Case study (4 Marks)

Question 14 Marks

Using matrix method, solve the following system of equations for x, y and z :
x - y + z = 4
2x + y - 3z = 0
x + y + z = 2

Answer

The matrix equation AX = B is
$\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$
|A| = 10
$\operatorname{adj} A=\left[\begin{array}{ccc}4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3\end{array}\right]^{\prime}=\left[\begin{array}{rrc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$
Here $A ^{-1}=\frac{1}{10}\left[\begin{array}{rrr}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$
So, $\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]=\left[\begin{array}{c}2 \\ -1 \\ 1\end{array}\right]$
Thus, $x =2, y =-1, z =1$

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Question 24 Marks

Find the inverse of the matrix :
$A=\left[\begin{array}{rrr}-1 & 1 & 2 \\ 3 & -1 & 1 \\ -1 & 3 & 4\end{array}\right]$
and hence show that $AA ^{-1}=1$.

Answer

Here, |A| = -(-4 - 3) - (12 + 1) + 2(9 - 1)
$=7-13+16=10 \neq 0$
$\begin{array}{l}\Rightarrow \operatorname{adj}( A )=\left[\begin{array}{rrr}-7 & -13 & 8 \\ 2 & -2 & 2 \\ 3 & 7 & -2\end{array}\right]^T=\left[\begin{array}{rrr}-7 & 2 & 3 \\ -13 & -2 & 7 \\ 8 & 2 & -2\end{array}\right] \\ \text { Hence } A ^{-1}=\frac{1}{10}\left[\begin{array}{rrr}-7 & 2 & 3 \\ -13 & -2 & 7 \\ 8 & 2 & -2\end{array}\right] \\ A A^{-1}=\frac{1}{10}\left[\begin{array}{ccc}-1 & 1 & 2 \\ 3 & -1 & 1 \\ -1 & 3 & 4\end{array}\right]\left[\begin{array}{ccc}-7 & 2 & 3 \\ -13 & -2 & 7 \\ 8 & 2 & -2\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\end{array}$

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Question 34 Marks

Find principal contained in 40th payment.

Answer

Given, $P =₹ 250000, i =\frac{6}{12 \times 100}=0.005$ and $n =5 \times 12=60$
Principal paid in 40th payment = EMI - Interest paid in 40th payment
= 4832.69 - 480.48 = ₹ 4352.21

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Question 44 Marks

An equated monthly installment (EMI) is a set monthly payment provided by a borrower to a creditor on a set day, each month. EMIs apply to both interest and principal each month, and the loan is paid off in full over some years.
How is EMI calculated?
There are two ways in which EMI can be calculated. These methods are:
•The flat rate method: When the loan amount is progressively being repaid, each interest charge is computed using the original principal amount in the flat rate method.
•The reducing balance method: The reducing balance technique, compared to the flat rate method, determines the interest payment according to the outstanding principal.
Example:
A loan of ₹250000 at the interest rate of 6% p.a. compounded monthly is to be amortized by equal payments at the end of each month for 5 years.
$\left(\right.$ Given $\left.(1.005)^{60}=1.3489,(1.005)^{21}=1.1104\right)$
(a) Find the size of each monthly payment.
(b) Find the principal outstanding at beginning of 40th month.
(c) Find interest paid in 40th payment.

