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10 Jan 2018, 01:10
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E
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70% (02:04) correct
30%
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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. \(\frac{1}{100}\)
A. \(\frac{1}{20}\)
B. \(\frac{1}{50}\)
C. \(\frac{1}{60}\)
D. \(\frac{1}{90}\)
E. \(\frac{1}{100}\)
12 Jan 2018, 00:52
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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 \(9 * 10 = 90\) 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 \(\frac{90}{9000} = \frac{1}{100}.\)
Therefore, the answer is E.
Answer: E
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 \(9 * 10 = 90\) 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 \(\frac{90}{9000} = \frac{1}{100}.\)
Therefore, the answer is E.
Answer: E
10 Jan 2018, 02:15
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Total 4-digit integers= 9999-1000+1= 9000.
There is only one palindromic number in evey 100 nos.
Therefore, total no.of palindromic= 90.
Probability= 90/9000=1/100.
Sent from my ONEPLUS A5010 using GMAT Club Forum mobile app
There is only one palindromic number in evey 100 nos.
Therefore, total no.of palindromic= 90.
Probability= 90/9000=1/100.
Sent from my ONEPLUS A5010 using GMAT Club Forum mobile app
