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Solving problems with equations II
Example 1
It would take Peter \latex{ 4 } hours to paint the fence. Andrew could do the same work in \latex{ 6 } hours. How much time does it take them to paint the fence together?
Solution
Peter alone finished the work in \latex{ 4 } hours, therefore in \latex{ 1 } hour he can paint \latex{\frac{1}{4}} of the fence, while Andrew finishes \latex{\frac{1}{6}} of it in \latex{ 1 } hour.
Let t denote the number hours of the joint work. In t hours Peter paints \latex{t\times \frac{1}{4} =\frac{t}{4}}, and Andrew paints \latex{t\times \frac{1}{6} =\frac{t}{6}} of the fence, and the sum of these two parts gives the whole work. Thus the equation that can be set up is:
\latex{\frac{t}{4} +\frac{t}{6} =1}
\latex{3t+2t=12}
\latex{5t=12}
\latex{t=2.4}.
\latex{3t+2t=12}
\latex{5t=12}
\latex{t=2.4}.
Therefore together they can finish the work in \latex{ 2.4 } hours, so \latex{ 2 } hours \latex{ 24 } minutes.
In \latex{ 2.4 } hours Peter finishes \latex{\frac{2.4}{4}=0,6}, whereas Andrew finishes \latex{\frac{2.4}{6}=0.4} of the work.
Example 2
A group of tourists goes up a hill in \latex{ 5 } hours, they cover the same distance in \latex{ 4 } hours downhill, since they can cover \latex{ 1 } \latex{ km } more every hour downhill. What is total distance covered during the trip?
Solution
Let the distance covered in one direction by \latex{ s }. It is practical to tabulate the data.
We know that the distance covered in an hour was \latex{ 1 } \latex{ km } longer downhill than uphill, i.e. the speed (velocity) was \latex{ 1 } \latex{ km/h } faster.
Since the measurement units correspond to each other, we can leave them out throughout the detailed calculation.
Thus the equation is: \latex{\frac{s}{5}+1=\frac{s}{4}}.
Uphill
Downhill
Time (h)
Distance (km)
Speed (km/h)
\latex{5}
\latex{4}
\latex{s}
\latex{s}
\latex{\frac{s}{5}}
\latex{\frac{s}{4}}
Example 3
From the end of a \latex{ 10 } metre long row of children walking straight ahead the accompanying teacher went to the start of this row, and once she got there, she turned around and went back to the end of the row. The teacher was walking three times as fast as the children. What distance did the children cover by the time the teacher reached the end of the row again?
Solution
Let us make a diagram to find the solution. (From a well-drawn figure we can often get very useful information about the quantities mentioned in the exercise.)
Figure 25
Towards the start of the row
\latex{ 10 } \latex{ m }
\latex{ 10 } \latex{ m }
\latex{ (x+10) } \latex{ m }
\latex{ x } \latex{ m }
row of children
row of children
teacher
teacher
\latex{ 1 }st case (Figure 25): \latex{ x } be the distance covered by the row, then \latex{ 10 + x } is the distance covered by the teacher. Let the speed of the children be \latex{ 1 }, then the speed of the teacher is \latex{ 3 }. Since the time taken must equate:
\latex{\frac{x}{1}=\frac{\left(x+10\right) }{3} \longrightarrow x=5}
Figure 25
To the end of the row
\latex{ 10 } \latex{ m }
\latex{ 10 } \latex{ m }
\latex{ (10-y) } \latex{ m }
\latex{ y } \latex{ m }
row of children
row of children
teacher
teacher
\latex{ 2 }nd case (Figure 26): \latex{ y } be the distance covered by the row, then \latex{ y – 10 } is the distance covered by the teacher. Let the speed of the children be \latex{ 1 }, then the speed of the teacher is \latex{ 3 }. Since the time taken must equate:
\latex{\frac{y}{1}=\frac{\left(10-y\right) }{3} \longrightarrow y=2.5}m.
So the row covered a total of x + y = 7.5 m.
Example 4
We mixed two types of orange juice. One of them was \latex{ 60 }% concentrated, the other was \latex{ 85 }%. How many litres did we buy from each if the mixture became \latex{ 18 } litres of \latex{ 70 }% concentrated orange juice?
