Twons A,B and C are on a straight line. Town C is between A and B. The distance A to B is 100 km. How far is A from C ?
(i) The distance from A to B is `25%` more than the distance from C to B.
(ii) The distance from A to C is `1/4` of the distance from C to B.

Answer

To solve the problem step by step, let's break down the information provided and analyze each condition separately.

### Given Information:
- Towns A, B, and C are on a straight line, with C between A and B.
- The distance from A to B is 100 km.

### Objective:
Find the distance from A to C (denote this distance as AC).

### Let’s denote:
- Distance from A to C = AC = X km
- Distance from C to B = CB = (100 - X) km

### Condition (i):
The distance from A to B is 25% more than the distance from C to B.

1. **Translate the condition into an equation:**
- We know that the distance from A to B is 100 km.
- According to the condition, 100 km is 25% more than the distance from C to B.
- Therefore, we can write the equation:
\[ 100 = CB + 0.25 \times CB \]
- This simplifies to:
\[ 100 = 1.25 \times CB \]
- Substituting CB with (100 - X):
\[ 100 = 1.25 \times (100 - X) \]

2. **Solve for X:**
- Distributing gives:
\[ 100 = 125 - 1.25X \]
- Rearranging gives:
\[ 1.25X = 125 - 100 \]
\[ 1.25X = 25 \]
- Dividing both sides by 1.25:
\[ X = \frac{25}{1.25} = 20 \text{ km} \]

### Conclusion for Condition (i):
The distance from A to C (AC) is **20 km**.

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### Condition (ii):
The distance from A to C is 1/4 of the distance from C to B.

1. **Translate the condition into an equation:**
- According to this condition, we can write:
\[ AC = \frac{1}{4} \times CB \]
- Substituting CB with (100 - X):
\[ X = \frac{1}{4} \times (100 - X) \]

2. **Solve for X:**
- Multiplying both sides by 4 gives:
\[ 4X = 100 - X \]
- Rearranging gives:
\[ 4X + X = 100 \]
\[ 5X = 100 \]
- Dividing both sides by 5:
\[ X = 20 \text{ km} \]

### Conclusion for Condition (ii):
The distance from A to C (AC) is **20 km**.

### Final Answer:
In both conditions, the distance from A to C is **20 km**.

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