Microsoft Word - FGN101006_Percentage, Profit and Loss.doc

Marks scored by A is 50 and marks scored

by B is 40. Then

(i) A is what % more than B?

(ii) B is what % less than A?

(iii) A is what % of B?

(iv) B is what % of A?

(i) x100 = 25%

40

(ii) x100 = 20%

50

(iii) x100 = 125%

(iv) x100 = 80%

50

What % of 60 is 40?

Keeping the above point in mind, answer can

be calculated as

Sol:

400 - I

I =

400

100

125

Please note that last two methods are based on the

same concept, only way of representation is different

student can follow any method base on his comfort.

But its better to practice last method.

(c) The quantity of jowar produced in the year 1997

was 21,000 tons. During the next year the

production increased by 33 %. What was the

33

production in 1998?

Sol:

1

1

(d) The production of tobacco in the year 2000 was

160 lakhs tons and this was 20% less than the

production of the year 1999. How much tobacco

was produced in 1999?

Sol: Let x be the production in

x 0.2x = 160

0. 8x = 160

160

Out of 80% given we want 100%.

x = 160 = 200

80

Explain to the students the concept of a number

becoming n times is equivalent to increase of

(n 1) x 100 % increase with examples as below.

For example if a value x is doubled over a period

of time then % change is

% increase = x100 = 100%

x

Parallely if x is tripled then

% increase = x100 = 200%

x

If x is made 4 times then

% Increase = x100 = 300%

x

if x is made n times then

% increase = x100 = (n -1)100%

x

Using this concept explain a few problems as

below

Ex 9:

Sol:

3x - x

2x

Ex 10: If the population of Delhi is 220% more than

Sol:

Ex 16: If the price of tea is decreased by 20% then

by what % should the consumption be

increased in order to maintain constant

expenditure?

Sol: Method I First of all explain to the

students that expenditure is equal to price of

tea x consumption of tea. Hence if expenditure,

price, consumption are represented as e, p, c

respectively. We can say

e = p x c = cost

p1, c1, = p2 c2

price of tea decreased by 20%

p2 = p1 20% p1 = 0.8 p1

p1 x c1 = 0.8 p1 x c2

c2 = = 1.25c1

0.8

% increase = x 100 = 25%

c1

Method II

Explain to the students that this can be solved

by using the formula also as

Req % inc = % = = 25%

(100 - 20) 80

Method III

As price becomes 80% i.e times,

5

consumption will become times to maintain

4

the expenditure

Now = 1 + and = 25%

4 4 4

Hence consumption must be increased by 25%.

Ex 17: The production of tea increases by 150%

from 1979 to 1980 by what % should the

production of tea increase from 1980 1981

so that the production of tea in 1981 recovers

becomes five times that of 1979.

6. Successive increase/decrease in percentages.

Many times we will come across a situation where

successive percentage changes are made and we

are asked to find overall percentage change.

For example, price of an article is increased by

20% and again it is increased by 30%. The overall

% change can be calculated as

Let initial price = 100

100 120

+20 +36

Hence overall % increase is 56%.

We can solve this question by using multiplication

factor also

Let initial price = P

Final price = P 1.2 1.3

= P 1.56

Hence overall 56% increase. As values 20% and 30%

are simple, we can use either of the methods. But if

values are 7% and 9% then it will involve a more

calculation. To avoid that we can use formula for

successive percentage change. If successive

changes of a% and b% are made, then overall

% change will be

ab

In this formula a and b will be positive for percentage

increase and negative for percentage decrease.

Same logic can be applied for the final answer. In the

example discussed above, using formula we can find

answer as

20 + 30 + = 56%

100

As final answer is positive, it is 56% increase.

In the above example if it is 20% increase I and then

30% decrease then

20 30 + = 16

100

As final answer is negative, it will be 16% decrease.

The above formula is applicable only for 2 successive

changes. For 3 successive change, first apply formula

for any 2 changes and then apply for resultant and

the third change.

Ex 20: The length of rectangle is increased by 20%

and its breadth is increased by 10%.

What is the percentage increase in its area?

Loss = C.P S.P

But just saying profit is Rs 2 will not give any idea

about how good that transaction is therefore

many times we express the profit in percentage

term. Now if you look at the value Rs 5 is

common and is not going to change. Hence while

calculating %profit or %Loss we will take C.P as

a base.

