FUnny DS question

[vishal205]

Whenever Martin has a restaurant bill with an amount between $10 and $99, he calculates the dollar amount of the tip as 2 times the tens digit of the amount of his bill. If the amount of Martin' most recent restaurant bill was between $10 and $99, was the tip calculated by Martin on this bill greater than 15 percent of the
amount of the bill?

(1) The amount of the bill was between $15 and $50
(2) The tip calculated by Martin was $8

[this problem was transcribed incorrectly in the original post. it has now been fixed. --ed.]

Its an easy problem..but the issue i have with it is that there is conflicting info in the two statements.
For the tip to be 8$ in (2)..the amt has to be in between 40 and 49

But acc to statement (1) the amt is between 15 and 30 which doesnt overlap with the above statement.

Is this a flawed question????

[tim]

As per the forum rules, we need a verifiable source for the problem before we can deal with this question. If the question is not one of ours or one that is copyright-free, we will have to delete the question..

[tim]

but yes, the question is flawed..

[jonathanblattner]

This is one of your questions. It is not flawed. When taking statement two by itself, the bill amount is between $10 and $99 since we do not take into consideration option one. The answer is B, option 2 is enough to answer the question, since even at $49, the tip is greater than 15% of the bill.

[RonPurewal]

Jonathan, you are correct that the statements don't have to be combined, but that's not the issue here.
The issue here is that the two statements will never, ever contradict each other, even in cases where you don't ever reach the point of combining them (like this one).

This problem is from GMAT PREP; the first statement has been transcribed incorrectly. The first statement is supposed to say "between $15 and $50."

[RonPurewal]

I realized that there's no actual solution posted on this thread, so here's a quick summary (feel free to reply if anything is unclear).

The goal here is to make both possible outcomes happen -- i.e., a tip that's greater than 15% of the bill, AND a tip that's not greater (i.e., equal to or less than 15% of the bill).
If we succeed in making that happen, then that's "not sufficient". If we fail -- if we're stuck with just one of those possibilities -- then that's "sufficient".

Statement one:
If the bill is $20, then the tip is $4, which is more than 15% of $20.
If the bill is $28, then the tip is still $4, which is not more than 15% of $28. (15% of $28 = 10% of $28 + 5% of $28, which is 2.80 + 1.40 = 4.20.)
Not sufficient.

Statement two:
This means that the bill was between $40 and $49.99. Therefore, 15% of the bill is anywhere from $6 to juuuuust barely less than $7.50 (= 15% of $50.00).
The tip is $8, which is always greater than anything between $6 and $7.50.
Sufficient.

[larissaho0907]

Below is my approach to this question:

assume the tens digit of the bill is a, the ones digit is b, so the amount of the bill =10a+b

From the question stem, we know the tip =2a. When the tip is greater than 15% of the amount of the bill, we have the following:

2a> .15(10a+b)

2a> .15*10a+ .15*b

2a>1.5a + .15b

2a-1.5a> .15b

.5a>.15b

a> .3b

So as long as the number of ten digit is great than 30% of the ones digit, the tip is greater than 15% of the amount of the bill. Is my approach correct?

Thank you! Any help is appreciated!

[jlucero]

Below is my approach to this question:

assume the tens digit of the bill is a, the ones digit is b, so the amount of the bill =10a+b

From the question stem, we know the tip =2a. When the tip is greater than 15% of the amount of the bill, we have the following:

2a> .15(10a+b)

2a> .15*10a+ .15*b

2a>1.5a + .15b

2a-1.5a> .15b

.5a>.15b

a> .3b

So as long as the number of ten digit is great than 30% of the ones digit, the tip is greater than 15% of the amount of the bill. Is my approach correct?

Thank you! Any help is appreciated!

Very nice. This goes to show that if it is sufficient, there should be a way to mathematically prove it. That said, don't think that one method or the other is the way you have to solve. The best way is always the fastest way for YOU to solve.

[larissaho0907]

Thank you so much Joe!

[jnelson0612]

Thank you so much Joe!

:-)