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# M61-02

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 7993
GMAT 1: 760 Q51 V42
GPA: 3.82

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18 Jun 2018, 00:36
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Difficulty:

55% (hard)

Question Stats:

53% (01:59) correct 47% (02:24) wrong based on 15 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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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7993 GMAT 1: 760 Q51 V42 GPA: 3.82 Re M61-02 [#permalink] ### Show Tags 18 Jun 2018, 00:36 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Joined: 18 Oct 2018
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29 Oct 2018, 12:31
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
Re: M61-02   [#permalink] 29 Oct 2018, 12:31
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# M61-02

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