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

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Official Answer

A

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

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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

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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!

Usernamevisible · Answer

You can use GP formula as well

Tn = a.r^(n-1)

Remember dont use

a(r^n-1)/r-1 because this is the formula for sum and not 19th term

NetOrb · Answer

How to become faster at such arithmetic? solving 1/2^20?

AndyRugger · Answer

Most of the other solutions I've seen are predicated on the assumption that you will recognize that 2^20 can be split into (2^10)^2, and immediately recognize 2^10 as 1,024.

Now, that's a reasonable assumption ... the GMAT expects you to know the powers of 2 up to 10, I think. But what if you didn't? Someone else mentioned that it was "obvious" that 2^20 would be bigger than a million, but that's actually not correct and it's a dangerous assumption. It's less than 5% bigger.

Here's an alternative: Comparing denominators. How do you compare 2^20 to 10^6? Well, you can split apart the million into 2's and 5's, in which case you realize that you're comparing 2^14 to 5^6, since 2^6 is shared.

Again, the answer isn't obvious ... but since you don't care about absolute amounts, only relative, you can compare the  square roots  of the two numbers! That turns into 2^7 vs. 5^3, which is easy to calculate even if you don't have any exponent values memorized. 125 is smaller than 128 ... work backwards, and you realize that d, the answer, must be smaller than 1 over 1 million.

Last year a company produced millions of widgets each week

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Last year a company produced millions of widgets each week

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