Case study (4 Marks)

Question 14 Marks

Rohan has completed his MBA and now he wants to start a new business. So, he approaches to many banks. One bank is agreed to give loan to Rohan. So, Rohan has borrowed ₹ 5 lakhs from a bank on the interest rate of 12 per cent for 10 years.
Q. 1. EMI stands for:
(A) Equated Monthly Instalments
(B) Emerging Monthly Instalments
(C) Easy Monthly Instalments
(D) None of the above
Q. 2. To calculate monthly instalment, we use the following formula:
(A) Instalment Amount $=\frac{(1+i)^n}{(1+i)^n} \times(P \times i)$
(B) Instalment Amount $=\frac{(1+i)^n}{(1+i)^n-1} \times(P \times i)$
(C) Instalment Amount $=\frac{(1+i)^n}{(1+i)^{n-1}} \times(P \times i)$
(D) None of the above
Q. 3. Calculate monthly instalment using $( 1 . 0 1 )^{120}$ $=3.300$
(A) ₹ 7100
(B) ₹ 7174
(C) ₹ 7147
(D) ₹ 7200
Q. 4. Find the amount of total payment made by Rohan.
(A) ₹ $8,60,88$
(B) ₹ $8,80,880$
(C) ₹ $8,60,000$
(D) ₹ $8,60,880$

Answer

(1) (A) Equated Monthly Instalments
(2) (B) Instalment Amount $=\frac{(1+i)^n}{(1+i)^n} \times(P \times i)$
Explanation: Formula to calculate monthly instalment is:
$
\text { Instalment Amount }=\frac{(1+i)^n}{(1+i)^n-1} \times(P \times i)
$
(3) (B)
Explanation:Given,
$
i=\left[\frac{\left(\frac{\text { annual rate }}{12}\right)}{100}\right]
$
$\begin{aligned} & =\left[\frac{\left(\frac{12}{12}\right)}{100}\right] \\ & =\frac{1}{100} \\ & =0.01 \\ n & =10 \times 12 \\ & =120\end{aligned}$
P = ₹ 5,00,000
$\begin{array}{l}\text { Instalment Amount }=\frac{(1+i)^n}{(1+i)^n-1} \times(P \times i) \\ \text { Instalment Amount }=\frac{(1+0.01)^{120}}{(1+0.01)^{120}-1}\end{array}$
$\begin{array}{l} \quad \times(5,00,000 \times 0.01) \\ = \frac{3.300}{3.300-1} \times 5,000 \\ = \frac{16,500}{2.300}\end{array}$
= ₹ 7173.91 ~ ₹7174
So, EMI that Rohan has to pay is ₹ 7174
(4) (D) ₹ 8,60,880
Explanation: Total payment made by Rohan to the bank in 10 years $=($ EMI $\times$ Total tenure in months $)$
= ₹ 8,60,880

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Question 24 Marks

Ajay is a service man. He lives in a joint family. There are 6 members in his family. He is planning to purchase a car so he is searching for a bank for a loan. He take a loan of ₹ 250000 at the interest rate of $6 \%$ p.a. compounded monthly is to be amortize by equal payment at the end of each month for 5 years. Given that $(1.005)^{60}=1.3489,(1.005)^{21}=1.1104$.

Q. 1. The size of each monthly payment is:
(A) ₹ 4833
(B) ₹ 8432
(C) ₹ 4383
(D) ₹ 4338
Q. 2. The principal outstanding at beginning of $40^{\text {th }}$ month is:
(A) ₹ 60069
(B) ₹ 96097
(C) ₹ 96000
(D) ₹ 98670
Q. 3. The interest paid in $40^{\text {th }}$ payment is:
(A) ₹ 480
(B) ₹ 408
(C) ₹ 840
(D) ₹ 482
Q. 4. Principal amount paid in $40^{\text {th }}$ payment is:
(A) ₹ 4325
(B) ₹ 4359
(C) ₹ 4352
(D) ₹ 4532

Answer

(1) (A) ₹ 4833
Explanation: Given, P=₹ 25,000
and $\quad \begin{aligned} i & =\frac{6}{12 \times 100} \\ & =0.005 \\ n & =5 \times 12 \\ & =60\end{aligned}$
$\begin{aligned} \therefore \quad EMI & =\frac{p \times i \times(1+i)^n}{(1+i)^n-1} \\ & =\frac{25000 \times 0.005 \times(1.005)^{60}}{(1.005)^{60}-1} \\ & =\frac{25000 \times 0.005 \times 1.3489}{0.3489}\end{aligned}$
=₹ 4832.69
=₹ 4833
(2) (B) ₹ 96097
Explanation: Principal outstanding at beginning of $40^{\text {th }}$ month
$
=\frac{\operatorname{EMI}\left[(1+i)^{n-40+1}-1\right]}{i(1+i)^{n-40+1}}
$
$\begin{array}{l}=\frac{4832.69\left[(1+0.005)^{60-40+1}-1\right]}{0.005(1+0.005)^{60-40+1}} \\ =\frac{4832.69\left[(1.005)^{21}-1\right]}{0.005 \times(1.005)^{21}} \\ =\frac{4832.69[1.1104-1]}{0.005 \times 1.1104} \\ =\frac{4832.69 \times 0.1104}{0.005 \times 1.1104}\end{array}$
=₹ 96096.72
= ₹ 96097
(3) (A) ₹ 480
Explanation: Interest paid in $40^{\text {th }}$ payment
$
\begin{array}{l}
=\frac{EMI\left[(1+i)^{n-40+1}-1\right]}{(1+i)^{n-40+1}} \\
=\frac{4832.69\left[(1+i)^{60-40+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}
\end{array}
$
= ₹ 480.48
= ₹ 480
(4) (C) ₹ 4352
$\begin{array}{l}\text { Explanation: Principal paid in } 40^{ th } \text { payment } \\ \quad= EMI \text { - Interest paid in } 40^{\text {th }} \text { payment } \\ =4832.69-480.48\end{array}$
= ₹ 4352.21
= ₹ 4352

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Applied Maths Case study (4 Marks)

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