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Manager  Joined: 02 Dec 2012
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How many of the 60 cars sold last month by a certain dealer  [#permalink]

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Question Stats: 86% (01:23) correct 14% (01:18) wrong based on 1012 sessions

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How many of the 60 cars sold last month by a certain dealer had neither power windows nor a stereo?

(1) Of the 60 cars sold, 20 had a stereo but not power windows.
(2) Of the 60 cars sold, 30 had both power windows and a stereo.
Math Expert V
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How many of the 60 cars sold last month by a certain dealer  [#permalink]

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How many of the 60 cars sold last month by a certain dealer had neither power windows nor a stereo?

Given: 60 = {Windows} + {Stereo} - {Both} + {Neither}.
Question: {Neither} = ?

(1) Of the 60 cars sold, 20 had a stereo but not power windows:
{Stereo} - {Both} = 20
60 = {Windows} + 20 + {Neither}. Not sufficient.

(2) Of the 60 cars sold, 30 had both power windows and a stereo: {Both} = 30. Not sufficient.

(1)+(2) We can find how many had stereo, but we still cannot get how many had neither power windows nor a stereo. Not sufficient.

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Manager  Joined: 13 Dec 2012
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Re: How many of the 60 cars sold last month by a certain dealer  [#permalink]

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Hi all. I am new to GmatClub and this is my first post ever. I just started preparations for the exams last week and I am targeting end-April for the exams. I seem to be able to get most of the questions on DS and Maths right without using elaborate equations and now I am worried. I have never liked equations plus English is my third language.

Must I learn to use all these esoteric equations to do well in Gmat? Call me crazy but my target is 770.

Posted from my mobile device
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Re: How many of the 60 cars sold last month by a certain dealer  [#permalink]

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Also using a chart is pretty straight Attachments screenshot.2.png [ 3.5 KiB | Viewed 13107 times ]

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Re: How many of the 60 cars sold last month by a certain dealer  [#permalink]

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knightofdelta wrote:
Hi all. I am new to GmatClub and this is my first post ever. I just started preparations for the exams last week and I am targeting end-April for the exams. I seem to be able to get most of the questions on DS and Maths right without using elaborate equations and now I am worried. I have never liked equations plus English is my third language.

Must I learn to use all these esoteric equations to do well in Gmat? Call me crazy but my target is 770.

Posted from my mobile device

if you think right away that gmat need somehing of esoteric to solve......I think is wrong.

reasoning has nothing to do with the arcane _________________
Manager  Joined: 13 Dec 2012
Posts: 186
Re: How many of the 60 cars sold last month by a certain dealer  [#permalink]

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carcass wrote:
knightofdelta wrote:
Hi all. I am new to GmatClub and this is my first post ever. I just started preparations for the exams last week and I am targeting end-April for the exams. I seem to be able to get most of the questions on DS and Maths right without using elaborate equations and now I am worried. I have never liked equations plus English is my third language.

Must I learn to use all these esoteric equations to do well in Gmat? Call me crazy but my target is 770.

if you think right away that gmat need somehing of esoteric to solve......I think is wrong.

reasoning has nothing to do with the arcane Thank you.

Now I can breathe easier.
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Manager  B
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Re: How many of the 60 cars sold last month by a certain dealer  [#permalink]

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Bunuel wrote:
How many of the 60 cars sold last month by a certain dealer had neither power windows nor a stereo?

Given: 60={Windows}+{Stereo}-{Both}+{Neither}.
Question: {Neither}=?

(1) Of the 60 cars sold, 20 had a stereo but not power windows --> {Stereo}-{Both}=20 --> 60={Windows}+20+{Neither}. Not sufficient.

(2) Of the 60 cars sold, 30 had both power windows and a stereo --> {Both}=30. Not sufficient.

(1)+(2) We can find how many had stereo, but we still cannot get how many had neither power windows nor a stereo. Not sufficient.

I did this question using Venn diagrams rather than equations or matrix, is that a fine approach too considering i was able to solve it under 2 minutes?
Thanks.
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: How many of the 60 cars sold last month by a certain dealer  [#permalink]

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Dienekes wrote:
Bunuel wrote:
How many of the 60 cars sold last month by a certain dealer had neither power windows nor a stereo?

Given: 60={Windows}+{Stereo}-{Both}+{Neither}.
Question: {Neither}=?

(1) Of the 60 cars sold, 20 had a stereo but not power windows --> {Stereo}-{Both}=20 --> 60={Windows}+20+{Neither}. Not sufficient.

(2) Of the 60 cars sold, 30 had both power windows and a stereo --> {Both}=30. Not sufficient.

(1)+(2) We can find how many had stereo, but we still cannot get how many had neither power windows nor a stereo. Not sufficient.

