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M19-10

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M19-10 [#permalink]

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New post 16 Sep 2014, 01:05
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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. \(15 \frac{2}{3}\)

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Re M19-10 [#permalink]

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New post 16 Sep 2014, 01:05
Official Solution:

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. \(15 \frac{2}{3}\)

If \(x\) is the price of the car without the mark-up, then its price with the mark-up is \(1.2x\). If the mark-up is cut in half, the price of the car will drop to \(1.1x\) and be reduced by \(\frac{1.2x - 1.1x}{1.2x} = \frac{1}{12}\) or \(\frac{100}{12}\% = 8 \frac{1}{3}\%\).

Answer: B
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M19-10 [#permalink]

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New post 04 Oct 2016, 17:43
how will the markup be 1.1x after reducing to half?
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Re: M19-10 [#permalink]

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New post 29 Oct 2016, 10:47
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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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M19-10 [#permalink]

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New post 04 Aug 2017, 06:25
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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New post 04 Aug 2017, 06:42
Assume the price to be x. Mark up becomes 0.2x. If the mark up is half , the amount becomes 1.1x. The reduction percentage is then (0.1x/1.2x)*100. So it becomes 100/12 or option B.
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Re: M19-10   [#permalink] 04 Aug 2017, 06:42
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