Wednesday, June 1, 2011

Percentages without Tears

In most of the exams like Bank P.O, SSC, CAT or NIFT, questions are asked on percentages. Here is a layman's guide to solving these problems.

A fraction whose denominator is 100 is called a percentage and the numerator is called the rate percent.

Thus 60% of x is 60/100 of x. Where 60 is the rate percent.

It will be useful to remember the following percentages and the corresponding fractions in order to solve the problems:

1=100%, 1/2= 50%, 1/3= 33.33%, 1/4= 25%, 1/5= 20%, 1/6= 16.66%, 1/7= 14.28%, 1/8= 12.5%, 1/9= 11.11%, 1/10= 10% and 1/11= 9.09% 

Example 1: A man Pays 33 1/3% of his salary in income tax. How much does he pay if his income is 3516. In this case his income will be 3516/3=1172

Then there are some usual problems in percentages in which we assume the base number to be 100 and solve the problems. eg. Two numbers are respectively 20% and 50% more than the third. What percentage is the first of the second.Here we assume that the third number is 100 so the required percentage is 120 x 100/150 = 80%. or if two numbers are respectively 20% and 25% of the third number than what % is the first of the second.It will be 20 x 100/25 = 80%

Of course one can use decision tree method to calculate the values. For example if a man loses 12 1/2 % of the money and, after spending 70% of the remainder, is left with Rs. 210. How much had he at first. To solve this problem we go like the following diagram.

Thus let us say he had M money at first he loses 12 1/2 % of the money. It means that he loses 1/8. Thus he is left with 7/8. Then he spent 70% of the remainder (i.e. 7/8) thus he is left with 30% or 3/10. Then our chain goes like this  M x 7/8 x 3/10 = 210.   

Then we have Problems in which one quantity is equal to the product of other two. eg. questions in which distance = speed x time, Expenditure = price x consumption. The questions are often asked as if price are increased by x% how much should be the reduction in consumption so as not to increase the expenditure etc. For example if price of a commodity increases by 20% what should be the reduction in consumption so as not to increase the expenditure. In such case we go like this 20% increase means +1/5  increase thus reduction in consumption will be derived from +1/5. To numerator will remain as such i.e. 1 , to find the denominator we take 5 add the + sign and take the numerator thus 5+1=6. Thus decrease in consumption is 1/6 or 16.66%.

The same method is used in solving such problems as if A's salary is 25% less than B, how much percentage is B's salary more than A. Thus A's Salary is -1/4 less than B so B's salary is +1/3 or 33.33% more than A.

If the questions involve such as if price are increased by x % and consumption is reduced by y% what should be the percentage increase (decrease) in expenditure we can use the formula (+(-)x+(-)y+(-) xy/100)% if the sign is positive there is an increase and if negative, there is a decrease.The sign of x and y vary if value are increasing or decreasing. For increase we take positive sign and for decrease we take negative sign.

We can use this method in many situations. For example if a shop keeper marks the price of the goods 12% higher than he allows a discount of 12% the percentage change in profit will be (12-12-12 x 12/100)% ie.1.44 percent decrease.

Similary in questions like A's salary first increases by 10% and then decreased by 5 % the net change in salary will be (10-5-10 x 5/100) percent i.e. 4.5% increase.

Or if one side of a rectangle is first increased by 20% and the other decreased by 5% then the net change in area will be (20-5-20 x 5/100) i.e. 14% increase. In fact for any two-d figure if all the sides are increased by x % and we are asked to find the change in area, we can assume that it is a rectangle and where length is also increased by x% and breadth is also increased by x% and solve the problem. For example if in a hexagon if all the sides are increased by 10% then change in area will be ( 10+10+10x10/100)% = 21% increase

We can also solve the questions on percentage by comparing directly. For example pass marks in an exam is 80%. If a candidate who scores 210 marks fails by 50 marks then to find the maximum marks we go like this Pass marks = 210+50 but these are 80% so 80--> 260 then 100--> (80/260 )x100= 325 marks. Similarly if a candidate scoring 25% in an examination fails by 20 marks while other candidate scoring 50% marks gets 30 marks more than the pass marks then to find the maximum marks, we have difference in percentage is 25% and difference in marks is 50 so 25--> 50 then 100--> (50/25) x 100 

Solving Problems in Averages without Tears

Average is defined as the Total of data/ Number of Data

Sometimes average of two groups of quantities is given and we are asked to find the combined average. For example The average age of 40 students in section A is 10 years and the average age of students in section B of 30 students is 12 years. Find the average age of the students in both the sections taken together. To solve such problems we go like this: Suppose the assumed average age is 10 years. Total excess over assumed average will be 30 x (12-10) = 60 years for second group.These 60 years will be divided in all 70 (40+30) students. So increase in average will be 60/70= .85 years. So actual average will be 10.85 years ( 10+.85 years).

