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26 Sep 2019, 04:46
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
59% (02:08) correct
41%
(02:05)
wrong
based on 325
sessions
History
Date
Time
Result
Not Attempted Yet
The original cost of a mobile was $799. Mr. Thomas increased the price by p% and gave a discount of d% on the increased price. Was the new cost less than the original cost?
(1) p = d
(2) 100p - 100d < pd
(1) p = d
(2) 100p - 100d < pd
26 Sep 2019, 09:10
Kudos
Bookmarks
Bunuel
Mr. Thomas increased the price by p% and gave a discount of d% on the increased price
--> New cost = \(799*(1 + \frac{p}{100})*(1 - \frac{d}{100})\)
--> New cost = \(799*(1 + \frac{p}{100} - \frac{d}{100} - \frac{pd}{10000})\)
(1) p = d
--> New cost = \(799*(1 + \frac{p}{100} - \frac{p}{100} - \frac{p^2}{10000})\)
--> New cost = \(799*(1 - \frac{p^2}{10000})\)
For any value of p, \((1 - \frac{p^2}{10000})\) is always less than 1
--> New cost is ALWAYS less than original cost --> Sufficient
(2) 100p - 100d < pd
--> New cost = \(799*(1 + \frac{p}{100} - \frac{d}{100} - \frac{pd}{10000})\)
--> New cost = \(799*(1 + \frac{(100p - 100d - pd)}{10000})\)
Since \(100p - 100d < pd\)
--> \(100p - 100d - pd < 0\)
New cost = 799*(fraction less than 1), which is less than original cost --> Sufficient
IMO Option D
Pls Hit kudos if you like the solution
26 Sep 2019, 06:09
Kudos
Bookmarks
The original cost of a mobile was $799. Mr. Thomas increased the price by p% and gave a discount of d% on the increased price. Was the new cost less than the original cost?
New cost = \(799*(\frac{100+p}{100})(\frac{100-d}{100}).\)...
So is \(799*(\frac{100+p}{100})(\frac{100-d}{100})>799\)....
\((100+p)(100-d)>10000\)....
\(10000+100p-100d-pd>10000\)......
\(100(p-d)>pd\)
(1) p = d
So the question is 100(p-d)>pd.....100(p-p)>p^2.......0>p^2...ans is NO
(2) 100p - 100d < pd
We are asked the opposite is 100(p-d)>pd..answer is NO
D
New cost = \(799*(\frac{100+p}{100})(\frac{100-d}{100}).\)...
So is \(799*(\frac{100+p}{100})(\frac{100-d}{100})>799\)....
\((100+p)(100-d)>10000\)....
\(10000+100p-100d-pd>10000\)......
\(100(p-d)>pd\)
(1) p = d
So the question is 100(p-d)>pd.....100(p-p)>p^2.......0>p^2...ans is NO
(2) 100p - 100d < pd
We are asked the opposite is 100(p-d)>pd..answer is NO
D