I did this question using Venn diagrams rather than equations or matrix, is that a fine approach too considering i was able to solve it under 2 minutes?
Thanks.

Absolutely. Choose whichever approach suits you best.
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Re: How many of the 60 cars sold last month by a certain dealer  [#permalink]

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How many of the 60 cars sold last month by a certain dealer had neither power windows nor a stereo?

Given: 60={Windows}+{Stereo}-{Both}+{Neither}.
Question: {Neither}=?

(1) Of the 60 cars sold, 20 had a stereo but not power windows --> {Stereo}-{Both}=20 --> 60={Windows}+20+{Neither}. Not sufficient.

(2) Of the 60 cars sold, 30 had both power windows and a stereo --> {Both}=30. Not sufficient.

(1)+(2) We can find how many had stereo, but we still cannot get how many had neither power windows nor a stereo. Not sufficient.

_________________

Bunuel,
Can we not get the answer if we use Statement 1 & 2?

We know total for Stereo & No Stereo and we know the total for Power Windows and No Power Windows.
By taking the above information.. 30+x + 20 + x = 60 --> x=5 Plug in 5 and you have No Power Windows at 25. Which mean's the cars with No Stereo and Power Window is 5. Am I thinking too much into this question?
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3074
Re: How many of the 60 cars sold last month by a certain dealer  [#permalink]

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roro555 wrote:
How many of the 60 cars sold last month by a certain dealer had neither power windows nor a stereo?

Given: 60={Windows}+{Stereo}-{Both}+{Neither}.
Question: {Neither}=?

(1) Of the 60 cars sold, 20 had a stereo but not power windows --> {Stereo}-{Both}=20 --> 60={Windows}+20+{Neither}. Not sufficient.

(2) Of the 60 cars sold, 30 had both power windows and a stereo --> {Both}=30. Not sufficient.

(1)+(2) We can find how many had stereo, but we still cannot get how many had neither power windows nor a stereo. Not sufficient.

_________________

Bunuel,
Can we not get the answer if we use Statement 1 & 2?

We know total for Stereo & No Stereo and we know the total for Power Windows and No Power Windows.
By taking the above information.. 30+x + 20 + x = 60 --> x=5 Plug in 5 and you have No Power Windows at 25. Which mean's the cars with No Stereo and Power Window is 5. Am I thinking too much into this question?

Dear roro555

Please refer to the statement in red above. It is not correct.

To explain why, have represented the information we get from St 1 + 2 in a matrix. Looking at this table, it's easy to see that

We know the total for Power Windows and No PW only in terms of X, the unknown value we had to find.

From this table, the only thing we can say for sure is that X lies between 0 and 10, inclusive. But since the question had asked us about a unique value of X, and we haven't been able to find one, the correct answer will be Option E.

Hope this helped! Japinder
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Re: How many of the 60 cars sold last month by a certain dealer  [#permalink]

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Is the following approach correct???

We know that

T-N = A + B - k

1) 60-N= A+20 (since b-k is given as 20 ) therefore not sufficient

2) 60-N= A+B - 30 ( k=30 is given)
Therefore not sufficient

Combining 1 and 2

60-N= A+50-30 again not sufficient
Therefore answer is E

Is this approach correct?

Waiting for a response!
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Re: How many of the 60 cars sold last month by a certain dealer  [#permalink]

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1
How many of the 60 cars sold last month by a certain dealer had neither power windows nor a stereo?

(1) Of the 60 cars sold, 20 had a stereo but not power windows.
(2) Of the 60 cars sold, 30 had both power windows and a stereo.

An easy way to solve this problem is to set up a double set matrix. In our matrix we have two main categories: power windows and stereo. More specifically, our table will be labeled with:

1) Power windows (PW)

2) No power windows (No PW)

3) Stereo

4) No stereo

(To save room on our table headings, we will use the abbreviations for some of these categories.)

We are given that a total of 60 cars were sold last month. We are trying to determine how many cars sold had neither power windows nor a stereo.

Let’s fill all this information into a table. Note that each row sums to a row total, and each column sums to a column total. These totals also sum to the grand total, designated by 60 at the bottom right of the table. We need to determine the value for the question mark in the table.

Statement One Alone:

Of the 60 cars sold, 20 had a stereo but not power windows.

Let’s fill the information from statement one into our matrix. Statement one does not provide enough information to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

Of the 60 cars sold, 30 had power windows and stereo.

Let’s fill the information from statement two into our matrix. Statement two is insufficient to answer the question. Eliminate answer choice B.

Statements One and Two Together:

From statements one and two we know that of the 60 cars sold, 20 had a stereo but not power windows and that 30 had power windows and stereo. We can fill all this into our matrix. We see that we still do not have enough information to determine how many cars sold had neither power windows nor a stereo.

