrac { 2 }{ n } \left( 1+\cfrac { 2r }{ n } \right) \)
= \(\underset { n\rightarrow \infty }{ lim } \sum _{ r=1 }^{ n }{ \left( \cfrac { 2 }{ n } +\cfrac { 4r }{ { n }^{ 2 } } \right) } \)
= \(\underset { n\rightarrow \infty }{ lim } \left( \cfrac { 2 }{ n } .\sum _{ r=1 }^{ n }{ .1+\cfrac { 4 }{ { n }^{ 2 } } .\sum _{ r=1 }^{ n }{ .r } } \right) \)
\(\left( \because \sum _{ r=1 }^{ n }{ 1=n } \right) \)

\(\left[ \sum _{ r=1 }^{ n }{ r } =\cfrac { n(n+1) }{ 2 } \right] \)

= \(2+\underset { n\rightarrow \infty }{ lim } \left( 1+\cfrac { 1 }{ n } \right) ^{ 2 }\)
= \(2+n\underset { n\rightarrow \infty }{ lim } \left( 2+\cfrac { 2 }{ n } \right) \)
= \(2+2\underset { n\rightarrow \infty }{ +lim } \left( 2+\cfrac { 2 }{ n } \right) \)

= 2 + 2 + 0
= 4

Let p be the probability of having a laptop
Given p = \(\frac { 40 }{ 100 } \) = 0.4 ⇒ q = 1-p = 1-0.4 = 0.6
n = 15
(i) P(X = 3)
=¹⁵C₃ (0.4)³ (0.6)¹²  ∵ p(x) =ⁿC_x p^xq^n-x, n = 15, x = 3

= 455 (0.064) (0.002176) = 0.0634
(ii) P (12 travellers not having laptop)
= P (3 travellers having laptop)
[∵ n = 15 and 15 - 12 = 3]
= P(X = 3)
= 0.0634.
(iii) P (atleast 3 travellers have laptop)
= P (X ≥ 3) = 1 - P(X < 3)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
=1-[¹⁵C₀ (0.4)⁰ (0.6)¹⁵ + ¹⁵C₁ (0.4)² (0.6)¹³ (0.6)¹⁴]

= 1 - (0.6)¹³ [(0.6)² + 6(0.6)¹ + 105(0.16)]
= 1 - (0.6)¹³ [0.36 + 3.6 + 16.8]
= 1 - (0.0013) (20.76) = 1 - 0.0270
= 0.9730.

Given demand function P_d = 25 - 3x and
Supply function P_s= 5 + 2x
At market equilibrium, P_d = P_s
⇒ 25-3x = 5+2x
⇒ 25-5 = 2x+3x
⇒ 20 = 5x
⇒ x = \(\frac{20}{5}\)
⇒ x₀ = 4
When x₀ = 4, p₀ = 25-3(4)
= 25-12 = 13
p₀ = 13
∴p₀x₀ = 13(4) = 52
∴ Consumer's surplus
\(CS=\int _{ 0 }^{ x }{ f(x) } dx-{ p }_{ 0 }{ x }_{ 0 }\)
\(\\ =\int _{ 0 }^{ 4 }{ (25-3x)dx-52 } \)
\(={ \left[ 25x-\frac { { 3x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 4 }-52\)
\(=25(4)-\frac { 3\left( { 4 }^{ 2 } \right) }{ 2 } -52\)
=100-24-52
=100-76
C.S = 24 units
Producer's Surplus
\((PS)={ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x }{ g(x)dx } \)
\(=52-\int _{ 0 }^{ 4 }{ (5+2x)dx } \)

= 52 - (5(4) + 4²}
= 52 - (20 + 16)
= 52 - 36
PS = 16 units

Given probability function is

Value of X = x	0	1	2	3	4	5	6	7
P(x)	0	k	2k	2k	3k	k2	2k2	7k2+k

i) Since the given function is a probability function, each ρ_i>0 and Σρ_i=1
⇒ 0+k+2k+2k+3k+k²+2k²+7k²+k = 1
⇒ 10k²+9k = 1
⇒10k²+9k-1 = 0
on factoring we get
(k+1)(10k-1) = 0
⇒k = -1 or k = \(\frac{1}{10}\)
k = -1 is not possible [each ρ_i>0]
we get k = \(\frac{1}{10}\)
ii) P(X<6) = P(X = 0)+P(X = 1)+P(X=2+P(X=3)+P(X=4)+P(X=5)
P(X<6) = 0 + k + 2k + 2k + 3k + k²
= 8k+k²
= \(8\left( \frac { 1 }{ 10 } \right) { \left( \frac { 1 }{ 10 } \right) }^{ 2 }\)
\(=\frac { 8 }{ 10 } +\frac { 1 }{ 100 } =\frac { 80+1 }{ 100 } =\frac { 81 }{ 100 } \)
\(\therefore P(X<6)=\frac { 81 }{ 100 } \)
Now P(X≥6) = P(X=6)+P(X=7)
= 2k²+7k²+k
= 9k²+k
\(=9{ \left( \frac { 1 }{ 10 } \right) }^{ 2 }+\frac { 1 }{ 10 } \)
\(=\frac { 9 }{ 100 } +\frac { 1 }{ 10 } =\frac { 9+10 }{ 100 } =\frac { 19 }{ 100 } \)
\(\therefore P(X\ge 6)=\frac { 19 }{ 100 } \)
And P(0X<5) = P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)
= k+2k+2k+3k
\(8k=8\left( \frac { 1 }{ 10 } \right) =\frac { 8 }{ 10 } \)
\( \therefore P(0\))
iii) Given P(X ≤ x) ≥ \(\frac{1}{2}\)
⇒P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)
= 0+k+2k+2k+3k
\(=8k=8\left( \frac { 1 }{ 10 } \right) =\frac { 8 }{ 10 } =\frac { 4 }{ 5 } >\frac { 1 }{ 2 } \)
\(\therefore P(X\le 4)\frac { 1 }{ 2 } \le x=4\)
∴ The minimum value of x is 4.

Let ‘x’ be the cost of a Business Mathematics book
Let ‘y’ be the cost of a Accountancy book.
Let ‘z’ be the cost of a Commerce book.
\(\therefore \) 3x + 2y + z = 840
2x + y + z = 570
x + y + 2z = 630
Here \({ \triangle }=\left| \begin{matrix} 3 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{matrix} \right| =-2\neq 0\)
\({ \triangle }_{ x }=\left| \begin{matrix} 840 & 2 & 1 \\ 570 & 1 & 1 \\ 630 & 1 & 2 \end{matrix}\begin{matrix} 1 \\ 1 \\ 2 \end{matrix} \right| =-240 \)
\({ \triangle }_{ y }=\left| \begin{matrix} 3 & 840 & 1 \\ 2 & 570 & 1 \\ 1 & 630 & 2 \end{matrix} \right| =-300 \)
\({ \triangle }_{ z }=\left| \begin{matrix} 3 & 2 & 840 \\ 2 & 1 & 570 \\ 1 & 1 & 630 \end{matrix} \right| =-360\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }x }{ { \triangle } } =\frac { -240 }{ -2 } =120 \)
\(y=\frac { { \triangle }y }{ { \triangle } } =\frac { 300 }{ -2 } =150 \)
\(z=\frac { { \triangle }z }{ { \triangle } } =\frac { 360 }{ -2 } =180\) 
\(\therefore \) The cost of a Business Mathematics book is Rs. 120,
the cost of a Accountancy book is Rs. 150 and
the cost of a Commerce book is Rs. 180.