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There are 2 decks of cards. The first deck has a 100 cards labelled...

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There are 2 decks of cards. The first deck has a 100 cards labelled... [#permalink]

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There are 2 decks of cards. The first deck has a 100 cards labelled with integers from 1 to 100. The second one has 150 cards labelled with integers 101 to 250. If we select one card at random from each deck, what is the probability that numbers on both selected cards will be multiples of 7?

A. \(\frac{7}{3000}\)

B. \(\frac{7}{2500}\)

C. \(\frac{7^2}{15000}\)

D. \(\frac{7^2}{7500}\)

E. \(\frac{7^2}{2500}\)

Could someone please explain the answer to this question. It will be much appreciated, thank you :)
[Reveal] Spoiler: OA

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There are 2 decks of cards. The first deck has a 100 cards labelled... [#permalink]

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kdatt1991 wrote:
There are 2 decks of cards. The first deck has a 100 cards labelled with integers from 1 to 100. The second one has 150 cards labelled with integers 101 to 250. If we select one card at random from each deck, what is the probability that numbers on both selected cards will be multiples of 7?

A. \(\frac{7}{3000}\)

B. \(\frac{7}{2500}\)

C. \(\frac{7^2}{15000}\)

D. \(\frac{7^2}{7500}\)

E. \(\frac{7^2}{2500}\)

Could someone please explain the answer to this question. It will be much appreciated, thank you :)


Hi
you have to first find the probability of getting seven in each selection..

there are 14 multiples of seven in 1 to 100... \(\frac{100}{7}=14.3\)...
so the probability of picking seven multiple will be \(\frac{14}{100}\)..

similarily there are 21 multiples of seven in 100 to 250...
so the probability of picking seven multiple will be \(\frac{21}{150}\)..

therefore getting seven multiple on both occasions will be \(\frac{14}{100}*\frac{21}{150}= \frac{7^2}{2500}\)...
ans E
pl feel free to ask any query ...
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Re: There are 2 decks of cards. The first deck has a 100 cards labelled... [#permalink]

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New post 03 Feb 2015, 22:26
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Hi kdatt1991,

The key to this question is in finding a quick way to figure out how many multiples of 7 you're dealing with. If you're comfortable doing basic division by hand, the "math" required isn't too bad.

I'm going to start by figuring out how many TOTAL multiples of 7 we're dealing with:

250/7 = 35 remainder 5, so we know that there are 35 multiples of 7 in total. NOW we have to figure out how many of those multiples are in the first group of 100 cards...

100/7 = 14 remainder 2, so we know....

35 TOTAL multiples:
There are 14 multiples of 7 in the first deck.
There are 21 multiples of 7 in the second deck.

Pulling 1 card from both decks, the probability of pulling two multiples of 7 is:

(14/100)(21/150)

While this might look "scary", you can reduce BOTH fractions before you multiply. The answer choices hint that the product will reduce anyway...

(14/100) = 7/50
(21/150) = 7/50
(7/50)(7/50) = (7^2)/2500

Final Answer:
[Reveal] Spoiler:
E


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Re: There are 2 decks of cards. The first deck has a 100 cards labelled... [#permalink]

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New post 03 Feb 2015, 22:46
was time consuming to find number of multiples in two intervals

1-100: 98-7/7+1=14

101-250: 245-105/7+1=21

14/100*21/150=2*7*3*7/15000=7^2*6/15000=7^2/2500

E

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Re: There are 2 decks of cards. The first deck has a 100 cards labelled... [#permalink]

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New post 03 Feb 2015, 22:57
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kdatt1991 wrote:
There are 2 decks of cards. The first deck has a 100 cards labelled with integers from 1 to 100. The second one has 150 cards labelled with integers 101 to 250. If we select one card at random from each deck, what is the probability that numbers on both selected cards will be multiples of 7?

A. \(\frac{7}{3000}\)

B. \(\frac{7}{2500}\)

C. \(\frac{7^2}{15000}\)

D. \(\frac{7^2}{7500}\)

E. \(\frac{7^2}{2500}\)

Could someone please explain the answer to this question. It will be much appreciated, thank you :)


Multiples of 7 in 1 - 100:
7, 14, ..., 98
7*1 to 7*14 (= 98) gives us 14 multiples.

Multiples of 7 in 101 - 250
105, 112, ..., 245
7*15 (=105) to 7*35 (= 245) gives us 35 - 15 + 1 = 21 multiples

Probability of picking a multiple of 7 in both cases = (14/100) * (21/150) = 7^2/2500

Answer (E)
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Re: There are 2 decks of cards. The first deck has a 100 cards labelled... [#permalink]

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Re: There are 2 decks of cards. The first deck has a 100 cards labelled... [#permalink]

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New post 12 Oct 2017, 18:06
kdatt1991 wrote:
There are 2 decks of cards. The first deck has a 100 cards labelled with integers from 1 to 100. The second one has 150 cards labelled with integers 101 to 250. If we select one card at random from each deck, what is the probability that numbers on both selected cards will be multiples of 7?

A. \(\frac{7}{3000}\)

B. \(\frac{7}{2500}\)

C. \(\frac{7^2}{15000}\)

D. \(\frac{7^2}{7500}\)

E. \(\frac{7^2}{2500}\)



Let’s determine the number of multiples of 7 from 1 to 100 and from 101 to 250.

From 1 to 100: (Note that the greatest multiple of 7 is 98 and the smallest multiple of 7 is 7.)

(98 - 7)/7+ 1 = 91/7 + 1 = 14

From 101 to 250: (Note that the greatest multiple of 7 is 245 and the smallest multiple of 7 is 105.)

(245 - 105)/7+ 1 = 140/7 + 1 = 21

Thus, the probability of selecting a multiple of 7 in the first set of cards is 14/100, and in the second set it is 21/150. We want the probability that both selected cards will contain multiples of 7, so we multiply the two probabilities to determine the total probability: 14/100 x 21/150 = (14 x 21)/(100 x 150) = (7 x 7)/(50 x 50) = 7^2/2500.

Answer: E
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Re: There are 2 decks of cards. The first deck has a 100 cards labelled...   [#permalink] 12 Oct 2017, 18:06
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