The answer is E.
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Joined: 02 Oct 2015
Posts: 11
How many of the 60 cars sold last month by a certain dealer  [#permalink]

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Bunuel wrote:
Dienekes wrote:
Bunuel wrote:
How many of the 60 cars sold last month by a certain dealer had neither power windows nor a stereo?

Given: 60={Windows}+{Stereo}-{Both}+{Neither}.
Question: {Neither}=?

(1) Of the 60 cars sold, 20 had a stereo but not power windows --> {Stereo}-{Both}=20 --> 60={Windows}+20+{Neither}. Not sufficient.

(2) Of the 60 cars sold, 30 had both power windows and a stereo --> {Both}=30. Not sufficient.

(1)+(2) We can find how many had stereo, but we still cannot get how many had neither power windows nor a stereo. Not sufficient.

I did this question using Venn diagrams rather than equations or matrix, is that a fine approach too considering i was able to solve it under 2 minutes?
Thanks.

Absolutely. Choose whichever approach suits you best.

hello bunuel,

I have encountered this problem today and I thought that the answer is C. please correct me if I am wrong. when checking both statements we have 4 different unknowns, I will presume that they are defined as a,b,c, and d. check the attached picture, we can derive 4 different equations. a+b=10 , b+d=20 , 30+a=c , c+d=60. I tried to solve them manually and the answer was a=10. I thought that maybe I missed something, and we know that the maximum number for a is 10, so I tried 9, which made c=39 and d=21, which is impossible. So the only value that a can take is actually 10. Am I missing something?
Attachments 11.jpg [ 15.18 KiB | Viewed 4440 times ]

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How many of the 60 cars sold last month by a certain dealer had neithe  [#permalink]

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How many of the 60 cars sold last month by a certain dealer had neither power windows nor a stereo?

(1) Of the 60 cars sold, 20 had a stereo but not power windows.
(2) Of the 60 cars sold, 30 had both power windows and a stereo.

This is a simple question actually, but we all know that the gmat can trick us into believing that a hard question has a simple answer. The answer in the Official guide and in the forum is E. I am not convinced, I think that the actual answer is C, my explanation is below.

when checking both statements we have 4 different unknowns, I will presume that they are defined as a,b,c, and d. check the attached picture, we can derive 4 different equations. a+b=10 , b+d=20 , 30+a=c , c+d=60. I tried to solve them manually and the answer was a=10. I thought that maybe I missed something, and we know that the maximum number for a is 10, so I tried 9, which made c=39 and d=21, which is impossible because the maximum value of d+b is 20. So the only value that a can take is actually 10. C is the correct answer.
Attachments 11.jpg [ 15.18 KiB | Viewed 4455 times ]

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Re: How many of the 60 cars sold last month by a certain dealer had neithe  [#permalink]

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tinayni552 wrote:
when checking both statements we have 4 different unknowns, I will presume that they are defined as a,b,c, and d. check the attached picture, we can derive 4 different equations. a+b=10 , b+d=20 , 30+a=c , c+d=60. I tried to solve them manually and the answer was a=10. I thought that maybe I missed something, and we know that the maximum number for a is 10, so I tried 9, which made c=39 and d=21, which is impossible because the maximum value of d+b is 20. So the only value that a can take is actually 10. C is the correct answer.

I think you're arriving at an incorrect conclusion because the equation I highlighted above is not right - it should read 20 + b = d.

You can write down four equations here in four unknowns, but the equations are not "independent", to use the technical term. You can derive one of the equations from the rest of them. If you take the first three equations and put the number alone on one side in each:

10 = a + b
20 = d - b
30 = c - a

and just add all three of those equations together, what you'll get is:

60 = c + d

which is the fourth equation. So the fourth equation isn't any new information, since you already know it using the first three equations, and you really only have 3 equations in 4 unknowns. You'll have an infinite number of solutions, though if you need the unknowns to represent non-negative integers, I think you'll find a can take any value between 0 and 10, inclusive, for eleven possible non-negative integer solutions.

The question does illustrate why counting equations and unknowns is generally not a successful strategy in DS questions, besides on the easiest questions. And for this question, you don't really need to use any algebra. We know about 50 of the cars using the two statements, but we don't know how the remaining 10 cars are divided between the two remaining categories - the cars with only power windows, and the cars with neither stereos nor power windows. So the answer is E.
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Re: How many of the 60 cars sold last month by a certain dealer  [#permalink]

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IanStewart wrote:
tinayni552 wrote:
when checking both statements we have 4 different unknowns, I will presume that they are defined as a,b,c, and d. check the attached picture, we can derive 4 different equations. a+b=10 , b+d=20 , 30+a=c , c+d=60. I tried to solve them manually and the answer was a=10. I thought that maybe I missed something, and we know that the maximum number for a is 10, so I tried 9, which made c=39 and d=21, which is impossible because the maximum value of d+b is 20. So the only value that a can take is actually 10. C is the correct answer.

