Question 322395: A boat goes up stream 30 miles and down stream 44 miles in 10 hours. Again it goes up stream 40 miles and down stream
55 miles in 13 hours. Find the rates of the stream and of the boat.
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! Distance(d) equals Rate (r) times Time(t) or d=rt; r=d/t and t=d/r
Let r1=rate of stream
And r2=rate of the boat
r2-r1=rate upstream
r2+r1=rate downstream
On the first run:
time upstream=30/(r2-r1)
time downstream=44/(r2+r1)
and we are told that the above times equals 10 hrs, so:
30/(r2-r1) + 44/(r2+r1)=10-----------------------eq1
On the second run:
time upstream=40/(r2-r1)
time downstream=55/(r2+r1)
And these times equals 13 hours, so:
40/(r2-r1)+55/(r2+r1)=13----------------------eq2
In eq1, multiply each term by (r2-r1)(r2+r1) and we get:
30(r2+r1)+44(r2-r1)=10(r2^2-r1^2) simplify
30r2+30r1+44r2-44r1=10r2^2-10r1^2
74r2-14r1=10r2^2-10r1^2-----------------eq1a
Let try to simplify eq2 in the same manner
40(r2+r1)+55(r2-r1)=13(r2^2-r1^2)
40r2+40r1+55r2-55r1=13r2^2-13r1^2
95r2-15r1=13r2^2-13r1^2------------eq2a
From eq1a, (r2^2-r1^2)=(74r2-14r1)/10. Substitute this into eq2a:
95r2-15r1=13(74r2-14r1)/10) multiply each side by 10
950r2-150r1=962r2-182r1 simplify
-150r1+182r1=962r2-950r2
32r1=12r2 or
8r1=3r2
r1=(3/8)r2 and r1^2=(9/64)r2^2; substitute these into eq1a
74r2-14(3/8)r2=10r2^2-10(9/64)r2^2 multiply each term by 64
4736r2-336r2=640r2^2-90r2^2 simplify
4400r2=550r2^2 divide each side by 10r2
440=55r2
r2=8 mph-----------------------------------------speed of boat
r1=(3/8)r2=(3/8)*8=3 mph ------------------------speed of stream
CK
30/(8-3)=6hr
44/(8+3)=4hr
10=10
40/5=8 hr
55/11=5 hrs
13=13
You can look it over. There may be an easier way.
Hope this helps---ptaylor
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