Or there may be questions where the combined average and average of one group is given and we are asked to find the average of other group. For example if average of 5 quantities is 6. The average of three of them is 4 then what will be average of remaining 2 numbers. Here average of 5 quantities is  and average of 3 quantities is 4. The total deficit in 3 quantities is 3 x (6-4)= +6. Total increase in the remaining 2 numbers is +6/2= 3. Therefore average expenditure of remaining 2 numbers is 6+3=9.

How to solve such problems in averages when some members are added or subtracted and the average increases or decreases. For example, the average weight of 29 students in a class is 48 kg. If the weight of teacher is included, the average weight rises by 500 gms. Find the weight of the teacher. Here total increase in weight after teacher is added is .500 x 30= 15 kg. So teacher's weight must be more than 48 kg by 15 kg. Because only teacher was responsible for increase in weight. So teacher's weight must be 48+15=63 kg.

Or a batsman has certain average runs for 20 innings. In the 21st innings, he made 107 runs, thereby increasing his average by 2. What is his average after 21 innings. In this case, after 21 innings his average increase by 2. Therefore total runs that are more than average will be 21 x 2 = 42. Therefore old average must be 107-42=65 and new average will be more than 2 ie. 65+2=67.

In questions where a group of quantities is replaced by another group of quantities and average we proceed as follows: (sum of) replacing quantities - ( sum of ) replaced quantities = change in average x total number of quantities in the group. For example the average temperature of June, July and August was 31 deg C. The average temperature of July, August and September was 30 deg C. If the temperature of June was 29 deg C, find the temperature of September.Here Temperature of September ( Replacing Qty)- 29 ( Replaced)= -1 x 3 which means Temperature of September= 29-3=26 deg C.

Finally the questions are asked about the mid term quantity. For example average of 11 observations is 50 and average of first 6 is 49 and average of last 6 is 52, what is the 6th quantity. In such questions the decrease in total for 1st 6 observations is -1 x 6= -6. and increase in the total for last 6 observations is +2 x 6= 12 from average therefore total change is 12-6=+6, therefore 6th quantity= 50+6=56.

I hope this theoretical discussion will be able to see you through from the first level of questions in average.

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How to Solve Problems on Ages without Tears--

Problems on age are similar to those of Ratio and Proportion. However in certain questions,calculations can be shortened using the concepts of Singapore Maths. For example the age of the father 3 years ago was 7 times that of his son. At present the father's age is 5 times that of his son than to find the ages of father and son we proceed as follows:

F: S
7:1  <-- 3 Years Ago
5:1  <-- Present

Here if the ratios were the actual ages then Present ratio should be more than that of 3 years ago. Also the same years would be added to father's age and son's age. So what we do here is we make the ratio of present age to be more than that of 3 years ago  and also make the difference of the ratio to be equal. To do that we find the difference of the ratios and cross multiply as:

F: S
7:1  <-- Difference is 6
5:1<--- Difference is 4 Thus the ratio of difference is 6:4 or 3:2, cross multiplying the upper ratio by 2 and lower by 3, we have:

F: S
14:2  <-- three years ago
15:3  < --Present
1 :1
Here we see that the difference is 1: 1 which is same and also the present age is more than that of 3 years ago.
Now the calculations is very simple :1 year difference in ratio--> actual 3 years difference so Father's age that is 15 present ---> actual will be 3/1 x 15=45 years and son's age will be 3/1 x 3 = 9 years

This concept becomes very handy when we solve similar problems in ratio and proportions.

Similarly if Three years earlier the father's age was 7 times as old as his son. Three years hence, the father's age would be four times that of his son. What are the present ages of the father and the son. We go like this:

F: S
7:1  <- Three years earlier
4:1  <-- Three years hence

To make the difference same we cross multiply by the difference

F: S
7:1  <--6~2 in ratio
       \  /  
        / \
4:1  <--3~1 in ratio

F: S
7:1 <-- -3
8:2<-- +3
Now if difference in ratio is 1--> Actual difference is 3+3=6 years. If the age 3 years earlier is 7 in ratio--> actual age is 6/1 x 7= 42. So present age is 42+3=45. similarly 1-->6 then 1-->6 so present age of son is 6+3=9 years.

The concept is very simple and gives the answer almost intuitively.

Sometimes the questions are asked like this : The sum of ages of a mother and her daughter is 50 years. Also 5 years ago, the mother's age was 7 times the age of the daughter. What are the present ages of the mother and the daughter. Here 5 years ago, the sum of the ages will be 40 years. Divide 40 in the ratio of 7:1, Thus 8~40 so 7~40/8 x 7 =35 years and 1~ 40/8= 5 years. So mothers age today is 35+5= 40 years and daughter's age today is 5+5=10 years.  

Most of the problems on ages fall into one or more of these categories

You can now evaluate yourself with this test.

How to Solve Problems of Ratio and Proportions

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


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:


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


0: 1 <-- Difference in ratios.


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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