%Profit = 100

C.P.

In the above example % Profit = 100=40%

5

Similarly % Loss = 100

C.P.

In the above example

%Loss = 100=20%

5

Please note that while calculating % profit and %loss

C.P is taken as a base 100%.

If someone gets 10% profit means profit = 10% of

C.P. In different words we can say if C.P = 100 the

S.P =110.

Similarly if someone suffers a loss of 30% means,

Loss = 30% of C.P

In different words we can say if C.P = 100 then S.P =70

These have to be said orally and only the final

formulae are to be written on the board.

Now explain to the students that generally we have

to calculate % Profit or %loss S. P or C. P accordingly

to the problem and these type of problems can either

be solved using the normal relations or direct

formulae as explained below:

If C.P is know and there is a P% Profit then S.P is

S. P = C.P +

100

If the above one is written as formula it is

S. P = C.P

100

Parallelly if C.P & L % are given then

S. P = C.P

100

If S.P & P% are given then

C.P = S. P

100 + P

if S. P & L% are given then

C.P = S.P

100 - L

Explain to the students that once if we understand

the relation between these parameters we need not

remember any formulae even to solve any problem.

Now explain some examples based on these

concepts.

If x is C.P of A, then C. P of B & C are 1.25x

and 1.5x respectively.

Hence C.P of C = 1200

1. 5x = 1200

C.P of A = x = = Rs.800

1.5

Note: Now explain to the students the concept of

two articles with same S. P & C. P when one sold

at P% profit and another at P% loss is then what

is the % profit or loss on the whole.

Explain to them that if we have two articles

with same C. P one sold at P% profit and other at

P% loss then we neither have a gain nor loss on

the whole.

Whereas if the S.P of the two articles is same

and one is sold at P% profit and other at P% loss

then we will have a overall loss and the loss %

= %

100

Ex.5: The C.P. of two articles is same. If one item

is sold at 10% p and other at 10% L, what is

% P or %L on the whole?

Explain to the students that one can solve

this problem even by conventional method as

below: If we assure C.P. of the articles are

100 each then the S. P of them are 110 & 90

respectively. Hence total C. P and S.P. is 200

and 200 respectively. Hence we neither have

a profit nor a loss.

Ex.11: Merchant sold a cycle at a loss of 20%.

Instead he sold it for Rs. 450 more he would

have made a 10% profit. What is his C. P?

Sol: Method :

Explain to this students that if we assume the

C.P. is x then difference in S. P when sold at

10% P and 20% L i.e., 1.1 x & 0.8x is Rs. 450

respectively.

1.1x 0.8x = Rs. 450

x = Rs. 1500

Method :

Explain to men that this problem can also be

solved in a different way. Since in the two

cases of P & L the C.P is same the profit of

10% can be accumulated as 10-(-20) = 30%

C.P. Hence. If 30% of C.P is 450 then clearly

we can say C.P is Rs. 1500

According to given data

Profit =30% of S.P

Let S.P = 100

Profit =30

C.P = 70

%Profit = 100 = 42.28%

Ex.12: Arvind sells a pen at a certain S.P had he

8 PARTNERSHIP:

Explain to the students the concept of

partnership as it is the case that occurs when two

or more persons involve in a business by

investing some amount respectively. At the end

of the year the profits are shared according to the

ratio of their investments. i.e.

If the time period for which all the persons

invested is same then the profit is divided in the

ratio of investments.

If the time period is not same then the profit is

divided in the ratio of their product of time &

Investment.

If one of the Investors involves in day-to-day

activities of the business men he is called

working partners, if not sleeping partners.

Now explain the difference types of problems

which examples as below:

Ex.1: A & B invests Rs. 8,000 & Rs. 7,000

respectively in a business. After one year

they make a profit of Rs. 4,500.

(a) Find the ratio of shares.

Sol: Since the time period of investment is same

for both the shares will be in the ratio of their

investment i.e., 8000 : 7000 8:7

(b) Find their respective shares

Sol: Since we know that total profit & the ratio of

share we can calculate

share of A = x4500 = Rs.2400

15

share of B = x4500 = Rs.2100

15

Ex.2: A started a business with an investment of

Rs. 20,000. After 2 months B joins him with

an investment of Rs. 40,000. If they make a

profit of Rs. 9,600 at end of the year. Find the

share of B.

