M47-163

Question

At the start of a year, Elena deposited $800 into Bank A and $800 into Bank B. Each deposit earned interest at its own fixed annual rate, compounded annually, and no additional deposits or withdrawals were made. What was the annual interest rate at Bank B?

(1) During the first year, the deposit at Bank B earned $40 more interest than the deposit at Bank A.
 
(2) During the second year, the deposit at Bank B earned $50 more interest than the deposit at Bank A.

Bunuel · Answer

Official Solution:

At the start of a year, Elena deposited $800 into Bank A and $800 into Bank B. Each deposit earned interest at its own fixed annual rate, compounded annually, and no additional deposits or withdrawals were made. What was the annual interest rate at Bank B? 
 
Let  a  be the annual interest rate at Bank A, and let  b  be the annual interest rate at Bank B.
 
(1) During the first year, the deposit at Bank B earned $40 more interest than the deposit at Bank A.
 
So:
 
 800b - 800a = 40 
 
 800(b - a) = 40 
 
 b - a = 0.05 
 
This gives only the difference between the two interest rates, not the value of  b .
 
Not sufficient.
 
(2) During the second year, the deposit at Bank B earned $50 more interest than the deposit at Bank A.
 
At the beginning of the second year, the deposit at Bank A was:
 
 800(1 + a) 
 
So the interest earned at Bank A during the second year was:
 
 800(1 + a) * a 
 
At the beginning of the second year, the deposit at Bank B was:
 
 800(1 + b) 
 
So the interest earned at Bank B during the second year was:
 
 800(1 + b) * b 
 
So, statement (2) gives:
 
 800b(1 + b) - 800a(1 + a) = 50 
 
This is one equation with two unknowns, so  b  cannot be determined.
 
Not sufficient.
 
(1)+(2) From statement (1):
 
 b - a = 0.05 
 
From statement (2):
 
 800b(1 + b) - 800a(1 + a) = 50 
 
Divide by 800:
 
 b(1 + b) - a(1 + a) = \frac{50}{800} 
 
 b + b^2 - a - a^2 = \frac{1}{16} 
 
 (b - a) + (b^2 - a^2) = \frac{1}{16} 
 
 (b - a) + (b - a)(b + a) = \frac{1}{16} 
 
 (b - a)(1 + a + b) = \frac{1}{16} 
 
Since  b - a = 0.05 = \frac{1}{20} :
 
 \frac{1}{20} * (1 + a + b) = \frac{1}{16} 
 
 1 + a + b = \frac{20}{16} = \frac{5}{4} 
 
 a + b = \frac{1}{4} 
 
Now we have:
 
 b - a = 0.05 
 
 a + b = 0.25 
 
So  b  is determined.
 
Sufficient.
 
Answer: C

M47-163

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