A bus carrying 50 tourists stopped at a roadside restaurant.

Question

A bus carrying 50 tourists stopped at a roadside restaurant. Each tourist ordered exactly one beverage: wine for €6, beer for €5, or water for €1. If the group’s total bill was €199, what is the greatest possible number of tourists who ordered wine?

(A) 28
(B) 29
(C) 30
(D) 31
(E) 33

(A) 28
(B) 29
(C) 30
(D) 31
(E) 33

sujoykrdatta · Answer

Let the number of people who order wine, beer and water be w, b and r respectively
w + b + r = 50
6w + 5b + r = 199
subtracting: 5w + 4b = 149
To maximize w, we need to minimise b. 
If b = 1: 5w = 149-4 = 145 => w = 29 - possible
(Complete solution: w = 29, b = 1, r = 50 - 29 - 1 = 20)

Answer B

3434will · Answer

You didn't say at least one of each was ordered.

kevincan · Answer

Would it matter if it were stated?

vasu1104 · Answer

total people= 50
each ordered exactly one drink
wine = 6$, beer= 5$ and water= 1$
total cost = 199

say people with wine = x
people with beer = y
people with water =z

x+y+z = 50

6x+5y + z = 199

we need maximum people with wine. so lets minimize the other two. 
x = 50-y- z

put into second equ,

6(50-y-z) + 5y + z = 199
300-6y-6z +5y+z = 199
101 = y+ 5z
z is multiple of 5 so with that in mind lets try value for y and then find for z.
y=1 , z=20 so x= 29
y=6, z= 19 so x= 25
y=11 , z= 18, so x= 21

x is decreasing with each rep. that means the greatest value for x can be 29

option B

rameezhasalife · Answer

if they mentioned atleast one of each was ordered, one wouldn't have put 0 for Beer, hence making 33 wines and 1 water bottle amounting to 199

kevincan · Answer

Remember that each tourist ordered exactly one beverage , so even though 33 tourists ordering wine and 1 ordering water would result in a total of 199, there would be 16 tourists with nothing to drink

A bus carrying 50 tourists stopped at a roadside restaurant.

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A bus carrying 50 tourists stopped at a roadside restaurant.

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