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There were 2 apples and 5 bananas in a basket. After
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Jcpenny
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There were 2 apples and 5 bananas in a basket. After [#permalink] New post  08 Nov 2008, 17:15
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There were 2 apples and 5 bananas in a basket. After additional apples and bananas were placed in the basket, the ratio of the number of apples to the number of bananas was 1/2. How many apples were added?

(1) The number of apples added was 2/3 the number of bananas added.
(2) A total of 5 apples and bananas were added.
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icandy
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Re: apple and banana---29 [#permalink] New post  08 Nov 2008, 17:49
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Jcpenny wrote:
There were 2 apples and 5 bananas in a basket. After additional apples and bananas were placed in the basket, the ratio of the number of apples to the number of bananas was 1/2 . How many apples were added?
(1) The number of apples added was 2/3 the number of bananas added.
(2) A total of 5 apples and bananas were added.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.



Given

A=2, B =5

2+X/ 5+Y = 1/2

(1) says X=2/3Y

2 eq's 2 unknowns. Solve for x to get the answer. Suff

(2) says X +Y =5

2 eq's 2 unknowns. Solve for x to get the answer. Suff

D
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rajareena
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Re: apple and banana---29 [#permalink] New post  26 May 2012, 04:06
I thought the answer should be C.

In case of statement 1, we can have multiple values of banana and apple. There can be 5 apples and 10 banana thus adding 3 apples and 5 bananas and there can also be 10 apples and 20 bananas giving ratio of 1/2 but apples added will be 8 and 15 banana.

Can someone clarify if I am missing something?


Jcpenny wrote:
There were 2 apples and 5 bananas in a basket. After additional apples and bananas were placed in the basket, the ratio of the number of apples to the number of bananas was 1/2 . How many apples were added?
(1) The number of apples added was 2/3 the number of bananas added.
(2) A total of 5 apples and bananas were added.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
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Re: apple and banana---29 [#permalink] New post  26 May 2012, 05:09
If A apples were added and B bananas were added:
(2+A)/(5+B)=1/2
Statement 1: is telling 2 bananas were added for every 3 apple added. i. e. A=2/3 B . plugging the value of B we can solve for A.
Statement 2: A+B=5. similar as statement 1. can solve for A.

D.



manjeet1972 wrote:
I thought the answer should be C.

In case of statement 1, we can have multiple values of banana and apple. There can be 5 apples and 10 banana thus adding 3 apples and 5 bananas and there can also be 10 apples and 20 bananas giving ratio of 1/2 but apples added will be 8 and 15 banana.

Can someone clarify if I am missing something?
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nevermoreflow
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Re: apple and banana---29 [#permalink] New post  10 Apr 2014, 16:03
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BDSunDevil wrote:
If A apples were added and B bananas were added:
(2+A)/(5+B)=1/2
Statement 1: is telling 2 bananas were added for every 3 apple added. i. e. A=2/3 B . plugging the value of B we can solve for A.
Statement 2: A+B=5. similar as statement 1. can solve for A.

D.



manjeet1972 wrote:
I thought the answer should be C.

In case of statement 1, we can have multiple values of banana and apple. There can be 5 apples and 10 banana thus adding 3 apples and 5 bananas and there can also be 10 apples and 20 bananas giving ratio of 1/2 but apples added will be 8 and 15 banana.

Can someone clarify if I am missing something?


Isn't (1) saying the # of apples to bananas is 2:3? So for every 2 apples, there are 3 bananas? Couldn't it be 4 to 6, 6 to 9, etc? We can't get an exact amount of apples that were added from this like this question is asking for.

(2) is clearly sufficient, but I'm having issues with (1).
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ind23
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Re: apple and banana---29 [#permalink] New post  10 Apr 2014, 22:24
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nevermoreflow wrote:
BDSunDevil wrote:
If A apples were added and B bananas were added:
(2+A)/(5+B)=1/2
Statement 1: is telling 2 bananas were added for every 3 apple added. i. e. A=2/3 B . plugging the value of B we can solve for A.
Statement 2: A+B=5. similar as statement 1. can solve for A.

D.



manjeet1972 wrote:
I thought the answer should be C.

In case of statement 1, we can have multiple values of banana and apple. There can be 5 apples and 10 banana thus adding 3 apples and 5 bananas and there can also be 10 apples and 20 bananas giving ratio of 1/2 but apples added will be 8 and 15 banana.

Can someone clarify if I am missing something?


Isn't (1) saying the # of apples to bananas is 2:3? So for every 2 apples, there are 3 bananas? Couldn't it be 4 to 6, 6 to 9, etc? We can't get an exact amount of apples that were added from this like this question is asking for.

(2) is clearly sufficient, but I'm having issues with (1).


Well the way I look at this question is that there are two unknowns.

