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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
&