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3 digit integer between 100 and 199, inclusive

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3 digit integer between 100 and 199, inclusive

by pareekbharat86 » Thu Nov 21, 2013 4:34 am
If a 3-digit integer is selected at random from the integers 100 through 199, inclusive, what is the probability that the first digit and the last digit of the integer are each equal to one more than the middle digit?

a.2/225
b.1/111
c.1/110
d.1/100
e.1/50

OA is D
[spoiler]
The no. that we get after interpreting the above is 101, right?[/spoiler]
Thanks,
Bharat.
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by riz_gmat » Thu Nov 21, 2013 5:12 am
Yes, 101 is the only possible no satisfying the conditions.

Total numbers = 199 - 100 + 1 (since bothr inclusive) = 100

[spoiler]P = 1 / 100[/spoiler]
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by Mathsbuddy » Thu Nov 21, 2013 6:23 am
I reckon the answer is not on the list.
Can somebody spot my misinterpretation of the question?

ABC = 3 digit number where A, B and C are the digits.

A = 1

So ABC = 1BC

Find P = p(1>=B<=C)

Q = p(B=0 and C>=0) = 1/10 * 1 = 1/10
R = p(B=1 and C>=1) = 1/10 * 9/10 = 9/100

P = p(Q or R) = Q + R = 1/10 + 9/100 = 19/100

Therefore, I conclude that P = 19/100
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by Mathsbuddy » Thu Nov 21, 2013 6:27 am
riz_gmat wrote:Yes, 101 is the only possible no satisfying the conditions.

Total numbers = 199 - 100 + 1 (since bothr inclusive) = 100

[spoiler]P = 1 / 100[/spoiler]
We are trying to find the probability that the first digit and the last digit of the integer are each EQUAL TO OR MORE THAN the middle digit,therefore 101 is surely not the only solution?

Here are some more: 102, 103, 104,... 111, 112, 113,... ...119.
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by Mathsbuddy » Thu Nov 21, 2013 6:27 am
pareekbharat86 wrote:If a 3-digit integer is selected at random from the integers 100 through 199, inclusive, what is the probability that the first digit and the last digit of the integer are each equal to one more than the middle digit?

a.2/225
b.1/111
c.1/110
d.1/100
e.1/50

OA is D
[spoiler]
The no. that we get after interpreting the above is 101, right?[/spoiler]
We are trying to find the probability that the first digit and the last digit of the integer are each EQUAL TO OR MORE THAN the middle digit,therefore 101 is surely not the only solution?

Here are some more: 102, 103, 104,... 111, 112, 113,... ...119.
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by [email protected] » Thu Nov 21, 2013 12:48 pm
Hi Mathsbuddy,

The question uses the phrase "the first digit and the last digit of the integer are EACH EQUAL to "ONE MORE" than the middle digit."

This means that the first and third digits are the SAME digit and that each is ONE MORE than the second digit.

Given the range that we have to work with (101 - 199), the ONLY number that fits this description is "101."

Since there are 100 total numbers in that range, the probability is 1/100

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by Mathsbuddy » Thu Nov 21, 2013 11:49 pm
[email protected] wrote:Hi Mathsbuddy,

The question uses the phrase "the first digit and the last digit of the integer are EACH EQUAL to "ONE MORE" than the middle digit."

This means that the first and third digits are the SAME digit and that each is ONE MORE than the second digit.

Given the range that we have to work with (101 - 199), the ONLY number that fits this description is "101."

Since there are 100 total numbers in that range, the probability is 1/100

GMAT assassins aren't born, they're made,
Rich
Thanks Rich,

That'll teach me to read the question more thoroughly! I had incorrectly read "...are each equal to OR more than...". That erroneous imaginary "OR" has cost me my dignity, but from it a good lesson is learned :)
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by sanju09 » Fri Nov 22, 2013 12:05 am
Mathsbuddy wrote:
[email protected] wrote:Hi Mathsbuddy,

The question uses the phrase "the first digit and the last digit of the integer are EACH EQUAL to "ONE MORE" than the middle digit."

This means that the first and third digits are the SAME digit and that each is ONE MORE than the second digit.

Given the range that we have to work with (101 - 199), the ONLY number that fits this description is "101."

Since there are 100 total numbers in that range, the probability is 1/100

GMAT assassins aren't born, they're made,
Rich
Thanks Rich,

That'll teach me to read the question more thoroughly! I had incorrectly read "...are each equal to OR more than...". That erroneous imaginary "OR" has cost me my dignity, but from it a good lesson is learned :)
No, it doesn't cost you your dignity my friend; but it could definitely cost you good score if you are not cheerfully careful while reading.
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by pareekbharat86 » Fri Nov 22, 2013 12:45 am
Mathsbuddy wrote:
[email protected] wrote:Hi Mathsbuddy,

The question uses the phrase "the first digit and the last digit of the integer are EACH EQUAL to "ONE MORE" than the middle digit."

This means that the first and third digits are the SAME digit and that each is ONE MORE than the second digit.

Given the range that we have to work with (101 - 199), the ONLY number that fits this description is "101."

Since there are 100 total numbers in that range, the probability is 1/100

GMAT assassins aren't born, they're made,
Rich
Thanks Rich,

That'll teach me to read the question more thoroughly! I had incorrectly read "...are each equal to OR more than...". That erroneous imaginary "OR" has cost me my dignity, but from it a good lesson is learned :)
I committed the exact same error in my first attempt. 19/100 was nowhere to be seen in the choices. ;-)
Thanks,
Bharat.
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