Wednesday, January 29, 2014
PERCENTAGE
IMPORTANT FACTS AND FORMULAE
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, ¼
=[(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 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]%.
SOLVED
EXAMPLES
Ex. 1. Express each of the
following as a fraction :
(i) 56% (ii) 4% (iii) 0.6% (iv) 0.008%
sol. (i) 56% = 56/100= 14/25. (ii) 4% =4/100 =1/25.
(iii) 0.6 =6/1000 = 3/500. (iv) 0.008 = 8/100 = 1/1250.
Ex. 2. Express each of the
following as a Decimal :
(i) 6% (ii)28% (iii) 0.2% (iv)
0.04%
Sol. (i) 6% = 6/100 =0.06. (ii) 28% = 28/100 =0.28.
(iii) 0.2% =0.2/100 = 0.002. (iv) 0.04%= 0.04/100 =0.004.
Ex. 3. Express each of the
following as rate percent :
(i) 23/36 (ii) 6 ¾ (iii) 0.004
Sol. (i) 23/36 = [(23/36)*100]% = [575/9]% = 63
8/9%.
(ii) 0.004 = [(4/1000)*100]% = 0.4%.
(iii) 6 ¾ =27/4 =[(27/4)*100]% = 675%.
Ex. 4. Evaluate :
(i)
28% of 450+ 45% of 280
(ii)
16 2/3% of 600 gm- 33 1/3%
of 180 gm
Sol. (i) 28% of 450 + 45% of 280 =[(28/100)*450 + (45/100)*280] =
(126+126) =252.
(iii)
16 2/3% of 600 gm –33 1/3% of 180 gm = [ ((50/3)*(1/100)*600) – ((100/3)*(1/3)*280)]gm
= (100-60) gm = 40gm.
Ex. 5.
(i) 2 is what percent of 50
?
(ii) ½ is what percent of
1/3 ?
(iii)What percent of 8 is 64
?
(iv)What percent of 2 metric
tones is 40 quintals ?
(v)What percent of 6.5
litres is 130 ml?
Sol.
(i)
Required Percentage = [(2/50)*100]% =
4%.
(ii)
Required Percentage = [ (1/2)*(3/1)*100]% = 150%.
(iii)Required
Percentage = [(84/7)*100]% = 1200%.
(iv)
1 metric tonne = 10 quintals.
Required percentage = [ (40/(2 * 10)) * 100]% =
200%.
(v)
Required Percentage = [
(130/(6.5 * 1000)) * 100]% = 2%.
Ex. 6.
Find the missing figures :
(i) ?% of 25 = 20125 (ii) 9% of ? = 63 (iii) 0.25% of ? = 0.04
Sol.
(i)
Let x% of 25 = 2.125. Then , (x/100)*25 = 2.125
X = (2.125 * 4) = 8.5.
(ii)
Let 9% of x =6.3. Then , 9*x/100 = 6.3
X =
[(6.3*100)/9] =70.
(iii)
Let 0.25% of x = 0.04. Then , 0.25*x/100 = 0.04
X=
[(0.04*100)/0.25] = 16.
Ex. 7.
Which is greatest in 16 (
2/3) %, 2/5 and 0.17 ?
Sol. 16 (2/3)% =[ (50/3)*
)1/100)] = 1/6 = 0.166, 2/15 = 0.133. Clearly, 0.17 is the greatest.
Ex. 8.
If the sales tax reduced from
3 1/2 % to 3 1/3%, then what
difference does it make to a person who purchases an article with market price
of Rs. 8400 ?
Sol. Required difference =
[3 ½ % of Rs.8400] – [3 1/3 % of Rs.8400]
= [(7/20-(10/3)]% of Rs.8400
=1/6 % of Rs.8400
= Rs. [(1/6)8(1/100)*8400] = Rs. 14.
Ex. 9. An inspector rejects 0.08% of the meters as defective. How many
will be examine to project ?
Sol. Let the number of
meters to be examined be x.
Then, 0.08% of x =2
[(8/100)*(1/100)*x] = 2
x = [(2*100*100)/8] = 2500.
Ex. 10. Sixty five percent of a number is 21 less than four fifth of that
number. What is the number ?
Sol. Let the number be x.
Then, 4*x/5 –(65% of x) = 21
4x/5 –65x/100 = 21
5 x = 2100
x = 140.
Ex.11. Difference of two numbers is 1660. If 7.5% of the
number is 12.5% of the other number , find the number ?
Sol. Let the numbers be x
and y. Then , 7.5 % of x =12.5% of y
X = 125*y/75 = 5*y/3.
Now, x-y =1660
5*y/3 –y =1660
2*y/3= 1660
y =[ (1660*3)/2] =2490.
One number = 2490, Second number
=5*y/3 =4150.
Ex. 12.
In expressing a length 810472 km as nearly as possible with three
significant digits , find the percentage error.
