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Perry buys a book that has pages numbered from 1 to 980. He then selec 26 Nov 2019, 01:56
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C
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58% (02:28) correct 42% (02:17) wrong based on 283 sessions

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Perry buys a book that has pages numbered from 1 to 980. He then selects one of the pages at random. What is the probability that the number of the page he selects is divisible by 7, given that it is divisible by 3?

A. \(\frac{1}{21}\)

B. \(\frac{1}{7}\)

C. \(\frac{23}{163}\)

D. \(\frac{23}{490}\)

E. \(\frac{29}{171}\)


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C
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Perry buys a book that has pages numbered from 1 to 980. He then selec 26 Nov 2019, 02:06
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pages divisble by 3 ; 980/3 ; ~ 326
and LCM of 3,7 = 21 ; ~ 46 pages
P = 46/326 ; 23/163
IMO C


Bunuel
Perry buys a book that has pages numbered from 1 to 980. He then selects one of the pages at random. What is the probability that the number of the page he selects is divisible by 7, given that it is divisible by 3?

A. \(\frac{1}{21}\)

B. \(\frac{1}{7}\)

C. \(\frac{23}{163}\)

D. \(\frac{23}{490}\)

E. \(\frac{29}{171}\)


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Perry buys a book that has pages numbered from 1 to 980. He then selec 26 Mar 2020, 19:58
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The denominator will be 326 and C is the only answer choice that has a factor of 326.
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Perry buys a book that has pages numbered from 1 to 980. He then selec 26 Mar 2020, 20:04
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So the question states that the page randomly selected is 100% a factor of 3. 326 pages are divisible by 3. To find the total number of pages divisible by 3, do total pages divided by 3 and do not consider the decimal (the decimal just tells you that there is/are numbers that aren't divisible by 3). 980/3 = 326.67 --> 326.

The denominator will be 326 and C is the only answer choice that has a factor of 326.
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Perry buys a book that has pages numbered from 1 to 980. He then selec 26 Mar 2020, 21:31
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Bunuel
Perry buys a book that has pages numbered from 1 to 980. He then selects one of the pages at random. What is the probability that the number of the page he selects is divisible by 7, given that it is divisible by 3?

A. \(\frac{1}{21}\)

B. \(\frac{1}{7}\)

C. \(\frac{23}{163}\)

D. \(\frac{23}{490}\)

E. \(\frac{29}{171}\)

The pages divisible by both 7 and 3 will have to divisible by 21. There are (980/21)=46 pages that are divisible by 21. The probability is 46/980 = 23/490.

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The pages divisible by both 7 and 3 will have to divisible by 21. There are (980/21)=46 pages that are divisible by 21. The probability is 46/980 = 23/490.
D is the answer.
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Perry buys a book that has pages numbered from 1 to 980. He then selec 19 Dec 2024, 01:59
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