I have a tin of chocolates to share with some friends. In the tin are four types of chocolate — one type in red wrappers, one type in blue, one type in green and one type in yellow, with equal numbers of each.
The tin is passed around in turn from friend to friend. When the tin gets to them:
• John always takes three red chocolates.
• Peter always takes one of each colour.
• Jane always takes one yellow chocolate.
• I always take two green chocolates.
After six passings around the tin, half of the chocolates remaining in the tin are blue.
How many chocolates were there in the tin when it was full?
Full explanation
Let N stand for how many chocolates of each colour the tin started with – it’s the same number for red, blue, yellow and green, since the tin has equal amounts of each.
Track what one full round of passing removes. John takes 3 red. Peter takes one of each colour (1 red, 1 green, 1 yellow, 1 blue). Jane takes 1 yellow. And the tin’s owner takes 2 green. Add that up and a single round removes 4 red, 2 yellow, 1 blue and 3 green.
After six rounds, six times those amounts are gone: 24 red, 12 yellow, 6 blue, 18 green. So what’s left in the tin is (N − 24) red, (N − 12) yellow, (N − 6) blue and (N − 18) green – a grand total of 4N − 60 chocolates remaining.
We’re told half of what’s left is blue, so the blue amount has to equal half that total:
N − 6 = (4N − 60) / 2 = 2N − 30
Rearranging: 30 − 6 = 2N − N, so N = 24.
The tin started with 24 chocolates of each colour, so 96 in total (4 × 24) – matching option D. As a check: with N = 24, the red runs out exactly (24 − 24 = 0), leaving 12 yellow, 18 blue and 6 green – 36 chocolates in total – and 18 is indeed exactly half of 36, just as the question says.
Word problems like this one turn into algebra once you name the unknown and track it through the story. Let a variable stand for the starting quantity, then write an expression for how much of it is removed by each repeating step, and multiply by however many times that step happens. The question’s real condition (here, that one colour ends up as exactly half of what’s left) becomes an equation you can solve for the unknown — and once solved, it’s always worth checking the answer against the original condition to make sure it actually holds.