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Ratio & Proportion
A ratio is a comparison of two numbers. We generally separate the two numbers in the ratio with a colon ":". Suppose we want to write the ratio of 8 and 12.
We can write this as 8:12 or as a fraction 8/12, and we say the ratio is eight to twelve.
- In a certain class, the ratio of passing grades to failing grades is 7 to 5. How many of the 36 students failed the course?
The ratio, "7 to 5" (or 7 : 5 or 7/5), tells you that, of every 7 + 5 = 12 students, five failed. That is, 5/12 of the class flunked. Then ( 5/12 )(36) = 15 students failed.
- Express the ratio in simplest form: $10 to $45
This means that you should write the ratio as a fraction, and you should then reduce the fraction:
.10/45 = 2/9.
This reduced fraction is the ratio's expression in simplest form. Note that the units "canceled" on the fraction, since the units, "$", were the same on both values. So there is no unit on the answer.
- Express the ratio in simplest form: 240 miles to 8 gallons
Depending on the text (or instructor), you may be supposed to keep the units on a ratio. In this case, you would have (240 miles)/(8 gallons) = (30 miles)/(1 gallon), or, in more common language, 30 miles per gallon.
To compare ratios, write them as fractions. The ratios are equal if they are equal when written as fractions.
Example: Are the ratios 3 to 4 and 6:8 equal?
The ratios are equal if 3/4 = 6/8.
These are equal if their cross products are equal; that is, if 3 x 8 = 4 x 6.
Since both of these products equal 24, the answer is yes, the ratios are equal.
Remember to be careful! Order matters!A ratio of 1:7 is not the same as a ratio of 7:1
Let x : y be a ratio, which can also be written as x/y. We will try to find out what will happen when a constant a is added both to the numerator and denominator
CASE I (x/y < 1 )
If x/y<1, then addition of a constant positive number to numerator and denominator leads to a bigger ratio than the ratio itself, i.e.
x/y < (a+x)/(a+y) for x/y<1
where a is a constant positive number. e.g. 1/2 is less than 1 and when we add 2 to both numerator and denominator we get 3/4 and 3/4 is grater than 1/2 . Similarly, Subtraction leads to a simillar ratio, i.e., x/y > (x-a)/(y-a) for x/y<1 Lets consider a fraction 5/10, if 5 is subtracted from numerator as well as denominator, we get 0 and it is less than 5/10(i.e. 1/2) Thus the rule for the case of subtraction is the reverse of the case of addition, as can be easily seen by the given example.
CASE II (x/y > 1 )
The above rule gets totally and exactly reversed for x/y >1. Therefore,
x/y > (x+a)/(y+b) and x/y < (x-a)/(y-b).
A proportion is an equation with a ratio on each side. It is a statement that two ratios are equal.
3/4 = 6/8 is an example of a proportion. When one of the four numbers in a proportion is unknown, cross products may be used to find the unknown number. This is called solving the proportion. Question marks or letters are frequently used in place of the unknown number.
Example: Solve for n: 1/2 = n/4.
Using cross products we see that 2 x n = 1 x 4 =4, so 2n = 4. Dividing both sides by 2, n = 4 / 2 so that n = 2.
1.Compounded Ratio of two ratios a/b and c/d is ac/bd, i.e., ac : bd.
Duplicate ratio of a : b is a2 : b2
Triplicate ratio of a : b is a3 : b3
Sub-duplicate ratio of a : b is a : b
Sub-triplicate ratio of a : b is ³a : ³b
Reciprocal ratio of a : b is b : a
2.Proportion. Four (non-zero) quantities of the same kind a, b, c, d are in proportion,
written as a : b :: c : d iff a/b = c/d
3.The non-zero quantities of the same kind a, b, c, d, ... are in continued proportion iff
a/b = b/c = c/d = ...
In particular, a, b, c are in continued proportion iff a/b = c/d. In this case b is called the mean proportion; b = ac; c is called third proportional. If a, b, c, d are in proportion, then d is called fourth proportional.
4.Invertendo. If a : b :: c : d then b : a :: d : c
Alternendo. If a : b :: c : d then a : c :: b : d
Componendo. If a : b :: c : d then (a +b) : b :: (c +d) : d
Dividendo. If a : b :: c : d then (a -b) : b :: (c -d) : d
Componendo and dividendo.
If a : b :: c : d then (a +b) : (a -b) :: (c +d) : (c -d)
i.e., a/b = c/d => (a +b)/(a - b) = (c +d)/(c +d)
5.If a/b = c/d = e/f = ..., then each ratio = (a +c +e +...)/(b +d +f +...)
Which is same as ( pa + qc + re +. . . )/( pb + qd + rf + . . . )
a/b = c/d = e/f
=(a +c +e +...)/(b +d +f +...)
