Percentage : Percentage By percentage we many hundredths. Thus, x per cent means x hundredths, written as x%
For eg:
9%=9/100
16%=16/100=4/25
To convert a fraction in percentage
3/4=(3/4*100)%=75%
0.2=2/10=(2/10*100)%=20%
Various types of sums : Various types of sums Express as fraction
1)64%
2)0.5%
Express as decimal
1)36%
=36/100=0.36
2)0.3%
Express as rate per cent : Express as rate per cent For eg:
1)2/3
2/3=(2/3*100)%=66.66%
2)0.002=? %
Find the value of:
1)70% of 70
70/100*70=49
2)3% of 6
Fill in the blanks : Fill in the blanks 1)(…?...)% of 64 is 8
we assume it as x
so, x/100*64=8
x=(8*100)/64=12.5
2)(…?...)% of 24 is 0.72
3)What per cent is 120 of 90?
Let x% is 120 of 90.
So, x/100*90=120
x=(120*100)/90=133.33%
Slide 5 : 4) What per cent is 5mg of 1 kg?
Queries based on income
Rules
1)If a’s income is r%more than b’s income,then b’s income is less than a’s income by [r/(100+r)*100]%
2)If a’s income is r% less than b’s income,then b’s income is more than a’s income by [r/(100-r)*100]%
er : er For eg:
1)If A’s salary is 50% more than that of B,then how much percent is B’s salary less than that of A?
Sol: B’s salary is less than that of A by
[r/(100+r)*100]%
=[(50/150)*100]%
=33.33%
For you to solve:
2)If A’s salary is 30% less than that of B, then how much percent is B’s salary more than that of A?
l : l Rule: If the price of a commodity increases by r%,then reduction in consumption so as not to increase the expenditure, is [r/(100+r)*100]%
For eg:
1)If the price of tea is increased by 20%,find by how much percent a householder must reduce her consumption of tea so as not to increase the expenditure?
Sol: reduction in consumption=[r/(100+r)*100]%
=[20/(120)*100]%
=16.66%
Slide 8 : 2)If the price of sugar falls down by 10%, by how much per cent must a householder increase its consumption, so as not to decrease expenditure on this item?
Promblems on population
Formulae:
1) If the population of town is P and its annual increase is r%,then:
t : t Population after n years= P(1+(r/100))^n
Population n years ago=p/(1+(r/100))^n
If the population is P and annual decrease is r%,then:
Population after n years=P(1+(r/100))^n
Population n years ago=P/(1+r/100))^n
For eg:
1) The population of a town is 176400.It increases annually at the rate of 5% per annum. What will be its population after 2 years? What it was 2 years ago?
Slide 10 : Population after 2 years=[17640*(1+5/100)^2]
=17640*21/20*21/20
=194481
Population 2 years ago=17640/(1+5/100)^2
=17640*20/21*20/21
=16000
For you to solve:
The value of a machine depreciates at the rate of 10% per annum. If its present value is Rs.81000,what will be its worth after 3 years? What was its value 2 years ago?