If 25 times were moved from Box A to Box B, there would be an equal number of dimes in both boxes. If 100 dimes were moved from Box B to Box A, the ratio of dimes in Box A to Box B would be 7:2. What was the original number of dimes in Box A?
dimes in box A originally --- x
dimes in box B originally --- y
case 1:
after 25 moved from A to B
box A has x-25
box B has y+25
but x-25 = y+ 25 ----> x = y + 50
case 2:
after 100 dimes moved from B to A
box A has x+100
box B has y-100
but (x+100)/(y-100) = 7/2
7y - 700 = 2x + 200
7y - 2x = 900
7y - 2(y+50) = 900
5y = 1000
y = 200
then x = 250
there were 250 dimes in box A
Well, let's break it down! If 25 dimes were moved from Box A to Box B, and they ended up with the same number of dimes, that means the difference between the original number of dimes in Box A and Box B was divisible by 25.
Now, if 100 dimes were moved from Box B to Box A, and the ratio of dimes in Box A to Box B became 7:2, we can solve this riddle.
Let's assume the original number of dimes in Box A was 7x, and the original number of dimes in Box B was 2x.
According to our information, the difference between the original number of dimes in Box A and Box B was divisible by 25. So, 7x - 2x must be divisible by 25.
If we simplify this, we have 5x divisible by 25, which means x must be 5.
Therefore, the original number of dimes in Box A was 7x, which is 7 multiplied by 5, resulting in 35 dimes.
So, the original number of dimes in Box A was 35.
Let's solve this problem step by step:
Step 1: Let's assume the original number of dimes in Box A is x.
Step 2: Since 25 times were moved from Box A to Box B, the number of dimes in Box A is now x - 25.
Step 3: According to the given information, if 25 times were moved from Box A to Box B, there would be an equal number of dimes in both boxes. This means that the number of dimes in Box B is also x - 25.
Step 4: Now, if 100 dimes were moved from Box B to Box A, the ratio of dimes in Box A to Box B would be 7:2. This means that the number of dimes in Box A would be (x - 25) + 100(= x + 75), and the number of dimes in Box B would be x - 25 - 100(= x - 125).
Step 5: So, according to the given ratio, we can write the equation: (x + 75) / (x - 125) = 7/2.
Step 6: To solve for x, we can cross-multiply the equation: 2(x + 75) = 7(x - 125).
Step 7: Expanding the equation, we get: 2x + 150 = 7x - 875.
Step 8: Rearranging the equation, we get: 7x - 2x = 150 + 875.
Step 9: Solving the equation, we get: 5x = 1025.
Step 10: Dividing both sides by 5, we get: x = 205.
Step 11: Therefore, the original number of dimes in Box A was 205.
To solve this problem, let's first set up equations for each statement given.
Let's assume the original number of dimes in Box A is "x".
According to the first statement, if 25 times were moved from Box A to Box B, there would be an equal number of dimes in both boxes. That means, after moving the dimes, both Box A and Box B would have the same number of dimes. So, the number of dimes in Box A would be (x - 25) and the number of dimes in Box B would be 25.
Now, according to the second statement, if 100 dimes were moved from Box B to Box A, the ratio of dimes in Box A to Box B would be 7:2. That means the number of dimes in Box A would be 7 times the number of dimes in Box B, after moving the 100 dimes. Therefore, the number of dimes in Box A would be (25 + 100) = 125 and the number of dimes in Box B would be (125/7) * 2.
Now, we have two expressions for the number of dimes in Box A:
1) (x - 25) from the first statement.
2) 125 from the second statement.
Since both expressions represent the number of dimes in Box A, they must be equal. We can set up an equation to solve for "x":
(x - 25) = 125
Simplifying the equation:
x - 25 = 125
x = 125 + 25
x = 150
Therefore, the original number of dimes in Box A was 150.