Question

Betty and Charlie shared a packet of sweets. Betty would have 48 more sweets more than Charlie if Charlie gave 2/5 of his sweets to Betty. If Charlie gave 5/8 of his sweets to Betty, he would have 192 less sweets than Betty. How many sweets did Betty have?

Answer 1

Let's assume that Betty initially had B sweets, and Charlie had C sweets.
If Charlie gave 2/5 of his sweets to Betty, Betty would get 2/5*C = (2/5)*C = (2/5)*C sweets
And since Charlie gave Betty 48 sweets, we have 2C/5 = 48
Multiplying both sides of the equation by 5/2 gives us 5C/2 = 48
Multiplying both sides by 2/5, C = 48 * 2/5 = 19.2

This means Charlie initially had 19.2 sweets, which is impossible since the number of sweets cannot be in decimal values. Therefore, our assumption is wrong.

Now let's assume Charlie initially had c sweets, where c is a whole number.

If Charlie gave 5/8 of his sweets to Betty, Betty would get 5/8*c = (5/8)*c = (5/8)*c sweets
And since Charlie gave Betty 192 sweets, we have 5c/8 = 192
Multiplying both sides of the equation by 8/5 gives us 8c/5 = 192
Multiplying both sides by 5/8, c = 192 * 5/8 = 120

This means Charlie initially had 120 sweets.

So, if Charlie gave 5/8 of his sweets, Betty would get 5/8*120 = 75 sweets
In this case, Betty would have 75 + 192 = <<75+192=267>>267 sweets. Answer: \boxed{267}.