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RenB
Joined: 13 Jul 2022
Last visit: 02 Mar 2026
Posts: 389
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Location: India
Concentration: Finance, Nonprofit
Schools: Stanford '26 Wharton '26 (D) Booth '26 (D) Yale '26 (D) CBS '26 (A$$$) Darden '26 (WL) HBS '26 (D) Ross '26 (A$$$$) LBS '26 (D) Tuck '26 (A$$) Kellogg '26 (A)
GMAT Focus 1: 715 Q90 V84 DI82 
GPA: 3.74
WE:Corporate Finance (Consulting)
Schools: Stanford '26 Wharton '26 (D) Booth '26 (D) Yale '26 (D) CBS '26 (A$$$) Darden '26 (WL) HBS '26 (D) Ross '26 (A$$$$) LBS '26 (D) Tuck '26 (A$$) Kellogg '26 (A)
GMAT Focus 1: 715 Q90 V84 DI82
Posts: 389
30 Aug 2023, 18:14
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
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Question Stats:
42% (02:09) correct
58%
(02:17)
wrong
based on 258
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An investment grows at a fixed annual rate of 10%. If there are no withdrawals during the investment period, and the interest earned is reinvested in the same investment every year, what is the total amount at the end?
(1) At the end, the value of the investment has increased by 21%.
(2) If in the beginning of the last year of the investment, $500 had been withdrawn from the investment made, the investment worth at the end would be 10% less than its actual worth at the end.
(1) At the end, the value of the investment has increased by 21%.
(2) If in the beginning of the last year of the investment, $500 had been withdrawn from the investment made, the investment worth at the end would be 10% less than its actual worth at the end.
30 Aug 2023, 20:29
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An investment grows at a fixed annual rate of 10%. If there are no withdrawals during the investment period, and the interest earned is reinvested in the same investment every year, what is the total amount at the end?
The above means the interest is compounded annually, that is the principal amount keeps increasing every year with addition of interest.
(1) At the end, the value of the investment has increased by 21%.
First year is it 1.1x, and next it becomes 1.1*1.1x or 1.21x
So, we can say that the amount was put for 2 years but nothing about the amount.
(2) If in the beginning of the last year of the investment, $500 had been withdrawn from the investment made, the investment worth at the end would be 10% less than its actual worth at the end.
So what we have in the last year is 500 less and in the end of the year, it is 500 less plus the 10% interest on it.
Say actual worth is x.
Under the new scenario, it is less by \(500+\frac{10}{100}*500\) or 550
Thus, \(x-550=x-x*\frac{10}{100}…….x=5500\)
Sufficient
B
The above means the interest is compounded annually, that is the principal amount keeps increasing every year with addition of interest.
(1) At the end, the value of the investment has increased by 21%.
First year is it 1.1x, and next it becomes 1.1*1.1x or 1.21x
So, we can say that the amount was put for 2 years but nothing about the amount.
(2) If in the beginning of the last year of the investment, $500 had been withdrawn from the investment made, the investment worth at the end would be 10% less than its actual worth at the end.
So what we have in the last year is 500 less and in the end of the year, it is 500 less plus the 10% interest on it.
Say actual worth is x.
Under the new scenario, it is less by \(500+\frac{10}{100}*500\) or 550
Thus, \(x-550=x-x*\frac{10}{100}…….x=5500\)
Sufficient
B
31 Aug 2023, 14:28
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This is a good example of a problem on which it's worthwhile to stop and assess before you do too much math. It looks like we're going to have to calculate compound interest, but we never actually need that!
Statement 1 just tells us that the investment has been compounding for 2 years, but has no information on actual value, so it's useless by itself. A and D are out.
Statement 2 does something very common for DS: it gives us the numerical difference between one scenario and another. As long as we're confident that it is possible to express the final year-end amount in terms of just one variable, then we know this is a solvable equation, and we don't need to do it! Sure, we CAN. I think chetan2u's solution above is the most direct, because it uses only the needed variable, but we could also solve in terms of the initial principal at the beginning of the final year:
.9(1.1P) = 1.1(P-500)
.99P = 1.1P - 550
550 = 0.11P
5,000 = P
(It's important that we are just dealing with the FINAL year. If we were dealing with the whole investment period, we'd need to use Statement 1, as well, since we might have 0.9(1.21P) = 1.21(P-500) or 0.9(1.331P) = 1.331(P-500), etc.)
The main point is that in either case, we have one linear equation for one variable, so we don't want to spend time actually solving! Imagine we'd been dealing with something much more involved, such as a 10-year investment in which $500 were withdrawn at the start of each year after the first. Sure, we could write that out:
.9(1.1^10)(P) = 1.1P + 1.1(1.1P - 500) + 1.1((1.1P-500) - 500) . . . Oh wait, did I say I could write that out???
That looks like a mess, but it's still a linear equation with one variable. There will be lots of powers of 1.1, but no extra powers of P, so I can still solve and I don't have to write any of this!
In short, if we are confident that there is only one value that would produce the exact outcome described, we can rule the statement sufficient and move on. In this case, there's only one final value that would be reduced by 10% if we had withdrawn $500 at the beginning of the year. If the value were any higher or lower, the difference would be correspondingly less or more than 10%.
Statement 1 just tells us that the investment has been compounding for 2 years, but has no information on actual value, so it's useless by itself. A and D are out.
Statement 2 does something very common for DS: it gives us the numerical difference between one scenario and another. As long as we're confident that it is possible to express the final year-end amount in terms of just one variable, then we know this is a solvable equation, and we don't need to do it! Sure, we CAN. I think chetan2u's solution above is the most direct, because it uses only the needed variable, but we could also solve in terms of the initial principal at the beginning of the final year:
.9(1.1P) = 1.1(P-500)
.99P = 1.1P - 550
550 = 0.11P
5,000 = P
(It's important that we are just dealing with the FINAL year. If we were dealing with the whole investment period, we'd need to use Statement 1, as well, since we might have 0.9(1.21P) = 1.21(P-500) or 0.9(1.331P) = 1.331(P-500), etc.)
The main point is that in either case, we have one linear equation for one variable, so we don't want to spend time actually solving! Imagine we'd been dealing with something much more involved, such as a 10-year investment in which $500 were withdrawn at the start of each year after the first. Sure, we could write that out:
.9(1.1^10)(P) = 1.1P + 1.1(1.1P - 500) + 1.1((1.1P-500) - 500) . . . Oh wait, did I say I could write that out???
That looks like a mess, but it's still a linear equation with one variable. There will be lots of powers of 1.1, but no extra powers of P, so I can still solve and I don't have to write any of this!
In short, if we are confident that there is only one value that would produce the exact outcome described, we can rule the statement sufficient and move on. In this case, there's only one final value that would be reduced by 10% if we had withdrawn $500 at the beginning of the year. If the value were any higher or lower, the difference would be correspondingly less or more than 10%.
