SOLUTION: Three brothers shared a box of biscuits. Roy ate {{{2/3}}} of the box of biscuits and 1/3 of a biscuit. Sam ate {{{2/3}}} of the remaining biscuits and 1/3 of a biscuit. Tom ate

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Three brothers shared a box of biscuits. Roy ate {{{2/3}}} of the box of biscuits and 1/3 of a biscuit. Sam ate {{{2/3}}} of the remaining biscuits and 1/3 of a biscuit. Tom ate      Log On


   

Question 1185956: Three brothers shared a box of biscuits.
Roy ate 2%2F3 of the box of biscuits and 1/3 of a biscuit.
Sam ate 2%2F3 of the remaining biscuits and 1/3 of a biscuit.
Tom ate 2%2F3 of the rest of the biscuits and 1/3 of a biscuit.
There were no biscuits left after this. How many biscuits did Roy a eat?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Box began with 13 biscuits.
Roy at 4 biscuits.

p, for how many bisquits starting in the box,

After Roy,
p%2F3-1%2F3

After Sam,
.
.
p%2F9-4%2F9

After Tom,
.
.
%281%2F3%29%28p%2F9-4%2F9%29-1%2F3=0
.
.
p=13

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


When each brother eats 2/3 of the biscuits, 1/3 of them are left. So when each brother eats some of the biscuits, the number of biscuits remaining gets multiplied by 1/3 and the reduced by 1/3. So

original number of biscuits: x

number after Roy eats his: %281%2F3%29x-1%2F3

number after Sam eats his: %281%2F3%29%28%281%2F3%29x-1%2F3%29-1%2F3

number after Tom eats his: %281%2F3%29%28%281%2F3%29%28%281%2F3%29x-1%2F3%29-1%2F3%29-1%2F3

The number of biscuits left at the end is 0, so

%281%2F3%29%28%281%2F3%29%28%281%2F3%29x-1%2F3%29-1%2F3%29-1%2F3=0

That equation is actually not too hard to solve, although it is easy to get lost along the way....

This kind of problem is quite often easier to work backwards, especially if there are a large number of steps. (Imagine what the final equation above would look like if there had been five (or more) brothers.)

At each step working the problem from beginning to end, the number of biscuits remaining is multiplied by 1/3 and then 1/3 is subtracted. Working that backwards, each step consists of adding 1/3 to the number remaining (the opposite of subtracting 1/3) and multiplying that by 3 (the opposite of dividing by 3).

Working the problem backwards is then simple:

biscuits remaining at the end: 0
biscuits remaining before Tom eats his: 3(0+1/3) = 3(1/3) = 1
biscuits remaining before Sam eats his: 3(1+1/3) = 3(4/3) = 4
biscuits remaining before Roy eats his: 3(4+1/3) = 3(13/3) = 13

There were 13 biscuits originally; Roy ate 9.

ANSWER: Roy ate 9 biscuits