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Raffle tickets numbered consecutively from 101 through 350 are placed
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Updated on: 20 Jul 2022, 09:36
Updated on: 20 Jul 2022, 09:36
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Question Stats:
50% (00:06) correct
50% (01:50) wrong based on 2 sessions
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Raffle tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random will have a number with a hundreds digit of 2?
(A) \(\frac{2}{5}\)
(B)\(\frac{2}{7}\)
(C) \(\frac{33}{83}\)
(D) \(\frac{99}{250}\)
(E) \(\frac{100}{249}\)
(A) \(\frac{2}{5}\)
(B)\(\frac{2}{7}\)
(C) \(\frac{33}{83}\)
(D) \(\frac{99}{250}\)
(E) \(\frac{100}{249}\)
Originally posted by SeaOyster1 on 20 Jul 2022, 06:51.
Last edited by Carcass on 20 Jul 2022, 09:36, edited 1 time in total.
Last edited by Carcass on 20 Jul 2022, 09:36, edited 1 time in total.
Formatted
Re: Raffle tickets numbered consecutively from 101 through 350 are placed
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20 Jul 2022, 06:57
20 Jul 2022, 06:57
Arithmetic Probability
There are 250 integers from 101 to 350 inclusive, 100 of which ( that is, 200 through 299) have a hundreds digit of 2. Therefore, the probability that a ticket selected from the box at random will have a hundreds digit of 2 can be expressed as -
\(\frac{100}{250}\)
This can be reduced as -
\(\frac{2}{5}\)
The Correct Answer is A
There are 250 integers from 101 to 350 inclusive, 100 of which ( that is, 200 through 299) have a hundreds digit of 2. Therefore, the probability that a ticket selected from the box at random will have a hundreds digit of 2 can be expressed as -
\(\frac{100}{250}\)
This can be reduced as -
\(\frac{2}{5}\)
The Correct Answer is A
gmatclubot
Re: Raffle tickets numbered consecutively from 101 through 350 are placed [#permalink]
20 Jul 2022, 06:57
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