Answer

Verified

449.1k+ 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

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

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

Write a letter to the principal requesting him to grant class 10 english CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

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

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE