The cost of an android mobile phone is Rs. 8,990. Joseph bought it by paying Rs. 500 cash down and the balance he agreed to pay in 10 monthly installments of Rs. 900 each. Nizam bought the same phone by initially paying Rs. 900 and the remaining balance in 8 installments of Rs. 1200 each. Who has paid more rate of interest?
Answer
381.9k+ views
Hint: In the above given question, we are asked to compare the rate of interest paid by Joseph and Nizam. Simply, calculate the rate of interests for both of them separately by using the formula of rate of interest and after obtaining the specific values for both the cases, you can easily compare the two results.
Complete step-by-step answer:
We are given that,
The cost of an android mobile phone is Rs. 8,990.
In case of Joseph:
The cash down payment by Joseph = Rs.500
The balance left=Rs. (8990−500) = Rs. 8490
The number of installments (n)=10
The amount of each installment (I)= Rs.900
The amount paid in installments = 10×900
= Rs.9000
Now, the extra amount paid (E)= Rs. (9000−8490)
=510
Now, the rate of interest is given as
$ = \dfrac{{2400E}}{{n((n + 1)I - 2E)}}$
After substituting the value of E, I and N, we get the value of rate of interest as,
$ = \dfrac{{2400 \times 510}}{{10((10 + 1)900 - 2 \times 510)}}$
$ = \dfrac{{2400 \times 510}}{{10(11 \times 900 - 1020)}}$
$ = \dfrac{{1224000}}{{10(9900 - 1020)}}$
$ = \dfrac{{1224000}}{{10(8880)}}$
$ = 13.78$
Thus, the rate of interest is 13.78%.
In case of Nizam:
The cash down payment by Nizam = Rs.900
The balance = Rs. (8990−900)
= Rs.8090
The number of installments (n)=8
The amount of each installment (I)= Rs.1200
Now, the amount paid in installments =8×1200
= Rs.9600.
Extra amount paid (E)= Rs. (9600−8090) = Rs.1510.
Now, we know that, the rate of interest is given as
$ = \dfrac{{2400E}}{{n((n + 1)I - 2E)}}$
After substituting the value of E, I and N, we get the value of rate of interest as,
$ = \dfrac{{2400 \times 1510}}{{8((8 + 1)1200 - 2 \times 1510)}}$
$ = \dfrac{{2400 \times 1510}}{{8(9 \times 1200 - 3020)}}$
$ = \dfrac{{2400 \times 1510}}{{8(10800 - 3020)}}$
$ = \dfrac{{2400 \times 1510}}{{8(7780)}}$
$ = 58.226$
Hence, the rate of interest is $ = 58.226\% $.
Therefore, Nizam has paid more rate of interest than Joseph.
Note: When we face such types of problems, the key point is that we must have a good understanding of the rate of interest. We have to simply substitute all the given values in the formula for calculating the simple interest and further evaluate it to reach the appropriate solution.
Complete step-by-step answer:
We are given that,
The cost of an android mobile phone is Rs. 8,990.
In case of Joseph:
The cash down payment by Joseph = Rs.500
The balance left=Rs. (8990−500) = Rs. 8490
The number of installments (n)=10
The amount of each installment (I)= Rs.900
The amount paid in installments = 10×900
= Rs.9000
Now, the extra amount paid (E)= Rs. (9000−8490)
=510
Now, the rate of interest is given as
$ = \dfrac{{2400E}}{{n((n + 1)I - 2E)}}$
After substituting the value of E, I and N, we get the value of rate of interest as,
$ = \dfrac{{2400 \times 510}}{{10((10 + 1)900 - 2 \times 510)}}$
$ = \dfrac{{2400 \times 510}}{{10(11 \times 900 - 1020)}}$
$ = \dfrac{{1224000}}{{10(9900 - 1020)}}$
$ = \dfrac{{1224000}}{{10(8880)}}$
$ = 13.78$
Thus, the rate of interest is 13.78%.
In case of Nizam:
The cash down payment by Nizam = Rs.900
The balance = Rs. (8990−900)
= Rs.8090
The number of installments (n)=8
The amount of each installment (I)= Rs.1200
Now, the amount paid in installments =8×1200
= Rs.9600.
Extra amount paid (E)= Rs. (9600−8090) = Rs.1510.
Now, we know that, the rate of interest is given as
$ = \dfrac{{2400E}}{{n((n + 1)I - 2E)}}$
After substituting the value of E, I and N, we get the value of rate of interest as,
$ = \dfrac{{2400 \times 1510}}{{8((8 + 1)1200 - 2 \times 1510)}}$
$ = \dfrac{{2400 \times 1510}}{{8(9 \times 1200 - 3020)}}$
$ = \dfrac{{2400 \times 1510}}{{8(10800 - 3020)}}$
$ = \dfrac{{2400 \times 1510}}{{8(7780)}}$
$ = 58.226$
Hence, the rate of interest is $ = 58.226\% $.
Therefore, Nizam has paid more rate of interest than Joseph.
Note: When we face such types of problems, the key point is that we must have a good understanding of the rate of interest. We have to simply substitute all the given values in the formula for calculating the simple interest and further evaluate it to reach the appropriate solution.
Recently Updated Pages
Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Find the values of other five trigonometric ratios class 10 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts
Which one of the following places is unlikely to be class 8 physics CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is 1 divided by 0 class 8 maths CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

What is pollution? How many types of pollution? Define it

Difference Between Plant Cell and Animal Cell

Find the HCF and LCM of 6 72 and 120 using the prime class 6 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers
