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Thirty airmail and 40 ordinary envelops

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Thirty airmail and 40 ordinary envelops [#permalink] New post 30 Apr 2016, 20:43
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Thirty airmail and 40 ordinary envelops are the only envelop in a bag. Thirty-five 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) \(\frac{23}{70}\)
[Reveal] Spoiler: OA
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Re: Thirty airmail and 40 ordinary envelops [#permalink] New post 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] New post 01 May 2016, 12:32
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happy1992 wrote:
Thirty airmail and 40 ordinary envelops are the only envelop in a bag. Thirty-five 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 un-stamped 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] New post 25 Nov 2016, 02:54
Thirty airmail and 40 ordinary envelopes are the only envelopes in a bag. Thirty-five 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.
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Re: my answer was not in the answer [#permalink] New post 25 Nov 2016, 12:22
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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] New post 05 Dec 2016, 02:37
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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] New post 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] New post 03 Dec 2017, 11:46
Expert's post
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] New post 17 May 2018, 15:51
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happy1992 wrote:
Thirty airmail and 40 ordinary envelops are the only envelop in a bag. Thirty-five 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] New post 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] New post 25 Nov 2019, 21:08
Thirty airmail and 40 ordinary envelopes are the only envelopes in a bag. Thirty-five 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] New post 25 Nov 2019, 21:08
Need explanation.
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Re: Thirty airmail and 40 ordinary envelopes are the only envelo [#permalink] New post 27 Nov 2019, 02:51
huda wrote:
Thirty airmail and 40 ordinary envelopes are the only envelopes in a bag. Thirty-five 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] New post 27 Nov 2019, 02:54
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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] New post 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
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