Ratio and Proportion Learning Outcome When you complete this module you will be able to: Describe the concepts of ratio and proportion.

Learning Objectives Here is what you will be able to do when you complete each objective: 1. Find ratios of one quantity to another quantity. 2. Solve word problems involving ratios and proportions.

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RATIO A ratio is the number of times that one number is contained in another number. It is a way to express the relative sizes of two or more quantities with the same units. For example, if the stroke of a pump is 300 mm and the diameter of the pump cylinder is 150 mm, the ratio of stroke to diameter is 300 to 150, or 2 to 1, reducing it to its lowest terms. The ratio could be left in its original figures but reducing it to its lowest terms conveys a much better idea of the comparative size of the two dimensions. In order to get a ratio between two quantities they must be in the same kind of units. Tonnes and kilometers cannot be compared nor can grams and metres, as they are entirely different kinds of measurements. Neither can we compare tonnes and kilograms, or metres and millimetres, unless we bring both masses to either tonnes or kilograms in the first case, and both measurements to either metres or millimetres in the second case.

Example 1: What is the ratio between a mass of 3 tonnes and a mass of 200 kg? Solution: First the values must be converted to the same units. Either number can be converted to the other’s units. 3 tonnes x 1000 kg/tonne

= 3000 kg

Now the ratio between the two numbers with the same units can be found. 3000 kg 200 kg 15 1 = 15 to 1 (Ans.)

Example 2: What is the ratio between a measurement of 3 m and a measurement of 80 mm?

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Solution: Convert 3 m to millimetres. 3 m x 1000 mm/m The ratio

= 3000 mm

= 3000 mm 80 mm = 75 2 = 75 to 2 (Ans.)

The rule then for finding the ratio between two quantities is: REDUCE BOTH QUANTITIES TO THE SAME UNITS AND DIVIDE THE FIRST QUANTITY BY THE SECOND QUANTITY.

PROPORTION When two ratios have the same value (are equal), the four quantities composing the ratios are said to be in proportion. Thus, the ratio 3 to 1 is of the same value as the ratio 9 to 3. This could be stated as: 3 is to 1 as 9 is to 3 but it is customary to replace the words by dots which have the same meaning. The last statement can be rewritten as: 3 : 1 :: 9 : 3 A proportion can also be stated in the fractional form: 3 1

=

9 3

or with an equal sign: 3:1

=

9:3

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Application of Proportion The principle of proportion is applied where two quantities are known to be in the same ratio but only one of the quantities is given and we wish to find the other. The proportion is set down in the form shown above, the missing quantity being denoted by x or some suitable symbol. The two outer terms (1st and 4th) are called the Extremes, and the two inner terms (2nd and 3rd) are called the Means. extreme

extreme

3 : 1

::

mean

9 : 3 mean

The product of the means is always equal to the product of the extremes. Therefore in the above proportion, 3 x 3 = 1 x 9 From this we get the rules: THE PRODUCT OF THE TWO MEANS DIVIDED BY EITHER EXTREME GIVES THE OTHER EXTREME. THE PRODUCT OF THE TWO EXTREMES DIVIDED BY EITHER MEAN GIVES THE OTHER MEAN.

Example 3: If 10 tonnes of coal cost $85.00, what will 3 1/2 tonnes cost if the cost per tonne is the same? Solution: The ratio of the prices must be the same as the ratio between the masses. Let x = the price of 3 1/2 tonnes. The proportion can be written as follows: x : 85 :: 3 1/2 : 10 x

= 3 1/2 tonnes x $85.00 10 tonnes = $29.75 (Ans.)

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Example 4: Brass is a mixture of copper and zinc. How much copper must be mixed with 80 kg of zinc if the ratio of zinc to copper is 3 to 7? Solution: Zinc : Copper :: 3 : 7 80 : Copper :: 3 : 7 Copper

= 80 kg x 7 3 = 186.7 kg (Ans.)

