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# Among the customers who bought either tables or chairs at a certain fu

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Math Expert
Joined: 02 Sep 2009
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Among the customers who bought either tables or chairs at a certain fu [#permalink]

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17 Feb 2017, 03:01
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95% (hard)

Question Stats:

38% (01:03) correct 62% (01:16) wrong based on 297 sessions

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Among the customers who bought either tables or chairs at a certain furniture store, the ratio of customers who bought tables to customers who bought chairs is 3:2. If half of those who bought chairs also bought tables, what percent of customers bought tables?

A. 60%
B. 65%
C. 70%
D. 75%
E. 80%
[Reveal] Spoiler: OA

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Re: Among the customers who bought either tables or chairs at a certain fu [#permalink]

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17 Feb 2017, 04:06
lets say total was 100
so only table was 60 and only chair 40
half of chair buyers bought table also, so 20 of them
total table 80
thats 80%.

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Re: Among the customers who bought either tables or chairs at a certain fu [#permalink]

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17 Feb 2017, 12:10
giobas wrote:
lets say total was 100
so only table was 60 and only chair 40
half of chair buyers bought table also, so 20 of them
total table 80
thats 80%.

But if now the total table buyers is 80, then it doesn't fit the 3:2 table:chair ratio any more, does it?
Intern
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Re: Among the customers who bought either tables or chairs at a certain fu [#permalink]

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18 Feb 2017, 13:06
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Lets say 6 people buy tables: 4 chairs. If half the people who bought chairs also bought tables, then 2 people bought chairs and tables. This means that 6 people bought tables, and Total # people= 6+4 -2 (to account for those that bought both) yields 8 people total. 6/8*100= 75%. D
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Re: Among the customers who bought either tables or chairs at a certain fu [#permalink]

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19 Feb 2017, 03:54
1
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Bunuel wrote:
Among the customers who bought either tables or chairs at a certain furniture store, the ratio of customers who bought tables to customers who bought chairs is 3:2. If half of those who bought chairs also bought tables, what percent of customers bought tables?

A. 60%
B. 65%
C. 70%
D. 75%
E. 80%

Using set theory to solve this sum. Given the following

n(cust buying tables) = n(a) = 3x

We are asked to find n(A)/n(AuB).

n(AuB)=n(A)+n(B)-n(AnB)

Given n(AnB)=x - half of those who bought chairs also bought tables
n(AuB) = 3x+2x - x = 4x.

So ratio is 3x/4x = 75% - D

Cheers!
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Among the customers who bought either tables or chairs at a certain fu [#permalink]

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21 Feb 2017, 23:27
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If the total was 100. Tables : Chair = 60 : 40 (3 : 2)

0.5 (40) = 20 buy tables.

$$=\frac{(40+20)}{(40+20+20)}*{100%}$$

$$={75%}$$
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Director
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Posts: 618
Re: Among the customers who bought either tables or chairs at a certain fu [#permalink]

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22 Feb 2017, 02:13
Bunuel wrote:
Among the customers who bought either tables or chairs at a certain furniture store, the ratio of customers who bought tables to customers who bought chairs is 3:2. If half of those who bought chairs also bought tables, what percent of customers bought tables?

A. 60%
B. 65%
C. 70%
D. 75%
E. 80%

I think answer should be A 60%. If # of people who bought tables=300 and chairs=200 and , then both= 200/2=100, but this does not change anything about what is given
So, %of customers who bought tables=300/500=60%
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Among the customers who bought either tables or chairs at a certain fu [#permalink]

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23 Feb 2017, 06:45
Bunuel wrote:
Among the customers who bought either tables or chairs at a certain furniture store, the ratio of customers who bought tables to customers who bought chairs is 3:2. If half of those who bought chairs also bought tables, what percent of customers bought tables?

A. 60%
B. 65%
C. 70%
D. 75%
E. 80%

Part of the questions ->half of those who bought chairs also bought tables is seemingly irrelevant; since at the end of the day what is being asked is how many people bought tables AND NOT how many people bought ONLY tables (and not chairs).
Therefore, only part relevant is : $$\frac{tables}{chairs} = \frac{3}{2}$$
inverting and adding 1 on both the sides: $$\frac{(tables + chairs)}{(tables)} = \frac{5}{3}$$
Anse -> 60%

Hope this is correct?
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Re: Among the customers who bought either tables or chairs at a certain fu [#permalink]

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23 Feb 2017, 10:31
Expert's post
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Bunuel wrote:
Among the customers who bought either tables or chairs at a certain furniture store, the ratio of customers who bought tables to customers who bought chairs is 3:2. If half of those who bought chairs also bought tables, what percent of customers bought tables?

