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# Profit and Loss Questions

Home > Quantitative Aptitude > Profit and Loss > General Questions
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A man buys oranges at Rs 5 a dozen and an equal number at Rs 4 a dozen. He sells them at Rs 5.50 a dozen and makes a profit of Rs 50. How many oranges does he buy ?

A30 dozens

B40 dozens

C50 dozens

D60 dozens

Explanation:

Avg. of both = 4.5 (Rs. 5 and same at Rs. 4 )
so he gain 1 Re.
So, for profit of 50 is 50

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A man gains 10 % by selling a certain article for a certain price. If he sells it at double the price, the profit made is :

A20%

B120%

C100%

D140%

Explanation:

Suppose the cost price of articel is 100,
then selling price is 100+10% = 100+10= 110.
If he sells at double price it becomes 110*2=220
So, [220/100]*100=220 [ selling price ]
so profit = (SP - CP)
= 220-100 = 120

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The price of an article was increased by p%. Later the new price was decreased by p%. If the latest price was Re. 1, the original price was :

ARe. 1

BRs. (1 - p2 / 100)

CRs. (10000 / 10000 - p2)

DRs. (?1 - p2 / 100)

Explanation:

If ooriginal price was x, then
x* (1+p/100)*(1-p/100) =1
x*(1-p^2/10000) =1
x*((10000-p^2)/10000
x= Rs. (10000 / (10000 - p2))

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Monika purchased a pressure cooker at 9/10th of its selling price and sold it at 8% more than its selling price. Her gain is :

A20%

B18%

C10%

DNone of these

Explanation:

Let the S.P. of pressure cooker = Rs. X .
So, C.P of pressure cooker = Rs. 9x/10.
Receipt=108% of Rs. X = Rs 27x/25

Gain=Rs (27x/25*9x/10) = Rs (108x-90x/100) = Rs18x/100

Gain %=( 18x/100*10/9x*100) %=20%

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A man sells two houses at the rate of Rs. 1.995 lakhs each. On one he gains 5% and on the other, he loses 5%. His gain or loss percent in the whole transaction is :

A0.25% loss

B0.25% gain

C2.5% loss

D25% loss

Explanation:

Note: If a person sells two items at the same price; one at a gain of x% and another at a loss of x%, then the seller always incurs a loss expressed as

Loss%=[(Common Loss and Gain%)/10]^2 = (x/10)^2

Loss % = (5/10)^2= 0.5^2= 0.25

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Toffees are bought at the rate of 3 for a rupee. To gain 50 %, they must be sold at :

A2 for a rupee

B1 for a rupee

C4 for a rupee

D5 for a rupee

Explanation:

Solution: 3 pcs for 1 Rs  so 1 pc for 1/3 Rs

For a profit of 50% S.P= 150/100* 1/3 =1/2 Rs for 1 pc

Therefore for a rupee 2 pcs can be sold.

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The printed price on a book is 400, a bookseller offers a 10% discount on it. If he still earns a profit of 12%, the CP of the book is:

A280

B352

C360

D300

Explanation:

Lets assume book CP = Rs. X

Given printed price = Rs 400
Bookseller offer discount of 10% = 0.1*400= Rs. 40
Therefore new S.P = Rs.360
Condition: He still earn profit of 12%
x*120/100 = Rs. 360
=>x = Rs.360*100/120= Rs.300

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Nitish sold his watch and sunglasses at a loss of 4% and gain of 4% respectively for 2600 to Kamal. Kamal sold the same sun glasses and watch at a loss of 4% and gain of 4% respectively for 2700. The price of watch and sun glasses to Nitish were.

ARs.1960, Rs.700

BRs.2000, Rs.1000

CRs.1500, Rs.700

DRs.1500, Rs.700

E(Rs.800, Rs.2000)

Explanation:

Lets assume the CP of watch = Rs x
and sunglasses = Rs y.
2600=96x/100 + 104y/100
2700= 104x/100 + 96y/100
By solving,
y=700
x=1960

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A man purchased a waten for Rs. 400 and sold it at a gain of 20% of the selling price. The selling price of the watch is:

A340

B400

C480

DNone of these

Explanation:

S.P =((100+ GAIN%)/100) *C.P
= (120/100 ) *400
= 480

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Cost of an item is x. It's value increases by p% and decreases by p% Now the new value is 1 rupee, what is the actual value ?

A10000/[10000-(p^2)]

B1000/[1000-(p^2)]

C100/[100-(p^2)]

DNone of these

Explanation:

Given cost of an item = Rs. X
Item cost by increasing p% will be = x(100+p)/100
After p% decrement, value = {x+(xp/100)}-[{x+(xp/100)}p/100]
= [x+(xp/100)] [1-(p/100)]
= x[1+(p/100)][1-(p/100)]
Givne new value is 1
x[1-((p^2)/10000)] = 1
x = 1/[1-((p^2)/10000)] = 10000/[10000-(p^2)]

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