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Four buses carry tourists, so that the number of tourists in the first

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Four buses carry tourists, so that the number of tourists in the first  [#permalink]

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New post 21 Jun 2019, 00:38
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  45% (medium)

Question Stats:

73% (02:37) correct 27% (02:41) wrong based on 26 sessions

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Four buses carry tourists, so that the number of tourists in the first bus is 40 percent more than that in the second bus, but 30 percent less than that in the third bus. The fourth bus carries twice as many tourists as the second bus. The average (arithmetic mean) of the number of tourists in each bus is what percentage of the mode of these numbers?


A. 16 \(\frac{2}{3}\)

B. 33 \(\frac{1}{3}\)

C. 60

D. 75

E. 80


Source: examPAL
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Re: Four buses carry tourists, so that the number of tourists in the first  [#permalink]

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New post 21 Jun 2019, 01:19
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Bunuel wrote:
Four buses carry tourists, so that the number of tourists in the first bus is 40 percent more than that in the second bus, but 30 percent less than that in the third bus. The fourth bus carries twice as many tourists as the second bus. The average (arithmetic mean) of the number of tourists in each bus is what percentage of the mode of these numbers?


A. 16 \(\frac{2}{3}\)

B. 33 \(\frac{1}{3}\)

C. 60

D. 75

E. 80

Source: examPAL


Let number of tourists in Bus 2 = 100

number of tourists in the first bus is 40 percent more than that in the second bus
--> Bus 1 = 140

number of tourists in the first bus is 30 percent less than that in the third bus
--> Number of tourists in Bus 1 = 70% of Bus 3
--> 70/100*(Bus 3) = 140
--> Bus 3 = 140*100/70 = 200

The fourth bus carries twice as many tourists as the second bus
--> Bus 4 = 2*100 = 200

{Bus 1, Bus 2, Bus 3, Bus 4} = {140, 100, 200, 200}
--> Mode = 200

Average = Total/4 = 640/4 = 160

The average (arithmetic mean) of the number of tourists in each bus is what percentage of the mode of these numbers
--> Average = x% of Mode
--> 160 = x/100*200
--> x = (160*100)/200 = 80

IMO Option E

Pls Hit Kudos if you like the solution
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Re: Four buses carry tourists, so that the number of tourists in the first  [#permalink]

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New post 21 Jun 2019, 01:34
This is PS and in any case, you will get one definite answer, so lets plug some noble integers
1) Let's name the buses I, II, III, IV. Let the number of tourists in II bus be 100.
According to stem in I bus there will be: 140
III bus: 200 (140 is 30% less, x*0.7=140)
IV bus: 200 (twice as many as the II bus)

Calculate average: (140+100+200+200)/4=160
Observe the mode: 200

160/200=0.8

We could use algebra
I buss: 1.4x
II buss: x (we have assumed that)
III buss: 0.7*y=1.4x, y=2x. Assumed y is the number of tourists in III bus)
IV bus: 2x

Average: (1.4x+x+2x+2x)/4=1.6x
mode: 2x

1.6x/2x=0.8

IMO
Ans: E
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Re: Four buses carry tourists, so that the number of tourists in the first  [#permalink]

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New post 21 Jun 2019, 03:43
Bunuel wrote:
Four buses carry tourists, so that the number of tourists in the first bus is 40 percent more than that in the second bus, but 30 percent less than that in the third bus. The fourth bus carries twice as many tourists as the second bus. The average (arithmetic mean) of the number of tourists in each bus is what percentage of the mode of these numbers?


A. 16 \(\frac{2}{3}\)

B. 33 \(\frac{1}{3}\)

C. 60

D. 75

E. 80


Source: examPAL


Bus#

1. 140x/100

2. x

3. y

4. 2x

y(7/10) = 140x/100

So, y=2x, Hence Mode= 2x

To find: [{(7x/5 + 5x)/4} / 2x ] *100
E 80%
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Re: Four buses carry tourists, so that the number of tourists in the first  [#permalink]

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New post 21 Jun 2019, 10:39
2nd bus = 100
1st bus = 140
3rd bus = 200
4th bus = 200
total = 640
avg = 160
mode = 200
so % ; 160/200 ; 80%
IMO E



Bunuel wrote:
Four buses carry tourists, so that the number of tourists in the first bus is 40 percent more than that in the second bus, but 30 percent less than that in the third bus. The fourth bus carries twice as many tourists as the second bus. The average (arithmetic mean) of the number of tourists in each bus is what percentage of the mode of these numbers?


A. 16 \(\frac{2}{3}\)

B. 33 \(\frac{1}{3}\)

C. 60

D. 75

E. 80


Source: examPAL
GMAT Club Bot
Re: Four buses carry tourists, so that the number of tourists in the first   [#permalink] 21 Jun 2019, 10:39
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