A. 60%
B. 65%
C. 70%
D. 75%
E. 80%

We are given that the ratio of customers who bought tables to customers who bought chairs is 3:2. In other words, the ratio is 3x : 2x, in which 3x is the number of customers who bought tables and 2x is the number of customers who bought chairs. We are also given that half of those who bought chairs also bought tables. Thus (½)(2x) = x customers bought both chairs and tables, and hence, there is a total of 2x + 3x - x = 4x customers who bought one or both pieces of furniture. Since there are 3x customers who bought tables, the percentage of customers who bought tables is 3x/4x = ¾ = 75%.

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Posts: 191
Re: Among the customers who bought either tables or chairs at a certain fu [#permalink]

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11 Mar 2017, 13:20
Overlapping problem:
ration is 3 to 2, but 2 includes those who bought tables (1/2), so 3x bought tables, total people are 3x + 0,5*2x=4x. 3/4=75%. Hence answer is D
Manager
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Among the customers who bought either tables or chairs at a certain fu [#permalink]

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07 May 2017, 11:56
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As stated by Ron from MGMAT, it's very important to organize the data in some form or the other.
In this case, lets organize it in form of a table, as follows.
Ratio of Tables to Chairs inserted as 3a to 2a respectively.

-----------------------------Tables / Yes. -------Tables / No --------Total
Chairs / Yes.--------------a---------------------------a-----------------------2a
Chairs / No----------------2a----------------------- Nil----------------------2a
Total--------------------------3a-------------------------a-----------------------100

4a = 100 , means a = 25
Therefore, percent of Tables i.e. 3a = 75%. / Ans : Option D
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Re: Among the customers who bought either tables or chairs at a certain fu [#permalink]

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07 May 2017, 12:50
table=30
chair=20
table+chair=20/2=10
total=30+20-10=40
table customer=30/40=75%
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Re: Among the customers who bought either tables or chairs at a certain fu [#permalink]

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08 May 2017, 05:35
Bunuel

Sir I have approached the problem in following way

Table/Toatal *100

$$\frac{x+x/2}{x+x/2+y}$$

Now using x/y=3/2 I'm getting 70%

What did I miss??
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Director
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Re: Among the customers who bought either tables or chairs at a certain fu [#permalink]

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12 Jun 2017, 21:43
Bunuel wrote:
Among the customers who bought either tables or chairs at a certain furniture store, the ratio of customers who bought tables to customers who bought chairs is 3:2. If half of those who bought chairs also bought tables, what percent of customers bought tables?

A. 60%
B. 65%
C. 70%
D. 75%
E. 80%

t: c= 3:2 -
2(1/2) = 1
3+2 = 5
3 +4/5 = .8

Thus
"E"
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Re: Among the customers who bought either tables or chairs at a certain fu [#permalink]

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13 Jun 2017, 00:58
1
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techiesam wrote:
Bunuel

Sir I have approached the problem in following way

Table/Toatal *100

$$\frac{x+x/2}{x+x/2+y}$$

Now using x/y=3/2 I'm getting 70%

What did I miss??

Hi

What you have done is x/y = 3/2. So you have taken the ratio of those people who bought Only Tables to those people who bought Only Chairs to be 3:2.

But that is NOT what is given in the question. The question says the ratio of People who bought Tables (these include those who bought chairs also along with tables) TO People who bought Chairs (these include those who bought tables also along with chairs) to be 3:2

So what you should do instead is (x + x/2)/(y + x/2) = 3/2. This will give you y=x/2, which you can then replace in the original venn diagram. Then number of customers with tables will become = x + x/2 = 3x/2. Number of customers with chairs will become = x/2 + x/2 = x. And Total customers = x + x/2 + x/2 = 2x

Therefore, percentage of people who bought tables = (3x/2)/(2x) * 100 = 75%

Hope this helps.
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Re: Among the customers who bought either tables or chairs at a certain fu [#permalink]

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16 Jun 2017, 02:15
Fabulous question.

It is essentially a overleaping sets question.

Chairs => 200
Tables => 300

Only chairs => 100
Only tables => 200
Both => 100

Hence p/100 * 400 = 300

p=> 75

Hence D.
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Re: Among the customers who bought either tables or chairs at a certain fu [#permalink]

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01 Oct 2017, 06:24
What is confusing here is the part in the problem statement "who bought either tables or chairs" is 3:2 but not both. I am sure most of us are confused here as I am when it comes to accepting the answer choice D as the correct answer.

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Re: Among the customers who bought either tables or chairs at a certain fu [#permalink]

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02 Oct 2017, 10:25
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Expert's post
amol143 wrote:
What is confusing here is the part in the problem statement "who bought either tables or chairs" is 3:2 but not both. I am sure most of us are confused here as I am when it comes to accepting the answer choice D as the correct answer.

The question doesn't specify in this statement that the customers bought ONLY tables or ONLY chairs. And without such specification the GMAT always assumes inclusivity. That is, "people who bought tables or chairs" means "people who bought tables or chairs OR BOTH" unless explicitly stated to the contrary.
Re: Among the customers who bought either tables or chairs at a certain fu   [#permalink] 02 Oct 2017, 10:25
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