If a number between 10 and 50 (inclusive) at random. What is the probability that the tens digit is greater than one digit?
A. 31/41
B. 1041
C. 14/41
D. 27/41
E. 11/41
The OA is E.
Is there a strategic approach to this question? Can any experts help?
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If a number between 10 and 50 (inclusive) at random.
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Brent@GMATPrepNow
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I think that listing (while looking for a pattern) and counting is the best/fastest approach here.AAPL wrote:If a number between 10 and 50 (inclusive) at random. What is the probability that the tens digit is greater than one digit?
A. 31/41
B. 1041
C. 14/41
D. 27/41
E. 11/41
Let's list all of the numbers (from 10 to 50 inclusive) such that the tens digit is greater than one digit
Numbers with tens digit 1: 10
Numbers with tens digit 2: 20, 21
Numbers with tens digit 3: 30, 31, 32 [can you see the pattern yet?]
Numbers with tens digit 4: 40, 41, 42, 43
Numbers with tens digit 5: 50
TOTAL values that meet the given condition = 1 + 2 + 3 + 4 + 1 = 11
Number of integers from 10 to 50 inclusive = 50 - 10 + 1 = 41
So, P(tens digit is greater than one digit) = 11/41
Answer: E
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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Jake@ThePrincetonReview
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You could also solve this problem with what's called a probability tree. It's basically the same idea as counting but with a handy tree visualization!
From the "top" of the tree, draw 5 branches going down, labelled 1,2,3,4 and 5 at the bottom of the "branch". These represent your choices for the first digit.
From each of the 1-4 branches, draw 10 more branches labelled 0-9. These represent your choices for the 2nd digit. From the 5 branch, draw one branch, labelled 0, as that's your only choice from that branch.
Now count the total branches at the "bottom". 10+10+10+10+1= 41. That's the denominator of your answer.
Now count the branches that meet your goal (ones digit greater than tens digit): 10, 20, 21, 30, 31, 32, 40, 41, 42, 43, 50. That's 11 numbers and that's your numerator.
Final answer: 11/41.
I use the tree whenever I get stuck on a probability question.
From the "top" of the tree, draw 5 branches going down, labelled 1,2,3,4 and 5 at the bottom of the "branch". These represent your choices for the first digit.
From each of the 1-4 branches, draw 10 more branches labelled 0-9. These represent your choices for the 2nd digit. From the 5 branch, draw one branch, labelled 0, as that's your only choice from that branch.
Now count the total branches at the "bottom". 10+10+10+10+1= 41. That's the denominator of your answer.
Now count the branches that meet your goal (ones digit greater than tens digit): 10, 20, 21, 30, 31, 32, 40, 41, 42, 43, 50. That's 11 numbers and that's your numerator.
Final answer: 11/41.
I use the tree whenever I get stuck on a probability question.
Jake Schiff
GMAT Instructor and Master Trainer
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Scott@TargetTestPrep
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From 10 to 19, there is 1 number (10) in which the tens digit is greater than the ones digit.AAPL wrote:If a number between 10 and 50 (inclusive) at random. What is the probability that the tens digit is greater than one digit?
A. 31/41
B. 1041
C. 14/41
D. 27/41
E. 11/41
From 20 to 29, there are 2 numbers (20, 21) in which the tens digit is greater than the ones digit.
From 30 to 39, there are 3 numbers (30, 31, 32) in which the tens digit is greater than the ones digit.
From 40 to 49, there are 4 numbers (40, 41, 42, 43) in which the tens digit is greater than the ones digit.
And 50 also works.
Since there are 50 - 10 + 1 = 41 numbers from 10 to 50 (inclusive), then the probability is 11/41.
Answer: E
Scott Woodbury-Stewart
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