Answer

An equated monthly installment (EMI) is a set monthly payment provided by a borrower to a creditor on a set day, each month. EMIs apply to both interest and principal each month, and the loan is paid off in full over some years.
How is EMI calculated?
There are two ways in which EMI can be calculated. These methods are:
The flat rate method: When the loan amount is progressively being repaid, each interest charge is computed using the original principal amount in the flat rate method.
The reducing balance method: The reducing balance technique, compared to the flat rate method, determines the interest payment according to the outstanding principal.
Example:
A loan of ₹250000 at the interest rate of 6% p.a. compounded monthly is to be amortized by equal payments at the end of each month for 5 years.
$\left(\right.$ Given $\left.(1.005)^{60}=1.3489,(1.005)^{21}=1.1104\right)$
(i) Given, P =₹ 250000,$i =\frac{6}{12 \times 100}=0.005$ and $n =5 \times 12=60$
$EMI =\frac{250000 \times 0.005 \times(1.005)^{60}}{(1.005)^{60}-1}$
$=\frac{250000 \times 0.005 \times 1.3489}{0.3489}=₹ 4832.69$
(ii) Given, $P =₹ 250000, i =\frac{6}{12 \times 100}=0.005$ and $n =5 \times 12=60$
Principal outstanding at beginning of 40th month
$=\frac{ EMI \left[(1+i)^{60-40+1}-1\right]}{i(1+i)^{60-40+1}}=\frac{4832.69 \times\left[(1.005)^{21}-1\right]}{0.005 \times(1.005)^{21}}$
$=\frac{4832.69 \times[1.1104-1]}{0.005 \times 1.1104}=\frac{4832.69 \times 0.1104}{0.005 \times 1.1104}=₹ 96096.72$
(iii)Given, $P =₹ 250000, i =\frac{6}{12 \times 100}=0.005$ and $n =5 \times 12=60$
Interest paid in 40th payment $=\frac{\operatorname{EMI}\left[(1+i)^{60-40+1}-1\right]}{(1+i)^{60-40+1}}$
$=\frac{4832.69 \times\left[(1.005)^{21}-1\right]}{(1.005)^{21}}=\frac{4832.69 \times 0.1104}{1.1104}=₹ 480.48$

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Question 54 Marks

Find the least cost of construction of the tank.

Answer

The least cost of the tank = ₹ $\left[1120+720\left(2+\frac{4}{2}\right)\right]$
= F(1120 + 2880) =F 4000

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Question 64 Marks

A factory owner wants to construct a tank with rectangular base and rectangular sides, open at the top, so that its depth is 2 m and capacity is $8 m^3$. The building of the tank costs ₹280 per square metre for the base and ₹180 per square metre for the sides.

(a) If the length and the breadth of the rectangular base of the tank are x metres and y metres respectively, then find a relation between x and y.
(b) If C (in ₹) is the cost of construction of the tank, then find C as a function of x.
(c) Find the value of x for which the cost of construction of the tank is least.

Answer

A factory owner wants to construct a tank with rectangular base and rectangular sides, open at the top, so that its depth is 2 m and capacity is $8 m^3$. The building of the tank costs ₹280 per square metre for the base and ₹180 per square metre for the sides.

(i) Given volume $=8 m^3$ and height $=2 m$ So, $x \times y \times 2=8 \Rightarrow x y=4$
(ii) Area of sides of the tank $=2(x+y) \times h=4(x+y) m^2$
$\therefore$ The cost of construction of the sides of the tank $=₹ 180 \times 4(x+y)=₹ 720(x+y)$ The cost of construction of the base of the tank $=₹ x \times y \times 280=₹ 4 \times 280=₹ 1120$ (using part ( $x \times y \times 2=8 \Rightarrow x y$ $=4)$ )
$\begin{array}{l}\text { So, } C =₹[1120+720( x + y )] \\ \Rightarrow C =1120+720\left(x+\frac{4}{x}\right)\end{array}$
(iii) $\frac{d C }{d x}==0+720\left(1-\frac{4}{x^2}\right)$ and $\frac{d^2 C}{d x^2}=\frac{5760}{x^3}$.
Now, $\frac{d C }{d x}=0 \Rightarrow\left(1-\frac{4}{x^2}\right)=0 \Rightarrow x ^2=4 \Rightarrow x =2$
$\left[\frac{d^2 C }{d x^2}\right]_{x=2}=\frac{5760}{8}>0$
$\Rightarrow C$ is minimum when $x =2$

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Applied Maths Case study (4 Marks)

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