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July 29, 2023
Profit and Loss is the most important topic in quantitative section of all the job entrance exams. This page here on contains Tips, Tricks and Shortcuts for Profit and Loss questions.
Type 1 Problem- Seller has two Articles for same price, but first article is sold at x% profit and other at x% loss. Total Profit/Loss incurred by him is not 0%
Way to solve this question is –
Apply direct formula Loss = ( \frac{x}{10})2%
Proof with example -> Let us assume the articles were sold at Rs1200, and 20% profit in case 1 is made and 20% loss in case 2 is made.
SP in case 1(Profit) – 1200
Thus CP = ( \frac{100}{100 + Gain} )× SP = ( \frac{100}{120}) × 1200 = ( \frac{5}{6}) × 1200 = 1000
SP in case 2(Loss) – 1200
Thus CP = ( \frac{100}{100 – Loss}) × SP = ( \frac{100}{80}) × 1200 = ( \frac{5}{4}) × 1200 = 1500
Total SP = 1200 + 1200 = 2400
Total CP = 1000 + 1500 = 2500
Loss = ( \frac{CP – SP}{CP}) × 100 = ( \frac{100}{2500}) × 100 = \frac{100}{25}= 4%
Also from direct formula above = ( \frac{20}{10})2
In such cases always, loss is incurred.
Type 2 Problem- Where no CP or SP is given. But whole concept is about Percentages.
Way to Solve Type 2 Questions
Assume the CP to be 100 and then solve the whole problem.
Example. In a transaction, the profit percentage is 80% of the cost. If the cost further increases by 20% but the selling price remains the same, how much is the decrease in profit percentage?
Let us assume CP = Rs. 100.
Then Profit = Rs. 80 and selling price = Rs. 180.
The cost increases by 20% → New CP = Rs. 120, SP = Rs. 180.
Profit % = \frac{60}{120} * 100 = 50%.
Therefore, Profit decreases by 30%.
Type 3 Problem- There are two Articles and you have to calculate total loss or profit.
Way to solve type 3 Problem
Now these problems are generally easy. But the whole point of solving is not to even use a pen and solve in 20 seconds.
Example. A man bought some toys at the rate of 10 for Rs. 40 and sold them at 12 for Rs. 60. Find his gain or loss percent
Cost price of 10 toys = Rs. 40 → CP of 1 toy = Rs. 4.
Selling price of 12 toys = Rs. 60 → SP of 1 toy = Rs. \frac{160}{12} = 5
Therefore, Gain = 5 – 4 = 1.
Gain percent = \frac{1}{4}\times 100 = 25%
Now in your mind you must do value 4 and 5 and \frac{1}{4} = 25%.
Type 4 Problem- CP of y items is same as SP of x items and Profit or Loss of some percentage is made.
Way to solve type 4 Question
The cost price of 10 pens is the same as the selling price of n pens. If there is a loss of 40%, approximately what is the value of n?
Solution:
Let the price of each pen be Re. 1.
Then the cost price of n pens is Rs. n and
the selling price of n pens is Rs. 10.
Loss = n-10.
Loss of 40% → ( \frac{loss}{CP}) × 100 = 40
Therefore, \frac{n – 10}{n} × 100 = 40 → n = 17 (approx)
Type 5 Problem-
If the price of an item increases by r% , then the reduction in consumption so that expenditure remains the same is
or
If the price of a commodity decreases by r% then increase in consumption , so as not to decrease expenditure on this item is
Way to solve type 5 Questions
Just apply the following two formulas
Case 1
Case 2
Type 6 Problem(IMP)- A dishonest dealer claims to sell his goods at cost price ,but he uses a weight of lesser weight .Find his gain%.
Way to solve type 6 Problem
Apply the following formula directly
Gain % = \frac{True \: Weight – False\: Weight}{False \: Weight} × 100
Example. Shopkeeper bought a product for Rs1000 per kg and is selling that at the same price. However he uses, a weighing scale that gives scale of 1kg for every 800gms. What is his profit?
Answer will be( \frac{1000 – 800}{800})\times 100 = ( \frac{2}{8}) × 100 = 25% profit.
Type 7 Problem(IMP)-
These questions will not be there for exams like AMCAT and Cocubes etc but for eLitmus.
A shopkeeper sells an item at a profit of x % and uses a weight which is y % less .find his total profit
Use Formula: Gain% =
\left ( \frac{\%Profit +\%{ \text Less \: in \: weight}}{100 – \% { \text Less \: in \: weight}} \right )\times 100
When dealer sells goods at loss on cost price but uses less weight .
Profit% or Loss% = \left ( \frac{\%Less\: weight -\% Loss}{100 – \% Less\: in \: weight} \right )\times 100
A dishonest dealer sells goods at x % loss on cost price but uses a gms instead of b gms to measure as standard, his profit or loss percent :-
Use Formula: Profit% or Loss% = \left ( 100 – Loss\% \right )\left ( \frac{Original\: Weight}{Altered\: Weight} \right )- 100
Note :- profit or loss will be decided according to sign .if +ive it is profit ,if –ve it is loss .
Case-1: When dealer sells product at profit but alters weight
Profit% or loss% = [100+gain%][ \frac{1000}{Altered \: Weight} ] – 100
Case-2: When dealer reduces weight in terms of percentage and earns profit
Example: A shopkeeper sells an item at a profit of 20 % and uses a weight which is 20% less. Find his total profit.
Applying the first formula
( \frac{20+20}{100 – 20}) × 100 = 50%
Case-3: When dealer sells goods at loss on cost price but uses less weight.
Note :- profit or loss will be decided according to sign . If +ive it is profit ,if –ve it is loss.Example: A dishonest dealer sells goods at 10% loss on cost price but uses 20% less weight. Calculate profit or loss percent.
Solution:Apply formula: Case 2 Formula
\left ( \frac{20-10}{100-20} \right )x 100
= \frac{25}{2}%
Here sign is positive so there is a profit of 12.5%
Case 4
Example: A dishonest dealer sells products at 10% loss on cost price but uses 2 gm instead of 4 gm . what is his profit or loss percent?
Apply formula :
[100-10] \frac{4}{2}-100 = 80%
Note :- Profit or loss will be decided according to sign. If +ive it is profit ,if –ve it is loss .
Example: A shopkeeper uses 940 gm in place of one kg. He sells it at 4% profit. What will be the overall profit or loss?
Solve this on your own, answer is 10.6%
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