# Area and Perimeter Formulas with Examples Part- 2

In this article, we will learn about successive percentage change, it deals with two or more percentage changes in a quantity consecutively.
Why this isn’t the simple addition of two percentage changes?
Successive Percentage Change: If there are percentage changes of a% and b% in a quantity consecutively, then total equivalent percentage change will be equal to the (a + b + \frac { ab }{ 100 } )%.

### Example1:

There is two outlet, one is offering a discount of 50%+ 50% and other is offering a discount of 60% + 40%. At which outlet, one must visit so that she gets more discount?
Solution: Case1: 50%+ 50%
Total discount = -50 + -50 + \frac {-50\times -50  }{ 100 }  = -100 +25 =-75% ⇒ 75% discount

Case2: 60% + 40%. Total discount = (-60)+(-40)+(-60)×\frac { -40 }{ 100 } =-100+24=-76% ⇒ 76% discount
Therefore, she must visit outlet offering a discount of 60% + 40%.

### Example2:

The length & breadth of a rectangle have been increase by 30% & 20% respectively. By what percentage its area will increase?
Total Percentage change=( a+b+\frac { ab }{ 100 } ) %
= P(equivalent)= 30+20+\frac { 30 \times 20 }{ 100 }= 60%

### Example3:

The length of the rectangle has been increased by 30% & breadth has been decreased by 20%. By what percentage its area will change?
Total Percentage change= ( a+b+\frac { ab }{ 100 } ) %
= P(equivalent)= 30+(-20)+\frac { 30 \times (-20) }{ 100 }= 4%

### Example4:

There is 10%, 15% & 20% depreciation in the value of mobile phone in 1st, 2nd & 3rd month after sale if the price at beginning was 10,000R, then price of mobile after 3rd month will be:-
Solution: Total Percentage change= ( a+b+\frac { ab }{ 100 } ) %
Take 10% & %20, Percentage Equivalent = -10 -20 +(-10) \times \frac { -20 }{ 100 } = -28%
Now, taking 28% & 15%. Percentage Equivalent =  -28 -15 +(-28) \times \frac { -15 }{ 100 }= 38.8%
Therefore, Price after 3rd month = 10000× (100-38.8) % = 6120Rs
Aliter: Final Price= Original Price× MF _{ 1 }× MF _{ 2 }× MF _{ 3 }
⇒ = 10000×0.9×0.85×0.8 = 6120 D

### Example5:

Price of an item is increased by 40% and its sales decrease by 20%, what will be tha percentage effect on income of shopkeeper?
Solution: Income= Price × sales
⇒ Percentage (effect) = 40 +(-20) \times \frac { (-40)(-20) }{ 100 } = 12%↑se

### Example6:

The radius of the circle has increased by 15%. By what percent its area will be increased?
Solution: Area of circle = \pi e ^{ 2 }
Area (eq.) = ( a+b+\frac { ab }{ 100 } ) % = r + r+   frac { (r)\times(r) }{ 100 }= 2r+ \frac { (r)(2) }{ 100 }
⇒ 2×15+ \frac { { 15 }^{ 2 } }{ 100 } = 32.25%

Note: Effect on Area= 2P +(P×P)/100 (where, P is %change in variable) this is valid for Circle, Square & Equilateral triangle.

## Faulty Balance:

### Example1:

A milkman mixes 100 litres of water with every 800lit. Of milk and sells at a markup of 11.11%. Find the percentage profit?
Solution: Total Profit = Adding water + Mark-Up
⇒ Profit = \frac { 100 }{ 800 } + \frac { 1 }{ 9 } + \frac { 100 }{ 800 } \times \frac { 1 }{ 9 } = \frac { 9+8+1 }{ 72 }= \frac { 1 }{ 4 }⇛ 25% Profit
Aliter: 100lit water+ 800lit milk=900lit milk
Let, CP= Rs1/lit⇒ total CP= 800Rs
& SP= 900Rs & there is mark-up as well. So Mark Up=\frac { 1 }{ 9 } \times900= 100Rs
⇒ total SP= 900+100=1000Rs
⇒\frac { SP }{ CP }= \frac { 1000 }{ 800 }= \frac { 5 }{ 4 }= 1.25 ⇛25%Profit

## Compound Interest:

Compound Interest in simple terms, is successive percentage equivalent of simple interest.

### Example1:

The difference between compound interest and simple interest on a sum for two years at 8% per annum, where the interest is compounded annually is Rs.16. Find the principal amount?
Solution: Simple Interest for 2years = 2×8%=16%
Compound Interest for 2years =8+8+\frac { 8\times 8 }{ 100 } =16.64%
Therefore, difference = 0.64%
= Principal \times  0.64%= 16
= Principal = 16+\frac { 100 }{ 0.64 }= 2500 Rs

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