APTITUDE PREPARATION: Profit and Loss


Introduction:

Selling Price is represented by SP; Cost Price is represented by CP
Percentage refers to ‘for 100’; Percentage is represented by %
Gain (or) Profit = (SP-CP) … SP > CP
Gain % =(Gain x 100)/CP … CP will be in the denominator
Loss = (CP-SP) … CP > SP
Loss % =(Loss x 100)/CP … CP will be in the denominator
Percentage change = (Actual Change x 100)/ Initial Value
In general, the initial value will be in the denominator
A fraction (1/2) is represented as 0.5 in decimal form and (0.50×100) = 50% in percentage form
If a number is increased by (1/n) times, it should be decreased by (1/n+1) times to bring back to the original value and vice versa
If a value increases successively by p%, q% and decreases by r%… then, the net multiplication factor
= [(100+p)/100]x[(100+q)/100]x[(100-r)/100]…
That is, ‘+’ if increases and ‘-‘ if decreases

Example 1:

A person bought some apples at a price of 6 for Rs. 20 and sold them at a rate of 8 for Rs. 30. Find the gain or loss percentage.

Solution:

To avoid fractions, we can make the number of apples to the LCM of 6 and 8 = 24.
CP of 24 apples = Rs. 20(24/6) =80
SP of 24 apples = Rs. 30(24/8) = 90
Gain = SP-CP = 90-80 = 10
Gain % = (Gain x 100)/CP = (10×100)/80 = 12.5 %

Example 2:

A merchant offers a discount of 10% on cash payment. At how much percent above CP must he mark the goods (MRP) to make a profit of 8%?

Solution:

Assume the CP = Rs. 100
SP = Rs. 108 = 90% of MRP … (considering 10% discount)
108 = 0.9 x MRP … (90% = 90/100 = 0.9)
MRP =(108/0.9) = Rs.120
That is, the MRP should be 20% above the CP …(since Rs. 20 above Rs. 100 equals 20%)

Example 3:

If the price of wheat is reduced by 10%, a person can purchase 25 kg more for Rs. 2250. What is the reduced price of wheat per kg?

Solution:

If the price is reduced by 10%, he can save 10% of 2250 = Rs. 225.
For this Rs. 225, he can get 25 kg more wheat.
Reduced price = 225/25 = Rs. 9


Example 4:

A man sold his scooter for Rs. 25000 and incurred some loss. Had it been sold at Rs. 28000, the gain would have been double the former loss amount. Find the cost price.

Solution:

SP at Rs. 25000:
Loss Amount = (CP – SP) = CP – 25000
SP at Rs. 28000:
Gain = (SP – CP) = (28000 – CP) = 2(Loss Amount)
= 2(CP – 25000)
(28000 – CP) = 2CP – 50000
3CP = Rs. 78000.
CP = Rs. 26000.

Example 5:

A person buys milk at some price per litre. He mixes some water with the milk and sells again for the same price. If he makes a profit of 20%, what is the quantity of water added for every litre of milk? Also, find the quantity of water present in 1 litre of the mixture.

Solution:

Let the CP of milk = Rs. 100 per litre
For making 20% profit, he should earn Rs. 120
For this Rs. 20, he should add 200 ml or 0.2 litre of water for every litre of milk.
Quantity of water in one litre of mixture:
In 1.2 litre of the mixture, 0.2 litre water is present
In 1 litre of the mixture, (0.2 x 1)/1.2 = (1/6) litre of water is present

Example 6:

The price of cooking oil rises from Rs. 240 in October to Rs. 300 in November. How much percentage a family should reduce in the monthly consumption to keep the monthly expenses same.

Solution:

Fraction increase in price = (300-240)/240 = 60/240 = 1/4
Fraction reduction in consumption = 1/(4+1) = 1/5, to keep the expenses same
Percentage reduction in consumption = 100 x (1/5) = 20 %


Example 7:

The speed of a car increases by 25% for some distance and then by 30%. Afterwards, the speed is reduced by 20%. What is the net percentage of speed with respect to its initial speed?

Solution:

Let the initial speed = V
Final speed = V(1.25)(1.3)(0.8) = 1.3V
Net percentage increase in speed = (Final speed – Initial speed) x 100/ Initial speed
= [(1.3V – V) x 100]/ V = (0.3 V x 100)/ V
= 30 %

Example 8:

A merchant buys apples at certain price per dozen and sells them at a price 10 times that price for 100 apples. What is the percentage gain (or) loss?

Solution:

Let the cost price of 1 apple = CP
Cost price of 12 apples (one dozen) = 12CP
Cost price of 100 apples = 100CP
Selling price of 100 apples = 10 times of cost price of 12 apples = 10(12CP) = 120CP
Considering 100 apples;
Gain % = (SP-CP)x100/CP
= (120CP – 100CP)x100/100CP
= 20 %.

Example 9:

P’s salary is 20% less than Q’s salary which is 15% less than R’s salary. By how much percentage R’s salary is higher than P’s salary?

Solution:

Let the salaries are P, Q and R respectively
P = 0.8Q and Q = (0.85R)
We have to relate P and R
P = 0.8(0.85R)
R = P/(0.8 x 0.85) = 1.4706P (or)
R’s salary is 47. 06 % higher than P’s

Example 10:

A man sells an article at 25% profit level. Had he bought it at 25% less and sold at Rs. 25 less, he would have still gained 25%. Find the cost the article.

Solution:

Assume the actual cost price = CP
First selling price, SP1= CP + CP(25/100) = 1.25CP
Second CP, CP2 = CP- CP(25/100) = 0.75CP
Second selling price, SP2 = (SP1-25) = 1.25CP-25
Gain % = 25% = (SP2-CP2)x100/CP2 = [(1.25CP-25) – (0.75CP)]x100/0.75CP
[(0.5CP-25)]x100 = 25 x (0.75CP)
50CP – 2500 = 25 x 0.75 CP = (25 x ¾)CP = (75CP/4)
50CP – (75CP/4) = 2500
(200CP-75CP)/4 = 2500
125CP = 10000
CP =(10000/125) = Rs. 80.

About Author

Dr. BASKAR .A,

a former Research Scientist Government of India, presently working as a Professor in the Dept of Mechanical Engineering, Authored books on Kinematics and Dynamics of Machinery, Also Published Numerous Research papers moreover a Geek in Mathematics

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