I think you're arriving at an incorrect conclusion because the equation I highlighted above is not right - it should read 20 + b = d.

You can write down four equations here in four unknowns, but the equations are not "independent", to use the technical term. You can derive one of the equations from the rest of them. If you take the first three equations and put the number alone on one side in each:

10 = a + b
20 = d - b
30 = c - a

and just add all three of those equations together, what you'll get is:

60 = c + d

which is the fourth equation. So the fourth equation isn't any new information, since you already know it using the first three equations, and you really only have 3 equations in 4 unknowns. You'll have an infinite number of solutions, though if you need the unknowns to represent non-negative integers, I think you'll find a can take any value between 0 and 10, inclusive, for eleven possible non-negative integer solutions.

The question does illustrate why counting equations and unknowns is generally not a successful strategy in DS questions, besides on the easiest questions. And for this question, you don't really need to use any algebra. We know about 50 of the cars using the two statements, but we don't know how the remaining 10 cars are divided between the two remaining categories - the cars with only power windows, and the cars with neither stereos nor power windows. So the answer is E.

but "a" cannot take any value smaller than 10, check it, if a was 9 then c would be 39 and d would be 21 which is impossible because d is equal to 20+b. Explain this issue please.
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Re: How many of the 60 cars sold last month by a certain dealer  [#permalink]

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tinayni552 wrote:
IanStewart wrote:
tinayni552 wrote:
when checking both statements we have 4 different unknowns, I will presume that they are defined as a,b,c, and d. check the attached picture, we can derive 4 different equations. a+b=10 , b+d=20 , 30+a=c , c+d=60. I tried to solve them manually and the answer was a=10. I thought that maybe I missed something, and we know that the maximum number for a is 10, so I tried 9, which made c=39 and d=21, which is impossible because the maximum value of d+b is 20. So the only value that a can take is actually 10. C is the correct answer.

I think you're arriving at an incorrect conclusion because the equation I highlighted above is not right - it should read 20 + b = d.

You can write down four equations here in four unknowns, but the equations are not "independent", to use the technical term. You can derive one of the equations from the rest of them. If you take the first three equations and put the number alone on one side in each:

10 = a + b
20 = d - b
30 = c - a

and just add all three of those equations together, what you'll get is:

60 = c + d

which is the fourth equation. So the fourth equation isn't any new information, since you already know it using the first three equations, and you really only have 3 equations in 4 unknowns. You'll have an infinite number of solutions, though if you need the unknowns to represent non-negative integers, I think you'll find a can take any value between 0 and 10, inclusive, for eleven possible non-negative integer solutions.

The question does illustrate why counting equations and unknowns is generally not a successful strategy in DS questions, besides on the easiest questions. And for this question, you don't really need to use any algebra. We know about 50 of the cars using the two statements, but we don't know how the remaining 10 cars are divided between the two remaining categories - the cars with only power windows, and the cars with neither stereos nor power windows. So the answer is E.

but "a" cannot take any value smaller than 10, check it, if a was 9 then c would be 39 and d would be 21 which is impossible because d is equal to 20+b. Explain this issue please.

Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: How many of the 60 cars sold last month by a certain dealer  [#permalink]

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tinayni552 wrote:
tinayni552 wrote:
IanStewart wrote:

I think you're arriving at an incorrect conclusion because the equation I highlighted above is not right - it should read 20 + b = d.

You can write down four equations here in four unknowns, but the equations are not "independent", to use the technical term. You can derive one of the equations from the rest of them. If you take the first three equations and put the number alone on one side in each:

10 = a + b
20 = d - b
30 = c - a

and just add all three of those equations together, what you'll get is:

60 = c + d

which is the fourth equation. So the fourth equation isn't any new information, since you already know it using the first three equations, and you really only have 3 equations in 4 unknowns. You'll have an infinite number of solutions, though if you need the unknowns to represent non-negative integers, I think you'll find a can take any value between 0 and 10, inclusive, for eleven possible non-negative integer solutions.

The question does illustrate why counting equations and unknowns is generally not a successful strategy in DS questions, besides on the easiest questions. And for this question, you don't really need to use any algebra. We know about 50 of the cars using the two statements, but we don't know how the remaining 10 cars are divided between the two remaining categories - the cars with only power windows, and the cars with neither stereos nor power windows. So the answer is E.

but "a" cannot take any value smaller than 10, check it, if a was 9 then c would be 39 and d would be 21 which is impossible because d is equal to 20+b. Explain this issue please.

Why not?

a = 9;
c = 39;
b = 1;
d = 21.

Above values fit perfectly.
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