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 ?
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 :
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 :
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 :
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.
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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