Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
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.
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.