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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
74% (00:49) correct
26%
(01:21)
wrong
based on 85
sessions
History
Date
Time
Result
Not Attempted Yet
If w widgets cost d dollars, then at this rate how many dollars will 2000 widgets cost?
(A) wd/2000
(B) 2000w/d
(C) 2000d/w
(D) d/(2000w)
(E) 2000/(wd)
(A) wd/2000
(B) 2000w/d
(C) 2000d/w
(D) d/(2000w)
(E) 2000/(wd)
20 May 2017, 06:49
Kudos
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Cost of w widgets = d dollars
Cost of 1 widget = d/w
Thus, cost of 2000 widgets = 2000d/w
Hence C
Cost of 1 widget = d/w
Thus, cost of 2000 widgets = 2000d/w
Hence C
21 May 2017, 08:07
Kudos
Bookmarks
Bunuel
Probably not as quick as algebraic method but perhaps helpful to some: Choose smart numbers and plug into answer choices.
Let original number of widgets, w = 2
Let original total cost, d = $10
1. Do the mental math. $10 for 2 items means $5 per item. $5 per item * 2000 items = $10,000; OR
2. Unit cost for w is total cost / # of items = \(\frac{$10}{2}\)= $5 per w or \(\frac{$5}{w}\)
\(\frac{$5}{w}\) * 2000w = $10,000
3. Plug w = 2 and d = 10 into answer choices, don't worry about the $5 unit cost, just look for $10,000 as the answer*:
(A) wd/2000 =2*10/2000 = 20/2000 ... no need to calculate. Too small. Eliminate
(B) 2000w/d = 2000*2/10 = 4000/10 = $400. Eliminate
(C) 2000d/w = 2000*10/2 = 20,000/2 = $10,000. Correct
(D) d/(2000w) = 10/2000*2 = 10/4000 . . . no need to calculate. Too small. Eliminate
(E) 2000/(wd) = 2000/2*10 = 2000/20 = $100. Eliminate
Answer C
*If you scan the answers quickly, with these chosen small values for w and d, it's immediately clear that A and D, with small values in numerator and large values in denominator, will yield an answer that is too small. Similarly, with E, you might notice that the denominator is too big, even with the relatively small numbers chosen for w and d. Now you're down to (B) and (C).
