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**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.

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