Question:Express each of the following as a fraction:

(i) 56%
(i) 4%
(i) 0.6% 

(i) 56% =`56/100` =`14/25`
(ii) 6% = `6/100` = 0.06
(iii) 0.6% = `0.6/100` =`3/500`

Question:Express each of the following as a decimal:

(i) 28%
(ii) 6%
(iii) 0.2%
(iv) 0.04% 

(i) 28% = `28/100`=0.28

(ii) 6% = `6/100` =0.06

(iii) 0.2% = `0.2/100` =0.002

(iv) 0.04% = `0.04/100` = 0.0004

Question:Express each of following as rate per cent:

(i) `23/36`  (ii) .004  (iii) `6 3/4` 

(i) `23/36` = `(23/36 xx 100)`% = `(575/9)`% = `63 8/9` %.

(ii) .004 = `4/1000` = `(4/1000 xx 100)`% = .4 %.

(iii) `6 3/4` = `27/4` = `(27/4 xx 100)`% = 675 %.

Question:(i) What percent of 72 is 6 ?
(ii) What percent of 7 is 84 ?
(iii) What percent of 2.4 kg is 8 gms ?
(iv) What percent of 6.5 litres is 130 ml ? 

(i) Required percentage = `(6/72 xx 100)`% = `8 1/3`%.

(ii) Required percentage = `(84/7 xx 100)`% = 1200%.

(iii) Required percentage = `(8/(2.4 xx 1000) xx 100)`% = `1/3`%.

(iv) Required percentage = `(130/(6.5 xx 1000) xx 100)`% = 2%.

Question:If A's income is 30% more than that of B, then how much percent is B's income less than that of A ? 

Required percentage = `{33/(100 + 33) xx 100}` % = 24.8%.

Question:If A's height is 24% less than that of B, then how much percent is B's height more than that of A ? 

Required percentage 

   = `{24/(100-24) xx 100}`% 

   = 31.57%.

Question:If the price of tea is increased by 20%, find by how much percent must a householder reduce her consumption of tea so as not to increase the expenditure ? 

Reduction in Consumption 

 = `{ R/(100+R) xx 100}` % 

 =`(20/120 xx 100)` % 

 = `16 2/3` %.

Question:If the price of sugar decreases by 20%,find by how much percent must a house holder increase her consumption of sugar so as not to decrease the expenditure ? 

Increase in consumption 

 = `{R/(100-R xx 100)}` %

 = `(20/80 xx 100)` % 

 = 25 %.

Question:The population of a town is 176400. If it increases at the rate of 5% per annum, what will be its population 2 years hence ?
What was it 2 years ago ? 

Population after 2 years

 = `176400 xx (1 +5/100)^2`

 = `(176400 xx 21/20 xx 21/20)` 

 = 194481.

 Population 2 years ago 

 = `176400/(1 +5/100)^2` 

 = `(176400 xx 20/21 xx 20/21)`

 = 160000.

Question:The value of a machine depreciates at the rate of 10% per annum. If its present value is Tk. 162000,What will be its worth after 2 years ?What was the value of the machine 2 years ago ? 

Value of the machine after 2 years  

 = `Tk. [162000 xx (1 -10/100)^2]`

 = `Tk.(162000 xx 9/10 xx 9/10)` 

 = Tk. 131220.

 Value of the machine 2 years ago 

 = `Tk.{162000/(1 -10/100)^2}` 

 = `Tk (162000 xx 10/9 xx 10/9)` 

 = Tk. 200000.