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Sagot :
Sure, let's solve these problems step by step.
### (b) Cargo Problem
A ship's hold, A, contains 250 tonnes of cargo, and another hold, B, contains 620 tonnes of cargo. We want to find how much cargo must be taken from B and put into A so that A will contain five times as much as B.
1. Define the unknown variable:
Let \( x \) be the amount of cargo (in tonnes) taken from B and put into A.
2. Write the expression for the new amounts of cargo after transferring x tonnes:
- Cargo in A after the transfer: \( 250 + x \)
- Cargo in B after the transfer: \( 620 - x \)
3. Set up the equation modeling the condition that A will contain five times as much as B:
[tex]\[ 250 + x = 5 \times (620 - x) \][/tex]
4. Solve the equation step by step:
[tex]\[ 250 + x = 5 \times 620 - 5x \][/tex]
[tex]\[ 250 + x = 3100 - 5x \][/tex]
[tex]\[ 250 + 6x = 3100 \][/tex]
[tex]\[ 6x = 3100 - 250 \][/tex]
[tex]\[ 6x = 2850 \][/tex]
[tex]\[ x = \frac{2850}{6} \][/tex]
[tex]\[ x = 475 \][/tex]
So, 475 tonnes of cargo must be taken from hold B and put into hold A to satisfy the condition.
### (c) Motor Boat Problem
A motor boat travels up-river against the current from one point to another at a speed of 6 knots, and then down-river with the current back to the original point at a speed of 9 knots, taking a total time of 2.5 hours. We need to find the distance between the two points.
1. Define the unknown variable:
Let \( d \) be the distance (in nautical miles) between the two points.
2. Write the expressions for the time taken to travel up-river and down-river:
- Time taken to travel up-river: \( t_1 = \frac{d}{6} \) hours
- Time taken to travel down-river: \( t_2 = \frac{d}{9} \) hours
3. Set up the equation modeling the total travel time:
[tex]\[ t_1 + t_2 = 2.5 \][/tex]
Substitute the expressions for \( t_1 \) and \( t_2 \):
[tex]\[ \frac{d}{6} + \frac{d}{9} = 2.5 \][/tex]
4. Solve the equation step by step:
First, find a common denominator for the fractions:
[tex]\[ \frac{d}{6} + \frac{d}{9} = 2.5 \][/tex]
[tex]\[ \frac{3d + 2d}{18} = 2.5 \][/tex]
[tex]\[ \frac{5d}{18} = 2.5 \][/tex]
Multiply both sides by 18 to clear the fraction:
[tex]\[ 5d = 2.5 \times 18 \][/tex]
[tex]\[ 5d = 45 \][/tex]
Divide both sides by 5:
[tex]\[ d = \frac{45}{5} \][/tex]
[tex]\[ d = 9 \][/tex]
So, the distance between the two points is 9 nautical miles.
### (b) Cargo Problem
A ship's hold, A, contains 250 tonnes of cargo, and another hold, B, contains 620 tonnes of cargo. We want to find how much cargo must be taken from B and put into A so that A will contain five times as much as B.
1. Define the unknown variable:
Let \( x \) be the amount of cargo (in tonnes) taken from B and put into A.
2. Write the expression for the new amounts of cargo after transferring x tonnes:
- Cargo in A after the transfer: \( 250 + x \)
- Cargo in B after the transfer: \( 620 - x \)
3. Set up the equation modeling the condition that A will contain five times as much as B:
[tex]\[ 250 + x = 5 \times (620 - x) \][/tex]
4. Solve the equation step by step:
[tex]\[ 250 + x = 5 \times 620 - 5x \][/tex]
[tex]\[ 250 + x = 3100 - 5x \][/tex]
[tex]\[ 250 + 6x = 3100 \][/tex]
[tex]\[ 6x = 3100 - 250 \][/tex]
[tex]\[ 6x = 2850 \][/tex]
[tex]\[ x = \frac{2850}{6} \][/tex]
[tex]\[ x = 475 \][/tex]
So, 475 tonnes of cargo must be taken from hold B and put into hold A to satisfy the condition.
### (c) Motor Boat Problem
A motor boat travels up-river against the current from one point to another at a speed of 6 knots, and then down-river with the current back to the original point at a speed of 9 knots, taking a total time of 2.5 hours. We need to find the distance between the two points.
1. Define the unknown variable:
Let \( d \) be the distance (in nautical miles) between the two points.
2. Write the expressions for the time taken to travel up-river and down-river:
- Time taken to travel up-river: \( t_1 = \frac{d}{6} \) hours
- Time taken to travel down-river: \( t_2 = \frac{d}{9} \) hours
3. Set up the equation modeling the total travel time:
[tex]\[ t_1 + t_2 = 2.5 \][/tex]
Substitute the expressions for \( t_1 \) and \( t_2 \):
[tex]\[ \frac{d}{6} + \frac{d}{9} = 2.5 \][/tex]
4. Solve the equation step by step:
First, find a common denominator for the fractions:
[tex]\[ \frac{d}{6} + \frac{d}{9} = 2.5 \][/tex]
[tex]\[ \frac{3d + 2d}{18} = 2.5 \][/tex]
[tex]\[ \frac{5d}{18} = 2.5 \][/tex]
Multiply both sides by 18 to clear the fraction:
[tex]\[ 5d = 2.5 \times 18 \][/tex]
[tex]\[ 5d = 45 \][/tex]
Divide both sides by 5:
[tex]\[ d = \frac{45}{5} \][/tex]
[tex]\[ d = 9 \][/tex]
So, the distance between the two points is 9 nautical miles.
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