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M61-02

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Math Revolution GMAT Instructor
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M61-02  [#permalink]

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New post 17 Jun 2018, 23:36
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E

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  65% (hard)

Question Stats:

50% (02:01) correct 50% (02:27) wrong based on 8 sessions

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A palindromic number is a number that remains the same when its digits are reversed. For example, 16461 is a palindromic number. If a 4-digit integer is selected randomly from the set of all 4-digit integers, what is the probability that it is palindromic?

A. \(\frac{1}{20}\)
B. \(\frac{1}{50}\)
C. \(\frac{1}{60}\)
D. \(\frac{1}{90}\)
E. \(\frac{1}{100}\)

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Re M61-02  [#permalink]

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New post 17 Jun 2018, 23:36
Official Solution:

A palindromic number is a number that remains the same when its digits are reversed. For example, 16461 is a palindromic number. If a 4-digit integer is selected randomly from the set of all 4-digit integers, what is the probability that it is palindromic?

A. \(\frac{1}{20}\)
B. \(\frac{1}{50}\)
C. \(\frac{1}{60}\)
D. \(\frac{1}{90}\)
E. \(\frac{1}{100}\)


4-digit palindromic numbers have the form \(xyyx\), where \(x\) is one of values 1, 2, …, 9 and \(y\) is one of values 0, 1, 2, …, 9.

So, there are 9 x 10 = 90 four-digit palindromic numbers.

The total number of 4-digit numbers between 1000 and 9999, inclusive, is 9000 ( = 9999 – 1000 + 1 ).

Therefore, the probability that the selected 4-digit is palindromic is \(\frac{90}{9000} = \frac{1}{100}\).


Answer: E
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Re: M61-02  [#permalink]

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New post 29 Oct 2018, 11:31
Here is a better explanation.

Total no. of outcomes= 9999-999=9000 numbers of 4 digits

Favorable outcomes= 9C1 x 10C1 x 1C1 x 1C1=90
Hence , probability = 90/9000=1/100
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Re: M61-02 &nbs [#permalink] 29 Oct 2018, 11:31
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