200 coins are sorted and kept 3 bags. The first bag has 20 more coins than the second bag. The second bag has twice as many coins as the third bag. How many coins are there in the third bag?
second bag --- x
first bag ----- x+20
third bag = 200 - (x + x+20)
= 200 - 2x - 20
= 180 - 2x
Let's break down the information given step by step:
1. We have three bags. Let's call them bag A, bag B, and bag C.
2. In total, there are 200 coins.
3. The first bag, A, has 20 more coins than the second bag, B. So the number of coins in A is equal to the number of coins in B plus 20.
4. The second bag, B, has twice as many coins as the third bag, C. So the number of coins in B is equal to the number of coins in C multiplied by 2.
Now, let's solve the problem:
Let's assume that the number of coins in the third bag, C, is 'x'.
According to point 4, the number of coins in the second bag, B, is '2x'.
According to point 3, the number of coins in the first bag, A, is '2x + 20'.
The total number of coins is given as 200, so we can write the equation:
x + 2x + (2x + 20) = 200.
Simplifying the equation:
5x + 20 = 200,
5x = 200 - 20,
5x = 180,
x = 36.
Therefore, there are 36 coins in the third bag (C).