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"a" Adults and "k" students are trying to cross a river with a raft. However the raft can carry only 1 adult at a time, or 2 students, or 1 students. What is the minimum number of trips we need to take to bring everyone across? Express it as a general equation.

Mathematics
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Koala Bird has put up a difficult question. I dunno how to do.
x+2k=Answer? Or; x+k=Answer? I really dunno.
If moneybird asks a question, you know it isn't going to be easy

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Other answers:

a+k/2 = number of trips.
actually--thats not right, because you need to get the raft back over
a+k = number of trips.
x = trips 2kx(2) = total number of trips or ax(2) = total number of trips or kx(2) = total numer of trips or as one equation kx(2) + 2kx(2) + ax(2) = total number of trips
Here is my attempt at this: A == Adults S == Students Left Right ---- ----- a, k 0A, 0S a, k-2 0A, 2S -> 2 students take raft from Left to Right a, k-1 0A, 1S -> 1 student takes raft back to Left a-1, k-1 1A, 1S -> 1 adult takes raft from Left to Right a-1, k 1A, 0S -> 1 student takes raft back to Left a-1, k-2 1A, 2S -+ a-1, k-1 1A, 1S |__ 4 steps to tranfer each adult from Left to Right a-2, k-1 2A, 1S | a-2, k 2A, 0S -+ a-2, k-2 2A, 2S a-2, k-1 2A, 1S a-3, k-1 3A, 1S a-3, k 3A, 0S ... 1, k-2 (a-1)A, 2S 1, k-1 (a-1)A, 1S 0, k-1 aA, 1S 0, k aA, 0S --> we get here after 4a trips 0, k-2 aA, 2S -> 2 students take raft from Left to Right 0, k-1 aA, 1S -> 1 student takes raft back to Left 0, k-3 aA, 3S -+ 0, k-2 aA, 2S -+-> 2 steps to tranfer each student from Left to Right 0, k-5 aA, 4S 0, k-4 aA, 3S ... 0, 1 aA, (k-1)S 0, 2 aA, (k-2)S --> we get here after 2(k-2) trips 0, 0 aA, kS --> one last trip to get last 2 students across so total trips = 4a + 2(k-2) + 1 = 4a + 2k -3

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