Boat Speed
Three boats, X, Y and Z are there to let the touristas tour a lake. They leave at different times and take 30 minutes to tour the lake. The time difference between X and Z is 20 minutes and that between Y and Z is 15 minutes. Given that the sum of time (in minutes) is 40 for all three to leave, after starting the countdown, at what time did each boat leave?
Labels: mathemagic, mathschallenge
posted by Rajesh Lal at
11:57 AM
8 Comments:
Being thick, I don't understand the question, but my best guess is...
The countdown starts.
Boat Y's skipper struggles to load his unruly touristas
T + 5 mins Boat Y starts out
No such delay for the other boats...
T + 20 Mins Boat Z starts out
T + 40 Mins Boat X starts out
Karl ... I came up with the same solution ... if we interpreted the question correctly ...
x-20
y-25
z-40
where's my cookie!
Interesting ... 2 different solutions meet the criteria.
5mins - x
10mins - y
25 mins - z
Boat Speed
(1A) (Assume z is less than x) x-z=20 OR
(1B) (Assume x is less than z) z-x=20
(2A) (Assume z is less than y) y-z=15 OR
(2B) (Assume y is less than z) z-y=15
"Given that the sum of time (in minutes) is 40 for all three to leave," interpreted as after the clock has started all boats will have left by 40 minutes later
(3AA)(Assume z is less than y is less than x)
(x-z)+z=40 OR 20+z=40 OR z=20
OR
(3AB)(Assume y is less than z is less than x) (x-z)+(z-y)+y =40 OR 20+15+y=40 OR y=5
OR
(3BA)(Assume x is less than z is less than y)
(z- x)+(y-z)+ x =40 OR 20+15+ x=40 OR x=5
OR
(3BB)(Assume x is less than y is less than z)
(z- x)+ x=40 OR 20+ x=40 OR x=20
Now matching up consistent equations....
We have 4 possible solutions.
ANSWERS:
*1A,2A, 3AA solution
z=20,y=35, x=40
*1A,2B,3AB solution
y=5,z=20,x=40
*1B,2A,3BA solution
x=5,z=25,y=40
*1B,2B,3BB solution
x=20, y=25, z=40
ALTERNATE SOLUTIONS:
"Given that the sum of time (in minutes) is 40 for all three to leave," interpreted as:
(4) x+y+z=40
*1A,2A,4 solution
(5) *1A-*2A= x-y=5 OR x=y+5
(6) from *2A, z=y-15
*4,*5,*6 x+y+z=40, (y+5)+(y)+(y-15)=40 OR 3y-10=40 OR y=50/3
x=65/3 ~= 21.667, y=50/3 ~= 16.667, z= 5/3 ~= 1.667
*1A,2B,4 solution
(7)*1A+*2B= x-y=35 OR x=y+35
(8) from *2B, z=15+y
*4,*7,*8 x+y+z=40, (y+35)+y+(15+y)=40, OR 3y+50=40 OR y=-10/3
Since the time must be positive the solution is invalid
*1B,2A,4 solution
(9) *1B+*2A= y- x=35 OR x=y-35
*4,*6,*9, x+y+z=40, (y-35)+y+(y-15)=40 OR 3y-50=40 OR y=30
Then subbing y into *9, x=-5
Since the time must be positive the solution is invalid
*1B,2B,4 solution
(10) *1B-*2B= y-x=5 OR x=y-5
*4,*8,*10, x+y+z=40, (y-5)+y+(15+y)=40, OR 3y+10=40 OR y=10
x=5, y=10, z=25
ALTERNATE SOLUTION ANSWERS:
x=5, y=10, z=25
OR
x=65/3 ~= 21.667, y=50/3 ~= 16.667, z= 5/3 ~= 1.667
Hopefully no mistakes above.
Cam
If x, y and z are the times taken to leave, then I interpret the sum of the times as x + y + z = 40. i.e. I think vishav has a correct answer.
boat x left after the countdown by 5 minutes.
boat y left after the countdown by 10 minutes.
boat z was late 25 minutes
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