Percentage

x percent means x hundredths written as x%.

 

Formula:

1. If the price of a thing increases by X% , then the reduction in consumption so as not to increase the expenditure is 100*X/(100+X)%.


2. If the price of a thing decreases by X% , then the increases and consumption so as not to decreases the expenditure is 100*X/(100-X)%.


3. Value of a thing after n years = p(1-R/100)ⁿ .


4. If A is X% more than B , then B is less than A by 100*X/(100+X)%.


5. If A is X% less than B , then B is more than A by 100*X/(100-X)%.


6. If A is increased by X% and again decreased by Y% then

% change = X + Y + X*Y/100

Example:

Ex. 1. 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. 2. 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 3. 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.

Profit & Loss

Cost Price: The price at which article is purchased.


SELLING PRICE: The price at which article is sold.

PROFIT OR GAIN: If SP is more than CP, The SELLING PRICE is said to have PROFIT or GAIN.

LOSS: If SP is Less than CP, The S.P. is said to have incurred a LOSS.

FORMULA

1.GAIN=(SP)-(CP).

2.LOSS=(CP)-(SP).

3.LOSS OR GAIN IS ALWAYS RECKONED ON CP

4. GAIN %={GAIN*100}/CP.

5.LOSS%={LOSS*100}/CP.

6.SP={(100+GAIN%) /100}*CP.

7.SP={(100-LOSS%)/100}*CP.

8.{100/(100+GAIN%)} *SP

9.CP=100/(100-LOSS%)}*SP

10.IF THE ARTICLE IS SOLD AT A GAIN OF SAY x%, THEN SP = (100+x)% OF CP

11.IF A ARTICLE IS SOLD AT A LOSS OF SAY x%. THEN SP= (100-x)% OF CP.

12.WHEN A PERSON SELLS TWO ITEMS,ONE AT A GAIN OF X% AND OTHER AT A LOSS OF X%.THEN THE SELLER ALWAYS INCURES A LOSS GIVEN:

{LOSS%=(COMON LOSS AND GAIN ) ²}/10.=(X/10) ²

13.IF THE TRADER PROFESSES TO SELL HIS GOODS AT CP BUT USES FALSE WEIGHTS,THEN

GAIN=[ERROR/(TRUE VALUE)-(ERROR)*100]%

Examples:

Ex 1. If the cost price is 96% of sp then whqt is the profit %

Sol. sp=Rs100 : then cp=Rs 96:profit =Rs 4.

Profit={(4/96)*100}%=4.17%

Ex. 2. The cp of 21 articles is equal to sp of 18 articles.find gain or loss %

Sol: CP of each article be Rs 1

CP of 18 articles =Rs18 ,sp of 18 articles =Rs 21.

Gain%=[(3/18)*100]%=50/3%

Ex.9 By selling 33 metres of cloth , one gains the selling price of 11 metres . Find the gain percent .

Sol: (SP of 33m)-(CP of 33m)=Gain=SP of 11m

SP of 22m = CP of 33m

Let CP of each metre be Re.1 , Then, CP of 22m= Rs.22,SP of 22m=Rs.33.

Gain%=[(11/22)*100]%=50%

ALLIGATION OR MIXTURE

IMPORTANT FACTS AND FORMULAE

1. Alligation: It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a desired price.

2. Mean Price: The cost price of a unit quantity of the mixture is called the mean price.

3. Rule of Alligation: If two ingredients are mixed, then

• (Quantity of cheaper) = (C.P. of dearer) - (Mean price)


• (Quantity of dearer) = (Mean price) - (C.P. of cheaper)


• (Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c).

4. Suppose a container contains x units of liquid from which y units are taken out and replaced by water. After n operations the quantity of pure liquid=[ x(1-y/x)^n]units.

Example:

Ex. 1. In what ratio must rice at Rs. 9.30 per kg be mixed with rice at Rs. 10.80 per kg so that the mixture be worth Rs. 10 per kg ?

Sol. By the rule of alligation, we have:

C.P. of 1 kg rice of 1st kind (in paise) = 930
C.P. of 1 kg rice of 2nd kind (in paise) = 1080

Mean price = 1000 (in paise)

(Cheaper quantity) : (Dearer quantity) = (1080 - 1000) : (1000 - 930)
:. Required ratio = 80 : 70 = 8 : 7.

