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07 Dec 2023, 00:37
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
65% (02:33) correct
35%
(02:55)
wrong
based on 564
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Last year a company produced millions of widgets each week. Last year the ratio of the number of defective widgets to the number of widgets produced was \(\frac{1}{4}\) for the first week, \(\frac{1}{8}\) for the second week, \(\frac{1}{16}\) for the third week, and so on for 19 weeks, where the ratio for each week after the first week was half of the ratio for the preceding week. If last year the ratio of the number of defective widgets to the number of widgets produced was \(d\) for the 19th week, then \(d\) satisfies which of the following inequalities?
A. \(d < \frac{1}{1,000,000}\)
B. \( \frac{1}{1,000,000} \leq d < \frac{1}{100,000} \)
C. \( \frac{1}{100,000} \leq d < \frac{1}{10,000} \)
D. \( \frac{1}{10,000} \leq d < \frac{1}{1,000} \)
E. \(d \geq \frac{1}{1,000}\)
Screenshot 2023-12-07 130523.png [ 121.56 KiB | Viewed 10862 times ]
A. \(d < \frac{1}{1,000,000}\)
B. \( \frac{1}{1,000,000} \leq d < \frac{1}{100,000} \)
C. \( \frac{1}{100,000} \leq d < \frac{1}{10,000} \)
D. \( \frac{1}{10,000} \leq d < \frac{1}{1,000} \)
E. \(d \geq \frac{1}{1,000}\)
Attachment:
Screenshot 2023-12-07 130523.png [ 121.56 KiB | Viewed 10862 times ]
11 Jan 2024, 06:54
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AnkurGMAT20
Bunuel, I'm having difficulty understanding this question and the solution. Can you please help?
Last year a company produced millions of widgets each week. Last year the ratio of the number of defective widgets to the number of widgets produced was \(\frac{1}{4}\) for the first week, \(\frac{1}{8}\) for the second week, \(\frac{1}{16}\) for the third week, and so on for 19 weeks, where the ratio for each week after the first week was half of the ratio for the preceding week. If last year the ratio of the number of defective widgets to the number of widgets produced was \(d\) for the 19th week, then \(d\) satisfies which of the following inequalities?
A. \(d < \frac{1}{1,000,000}\)
B. \( \frac{1}{1,000,000} \leq d < \frac{1}{100,000} \)
C. \( \frac{1}{100,000} \leq d < \frac{1}{10,000} \)
D. \( \frac{1}{10,000} \leq d < \frac{1}{1,000} \)
E. \(d \geq \frac{1}{1,000}\)
We are given that the ratio of the number of defective widgets to the number of widgets produced halves each week.
- The ratio for 1st week was \(\frac{1}{4} = \frac{1}{2^2}\);
- The ratio for 2nd week was \(\frac{1}{8} = \frac{1}{2^3}\);
- The ratio for 3rd week was \(\frac{1}{16} = \frac{1}{2^4}\);
and so on.
Hence, the ratio for the nth week would be \(\frac{1}{2^{n+1}}\), making the ratio for the 19th week equal to \(\frac{1}{2^{20}}\).
Essentially now we need to evaluate the value \(\frac{1}{2^{20}}\) to answer the question. Since \(2^{20}=(2^{10})^2=1,024^2 > (1,000^2=1,000,000)\), then \(\frac{1}{2^{20}}< \frac{1}{1,000,000}\).
Answer: A.
Hope it helps.
General Discussion
07 Dec 2023, 07:12
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gmatophobia
Last year a company produced millions of widgets each week. Last year the ratio of the number of defective widgets to the number of widgets produced was \(\frac{1}{4}\) for the first week, \(\frac{1}{8}\) for the second week, \(\frac{1}{16}\) for the third week, and so on for 19 weeks, where the ratio for each week after the first week was half of the ratio for the preceding week. If last year the ratio of the number of defective widgets to the number of widgets produced was \(d\) for the 19th week, then \(d\) satisfies which of the following inequalities?
A. \(d < \frac{1}{1,000,000}\)
B. \( \frac{1}{1,000,000} \leq d < \frac{1}{100,000} \)
C. \( \frac{1}{100,000} \leq d < \frac{1}{10,000} \)
D. \( \frac{1}{10,000} \leq d < \frac{1}{1,000} \)
E. \(d \geq \frac{1}{1,000}\)
A. \(d < \frac{1}{1,000,000}\)
B. \( \frac{1}{1,000,000} \leq d < \frac{1}{100,000} \)
C. \( \frac{1}{100,000} \leq d < \frac{1}{10,000} \)
D. \( \frac{1}{10,000} \leq d < \frac{1}{1,000} \)
E. \(d \geq \frac{1}{1,000}\)
Attachment:
Screenshot 2023-12-07 130523.png
d = (1/4)*(1/2)^(19-1) = (1/2)^20 = (1/2^10)^2 = (1/1024)^2
1/1024 < 1/1000
Since both numbers are positive, we can write,
(1/1024)^2 < (1/1000)^2
=> d<1/1000000
Option A.
