## CHAPTER – 10 : PROFIT AND LOSS

INTRODUCTION

In any competitive examination the quant section off the paper will always consist 2-3 problems based on this concept because profit and loss this concept is used everywhere weather it is our day to day life or in business.
Now to understand this we will define this terms related to it which are as follows:-

1.     Cost price (CP)
2.     Selling price (SP)
3.     Profit (P)
4.     Loss (L)
5.     Marked price
6.     Discount (D)

Cost price:
Cost price is the price at which shopkeeper bought the product.
Selling price:
Selling price is the price at which the shopkeeper sold the product.
Profit:
If CP < SP then the profit has been occurred in whole transaction.
Loss:
If CP > SP then in this transaction loss has been found
Marked price:
The price at which the price is marked with the mark up to and its relation with CP and SP is,
MP = CP + (% markup on CP)
MP = SP + discount
Discount = d % on MP
= ((d×MP)/100)  ………(d = Discount percentage)
Note: concept of discount is explained briefly later in this chapter.

IMPORTANT FORMULA

1.       Profit = SP – CP
2.     Loss = CP – SP
3.     Profit Percentage = ((SP-CP)/CP)× 100 = (Profit/CP) × 100
4.     Loss Percentage = ((CP-SP)/CP)× 100 = (Loss/CP) × 100
5.     SP = ((100+Profit%)/100) × CP  = ((100-Loss%)/100) × CP
6.     CP = (100/(100+Profit%)) × SP  = (100/(100-Loss%)) × SP

Note: Profit and Loss are always reckoned on cost price

SOLVED EXAMPLES

Q.1. A book seller buys a lot of book at RS 700 and later he sells the same lot at RS 900 Find his profit OR Loss if any? And also find Profit/Loss Percentage?
Solution:
Here CP = 700
SP = 900
SP > CP     ∴ Profit will occur in whole transaction.
∴ Profit = SP – CP
= 900 – 700
= 200
Now, to find profit ‘%’ we know that it is always counted on the cost price.
∴  Profit Percentage = ((SP-CP)/CP)× 100, = (Profit/CP) × 100
= 200/700×100
= 200/7
= 28.57%

Q.2. By selling a refrigerator at RS 16000 a wholesaler suffers a loss of 4% what is the cost price of Refrigerator? At what price he should sell in order to gain 8%?
Solution:
Here SP = 16000
Loss = 4%
CP =?
By formula,
SP = ((Loss-Loss%)/100) × CP
16000 = ((100-4)/100) × CP
(16000×100)/96 = CP
CP = 16666 Rs.

Q.3. By selling 16 bananas a vendor losses the selling price of two bananas find the loss percent?
Solution:
Let the CP of one banana be 1 RS.
CP of 16 Bananas = 16 RS
Now, Loss suffered = (2)× 1
= 2 RS
Loss = CP – SP
2 = CP – 16
CP = 18 Rs.
Percentage Loss = (Loss/CP)× 100
= 2/18× 100
= 100/9 = 111/9 %

Q.4. A trader marked up the goods by 15% and then gives a discount of 10%, what is his profit/loss Percentage?
Solution:
Let CP = 100
MP = 100 + 15% 100
= 100 + 15/100×100
= 100 + 15
= 115
Note:
DISCOUNT = D% × MP
DISCOUNTED PRICE = SP
= MP – (D% MP)
= MP – DISCOUNT
= SP
Now Discount is always given on marked price
Discounted Price = ((100-D)/100) MP
= ((100-10)/100)×115
= (90×115)/100
= 103.5
= Selling Price

Profit % = ((103.5-100)/100)×100
= 3.5%

METHOD ‘100’

This method is very much useful when in the question Profit/Loss Percentage is asked because rather than assuming the unknown by ‘x’ and then finding the CP, SP, MP and then finding the Profit/Loss/Discount% will be the tedious
e.g. 1.  CP = 100 and Profit% = 10
Then SP = 110
e.g. 2.  If SP increases by 2.5% Due to which Profit % increase from 20% to 25% what is the % increase in CP?
CP = 100
SP 120 -> 150         (by increase 25%)
Profit 20% -> 25%       (from 20 to 25%)
SP = (100 + 25)×18/100
= 150/1.25
= 120
% change in CP = ((120-100)/100)×100
= 20%

