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?

Answer Choices

Correct answer marked below

d < \frac{1}{1,000,000}

Correct

\frac{1}{1,000,000} \le d < \frac{1}{100,000}

\frac{1}{100,000} \le d < \frac{1}{10,000}

\frac{1}{10,000} \le d < \frac{1}{1,000}

d \ge \frac{1}{1,000}

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Comments

ChrisApr 26

Took me a second to realize the ratios just keep halving, so week 19 is 1/4 divided by 2 eighteen times. Once I saw that it was just powers of 2 it got much easier.

SathwikaMay 4

Yeah same, I initially tried to sum the ratios which was totally wrong lol.

LilianApr 30

Wait, why 18 times and not 19? That's where I slipped up.

ManelApr 21

This one looks scary with all those fractions but it's really just counting the doublings. Definitely more of a 650-level question once you see through the setup.

NaveenApr 22

Agreed, the wording is intimidating but the math is straightforward.

JakubJul 9

Can someone explain why the answer compares against those specific powers of 10? I got the denominator as 2^something but then had to estimate where it falls.

SeharJul 14

You just need to know 2^10 is about 1000, so you can ballpark the exponent.

MatildeAug 2

Got the right range but only after approximating 2^20 ~ 1,000,000. Nice question, the trap is thinking you need to compute the exact fraction.

What this question tests

What it tests

Your fluency with the order of operations, fractions, decimals, and basic number sense — the foundation every quant question leans on.

Common trap

Applying the order of operations out of sequence or rounding intermediate values before the final step.