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