Two boats go downstream from A to B. The faster boat covers the distance from A to B, 1.5 times as fast as the slower boat. It is known that for every hour the slower boat

lags behind the faster boat by 8 km. However, if they go upstream, then the faster boat covers the distance B to A in half the time as the slower boat. Find the speed of the

faster boat in still water.

A. 12 k/h

B. 20 k/h

C. 24k/h

D. 25k/h

Let’s break this problem down step-by-step.

---

### Step 1: Define variables  
- Let \( v_f \) be the speed of the **faster boat in still water** (this is what we need to find).
- Let \( v_s \) be the speed of the **slower boat in still water**.
- Let \( c \) be the speed of the **current**.

---

### Step 2: Downstream case
- **Downstream speed** for a boat = Boat speed + Current speed.

Thus:
- Speed of the faster boat downstream = \( v_f + c \).
- Speed of the slower boat downstream = \( v_s + c \).

The problem states:
1. **The faster boat is 1.5 times as fast as the slower boat downstream:**
   \[
   v_f + c = 1.5 \times (v_s + c)
   \]

2. **The faster boat is ahead by 8 km for every hour:**
   This means the difference in their downstream speeds is 8 km/h:
   \[
   (v_f + c) - (v_s + c) = 8
   \]
   Simplifying:
   \[
   v_f - v_s = 8
   \tag{1}
   \]

---

### Step 3: Upstream case
- **Upstream speed** for a boat = Boat speed - Current speed.

Thus:
- Speed of the faster boat upstream = \( v_f - c \).
- Speed of the slower boat upstream = \( v_s - c \).

The problem states:
1. **The faster boat covers the upstream distance (B to A) in half the time as the slower boat:**
   \[
   \frac{v_f - c}{v_s - c} = 2
   \]

---

### Step 4: Solve the system of equations

#### Use Equation (1):  
\[
v_f = v_s + 8
\]

#### Substituting \( v_f = v_s + 8 \) into the ratio equation:  
\[
\frac{(v_s + 8) - c}{v_s - c} = 2
\]
Simplify the numerator:
\[
\frac{v_s + 8 - c}{v_s - c} = 2
\]
Cross-multiply:
\[
v_s + 8 - c = 2(v_s - c)
\]
Distribute on the right-hand side:
\[
v_s + 8 - c = 2v_s - 2c
\]
Rearrange terms:
\[
8 = 2v_s - v_s - 2c + c
\]
\[
8 = v_s - c
\]
Thus:
\[
v_s = c + 8
\tag{2}
\]

---

### Step 5: Substitute into Equation (1)
From Equation (2), \( v_s = c + 8 \). Substituting into Equation (1):
\[
v_f = (c + 8) + 8 = c + 16
\]

---

### Step 6: Determine the speed in still water

We need to determine \( v_f \) (the speed of the faster boat in still water). Notice that \( c \) (the current speed) cancels out across different steps. 

If

Two boats go downstream from A to B. The faster boat covers the distance from A to B, 1.5 times as fast as the slower boat. It is known that for every hour the slower boat lags behind the faster boat by 8 km. However, if they go upstream, then the faster boat covers the distance B to A in half the time as the slower boat. Find the speed of the faster boat in still water. A. 12 k/h B. 20 k/h C. 24 k/h D. 25 k/h

Solve a boat speed problem where two boats move downstream and upstream, and the faster boat covers the distance 1.5 times quicker than the slower boat. Learn to use algebraic equations and ratios to determine the speed of the faster boat in still water. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation