COMPOUND PROPORTIONS AND RATES OF WORK
Compound Proportions
Consider the numbers 4, 10, 20 and 50. 4:10 = 2:5 and 20:50 =2:5 . The two ratios are equal ( 4:10 = 20:50 ). Therefore, these 4 numbers are said to be proportional.
Generally, any 4 numbers a, b, c and d are said to be proportional if:
\frac{a}{b}=\frac{c}{d} OR \frac{a}{c}=\frac{b}{d}
Continued Proportions
Consider the numbers 2, 4, 8 and 16. You may notice that each number is equal to the previous number multiplied by 2. The 4 numbers are therefore said to be in continued proportion.
Generally, if 4 numbers, a, b, c and d are in continued proportion:
\frac{a}{b}=\frac{b}{c}=\frac{c}{d}If numbers are in continued proportion, they form a geometric progression since they have a common ratio.
Example
Given that the ratio x:y=2:3 , find the ratio (5x-2y):(x+y) .
We are told that x:y=2:3 . This can alternatively be expressed as \displaystyle \frac{x}{y}=\frac{2}{3} To eliminate the y terms in the second ratio, we can express y in terms of x. 2y=3x y=\frac{3}{2}x Substituting this into the second ratio, we get: \left( 5x-2 \left( \frac{3}{2}x \right) \right): \left( x+\frac{3}{2}x \right) = \left( 5x-3x \right): \left(\frac{5}{2}x \right) = 2x: \frac{5}{2}x We then multiply both sides by \frac{2}{x} to simplify. = \frac{2}{\cancel{x}}\times2\cancel{x}: \frac{5}{\cancel{2}}\cancel{x} \times \frac{\cancel{2}}{\cancel{x}} =4:5
Proportions are everywhere in our daily life, and are very useful. For example, when preparing a meal for many people you would need to increase the amount of ingredients. To do this without altering the taste, you would have to maintain the proportions between various ingredients. They are also important for determining relationships. If you were at a shop and told an item has a discount of sh 50 but you do not know its original price, this would not be useful information. However, if you are instead told it is 15% off, it has a larger effect as you now know the proportion of the new price versus the old price.
Proportional Parts
Generally, if an amount n is to be divided in the ratio a:b:c , a will get \displaystyle \frac{a}{a+b+c}\times n , b will get \displaystyle\frac{b}{a+b+c}\times n and c will get \displaystyle \frac{c}{a+b+c}\times n .
Example
An alloy is made of copper, tin and aluminium in the ratio 5 : 14 : 1 . Find the mass of copper in 185 kg of the alloy.
The mass of copper is given by the expression: \frac{5}{5+14+1}\times 185 =\frac{5}{20}\times 185 =46.25\ \text{kg}
Rates of Work
When dealing with rates of work, the key is to find out how much work a person/thing can do in one hour. Take a look at an example.
Example
A water tank has a volume of 7.2 \text{m}^3 . An inlet pipe was opened and water let to flow into the tank at a rate of 10L per minute. After 1 hour, a drain pipe was opened and water allowed to flow out of the tank at a rate of 4L per minute. Calculate the total time taken to fill up the tank.
The volume of the tank expressed in litres is 7.2\times1000=7200\text{L} . This is the amount of water we need to fill it.
Let us find out how much work is done by each of the pipes in 1 hour. The inlet pipe adds 10 \times60=600\text{L} of water every hour. The drain pipe removes 4\times60=240\text{L} of water every hour.
In the first hour, the inlet pipe acts alone. Thus, 600L of water is added.
For the rest of the time, both pipes act at the same time. Therefore, we can subtract the amount of water that is removed from the amount that is added to find the net amount of water added to the tank.
600-240=360 \text{L added every hour}
The amount of water remaining to be added is 7200-600=6600L . Thus, the time taken after the first hour is: \frac{6600}{360}=18.33h = \text{18h 20 min}
Our total time is therefore 1h+\text{18h 20 min} = \text{19h 20 min} .
In questions with 3 or more factors to consider and 1 unknown, think of it like this: If I increase this value, will the unknown increase or decrease? If the unknown will increase, multiply by a fraction greater than 1. If the unknown will decrease, multiply by a fraction less than 1. Let us look at an example.
Example
Three workers, working 8 hours per day, can complete a task in 5 days. Each worker is paid Ksh 40 per hour. Calculate the cost of hiring 5 workers if they work for 6 hours per day to complete the same task.
In this question, the unknown is the number of days for the second situation. We can thus start our expression with the original number of days. This is what will be manipulated to determine the final number of days. 5\times Now, let us look at the number of workers. We originally had 3 workers, but we now have 5. Therefore, we expect the number of days to decrease, so we will multiply the by a fraction less than 1. The number of days decreases in the ratio 3:5 because instead of 3 workers we now have 5. 5\times\frac{3}{5}\times Now we look at the number of hours. The number of hours a day decreases, thus we expect the number of days to increase. We therefore multiply by a fraction greater than 1. The number of days will increase in the ratio 8:6 because we have gone from 8 hours a day to 6. Thus, our final expression will be: 5\times\frac{3}{5}\times\frac{8}{6}=4\text{ days}The cost of hiring the workers is 5\times6\times4\times40=\text{Ksh }4800 .
Mixtures
Example 1
In what proportion should grades of sugar costing sh 45 and sh 50 per kilogram be mixed in order to produce a mixture worth sh 48 per kilogram?
Let x kg of the sugar costing sh 45 be mixed with y kg of the sugar costing sh 50 per kg. The total cost of this mixture would be 45x+50y . The cost of 1 kilogram of this mixture would be the total cost divided by the total mass, i.e \displaystyle\frac{45x+50y}{x+y} .
We have that the cost per kg is Sh 48. We then equate this with the expression above. \frac{45x+50y}{x+y}=48 45x+50y=48(x+y) 45x+50y=48x+48y 2y=3x \frac{x}{y}=\frac{2}{3} The proportion is x:y=2:3 .
Example 2
An industrialist has 450L of a chemical which is 70% pure. He mixes it with a chemical of the same type but 90% pure to obtain a mixture that is 75% pure. Find the amount of the 90% pure chemical used.
If we let the volume of the 90% pure chemical be x, the total volume of the mixture is 450+x . The amount of chemical in the mixture is (450\times0.7)+0.9x . To get the purity of the new mixture, we must divide the amount of chemical by the total volume. \displaystyle\frac{(450\times0.7)+0.9x}{450+x}
We are given that the new purity is 75%. We can now equate this to the expression above to get: \frac{(450\times0.7)+0.9x}{450+x}=0.75 315+0.9x=0.75(450+x) 315+0.9x=337.5+0.75x 0.15x=22.5 x=150L