Number of apples added = x
Number of bananas added = y

If we have two different equation, we can find unique values of x & y

One equation can be made using the information in question => (2+x)/(5+y) = 1/2

Argument 1 gives us another equation => x = 2/3y
If you put x = 2/3y in first equation you will get a unique solution.
So, it is sufficient.

Argument 2 also gives us on equation => x+y = 5
Again a unique solution is possible. It
So, it is sufficient.

Hence, the answer is D.

----------------
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Re: There were 2 apples and 5 bananas in a basket. After [#permalink] New post  18 Jul 2015, 11:04
nevermoreflow wrote:
BDSunDevil wrote:
If A apples were added and B bananas were added:
(2+A)/(5+B)=1/2
Statement 1: is telling 2 bananas were added for every 3 apple added. i. e. A=2/3 B . plugging the value of B we can solve for A.
Statement 2: A+B=5. similar as statement 1. can solve for A.

D.



manjeet1972 wrote:
I thought the answer should be C.

In case of statement 1, we can have multiple values of banana and apple. There can be 5 apples and 10 banana thus adding 3 apples and 5 bananas and there can also be 10 apples and 20 bananas giving ratio of 1/2 but apples added will be 8 and 15 banana.

Can someone clarify if I am missing something?


Isn't (1) saying the # of apples to bananas is 2:3? So for every 2 apples, there are 3 bananas? Couldn't it be 4 to 6, 6 to 9, etc? We can't get an exact amount of apples that were added from this like this question is asking for.

(2) is clearly sufficient, but I'm having issues with (1).


Whatever value you take, the ratio remains same and so our answer also. By taking 4/6 you are taking extra 2 multiplied in both numerator and denominator, so whatever you take 4/6 or 6/9 or 8/12 answer will always be same.
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Kekki
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Re: There were 2 apples and 5 bananas in a basket. After [#permalink] New post  15 Sep 2015, 03:06
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nevermoreflow wrote:
BDSunDevil wrote:
If A apples were added and B bananas were added:
(2+A)/(5+B)=1/2
Statement 1: is telling 2 bananas were added for every 3 apple added. i. e. A=2/3 B . plugging the value of B we can solve for A.
Statement 2: A+B=5. similar as statement 1. can solve for A.

D.



manjeet1972 wrote:
I thought the answer should be C.

In case of statement 1, we can have multiple values of banana and apple. There can be 5 apples and 10 banana thus adding 3 apples and 5 bananas and there can also be 10 apples and 20 bananas giving ratio of 1/2 but apples added will be 8 and 15 banana.

Can someone clarify if I am missing something?


Isn't (1) saying the # of apples to bananas is 2:3? So for every 2 apples, there are 3 bananas? Couldn't it be 4 to 6, 6 to 9, etc? We can't get an exact amount of apples that were added from this like this question is asking for.

(2) is clearly sufficient, but I'm having issues with (1).


If you add 4 apples and 6 bananas then you have 6 apples and 11 bananas which isn't a ratio of 1:2.
Only 2 apples + 3 banans lead to a total of 4 apples and 8 bananas -> Ratio of 1:2. No other multiple of 2:3 would lead to a total ration of 1:2.

Therefore (1) is sufficient as we get a clear # value for apples and bananas.
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Re: There were 2 apples and 5 bananas in a basket. After [#permalink] New post  01 Nov 2015, 04:35
Can someone please post the solution to statement 2?

I know it's sufficient, but I can't understand why?
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Re: There were 2 apples and 5 bananas in a basket. After [#permalink] New post  01 Nov 2015, 08:03
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Akshobh wrote:
Can someone please post the solution to statement 2?

I know it's sufficient, but I can't understand why?


There were 2 apples and 5 bananas in a basket. After additional apples and bananas were placed in the basket, the ratio of the number of apples to the number of bananas was 1/2. How many apples were added?

From the stem: \(\frac{2+a}{5+b}=\frac{1}{2}\)]

(2) A total of 5 apples and bananas were added --> \(a+b=5\).

Solve these two equations to get a=2 and b=3.
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Re: There were 2 apples and 5 bananas in a basket. After [#permalink] New post  22 Jul 2017, 17:24
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Jcpenny wrote:
There were 2 apples and 5 bananas in a basket. After additional apples and bananas were placed in the basket, the ratio of the number of apples to the number of bananas was 1/2. How many apples were added?

(1) The number of apples added was 2/3 the number of bananas added.
(2) A total of 5 apples and bananas were added.


(2+A) / (5+B) = 1/2

4+2A = 5+B
2A = 1+B

The goal is to find A.

Statement 1) A = 2/3*B

2A = 1+B
A = 2B/3

3A = 2B
4A = 2+2B
-A = -2
A = 2
B = 3

Sufficient.

