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555-605 Level|   Percent and Interest Problems|                              
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Bunuel
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Alex deposited x dollars into a new account that earned 8 percent annual interest, compounded annually. One year later Alex deposited an additional x dollars into the account. If there were no other transactions and if the account contained w dollars at the end of two years, which of the following expresses x in terms of w ?

A. w/(1+1.08)
B. w/(1.08+1.16)
C. w/(1.16+1.24)
D. w/(1.08+1.08^2)
E. w/(1.08^2+1.08^2)

I thought as
1.08x+2x(1.08) = w

Account at the end of the first year would be 1.08x dollars. At this time x dollars was deposited, hence the account at the beginning of the second year would be (1.08x+x) dollars. Account at the end of the second year would be (1.08x+x)*1.08=w --> x(1.08^2+1.08)=w --> x=w/(1.08+1.08^2).

Answer: D.

I did the math, 1.08x + x = 2.08x
2.08x * 1.08 = 2.2464
couldn't spot the answer after 2+ mins.

How are we supposed to know to leave (1.08x + x) in order to see the cube to 1.08^2 x?
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kajolnb
Alex deposited x dollars into a new account that earned 8 percent annual interest, compounded annually. One year later Alex deposited an additional x dollars into the account. If there were no other transactions and if the account contained w dollars at the end of two years, which of the following expresses x in terms of w ?

A. w/(1+1.08)
B. w/(1.08+1.16)
C. w/(1.16+1.24)
D. w/(1.08+1.08^2)
E. w/(1.08^2+1.08^2)

I thought as
1.08x+2x(1.08) = w

Account at the end of the first year would be 1.08x dollars. At this time x dollars was deposited, hence the account at the beginning of the second year would be (1.08x+x) dollars. Account at the end of the second year would be (1.08x+x)*1.08=w --> x(1.08^2+1.08)=w --> x=w/(1.08+1.08^2).

Answer: D.

I did the math, 1.08x + x = 2.08x
2.08x * 1.08 = 2.2464
couldn't spot the answer after 2+ mins.

How are we supposed to know to leave (1.08x + x) in order to see the cube to 1.08^2 x?

On the PS section always look at the answer choices before you start to solve a problem. They might often give you a clue on how to approach the question.

For this question this would give you a hint that you shouldn't calculate 1.08^2+1.08.
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kajolnb
Alex deposited x dollars into a new account that earned 8 percent annual interest, compounded annually. One year later Alex deposited an additional x dollars into the account. If there were no other transactions and if the account contained w dollars at the end of two years, which of the following expresses x in terms of w ?

A. w/(1+1.08)
B. w/(1.08+1.16)
C. w/(1.16+1.24)
D. w/(1.08+1.08^2)
E. w/(1.08^2+1.08^2)

I thought as
1.08x+2x(1.08) = w

Account at the end of the first year would be 1.08x dollars. At this time x dollars was deposited, hence the account at the beginning of the second year would be (1.08x+x) dollars. Account at the end of the second year would be (1.08x+x)*1.08=w --> x(1.08^2+1.08)=w --> x=w/(1.08+1.08^2).

Answer: D.


I did quick math (1.08)^2 = 1.16 and selected option B.

I know option D is more precise, but can GMAC give two option different only by third decimal digit (1.16 Vs 1.1664)?
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idinuv
I did quick math (1.08)^2 = 1.16 and selected option B.

I know option D is more precise, but can GMAC give two option different only by third decimal digit (1.16 Vs 1.1664)?
Dear idinuv,
I'm happy to respond. :-)
The short answer to your question is: "absolutely." Math is all about precision. Yes, in many Quant questions, GMAC spreads out the answer choices and allows for estimation and quick approximations, but that is not always the case. One way to think about it is that, for a pure mathematician, there is a continuous infinity of decimals between 1.16 and 1.1664 --- more decimals in that separation than the number of grains of sand it would take to fill the Universe. For a pure mathematician, there is just equal or completely unequal, and any inequality, no matter how small, is vast beyond all reckoning. Another perspective is what business people care about. Suppose, for the sake of argument, that x = $100,000,000 --- then, whether we divide by 1.16 or 1.1664 results in a difference of $437,014.52 : do you want that discrepancy to come out of your paycheck, because you were the person who rounded to two decimal places? Small decimal difference get very big in a hurry when one starts dealing with numbers in the millions & billions --- which values, of course, are typical in some industries. ------ Both the perspective of the pure mathematician and the perspective of big business are very important in informing the design of GMAT Quant questions, and from the point of view of both of these perspectives, the difference between 1.16 and 1.1664 could be tremendously important, not something to overlook.
Does all this make sense?
Mike :-)
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See attachment for the solution
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kajolnb
Alex deposited x dollars into a new account that earned 8 percent annual interest, compounded annually. One year later Alex deposited an additional x dollars into the account. If there were no other transactions and if the account contained w dollars at the end of two years, which of the following expresses x in terms of w ?

