If a 3-digit integer is selected at random from the integers 100 throu

Question

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

(A) 2/225 
 (B) 1/111
 (C) 1/110
 (D) 1/100
 (E) 1/50

Skywalker18 · Answer

Hundreds digit has to be 1 . So units digit needs to 1 as well
And both Hundreds and Units digit needs to be 1 more than ten's digit
The number is 101
Total numbers between 100 and 199 = (199-100)+1 = 100

Probability = 1/100

Answer D

sahil7389 · Answer

Answer-D
Total number of digits=199-100=99+1=100= Number of outcomes
Numbers which fulfill given criteria= only 1 number=101
Probability= Favourable Outcome/Total number of outcomes= 1/100

EMPOWERgmatRichC · Answer

Hi All,

The answer choices to this question provide some significant hints as to the possible outcomes that 'fit' the restrictions mentioned in the prompt.

We're told to consider the integers from 100 to 199, inclusive. That is a group of 100 numbers, so any probability question that is based on that range will have a denominator that is either 100 or reduced from 100. Only Answers D and E fit that pattern. With those answers, we know that there is either one number that fits the restrictions (1/100) or two numbers (2/100 = 1/50).

It's actually not too hard to find the integer that fits the description (since we know that the first digit has to be 1): 101. Since there's no other option, we have the answer.

Final Answer:  D

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

EMPOWERgmatRichC · Answer

Hi ALL,

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

Final Answer:  D

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

gota900 · Answer

I don't know if I am the only one, but I think the wording here is very weird... would the GMAC provide such a weird phrasing or is it just me?

Best, gota900

EMPOWERgmatRichC · Answer

Hi gota900,

When you approached this prompt, did you interpret the wording differently (and if you did, then what 'result' did you end up with?)? One of the great aspects about Quant questions on the Official GMAT is that they are written so that there is little-to-no chance for "interpretational bias." In simple terms, if you interpret a question in a way that is not correct, then the end result that you get to will NOT be among the 5 choices... meaning that you will then be able to deduce that you misinterpreted something.

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

gota900 · Answer

EMPOWERgmatRichC

Hey Rich,

the prompt says:

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

If a 3-digit integer is selected at random from the integers 100 through 199, inclusive

Totally fine. We select a 3-digit integer, which will be a random integer that is located between 100 and 199. No problem so far. I'm cool as a cucumber.

Next bit:

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 ?

I underlined everything I have problems with.

What this means to me:

the first digit  (needless to say that has to be 1, bc every integer between 100 up to 199 will be in the hundreds, so that's that)

reading on: ...  and the last digit are each equal to one or more than the middle digit.

This bit is what gets me confused.

If both have to be equal to  one or more than the middle digit  , why can't the last digit be 2? or 3? or e.g. 6?

In my understanding, if we had, say, 1X2, then both, the first and the last digit could indeed be 1 OR MORE than the middle digit for the case 102, same would apply to 1X3 and 1X4 and so on...

Nothing in the prompt gives me the definitive information (at least from my point of view) that the first and the last integer should be equal to 1.

Best, gota900

EMPOWERgmatRichC · Answer

Hi gota900,

The prompt does NOT state that the two digits are each "one OR more greater"... it states that those digits are "ONE MORE greater" than the middle digit.

What is "one more" than 0?
If there are 5 people in a room, then what would "one more person" be?
If you have $10, but the shirt you want to buy costs "one more dollar" than that, then what is the cost of the shirt?

In all of these examples, the phrase "one more" clearly means "add 1"... and that same meaning applies to this question.

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

gota900 · Answer

Thanks mate! I must have been too exhausted from the studying... I don't know where I took that OR from

Best, gota900

energetics · Answer

This problem could be made harder in various ways by changing the conditions.

For example, "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?" In this case, p=10/100 = 1/10 {100, 111, 122, 133 ... 199}

An additional change would be increasing the range to something like 100-399. Now we have more than one possibility for the first digit, set with First=2 would be one less {210, 221, 232 ... 298} and First=3 would be 2 less. p=10+9+8/300 = 27/300 = 9/100. It would be even more cumbersome if the range was something like 100-250 because we'd have a partial set of numbers to choose for 200-250.

Another change would be to ask, "what is the probability that the first digit and the last digit of the integer are equal to  two or more  than the middle digit?" Now p=45/100= 9/20 {101, 102, 103 ... 109} {112, 113, 114 ... 119} etc. Each time First increases by 1 the number of possibilities is reduced by 1, so 9+8+7...+1. You could change the range here too and make it more calculation intensive.

EMPOWERgmatRichC  Hope that's all correct.

ScottTargetTestPrep · Answer

Since the first digit (the hundreds digit) must be 1, the second digit (the tens digit) must be 0 and the third digit (the ones digit) must be 1. That is, the number must be 101. The probability of selecting one number (101) from 100 numbers (100 through 199, inclusive) is 1/100.

Answer: D

MBAHOUSE · Answer

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

(A) 2/225 
 (B) 1/111
 (C) 1/110
 (D) 1/100
 (E) 1/50

bumpbot · Answer

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