**1.**

**Concept of percentage or Percentage formula**

To express x% as a fraction: We
have, x%

x/100 Thus, 30%=30/100= 3/10

x/100 Thus, 30%=30/100= 3/10

**In simple terms**we can conclude that percentage symbol

**"%"**means

**1/100**

that is,

**a/b=(a/b)×100)%**

**2.**

**Increase or decrease in percentage with price**

(

**1**) If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure will be**= [R/ (100+R)×100]%**
(

**2**) If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure will be**= [R/(100−R)×100]**

**3.**

**Results on population**

Formula's on result of population are very important when we have to calculate the population n years after or n years before,

Let the population of a town be P now and suppose it increases at the rate of R% per annum, then

(

**1**) Population

**after**

**n**years

**=P(1+R/100)**

^{n}
(

**2**) Population**before****n**years**=P/(1+R/100**)^{n}^{}

**4.**

**Results on Depreciation**

We know that value of a machine depreciate with time, so it will decrease with the time. To calculate the value of machine after n years or before n years, we use these formula's

Let the present value of a machine
be P. Suppose it depreciates at the rate of R% per annum.

**(1**) Population after n years**= P(1−R/100)**^{n}**(2**) Population before n years

**=P/(1−R/100)**

^{n}

^{}**5.**

**Important results for Percentage**

The formulas we are going to mention below is same as of increase or decrease in consumption with increase or decrease in the commodity price, just here we are in a bit different context.

(

**1**) If A is R% more than B, then B is less than A by

**= [R/(100+R)×100]%**

(

**2**) If A is R% less than B, then B is more than A by**= [R/(100−R)×100]%**

**QUESTIONS**:

**1.**Two numbers are less than third number by 30% and 37% respectively. How much percent is the second number less than by the first

(1) 20% (2) 30 %

(3) 10% (4) 40%

**2.**In an examination, 34% of the students failed in mathematics and 42% failed in English. If 20% of the students failed in both the subjects, then find the percentage of students who passed in both the subjects.

(1) 40% (2) 41%

(3) 43% (4) 44%

**3.**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 ?

(1) 20% (2) 30%

(3) 22% (4) 24%

**4.**The value of a machine depreciates at the rate of 10% per annum. If its present is Rs.1,62,000.? What was the value of the machine 2 years ago ?

(1) 200000 (2) 20000

(3) 2000 (4) None of these

**5.**A student multiplied a number by 3/5 instead of 5/3. What is the percentage error in the calculation?

(1) 60% (2) 62%

(3) 64% (4) 65%

**ANSWERS WITH EXPLANATIONS:**

**1.**(3) : Let the third number is x.

then first number = (100-30)% of x

= 70% of x = 7x/10

Second number is (63x/100)

Difference = 7x/10 - 63x/100 = 7x/10

So required percentage is, difference is what percent of first number

=> (7x/100 * 10/7x * 100 )% =

Difference = 7x/10 - 63x/100 = 7x/10

So required percentage is, difference is what percent of first number

=> (7x/100 * 10/7x * 100 )% =

**10%**

**2.**(4) : Failed in mathematics, n(A) = 34

Failed in English, n(B) = 42

n(AUB) = n(A) + n (B) – n (A∩ B )

= 34 + 42- 20 = 56

Failed in either or both subjects are 56

Failed in either or both subjects are 56

Percentage
passed = (100−56)%=

**44%**

**3.**(1) : Reduction in consumption = [((R/(100+R))*100]%

? [(25/125)*100]%=

**20%**.

**4.**(1): Value of the machine 2 years ago

= Rs.[162000/(1-(10/100)

^{2})]=Rs.[162000*(10/9)*(10/9)]=

**Rs.200000**

**5.**(3) : Let the number be x

Then, error = (5/3)x – (3/5)x =(16/15)x

Error% = [(16x/15) /(5x/3)] * 100% =

**64%**

## 20 comments:

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