IMPORTANT
FACTS AND FORMULAE
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%.
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.
express x% as a fraction : We have , x% = x/100.
Thus,
20% =20/100 =1/5; 48%
=48/100 =12/25, etc.
=48/100 =12/25, etc.
To
express a/b as a percent : We have, a/b =((a/b)*100)%.
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%.
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
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
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
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.
Population after nyeras = P
[1+(R/100)]^n.
2.
Population n years ago = P
/[1+(R/100)]^n.
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
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.
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.
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
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
A is R% less than B , then B is more than A by
[(R/(100-R))*100]%.
SOLVED
EXAMPLES
EXAMPLES
Ex.
1. Express each of the
following as a fraction :
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 :
following as a Decimal :
(i)
6% (ii)28% (iii) 0.2% (iv)
0.04%
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.
0.002. (iv) 0.04%= 0.04/100 =0.004.
Ex.
3. Express each of the
following as rate percent :
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%.
8/9%.
(ii)
0.004 =
[(4/1000)*100]% = 0.4%.
[(4/1000)*100]% = 0.4%.
(iii)
6 ¾ =27/4
=[(27/4)*100]% = 675%.
=[(27/4)*100]% = 675%.
Ex.
4. Evaluate :
(i)
28% of 450+ 45%
of 280
28% of 450+ 45%
of 280
(ii)
16 2/3% of 600
gm- 33 1/3% of 180 gm
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.
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.
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 ?
1/3 ?
(iii)What
percent of 8 is 64
?
?
(iv)What
percent of 2 metric
tones is 40 quintals ?
tones is 40 quintals ?
(v)What
percent of 6.5
litres is 130 ml?
litres is 130 ml?
Sol.
(i)
Required Percentage =
[(2/50)*100]% = 4%.
[(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.
1 metric tonne = 10 quintals.
Required
percentage = [
(40/(2 * 10)) * 100]% = 200%.
(40/(2 * 10)) * 100]% = 200%.
(v)
Required Percentage
= [ (130/(6.5 * 1000)) * 100]% = 2%.
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
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
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
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 ?
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.
(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 ?
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]
1/3 % of Rs.8400]
=
[(7/20-(10/3)]% of Rs.8400
=1/6 % of Rs.8400
=1/6 % of Rs.8400
=
Rs. [(1/6)8(1/100)*8400] = Rs.
14.
14.
Ex.
9. An inspector rejects 0.08% of the meters as defective.
How many will be examine to project ?
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 ?
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 ?
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
=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.
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.
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%.
0.034%.
Ex.
13.
In
an election between two candidates, 75% of the voters cast
their 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.
their 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).
x. Valid votes = 98% of (75% of x).
75%
of (98% of (75%of x)) =9261.
[(75/100)*(98/100)*(75/100)*x]
=9261.
=9261.
X
= [(9261*100*100*100)/(75*98*75)]
=16800.
=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?
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)
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)
(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%
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?
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.
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.
Rs(3X((20/200) X (3x/5)) = Rs.9x/5.
Balance
= Rs.((3x/5) –
(9x/25))=Rs.6x/25.
(9x/25))=Rs.6x/25.
Amount
deposited in bank=
Rs(1/2 X 6x/25)=Rs.3x/25.
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.
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.
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.
total number of orginal inhabitants be x.
((75/100))*(90/100)*x)=4050
ó (27/40)*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.
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
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
+ (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.
ó 24x/25=31200óx=[(31200*25)/24)=32,500.
Total
sales=Rs.32,500
sales=Rs.32,500
Ex
.19 Raman`s salary was decreased by 50% and subsequently
increased by 50%.How much percent does he lose?
increased by 50%.How much percent does he lose?
Sol:
Let
the
original salary = Rs.100
original salary = Rs.100
New
final
salary=150% of (50% of Rs.100)=
salary=150% of (50% of Rs.100)=
Rs.((150/100)*(50/100)*100)=Rs.75.
Decrease
=
25%
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 .
