05 December, 2015

Percentage Formulas, Shortcuts and Problems

Percentage Basic Concepts:

  • Percentage means per 100 or for every 100. i.e. X % of Y = (X/Y)*100 
          Ex:  1. 20% = 20/100 = 1/5
  • Let the population of a town be P now and suppose it increases at the rate of R% per annum, then:
          Population after 'n' years = P[1 + (R/100)]
          Population 'n' years ago = P/[1 + (R/100)]
  • If the price of the commodity increases by R %, then the reduction in consumption so as not to increase the expenditure is:  {[R/(100 + R)] * 100}%

  • If the price of the commodity decreases by R %, then the increase in consumption so as not to decrease the expenditure is:  {[R/(100 - R)] * 100}%

  • Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum.
          Value of the machine after 'n' years = P[1 - (R/100)]n
          Value of the machine n years ago    = P/[1 - (R/100)]n
Problems on Percentages:
1. Find 20 % of 6200 ?
A) 1200                 B) 1220                   C) 1240                 D) 1250


Ans.C
Solution:
20 % of 6200 = (20/100)*6200 = 1240.

2. 40 is what percentage of 20 ?
A) 50                    B) 150                     C)200                     D)250


Ans.C
Solution:
Let required percentage is 'X';
20*(X/100) = 40;
X = (40 * 100)/20;
X = 200 %.

3. 12 % of a certain sum of money is Rs. 43.5. Find the sum ?
A) Rs. 362.5         B) 364.5                 C) 365.0                 D) 366.5


Ans.A
Solution:
Let required sum is 'X';
X*(12/100) = 43.5;
X = (43.5 * 100)/12;
X = Rs. 362.5.

4. 16 % of 40 % of a value is 8. Then find the value ?
A) 75                    B) 100                    C) 125                    D) 150


Ans.C
Solution:
Let required Number is 'X';
(16/100)*(40/100)* X = 8;
X = (8 * 100 * 100)/(16 * 40);
X = 125.

5. The population of a town has increased from 60,000 to 65,000. Find the increase percent ?
A) 22/3                B) 25/3                    C) 50/7                  D) 52/7


Ans.B
Solution:
Increased Percentage = [5,000/6,0000]*100
Increased Percentage = 100/12
Increased Percentage = 25/3.

6. In an election of two candidates, the candidate who gets 41 % is rejected by a majority of 2412 votes. Find the total no. of votes polled ?
A) 13,100            B) 13,200                C) 13,300              D) 13,400


Ans.D
Solution:
59 % - 41 % = 18 % = 2412;
(18/100) * X = 2412;
X = (2412 * 100)/18;
X = 13,400.

7. If the income of Rajesh is more than Suresh by 25 %. Then how much percent of Suresh income is less than that of Rajesh ?
A) 5 %                B) 10 %                   C) 15 %                 D) 20 %


Ans.D
Solution:
Let the income of suresh = Rs. 100
Then Rajesh income = Rs. 125
X = (25/125)*100;
X = 20 %.

8. Prakash scored 60 % marks in an exam and Raghu scored 62 % of marks. If one of them has got 5 marks more than other. Find maximum marks ?
A) 150                B) 200                     C) 250                   D) 300


Ans.C
Solution:
62 % - 60 % = 5;
2 % = 5;
100 % = Maximum Marks = X;
X = (100/2)*5;
X = 250.

9. An ore contain 15 % of iron. How many kg of the ore are required to get 72 kg of iron ?
A) 480                B) 460                     C) 380                   D) 490


Ans.A
Solution:
(72/X)*100 = 15;
X = (72 * 100)/15;
X = 480 Kg.

10. If the length of a rectangle is increased by 30 % and its breadth is decreased by 20 %. Then what is effected of its area ?
A) 2 % Increase                         B) 4 % Increase
C) 2 % Decrease                        D) 4 % Decrease


Ans.B
Solution:
Consider that length = 100, Breadth = 50;
Area 1 = length * breadth;
Area 1 = 100 * 50;
Area 1 = 5,000;
If the length is increased by 30 % then length = 130;
If the breadth is decreased by 20 % then breadth = 40;
Area 2 = 130 * 40;
Area 2 = 5,200;
Effected Area = (200/5000)*100;
Effected Area = 4 % Increase.

           [OR]
Short Cut Method:
Effected Area = X + Y + (XY/100);
Effected Area = 30 - 20 + (30(-20))/100;
Effected Area = 10 - (600/100);
Effected Area = 4 % Increase.

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