­SIMPLE INTEREST

Important Facts and Formulae

1.. Principal: The money borrowed or lent out for a certain period is called the

principal or the sum.

2. Interest: Extra money paid for using other's money is called interest.

3. Simple Interest (S.I.) : If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest.

Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then,

(i) S.I. = (P*R*T )/100

(ii) P=(100*S.I)/(R*T) ;R=(100*S.I)/(P*T) and T=(100*S.I)/(P*R)



Examples:

Ex. 1. Find the simple interest on Rs. 68,000 at 16 2/3% per annum for 9 months.

Sol. P = Rs.68000,R = 50/3% p.a and T = 9/12 years = 3/4years.

S.I. = (P*R*T)/100 = Rs.(68,000*(50/3)*(3/4)*(1/100))= Rs.8500

Ex. 2. Find the simple interest on Rs. 3000 at 6 1/4% per annum for the period from 4th Feb., 2005 to 18th April, 2005.

Sol. Time = (24+31+18)days = 73 days = 73/365 years = 1/5 years.

P = Rs.3000 and R = 6 ¼ %p.a = 25/4%p.a

S.I. = Rs.(3,000*(25/4)*(1/5)*(1/100))= Rs.37.50.

Remark : The day on which money is deposited is not counted while the day on which money is withdrawn is counted .

Ex. 3. A sum at simple interests at 13 ½ % per annum amounts to Rs.2502.50 after 4 years find the sum.

Sol. Let sum be Rs. x then , S.I.=Rs.(x*(27/2) *4*(1/100) ) = Rs.27x/50

:.amount = (Rs. x+(27x/50)) = Rs.77x/50

:. 77x/50 = 2502.50 <=> x = (2502.50 * 50)/77 = 1625

Hence , sum = Rs.1625.

Ex. 4. A sum of Rs. 800 amounts to Rs. 920 in 8 years at simple interest, interest rate is increased by 8%, it would amount to bow much ?

Sol. S.l. = Rs. (920 - 800) = Rs. 120; p = Rs. 800, T = 3 yrs

. R = ((100 x 120)/(800*3) ) % = 5%.

New rate = (5 + 3)% = 8%.

New S.l. = Rs. (800*8*3)/100 = Rs. 192.

: New amount = Rs.(800+192) = Rs. 992.

COMPOUND INTEREST

Compound Interest: Sometimes it so happens that the borrower and the lender agree to fix up a certain unit of time, say yearly or half-yearly or quarterly to settle the previous account.

In such cases, the amount after first unit of time becomes the principal for the second unit,the amount after second unit becomes the principal for the third unit and so on.

After a specified period, the difference between the amount and the money borrowed is called the Compound Interest (abbreviated as C.I.) for that period.

IMPORTANT FACTS AND FORMULAE

Let Principal = P, Rate = R% per annum, Time = n years.

I. When interest is compound Annually:

Amount = P(1+R/100)ⁿ

II. When interest is compounded Half-yearly:

Amount = P[1+(R/2)/100]²ⁿ­

III. When interest is compounded Quarterly:

Amount = P[ 1+(R/4)/100]⁴ⁿ­



IV. When interest is compounded AnnuaI1y but time is in fraction, say 3(2/5) years.



Amount = P(1+R/100)³ x (1+(2R/5)/100)

V. When Rates are different for different years, say Rl%, R2%, R3% for 1st, 2nd and 3rd year respectively.

Then, Amount = P(1+R₁/100)(1+R₂/100)(1+R₃/100)



VI. Present worth of Rs.x due n years hence is given by :

Present Worth = x/(1+(R/100))ⁿ

Examples:


Ex.1. Find compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annually.

Sol: Amount = Rs [7500*(1+(4/100)²] = Rs (7500 * (26/25) * (26/25)) = Rs. 8112.

therefore, C.I. = Rs. (8112 - 7500) = Rs. 612.

Ex. 2. Find the compound interest on Rs. 10,000 in 2 years at 4% per annum, the

interest being compounded half-yearly.

Sol. Principal = Rs. 10000; Rate = 2% per half-year; Time = 2 years = 4 half-years.

Amount

= Rs [10000 * (1+(2/100))⁴] = Rs(10000 * (51/50) * (51/50) * (51/50) * (51/50))

= Rs. 10824.32.

:. C.I. = Rs. (10824.32 - 10000) = Rs. 824.32.