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A friend of you wants to buy an electric car for which he plans to take a loan from a bank and plans to pay the total loan and the interest = ₹$5,90,000$, by paying every month starting with the first instalment of ₹$5,000$. He increases the instalment by ₹$500$ every month.
Based on the above, answer the following questions :
(a) What are the first three instalments paid by him ?
(b) Find the amount to be paid by him in $11^{th}$ instalment.
(c) Find the number of instalments in which he would clear his total loan.
OR
(c) After paying the $31^{st}$ instalment, find how much money he still has to pay.
Based on the above, answer the following questions :
(a) What are the first three instalments paid by him ?
(b) Find the amount to be paid by him in $11^{th}$ instalment.
(c) Find the number of instalments in which he would clear his total loan.
OR
(c) After paying the $31^{st}$ instalment, find how much money he still has to pay.
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Total amount = ₹$5,90,000$
First instalment = ₹$5,000$ and Increase in the instalment = ₹$500$
(a) ₹$5,000$ ; ₹$5,500$ ; ₹$6,000$ (1 Mark)
(b) $11^{th}$ instalment ($a_{11}$) = $5000 + 10 \times 500$ (1 Mark)
$= 10000$
Amount paid in $11^{th}$ instalment is ₹$10,000$
(c) Let '$n$' be the number of instalments to clear the loan.
$590000 = \frac{n}{2} [2 \times 5000 + (n - 1)500]$ (1 Mark)
$\Rightarrow n^2 + 19n - 2360 = 0$ ($\frac{1}{2}$ Mark)
$\Rightarrow (n + 59)(n - 40) = 0$ ($\frac{1}{2}$ Mark)
Thus, $n = -59, 40$
rejecting $n = -59$,
$\therefore n = 40$ ($\frac{1}{2}$ Mark)
Thus, the loan will be cleared in $40$ instalments. ($\frac{1}{2}$ Mark)
OR
(c) Amount to be paid upto $31^{st}$ instalment is $S_{31}$ ($\frac{1}{2}$ Mark)
$S_{31} = \frac{31}{2} [2 \times 5000 + (31 - 1)500]$ (1 Mark)
$= \frac{31}{2} \times 25000$
$= 387500$ ($\frac{1}{2}$ Mark)
Amount still to be paid = $5,90,000 - 3,87,500$ ($\frac{1}{2}$ Mark)
$= \text{Rs}2,02,500$ ($\frac{1}{2}$ Mark)
First instalment = ₹$5,000$ and Increase in the instalment = ₹$500$
(a) ₹$5,000$ ; ₹$5,500$ ; ₹$6,000$ (1 Mark)
(b) $11^{th}$ instalment ($a_{11}$) = $5000 + 10 \times 500$ (1 Mark)
$= 10000$
Amount paid in $11^{th}$ instalment is ₹$10,000$
(c) Let '$n$' be the number of instalments to clear the loan.
$590000 = \frac{n}{2} [2 \times 5000 + (n - 1)500]$ (1 Mark)
$\Rightarrow n^2 + 19n - 2360 = 0$ ($\frac{1}{2}$ Mark)
$\Rightarrow (n + 59)(n - 40) = 0$ ($\frac{1}{2}$ Mark)
Thus, $n = -59, 40$
rejecting $n = -59$,
$\therefore n = 40$ ($\frac{1}{2}$ Mark)
Thus, the loan will be cleared in $40$ instalments. ($\frac{1}{2}$ Mark)
OR
(c) Amount to be paid upto $31^{st}$ instalment is $S_{31}$ ($\frac{1}{2}$ Mark)
$S_{31} = \frac{31}{2} [2 \times 5000 + (31 - 1)500]$ (1 Mark)
$= \frac{31}{2} \times 25000$
$= 387500$ ($\frac{1}{2}$ Mark)
Amount still to be paid = $5,90,000 - 3,87,500$ ($\frac{1}{2}$ Mark)
$= \text{Rs}2,02,500$ ($\frac{1}{2}$ Mark)