Solution
Let \latex{ x } (litre) denote the amount used from the \latex{ 60 }% orange juice. Thus we can create the following table:
Amount of solution \latex{ (l) }
Concentration
\latex{ ( }%\latex{ ) }
\latex{ ( }%\latex{ ) }
Concentrate
\latex{ (l) }
\latex{ (l) }
Orange juice 1
Orange juice 2
Mixture
\latex{ x }
\latex{ 18-x }
\latex{ 18 }
\latex{ 60 }%
\latex{ 85 }%
\latex{ 70 }%
\latex{ 0.6x }
\latex{ 0.85\times(18-x) }
\latex{ 0.7 × 18 }
The amount of concentrate in the mixture is the sum of the concentrates found in the two types of orange juice. Thus:
\latex{ 0.6x + 0.85(18 - x) = 12.6, }
\latex{ 0.6x + 15.3 - 0.85x = 12.6, }
\latex{ -0.25x = -2.7, }
\latex{ x = 10.8. }
\latex{ 0.6x + 15.3 - 0.85x = 12.6, }
\latex{ -0.25x = -2.7, }
\latex{ x = 10.8. }
Therefore to get \latex{ 18 } litres of \latex{ 70 }% orange juice we have to mix \latex{ 10.8 } litres of \latex{ 60 }% and \latex{ 7.2 } litres of \latex{ 85 }% orange juice.
Indeed the amount of concentrate found in \latex{ 10.8 } litres of \latex{ 60 }% orange juice is
\latex{ 0.6 × 10.8 } litres \latex{ = 6.48 } litres,
the amount of concentrate found in \latex{ 7.2 } litres of \latex{ 85 }% orange juice is
\latex{ 0.85 × 7.2 } litres \latex{ = 6.12 } litres.
And the sum of these is
\latex{ 0.6 × 10.8 } litres \latex{ = 6.48 } litres,
\latex{ 0.85 × 7.2 } litres \latex{ = 6.12 } litres.
\latex{ 6.12 } litres \latex{ + \,6.48 } litres \latex{ = 12.6 } litres.
Example 5
A bookseller bought two books, and then he sold them at equal prices. He made \latex{ 20 }% profit on one of them and lost \latex{ 20 }% on the other one, thus he got \latex{ 1 } EUR less than he bought them for. For how much did he buy and sell the books?
Solution
Let the selling price of the books be \latex{ x } EUR.
Thus he bought the first book for \latex{\frac{x}{1.2}} EUR, and the second one for \latex{\frac{x}{0.8}} EUR. The sum of these is \latex{ 1 } EUR more than he sold the books for, therefore:
\latex{\frac{x}{1.2} +\frac{x}{0.8}=2x+1}
\latex{0.8x+1,2x=1,92x+0.96}
\latex{0.08x=0.96}
\latex{x=12}
\latex{0.8x+1,2x=1,92x+0.96}
\latex{0.08x=0.96}
\latex{x=12}
So he sold the books for \latex{ 12 } EUR, and he bought them for \latex{ 10 } EUR and \latex{ 15 } EUR respectively. Thus he won \latex{ 20 }% on the first book, while he lost \latex{ 20 }% on the second one.
Exercises
{{exercise_number}}. Changing the wallpaper in a flat takes one worker \latex{ 24 } hours, and it takes the other worker \latex{ 30 } hours. In how much time can they finish if they work together?
{{exercise_number}}. A bathtub gets full in \latex{ 20 } minutes from one tap and in \latex{ 15 } minutes from the other tap. When opening the plughole it gets empty in \latex{ 16 } minutes. How long does it take to fill the bathtub if we open both taps but leave the plughole open too?
{{exercise_number}}. A ship can cover the distance between two harbours in \latex{ 3.5 } hours downstream and in \latex{ 5 } hours upstream. The speed of the river is \latex{ 3 } \latex{ km/h }. How many kilometres away are the two harbours?
{{exercise_number}}. A greyhound starts chasing a fox who is \latex{ 30 } \latex{ m } away. One leap of the greyhound is \latex{ 2 } \latex{ m } long, and one leap of the fox is \latex{ 1 } \latex{ m } long. For every three fox leaps the greyhound takes two leaps. What distance shall the greyhound cover to catch up to the fox?
{{exercise_number}}. How much water should be evaporated from \latex{ 10 } litres of \latex{ 40 }% salt solution to get a \latex{ 60 }% salt solution?
{{exercise_number}}. The price of an item was reduced by \latex{ 20 }%, then it was raised by \latex{ 20 }%, thus its price became \latex{ 1 } EUR less than the original price. What was the original price of the item?
Puzzle
Three hens lay three eggs in three days.
- How many eggs do nine hens lay in nine days?
- How many hens lay \latex{ 21 } eggs in nine days?
- In how many days do five hens lay \latex{ 10 } eggs?