Looking for help on finding Profit and Loss Concepts? Then, you have come the right way. Here let us observe some fully solved example problems on calculating profit or loss. You can find step by step solutions to all the Profit and Loss Questions available here. Try Practicing from the Profit and Loss Problems and get acquainted with the concepts better. Learn various methods for Calculating Profit and Loss and solve related problems easily. Assess your preparation standards on the concept and concentrate on the areas you are lagging in accordingly.

**Question 1:**

**If a manufacturer allows 40% commission on the retail price of his product, he earns a profit of 9%. What would be his profit percent if the commission is reduced by 25 percent?**

Solution:

We need to find out the profit % when the given commission is reduced by 25 percent.

Given data:

According to the question consider

Cost price (C.P.) of the product = $ 100

Then, a commission of the product = $ 40

Therefore selling price (S.P.) = $ (cost price (C.P.) – commission)

= $ (100 – 40)

= $ 60

Given that profit = 9%

Therefore Cost price (C.P.) = \(\frac { 100 }{ 100+gain%} \)* S.P

So,

C.P. = $ \(\frac { 100 }{ 100+9 } \)* 60

= $ \(\frac { 6000 }{109 } \)

Now new commission = $ 15

Therefore new selling price (S.P.) = $ 100 – 15

= $ 75

Gain = S.P. – C.P.

= $ (75 – \(\frac { 6000 }{109 } \))

= $ \(\frac { 2175 }{109 } \)

Gain% = (\(\frac { Profit }{C.P. } \)*100)%

=(\(\frac { 2175 }{109 } \)*\(\frac { 109 }{6000 } \)*100)%

= 36.25 %

Hence, gain % is 36.25.

**Question 2:**

**After getting two successive discounts, a pant with the least price of $ 200 is available at $ 125. If the second discount is 14%, find the first discount.**

Solution:

Let the first discount be ‘P%’

Then, 86% of (100 – P) % of 200 = 125

\(\frac { 86 }{ 100 } \)*\(\frac { (100 – P) }{ 100 } \)*200 = 125

100-P = \(\frac {(125*100*100) }{ 200*86 } \)

100 – P = 72.67

P = 100 – 72.67

P = 27.32%

Therefore, first discount price of pant is 27.32%.

**Question 3:**

**A women sells an article at a profit of 20%. If he had bought it at 15% less and sold it for $ 11.50 less, he would have gained 25%. Find the cost price of the article.**

Solution:

Given data:

Consider cost price (C.P.) of article be ‘X’

First selling price of article ‘X’ = 120% of ‘X’

= \(\frac { 120 }{ 100 } \)*X

= \(\frac { 6 }{ 5 } \)*X

Cost price of article for ‘X’ at 75% = 75% of ‘X’

=\(\frac { 75 }{ 100 } \)*X

=\(\frac { 3 }{ 4 } \)*X

Second selling price of article ‘X’ = 125% of 3/4 * X

= \(\frac { 125 }{ 100 } \)*\(\frac { 3x }{ 4 } \)

= \(\frac { 15x }{ 16 } \)

As given the article is sold at $ 11.50 less

Therefore, selling prices are equalized to a reduced price

\(\frac { 6x }{5 } \) –\(\frac { 15x }{ 16 } \) = 11.50

\(\frac { 21x }{80 } \) = 11.50

X = $ 43.8

Almost equal to $ 44

Hence, the cost price of an article is given as $ 44.

**Question 4:**

**A dealer sold three – fourth of his articles at a gain of 25% and the remaining at cost price. Find the profit earned by him in the whole transaction.**

Solution:

A dealer sold his ¾ th quantity with a gain of 25% and the remaining ¼ that its cost price.

Given data:

Consider cost price (C.P.) of whole articles be ‘X’

Cost price (C.P.) of \(\frac { 3}{ 4} \)th quantity = $ \(\frac { 3x}{ 4} \)

Cost price (C.P.) of \(\frac { 1}{ 4} \)th quantity = $ \(\frac { x}{ 4} \)

Total selling price (S.P.) = $ ((125% of \(\frac { 3x}{ 4} \)) + \(\frac { x}{ 4} \))

= $ (\(\frac { 15x}{ 16} \) + \(\frac { x}{ 4} \))

= $ (\(\frac { 19x}{ 16} \))

Profit / Gain = S.P. – C.P.

= $ (\(\frac { 19x}{ 16} \) – x)

= $ \(\frac { 3x}{ 16} \).

Gain % = (\(\frac { gain}{ C.P. } \)*100)%

= (\(\frac {3x}{ 16 } \)*\(\frac {1}{ x } \)*100)%

= 18.75%.

Hence, the gain % of the article is 18.75%.

**Question 5:**

**A man sold two flats for $ 775,000 each. On one he gains 18% while on the other he losses 18%. How much does he gain or lose in the whole transaction?**

Solution:

In this problem he gets an equal amount of profit and loss such cases there is always a loss. Therefore the selling price (S.P.) is immaterial.

Loss % = (\(\frac {common loss and gain %}{ 10 } \))^{2}

= (\(\frac {18 }{ 10 } \))^{2}

= (\(\frac {324 }{ 100 } \))

= 3.24%

The total loss incurred by the person is 3.24%.

**Question 6:**

**Pure petrol costs $ 100 per lit. After adulterating it with kerosene costing $ 50 per lit, a shopkeeper sells the mixture at the rate of $ 96 per lit, thereby making a profit of 20%. In what ratio does he mix the two?**

Solution:

Here, we have two different cost prices for different mixtures and one selling price (S.P.).

Given data:

Cost price (C.P.) of petrol = $ 100 per lit

Cost price (C.P.) of kerosene = $ 50 per lit

Selling price (S.P.) of mixture = $ 96 per lit

As we have two cost prices,

Mean cost price = $(\(\frac {100 }{ 120 } \))* 96)

= $ 80 per lit.

Since they asked us to find a ratio it is easy to find out by the allegation rule

Cost price (C.P.) of a unit Cost price (C.P.) of a unit quantity of $ X item quantity of $ Y item

Mean cost

$ M

(M – Y) (X – M)

Similarly using this concept here,

Cost price (C.P.) of a unit Cost price (C.P.) of a unit quantity of $ 100 item quantity of $ 50 item

Mean cost

$ 80

(80 – 50) (100 – 80)

Therefore, required ratio = 30 : 20

= 3 : 2.

**Question 7:**

**Find cost price (C.P.), when**

**1. Selling price (S.P.) = $ 50, Gain = 18%**

**2. Selling price (S.P.) = $ 51, Loss = 14%**

Solution:

Here, we need to find cost price (C.P.) using below formulae

1. Given data

Selling price (S.P.) = $ 50 & Gain = 18%

C.P. = \(\frac { 100 + Gain%) }{ 100 } \)*S.P.

=$ \(\frac { 100 + 18 }{ 100 } \)*50= $ 59.

2. Given data

Selling price (S.P.) = $ 51, Loss = 14%

C.P. = \(\frac { 100 – Loss% }{ 100 } \)*S.P.

= $ \(\frac { 100 – 14 }{ 100 } \)*51

= $ 43.86.