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
original income=Rs.100 . Then , expenditure=Rs.75 and savings =Rs.25
New
income
=Rs.120 , New expenditure =
=Rs.120 , New expenditure =
Rs.((110/100)*75)=Rs.165/2
New
savings
= Rs.(120-(165/2)) = Rs.75/2
= Rs.(120-(165/2)) = Rs.75/2
Increase
in
savings = Rs.((75/2)-25)=Rs.25/2
savings = Rs.((75/2)-25)=Rs.25/2
Increase
%=
((25/2)*(1/25)*100)% = 50%.
((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 ?
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.
original salary be Rs.100 . New salary = Rs.90.
Increase
on
90=10 , Increase on 100=((10/90)*100)%
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 ?
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.
price of the product be Rs.100 and let original sale be 100 pieces.
Then
,
Total Revenue = Rs.(100*100)=Rs.10000.
Total Revenue = Rs.(100*100)=Rs.10000.
New
revenue
= Rs.(90*130)=Rs.11700.
= Rs.(90*130)=Rs.11700.
Increase
in
revenue = ((1700/10000)*100)%=17%.
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.
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.
original fraction be x/y.
Then
(115%of x)/(92% of y)=15/16 => (115x/92y)=15/16
(115%of x)/(92% of y)=15/16 => (115x/92y)=15/16
ð
((15/16)*(92/115))=3/4
((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 ?
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%.
[(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 ?
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
after 2 years = 176400*[1+(5/100)]^2
=[176400*(21/20)*(21/40)]
=
194481.
Population
2 years ago = 176400/[1+(5/100)]^2
2 years ago = 176400/[1+(5/100)]^2
=[716400*(20/21)*(20/21)]=
160000.
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
?
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
the machine after 2 years
=Rs.[162000*(1-(10/100))^2]
= Rs.[162000*(9/10)*(9/10)]
= Rs.[162000*(9/10)*(9/10)]
=Rs.
131220
Value
of
the machine 2 years ago
the machine 2 years ago
=
Rs.[162000/(1-(10/100)^2)]=Rs.[162000*(10/9)*(10/9)]=Rs.200000
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 ?
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
in the beginning of the first year
=
9975/[1+(5/100)]*[1-(5/100)] = [9975*(20/21)*(20/19)]=10000.
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 ?
than B,how much percent does B earn less then A ?
Sol:
Required
Percentage = [((100/3)*100)/[100+(100/3)]]%
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 ?
how much percent is B`s salary more than A`s ?
Sol:
Required
percentage = [(20*100)/(100-20)]%=25%.
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 ?
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
salt in 30kg solution = [(20/100)*30]kg=0.6kg
Let
x kg of pure salt be
added
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.
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.
Rs.x per kg.
Reduced
rate =
Rs.[(100-(25/4))*(1/100)*x}]=Rs.15x/16per kg
Rs.[(100-(25/4))*(1/100)*x}]=Rs.15x/16per kg
120/(15x/16)-(120/x)=1
ó
(128/x)-(120/x)=1
(128/x)-(120/x)=1
ó
x=8.
So,
the original rate = Rs.8
per kg
per kg
Reduce
rate =
Rs.[(15/16)*8]per kg = Rs.7.50 per kg
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 .
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 .
students who failed in Hindi and English respectively .
Then
, n(A) = 35 , n(B)=45 ,
n(AÇB)=20.
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%
and English or both=60%
Hence
, percentage passed =
(100-60)%=40%
(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.
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 .
students be x .
Let
A and B represent the
sets of students who passed in English and Mathematics respectively .
sets of students who passed in English and Mathematics respectively .
Then
, number of students
passed in one or both the subjects
passed in one or both the subjects
=
n(AÈB)=n(A)+n(B)- n(AÇB)=80% of x + 85% of x –75%
of x
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.
the subjects = [x-(9x/10)]=x/10.
So,
x/10=40 of x=400 .
Hence ,total number of
students = 400.
students = 400.
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