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M19-13

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M19-13 [#permalink]

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1000 dollars were converted into pounds and then the pounds were converted back into dollars at the same exchange rate of \(x\) pounds per dollar. If a commission of \(y\%\) is levied on any exchange operation, what dollar amount was left after the exchanges?


(1) \(x = 0.6\)

(2) \(y = 5\)
[Reveal] Spoiler: OA

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Official Solution:

Whatever the value of \(x\), the first exchange turned 1000 dollars into the equivalent of \(1000(1 - \frac{y}{100})\) dollars. This amount, in turn, became \(1000(1 - \frac{y}{100})^2\) dollars after the second exchange. To answer the question we need to know the value of \(y\).
Statement (1) by itself is insufficient. S1 tells us the value of\(x\), not \(y\).

Statement (2) by itself is sufficient. S2 tells us the value of \(y\).


Answer: B
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Re: M19-13 [#permalink]

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HI BRUNEL

CAN U PLS ELABORATE THE LOGIC BEHIND

1000(1−y/100) dollars

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Re: M19-13 [#permalink]

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New post 26 May 2016, 02:32
sidagar wrote:
HI BRUNEL

CAN U PLS ELABORATE THE LOGIC BEHIND

1000(1−y/100) dollars


sidagar: the question said y% means y% = y/100.

Also, final value you receive after exchange = begin - commission (on begin value) = 1000 - 1000*y% = 1000 (1-y%) = 1000(1-y/100).

Cheers.
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Re M19-13 [#permalink]

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I think this is a high-quality question and I agree with explanation.

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New post 10 Oct 2016, 07:42
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. I agree with the impact of Y on exchange but what about x, once the exchange operation is done what about the loss through exchange rate, I believe that should also be converted.

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chetanyasahu1 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. I agree with the impact of Y on exchange but what about x, once the exchange operation is done what about the loss through exchange rate, I believe that should also be converted.

CASE #1 Exchange Rate X = 0.6 (£0.6 = $1)
• $1000 --> £600
• (-) Commission 5%  £570
• £570--> $950
• (-) Commisson 5%  $902.5

CASE #2 Exchange Rate X = 2 (£2 = $1)
• $1000 --> $2000
• (-) Commission 5% --> £1900
• £1900 --> $950
• (-) Commisson 5% --> $902.5

SO, exchange rate does not matter.

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New post 09 Aug 2017, 08:46
neither of these two answers are logic and do not answer the question. The question was how much dollars you have after two exchanges. These two answers are more like conditions of task.

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Re: M19-13 [#permalink]

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New post 09 Aug 2017, 09:47
+1 for option B. We need value of "y". Only option B gives us the amount of dollars to be paid. The answer is B.
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sidagar wrote:
HI BRUNEL

CAN U PLS ELABORATE THE LOGIC BEHIND

1000(1−y/100) dollars


Hi,
Its simple math..
1000-y% of 1000 = 1000 - \(\frac{y}{100}*1000\) is what would remain for conversion to pounds.

Hope this answers your question.

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Re: M19-13 [#permalink]

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New post 17 Aug 2017, 22:52
Bunuel wrote:
Official Solution:

Whatever the value of \(x\), the first exchange turned 1000 dollars into the equivalent of \(1000(1 - \frac{y}{100})\) dollars. This amount, in turn, became \(1000(1 - \frac{y}{100})^2\) dollars after the second exchange. To answer the question we need to know the value of \(y\).
Statement (1) by itself is insufficient. S1 tells us the value of\(x\), not \(y\).

Statement (2) by itself is sufficient. S2 tells us the value of \(y\).


Answer: B


Adding to bunuel's solution
exchange rate X must be equal to a fraction i.e. [pound][/dollar] or [p][/d] .
Now after first conversion the amount would be = \(1000(1 - \frac{y}{100})\) . [p][/d]

Now when we convert again the pound to dollar the fraction would be inverted to [d[/p]
and the new dollar value that would be remaining = \(1000(1 - \frac{y}{100})^2\) . [p][/d] . [d[/p] = \(1000(1 - \frac{y}{100})^2\)

=> the Answer is B, as we just need to know the value of Y .

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Re: M19-13   [#permalink] 17 Aug 2017, 22:52
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