Problem Solving

Good Day,

This seems to be an easy problem but for some reason I just cannot get the right answer..appreciate help from some the geniuses here :

If $10,000 is invested at the rate of x% simple interest for n years, which of the following represents the total amount of interest in dollars that will be earned by the investment in the 'n' years :

a) 10,000 (x^n)

b) 10,000 (x/100)^n

c) 10,000n ( x/100)

d) 10,000 ( 1 + x/100) ^n

e) 10,000n ( 1 + x/100)

OA : C

[/spoiler]

Simple Interest = p * i * n

where:
p = principal (original amount borrowed or loaned)
i = interest rate for one period
n = number of periods

Example: You borrow $10,000 for 3 years at 5% simple annual interest.

interest = p * i * n = 10,000 * (5/100) * 3 = 1,500

Looks Easy...this was my method :

Principal : 10,000

SI : x% per year.

After one year, principal : 10,000 + 10,000x/100 = 10,000 ( 1+ x/100)
After second year, principal : 10,000 ( 1+ x/100) + 10,000 ( 1+ x/100)* x/100
= 10,000 (1+x/100) ^2

Similarly, after 'n' years, the principal = 10,000 (1+x/100)^n

What is wrong with this approach?

everything's eventual wrote:Good Day,

This seems to be an easy problem but for some reason I just cannot get the right answer..appreciate help from some the geniuses here :

If $10,000 is invested at the rate of x% simple interest for n years, which of the following represents the total amount of interest in dollars that will be earned by the investment in the 'n' years :

a) 10,000 (x^n)

b) 10,000 (x/100)^n

c) 10,000n ( x/100)

d) 10,000 ( 1 + x/100) ^n

e) 10,000n ( 1 + x/100)

OA : C

Let x=10 and n=2.
Each year, the amount of interest earned = x% of 10,000 = 10% of 10,000 = 1000.
Over 2 years, the total amount earned = 2*1000 = 2000. This is our target.

Now we plug x=10 and n=2 into the answers to see which yields our target of 2000.
Only answer choice C works:
10,000n(x/100) = 10,000*2(10/100) = 2000.

The correct answer is C.

Hello Sir...Can we use two values for variables in the same problem( in this case x and n)? With your method it is surely much easier to solve.

Just to understand, what was wrong with my approach?

Regards.

everything's eventual wrote:Hello Sir...Can we use two values for variables in the same problem( in this case x and n)? With your method it is surely much easier to solve.

Just to understand, what was wrong with my approach?

Regards.

When the answer choices contain variables, we can plug in our own numbers, regardless of the number of variables.
Type into the search bar "gmatguruny" and "this is our target" to find other problems that I've solved by plugging in numbers.

Your solution calculates the principal after the first year.
There is no reason to perform this calculation: with simple interest, the principal for each year remains the same.
Interest in the first year = (x/100)10,000.
Interest in the second year = (x/100)(10,000).
And so on.
Thus, the interest earned after n years = n(x/100)(10,000).

Wow...my basic understanding of the subject was wrong...thank you very much...