= ( pa + qc + re +. . . )/( pb + qd + rf + . . . )
= ( pna + qnc + rne +. . . )1/n/( pnb + qnd + rnf + . . . )1/n
where p, q, r, etc. are constants such that all of them are simultaneously not equal to zero
Examples 1. If a doctor prescribes 30 units of insulin in 500 ml to be administered over 2 hours, how many drops per minute should be administered if the set is calibrated to deliver 20 drops per ml?
The rate we want is 500 milliliters over two hours, which we can write as a fraction: 500 ml / 2 hr We know that there are 20 drops per milliliter, so that 20 drops/1ml = 1 since the numerator is the same amount as the denominator. If we multiply the rate by this fraction, we won't change the rate. We then have ml in the numerator and the denominator, so they will cancel:
(500 ml /2hr)*(20 drops /1ml) = (500*20) drops / (2*1hr) = 5000 drops/hr
We're not done yet, because we need to know the number of drops per MINUTE. But that's easy using another "unit multiplier". Since 1 hour = 60 minutes, we can write 1 hour / 60min = 1
I chose to put the hour on top to cancel the hours in the bottom of our fraction, 5000 drops/hour, when we multiply:
( 5000 drops/ 1 hour) *(1hour/60 mins) = 5000/60 drops/min = 83.333 drops/min
2. A petroleum distributor has two gasohol storage tanks, the first containing 9 percent alcohol and the second containing 12 percent alcohol. They receive an order for 300,000 gallons of 10 percent alcohol. How can they mix alcohol from the two tanks to fill this order?
Let x = volume of type(1) 100-x = volume of type(2) in 100 gallons of mixture. Then 0.09x + 0.12(100-x) = 10 gallons -0.03x + 12 = 10 2 = 0.03x and so x = 66.6667 gallons So we use 66+(2/3) gallons type (1) and 33+(1/3) gallons type (2) per 100 gallons of mixture. For 300,000 gallons, we multiply up by 3000 to get 200,000 and 100,000 of type (1) and type (2) respectively.
3.A cable television company plans to begin operations in a small town. The company knows that about 600 people will subscribe to the service if the price per subscriber is $5 per month, but that for each 5-cent increase in the monthly subscription price, 4 of the original 600 people will decide not to subscribe. The office begins operations, and its total revenue for the first month is $1500. How many people have subscribed to the cable television service?
Let x = number of increments of 5-cents added to the price. So the number of subscribers = 600-4x, and subscription = 5+.05x So we have: (600-4x)(5+.05x) = 1500 3000 - 20x + 30x - 0.2x^2 = 1500 -0.2x^2 + 10x + 1500 = 0 0.2x^2 - 10x - 1500 = 0 so x = -65.1387 or x = 115.13 The negative value we ignore, as the question does not say that they would gain subscribers by reducing the price. So they have increased the price by 115 increments of 5 cents = $5.75 Total cost of subscribing = $5 + $5.75 = $10.75 Total number of subscribers = 600-460 = 140 Total income = 140*10.75 = $1505 - which checks approximately.
4.Milk that has 5% butterfat is mixed with milk that has 2% butterfat. How much of each is needed to obtain 60 gallons of milk that has 3% butterfat0.05x + 0.02y = 0.03*60 = 1.8 You might like to avoid decimals by multiplying the whole equation by 100: 5x + 2y = 180 Now you have a system of simultaneous equations: x + y = 60 5x + 2y = 180 You have probably learned some method of solving this. If not, the easiest way is to solve the first equation for y: y = 60 - x and then replace y in the second equation with (60-x). This gives you a single equation to solve to find x.
5.Sterling Silver is 92.5% pure silver. How many grams of pure silver and sterling silver must be mixed to obtain 100g of a 94% Silver alloy?
The amounts of the two input mixtures are unknown; the amount of the output mixture is 100g. Let x = grams of 92.5% silver alloy Then, since the total weight is 100g, we have (100-x) = grams of 100% silver We now write an equation relating the amounts of pure silver in the two "input" mixtures and in the "output" mixture: "x" grams at 92.5% silver + (100-x) grams at 100% = 100g at 94% (x)(0.925) + (100-x)(1.0) = (100)(0.94) 0.925x + 100 - x = 94 6 = 0.075x 6000 = 75x 6000/75 = x 80 = x So to make 100g of an alloy of 94% silver, you need to mix 80g of 92.5% alloy and 20g of pure silver.
And now here is a completely different approach to the same problem. I prefer this method, because I find the calculations are usually easier. Understanding why this method works is probably a bit more difficult than understanding the traditional method, but it works for me, so I use it. Take a look at this alternative method and see if you like it. We have two "input" mixtures; one of 92.5% silver and the other of 100% silver. We want to make a mixture of 94% silver. If I think of plotting these percentages on a number line, I see that the "distance" from 92.5% to 94% is 1.5%, while the "distance" from 94% to 100% is 6%. And now here is the key to my method: The distances from 92.5% to 94% and from 94% to 100% are 1.5% and 6%; these two distances are in the ratio 1:4. This means that the two "input" mixtures must be mixed in the ratio 1:4 to get the 94% alloy. If there are to be 100g of the 94% alloy and the two input mixtures are in the ratio 1:4, then there must be 20g of one input and 80g of the other. Because the resulting alloy is closer to 92.5% than 100%, the required amounts of the inputs are 80g of the 92.5% alloy and 20g of the 100% silver.