Sol. Error = (81.5 –
81.472)km = 0.028.
Required percentage = [(0.028/81.472)*100]% = 0.034%.
Ex. 13.
In an election between two candidates, 75% of the voters cast thier thier
votes, out of which 2% of the votes were declared invalid. A candidate got 9261
votes which were 75% of the total valid votes. Find the total number of votes
enrolled in that election.
Sol.
Let the number of votes enrolled be x. Then ,
Number of votes cast =75% of x.
Valid votes = 98% of (75% of x).
75% of (98% of (75%of x)) =9261.
[(75/100)*(98/100)*(75/100)*x]
=9261.
X = [(9261*100*100*100)/(75*98*75)]
=16800.
Ex.14. Shobha’s mathematics
test had 75 problems i.e.10 arithmetic, 30 algebra and 35 geometry problems.
Although she answered 70% of the arithmetic ,40% of the algebra, and 60% of the
geometry problems correctly. she did not
pass the test because she got less than
60% of the problems right. How many more questions she would have to
answer correctly to earn 60% of the passing grade?
Sol. Number of questions attempted correctly=(70% of 10 +
40% of 30 + 60% 0f 35)
=7 + 12+21= 45
questions
to be answered correctly for 60% grade=60% of 75 = 45
therefore
required number of questions= (45-40) = 5.
Ex.15.
if 50% of (x-y) = 30% of (x+y) then what percent of x is y?
Sol.50% of (x-y)=30% of(x+y) ó (50/100)(x-y)=(30/100)(x+y)
ó5(x-y)=3(x+y) ó 2x=8y ó x=4y
therefore required percentage
=((y/x) X 100)% = ((y/4y) X 100) =25%
Ex.16.
Mr.Jones gave 40% of the money he had to his wife. he also gave 20% of the
remaining amount to his 3 sons. half of the amount now left was spent on
miscellaneous items and the remaining amount of Rs.12000 was deposited in the
bank. how much money did Mr.jones have initially?
Sol.
Let the initial
amount with Mr.jones be Rs.x then,
Money given to wife=
Rs.(40/100)x=Rs.2x/5.Balance=Rs(x-(2x/5)=Rs.3x/5.
Money given to 3 sons= Rs(3X((20/200) X
(3x/5)) = Rs.9x/5.
Balance = Rs.((3x/5) –
(9x/25))=Rs.6x/25.
Amount deposited in bank= Rs(1/2 X
6x/25)=Rs.3x/25.
Therefore 3x/25=12000 ó x= ((12000 x 35)/3)=100000
So Mr.Jones initially had Rs.1,00,000
with him.
Short-cut Method : Let the initial amount with Mr.Jones be Rs.x
Then,(1/2)[100-(3*20)]% of x=12000
ó (1/2)*(40/100)*(60/100)*x=12000
óx=((12000*25)/3)=100000
Ex 17 10% of the inhabitants of village having died
of cholera.,a panic set in , during which 25% of the remaining inhabitants left
the village. The population is then reduced to 4050. Find the number of
original inhabitants.
Sol:
Let the total number of orginal
inhabitants be x.
((75/100))*(90/100)*x)=4050 ó
(27/40)*x=4050
óx=((4050*40)/27)=6000.
Ex.18 A salesman`s commission is 5% on all sales
upto Rs.10,000 and 4% on all sales exceeding this.He remits Rs.31,100 to his
parent company after deducing his commission . Find the total sales.
Sol:
Let his total sales be
Rs.x.Now(Total sales) – (Commission )=Rs.31,100
x-[(5% of 10000 + 4% of (x-10000)]=31,100
x-[((5/100)*10000 +
(4/100)*(x-10000)]=31,100
óx-500-((x-10000)/25)=31,100
óx-(x/25)=31200 ó 24x/25=31200óx=[(31200*25)/24)=32,500.
Total sales=Rs.32,500
Ex .19 Raman`s salary was decreased by 50% and subsequently increased by
50%.How much percent does he lose?
Sol:
Let the original salary = Rs.100
New final salary=150% of (50% of Rs.100)=
Rs.((150/100)*(50/100)*100)=Rs.75.
Decrease = 25%
Ex.20 Paulson spends 75% of his income. His income is increased by 20% and
he increased his expenditure by 10%.Find the percentage increase in his savings
.
Sol:
Let the original income=Rs.100 .
Then , expenditure=Rs.75 and savings =Rs.25
New income =Rs.120 , New
expenditure =
Rs.((110/100)*75)=Rs.165/2
New savings = Rs.(120-(165/2))
= Rs.75/2
Increase in savings =
Rs.((75/2)-25)=Rs.25/2
Increase %= ((25/2)*(1/25)*100)% =
50%.
Ex21. The salary of a person was reduced by 10% .By
what percent should his reduced salary be raised so as to bring it at par with
his original salary ?
Sol:
Let the original salary be Rs.100 .
