Question based on permutation and combination coming soon

Simple interest and Compound interest

Formula 1:

 S.I =  P * R * N / 100 and A = S.I + P

Where,

P= principal amount

R= rate of interest

N= number of years &

A= amount

Formula 2:

A = P * ( 1 + R / 100n ) ^ nt

A = amount

P = principal

R = rate of interest

n = how the interest is compounded

e.g. n=1…when interest is compounded yearly

n=2… interest is compounded half yearly

n=4… interest is compounded quarterly

n=12… interest is compounded monthly

Exercise:-

1)      What will be the amount on Rs. 6400 for three years at the rate of interest 5% per annum ?

Solution:- here A=? , P=640 , R=5% and N=3

S.I=P*R*N/100

S.I=6400*5*3/100

S.I=960

A=960+6400=7360

2)      A sum of money amount to Rs. 5017 in 2 years at the rate of 8% per annum, what was the sum invested?

Solution:- here P=? , A=4500 , N=2 , R=8%

A=S.I+P=[P*R*N/100]+P

5017=[P*8*2/100]+P

5017=[4P/25]+P è 5017*25=29P èP=4325

3)       Find the time in which Rs. 5000 amount to Rs. 7000 at the rate of 2% p.a.

Solution:- here N=? , P=5000 , A=7000 , R=2%

A=S.I+P è S.I=2000

P*R*N/100=2000

5000*2*N=2000*100

N=20 years

4)      What would be the rate of interest when the sum of Rs. 6200 amount to Rs. 7440 in 5 years.

Solution:- here R=? , P=6200 , A=7440 , N=5

A=P+S.I  è S.I=1240

P*R*N/100=1240

6200*R*5=1240*100

R= 4%

5)      Find the amount when the sum of Rs. 20,000 is compounded half yearly at the rate of 5% p.a for 1 year.

Solution:- here A=? , P=20,000 , R=5% , n=2 , t=1

A = P * ( 1 + R / 100n ) ^ nt

A = 20,000 ( 1 + 5/100 ) ^ 2

A = 20,000 ( 21 / 20 ) ^2

A = 20,000 * 441/400 = 22050

Percentage

Basic concepts:-

1)      X% = x/100

e.g. 20% = 20/100 = 1/5

25% = 25/100 = 1/4

50% = 50/100 = 1/2

75% = 75/100 =3/4

80% = 80/100 = 4/5

When we want to convert a % value into fraction then

divide the given by 100

2)      x/5 = 100 x /5 = 20% of x

e.g.1/5= 100/ 5= 20%

1/4 = 100/4 = 25%

3/4 = 3*100/4 =75%

3)      Use ‘ + ‘ sign when the word ‘more’ comes  ,  ‘ – ‘

sign when the word ‘less’ comes and ‘ x ‘ when the word

‘ of ’ comes in word problem

When we want to convert a fraction into % value then

multiply the given fraction by 100

Exercise:-

1)      The price of an article is reduced by 20%. If the new

price is 825. Find the original price of the article.

Solution:- Assume that the original price of the article is

x Rs.

New price= x- 25% of X = 75% of x

75% of x = 825……………….given that the new price is 825

(75/100) * x=825

X=825*100/75 = 1100

2)      The price of an article is increased by 25%. If the new

price is 1200. Find the original price of the article.

Solution:- Assume that the original price is x Rs.

New price= x + 25% of x = 125% of x

125% of x = 1200………………….given that the new price is 1200

(125/100)*x = 1200

X=1200*100/125 = 960

Formula 1 :-

Suppose that A be a value / price / amount / quantity

which changes twice. First time increases by x% and

then by y%. Then the net % change in amount will be

( 100% of A + x% of A ) + y% of ( 100% of A + x% of A )

(100 + x)% of A + y% of [(100 + x)% of A ]

[( 100+x )% of A ]( 1+ y%)

[(100 + x)/100 ]*[1 + y/100] * A

1/100 [(100 + x)*(100 + y)] * A/100

1/100[10,000 + 100x + 100y + xy] * A/100

[100 + x + y + xy/100] * A/100

A + (x + y + xy/100)% of A

Net % increment = (x + y + xy/100)%

Remark:-

·       

if both x an y increases then formula will be

(x + y + xy/100)%

·       

if x increases and y decreases then formula will be

(x - y - xy/100)%

·       

If x and y both decreases then formula will be

( - x - y + xy/100)%

3)      The price of an article is first increased by 25% and

then reduced by 20%. If the new price is 1540.

Find the original price of the article.

Solution:- Here x=25% and y=20%

Net % change = (x + y + xy/100)%

Net % change = (25 – 20 – 500/100)%

= (5 – 5)% = 0%

There will not be any change in the amount.

Original price will remain same i.e. 1540

4)      Suppose a person gives 25% of his amount to his

elder son and 40% of the remainder to his younger

son if he is left with 150 Rs.then how  much money

he was having earlier?

Solution:- Assume that he has x Rs

Net % change in x=(25+40+25*40/100)%

Net % change in x=(65+1000/100%

Net % change in x=75%

He is left with=100-75=25% of x

But he is left with 150 Rs

150=25% of x

150*100/25=x

600=x

Earlier he was having 600 Rs.

