A farmer buys a used tractor for Rs. 12000. He pays Rs. 6000 cash and agrees to pay the balance in annual instalments of Rs. 500 plus 12% interest on the unpaid amount. How match the tractor cost him?

Answer

N/a

Explanation

Cost price = Rs. 12000
Cash payment = Rs. 6000
Remaining amount = Rs. 12000 - Rs. 6000 = Rs. 6000
Annual instalment = Rs. 500
Number of instalment `= (6000)/(500) = 12`
For first instalment
Principal = Rs. 6000
`therefore` Insterest `= Rs. (6000 xx 12 xx 1)/(100) = Rs. 720`
For second instalment
Principal = Rs. 6000 - Rs. 500 = Rs. 5500
`therefore` Interest `= Rs. (5500 xx 12 xx1)/(100) = Rs. 660`
For third instalment
Principal = Rs. 5500 - Rs. 500 = Rs. 5000
`therefore` Interest `= Rs. (5000 xx 12 xx 1)/(100) = Rs. 600`
Progression formed by the interests :
720 + 660 + 600 + ... 12 terms
It is an A.P.
Here a = 720
d = 660 - 720 = 600 - 660 = -60
n = 12
sum `S = (n)/(2)[2a + (n - 1)d]`
`= (12)/(2)[2 xx 720 + (12 - 1)(-60)]`
= Rs. 4680
Now, total cost of tractor `= Rs. 12000 + Rs. 4680`
= Rs. 16680