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# Of the 25 cars sold at a certain dealership yestarday, some

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VP
Joined: 22 Nov 2007
Posts: 1079
Of the 25 cars sold at a certain dealership yestarday, some [#permalink]

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28 Dec 2007, 09:29
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Of the 25 cars sold at a certain dealership yestarday, some had automatic transmission and some had antilock brakes. How many of the cars had automotive transmission but not antilock brakes?

1. all of the cars that had antilock brakes also had automatic transmission
2. 2 of the cars had neither automatic transmission nor antilock brakes
Director
Joined: 26 Mar 2006
Posts: 631

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28 Dec 2007, 09:33
marcodonzelli wrote:
Of the 25 cars sold at a certain dealership yestarday, some had automatic transmission and some had antilock brakes. How many of the cars had automotive transmission but not antilock brakes?

1. all of the cars that had antilock brakes also had automatic transmission
2. 2 of the cars had neither automatic transmission nor antilock brakes

I think it is 'C'. Will leave it to the experts to confirm.
VP
Joined: 09 Jul 2007
Posts: 1100
Location: London

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28 Dec 2007, 09:40
marcodonzelli wrote:
Of the 25 cars sold at a certain dealership yestarday, some had automatic transmission and some had antilock brakes. How many of the cars had automotive transmission but not antilock brakes?

1. all of the cars that had antilock brakes also had automatic transmission
2. 2 of the cars had neither automatic transmission nor antilock brakes

Total 25,
1. no info about automatic transmission supplied cars. there might be 20 cars with antilock brakes and the rest might have only auto transmission. or the other way around...

2. OK. we are left with 23 cars which we do not know what each of them has.

combined.
we know that 2 cars do not have either. so there are 23 cars have which have automatic transmission only, antilock brakes only or both so not suff.

E for me
VP
Joined: 22 Nov 2007
Posts: 1079

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28 Dec 2007, 09:50
OA is E

follow me...25=A+B-AB+N where A is the set of automatic trans. (but not only autom.trans), B the set of antilock brakes, AB the subset of cars with both of them and N the other cars...

1. B=AB....so 25=A+N...insuff
2. N=2.....insuff

Let's see C... 25=A+2...A=23....but A comprehends also the cars with antilock brakes....so we do not know how many cars have only automatic brakes...
Director
Joined: 12 Jul 2007
Posts: 858

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28 Dec 2007, 09:55

We have 25 cars in question. Per the second statement we know that 2 have neither and are thus off the table. Leaving us with 23 cars.

From statement 1 we know that all cars with B (anti-lock Brakes) have A (auto-transmission)...but that's all we know.

All 23 remaining cars could have B, and thus have A. Leaving us with 0 cars with A only.

OR

Of the 23 remaining, 10 could have B and the other 13 could A without B.

The equation looks like this:

X = cars with both
Y = cars with automatic transmission, but not brakes
Z = cars with neither

and we know that everything with brakes also has automatic, so there is no category for cars with only brakes (they don't exist!)

total cars = 25

25-z = 23 (remaining cars that have one or both)

X+Y = 23

and that's it. two variables with 1 equation. impossible to narrow it down further.

E
Director
Joined: 26 Mar 2006
Posts: 631

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28 Dec 2007, 10:14
eschn3am wrote:

We have 25 cars in question. Per the second statement we know that 2 have neither and are thus off the table. Leaving us with 23 cars.

From statement 1 we know that all cars with B (anti-lock Brakes) have A (auto-transmission)...but that's all we know.

All 23 remaining cars could have B, and thus have A. Leaving us with 0 cars with A only.

OR

Of the 23 remaining, 10 could have B and the other 13 could A without B.

The equation looks like this:

X = cars with both
Y = cars with automatic transmission, but not brakes
Z = cars with neither

and we know that everything with brakes also has automatic, so there is no category for cars with only brakes (they don't exist!)

total cars = 25

25-z = 23 (remaining cars that have one or both)

X+Y = 23

and that's it. two variables with 1 equation. impossible to narrow it down further.

E

now it makes sense...

Code:
all of the cars that had antilock brakes also had automatic transmission

All that had AB had AT but not the other way round.. there could be some only with AT and we don't know the number.. Thanks...
Re: counting   [#permalink] 28 Dec 2007, 10:14
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