Sol: Since the time of investment is not same their

ratio of shares will be equal to ratio of their

product of time, interest.

i.e. ratio of shares of A & B is

A : B = 20000 x 12 : 40000x10 = 3:5

The total profit is 9600 hence share of

B = x9600 = 86000

Ex.3: A starts a business with Rs. 30,000 after two

months B joined him with Rs. 20,000. A

leaves after 6 months while B stays on till

end of year.

(b) By what % does B get more/less than A?

Sol: with the ratio of shares we can calculate the

above as = x100 = 11.11% more than A.

9

respective investments of Rs. 10,000 & Rs.

15,000. The managing partner Aman

receives a salary of Rs. 2,000 for an amount

out of total profit of Rs. 9,500 at year end.

Find total amount received by Aman.

Sol: Out of profit the salary of Aman is Rs. 2000 &

the left out Rs.7,500 is shared between them.

The ratio of shares is the ratio of investments

as the time period is same

i.e, = 10,000 : 15,000

= 2 : 3

share of Aman = x7500 = Rs.3000

5

Total amount received by Aman

= Rs. 3000 + Rs. 2,000 = Rs. 5,000

Ex.5: Anitha starts a Business with Rs. 20,000.

Which is the better investment, 6% stock at

108 or 7% stock at 109.

Explain to them to compare investments we

have to compare rate of returns.

i.e., x100 & x100

108 109

i.e., &

Hence the rate of return on 7% stock at 109

is better and hence it is the better investment.

Which is better investment, 11% stock at 100

or 12% stock at 109

As in above problem

11 12

100 109

100 100

9.09 9.08

Hence 12% stock at 109 is better. Ex 6:

Sol:

Ex 7:

Sol:

Ex 1:

Sol:

Ex 2:

Sol:

Ex 3:

Sol:

Ex 4:

Sol:

Ex 5:

This successive increase model can be used only

when same quantity is multiplied with multiplication

factors.

Area = Length Breadth

If Length increases by 10% and breadth

increases by 20% then Area gets multiplied by

1.1 and 1.2 so it can be used.

Distance = Speed time

Expenditure = Price Consumption etc

4. Percentage Change:

% Increase = 100

Initial Value

What % of 40 is 60?

Similarly we get x100 = 150%

40

The ratio of As salary to Bs salary is 6 : 5.

(a) By what % does A earn more than B?

(b) By what % does B earn less than A?

Explain to the students, that even for the

problem of ratios the same concept explained

above can be applied and it is also not

necessary even to take the common term for

the ratio i.e. let salary of A & B be 6, 5.

(a) x100 = 20%

5

(b) x100 = 16.67%

6

A is taller than B by 25%, by what percentage

B is shorter than A?

1

4

Method II: We know that 25% =

If B = 4 then A = 5

1

5

Required % = = 20%.

Income of A is more than B. A spends 30% of

income on food and B spends 25% of his

income on food. Who spends more on food?

Income of A is more than B and he spends

higher % of income. (as 30% > 20%) on food.

Hence A spends more on food than B.

In the above example if income of A is less

than B then who spends more on food?

As a percentage, 30% > 25%. But here we

are taking. higher % (ie. 30%) of lower base

(i.e A) may be more or less than 25% of B.

Hence, answer is cannot be determined.

50 - 40

10

50

40

40

40

60

x100 = 66.67%

Actual Increase

Actual Decrease

Initial Value

% Decrease =

100

For example, present population of city A is 45000

and last year it was 35000. Then we can find %

increase.

Actual Increase = 45000 35000

= 10000

Initial population = 35000

% Increase = 100 = 100 = 28.56%.

35000 7

60

10000 2

6 - 5

6 - 5

Sol: Method I: As it is "than B"

Let B = 100 then A = 125

Required % = = = 20%.

125 5

25 1

Required Data

Given Data

60 + (20% of 60) = 60 + 12 = 72

We can do the same question in another way by

using concept of Multiplication factor.

