Question 1184054
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The solution from the other (new) tutor is fine.... But it involves a lot of computations with "ugly" fractions.<br>
Since there are many different paths to the solution to the problem, let's try something different and see if we find an easier path to the solution.<br>
When the statement of the problem gives ratios, it is often easiest to set up the problem using those ratios.<br>
In this problem, we have...
(1) K:L = 3:2
(2) M:K = 3:4<br>
We can use a single variable to solve the problem by finding the ratio M:K:L.  Do that by multiplying each of the given ratios by a constant so that the "K" in both ratios is the same number.
(1) K:L = 3:2 = 12:8
(2) M:K = 3:4 = 9:12
(3) M:K:L = 9:12:8<br>
Having done that manipulation with the given ratios...
Let 9x = number Mike starts with
Let 12x = number Ken starts with
Let 8x = number Lionel starts with<br>
Mike ends up with twice as many as he had to start, so
18x = number Mike ends with<br>
Between them, the number Ken and Lionel gave Mike was 9x.  Ken gave Mike 3 times as many as Lionel gave, so Ken gave Mike 3/4 of the 9x and Lionel gave Mike 1/4 of the 9x:
(3/4)9x = (27/4)x = number Ken gave Mike
(1/4)9x = (9/4)x = number Lionel gave Mike<br>
Then...
the number Ken ends up with is 12x-(27/4)x = (21/4)x
the number Lionel ends up with is 8x-(9/4)x = (23/4)x<br>
Lionel ends up with 32 more than Ken:
(23/4)x-(21/4)x = 32
(2/4)x = (1/2)x = 32
x = 64<br>
We could figure out all the numbers in the problem -- the numbers each of them started with and the numbers each of them ended with.  But the problem only asks for the number Ken gave to Mike.  That number is<br>
(27/4)x = (27/4)(64) = 27*16 = 432<br>
ANSWER: 432<br>
If this forum works the way it is supposed to, you will get another response or two showing still different ways to set up and solve the problem.<br>