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Re: M19-10 [#permalink]
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manchitkapoor wrote:
how will the markup be 1.1x after reducing to half?


I like to use numbers in cases like this. I chose 100 as the original car price since it makes things easy. A 20% markup would make the car $120. If the mark-up is reduced by half (20% to 10%), the car is now priced at $110. So percent change is clearly less than 10% but greater than 5% so you don't really need to do much more math then that.
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Re: M19-10 [#permalink]
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Bunuel wrote:
If the price of a car includes a 20% mark-up, by what percent will the price of the car be reduced if the mark-up is cut in half?

A. \(5\)
B. \(8 \frac{1}{3}\)
C. \(12\)
D. \(14 \frac{1}{3}\)
E. \(15 \frac{2}{3}\)


Let the initial price of car be \(= 100\)

Price of car after \(20\)% mark-up \(= 120\)% of \(100 = \frac{120}{100} * 100 = 120\)

If the mark-up is cut in half, ie; \(10\)%, price of car \(= 110\)% of \(100 = \frac{110}{100} * 100 = 110\)

Required percentage \(=\) ( Difference of price after mark-up cut \(/\) Price of car after \(20\)% mark-up ) \(* 100\)

Required percentage \(= \frac{120 - 110}{120} * 100 = \frac{10}{120} * 100 = \frac{25}{3}\) or \(8 \frac{1}{3}\)

Answer (B)...
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Re: M19-10 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M19-10 [#permalink]
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