Let Ajay's initial marks = 100% then his final exam

marks

= 100% + 20%

= 120%

= = 1.2

The number "1.2" is the multiplication factor.

Hence final exam marks can be obtained by simply

multiplying 60 by 1.2

i.e 60 1.2 = 72 marks.

Multiplication factor for p% increase = 1+

100

Multiplication factor for p% decrease = 1- .

100

Let's quickly see M.F for some P% changes.

% Change M. F

30% Increase 1.3

70% Increase 1.7

8% Increase 1.08 (not 1.8)

10% Decrease 0.9

40% Decrease 0.6

7% Decrease 0.93

As we discussed earlier if his semester marks are

100% then his final exam marks will be 120%. But we

know his semester exam marks as 60. Hence doing

cross multiplication

100 60

120 ?

? = = 72

100

The above process we can do in one step as out of

given 100% we want 120%. which can be obtained as

120

100

P

P

Explain to the students that this problem can

either be solved using formula or by

summation as follows

25 = x100

I

25 I = 40000 100 I 125 I = 40000

40000

125

I = 320 thousand tons

OR

Using multiplication factor,

(prod. in 1998) 1.25 = prod. in 1999

(prod. in 1998) 1.25 = 400

prod. in 1998 = = 320

1.25

OR

If production in 1998 = 100%.

then production in 1999 = 125%.

out of given 125%. We want 100%. Hence,

prod. in 1998 400 = 320

1

Explain to the students that reciprocal value

of 33 % is 1/3 and hence the problem can

33

be solved as

= 21,000 + [21,000]

3

= 21000 + 7,000 = 28,000 tons

x 20% of x = 160

x = 200 lakh tons.

OR

Using multiplication factor

x = = 200

0.8

OR

100

60 120

120

100

60 = 72

Ex 8: (a) Production of wheat in 1996 was

60,000 tons and in 1997, it was 75,000 tons.

By what percentage did the production

increase?

Sol: % change = = 25%

60,000

(b) In the year 1999, the production of

pulses increases by 25% over that of the

preceding year and reached 400

thousand tons. What was the quantity of

pulses produced in 1998?

75,000 - 60,000

Ex 11: If the population of India is 4.5 times that of

U. S. A, what % of Indias population is

population of U. S. A?

Sol: Assume population of U. S. A = x

population of India = 4.5 x

population of U. S. A as a % of population

of India is

= x100 = 100 = 100 = 22.22%

4.5x 9

2

Ex 12: If salary of A is 20% greater than that of

B then by what % is Bs salary less than that

of A?

Sol: Explain to the students that since the

salary of A is 20% more than B let us

assume, Bs salary = 100 thus making As

salary = 120.

% by which Bs salary is less than As

salary = x100 = 16.67%

120

Note: Explain to the students that these problems

can also be solved by a formula as given

below.

If A is x % more/less than B then B is

Make a note to the students that in

course of time if it happens that %

change is negative then it indicates that it

is a case of decrease.

Note:

Also say to them that same formulae can be used in

concepts like finding out % reduction/increase in a

value which has already increased/reduced by x %

and the reduction/increment in the consumption of an

item when its price is increased/reduced by x % to

make expenditure constant.

Ex 13: If As salary is 10% less than Bs salary then

by what % is Bs salary greater than that

of A?

Sol: Explain to the students that this problem

can be solved using the convention referred

ie., by assuming some values as in before

problem and also by application of formulas

as below:

Since here the initial product is 10% less than

next one then it is

100 - x

initial one

i.e., = % = = 11.11%

100 - 10 90

Ex 14: If the price of an item is increased by 25%, by

what % should it be brought down to bring

back to the original level?

Sol: Explain to the students that since the price

of an item is increased by 25% let us assume

that before increase it was 100. An increase

of 25% makes it 125. Now this 125 has to be

Or solving by conventional method if

population of Hyderabad is x then

population of Delhi is

X + 220% of x = 320% of x

Required ratio of population of Delhi to

Hyderabad

320 x

= = 3.2times

x

x 1 2

9

2x - x

3x - x

4x - x

nx - x

If the production of sugar tripled from

1989 to 1990 then find % increase in

production of sugar from 1989 to 1990

Explain to the students, that if we remember

the concept above straight away we can say

% increase is 200% or solving it in

conventional method we get. If x is

production in 1989.

