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Thirty airmail and 40 ordinary envelops [#permalink]
30 Apr 2016, 20:43
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Question Stats:
27% (01:53) correct
72% (01:37) wrong based on 11 sessions
Thirty airmail and 40 ordinary envelops are the only envelop in a bag. Thirtyfive envelops in the bag are unstamped, and \(\frac{5}{7}\) of the unstamped envelops are airmail. What is the probability that an envelope picked randomly from the bag is an unstamped airmail envelope? (A) \(\frac{1}{7}\) (B) \(\frac{1}{3}\) (C) \(\frac{13}{35}\) (D) \(\frac{17}{38}\) (E) \(\frac{23}{70}\)




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Re: Thirty airmail and 40 ordinary envelops [#permalink]
01 May 2016, 08:21
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Hi, are you sure the answers are correct ??
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Re: Thirty airmail and 40 ordinary envelops [#permalink]
01 May 2016, 12:32
happy1992 wrote: Thirty airmail and 40 ordinary envelops are the only envelop in a bag. Thirtyfive envelops in the bag are unstamped, and 5/7 of the unstamped envelops are airmail. What is the probability that an envelope picked randomly from the bag is an unstamped airmail envelope?
(A) 1/7 (B) 1/3 (C) 13/35 (D) 17/38 (E) 23/70 Here we have 30 airmail and 40 ordinary envelops. Hence a total of 70 envelops. Now 35 envelops are unstamped a 25 of them are airmail. Hence the new distribution is : 1. Unstamped airmail 25 2. Stamped airmail 5 3. Unstamped normal envelop 10 4. Stamped normal envelop 30 Hence probability of unstamped airmail envelope should be 25/70.
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Thirty airmail and 40 ordinary envelopes are the only envelo [#permalink]
25 Nov 2016, 02:54
Thirty airmail and 40 ordinary envelopes are the only envelopes in a bag. Thirtyfive envelopes in the bag are unstamped, and 5/7 of the unstamped envelopes are airmail letters. What is the probability that an envelope picked randomly from the bag is an unstamped airmail envelope?
(A) 1/7 (B) 1/3 (C) 13/35 (D) 17/38 (E) 23/70
25 which is 5/7 of 35 unstamped airmail envelopes.
so , P(unstamped airmail) = 25/70 => 5/14 is not in the options . am i missing anything?
Last edited by Carcass on 05 Dec 2016, 02:36, edited 1 time in total.
Edited the post



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Re: my answer was not in the answer [#permalink]
25 Nov 2016, 12:22
I think you are right. Unstamped ordinary is 10/70 or 1/7. Probably a typo.
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Re: Thirty airmail and 40 ordinary envelopes are the only envelo [#permalink]
05 Dec 2016, 02:37
Please follow the rules for posting in quant section. They are as announcement in main quant section Thank you for your collaboration
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Re: Thirty airmail and 40 ordinary envelopes are the only envelo [#permalink]
20 Dec 2016, 21:07
Yes, none of the answers are correct. Unstamped aitmail letters are 25 letters out of 70, so the answer should be 25/70 or 5/14.



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Re: Thirty airmail and 40 ordinary envelops [#permalink]
03 Dec 2017, 11:46
Merged similar topic. The OA is A. Quote: We have that 30 airmail and 40 ordinary envelopes are the only envelopes in the bag. Hence, the total number of envelopes is 30 + 40 = 70. We also have that 35 envelopes in the bag are unstamped, and 5/7 of these envelopes are airmail letters. Now, 5/7 * 35 = 25. So the remaining 35 – 25 = 10 are ordinary unstamped envelopes. Hence, the probability of picking such an envelope from the bag is (Number of unstamped ordinary envelopes) / (Total number of envelopes) =10/70 =1/7
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Re: Thirty airmail and 40 ordinary envelops [#permalink]
17 May 2018, 15:51
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happy1992 wrote: Thirty airmail and 40 ordinary envelops are the only envelop in a bag. Thirtyfive envelops in the bag are unstamped, and \(\frac{5}{7}\) of the unstamped envelops are airmail. What is the probability that an envelope picked randomly from the bag is an unstamped airmail envelope?
(A) \(\frac{1}{7}\) (B) \(\frac{1}{3}\) (C) \(\frac{13}{35}\) (D) \(\frac{17}{38}\) (E) \(\frac{23}{70}\) We have 30 airmail and 40 originary envelopes for a total of 70 envelopes. Since 5/7 of the 35 unstamped envelopes are airmail, 5/7 x 35 = 25 envelopes are unstamped airmail. So the probability of selecting an unstamped airmail envelope is 25/70 = 5/14 (not in the answer choices)
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Re: Thirty airmail and 40 ordinary envelops [#permalink]
17 May 2018, 16:33
it looks very complicated question



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Thirty airmail and 40 ordinary envelopes are the only envelo [#permalink]
25 Nov 2019, 21:08
Thirty airmail and 40 ordinary envelopes are the only envelopes in a bag. Thirtyfive envelopes in the bag are unstamped, and \(\frac{5}{7}\) of the unstamped envelopes are airmail letters. What is the probability that an envelope picked randomly from the bag is an ordinary airmail envelope?
(A) \(\frac{1}{7}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{5}{14}\)
(D) \(\frac{17}{38}\)
(E) \(\frac{23}{70}\)



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Re: Thirty airmail and 40 ordinary envelopes are the only envelo [#permalink]
25 Nov 2019, 21:08
Need explanation.



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Re: Thirty airmail and 40 ordinary envelopes are the only envelo [#permalink]
27 Nov 2019, 02:51
huda wrote: Thirty airmail and 40 ordinary envelopes are the only envelopes in a bag. Thirtyfive envelopes in the bag are unstamped, and \(\frac{5}{7}\) of the unstamped envelopes are airmail letters. What is the probability that an envelope picked randomly from the bag is an ordinary airmail envelope?
(A) \(\frac{1}{7}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{5}{14}\)
(D) \(\frac{17}{38}\)
(E) \(\frac{23}{70}\) Need explation, because I got 3/7 ans



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Re: Thirty airmail and 40 ordinary envelops [#permalink]
27 Nov 2019, 02:54
Merged similar topic. See the explanations above and use the search tool https://greprepclub.com/forum/search.php before to post a new question. Regards
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Re: Thirty airmail and 40 ordinary envelops [#permalink]
27 Nov 2019, 02:57
Carcass wrote: Merged similar topic. See the explanations above and use the search tool https://greprepclub.com/forum/search.php before to post a new question. Regards Sorry do not understand. And please mention the one whom you to write.




Re: Thirty airmail and 40 ordinary envelops
[#permalink]
27 Nov 2019, 02:57





