(i) What quantity must be added to each term of the ratio `8 : 15` so that it becomes equal to `3 : 5`?
(ii) What quantity must be subtracted from each term of the ratio `a : b` so that it becomes `c : d` ?

Answer

Let's solve the given problems step by step.

### Part (i): What quantity must be added to each term of the ratio `8 : 15` so that it becomes equal to `3 : 5`?

1. **Let the quantity to be added be \( x \)**.
- We can express the new ratio after adding \( x \) to both terms:
\[ \frac{8 + x}{15 + x} = \frac{3}{5} \]

2. **Cross-multiply to eliminate the fraction**:
- This gives us:
\[ 5(8 + x) = 3(15 + x) \]

3. **Distribute on both sides**:
- Expanding both sides results in:
\[ 40 + 5x = 45 + 3x \]

4. **Rearranging the equation**:
- Move all terms involving \( x \) to one side and constants to the other:
\[ 5x - 3x = 45 - 40 \]
- This simplifies to:
\[ 2x = 5 \]

5. **Solve for \( x \)**:
- Dividing both sides by 2 gives:
\[ x = \frac{5}{2} \]

6. **Final result**:
- Therefore, the quantity that must be added to each term of the ratio \( 8 : 15 \) is \( \frac{5}{2} \) or \( 2.5 \).

### Part (ii): What quantity must be subtracted from each term of the ratio `a : b` so that it becomes `c : d`?

1. **Let the quantity to be subtracted be \( t \)**.
- We can express the new ratio after subtracting \( t \) from both terms:
\[ \frac{a - t}{b - t} = \frac{c}{d} \]

2. **Cross-multiply to eliminate the fraction**:
- This gives us:
\[ d(a - t) = c(b - t) \]

3. **Distribute on both sides**:
- Expanding both sides results in:
\[ da - dt = cb - ct \]

4. **Rearranging the equation**:
- Move all terms involving \( t \) to one side and constants to the other:
\[ da - cb = dt - ct \]
- Factor out \( t \):
\[ da - cb = t(d - c) \]

5. **Solve for \( t \)**:
- Dividing both sides by \( (d - c) \) gives:
\[ t = \frac{da - cb}{d - c} \]

6. **Final result**:
- Therefore, the quantity that must be subtracted from each term of the ratio \( a : b \) to make it equal to \( c : d \) is \( \frac{da - cb}{d - c} \).