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# There are c chocolate chip cookies and r oatmeal raisin cookies in a j

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Senior DS Moderator
Joined: 27 Oct 2017
Posts: 1205
Location: India
GPA: 3.64
WE: Business Development (Energy and Utilities)

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20 Oct 2018, 09:14
1
00:00

Difficulty:

25% (medium)

Question Stats:

80% (01:47) correct 20% (01:53) wrong based on 61 sessions

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There are c chocolate chip cookies and r oatmeal raisin cookies in a jar. If there are no other cookies in the jar, is the probability of randomly selecting an oatmeal raisin cookie greater than the probability of selecting a chocolate chip cookie?
1)$$\frac{(r^2-rc)}{(r^2-c^2)}$$ > $$\frac{c(r+c)}{(r+c)^2}$$
2) If p peanut butter cookies were added to the jar then $$\frac{r}{(r+c+p)} > \frac{c}{(r+c+p)}$$

GMATbuster's Weekly GMAT Quant Quiz #5 Ques No 4

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20 Oct 2018, 09:20
Both are sufficient.

statement 1 can be solved using distributive law

Statement 2 is sufficient
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Joined: 02 Apr 2018
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20 Oct 2018, 09:22
D:

Statement 1:
the question stem asks if r / (r+c) > c / (r+c)
Statement 1 states the same thing except LHS is multiplied by (r-c)/(r-c) and RHS is multiplied by (r+c)/(r+c)

Statement 2:
adding in a constant does not change the proportion therefore also sufficient

Therefore D
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Joined: 16 Sep 2011
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20 Oct 2018, 09:23
1
D is sufficient

Option A, r(r-c)/(r-c)(r+c) >. c(r+c)/(r+c)^2
if (r-c) is not equal to 0 and (r+c) is not equal to 0,

r/(r+c) > c/(r+c)
which is 1/(r+c) (r-c)>0.... which gives r>c

sufficient

Option B, r/(r+c+p)> c/(r+c+p)
if r+c+p is not equal to 0,

1/(r+p+c) (r-c)>0,
which means r>c,
sufficient
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Joined: 08 Feb 2018
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20 Oct 2018, 09:23
D.

Since we can assume r and c would be positive, solving statement 1 gives r>c. The second statement introduces a constant to both sides and it can be concluded that r>c.
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20 Oct 2018, 09:30
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Manager
Joined: 21 Jul 2018
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18 Dec 2018, 14:47
Hi gmatbusters,

I am not able to get how we can get answer from option 2, could you please assist me.
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Re: There are c chocolate chip cookies and r oatmeal raisin cookies in a j   [#permalink] 18 Dec 2018, 14:47
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