## TYPES  OF PROBLEMS IN PROFIT AND LOSS

1.       CP of x items is equal to SP of y items and vice-versa.

2.     Discount based

3.     Successive Discount/Profit/Loss

## Type – 1  WHEN CP OF ‘X’ ITEMS IS EQUAL TO THE SP OF ‘Y’ ITEMS.

In these category sum no. of exact CP and SP will be given they will be given relatively like CP of 5 items is equal to SP of 3 items, CP of 6 items is equal to SP of 8 items etc and from the question we have to also deduct weather profit or loss occurring in entire transaction.
Q.1. If the cost price of 16 pencil is same as the selling price of 20 pencil then find gain or loss?
Solution:
CP of 16 pencil = SP of 20 pencil
CP × 16 = SP × 20
CP/SP = 20/16  = 5/4                      …here CP > SP (5 > 4) loss is Occurring in entire transaction
CP = 5x, SP = 4x
Loss % = ((CP-SP)/CP)× 100
= ((5x-4x)/5x)× 100
= 1/5 × 100 = 20%

Q.2. A retailer buys 30 lamps at the marked price of 25 pens from wholesaler if he sells lamps giving discount of 5% what is the profit/loss%?
Solution:
Let the marked price of each lamp =1 RS
CP of 30 lamps = MP of 25 lamps
= 25 × 1
= 25 RS
Now SP of 30 lamps = (100 – D) MP of 30 lamps
= (100 – 5) × 30
= 28.5 RS
Here SP > CP
Profit will Occur.
Profit Percentage = ((SP-CP)/CP)× 100
= ((28.5-25)/25)× 100 = 14%.

## Type – 2  DISCOUNT RELATED

The purpose of discount is to increase the selling price of product by some amount which will be further reduced price from the marked price of the product this is generally done to attract new customers by showing discount avoid loss because of bargaining by the customer empty the stock etc.
Formula:-
Marked price/Printing price/List price = CP + Mark up
MP = SP + Discount
Discount = D% on MP
SP = MP – Discount
Note:
Profit and loss are always counted on cost price unless given otherwise.
Discount is always counted on marked price unless given otherwise.

Q.1. If the marked price of the article is 500 RS and the discount given is 15% then what is the selling price?
Solution:
MP = 500
D = 15%
By Formula
SP = MP – Discount
= 500 – (15% 500)
= 500 – 75
= 425 Rs.

Q.2. If markup percentage of an article is 40% and discount Percentage is 18% then what is the profit?
Solution:
BY 100 Method
CP = 100
MP = 150           …[CP + % markup on once]
SP = 135             …[MP – D% on MP]
Profit = ((135-100)/100) ×  100
= 35%

Q.3. What is the percentage profit in selling an article at discount of 15% which was earlier being sold at 30% profit?
Solution:

By method 100
Case – 1                                                    Case – 2
CP = 100                                                   CP = 100
SP = 130                                                   SP = 130 – 130 × 0.15
…[=MP – D%MP]
MP = SP                                                                    = 130 × 0.15 = 10.5 %
Note: When discount is not given then selling price is marking price
Percentage Profit = (SP-CP)/CP×100
= (110.5-100)/100×100
= 10.5%

Q.4. A shopkeeper marks his goods such that he can make 30% profit after giving 30% discount. However instead of 15% customer availed 20% Discount find the new profit percentage?
Solution:

Let CP = 100
SP = 130
and SP = 0.85 MP
MP = 130/85×100
= 152.9≈ 153
Now discount availed by customer is 20%
SP = 0.8 × MP
= 0.8 × 153 = 122.4
New Profit = ((122.4-100)/100) × 100 = 22.4%

## Type – 3  SUCCESSIVE DISCOUNT / PROFIT / LOSS

In the whole supply chain product moves from manufacturer to wholesaler, wholesaler to retailer, retailer to shopkeeper, shopkeeper to customer. So at every stage same amount of profit or loss is incused Depending upon the situation similarly on discount as well one two or more than two discount are given on the same product.
e.g. on a product A 10 %  discount is given and after bargaining customer avails 4% discount
Total discount in this case will be
Let MP = 100            D1 = 10
SP = 90
Again 4% discount is given
D2 = 0.04 × 90
= 3.6
SP = 90 – 3.6 = 86.4
Overall discount after successive discount of 10%
And 4% is = ((100-86.4)/100)×100
= 13.6%
Note: successive discount of 10% and 4% ≠ 10 + 4
≠ 14%

Q.1. If the manufacturer gains 8%, the whole seller gains 10% and the Selling price  if cost price for manufacturer is 1200 RS?
Solution:
Here there is no need to calculate CP and SP in each transaction. We will apply one direct formula,
SP = ((100+gain)/100)×CP
= ((100+8)/100)     ×    ((100+10)/100)     ×    ((100+20)/100)   ×   CP
Manufacturer               wholesaler                    retailer
SP = 1.8 × 1.1 × 1200
= 2851.2 Rs.

