Bill brought four times as may apples as harry and three times as may apples as martin. If Bill, and martin purchased less than a total of 190, apples, what is the greatest number of apples that bill could have purchased?

1 answer

apples for Harry -- x
apples for Bill --- 4x
apples for Martin -- y
but Bill's apples = 3 times Martin
4x = 3y
y = 4x/3

Bill + martin < 190
4x + 4x/3 < 190
times 3
12x + 4x < 570
16x < 570
x <35.6
4x < 142.5
greatest number of apples for Bill is 142

but Bill bought 4 times as many as Harry, and we will assume that there were no fractional apples .
Thus Bill's number of apples should be a multiple of 4
and the greatest such multiple which is less than 142
is 140
BUT, that would still give Martin partial apples
how about 136?
then y = (4/3)(136) --- still not a whole number
how about 132 ???
then y = (4/3)(132) = 44 and x = 3y/4 = 33

yeah, whole apples all around

Harry --- 33 apples
Bill ----- 132 apples
Martin --- 44 apples

Notice ratio of
Harry : Bill : Martin = 33 : 132 : 44 , where 132 + 44 < 190
= 3: 12: 4

if we went for the ratio of
36 : 144 : 48 , the sum of apples for Bill and Martin > 190
even though the relationship of apples between them would be correct