There are a total of 100 animals: cows, sheep and buffaloes. These 100 animals ate 100 bunches of grass.

Every cow ate 5 bunches, every buffalo ate 3 bunches and every sheep ate only 1/3 bunch.

How many cow, sheep and buffalo are there? You only know that there is at least one of every kind of animal.

Allow me to give a more thorough response. I acknowledge there are better ways to solve this problem.

Let c = cows, b = buffaloes, and s = sheep.
From condition 1 (100 total animals): c + b + s = 100
From condition 2 (100 bunches eaten): 5c + 3b + s/3 = 100

Since the left hand sides of both equations equal 100, we can set both left hand sides equal to each other: c + b + s = 5c + 3b + s/3
Simplifying: 3*(c+b+s) = 3*(5c+3b+s/3)
3c+3b+3s = 15c+9b+s
(3c+3b+3s)+(-3c-3b-s) = (15c+9b+s)+(-3c-3b-s)
2s = 12c+6b
(2s)/2 = (12c+6b)/2
s = 6c + 3b

Subsituting into the first equation: c + b + (6c + 3b) = 100
7c + 4b = 100

Using the 'real-world' constraints of no partial or 'negative' animals, plus the stated requirement that there is at least one of each animal, we can limit c, b, and s to be positive integers. Thus, C must be at least 1 and cannot be larger than 14 (7*15 is 105, which would force b to be negative). So solving 7c + 4b = 100, and s = 6c + 3b, with the constraint of s, b, and c being positive integers yields three possible sets: (c=4, b=18, s=78); and (c=8, b=11, s=81); and (c=12, b=4, s=84).

These answers all check:
4 + 18+ 78 = 100; 5*4 + 3*18 + 78/3 = 20 + 54 + 26 = 100
8 + 11+ 81 = 100; 5*8 + 3*11 + 81/3 = 40 + 33 + 27 = 100
12+ 4+ 84 = 100; 5*12+ 3*4 + 84/3 = 60 + 12 + 28 = 100

Posted by Ender on 2002-06-12 04:03:14