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At the start of 2015, Jane opened two new accounts – X and Y – to inve

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Concentration: General Management, Finance
GMAT Date: 02-18-2019
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WE: Engineering (Consulting)
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At the start of 2015, Jane opened two new accounts – X and Y – to inve  [#permalink]

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New post 15 Jun 2019, 10:38
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Question Stats:

69% (02:56) correct 31% (02:33) wrong based on 32 sessions

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At the start of 2015, Jane opened two new accounts – X and Y – to invest her total savings of $1000. She invested p percent of her savings in account X, which yielded a simple interest of 8 percent per annum, and the rest of the savings in account Y, which yielded a simple interest of 6 percent per annum. Was the amount of interest earned by account X greater than the amount of interest earned by account Y during the year 2015?

(1) The value of p was between 25 and 30, inclusive

(2) The total interest earned by the two accounts during 2015 was $66
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At the start of 2015, Jane opened two new accounts – X and Y – to inve  [#permalink]

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New post 15 Jun 2019, 21:42
mangamma wrote:
At the start of 2015, Jane opened two new accounts – X and Y – to invest her total savings of $1000. She invested p percent of her savings in account X, which yielded a simple interest of 8 percent per annum, and the rest of the savings in account Y, which yielded a simple interest of 6 percent per annum. Was the amount of interest earned by account X greater than the amount of interest earned by account Y during the year 2015?

(1) The value of p was between 25 and 30, inclusive

(2) The total interest earned by the two accounts during 2015 was $66




She invested p percent of her savings in account X, which yielded a simple interest of 8 percent per annum,----
\(X=1000*\frac{p}{100}=10p\)
Interest = \(\frac{10p*8*1}{100}=\frac{4p}{5}\)

the rest of the savings in account Y, which yielded a simple interest of 6 percent per annum.
\(Y=1000-X=1000-10p\)
Interest = \(\frac{(1000-10p)*6*1}{100}=60-\frac{3p}{5}\)

We are looking for
Was the amount of interest earned by account X greater than the amount of interest earned by account Y during the year 2015?
OR \(\frac{4p}{5}>60-\frac{3p}{5}.......7p>60*5....p>\frac{300}{7}.....p>42.9\)

(1) The value of p was between 25 and 30, inclusive
So answer for -- Is p>42.9 is NO
sufficient

(2) The total interest earned by the two accounts during 2015 was $66
[m]\frac{4p}{5}+60-\frac{3p}{5}=66.........\frac{p}{5}=6....p=30[\m]
So answer for -- Is p>42.9 is NO
sufficient

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At the start of 2015, Jane opened two new accounts – X and Y – to inve   [#permalink] 15 Jun 2019, 21:42
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At the start of 2015, Jane opened two new accounts – X and Y – to inve

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