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# A total of n trucks and cars are parked in a lot. If the num

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A total of n trucks and cars are parked in a lot. If the num [#permalink]  24 Oct 2009, 16:11
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A total of n trucks and cars are parked in a lot. If the number of cars is 1/4 the number of trucks, and 2/3 of the trucks are pickups, how many pickups, in terms of n, are parked in the lot?

A. 1/6 n
B. 5/12 n
C. 1/2 n
D. 8/15 n
E. 11/12 n

[Reveal] Spoiler:
I decided to use 100 for the total of trucks + cars. So n = 100.
Number of cars = 1/4 # of trucks, then there are 25 cars and 75 trucks
2/3 of the trucks are pickups, so 2/3 of 75 is 50.

Therefore, the number of pickups (50) in terms of n would be (1/2)n, which equals 100 * 1/2 = 50, the number of pickups. = C

However, this is not the correct answer- the correct answer is D, 8/15. Can someone explain???
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Oct 2013, 01:03, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Director
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Re: Picking numbers on algebra question- Something is wrong! [#permalink]  25 Oct 2009, 01:20
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n = t+c.
c=1/4 *t
so the total number of vehicles in terms of t is n=5/4 *t
Now, 2/3 * t are pickups

Q asks: 2/3 * t = n * F ( F is the required fraction )
2/3 * t = 5/4 * F => F = 8/15.
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A total of n trucks and cars [#permalink]  05 Jan 2011, 17:19
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A total of n trucks and cars are parked in a slot. If the number of cars is 1/4 the number
of trucks, and 2/3 of the trucks are pickups, how many pickups, in terms of n, are parked
in the slot?

A. 1/6 n
B. 5/12 n
C. 1/2 n
D. 8/15 n
E. 11/12 n

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Re: A total of n trucks and cars [#permalink]  05 Jan 2011, 17:23
D.
x+x/4=n => x=4n/5 => 2x/3=8n/15
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Re: A total of n trucks and cars [#permalink]  05 Jan 2011, 23:31
ajit257 wrote:
A total of n trucks and cars are parked in a slot. If the number of cars is 1/4 the number
of trucks, and 2/3 of the trucks are pickups, how many pickups, in terms of n, are parked
in the slot?

A. 1/6 n
B. 5/12 n
C. 1/2 n
D. 8/15 n
E. 11/12 n

Let there be c cars and t trucks and p pickup trucks
n = t+c
t=4c
So n = 5c
(2/3)t = p
p = (2/3)t = (2/3)(4c) = (8/3)c = (8/3)(n/5) = (8/15)n

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Re: A total of n trucks and cars [#permalink]  06 Jan 2011, 21:30
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Expert's post
ajit257 wrote:
A total of n trucks and cars are parked in a slot. If the number of cars is 1/4 the number
of trucks, and 2/3 of the trucks are pickups, how many pickups, in terms of n, are parked
in the slot?

A. 1/6 n
B. 5/12 n
C. 1/2 n
D. 8/15 n
E. 11/12 n

Try and use ratios where possible. They make your life very easy.

No of cars : No of trucks = 1:4 (Since for every 4 trucks, there is 1 car)
So total cars and trucks (n) in ratio terms = 1+4 = 5
Now out of 4 trucks, 2/3 are pickup (p) i.e. 4*2/3 = 8/3 are pick up.
So p/n = (8/3)/5 or p = 8/15 n
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Manager
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Concentration: General Management, Finance
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Re: A total of n trucks and cars are parked in a lot. If the num [#permalink]  21 Mar 2014, 06:36
Lets say Total truck is 12. So car is 3.
Now pickups are =2/3rd of 12= 8
so total fraction is 8/15 ..ans
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Re: A total of n trucks and cars are parked in a lot. If the num [#permalink]  28 May 2014, 05:43
cronkey7 wrote:
A total of n trucks and cars are parked in a lot. If the number of cars is 1/4 the number of trucks, and 2/3 of the trucks are pickups, how many pickups, in terms of n, are parked in the lot?

A. 1/6 n
B. 5/12 n
C. 1/2 n
D. 8/15 n
E. 11/12 n

[Reveal] Spoiler:
I decided to use 100 for the total of trucks + cars. So n = 100.
Number of cars = 1/4 # of trucks, then there are 25 cars and 75 trucks
2/3 of the trucks are pickups, so 2/3 of 75 is 50.

Therefore, the number of pickups (50) in terms of n would be (1/2)n, which equals 100 * 1/2 = 50, the number of pickups. = C

However, this is not the correct answer- the correct answer is D, 8/15. Can someone explain???

C = Cars, T = Trucks, and P = Pickups.

$$C + T = n$$

$$C = \frac{1}{4} T$$
$$\frac{2}{3}T = P$$

We'd like to write everything in terms of Pickups, since that's what the question is asking us about. Thus,

$$T = \frac{3}{2}P$$,
$$C = \frac{1}{4}T = \frac{1}{4} * \frac{3}{2} P = \frac{3}{8}P$$

Therefore,

$$n = C + T = \frac{3}{8} P + \frac{3}{2} P = \frac{30}{16}P = \frac{15}{8}P.$$

Solving for P, we see that $$P = \frac{8}{15} n.$$

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Re: A total of n trucks and cars [#permalink]  09 Feb 2015, 23:04
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Re: A total of n trucks and cars   [#permalink] 09 Feb 2015, 23:04
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