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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 Updated on: 20 Feb 2018, 07:59
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Difficulty:

  35% (medium)

Question Stats:

74% (02:56) correct 26% (02:29) wrong based on 29 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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Originally posted by Jiggy11 on 20 Feb 2018, 00:36.
Last edited by Bunuel on 20 Feb 2018, 07:59, edited 1 time in total.
Renamed the topic.
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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 20 Feb 2018, 02:39
Jiggy11 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



hello :)


let bus one be x
let bus two be y
third bus = z

x= 1.40 %
y = 100 %
z= 1.30 %
Fourh bus = 2y --> 200

140+100+130+200/4 = 142.5

mode is number that occurs most ... frm this set of numbers i dont see such number... all numbers are unique 140+100+130+200

so what did i do wrong ? :? :)
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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 20 Feb 2018, 03:42
1
dave13 wrote:

hello :)


let bus one be x
let bus two be y
third bus = z

x= 1.40 %
y = 100 %
z= 1.30 %
Fourh bus = 2y --> 200

140+100+130+200/4 = 142.5

mode is number that occurs most ... frm this set of numbers i dont see such number... all numbers are unique 140+100+130+200

so what did i do wrong ? :? :)


Well Dave,

If you see all values are rotating around the Second Bus and the other three buses are expressed in percentage terms of the second bus.

So we assume the number of tourists in the Second Bus = 100
That makes the number of tourists in the First Bus = 140
It also makes the number of tourists in the Third Bus = 200
Fourth Bus has double passengers of that of Second. So it will be 200 in Fourth Bus as well.

Total Passengers = 100 + 140 + 200 + 200 = 640
Arithmetic Mean = 640/4 = 160

Mode of (100,140,200,200) is = 200

So the Mean is nothing but 80% of the Mode.

Hence Option E = 80 is the OA
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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 20 Feb 2018, 03:54
1
dave13 wrote:
Jiggy11 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



hello :)


let bus one be x
let bus two be y
third bus = z

x= 1.40 %
y = 100 %
z= 1.30 %
Fourh bus = 2y --> 200

140+100+130+200/4 = 142.5

mode is number that occurs most ... frm this set of numbers i dont see such number... all numbers are unique 140+100+130+200

so what did i do wrong ? :? :)



Hi dave13

it is given that number of passengers in 1st bus is 30% less than 3rd bus. so if you have got no of passenger in 1st bus as 140, then it means

70% of number of passenger in 3rd bus =140

so number of passengers in 3rd bus = \(\frac{140}{70}\)% \(= \frac{140*100}{70}=200\)

once you have got number of passengers in each bus, then its a simple mean & mode question

the highlighted part above is incorrect.
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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 20 Feb 2018, 05:34
Jiggy11 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 a b c d be the four buses . Let b = 100 d = 200 ( twice) a = 40% more and 30% less of c

a= 140 b= 100
c= 200
D= 200

mean = 640/4 = 160

160 is what percent of 200 = 80

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 20 Feb 2018, 06:16
Jiggy11 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


From the given question, Let a,b,c,d be the number. Set S can be formed as -
\(a = 1.4b = 0.7c\)
\(c = 2b, d =2b\)
Set S = \({1.4b, b, 2b, 2b}\)
Mode = \(2b\), Average = \(1.6b\)

Answer = \(0.8\)
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Re: Four buses carry tourists, so that the number of tourists in the first &nbs [#permalink] 20 Feb 2018, 06:16
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