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Question: Shamshad Ali buys a scooter for \(Rs22000.\) He pays \(4000\) cash and agrees to pay the balance in ...

Shamshad Ali buys a scooter for Rs22000.Rs22000. He pays 40004000 cash and agrees to pay the balance in annual instalment of Rs1000Rs1000 plus 1010% interest on the unpaid amount. How much will the scooter cost him?

Explanation

Solution

Hint: In order to solve this type of question, first we have to calculate the remaining balance and then we have to calculate the interest on the unpaid amount. After that we have to calculate the number of instalments.

Complete step-by-step answer:
Given-
Cost of the scooter = Rs22000Rs22000
Down payment made by cash = Rs4000Rs4000
Remaining balance =Rs22000Rs4000 =Rs18000  Rs22000 - Rs4000 \\\ = Rs18000 \\\
Annual instalment =1000+= 1000 + Interest on unpaid amount @10@10%
1st{1^{st}} Instalment, unpaid amount =18000 = 18000
Interest on unpaid amount =10100×18000=1800 = \dfrac{{10}}{{100}} \times 18000 = 1800
Therefore, amount of instalment =1000+1800=2800(1) = 1000 + 1800 = 2800 - - - - \left( 1 \right)
2nd{2^{nd}} Instalment, unpaid amount=180001000=17000 = 18000 - 1000 = 17000
Interest on unpaid amount ==10100×17000=1700 = = \dfrac{{10}}{{100}} \times 17000 = 1700
Therefore, amount of instalment =1000+1700=2700(2) = 1000 + 1700 = 2700 - - - - - \left( 2 \right)
3rd{3^{rd}} instalment, unpaid amount =170001000=16000 = 17000 - 1000 = 16000
Interest on unpaid amount ==10100×16000=1600 = = \dfrac{{10}}{{100}} \times 16000 = 1600
Therefore, amount of instalment =1000+1600=2600(3) = 1000 + 1600 = 2600 - - - - - \left( 3 \right)
Thus, from (1), (2) and (3) our instalments are 2800,2700,2600.2800,2700,2600.
Number of instalments =Remaining balance leftbalance cleared per instalment=180001000=18 = \dfrac{{\operatorname{Remaining} {\text{ }}balance{\text{ }}left}}{{balance{\text{ cleared per instalment}}}} = \dfrac{{18000}}{{1000}} = 18
So, our instalments are 2800,2700,2600,......to 18terms.2800,2700,2600,......to{\text{ 18terms}}{\text{.}}
We can observe that this is an APAP as difference between consecutive terms is an AP.AP.
Hence,
First term (a)=2800\left( a \right) = 2800
Common difference= (d)=27002800=100\left( d \right) = 2700 - 2800 = - 100
Number of terms=n=18 = n = 18
We need to calculate total amount paid in 1818 instalments i.e.(2800+2700+2600......to 18terms)\left( {2800 + 2700 + 2600......to{\text{ 18terms}}} \right)
We have to apply the formula
Sn=n2(2a+(n1)d){S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)
Where,
Sn=sum of n terms of A.P. n = number of terms. a = first term and d = common difference  {S_n} = sum{\text{ of n terms of A}}{\text{.P}}{\text{.}} \\\ {\text{n = number of terms}}{\text{.}} \\\ {\text{a = first term and d = common difference}} \\\
Putting the value of n=18,a=2800 and d = - 100n = 18,a = 2800{\text{ }}and{\text{ d = - 100}}
Sn=182(2(2800)+(181)(100)){S_n} = \dfrac{{18}}{2}\left( {2\left( {2800} \right) + \left( {18 - 1} \right)\left( { - 100} \right)} \right)
Or Sn=9(5600+17(100)){S_n} = 9\left( {5600 + 17\left( { - 100} \right)} \right)
Or Sn=9(56001700){S_n} = 9\left( {5600 - 1700} \right)
Or Sn=9(3900){S_n} = 9\left( {3900} \right)
Or Sn=35100{S_n} = 35100
Hence, the amount paid in 1818 instalments=Rs 35100 = Rs{\text{ }}35100

Note: Whenever we face these type of question the key concept is that we have to calculate the annual instalment with annual interest and then total number of instalment and simply substituting the value of (a)\left( a \right) and (d)\left( d \right) in the equation Sn=n2(2a+(n1)d){S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right) .