% increase = x100

x

= 100 = 200%

x

that of Hyderabad then the population of

Delhi is how many times of Hyderabad.?

Explain to the students, that this problem is a

reverse way of the concept explained before

and since the % increase is 220% the

number of times the value is would be

100x

% more than

10010 1000

120 - 20

100x

(100 x)

% less/more than A.

220

100

+ i.e 3.2

1

100

Sol:

Sol: Explain to the students that if we assume

production of tea is x in 1979 then its

production in 1980 would be x + 150% of

x = 2.5 x. Now the production in 1981 has to

be 5x. Hence the % change such that

2. 5 x becomes 5x is nothing by doubling the

value i.e., 100%.

5. Percentage point change:

Consider population of city X.

Year Males Females Total Population

1998 40 60 100

1999 50 100 150

From 1998 to 1999

% increase in no.of males = 100 = 20%

50

In 1998 Males are = 40% of total population.

100

and 1999 Males are = 33.33% of total

150

population.

From the above data we can say that number of

males are increase by 20% but number of males are

a percentage of total population is decreased by

approx 6.67 (i.e.40 33.33) percentage points.

Ex 18: The production of wheat formed 20% of the

total food grain production in 1996. In the

next year, the share of wheat in the total food

grain production went up by 5 percentage

points. Find the % change in production of

what from1996 to 1997 if the total food grain

production increased is by 20% from

1996 to 1997

1996 1997

Total Food

grain

production

Wheat

Production

100 120

20 30

10

40

50

brought back to 100 i.e., the % by which it

should be reduced is = 100

125

explain to the students that this problem can

be solved using the formula also.

Ex 15: If the price of an item is decreased by 25%

by what % should it be raised to bring it back

to original level?

125 - 100

Explain to the students that this can also

be solved using the formula as below:

100x25

Required % increase =

100 - 25

%

2500 100

75 3

= = 33.33%

c1

1.25c1 - c1

100x20 2000

105 - 75 30

75 75

4

5

5 1 1

30 - 20

20

% increase in x100 = 50%

Ex 19: Andy scored 75 out of 150 marks in his mid

term exam and 105 out of 150 in end term

exam.

(a) Find the % increase/decrease in his

marks.

Sol: = x100 = x100 = 40%

(b) Find the increase in his marks in terms of

% points.

However it cannot be used to find increase in

perimeter as in that case does not get multiplied as

the case becomes

Now perimeter = 1.1 L + 1.2 B

Thus answer will become cannot be determined if we

do not know exact values of L and B.

Ex 21: If A's income is 16% more than B's income

and B's income is 25% more than C's

income, by what percentage is A's income

more than C's income?

Sol: Method I

C B A

+20

100 125 145

Hence 45% more

Method II

Using successive percentage formula.

25 + 16 + = 45%.

100

7. PROFIT AND LOSS:

I appeared for Last CAT exam. Before going for

exam I purchased 3 pencils. each at Rs.5. In my

exam hall there were 2 students who were not

carring pencils and they were knowing the fact I

was carring extra pencils. One of them offered

me Rs 7 and other offered Rs 4 for one pencil. I

accepted their offer and sold one pencil to each

at the above mentioned prices. Now if you look at

there are three prices available Rs 5, Rs 7 and

Rs 4. If we compare Rs 5 and Rs 7 we realise

that my selling price (S.P) is more than my cost

price (C.P). Hence this transaction gave me a

profit of Rs.2 (i.e. 75). If we compare Rs 5 and

Rs 4, we realise that my S.P is less than C.P.

Hence this transaction resulted in a loss Rs.1

(i.e.5-4).So we can define profit and Loss as.

Profit:

One has a profit when selling price (S.P) is more

than cost price (C.P). Hence

Profit = S.P C.P

Loss:

One has a loss when selling price is less than

cost price .Hence

Method I: A = l b

New Area = 1.2l 1.1b

= 1.32 (lb)

Hence 32% increase in area.

Method II

Using successive percentage formula.

20 + 10 + = 32%

100

Sol: % marks in mid term exam Sol:

= x100 = 50%

150

% marks in end term exam

= x100 = 70%

Change in % points is 70 50 = 20%

75

105

150

20 10

25 16

20% 30%

156

25% 16%

a

+ b + %.

