**Ratio and Proportions**

The problems and ratio and proportions can be on the basis of unitary methods. For example if A can do a piece of work in 12 days. B is 20% more efficient than A. Find the number of days it takes B to do the same piece of work. It means If A is 100, B is 120. Then we do the calculations like this. We want to find out the number of days. So the days which are given are taken in the numerator. So we write

12

Then one quantity is 100 and 120. Now if A is 100 and B is 120, what will be the impact on the quantity we want to find out i.e. number of days. Definitely B will take less days. So we take 100 in the numerator and 120 in the denominator. And write

12 x (100/120) and calculate the required value.

Taking one more example, if 30 men working 7 hours a day can do a piece of work in 18 days, in how many days will 21 men working 8 hours a day do the same piece of work. Again we think like this: the quantity(days) that we want to find out, we take the corresponding quantity(18 days) in the numerator. Thus:

18

Now we find the impact of individual element on the number of days. Taking men, first there were 30men, now there are 21, So number of men have decreased so number of days will be more so we take greater quantity (30) in the numerator and 21 in the denominator as:

18 x (30/21)

Taking hours a day, first there were 7 hours per day, now there are 8 hours per day. So number of days to do a particular work will become less as the number of hours per day are increasing. So we take 7 ( less qty) in the numerator and the other in the denominator as:

18 x (30/21) x (7/8) Solving we can get the answer.

The questions can be asked about specific ratios e.g. Divide 581 into three parts such that 4 times the first may be equal to 5 times the second and 7 times the third.

In such cases the ratio will be ¼ : 1/5:1/7, Then divide the amount in this ratio.

In any two-d figure if the corresponding sides are in the ratio a: b, then there areas in the ratio a^2 :b^2. Eg. Sides of a hexagon becomes three times. Find the ratio of the areas of the new and the old hexagons. The ratio will be 9:1

The questions are also asked of Mixtures eg. A mixture contains milk and water in the ratio 8:3. On adding 3 liters of water, the ratio of milk to water becomes 2: 1. Find the quantity of milk and water in the mixture.

Here we proceed like this:

M : W

8:3 <-- Initial ratio

2:1<-- After adding three liters of water.

Now since there is no change in the milk , we make the ratio of both the quantities equal by multiplying the second ratio by 4. Then it looks like:

M: W

8:3 <-- Initial Ratio

8:4 <-- After adding 3 liters of water

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0: 1 <-- Difference in ratios.

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Thus increase in qty of water in ratio =1 ~ 3 liter actually, so Initial qty of milk must be (3/1) x 8= 24 liters and water (3/1) x 3 = 9 liters

This method is very powerful and can be used in such situations under varying exam conditions.

Or the questions can be asked like this: How many rupees, fifty paise coins and twenty-five paise coins of which the numbers are proportional to 2 ½, 3 and 4 are together worth Rs. 210. Here the ratio is 2 ½:3:4 = 5:6:8 Their proportional value= 5 x 1: 6/2: 8/4= 5:3:2, Now Divide 210 Rs. In this ratio we have value of rupees= 5/10 x 210= 105 Rs. So there are 105 coins of one rupee, Similarly value of 50 paise coins will be 3/10 x 210= 63 Rs. So there are 126 coins of 50 paise and so on.

Or it can be find the numbers which when added to the terms of the ratio 11:23 makes it equal to the ratio 4:7. Here we find a number that we add to the terms so that it becomes a multiple of 4 and then check. Adding 1 to 11 and 23 makes it equal to 12/24 but the ratio is not 1:2 . Then we add 5 to 11:23 and find that the numbers become 16:28 or the ratio becomes 4:7 so the number that should be added is 5.

Most of the questions in ratio and proportions can be solved by this method.

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