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25 May 2011, 11:55
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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Grace makes an initial deposit of x dollars into a savings account with a z percent interest rate, compounded annually. On the same day, Georgia makes an initial deposit of y dollars into a savings account with a z percent annual interest rate, compounded quarterly. Assuming that neither Grace nor Georgia makes any other deposits or withdrawals and that x, y, and z are positive numbers no greater than 50, whose savings account will contain more money at the end of exactly one year?
(1) z = 4
(2) 100y = zx
OPEN DISCUSSION OF THIS QUESTION IS HERE: grace-makes-an-initial-deposit-of-x-dollars-into-a-savings-a-123186.html
(1) z = 4
(2) 100y = zx
OPEN DISCUSSION OF THIS QUESTION IS HERE: grace-makes-an-initial-deposit-of-x-dollars-into-a-savings-a-123186.html
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Grace savings by the end of the year = x(1 +(z/100))
Georgia savings by the end of the year = y(1+(z/100))
So this can be easily found if we know relation between x and y .
1. Not sufficient
as we dont know anything about x and y.
2. Sufficient
(1+z/100)x vs (1+z/100)y
100y/z vs y
100/z vs 1
as z varies from 1 to 50 ,100/z will always be greater than 1
Hence grace will have more in her savings account by the end of the year.
Answer is B.
Georgia savings by the end of the year = y(1+(z/100))
So this can be easily found if we know relation between x and y .
1. Not sufficient
as we dont know anything about x and y.
2. Sufficient
(1+z/100)x vs (1+z/100)y
100y/z vs y
100/z vs 1
as z varies from 1 to 50 ,100/z will always be greater than 1
Hence grace will have more in her savings account by the end of the year.
Answer is B.
25 May 2011, 17:54
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I think answer could be B here.
Grace savings G1 = x (1+z/100)
Georgia's savings G2 = y (1+ z/q*100)^nq
st - 1 ==> z = 4; not sufficient
st - 2 ==> 100 * y = zx
==> x = 100*y/z and given x, y and z are >0 and <=50
plugging in the extreme possible values of x, y and z in above equation -
x = 50, y =1 and z = 2
x = 50, y = 25 and z = 50
x = 2, y =1 and z = 50
notice that y cannot have value greater than 25 and x can have value of 50.
if you plug these values in G1 and G2 equations; G1 > G2 for all possibilities.
So st -2 is sufficient alone.
Grace savings G1 = x (1+z/100)
Georgia's savings G2 = y (1+ z/q*100)^nq
st - 1 ==> z = 4; not sufficient
st - 2 ==> 100 * y = zx
==> x = 100*y/z and given x, y and z are >0 and <=50
plugging in the extreme possible values of x, y and z in above equation -
x = 50, y =1 and z = 2
x = 50, y = 25 and z = 50
x = 2, y =1 and z = 50
notice that y cannot have value greater than 25 and x can have value of 50.
if you plug these values in G1 and G2 equations; G1 > G2 for all possibilities.
So st -2 is sufficient alone.
