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Alex deposited x dollars into a new account that earned 8
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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)
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Originally posted by kajolnb on 23 Jan 2012, 13:12.
Last edited by MikeScarn on 03 Jul 2019, 04:58, edited 2 times in total.
Edited the answer choices




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Re: Alex deposited x dollars into a new account that earned 8
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23 Jan 2012, 16:37
Hi there! I'm happy to contribute to this one! The question: 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?So first, Alex puts in x dollars. One year goes by, and the x dollar accrues interest > x(1.08) Then, Alex adds another x dollars > x + x(1.08) Then the second year goes by, and that whole amount gets multiplied by 1.08 > [x + x(1.08)]*(1.08) = x(1.08) + x(1.08)^2 = x[1.08 + (1.08)^2] We are told this amount, the sum total after two years, equals w, so w = x[1.08 + (1.08)^2] Dividing by the brackets to solve for x, we get x = w/(1.08 + (1.08)^2) The answer choices as they appear in your post are technically incorrect, because they are lacking parentheses. If you underestimate the importance of parentheses, they will bite you in the butt over and over again on the real GMAT. Assuming the parentheses were in the right places, the answer would be . The key idea is: the x dollar amount that was in there for both years is multiplied twice by the multiplier. That's why there has to be a factor of (1.08)^2 floating around somewhere. Does this make sense? Please let me know if you have any questions on what I've said. Mike
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Re: Alex deposited x dollars into a new account that earned 8
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23 Jan 2012, 18:38
kajolnb wrote: 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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Re: Alex deposited x dollars into a new account that earned 8
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22 Sep 2013, 05:58
Bunuel wrote: kajolnb wrote: 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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Re: Alex deposited x dollars into a new account that earned 8
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22 Sep 2013, 06:07
Skag55 wrote: Bunuel wrote: kajolnb wrote: 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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Re: Alex deposited x dollars into a new account that earned 8
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22 Sep 2013, 08:51
Got it, wasn't aware of this. Thanks!



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Re: Alex deposited x dollars into a new account that earned 8
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05 Feb 2014, 02:45
Bunuel wrote: kajolnb wrote: 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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Re: Alex deposited x dollars into a new account that earned 8
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05 Feb 2014, 11:11
idinuv wrote: 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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Re: Alex deposited x dollars into a new account that earned 8
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05 Feb 2014, 11:48
mikemcgarry wrote: idinuv wrote: 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 Thanks for your clear explanation Mike ! I totally concede with both the perspectives you have putforth. My perspective about the design of incorrect answers has been that the incorrect answers generally are Partial answers, Wrong path answers, Simple manipulation answers etc. As suggested, I would now also lookout for 'Precision' based on the range of answer choices given.



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Re: Alex deposited x dollars into a new account that earned 8
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18 Sep 2015, 01:46
See attachment for the solution
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Re: Alex deposited x dollars into a new account that earned 8
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03 Aug 2016, 15:38
BrainLab wrote: See attachment for the solution might be a dumb question but at the end of 2nd year, why is it being multiplied by 1.08? should not it be multiplied by 0.08 since it is 8 %? then the resulting amount can be added back to the amount at the end of first year



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Re: Alex deposited x dollars into a new account that earned 8
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14 Oct 2016, 05:48
kajolnb wrote: 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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Re: Alex deposited x dollars into a new account that earned 8
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22 Mar 2017, 01:16
1.08(1.08x + x) = (1.08^2)x + 1.08x
The new total value obtained is equal to w, we can set up the following equation:
w = (1.08^2)x + 1.08x
So, 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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Re: Alex deposited x dollars into a new account that earned 8
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01 Jun 2017, 15:36
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 wrote: kajolnb wrote: 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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Re: Alex deposited x dollars into a new account that earned 8
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01 Jun 2017, 16:57
cuhmoon wrote: 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 GMATMy 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 xthat 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 onetime 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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Re: Alex deposited x dollars into a new account that earned 8
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06 Jul 2017, 05:58
suppose X = 800 and given R= 8 compounded annually now 800(1.08) ....... for first year so 800(1.08)^2.......for 2nd year and now as he added 800 after 1 year then calculate that separately as earlier 800(1.08) W= 800(1.08)^2+ 800(1.08) as we need to calculate X in terms of W , so look for the option now Check E first which will not give us 800 now look for option D , in which by solving we get .... 800{(1.08)^2+(1.08)}/ (1.08)^2+ (1.08) = 800=X hence D hit kudos to appreciate .



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Re: Alex deposited x dollars into a new account that earned 8
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28 Feb 2018, 03:08
BrainLab wrote: 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?



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Re: Alex deposited x dollars into a new account that earned 8
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11 May 2018, 09:27
I forgot adding the interest earned in the second year, and thus the result was option A  I did this silly mistake..
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Re: Alex deposited x dollars into a new account that earned 8
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27 Jan 2019, 16:18
kajolnb wrote: 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)
"Alex deposited x dollars into a new account that earned 8 percent annual interest, compounded annually" Means that w = x * 1.08 "One year later Alex deposited an additional x dollars into the account" w = x*1.08^1 + x "If there were no other transactions and if the account contained w dollars at the end of two years" w = (x*1.08^1) + (x*1.08^2) Express x in terms of w = Basically means rearrange the formula so that it equals to x w = x(1.08) + (1.08^2) (factor out an x) x = w / [(1.08) + (1.08^2)] Answer is D



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Re: Alex deposited x dollars into a new account that earned 8
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19 Sep 2019, 23:46
yonseiglobalstudent wrote: BrainLab wrote: 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?




Re: Alex deposited x dollars into a new account that earned 8
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