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# One week a certain vehicle rental outlet had a total of 40

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Director
Joined: 12 Jun 2006
Posts: 532

Kudos [?]: 162 [0], given: 1

One week a certain vehicle rental outlet had a total of 40 [#permalink]

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12 Jun 2006, 03:05
00:00

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(N/A)

Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 2 sessions

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One week a certain vehicle rental outlet had a total of 40 cars, 12 trucks, 28 vans and 20 SUVs available. Andre and Barbara went to the vehicle rental outlet and chose 2 vehicles at random, with the condition that Andre and Barbara would not select two of the same type of vehicle (in other words, if one of them has an SUV, the other won't take an SUV, so the second person doesn't even consider the SUVs). What is the probability that, of the two vehicles, one of them is a car of van?

OA:
[Reveal] Spoiler:
1037/1100

[Reveal] Spoiler:
In order to solve, I tried:

40/100*28/60 + 28/100+40/72
This is incorrect.

I was told that I should find the probability of them not choosing a car or van:

20/100*12/80 + 12/100*20/88
This equals 63/1100.
Now, subtract 63/1100 from 1.
Your answer is 1037/1100.

Can anyone explain why I should find the probability of them not choosing a car or van in order to solve. I was told this is the "1-x probability method". Is anyone familiar with this?

Hello by the way.

Kudos [?]: 162 [0], given: 1

Manager
Joined: 01 Jun 2006
Posts: 139

Kudos [?]: 6 [0], given: 0

Re: One week a certain vehicle rental outlet had a total of 40 [#permalink]

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12 Jun 2006, 04:12
I think that way is not correct. I likely solve the problem this way
The number of ways you can collect 2 car depends on the condition is:
(40*60+12*88+28*72+20*80)=7072
And the ways you collect 2 cars without any car or van is
12*20=240
So the needed probability must be
1-240/7072=427/442
What do u think?

Kudos [?]: 6 [0], given: 0

Manager
Joined: 10 May 2006
Posts: 186

Kudos [?]: 5 [0], given: 0

Location: USA
Re: One week a certain vehicle rental outlet had a total of 40 [#permalink]

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12 Jun 2006, 15:44
ggarr wrote:
One week a certain vehicle rental outlet had a total of 40 cars, 12 trucks, 28 vans and 20 SUVs available. Andre and Barbara went to the vehicle rental outlet and chose 2 vehicles at random, with the condition that Andre and Barbara would not select two of the same type of vehicle (in other words, if one of them has an SUV, the other won't take an SUV, so the second person doesn't even consider the SUVs). What is the probability that, of the two vehicles, one of them is a car of van?

Prob that A picks a van and B doesn't if A picks first is 28/100 = 7/25. Doesn't matter what B picks.

Prob that A does not pick a van and B does = 72/100 * 28/99 =56/275

Prob that either A or B picks a van = 7/25 + 56/275 = 133/275

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VP
Joined: 02 Jun 2006
Posts: 1258

Kudos [?]: 106 [0], given: 0

Re: One week a certain vehicle rental outlet had a total of 40 [#permalink]

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12 Jun 2006, 15:51
Is the question "... one of them picks a car or van?"

or

" .. one of them picks a car of type van"?

Kudos [?]: 106 [0], given: 0

Director
Joined: 12 Jun 2006
Posts: 532

Kudos [?]: 162 [0], given: 1

Re: One week a certain vehicle rental outlet had a total of 40 [#permalink]

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12 Jun 2006, 16:00
One week a certain vehicle rental outlet had a total of 40 cars, 12 trucks, 28 vans and 20 SUVs available. Andre and Barbara went to the vehicle rental outlet and chose 2 vehicles at random, with the condition that Andre and Barbara would not select two of the same type of vehicle (in other words, if one of them has an SUV, the other won't take an SUV, so the second person doesn't even consider the SUVs). What is the probability that, of the two vehicles, one of them is a car or van?

In order to solve, I tried:

40/100*28/60 + 28/100+40/72
This is incorrect.

I was told that I should find the probability of them not choosing a car or van:

20/100*12/80 + 12/100*20/88
This equals 63/1100.
Now, subtract 63/1100 from 1.
Your answer is 1037/1100.

Can anyone explain why I should find the probability of them not choosing a car or van in order to solve. I was told this is the "1-x probability method". Is anyone familiar with this?

Hello by the way.

Kudos [?]: 162 [0], given: 1

Director
Joined: 12 Jun 2006
Posts: 532

Kudos [?]: 162 [0], given: 1

Re: One week a certain vehicle rental outlet had a total of 40 [#permalink]

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21 Jun 2006, 03:22
Quote:
One week a certain vehicle rental outlet had a total of 40 cars, 12 trucks, 28 vans and 20 SUVs available. Andre and Barbara went to the vehicle rental outlet and chose 2 vehicles at random, with the condition that Andre and Barbara would not select two of the same type of vehicle (in other words, if one of them has an SUV, the other won't take an SUV, so the second person doesn't even consider the SUVs). What is the probability that, of the two vehicles, one of them is a car or van?

According to Manhattan Gmat the correct answer is 1037/1100.
This is arrived at by:

1) finding the prob of them not choosing the car or van
20/100*12/80 + 12/100*20/88.

2) then subtracting the solution, 63/1100, from 1100/1100.

3) Your answer is 1037-1100.

Does anyone have any idea why I need to find the probability of them not choosing a car or van in order to solve? Again, Manhattan Gmat names this method the "1-x probability trick". Has anyone heard of this method?

Kudos [?]: 162 [0], given: 1

Intern
Joined: 12 Aug 2014
Posts: 2

Kudos [?]: [0], given: 0

Re: One week a certain vehicle rental outlet had a total of 40 [#permalink]

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17 Aug 2014, 13:55
Has anyone figured this out yet?

Why do you have to use the 1-x approach rather than solving it directly?

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Re: One week a certain vehicle rental outlet had a total of 40   [#permalink] 17 Aug 2014, 13:55
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# One week a certain vehicle rental outlet had a total of 40

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