BTGmoderatorDC wrote:In 2001, John invests x dollars in a special account that yields y% simple interest annually. If he has $250 in his account in 2006 and in 2008 he has $270 in his account, what is x + y?

(A) 5

(B) 25

(C) 200

(D) 205

(E) 210

OA D

Source: Magoosh

Given that the sum $x amounted to $250 in 2006 and to $270 in 2008, we can deduce that the simple interest on $x for 2 years = 270 - 250 = $20

Thus, 20 = x*y*2/100 => xy = 1,000

Again, since the sum $x invested in 2001 amounted to $250 in 2006, SI for 5 years is $(250 - x).

Thus, 250 - x = x*y*5/100

25,000 - 100x = 5xy

Plugging-in the value of xy = 1,000 in 25,000 - 100x = 5xy, we have x = $200 and y = 5. Thus, x + y = 200 + 5 = 205.

The correct answer: D

Hope this helps!

-Jay
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BTGmoderatorDC wrote:In 2001, John invests x dollars in a special account that yields y% simple interest annually. If he has $250 in his account in 2006 and in 2008 he has $270 in his account, what is x + y?

(A) 5

(B) 25

(C) 200

(D) 205

(E) 210

OA D

Source: Magoosh

Since he has $250 in his account in 2006 and $270 in 2008, it must be true that the amount of interest is $20 during these two years. Since it's simple interest, it must mean that the amount of interest is $10 per year.

Therefore, in 2006, 5 years since 2001, the account must have accrued $50 in interest, which means the amount in 2001 is $200. So x = 200. Since 10/200 = 5/100 = 5%, y = 5. Therefore, x + y = 205.

Answer: D