Last year a company produced millions of widgets each week

Question

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. \(d \geq \frac{1}{1,000}\)

Show Spoiler

Attachment:

Screenshot 2023-12-07 130523.png [ 121.56 KiB | Viewed 21494 times ]

Show Answer

Official Answer

A

Pro tip: Combine official questions for familiarity with GMAT Club Tests for analytics and tougher Focus-level practice so the real exam feels manageable. Explore →

Bunuel · Accepted Answer

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. \(d \geq \frac{1}{1,000}\)

Show Spoiler

Attachment:

Screenshot 2023-12-07 130523.png

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.

jm2k5 · Answer

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.

AnkurGMAT20 · Answer

Bunuel , I'm having difficulty understanding this question and the solution. Can you please help?

ruis · Answer

Could we do an aritmetic progression here?
Something like a19 = a1 + (n + 1) * d
a19 = 4 + (19+1) * 2
a19 = 4 + 40
a19 = 44
Why wouldn´t this work? (1/44 on week 19th)

Bunuel · Answer

This would not work for the simple reason that the given sequence is not an arithmetic progression; it is a geometric progression.

Oppenheimer1945 · Answer

Week_19 ratio is 1/2^20

Knowing from basics of binary conversion 2^10=1024
2^10>10^3
1/2^10<1/10^3
1/2^20<1/10^6

Posted from my mobile device

Fish181 · Answer

Denomenator doubles every week so geometric formula would be 1/2^20. It is such a small number you don't need to solve. It has to be smaller than 1/1,000,000. IMO this question tests your ability to recognize the "exponentialness" of exponents.

KarishmaB · Answer

The question gives a GP. 1st Term = \frac{1}{4} = \frac{1}{2^2} 2nd Term = \frac{1}{8} = \frac{1}{2^3} ... 19th Term = \frac{1}{2^{20}} = d d = \frac{1}{1024*1024} So d is less than \frac{1}{1000*1000} = \frac{1}{1,000,000} Answer (A) Check this post on GPs https://anaprep.com/algebra-benefits-of ... -concepts/

Kinshook · Answer

Given: 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.

Asked: 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?

d = 1/2^{n+1} = 1/2^{20} = 1/{1024}^2 < 1/10^6

IMO A

nisiat · Answer

Has anyone else found the stem missleading?

"If last year the ratio of the number of defective widgets to the numbers of widgets produced was d for the 19th week" --> It kind of feel like the question was going to be for the addition of the whole year and it ends asking only about the 19th week.

That's what got me... Not a tough question, though.

ADisHere · Answer

You have been an amazing guide, tomorrow is my exam and your ways have really helped me to elevate my score. Kudos to you!

Last year a company produced millions of widgets each week

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Last year a company produced millions of widgets each week

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards