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# Judith and Quincy started with the same amount of money to invest at t

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Manager
Joined: 17 Nov 2013
Posts: 91
Judith and Quincy started with the same amount of money to invest at t  [#permalink]

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01 Nov 2016, 04:26
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00:00

Difficulty:

85% (hard)

Question Stats:

53% (02:32) correct 47% (02:47) wrong based on 91 sessions

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Judith and Quincy started with the same amount of money to invest at the beginning of 2012. Judith's investments increased by P percent in the year 2012 and then increased by Q percent in the year 2013. From the start of 2012 to the end of 2013, Quincy's investment increased by R percent. Which one of them had more money by the end of 2013?

(1) P + Q < R

(2) R - P - Q < (PQ/100)
Senior Manager
Joined: 04 Aug 2010
Posts: 310
Schools: Dartmouth College
Re: Judith and Quincy started with the same amount of money to invest at t  [#permalink]

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30 Jun 2018, 02:39
2
lalania1 wrote:
Judith and Quincy started with the same amount of money to invest at the beginning of 2012. Judith's investments increased by P percent in the year 2012 and then increased by Q percent in the year 2013. From the start of 2012 to the end of 2013, Quincy's investment increased by R percent. Which one of them had more money by the end of 2013?

(1) P + Q < R

(2) R - P - Q < (PQ/100)

Let the original investment in each case = 100.

JUDITH:
In 2012, Judith's P% increase = $$\frac{P}{100}(100) = P$$.
Thus, Judith's amount at the end of 2012 = $$100+P$$.
In 2013, Judith's Q% increase = $$\frac{Q}{100}(100+P) = Q+\frac{PQ}{100}$$.
Thus, Judith's total increase = $$P+Q+\frac{PQ}{100}$$.

QUINCY:
Quincy's R% increase = $$\frac{R}{100}(100) = R$$.

Judith's final amount will be greater than Quincy's final amount if her total increase is greater than his increase:
Question stem, rephrased:
Is $$P+Q+\frac{PQ}{100} > R$$?

Statement 1: $$P+Q < R$$
Case 1: P=10, Q=10 and R=21
In this case, $$P+Q+\frac{PQ}{100} = 10+10+\frac{(10*10)}{100} = 21$$ and $$R=21$$, so the answer to the rephrased question stem is NO.
Case 2: P=40, Q=50 and R=100
In this case, $$P+Q+\frac{PQ}{100} = 40+50+\frac{(40*50)}{100} = 110$$ and $$R=100$$, so the answer to the rephrased question stem is YES.
Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.

Statement 2: $$R - P - Q < \frac{PQ}{100}$$
$$R < P+Q+\frac{PQ}{100}$$
$$P+Q+\frac{PQ}{100} > R$$
Thus, the answer to the rephrased question stem is YES.
SUFFICIENT.

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##### General Discussion
Manager
Joined: 17 Nov 2013
Posts: 91
Re: Judith and Quincy started with the same amount of money to invest at t  [#permalink]

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01 Nov 2016, 04:31
the first statement is a trap.

the second statement is more complex. At first we dont know if it can be resolved.
reading the stem you can use formulas such as ((1+P)/100)((1+Q)/100) for Judy and ((1+R)/100) for Quincy.
in order to prove whether the 2nd stmt is correct, i need to solve the above equation. Once you solve it. you will find that it matches stmt 2, not only that, stmt2 gives you the > sign to answer the question from the stem.
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Re: Judith and Quincy started with the same amount of money to invest at t  [#permalink]

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22 Nov 2016, 23:03
lalania1 wrote:
Judith and Quincy started with the same amount of money to invest at the beginning of 2012. Judith's investments increased by P percent in the year 2012 and then increased by Q percent in the year 2013. From the start of 2012 to the end of 2013, Quincy's investment increased by R percent. Which one of them had more money by the end of 2013?

(1) P + Q < R

(2) R - P - Q < (PQ/100)

Bunuel Can you give detailed explanation of this question? I got the point in statement 2 need some elaboration nonetheless
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Judith and Quincy started with the same amount of money to invest at t  [#permalink]

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22 Nov 2016, 23:53
2
1
lalania1 wrote:
Judith and Quincy started with the same amount of money to invest at the beginning of 2012. Judith's investments increased by P percent in the year 2012 and then increased by Q percent in the year 2013. From the start of 2012 to the end of 2013, Quincy's investment increased by R percent. Which one of them had more money by the end of 2013?

(1) P + Q < R

(2) R - P - Q < (PQ/100)

Judith's investments by the end of 2013: $$N(1+\frac{P}{100})(1+\frac{Q}{100})$$
Quincy's investments by the end of 2013: $$N(1+\frac{R}{100})$$

Comparing between Judith's investments and Quincy's investments is the same as comparing between $$(1+\frac{P}{100})(1+\frac{Q}{100})$$ and $$(1+\frac{R}{100})$$
We have
$$(1+\frac{P}{100})(1+\frac{Q}{100})-(1+\frac{R}{100})$$
$$=(\frac{100+P}{100})(\frac{100+Q}{100})-(\frac{100+R}{100})$$
$$=\frac{10^4+100P+100Q+PQ}{10^4}-\frac{100(100+R)}{10^4}$$
$$=\frac{100(P+Q-R)+PQ}{10^4}$$(*)

(1) $$P+Q<R$$ so $$P+Q-R<0$$. However, from (*) we cannot deduce anything. Insufficient.

(2) $$R - P - Q < \frac{PQ}{100} \iff 100(R-P-Q)<PQ \iff 100(P+Q-R)+PQ>0$$.

Hence, from (*) we have $$(1+\frac{P}{100})(1+\frac{Q}{100})>(1+\frac{R}{100})$$ or Judith's investments is higher. Sufficient.

The answer is B
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Re: Judith and Quincy started with the same amount of money to invest at t  [#permalink]

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30 Jun 2018, 05:13
How do you assume that its compound interest ?
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Re: Judith and Quincy started with the same amount of money to invest at t  [#permalink]

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30 Jun 2018, 05:42
1
How do you assume that its compound interest ?

Judith's investments increased by P percent in the year 2012 and THEN increased by Q percent in the year 2013.
The usage of then implies that the interest in 2013 is accrued on the value of the investment at the end of 2012.
That said, the GMAT would use wording to make this intent crystal clear.
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Re: Judith and Quincy started with the same amount of money to invest at t &nbs [#permalink] 30 Jun 2018, 05:42
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# Judith and Quincy started with the same amount of money to invest at t

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