1) Concept 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
Thus, 20% = 20/100 = 1/5
To express a/b as a percent: We have, (a/b)=((a/b)*100)%.
Thus, 1/4 = ((1/4)*100)% = 25%.
2. Percentage Increase/Decrease:
If the price of a 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 a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is:
[((R)/(100-R))*100]%
3. Results on Population:
Let the population of a town be P now and suppose it increases at the rate of R% per annum, then:
i) Population after n years = P(1+(R/100))^n
ii) Population n years ago = (P)/(1+(R/100))^n
4. Results on Depreciation:
Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. Then:
i) Value of the machine after n years = P(1-(R/100))^n
ii) Value of the machine n years ago = (P)/(1-(R/100))^n
iii) If A is R% more than B, then B is less than A by [((R)/(100+R))*100]%
iv) If A is R% less than B, then B is more than A by [((R)/(100-R))*100]%
Also read:
Percentage Problems Shortcut and Tricks for Competitive Exams Free PDF Download
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
Thus, 20% = 20/100 = 1/5
To express a/b as a percent: We have, (a/b)=((a/b)*100)%.
Thus, 1/4 = ((1/4)*100)% = 25%.
2. Percentage Increase/Decrease:
If the price of a 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 a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is:
[((R)/(100-R))*100]%
3. Results on Population:
Let the population of a town be P now and suppose it increases at the rate of R% per annum, then:
i) Population after n years = P(1+(R/100))^n
ii) Population n years ago = (P)/(1+(R/100))^n
4. Results on Depreciation:
Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. Then:
i) Value of the machine after n years = P(1-(R/100))^n
ii) Value of the machine n years ago = (P)/(1-(R/100))^n
iii) If A is R% more than B, then B is less than A by [((R)/(100+R))*100]%
iv) If A is R% less than B, then B is more than A by [((R)/(100-R))*100]%
Also read:
Percentage Problems Shortcut and Tricks for Competitive Exams Free PDF Download
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