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Percentage

Percentage -> IMPORTANT FACTS AND FORMULAE

Concepts of Percentage : By a certain percent, we mean that many hundredths. Thus, x percent means x hundredths, written as x%.
To express x% as a fraction : we have, x% = x/100.
To express a/b as a percent : we have, a/b = (a/b×100)% .
Results on population : 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]n.
Population n years = p / [1+R/100]n.

Percentage -> SOLVED EXAMPLES

1. Express each of the following as a decimal.
28%
  Sol. = 28/100 = 0.28%.
2. Express each of the following as a decimal.
0.04%
  Sol. = 0.04/100 = 0.0004%.
3. Express each of the following as a fraction.
4%
  Sol. = 4/100 = 1/25.

Percentage -> EXERCISE

4. Entry fee in an examination was Re. 1. Later, this was reduced by 25% which increased the sale by 20%. The percentage increase in the number of visitors is
 
  • A. 60
  • B. 50
  • C. 66
  • D. 54
Ans: A.
Sol.
Let the total original sale be Rs. 100. Then, original number of visitors = 100.
New number of visitors = 120 / 0.75 = 160.
∴ Increase% = 60%.
 
5. In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got was
 
  • A. 2700
  • B. 2900
  • C. 3000
  • D. 3100
Ans: A.
Sol.
Number of valid votes = 80% of 7500 = 6000.
Valid votes polled by other candidate = 45% of 6000
= [45/100 x 6000] = 2700.
 
 
6. From a container having pure milk, 20% is replaced by water and the process is repeated thrice. At the end of the third operation, the milk is
 
  • A. 40% pure
  • B. 50% pure
  • C. 51.2% pure
  • D. 58.8% pure
Ans: C.
Sol.
Let total quantity of original milk = 1000 gm.
Milk after first operation = 80% of 1000 = 800 gm.
Milk after second operation = 80% of 800 = 640 gm.
Milk after third operation = 80% of 640 = 512 gm.
∴ Strength of final mixture = 51.2%.