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.