Question 80947: From towns 363 miles apart, Pat and Mike set out to meet each other. If Pat travels 1 mile the first day, 3 miles the second, 5 miles the third, etc. and Mike travels 2 miles the first day, 6 miles the second and 10 miles the third, etc., when will they meet? - I got the answer from my teacher, I know it's 11 days, but I have no idea how to work it. PLEASE HELP!
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! each day, Pat travels 2 miles more than the previous day ... on the first day he traveled 1 mile
so the distance traveled on day x is 1+2(x-1) ... for Mike, 2+4(x-1)
let d=days traveled ... distance equals (average per day)*(days) ... ((first day+last day)/2)*(days)
so Pat's distance is (1+(1+2(d-1))/2)*d ... similarly, Mike's distance is (2+(2+4(d-1))/2)*d
the combined distance is 363 ... so (1+(1+2(d-1))/2)*d+(2+(2+4(d-1))/2)*d=363 ... (d)*d+(2d)*d=363
 ...  ... d=11
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