Grace makes an initial deposit of x dollars into a savings : GMAT Data Sufficiency (DS)
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# Grace makes an initial deposit of x dollars into a savings

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Grace makes an initial deposit of x dollars into a savings [#permalink]

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25 May 2011, 10:55
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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

OA is not known
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Re: Grace and Georgia [#permalink]

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25 May 2011, 15:46
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.

Last edited by Spidy001 on 25 May 2011, 17:59, edited 1 time in total.
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Re: Grace and Georgia [#permalink]

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25 May 2011, 16:54
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.
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Re: Grace and Georgia [#permalink]

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25 May 2011, 23:56
its is asking x(1+z/100) is > | < y(1+z/400)^4

a gives x(1+4/100) and y(1+4/400) ^ 4
so since we have no idea of x and y nothing can be concluded.
Not sufficient.

b x(1+z/100) = x + y

and y(1+z/400)^4 = y^4 + 4y^3 * (z/400) + 6 y^2 * (z/400)^2 + 4y * (z/400)^3 + (z/400)^4

gives y(1+z/400)^4 > than x+y. since x,y and z are all >0 but <50.

thus B.
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Re: Grace and Georgia   [#permalink] 25 May 2011, 23:56
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# Grace makes an initial deposit of x dollars into a savings

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