100

20 (-30)

20 30

Ex.1: (a) The C.P & S. P of an article are Rs.400 &

Rs.600 respectively. What is profit %?

Sol:

Sol:

Ex.2:

Sol:

Ex.3:

Sol:

Loss

Profit is difference between S. P & C. P and

is given by 600 400 = 200. And since

everyone knows that

P% = x100 = x100 = 50%

C.P 400

Note: Explain to the students that unless

specified in any problem always P% and L%

should be calculated on C. P

(b) If the price above given are interchanged

what is % of loss.

L % = x100

600

= x100 = 33.33%

An article brought at Rs. 700 was sold at a

profit of 20% what is the S.P?

Explain to the students that we know that

S.P = C. P + P

and P = = = 140

100 100

S.P = 700 + 140 =840

Also say to them that the same problem can

also be done with the help of

S. P = C. P

100

Note that is nothing but a

100

multiplication factor. For 20% increase M.F is

1.2. Hence S.P = 700 1.2= 840

An article is sold for Rs. 400 at profit of 25%.

What is C. P?

Method :

Explain to the students that we know C. P of

the article is given by

S.P = C. P + P

P = 25% of C.P = 0.25 C. P

S. P = C. P + 0.25 C. P

C.P = = = Rs.320

1.25 1.25

Say to them that this problem also can be

solved with the help of formula

C. P = S. P

100 + 8

Method :

As we discussed 25% Profit means if C.P

=100 then S.P =125 .S.P is given, hence

C.P. = 400 = 320

P 200

profit

2

600 - 400

200

600

C.PP

100 + P

- L

100

100

100

1

P%xC.P 20x700

100 + P

100 + P

S.P 400

100

100

125

Ex.4:

Sol:

Ex.6:

Sol:

Ex.7:

Sol:

Ex.8:

Sol:

Ex.9:

Sol:

A, sells an article to B at 25% profit B sells

same to C at 20% profit.

(a) By what % is the C.P of C more than

A?

Explain to the students that in such problem it

is better to take the initial value of the article

as a base value like 100 rather than x.

Hence if we assure the C.P of the article to

A is 100 then C.P of A =100, S. P of A = 125

C. P of B = S. P of A = 125, S.P. of B = C. P

of C = 125 + 20% of 125 = 150.

% by which C.P. of C more than A is

= x100 = 50%

100

Two articles sold at same price. On one

article there is a loss of 10% and on the other

profit of 10% what is P% or L% in the whole.

Using formula,

The percentage loss is given by

% Loss = . = =1.

Two articles which were brought at same

price were sold making 20% profit on one

and 10% loss on other. What %P or % L on

the whole?

Explain to the students that if we assume

C. P of each one is Rs. 100 then Total C.P =

100+100 =200

Total S.P = 120 +90 =210

P % = x100 = 5%

200

The S. P of 12 oranges is equal to C. P of 15

oranges. Find P/L %

Given:

12 SP = 15 CP

SP 15 5

CP 12 4

SP = 5k

CP = 4k

P% = 100 = 25%.

4k

The profit made on selling 20m of cloth is

equal to S. P of 5 m of cloth. Find P%.

20 SP20 CP = 5SP

15 SP= 20CP

SP 21 4

CP 15 3

SP = 4k

CP = 3k

P = k

Profit = 100 = 33 1/3%

3k

(b) If the C. P of A is 600. How much did it cost

to C.

Sol: Explain to the students that the C. Ps of A &

C are varying directly. We can say.

(c) If C brought it

price did A buy it.

for Rs. 1200 at what

Sol: Let cost Price per gm be Rs 1

Bills

1000gms w/

Price/gm 1 Price/gm

S.P 1000 C.P

P% = 100

900

= 11.11%

Ex.10: A merchant dishonestly weighs 100gm short

for 1 Kg that he sells. Find his gain % if his S.

P & C. P are same.

Gives

900

1

900

= =

k

150 - 100

P2 102

100 100

10

100 600

150 x

= x = Rs.900

C.P of C = Rs. 900.

= =

k

1200

P2

1000 - 900

M.P = S.P.

