# MAth problems?

If S=( 1/5, 1/4, 1/3, 2/3, 3/4, 4/5) and x and y are elements of S, what is the maximum possible value of 3x-3y?

On a given day there are 12 trains to city X with an average of 1400 commuters per train. if the number of trains were cut to 7 and the total number of commuters remained the same, there would be an average of how many more commuters per train?

i need the steps please!! thanks thanks very muchhh………..

Favorite Answer

15 + 18 + 21 = 54, but since they are pairs, divide by 2, 54/2 = 27,and 27/3 = 9.

9 is the average for the three numbers.

The maximum possible value means taking the smallest number from the greatest, therefore, 4/5 is the greatest number, and 1/5 is the smallest.

3x – 3y = ? Inserting these numbers gives

3(4/5) – 3(1/5) = ? or 12/5 – 3/5 = 9/5

9/5 is the maximum possible value

12 trains @ 1400 commuters per train gives

12 x 1400 = 16,800 average total commuters

reduce the trains to 7 and you get

16,800/7 = 2400 average commuters per train

The average increase is 1000 commuters per train.

x+z=18

y+z=21

Add together to get 2x+2y+2z=54 so x+y+z=27 so average=27/3=9.

#2 max value for x-y is biggest-smallest

max value= 4/5-1/5=3/5

max for 3x-3y=3(3/5)=9/5

#3 number of commuters=12 x 1400

average per train =1400

with 7 trains, average per train= (12×1400)/7=2400

increase in average per train= 2400-1400=1000 passengers

average=(6+9+12)/3

=9

2)max value = 3*(4/5) – 3*(1/5)

= 9/5

3)….later

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