6.Person A has a cup of coffee. Person B has a cup of tea. Person A takes a teaspoon of coffee and puts it in the tea. Person B mixes the teaspoon of coffee and tea, then takes a teaspoon of the mixture and returns it to the coffee. Is there more coffee in the tea, or more tea in the coffee, or are they the same?
The volume of each cup after the swap is the same. That means that whatever coffee is missing from the first cup must have been replaced with exactly the same amount of tea.
7.Chris has 3 gallons of a solution that is 30 percent antifreeze, which he wants to use to winterize his car. How much pure antifreeze should he add to this solution so that the new solution will be 65 percent antifreeze?
you start out with x gallons of antifreeze that's been mixed with y gallons of water, and the total is 3 gallons, so I'd say x + y = 3 is a good equation to start with. You also know that the solution is 30 percent antifreeze, so x/(x+y) = .30. Since you have two equations and two unknowns, these equations are enough to tell you both x and y. Maybe you should find x and y before going any further, and then you can write the next equation more simply. Now we want to add some more antifreeze until the total is 65 percent antifreeze. Say you're adding z gallons. The equation is (x+z)/(x+y+z) = 0.65 The x+z is the total amount of antifreeze, and the x+y+z is the total amount of solution, antifreeze+water. Since you already know x and y, you can put them into this equation and you'll just have a single equation with z in it. That should do the trick!
8.Suppose 30 liters of a solution with an unknown percentage of alcohol is mixed with 5 liters of a 90% alcohol solution. If the resulting mixture is a 62% alcohol solution, what is the percentage of alcohol in the first solution?
Think of the amount of alcohol in each solution. In the first one, it is 30*(x/100) liters. In the second one, it is 5*(90/100). In the last one, it is 35*(62/100) (because the total amount of the mixture is 35 = 30 + 5 liters). When the first two solutions are mixed, the total amount of alcohol is the sum of the amounts in the two ingredients, so you get an equation: 30*(x/100) + 5*(90/100) = 35*(62/100).
9.A group of chemists are conducting an experiment to produce a new liquid material. One chemical contains 15% sodium (Na) and the other chemical contains 30% sodium (Na). Once they mix the two samples the resulting chemical contains 22% sodium (Na). How many milliliters (ml) of each sample must be mixed to obtain 600 ml of the new chemical?
Since the first material is 15% sodium, it means that if we have some amount, A, of the material, we can determine the amount of sodium it contains by taking 15% of that. Does that make sense? amount of sodium from first material = 0.15*A For example, if we have 100 ml of the material, then 15 ml of it will be sodium, and 85 ml will be something else. Similarly, if B is the amount of the second material, amount of sodium from second material = 0.30*B Now, if we combine the two chemicals to get a new chemical, which is 22% sodium, then the total amount of combined sodium is amount of sodium from both sources = 0.22*(A+B) So we can use that to set up an equation: sodium from sodium from sodium from first material + second materal = both materials 0.15*A + 0.30*B = 0.22*(A+B) Now, this is kind of a problem, since we have two variables but only one equation. But in fact, we _have_ a second equation, because we know that A + B = 600 So now you have two equations, and you can use substitution or elimination to find the values of the variables:
10.An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice?
You have 5 quarts of "drink" that is 10% orange juice. Thus the 5 quarts of drink contain 5*10% = 5*.1 = .5 quarts of orange juice, and hence must contain 4.5 quarts of other stuff. Now we're going to add 100% orange juice and we want to know how much to add so that the new mixture will be 40% orange juice. Well, what if you add 1 quart of 100% orange juice? What would that do to the concentration of orange juice in the mixture? It would mean that we now have 1.5 quarts of pure orange juice mixed in a carton that now contains 6 quarts of liquid. Thus the concentration of orange juice in this 6-quart jug would be 1.5/6 = .25 = 25% So we haven't added enough, but maybe you see what's going on. In a more general situation: If we add x quarts of pure orange juice to the original mixture that contains .5 quarts of OJ and 4.5 quarts of something else, then what we have is a mixture of 5+x quarts that contains .5+x quarts of OJ. So after we've added x quarts of pure OJ, the concentration of OJ will be (.5+x)/(5+x) We want to know when this will be equal to .4 So solve for x: (.5+x)/(5+x) = .4 .5+x = 2 + .4x .6x = 1.5 x = 1.5/.6 = 2.5 Thus you should add 2.5 quarts of pure OJ to make the concentration of the new mixture 40% OJ