But if discount is given then M.P is considered as

base value (i.e. 100%).

So if some one offers a discount of 15 % then it is 15

% of M.P.

In other words we can say if M.P =100 then S.P = 85

In the case of profit we can have the following chart.

M.P

Discount

S.P

Profit

C.P

Ex.13: Two successive discounts of 20% each is

equivalent to a single discount of what %?

Sol: Method :

Explain to them that if we assume the actual

M.P is 100 then by successive discounts of

20% each the S. P would be 80 & further 64.

Hence the equivalent discount is

= x100 = 36%

100

Ex.14: A trader marks his S.P. at 30% above the

C.P. & later offers a discount. Find the

discount % if trader make a net profit of 17%.

Sol:

MARKED PRICE OR LIST PRICE

On Last Sunday I went in a shop to purchase a shirt. I

short listed one shirt and saw a price tag on that

which was Rs.1000. Now this price is nothing but a

marked price (M.P.). Before making a payment

shopkeeper told me that a festival discount is going

on and he offered me a discount of 15% .Now can we

find the discount amount?

Discount will be 15% of 1000 i.e. Rs.150. This

indicates that discount percentage is always applied

on M.P.

Sol:

Ex.15: By selling at a disc of 30% as trader makes a

profit of 26% by what % does he mark up the

C.P.

Sol: Method :

As 30 % Discount is given,

S.P = 70% of M.P

Similarly, profit is 26%

S.P = 126% of C.P.

70% of Mp = 126% of C.P

M.P 126

C.P 70

30

70

If we assume the C.P. be Rs.100 then the

marked price will be Rs.130.

The net profit is 17% i.e. the S. P will be

Rs.100 + 17% of 100 = Rs. 117. Now as we

know the S. P and M.P the discount % can

be calculated as

% discount = x100

M.P

= x100 = x100 = 10%

130 130

M.P - S.P

130 -117 13

100 - 64

2

3

sold the pen at of actual S.P he would

have made a loss of 10%. What was the

Arvind actually made.

Let C.P = 100

New S.P =90 (as 10% Loss)

But New S.P = Actual S.P

3

90 = Actual S.P

3

Actual S.P = 90 =135

2

Hence 35% Profit.

2

2

3

=

Discount If C.P = 70 then M.P = 126

%discount = 100

Marked Pr ice

If no discount is given then

56

70

Marked up % = 100 = 80%.

Now after the completion of the above

examples still if we find time then the

following example can be covered.

Ex.16: A brought a Cow for 40 gold coins and sold it

to B for 30% P. If B sold it to C for a loss

of 25% find his S. P (in terms of gold coins)

Sol:

Since the C. P of A 40 gold coins is S. P is

sum of C. P & 30% profit.

S.P = 40 + 30% of 40 = 52

The S. P of A and C. P of B is 52 gold

coins. Now B is selling the Cow to C to a

loss of 25%. Hence his selling price is S.P =

52 25% of 52

= 52 13 = 39 gold coins.

Ex.17: Andy sells a scooter to Bandy for a profit of

20% & Bandy sells it to candy for a profit of

6000 making 20% P. Find Andys C.P.

Sol:

A

25

Profit of C is 6 parts Rs 6000

C.P of a is 25 parts Rs 25000.

Ex.18: Mohan purchased a certain amount of rice. If

he sold one third of rice at Profit of 20% then,

find the profit % he needs to make on the

remaining rice to make an overall profit of 25%

Sol: Method :

Let Total C.P = 300

As overall Profit is 25%

Total S.P = 300 +75=375

Now CP1 =100 CP2 = 200

SP1 = 120 SP2 = 375 120 = 255

%profit = 100 =27.5%

200

Ex.19: If discount on an article is increased from

20% to 35% the money realized from the sale

decrease by Rs. 27. What is the total price of

the article.

Sol: Explain to the students that we know,

discount is calculated as % of M. P and

difference between 20% disc & 35% disc. i.e.

