Problem 1:
Let one part be \(x\), and the other part be \((184 - x)\).
The condition is \(\frac{x}{3} = \frac{1}{7}(184 - x) + 8\).
**Solution:**
\[ x = 72 \]
So, one part is 72, and the other part is \(184 - 72 = 112\).
Problem 2:
Let \(x\) be the number of notes of 5, and \((90 - x)\) be the number of notes of 10.
The equation is \(5x + 10(90 - x) = 500\).
**Solution:**
\[ x = 80 \]
5x + 10(90 - x) = 500
-5x + 900 = 500
-5x = -400
x = 80
So, there are 80 notes of 5 and \(90 - 80 = 10\) notes of 10.
Problem 3:
Let \(x\) be the number of 50-paise coins, then the number of 25-paise coins is \(2x\).
The equation is \(0.50x + 0.25(2x) = 34\).
**Solution:**
\[ x = 17 \]
0.50x + 0.25(2x) = 34
0.5x + 0.5x = 34
x = 34
So, there are 17 coins of 50 paise and \(2 \times 17 = 34\) coins of 25 paise.
