2. If the price of a commodity increases by R%, then the reduction in consumption so asnot 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 to decrease the expenditure is [(R/(100-R)*100]%.

3.Results on Population : Let the population of the town be P now and suppose it increases at the rate of R% per annum, then :

1.Population after nyeras = P [1+(R/100)]^n. 2.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 R% per annum. Then, 1.Value of the machine after n years = P[1-(R/100)]n. 2.Value of the machine n years ago = P/[1-(R/100)]n.

5.If A is R% more than B, then B is less than A by [(R/(100+R))*100]%. If A is R% less than B , then B is more than A by [(R/(100-R))*100]%.

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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; 48% =48/100 =12/25, etc.

To express a/b as a percent :We have, a/b =((a/b)*100)%.Thus, 1/4 =[(1/4)*100] = 25%; 0.6 =6/10 =3/5 =[(3/5)*100]% =60%.

2. If the price of a commodity increases by R%, then the reduction in consumption so asnot 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 to decrease the expenditure is

[(R/(100-R)*100]%.

3.Results on Population :Let the population of the town be P now and suppose it increases at the rate ofR% per annum, then :

1.Population after nyeras = P [1+(R/100)]^n.

2.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 rateR% per annum. Then,

1.Value of the machine after n years = P[1-(R/100)]n.

2.Value of the machine n years ago = P/[1-(R/100)]n.

5.If A is R% more than B, then B is less than A by

[(R/(100+R))*100]%.

If A is R% less than B , then B is more than A by

[(R/(100-R))*100]%.