The number of ten-cent coins in a jar was 1/3 the number of fifty-cent coins. Zane took out 19 fifty-cent coins and exchanged them for ten-cent coins. Then he put the money back into the jar. The number of fifty-cent coins became 5/8 the number of ten-cent coins. How much money was there in the jar?
Starting with t 10¢ coins and f 50¢ coins,
t = 1/3 f
after the switch,
f-19 = 5/8 (t + 19)
solving those two equations,
f-19 = 5/8 (f/3+19)
f = 39
Ten 20-cent coins exchanged for four 80-cent coins
A first, ratio = 3:2
20 cent = 34
50 cent = 24
Exchanged
20 cent = -10
50 cent = +4
In the end
20 cent = 34 - 10
50 cent = 24 + 4
Given ratio in the end is 7:10
7:10 = (34 - 10) : (24 + 4)
6 = 8
Sum of money = 4.8 + 8
= $12.80
To solve this problem, let's use algebraic equations to represent the given information and find the solution.
Let's say the number of ten-cent coins in the jar is 'x', and the number of fifty-cent coins is 'y'.
According to the first condition, the number of ten-cent coins is 1/3 the number of fifty-cent coins. So we can write an equation as:
x = (1/3)y -- (Equation 1)
Let's consider the second condition. Zane took out 19 fifty-cent coins and exchanged them for ten-cent coins. This means the number of fifty-cent coins remaining in the jar is 'y - 19'. Also, 19 fifty-cent coins were replaced with an equal value of ten-cent coins, which means that the number of ten-cent coins in the jar increased by 19/10 times the number of fifty-cent coins taken out. So the equation can be written as:
x + (19/10)y = y - 19 -- (Equation 2)
Finally, according to the third condition, the number of fifty-cent coins became 5/8 the number of ten-cent coins after Zane put the money back into the jar. This gives us the equation:
y = (5/8)x -- (Equation 3)
Now we have a system of three equations (Equations 1, 2, and 3) that we can solve to find the values of 'x' and 'y'.
First, let's simplify Equation 2 by multiplying both sides by 10 to eliminate the fraction:
10x + 19y = 10y - 190 -- (Equation 4)
We can rewrite Equation 3 as x = (8/5)y and substitute this expression for 'x' in Equation 1 and Equation 4.
Substituting x in Equation 1:
(8/5)y = (1/3)y
Multiply both sides by 15 to eliminate the fraction:
24y = 5y
Now, solve for 'y':
24y - 5y = 0
19y = 0
y = 0
However, it is not possible for the number of fifty-cent coins to be zero (y = 0) since Zane started by taking out 19 fifty-cent coins.
Therefore, there is no solution to this problem, and we cannot determine the amount of money in the jar based on the given conditions.