Example 5: A certain chemical mixture should be in the ratio of 3 parts A, 5 parts B, and 2 parts C. A 15 kg batch is to be mixed. What quantities of A, B, and C should be used in the batch? Solution: It can be seen that 3 parts A, 5 parts B, and 2 parts C add up to a total of 10 parts in this mixture. A

=

3 10

of the mixture

B

=

5 10

or

1 2

of the mixture

C

=

2 10

or

1 5

of the mixture

To find the amount of A needed in the batch: A : 15 :: 3 : 10 A

= 15 kg x 3 10 = 4.5 kg (Ans.) 5

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To find the amount of B needed for the batch: B : 15 :: 1 : 2 B

= 15 kg x 1 2 = 7.5 kg (Ans.)

To find the amount of C needed for the batch: C : 15 :: 1 : 5 C

= 15 kg x 1 5 = 3 kg (Ans.)

A quick check is to make sure the amounts calculated for A, B, and C add up to the specified total of 15 kg. 4.5 kg A + 7.5 kg B + 3 kg C = 15 kg.

Inverse Proportion The problems just worked out are examples of direct proportion. Thus, in Example 3, the price increased directly as the mass increased. An increase in mass means an increase in price and a decrease in mass means a decrease in price. There are some instances, however, where the opposite holds true; that is, an increase in one factor means a corresponding decrease in the other factor. Such cases are called inverse proportions. The most common examples of inverse proportion are problems involving size and speed of gear wheels that are meshed together, or size and speed of pulleys that are connected together by belts. The speeds of such pulleys and gear wheels are inversely proportional to their diameters. The ratio between the speeds will be the same as the ratio between the diameters or number of teeth, but the smaller pulley or gear will run at the faster speed and the larger pulley or gear will run at the slower speed. Example 6: A gear wheel 300 mm in diameter and revolving at a speed of 100 r/min drives a wheel 120 mm in diameter. What is the speed of the second gear? 6 MATH 6011

Solution: Speed of 2nd : Speed of 1st :: Dia. of 1st : Dia. of 2nd Speed : 100 :: 300 : 120 Speed

= 100 r/min x 300 mm 120 mm = 250 r/min (Ans.)

Example 7: A supply fan is driven by an electric motor by means of V-belts. The pulley on the motor shaft is 80 mm diameter and rotates at 1750 r/min. The diameter of the pulley on the fan shaft is 280 mm. What is the speed of the fan? Solution: Fan speed : Motor speed :: Dia. motor pulley : Dia. fan pulley Fan speed : 1750 :: 80 : 280 Fan speed

= 1750 r/min x 80 mm 280 mm = 500 r/min (Ans.)

Example 8: A driving gear has 55 teeth and turns at 600 r/min. The driven gear has 80 teeth; how fast does it turn? Solution: Let the speed of the 80 tooth (driven) gear = x Driven speed : Driver Speed :: Driver teeth : Driven teeth x : 600 :: 55 : 80 x

= 600 r/min x 55 teeth 80 teeth = 412.5 r/min (Ans.) 7

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Self Test After completion of the self-test, check your answers with the answer guide that follows. 1. A manufacturer employs 500 females and 1500 males. (a) What is the ratio of females to males? (b) What is the ratio of males to females? (c) What is the ratio of females to the total number of employees? (d) What is the ratio of males to total number of employees? 2. A gas plant receives product from three different wells, A, B, and C in the proportion of 3 parts, 7 parts, and 2 parts respectively. If the total amount of gas processed is 2 400 000 m3, how much gas comes from each well? 3. A small 150 mm diameter pulley drives a larger one measuring 400 mm. If the small one rotates at a speed of 240 r/min, how fast does the large one rotate? 4. Gear A has a 120 mm diameter. What is the diameter of gear B if gears A and B have radii in the ratio of 5:3?

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Self Test Answers 1.

(a) (b) (c) (d)

1:3 3:1 or 3 1:4 3:4

2. 600 000 m3 from Well A, 1 400 000 m3 from Well B and 400 000 m3 from Well C. 3. 90 r/min 4. 72 mm

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Notes:

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