New salary = Rs.90.
Increase on 90=10 , Increase on
100=((10/90)*100)%
= (100/9)%
Ex.22 When the price fo a product was decreased by
10% , the number sold increased by 30%. What was the effect on the total
revenue ?
Sol:
Let the price of the product be
Rs.100 and let original sale be 100 pieces.
Then , Total Revenue =
Rs.(100*100)=Rs.10000.
New revenue = Rs.(90*130)=Rs.11700.
Increase in revenue =
((1700/10000)*100)%=17%.
Ex 23 . If the numerator of a fraction be increased
by 15% and its denominator be diminished by 8% , the value of the fraction is
15/16. Find the original fraction.
Sol:
Let the original fraction be x/y.
Then (115%of x)/(92% of y)=15/16
=> (115x/92y)=15/16
ð
((15/16)*(92/115))=3/4
Ex.24 In the new budget , the price of kerosene oil
rose by 25%. By how much percent must a person reduce his consumption so that
his expenditure on it does not increase
?
Sol:
Reduction in consumption = [((R/(100+R))*100]%
ð
[(25/125)*100]%=20%.
Ex.25 The population of a town is 1,76,400 . If it increases at the rate
of 5% per annum , what will be its population 2 years hence ? What was it 2
years ago ?
Sol:
Population after 2 years =
176400*[1+(5/100)]^2
=[176400*(21/20)*(21/40)]
= 194481.
Population 2 years ago =
176400/[1+(5/100)]^2
=[716400*(20/21)*(20/21)]= 160000.
Ex.26 The value of a machine depreiates at the rate
of 10% per annum. If its present is Rs.1,62,000 what will be its worth after 2
years ? What was the value of the machine 2 years ago ?
Sol.
Value of the machine after 2 years
=Rs.[162000*(1-(10/100))^2] =
Rs.[162000*(9/10)*(9/10)]
=Rs. 131220
Value of the machine 2 years ago
=
Rs.[162000/(1-(10/100)^2)]=Rs.[162000*(10/9)*(10/9)]=Rs.200000
Ex27. During one year, the population of town increased by 5% . If the total population
is 9975 at the end of the second year , then what was the population size in
the beginning of the first year ?
Sol:
Population in the beginning of the
first year
= 9975/[1+(5/100)]*[1-(5/100)] =
[9975*(20/21)*(20/19)]=10000.
Ex.28 If A earns 99/3% more than B,how much percent
does B earn less then A ?
Sol:
Required Percentage =
[((100/3)*100)/[100+(100/3)]]%
=[(100/400)*100]%=25%
Ex. 29 If A`s
salary is 20% less then B`s salary , by how much percent is B`s salary more
than A`s ?
Sol:
Required percentage =
[(20*100)/(100-20)]%=25%.
Ex30 .How many kg of pure salt must be added to 30kg
of 2% solution of salt and water to increase it to 10% solution ?
Sol:
Amount of salt in 30kg solution =
[(20/100)*30]kg=0.6kg
Let x kg of pure salt be added
Then , (0.6+x)/(30+x)=10/100ó60+100x=300+10x
ó90x=240 ó x=8/3.
Ex 31. Due to reduction of 25/4% in
the price of sugar , a man is able to buy 1kg more for Rs.120. Find the
original and reduced rate of sugar.
Sol:
Let the original rate be Rs.x per kg.
Reduced rate =
Rs.[(100-(25/4))*(1/100)*x}]=Rs.15x/16per kg
120/(15x/16)-(120/x)=1 ó (128/x)-(120/x)=1
ó x=8.
So, the original rate = Rs.8 per kg
Reduce rate = Rs.[(15/16)*8]per kg = Rs.7.50 per kg
Ex.32 In an examination , 35% of total students failed in Hindi , 45% failed in English and 20% in both . Find the percentage of those who passed in both subjects .
Sol:
Let A and B be the sets of students who failed in
Hindi and English respectively .
Then , n(A) = 35 , n(B)=45 , n(AÇB)=20.
So , n(AÈB)=n(A)+n(B)- n(AÇB)=35+45-20=60.
Percentage failed in Hindi and English or both=60%
Hence , percentage passed = (100-60)%=40%
Ex33. In an
examination , 80% of the students passed in English , 85% in Mathematics and
75% in both English and Mathematics. If 40 students failed in both the subjects
, find the total number of students.
Sol:
Let the total number of students be x .
Let A and B represent the sets of students who
passed in English and Mathematics respectively .
Then , number of students passed in one or both the
subjects
= n(AÈB)=n(A)+n(B)- n(AÇB)=80% of x + 85% of x –75% of x
=[(80/100)x+(85/100)x-(75/100)x]=(90/100)x=(9/10)x
Students who failed in both the subjects =
[x-(9x/10)]=x/10.
So, x/10=40 of x=400 .
Hence ,total number of students = 400.
