A collector has a certain amount of coins, all of different weights. If you remove the 3 heaviest coins, the total weight of all the coins you had decreases by 35%. If you remove the lightest 3 of the remaining coins, the total weight of these remaining coins decreases by 5/13.
How many coins did the collector originally have?
The 3 heaviest coins are 35%, then the average (because they cannot have the same weight) is 11'67%
On the other hand, the three least heavy are 25% of the total (65% * 5/13), so the average is 8'33%
Then we have to look for a number of coins whose weight is 40% and that has an average between 8'33% and 11'67%
This makes us need 4 coins (with weights between the lightest of the heaviest and the heaviest of the lightest), with an average that would be around 10%
Let us call a, c, b respectively the weight of the lightest 3, the heaviest three, and the rest.
- From the given conditions it is easy to write two equations and put b and c according to a.
- If I was not mistaken in the accounts it comes out: b = 8a / 5; c = 7a / 5;
- Now the thing is to know how many coins make up the weight b. Let's call that number n. The key is that the less heavy currency of b has to weigh more than the light 3 and the heaviest less than the heaviest 3.
- In the lighter three at least there is a coin that weighs a / 3 or more. In the heaviest three at least there is a coin that weighs 3 / less. The same for "central" currencies.
- From there:
a / 3 <= 8a / 5n <= 7a / 15 As n is integer there will only be one solution and the requested number is n + 6. 10 coins