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

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Senior Manager
Joined: 25 Dec 2018
Posts: 422
Location: India
Concentration: General Management, Finance
GMAT Date: 02-18-2019
GPA: 3.4
WE: Engineering (Consulting)
At the start of 2015, Jane opened two new accounts – X and Y – to inve  [#permalink]

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15 Jun 2019, 10:38
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Difficulty:

45% (medium)

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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Joined: 02 Aug 2009
Posts: 7764
At the start of 2015, Jane opened two new accounts – X and Y – to inve  [#permalink]

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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

D
_________________
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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