Answer
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Hint: In order to solve first find out the number of installments to be made. Then find the sums to be paid as installments. Find some sort of series in the installments to be made. And finally find the sum of the series.
Complete Step-by-Step solution:
As we know that
Amount to be paid to buy the tractor = Rs. 12000
Amount paid in cash = Rs. 6000
Remaining balance to be paid \[ = Rs.12000-Rs.6000 = Rs.6000\]
Given that Annual installment is Rs. 500 plus 12% interest on the unpaid amount.
For the first installment
Unpaid amount = Rs. 6000
Interest on unpaid amount
$
= 12\% {\text{ of }}Rs.6000 \\
= \dfrac{{12}}{{100}} \times 6000 \\
= Rs.720 \\
$
Amount of installment to be paid
$
= Rs.500 + Rs.720 \\
= Rs.1220 \\
$
For the second installment
Unpaid amount
$
= Rs.6000 - Rs.500 \\
= Rs.5500 \\
$
Interest on unpaid amount
$
= 12\% {\text{ of }}Rs.{\text{5500}} \\
= \dfrac{{12}}{{100}} \times 5500 \\
= Rs.660 \\
$
Amount of installment to be paid
$
= Rs.500 + Rs.660 \\
= Rs.1160 \\
$
Similarly for the third installment
Unpaid amount
$
= Rs.5500 - Rs.500 \\
= Rs.5000 \\
$
Interest on unpaid amount
$
= 12\% {\text{ of }}Rs.{\text{5000}} \\
= \dfrac{{12}}{{100}} \times 5000 \\
= Rs.600 \\
$
Amount of installment to be paid
$
= Rs.500 + Rs.600 \\
= Rs.1100 \\
$
Thus our installments are $Rs.1220,Rs.1160,Rs.1100,........$
Total number of installments will be
$
= \dfrac{{{\text{Remaining balance left}}}}{{{\text{Balance cleared per instalment}}}} \\
= \dfrac{{Rs.6000}}{{Rs.500}} \\
= 12 \\
$
Thus our installments are $Rs.1220,Rs.1160,Rs.1100,........$ upto 12 terms.
We can observe that the terms are decreasing every time by Rs. 60. So these balances to be paid are in A.P.
We can find out the sum of 12 terms of this A.P. easily by the help of formula for A.P.
Here,
First term $\left( a \right) = 1220$
Common difference $\left( d \right) = 1160 - 1220 = - 60$
Number of terms $\left( n \right) = 12$
As we know the formula for sum of n terms of A.P. is given by:
${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
Where, ${S_n}$ is the sum of n terms of an A.P.
n = number of terms
a = first term
d = common difference.
Let us substitute all the values for the given problem
Sum of 12 terms of A.P. is:
$
\because {S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right) \\
\Rightarrow {S_{12}} = \dfrac{{12}}{2}\left( {2\left( {1220} \right) + \left( {12 - 1} \right)\left( { - 60} \right)} \right) \\
\Rightarrow {S_{12}} = 6\left( {2440 + 11 \times \left( { - 60} \right)} \right) \\
\Rightarrow {S_{12}} = 6\left( {2440 - 660} \right) \\
\Rightarrow {S_{12}} = 6 \times \left( {1780} \right) \\
\Rightarrow {S_{12}} = 10680 \\
$
Hence, total amount to be paid in 12 installments = Rs. 10680
So, the total cost of the tractor = Amount paid in cash + Amount paid in 12 installment
$
= Rs.6000 + Rs.10680 \\
= Rs.16680 \\
$
Hence, the tractor will cost Rs. 16680 to the farmer.
Note: In order to solve these types of problems where we need to find the sum of money for 12-14 numbers of times, the best way to proceed is to find some kind of series or relation between the terms and directly find the sum. The question would be very hard to solve if we would have found the money for each installment separately.
Complete Step-by-Step solution:
As we know that
Amount to be paid to buy the tractor = Rs. 12000
Amount paid in cash = Rs. 6000
Remaining balance to be paid \[ = Rs.12000-Rs.6000 = Rs.6000\]
Given that Annual installment is Rs. 500 plus 12% interest on the unpaid amount.
For the first installment
Unpaid amount = Rs. 6000
Interest on unpaid amount
$
= 12\% {\text{ of }}Rs.6000 \\
= \dfrac{{12}}{{100}} \times 6000 \\
= Rs.720 \\
$
Amount of installment to be paid
$
= Rs.500 + Rs.720 \\
= Rs.1220 \\
$
For the second installment
Unpaid amount
$
= Rs.6000 - Rs.500 \\
= Rs.5500 \\
$
Interest on unpaid amount
$
= 12\% {\text{ of }}Rs.{\text{5500}} \\
= \dfrac{{12}}{{100}} \times 5500 \\
= Rs.660 \\
$
Amount of installment to be paid
$
= Rs.500 + Rs.660 \\
= Rs.1160 \\
$
Similarly for the third installment
Unpaid amount
$
= Rs.5500 - Rs.500 \\
= Rs.5000 \\
$
Interest on unpaid amount
$
= 12\% {\text{ of }}Rs.{\text{5000}} \\
= \dfrac{{12}}{{100}} \times 5000 \\
= Rs.600 \\
$
Amount of installment to be paid
$
= Rs.500 + Rs.600 \\
= Rs.1100 \\
$
Thus our installments are $Rs.1220,Rs.1160,Rs.1100,........$
Total number of installments will be
$
= \dfrac{{{\text{Remaining balance left}}}}{{{\text{Balance cleared per instalment}}}} \\
= \dfrac{{Rs.6000}}{{Rs.500}} \\
= 12 \\
$
Thus our installments are $Rs.1220,Rs.1160,Rs.1100,........$ upto 12 terms.
We can observe that the terms are decreasing every time by Rs. 60. So these balances to be paid are in A.P.
We can find out the sum of 12 terms of this A.P. easily by the help of formula for A.P.
Here,
First term $\left( a \right) = 1220$
Common difference $\left( d \right) = 1160 - 1220 = - 60$
Number of terms $\left( n \right) = 12$
As we know the formula for sum of n terms of A.P. is given by:
${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
Where, ${S_n}$ is the sum of n terms of an A.P.
n = number of terms
a = first term
d = common difference.
Let us substitute all the values for the given problem
Sum of 12 terms of A.P. is:
$
\because {S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right) \\
\Rightarrow {S_{12}} = \dfrac{{12}}{2}\left( {2\left( {1220} \right) + \left( {12 - 1} \right)\left( { - 60} \right)} \right) \\
\Rightarrow {S_{12}} = 6\left( {2440 + 11 \times \left( { - 60} \right)} \right) \\
\Rightarrow {S_{12}} = 6\left( {2440 - 660} \right) \\
\Rightarrow {S_{12}} = 6 \times \left( {1780} \right) \\
\Rightarrow {S_{12}} = 10680 \\
$
Hence, total amount to be paid in 12 installments = Rs. 10680
So, the total cost of the tractor = Amount paid in cash + Amount paid in 12 installment
$
= Rs.6000 + Rs.10680 \\
= Rs.16680 \\
$
Hence, the tractor will cost Rs. 16680 to the farmer.
Note: In order to solve these types of problems where we need to find the sum of money for 12-14 numbers of times, the best way to proceed is to find some kind of series or relation between the terms and directly find the sum. The question would be very hard to solve if we would have found the money for each installment separately.
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