Q.2. The marked price of a gadget is RS 1000. A shopkeeper offers 20% Discount on this gadget and then again he offers 10% discount on new price and then again he offers 30% Discount on the new price. How much will you have to pay finally?
Solution:            Let the MP = 1000
After D1 = 20% = 0.8 × MP
SP=800
After D2 = 10% = 0.8 × 800
SP=640
After D3 = 30% = 0.7 × 720
SP=504
I will get gadget at 504 Rs.
Note: successive discount of x % and y % on the product will be calculated as total discount by the formula
Overall discount = (x + y - xy/100)

e.g. Successive Discount of 5% and 10% will be equal to overall discount = 5 + 10 – (5×10)/100
= 15 – 0.5
= 14.5%

when x %, y % and z % these successive discount is given it will be easy to find the overall discount by taking x % and y % and then overall of x % and y % with z %.
for e.g. in Q.2. Successive discount of 20%, 10% and 30%
x = 20%                 (x+y)overall = x+y - xy/100
y = 10%                                    = 20 + 10 - 200/100
z = 30%                                    = 30 – 2  = 28%
now, (x + y)overall = 28 + 30 - (28×30)/100
= 58 – 8.4
= 49.6%

## Type – 4  ADULTERATION / DISLOYAL SHOPKEEPER

Adulteration is the process in which the quality of food is lowered by either by addition of inferior/cheaper quality of material or by extraction of valuable ingredients with the purpose of gaining more profit.
e.g. A milkman selling a milk 70 Rs per litre adds 100 ml of water in it and claims at selling at cost price
His gain = ((70-63)/63)×100
= 11.11 %       …here   [70 = SP of 1L milk & 60 = CP of 900 ml milk]
Concept of false weight is used by disloyal shopkeeper to gain more profit by selling it at lower price or they can give discount on selling price.
e.g.  A shopkeeper at grocery store uses weight of 950g (any weight less than 1kg) marked with 1 kg to equate with goods of 1 kg.

Q.1. One liter of oil cost 200 RS. In order to get more profit shopkeeper mixes it with another oil costing with 100 RS per liter if shopkeeper sells the oil at the rate of 180 per liter and making profit of 25% then what is the ratio of expensive oil to cheap oil?
Solution:
With the help of Alligation method it can be solved easily
As SP = 180 RS
P% = 25%
SP = 125/100× CP
CP = 100/125× 180
= 144
Now 144 RS per liter is the price of mixture

200-144= 56

144-100= 44

(expensive oil)/(cheap oil) = 44/56

= 11/14
Required ratio = 11:14

Q.2. A dishonest shopkeeper sells rice at 16/kg which he has bought for 14/kg and at the time of weighing he uses 900g of weight instead of 1000g
Find his actual profit?
Solution:
CP of 1000g = 14 RS
CP of 900g   = 12.6 RS
SP of 900g = 16 RS
Now, % profit = (SP-CP)/CP×100
= (16-12.6)/12.6×100
= 3.4/12.6×100
= 26.98
≈ 27%

Q.3. A shopkeeper marks up the product by 60% and gives discount of 20%. Apart from it he also weight 10% less amount while selling the product. What is the net profit of shopkeeper?
Solution:

By method 100

After mark up
Let CP = 100                          MP = 160
CP = 90                        SP = 128
P% = ((SP-CP)/CP)×100
= ((128-90)/90)×100
= 42.2%

Q.4. A disloyal wholesaler cheat his manufacturer and his retailer by getting 20% more from his manufacturer and selling 10% less to his retailer by using false weight. What is his total profit if he sells product at CP?
Solution:
This type of sum can be easily solved Rationally,
Let’s say he has 100 Rs.
For 100 Rs he is getting product of worth 120 Rs
And for 100 Rs he is selling product worth of 90 RS
Profit = (120-90)/90× 100
= 1/3 × 100
= 33.33%

Note:

1.       Whenever only percentage/profit/loss/discount is asked, use method 100

2.     Whenever actual profit and loss is asked. do it by normal method by assuming the unknown as ‘x’ Posted By: Hardik Kaneriya      • Time Left

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