A. w/(1+1.08)
B. w/(1.08+1.16)
C. w/(1.16+1.24)
D. w/(1.08+1.08^2)
E. w/(1.08^2+1.08^2)

We start by determining the new value of the x dollars Alex deposited into his account that earned 8 percent annual interest. At the end of the first year the amount of money in the account was 1.08x and then he added another x dollars to the account, so the account then had a total value of 1.08x + x dollars. The 1.08x + x dollars earned another 8 percent interest for the year. Thus, the total value of his account at the end of the second year is:

1.08(1.08x + x) = (1.08^2)x + 1.08x

Since the new total value is equal to w, we can set up the following equation:

w = (1.08^2)x + 1.08x

Now we must get x in terms of w.

w = x(1.08^2 + 1.08)

w/(1.08^2 + 1.08) = x

Answer: D
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This maybe a foolish question but how does one know the amount was compounded for the second year also or that compounding needs to be done again on the amount?

Why isn't the solution just using the amount and adding x to the amount: (1.08x+x)



Bunuel
kajolnb
Alex deposited x dollars into a new account that earned 8 percent annual interest, compounded annually. One year later Alex deposited an additional x dollars into the account. If there were no other transactions and if the account contained w dollars at the end of two years, which of the following expresses x in terms of w ?

A. w/(1+1.08)
B. w/(1.08+1.16)
C. w/(1.16+1.24)
D. w/(1.08+1.08^2)
E. w/(1.08^2+1.08^2)

I thought as
1.08x+2x(1.08) = w

Account at the end of the first year would be 1.08x dollars. At this time x dollars was deposited, hence the account at the beginning of the second year would be (1.08x+x) dollars. Account at the end of the second year would be (1.08x+x)*1.08=w --> x(1.08^2+1.08)=w --> x=w/(1.08+1.08^2).

Answer: D.
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cuhmoon
This maybe a foolish question but how does one know the amount was compounded for the second year also or that compounding needs to be done again on the amount?

Why isn't the solution just using the amount and adding x to the amount: (1.08x+x)
Dear cuhmoon,

I'm happy to respond. :-)

You may find this blog article helpful:
Compound Interest on the GMAT

My friend, please don't be afraid of asking a question because it might appear "foolish." Every basic question you ask that gets answered by an expert will there to help dozens of other GC users who were afraid to ask that same question.

The big idea of compound interest is interest on interest. If we compounded in the first year, and then stopped, that wouldn't be compound interest. The action of compounding is when new interest starts to accrue on interest already there.

If x is the initial amount, notice that 0.08x is 8% of x--that would be just the interest after one cycle, if the interest rate were 8%. If we want interest plus principal, we add x + 0.08x = 1x + 0.08x = 1.08x. That latter form, 1.08x is a 8% increase on a starting value of x; it's the value of interest plus principal after one cycle. Notice that the question very specifically says: "One year later Alex deposited an additional x dollars into the account." In other words, this is not just a single one-time deposit of principal that is building up interest: instead, Alex put money in, let it build up interest, and then put more money in from the outside. After the first year, 1.08x was already in the account, after the interest was added, and then separate from the process of interest, Alex deposited another x of his own money into this same account. That's why we add 1.08x + x.

In order to understand all this, it's very helpful to understand percents and percent changes as multipliers.

Does all this make sense?
Mike :-)
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yonseiglobalstudent
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See attachment for the solution


I just want to clarify one point. I follow up until the start of the 2nd year, but then why do we multiply again by 1.08? (2.08x)(1.08)
Where did the 1.08 come from? I believe we need to represent 8% of 2.08x, but why don't we multiply by .08 instead?

I have the same question. Can someone please explain clearly why we are multiplying by 1.08 and not 0.08 at the end of the second year?
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yonseiglobalstudent
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See attachment for the solution


I just want to clarify one point. I follow up until the start of the 2nd year, but then why do we multiply again by 1.08? (2.08x)(1.08)
Where did the 1.08 come from? I believe we need to represent 8% of 2.08x, but why don't we multiply by .08 instead?

I have the same question. Can someone please explain clearly why we are multiplying by 1.08 and not 0.08 at the end of the second year?

Interest earned at the end of the second year would be 2.08x*0.08 but the balance at the end of the second year would be (amount at the beginning of the second year) + (interest earned in second year) = 2.08x + 2.08x*0.08 = 2.08x*1.08.

Hope it's clear.
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kajolnb
Alex deposited x dollars into a new account that earned 8 percent annual interest, compounded annually. One year later Alex deposited an additional x dollars into the account. If there were no other transactions and if the account contained w dollars at the end of two years, which of the following expresses x in terms of w ?


