Medium #5

let B = number of fish Billy caught

let L = number of fish Lenny caught

let P = number of fish Peter caught

With that information, we can write 3 equations:

(1) B + L + P < 100

(2) B = 3L

(3) B = 4P

Then, we can rearrange the 2nd and 3rd equation to isolate the

number of fish Lenny and Peter caught like so:

B/3 = L

B/4 = P

With that, we can plug B/3 and B/4 into our first equation

respectively:

B + B/3 + B/4 < 100

Now we try to combine and simplify so that B will have a coefficient

of 1 by finding the LCD which is 12:

(12B + 4B + 3B)/12 < 100

Combine the B's and simplify to get:

19B < 1200

Divide by 19 to get B < 63.16

However, even though 63 is our highest whole number, it's not the

answer because it can't be divided evenly by 4 (you will get 15 with

a remainder of 3), our next highest number that's evenly divisible by

both 3 and 4 is 60, which is our answer.

Let the number of fish Billy caught be $12x$ , then the number Lenny caught is $4x$ and the number Peter caught is $3x$ for a total that they all caught of $12x+4x+3x=19x<100$ , meaning $x<5.26$ . The largest Billy could've caught is then $12(5)=\boxed{60}$

This is quite simple.Convert Lenny(L) and Peter(P) caught in terms of Billy(B) caught.

L=B/3, P=B/4.

Then, B+B/3+B/4 = 19B/12.

Now, we should take B value such a way it should satisfy 19B/12 and should be max value less than 100 which is 12*5=60.

This value could be achieved easily. We have to keep in mind that we are looking for a whole number value since it is regarding number of fishes caught. So fraction becomes a whole number value when B has a factor 12 and now to maximize the total value we can take only 5 as another factor so that total value will be 95. Hence 60 is the answer.