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Grace makes an initial deposit of x dollars into a savings a

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Grace makes an initial deposit of x dollars into a savings a [#permalink] New post 15 Nov 2011, 14:42
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48% (02:39) correct 52% (01:46) wrong based on 104 sessions
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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 18 Apr 2014, 03:33, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Grace and its deposit [#permalink] New post 15 Nov 2011, 17:14
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B.

B is sufficient due to the constraints in the problem. Since no number can be greater than 50 and Z is the same for both.

100y =zx

Basically says that X is larger than Y. The differences between annual and quarterly compounding of an identical interest rate over only one year will be fairly negligible. Assuming Z = 50 which is the most it can equal we are left with 2y = x. So we know that Grace is going to have more money than Georgia at the end of year 1.

Statement 2 is sufficient.

Hope that helps.
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Re: Grace and its deposit [#permalink] New post 16 Nov 2011, 20:21
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enigma123 wrote:
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

Your help is desperately needed folks please.


Responding to a pm:

Actually, u0422811's solution is perfect. That is exactly what went through my mind when I read this question.
Quarterly compounding yields more than annual compounding but the difference is minuscule in % terms.
e.g. if you invest $10 at 10% annual compounding, you get $11 at the end of the year.
but if you invest $10 at 10% quarterly compounding, you get $11.038 at the end of the year.
You get a small fraction of interest extra.

So x is invested at annual compounding and y at quarterly compounding. If x=y, the amount received from y will be a little more.

Statement 1 tells us z = 4. We need to compare x with y so this is not sufficient.

Statement 2 tells us 100y = zx
x/y = 100/z
Since maximum value of z is 50, x is at least twice of y.
If z% = 50%, amount obtained from x is 1.5x (= 3y) and that obtained from y is a little more than 1.5y.
Definitely an investment of $x results in a higher amount at the end of the year.

Answer (B)
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Re: Grace and its deposit [#permalink] New post 18 Apr 2014, 01:54
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Re: Grace and its deposit [#permalink] New post 08 Jun 2014, 09:44
VeritasPrepKarishma wrote:
enigma123 wrote:
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

Your help is desperately needed folks please.


Responding to a pm:

Actually, u0422811's solution is perfect. That is exactly what went through my mind when I read this question.
Quarterly compounding yields more than annual compounding but the difference is minuscule in % terms.
e.g. if you invest $10 at 10% annual compounding, you get $11 at the end of the year.
but if you invest $10 at 10% quarterly compounding, you get $11.038 at the end of the year.
You get a small fraction of interest extra.

So x is invested at annual compounding and y at quarterly compounding. If x=y, the amount received from y will be a little more.

Statement 1 tells us z = 4. We need to compare x with y so this is not sufficient.

Statement 2 tells us 100y = zx
x/y = 100/z
Since maximum value of z is 50, x is at least twice of y.
If z% = 50%, amount obtained from x is 1.5x (= 3y) and that obtained from y is a little more than 1.5y.
Definitely an investment of $x results in a higher amount at the end of the year.

Answer (B)

Hi Karishma,

taking B into consideration, we can have x=2, y=1, z=50.

so X after one year will be 3.

Y after one year and 4 interest bumps will be over 3....

Can you explain this?
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Re: Grace and its deposit [#permalink] New post 08 Jun 2014, 21:54
Expert's post
ronr34 wrote:
Hi Karishma,

taking B into consideration, we can have x=2, y=1, z=50.

so X after one year will be 3.

Y after one year and 4 interest bumps will be over 3....

Can you explain this?


How did you get that "Y after one year and 4 interest bumps will be over 3"?

If initial investment is $1 with an annual rate of 50% compounded quarterly, at the end of the year, the amount is 1(1 + 50/400)^4 = 1.608 i.e. somewhat above 1.5. How is it 3? Note that 50% is annual rate. If it is compounded quarterly, the quarterly rate becomes 50/4%
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Re: Grace and its deposit   [#permalink] 08 Jun 2014, 21:54
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