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Maybe my English was bad.
I took away 10 50-cent coins and replaced with 10 20-cent coins, and just couldn’t get the answer, until…
I took away 10 50-cent coins of value $5, and replaced with equal value of 20-cent coins, ($5 ÷ $0.20) = 25 coins
20-cent | 50-cent | |
Quantity at first | 3u | 4u |
Quantity change | +25 | -10 |
Quantity in the end | 7p = 3u + 25 | 5p = 4u – 10 |
Multiply by 5 | Multiply by 7 | |
35p = 15u + 125 | 35p = 28u – 70 |
28u – 70 = 15u + 125
13u = 195
1u = 15
Value of 20-cent coins at first -> 3u × $0.20 = $9
Value of 50-cent coins at first -> 4u × $0.50 = $30
Sum of money in the box -> $9 + $30 = $39
hellowk
Thank you. It’s like solving simultaneous equations.

You are absolutely right.
With 2 unknowns and 2 equations, it is indeed solving simultaneous equations.
hellowk
Thanks alot for the answer. Appreciate it!

You are most welcome!