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Re: Grace and its deposit [#permalink]
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VeritasPrepKarishma
enigma123
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]
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ronr34
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 makes an initial deposit of x dollars into a savings a [#permalink]
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enigma123
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

look at choice 2
grace=x(1+Z/10)
Georga= y( 1+z/100*4)^4

from 2 we have, y=zx/100, plug this into condition 2.

Geogra= zx/100(1+z/100*4)^4
it is clear that this expression is smaller than x(1+z/100) even if z=50
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Re: Grace makes an initial deposit of x dollars into a savings a [#permalink]
One doubt : If we are assuming that z=50, for Statement 2, why can't we assume the same for Statement 1 :

I mean, at the final stage we need to compare "x" with "y", since given that "x", "y" and "z" are positive numbers no greater than 50, why can't "x" = "y" = 50?

(Even though Statement 1 doesn't tell us about "x" or "y", we can assume right? Just as we assume for "z" in Statement 2)

If we substitute as above, we find "y" > "x".

So either Statement "1" or Statement "2" is sufficient right?

Kindly enlighten me.
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Re: Grace makes an initial deposit of x dollars into a savings a [#permalink]
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What I did was:
A) clearly not sufficient because we don't know X and Y, what if X is 49 and Y is 1 or vice versa? Interest rates won't make a difference;

B) 100y=zx, y=zx/100, which means y principal amount equals to the interests of z;
and since z<50, then the interest of y, which is <y*50%, is definitely less than y,
therefore the interest of y is definitely is less than the interest of z
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Grace makes an initial deposit of x dollars into a savings a [#permalink]
KarishmaB
enigma123
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)

hello KarishmaB
Ma'am could you please explain how to decide after coming to this point
eq 1 : x(3/2)
eq 2 : x/2(9/8)^4
Going by the equation, using st-2, taking z = 50
I came up with this:

Grace's amount at the end of a year = x(1+ 50/100) ----- 1

and Georgia's amount at the end of a year = zx/100(1 + 50/400 )^4 ---- 2 "taking y=zx/100 from the st-2"

eq 1 : x(3/2)
eq 2 : x/2(9/8)^4

I could not just conclude equation is 1 > equation 2.... it took a lot of calculations like doing 9/8 and figuring out (9/8)^4

please help ESPECIALLY with how to come up with a conclusion in a situation such as this
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Grace makes an initial deposit of x dollars into a savings a [#permalink]
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deep1624
KarishmaB
enigma123
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)

hello KarishmaB
Ma'am could you please explain how to decide after coming to this point
eq 1 : x(3/2)
eq 2 : x/2(9/8)^4
Going by the equation, using st-2, taking z = 50
I came up with this:

Grace's amount at the end of a year = x(1+ 50/100) ----- 1

and Georgia's amount at the end of a year = zx/100(1 + 50/400 )^4 ---- 2 "taking y=zx/100 from the st-2"

eq 1 : x(3/2)
eq 2 : x/2(9/8)^4

I could not just conclude equation is 1 > equation 2.... it took a lot of calculations like doing 9/8 and figuring out (9/8)^4

please help ESPECIALLY with how to come up with a conclusion in a situation such as this


Once you do some problems on compound interest, you realise that in the initial years, interest on interest is a small amount.

Also, half yearly compounding or quarterly compounding do give extra interest but slightly. The rate of interest is getting halved or quartered as the case may be in every half year or every quarter. After all, we are not using the rate of interest as 50% in each quarter. It becomes 50/4 = 12.5% in each quarter.
The power of compounding is seen over long term.

Try to keep a more big picture approach in mind instead of getting lost in algebra and equations. Once I realised that x will be at least twice of y from statement 2, and that the rate of interest will be maximum 50%, I know that 50% of x itself will give me the entire y amount. So 50% of y even if compounded quarterly cannot give me double of y. The quarterly compounding will make the interest on y more than y/2 but not by a whole lot.

Check out this video for compound interest in consecutive years: https://youtu.be/pfjZE2xcf04
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