 

5)      Mr. Robert spends 25% of his income on food

and 20% of the remaining on house rent. If at the

end he is left with 30,000 then what is his salary?

Solution:- Net % percent change will be

=(25+20+500/100)%

Net % change=50%

This means he spends his 50% salary

èhe is left with 50% salary which is 30,000

ð 

His salary is 60,000 Rs.

Formula 2 :-

Case 1:- The population of a town is suppose 5000

and it increase by 10% annually then the population

after 2 years will be

5000 ( 1 + 10/100 )^2 = 5000 ( 11/10 )^2

= 6050

Generalization: - Assume that the population of a

town is P and it increase R% annually then the

population after n years will be

·        

P ( 1 + R/100 )^2

Case 2:- The population of a town is 5000 and it

decreases by 10% annually due to some disease

then population after 2 years will be

5000 ( 1 – 10/100 )^2 = 5000 ( 9/10 )^2 = 4050

Generalization:- Assume that the population of a

town is P and it decreases by 10% annually then

the population after n years will be

·        

P ( 1 – R/100 )^2

Case 3:- The population of a town is suppose 5000

and it increase by 10% annually then the population

before 2 years will be

5000 / ( 1 + 10/100 )^2 = 5000 / (11/10)^2

5000 *100 /121 = 4132

Generalization:- Assume that the population of a

town is P and it increase R% annually then the

population before n years will be

·        

P / ( 1 + R/100 )^2

Case 4:- The population of a town is suppose 5000 and

it decreases by 10% annually then the population before

2 years will be

5000 / ( 1 – 10/100 )^2 = 5000 / ( 9/10)^2

5000*100/81 = 6173

Generalization:- Assume that the population of a

town is P and it decrease R% annually then the

population before n years will be

·        

P / ( 1 - R/100 )^2

Exercise

6)      The population of a town is 84000 and it increases by

12% annually what will it be in 3 years?

Solution:-  here P=84000, R=5 & n=3. We have

P ( 1 + R/100 )^n = 84000 ( 1+ 5/100 )^3

= 84000 ( 21/20 )^3 =97241

Age problems

          1)      Father is aged three times more than his son. After 8 years he would be two and a half of his son’s age. After further 16 years how many times would he be of his son’s age?

Solution:-  assume that the son’s present age is x years and father’s present age

=x+3x=4x years

After 8 years their ages will be x+8 and 4x+8

4x+8=(5/2)*(x+8)…….(given)

4x-5x/2=20-8

3x/2=12

è x=8

Therefore, son’s present age=8 years and fathers present age=32

(8+16) years their ages will be,

Son’s age=32 and father’s age=56

èfather’s age=7/4 of son’s age

     2)      The sum of ages of 6 children born at the interval off 2 years is 60 year. What is age of the youngest child?

Solutions:- assume that the age of the youngest child is x years

Age of the 2^nd child=x+2

Age of the 3^rd child=(x+2)+2=x+4

 Age of the 4^th child=(x+4)+2=x+6

Age of the 5^th child=(x+6)+2=x+8

Age of the 6^th child=(x+8)+2=x+10

Therefore, x+x+2+x+4+x+6+x+8+x+10=60

6x+30=30 è 6x=30 è x=5

Hence the youngest child is 5 years old.

     3)      Karan’s father was 30 years old when he was born. And his mother was 26 years old when his sister was born who is 3 years younger than him.What is the difference between his parent’s ages?

Solution:- assume that karan’s age is x years

Today	4 years ago
Karan=4                         <==	Karan=0
Father=34                        <==	Father=30
Sister=0
Mother=26                        è	Mother=22

Difference=34-26=30-22=8 years

      4)      If we subtract 8 from grandfather’s present age and divide the remainder by 15 we get the present age of grandson who his 3 years younger than his brothers of 7 years. Find the present age the grandfather?

Solution;- assume that the present age of grandfather is x years.

(x-8)/15=4 è x=60+8=68

Grandfather’s age=68 years

      5)      If after 10 years A will be thrice as old as B was 10 years ago. If A is now 4 years elder than B then what is the present age of B?

Solution:- assume that the present age of B is x years.

	10 years ago	Present ages	10 years later
A	(x+4)-10=x-4	x+4	(x+4)+10=3(x-10)
B	x-10	x	x+10

X+14=3x-30

14+30=3x-x

44=2x

X=22 years

6) If 6 years is subtracted from A's age and the remainder is divided by 8 then B's age is obtained. If Jasmine is 2 years younger than C who is 9 then what is the age of A ? 

Solution;- Assume that the A's age is x years.

According to condition given, we have

(x-6)/8= B's age = 9-2 = 7 years

(x-6)/8=7 ==> x=7*8+6=60 years

7) Manish was as old as his daughter Jasmine at the time of his birth. If the present age of Manish is 32 years then what will be the age of Jasmine 4 years ago?

       Solution:- Assume that the present age of Jasmine is x years. Now we get
    

Jasmine's birth	Manish= x	Jasmine= 0
Present ages	Manish=32	Jasmine= x

From the above table we have,

 32-x=x-0

 32=2x

16=x

4 years ago she will be 12 years old.