A. \(\frac{w}{1+1.08}\)

B. \(\frac{w}{1.08+1.16}\)

C. \(\frac{w}{1.16+1.24}\)

D. \(\frac{w}{1.08+1.08^2}\)

E. \(\frac{w}{1.08^2+1.08^2}\)
Answer: Option D

Video solution by GMATinsight

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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Bunuel
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Video solution from Quant Reasoning:

Cannot view the YouTube Solution. The link is broken

It works just fine for me.

Why can the answer not be B? I solved (1.08)^2 and added it to 1.08 so I got 1.16 + 1.08 in the denominator. Please help here
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SrAD

Why can the answer not be B? I solved (1.08)^2 and added it to 1.08 so I got 1.16 + 1.08 in the denominator. Please help here

1.16 is 1.08*2, not (1.08)^2.

(1.08)^2 means 1.08*1.08 and equals to 1.1664.

Hope it's clear.
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Bunuel
kajolnb
Alex deposited x dollars into a new account that earned 8 percent annual interest, compounded annually. One year later Alex deposited an additional x dollars into the account. If there were no other transactions and if the account contained w dollars at the end of two years, which of the following expresses x in terms of w ?

A. \(\frac{w}{1+1.08}\)

B. \(\frac{w}{1.08+1.16}\)

C. \(\frac{w}{1.16+1.24}\)

D. \(\frac{w}{1.08+1.08^2}\)

E. \(\frac{w}{1.08^2+1.08^2}\)

I thought as
1.08x+2x(1.08) = w

Account at the end of the first year would be 1.08x dollars. At this time x dollars was deposited, hence the account at the beginning of the second year would be (1.08x+x) dollars. Account at the end of the second year would be \((1.08x+x)*1.08=w\) --> \(x(1.08^2+1.08)=w\) --> \(x=\frac{w}{(1.08+1.08^2)}\).

Answer: D.


" 8 percent annual interest, compounded annually."
Sorry how to decide whether its just a simple interest or compounded interest?
I got this one wrong because I tried to use a compounded formula.
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kajolnb
Alex deposited x dollars into a new account that earned 8 percent annual interest, compounded annually. One year later Alex deposited an additional x dollars into the account. If there were no other transactions and if the account contained w dollars at the end of two years, which of the following expresses x in terms of w ?

A. \(\frac{w}{1+1.08}\)

B. \(\frac{w}{1.08+1.16}\)

C. \(\frac{w}{1.16+1.24}\)

D. \(\frac{w}{1.08+1.08^2}\)

E. \(\frac{w}{1.08^2+1.08^2}\)

I thought as
1.08x+2x(1.08) = w

Account at the end of the first year would be 1.08x dollars. At this time x dollars was deposited, hence the account at the beginning of the second year would be (1.08x+x) dollars. Account at the end of the second year would be \((1.08x+x)*1.08=w\) --> \(x(1.08^2+1.08)=w\) --> \(x=\frac{w}{(1.08+1.08^2)}\).

Answer: D.


" 8 percent annual interest, compounded annually."
Sorry how to decide whether its just a simple interest or compounded interest?
I got this one wrong because I tried to use a compounded formula.

We are directly told that:

    Alex deposited x dollars into a new account that earned 8 percent annual interest, compounded annually.

So, at the end of the second year the balance would be \((1.08x+x)*1.08=2.2464 x\).

If instead it were:

    Alex deposited x dollars into a new account that earned 8 percent simple annual interest.

Then at the end of the second year the balance would be \((x+0.08x+0.08x) +(x+0.08x)=2.24 x\).

Hope it helps.
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kajolnb
Alex deposited x dollars into a new account that earned 8 percent annual interest, compounded annually. One year later Alex deposited an additional x dollars into the account. If there were no other transactions and if the account contained w dollars at the end of two years, which of the following expresses x in terms of w ?

A. w/(1+1.08)
B. w/(1.08+1.16)
C. w/(1.16+1.24)
D. w/(1.08+1.08^2)
E. w/(1.08^2+1.08^2)

We start by determining the new value of the x dollars Alex deposited into his account that earned 8 percent annual interest. At the end of the first year the amount of money in the account was 1.08x and then he added another x dollars to the account, so the account then had a total value of 1.08x + x dollars. The 1.08x + x dollars earned another 8 percent interest for the year. Thus, the total value of his account at the end of the second year is:

1.08(1.08x + x) = (1.08^2)x + 1.08x

Since the new total value is equal to w, we can set up the following equation:

w = (1.08^2)x + 1.08x

Now we must get x in terms of w.

w = x(1.08^2 + 1.08)

w/(1.08^2 + 1.08) = x

Answer: D

Why are multiplying with 1.08 and not 0.08
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