A man bought an article at Rs.600. He marks up the cost of article and

Question

A man bought an article at Rs.600. He marks up the cost of article and sells it in such a way that if he will give 46(2/3)% discount then loss occur to him is equal to profit earn by him if he will give 20% discount. Find the Mark price of article?
(a) Rs.1500
(b) Rs.800
(c) Rs.900
(d) Rs.1200
(e) Rs.1800

sa119 · Accepted Answer

Answer is C

Man bought the article for 600 (this is the cost price). 
Let us assume that he adds up a cost of x to mark up the selling price to (600+x)

So CP= 600 and SP= 600+x

Now if he offers a discount of 46(2/3)% his new SP= 53(1/3)% of 600+x

Loss= CP- SP
= 600- 53(1/3)% (600+x)

If he sells at a discount of 20%, his SP in this case= 80% (600+x)

Profit= SP- CP
= 80% (600+x) - 600

Now given, Loss= Profit

600- 56(1/3)%(600+x) = 80% (600+x) - 600

Solving this we get the answer as 900.

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stne · Answer

46\frac{2}{3}  = \frac{140}{3} -> so a discount of  \frac{140}{3} % on m.p -> S.P.=  \frac{160}{300} m.p

and a discount of 20% on m.p. ->S.P.=  \frac{80}{100} m.p.

600 - \frac{160}{300}m.p. = \frac{80}{100}m.p. -600

on solving above = m.p.= 900

Hope it's clear.

pryadao · Answer

Did not understand the solution, can someone please explain this ?

SaquibHGMATWhiz · Answer

Solution:

CP of the article = 600  
 Let the marked price be MP  
 Accroding to the question, loss while giving  46\frac{2}{3}  or  \frac{140}{3}\%  is equal to profit while giving 20% discount  
 Thus, we can write  600-MP(1-\frac{140}{300})=MP(1-\frac{20}{100})-600 
 ⇒MP(1-\frac{20}{100})+MP(1-\frac{140}{300})=600+600 
 ⇒MP(1-\frac{20}{100}+1-\frac{140}{300})=1200 
 ⇒MP(2-\frac{60}{300}-\frac{140}{300})=1200 
 ⇒MP(2-\frac{2}{3})=1200 
 ⇒MP \frac{4}{3}=1200 
 ⇒ MP=\frac{1200	imes 3}{4} 
 ⇒MP=900

Hence the right answer is    Option C

ajay0520 · Answer

is there any logic for us to land to understanding that 46(2/3) = 140/3 or 92/3. just got headache because of this.

46(2/3)% = (46*3+2)/3 = 140/3

and not (46*3)/2 = 92/3

luisdicampo · Answer

Deconstructing the Question

Let the marked price be   M  .

The cost price is

600

There are two discount cases.

With a discount of

46\frac{2}{3}\%

there is a loss.

With a discount of

20\%

there is a profit.

The problem states that the loss in the first case equals the profit in the second case.

First note that

46\frac{2}{3}\%=\frac{7}{15}

Step-by-step

Selling price with discount

46\frac{2}{3}\%

M\left(1-\frac{7}{15}ight)=M\left(\frac{8}{15}ight)=\frac{8M}{15}

Loss in this case

600-\frac{8M}{15}

Selling price with discount

20\%

M\left(1-\frac{20}{100}ight)=\frac{4M}{5}

Profit in this case

\frac{4M}{5}-600

Set loss equal to profit

600-\frac{8M}{15}=\frac{4M}{5}-600

Move terms

1200=\frac{4M}{5}+\frac{8M}{15}

Convert to denominator  15

\frac{4M}{5}=\frac{12M}{15}

So

1200=\frac{12M}{15}+\frac{8M}{15}

1200=\frac{20M}{15}=\frac{4M}{3}

Solve

M=1200	imes\frac{3}{4}=900

Answer C

A man bought an article at Rs.600. He marks up the cost of article and

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A man bought an article at Rs.600. He marks up the cost of article and

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