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Digit problem

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by madsport » Sun Sep 14, 2008 7:24 pm
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

The answer is D

So what's the strategy for problems like these?
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Re: Digit problem

by justpal » Sun Sep 14, 2008 8:01 pm
Is it 1/100 ????

Workout for this problem
1. 1 0 1 which has first and last digits 1 more that the middle one.
2. 2 1 2 which is not possible as integer should be less than 199....

which would be 101 and total of 100 integers....

I'll go with 1/100

What's the answer??
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by alescau » Sun Sep 14, 2008 8:04 pm
1/100. an easy question phrased to confuse.
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by nil4700 » Mon Sep 15, 2008 3:44 am
Hello all,


there are 10 numbers that satisfy the condition: 101, 102, 103, 104.... 109.

the sentence only says "first digit and the last digit of the integer are each equal to one more than the middle digit". It is not saying that the FIRST digit and the THIRD digit must be same. So, probability 9/100.


If i'm wrong, please explain where i missed the trick.

Thanks,

Anil.
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by nil4700 » Mon Sep 15, 2008 3:46 am
Hello all,


there are 10 numbers that satisfy the condition: 101, 102, 103, 104.... 109.

the sentence only says "first digit and the last digit of the integer are each equal to one more than the middle digit". It is not saying that the FIRST digit and the THIRD digit must be same. So, probability 9/100.


If i'm wrong, please explain where i missed the trick.

Thanks,

Anil.
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by justpal » Mon Sep 15, 2008 6:31 am
nil4700 wrote:Hello all,


there are 10 numbers that satisfy the condition: 101, 102, 103, 104.... 109.

the sentence only says "first digit and the last digit of the integer are each equal to one more than the middle digit". It is not saying that the FIRST digit and the THIRD digit must be same. So, probability 9/100.


If i'm wrong, please explain where i missed the trick.

Thanks,

Anil.
Well, Anil

You are rite....
Question says that ""first digit and the last digit of the integer are each equal to one more than the middle digit". Now in 102, is 2 one more of middle digit which is 0, same as is 3 one more than the middle digit which is again 0 ?????

Also 9/100 is not in the answer list....

Is anyone out there has a different explanation....i am getting confused now....
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by chris500 » Mon Sep 15, 2008 8:12 pm
The only integer from 100 to 199 that meets the stated requirements is 101. There are 100 possibilities from 100-199. The next number above 101 that would meet the stated requirements would be 212, which is higher than the highest possible number.

So D is the answer
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by manulath » Tue Sep 16, 2008 1:09 am
the answer should be 1/100. D

as for "nil4700" and "justpal"

why did you stopped at 101, 102,....,109
this should have expanded further to 112, 123, 134, 145, 156, 167, 178, 189
so 17/100

anyways the question says – "the first digit and the last digit of the integer are each equal to one more than the middle digit"

'are each equal to one more" - both are one more than middle digit.
so only 1 - 101 is possible

............................................
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