Statement 2) A+B = 5

A+B = 5
2A = 1+B

A = 5-B

2(5-B) = 1+B
10-2B = 1+B
9 = 3B
3=B
2=5

Sufficient.
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Re: There were 2 apples and 5 bananas in a basket. After [#permalink] New post  02 Feb 2020, 23:15
nevermoreflow wrote:
BDSunDevil wrote:
If A apples were added and B bananas were added:
(2+A)/(5+B)=1/2
Statement 1: is telling 2 bananas were added for every 3 apple added. i. e. A=2/3 B . plugging the value of B we can solve for A.
Statement 2: A+B=5. similar as statement 1. can solve for A.

D.



manjeet1972 wrote:
I thought the answer should be C.

In case of statement 1, we can have multiple values of banana and apple. There can be 5 apples and 10 banana thus adding 3 apples and 5 bananas and there can also be 10 apples and 20 bananas giving ratio of 1/2 but apples added will be 8 and 15 banana.

Can someone clarify if I am missing something?


Isn't (1) saying the # of apples to bananas is 2:3? So for every 2 apples, there are 3 bananas? Couldn't it be 4 to 6, 6 to 9, etc? We can't get an exact amount of apples that were added from this like this question is asking for.

(2) is clearly sufficient, but I'm having issues with (1).




I too approached the problem in the same manner. But you must remember that the question explicitly states that the new ratio is 1/2.
The only set that maintains this ratio is 4/8. Other values like 6/11, 8/14 violate this condition.
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There were 2 apples and 5 bananas in a basket. After [#permalink] New post  10 Aug 2020, 17:28
There are 2 apples (A) and 5 bananas (B) initially, and then let x apples, and y bananas be added to get to new ratio of apples (A) to bananas (B) of 1:2.

Statement (1):

\(x = \frac{2}{3} y\)
Sufficient, have the relation between x and y, so can plug into the ratio equation to solve for x and then y.
To illustrate this:
Apples : Bananas
\(\frac{(2 + x)}{(5 + y)} = \frac{1}{2}\)
\((2 + \frac{2y}{3})/(5 + y) = \frac{1}{2}\)
\(2(2+ \frac{2}{3} y) = 5 + y\)
\(4 + \frac{4}{3} y = 5 + y\)
\(y = 3\)
\(x = 2/3 y\)
\(x = 2\)

Statement (2):

\(x + y = 5\)
Also sufficient since have the total of x and y.
Still can solve for x to illustrate:
\(y = 5 - x\)
\(\frac{(2 + x)}{(5 + 5 - x)} = \frac{1}{2}\)
\(\frac{(2 + x)}{(10 - x)} = \frac{1}{2}\)
\(2(2 + x) = 10 - x\)
\(x = 2\)

Jcpenny wrote:
There were 2 apples and 5 bananas in a basket. After additional apples and bananas were placed in the basket, the ratio of the number of apples to the number of bananas was 1/2. How many apples were added?

(1) The number of apples added was 2/3 the number of bananas added.
(2) A total of 5 apples and bananas were added.
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Re: There were 2 apples and 5 bananas in a basket. After [#permalink] New post  11 Sep 2020, 04:31
icandy wrote:
Jcpenny wrote:
There were 2 apples and 5 bananas in a basket. After additional apples and bananas were placed in the basket, the ratio of the number of apples to the number of bananas was 1/2 . How many apples were added?
(1) The number of apples added was 2/3 the number of bananas added.
(2) A total of 5 apples and bananas were added.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.



Given

A=2, B =5

2+X/ 5+Y = 1/2

(1) says X=2/3Y

2 eq's 2 unknowns. Solve for x to get the answer. Suff

(2) says X +Y =5

2 eq's 2 unknowns. Solve for x to get the answer. Suff

D


Thanks! But can you expand on '2 eq's 2 unknowns.' a little bit please? I know how to substitute/eliminate linear equations, but understanding the rule could help me with speed.
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Re: There were 2 apples and 5 bananas in a basket. After [#permalink] New post  15 Dec 2020, 19:21
Jcpenny wrote:
There were 2 apples and 5 bananas in a basket. After additional apples and bananas were placed in the basket, the ratio of the number of apples to the number of bananas was 1/2. How many apples were added?

(1) The number of apples added was 2/3 the number of bananas added.
(2) A total of 5 apples and bananas were added.


\(\frac{2 + A }{ 5 + B} = \frac{1}{2}\)

What is A?

(1) \(\frac{A }{ B} = \frac{2 }{ 3}\)
\(2B = 3A\)
\(\frac{2B}{3} = A\)

2 + 2B/3 / 5 + B = 1/2

B = 3, A = 2

SUFFICIENT

(2) Directly gives us our answer; SUFFICIENT.

Answer is D.
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Re: There were 2 apples and 5 bananas in a basket. After [#permalink]
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