35% M.P 20% M.P= Rs. 27

15% M. P = 27

M. P = x100 = Rs.180

15

1

5

30

C

36

1

5

20% =

We have to increase value by 20 %

Let C.P of A = 25

1

5

25

B

30

8

7

255 - 200

5

8

27

(a)

Sol:

Find the ratio of their shares

Since the time of investment is not same ratio

of shares is = 30,000 x 6 : 20,000 x 10

= 1,80,000 : 2,00,000 = 9 : 10

= 120 : 120

= 1 : 1

Now hence the profit will be divided equally

but if we assume total profit to be x of it 10%

i.e. 0.1 x is paid to Abdul as salary hence the

remaining profit i.e. 0.9x is divided equally

among them i.e. 0.45x.

Amount received by Abdul

= 0.45x + 0.1x = 4400

0.55x = 4400

x =

55

Amount received by Basker

= 0.45x = 0.45 x 8000 = 3600

Ex.7: A & B run a business in a certain year the

salary of A turns out to be equal to 90% of

the balance of profit left after his salary was

paid. If A finally received a total of Rs. 35,000

after splitting the remaining profit equally,

then find the total profit made that year.

Sol: Let us assume the total profit to be x and the

salary paid to A is y therefore

y = 20%of (x-y)

y = 0.20x 0.2y

0.2x = 1.2y

x = 6y

1

Ex.4: Aman & Bhavan started a business with 4400x100

2

x

6

After 3 months Vanitha joins her with Rs. y =

30,000. After some more time Cynthia joins

them with Rs. 40,000. If at the end of the

year Vanitha gets Rs. 18,000 out of a total

profit of Rs. 42,000 then how many months

after Vinitha did Cynthia join the business.

Sol: Ratio of shares of these persons is

(Let Cynthia join after x months)

20,000x12 : 30,000 x 9 : 40,000x(12-x)

24 : 27: 4(12x)

Now of the total profit share of Vanitha is Rs.

after the salary is paid profit remaining is x

- =

6 6

this is divided equally among A & B i.e., each

getting a share of + = 35,000

12 6

= 35,000 x = 60,000

12

x 5x

5x x

7x

18,000 9 STOCKS & SHARES

18000 27

47000 24 + 27 + 4(12 - x)

63 =- 51 + 48 4x

4x = 99 63 = 36 x = 9

Cynthia joined after 6 months Vanitha

joined.

After the explanation for the above problem is

over and there is some more time available

then the following problems are to be

discussed.

Ex.6: Abdul started on business with Rs. 10,000&

4 months later Basker joined from with Rs.

15,000. At the end of the year, out of total

profit, Abdul recovers as total amount of Rs.

4,400 which includes a salary of 10% of the

profit what was the amount Basker received.

Sol: The ratio of their shares is

10000 x 12 : 15000 x 8

=

Please refer to the SM Booklet for theory.

Ex.1: How many shares at 10% premium can be

purchased for an amount of Rs. 1,21,000 if

the face value of share is Rs. 100.

Sol: Market value of the share

= Face value + Premium

= Rs. 100 + 10% of 100 = Rs.110

No. of shares that can be brought =

1,21,000

110

= 1100 shares.

Ex.2: What is the dividend income from a 4% stock

worth Rs.4,00,000.

Sol: The rate of return from a stock of 4% worth

Rs.4,00,000 = x4,00,000 = Rs.16,000

100

4

Ex.3:

Sol:

Ex.4:

Sol:

Ex.5:

Sol:

Mohan has 10,000 worth of 6% stock which

he sells at 120 and invests the proceeds in

7% stock at 150. His annual dividend income

now is 16.67% more than earlier income. In

this transaction was be left with a balance

from the sale money did he invest additional

incomes? And how much was the amount.

Assuming the face value to be Rs. 100 the

10,000 worth stock is equal to 100 the 10,000

worth stock is equal to 100 units. They were

sold at 120 thereby getting 120 x 100 = Rs.

12,000. Since the stock was 6% worth annual

income on it is 6 x 100 = 600 Rs.

Next year there was increase of 16.67

thereby making it Rs. 700.

To get an annual income of Rs. 700 he must

have brought units =100 units since

7

each unit costs 150 total investment is 100 x

150 = 15,000.

The amount be incurred was Rs.1,20,000

and spent Rs. 15,000. Hence he had

invested an additional amount of Rs.3,000.

Announce to the students that with this the

topic of PPL is completed.

700

6 7

100 100

18 15.5

x100 & x100

&