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hmm okay so the total bill is a 2 digit number. I'll start with a random option to understand the problem..

If the bill was 80, Fran would have left a $25 tip, and Gwen would have left a $30 tip...
15% - 24, 20% - 32
If the bill was 70, Fran would have left a $25 tip, and Gwen would have left a $25 tip...
15% - 21, 20% 28
If the bill was 75....Fran would have left a $25 tip, Gwen would have left $25 to be less than 20%...
15% - 22.50, 20% - 30
If it was 76..Fran would have left $25 and Gwen would have left $30

In order for the tips to round to the same number, the difference between 15% and 20% has to be equal to or less than 7.5? So the bill has to be below 76

But I will also think about the minimum number for a second
If the bill was $10, Fran would have left a $5 tip, and Gwen would have left...$0?
15% - 1.5, 20% 2
If the bill is $20, Fran would have left $5, Gwen would have left again, $0
If the bill is $50, Fran would have left $20, and Gwen would have left $15
So the difference also has to be above 5?

So the set is... {51, 52....74, 75}...

so....uhh counting...25?
That's..not an option......so I will arbitrarily multiply it by 2 to get 50?

I'm new to this math sorry
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The tip they could agree on could be 5 , 10 , or 15 dollars

Note that 15% is 3/20 and 20% =1/5

Case I : If the tip is 5 dollars and the bill is for x dollars ,
5 > (3/20)x and 5 < x/5
x < 100/3 and x> 25
x would have to be an integer from 26 to 33 (8 possibilities)
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These are American style tips you mention lol

If the bill was 80, Fran would leave 15 dollars , since 15% of 80 is 12 and the smallest multiple of 5 greater than 12 is 15. Gwen would leave 15 as well, as 20% of 80 is 16 and the smallest multiple of 5 less than 16 is 15.
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hmm okay so the total bill is a 2 digit number. I'll start with a random option to understand the problem..

If the bill was 80, Fran would have left a $25 tip, and Gwen would have left a $30 tip...
15% - 24, 20% - 32
If the bill was 70, Fran would have left a $25 tip, and Gwen would have left a $25 tip...
15% - 21, 20% 28
If the bill was 75....Fran would have left a $25 tip, Gwen would have left $25 to be less than 20%...
15% - 22.50, 20% - 30
If it was 76..Fran would have left $25 and Gwen would have left $30

In order for the tips to round to the same number, the difference between 15% and 20% has to be equal to or less than 7.5? So the bill has to be below 76

But I will also think about the minimum number for a second
If the bill was $10, Fran would have left a $5 tip, and Gwen would have left...$0?
15% - 1.5, 20% 2
If the bill is $20, Fran would have left $5, Gwen would have left again, $0
If the bill is $50, Fran would have left $20, and Gwen would have left $15
So the difference also has to be above 5?

So the set is... {51, 52....74, 75}...

so....uhh counting...25?
That's..not an option......so I will arbitrarily multiply it by 2 to get 50?

I'm new to this math sorry
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Ahaha you're right I entirely forgot to do those percentages correctly! Once I stop making silly mistakes I swear my GMAT score is going to go up by like 50. Sheesh! :lol:

kevincan
These are American style tips you mention lol

If the bill was 80, Fran would leave 15 dollars , since 15% of 80 is 12 and the smallest multiple of 5 greater than 12 is 15. Gwen would leave 15 as well, as 20% of 80 is 16 and the smallest multiple of 5 less than 16 is 15.

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can someone please explain
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This is my explanation: For me answer is 48.

We know that bill is from 10-99 (two digit integer)

Tip will be a multiple of 5n and let the bill be B,

We can write that 15B/100 <5n<20B/100
This is 15B<500n<20B
This is 3B<100n<4B

For N=1
We have B to be minimum 26 and Max 33---> 8 combinations

For N=2
We have B to be minimum 51 and max 66--> 16 combinations

For N=3
We have B to be minimum 76 and maximum 99---> 24 combinations

N=4 is not a possibility as that will make B a 3 digit integer

So total combinations are 24+16+8= 48 combinations.

kevincan: Why this answer is wrong?
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Your answer is indeed right ! Don’t know why it said C
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I got 48 as the answer, don't know why it is wrong
Here's my solution
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Perfect solution !
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Hi paragw,

48 is correct, and here that is choice A.

Very clear solve and I think you've done a good job with the solution. Your setup holds at every step:

- Both tips equal the same multiple of 5, call it 5n.
- Fran leaves the smallest multiple of 5 above 15% of the bill, so 5n > 0.15B.
- Gwen leaves the greatest multiple of 5 below 20% of the bill, so 5n < 0.20B.
- Together: 0.15B < 5n < 0.20B, which is your 3B < 100n < 4B.

Counting the integer bills in each band:

- n = 1: B runs 26 to 33, so 8 bills
- n = 2: B runs 51 to 66, so 16 bills
- n = 3: B runs 76 to 99, so 24 bills
- n = 4 would push B past 99, so it is out.

Total = 8 + 16 + 24 = 48.

One point worth pinning down, because it is what this question gets argued over: both rules use strict inequalities. That is what excludes B = 25, 50 and 75, where 20% of the bill lands exactly on a multiple of 5. At B = 50, Fran leaves 10, but Gwen has to stay below 10 and leaves 5, so the tips do not match. Count those three in and you get 51. The wording keeps them out.

Answer: A
paragw
I got 48 as the answer, don't know why it is wrong
Here's my solution
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Sorry, what am I missing?

20% of 26 is 10.xx. So the greatest multiple of 5 less than 20% of 26 is 10, isn’t it?
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20% = 1/5
20% of 26 is a little more 5 (26/5=52/10=5.2)
So Gwen would a 5 dollar tip on a $26 restaurant bill
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Uff i need a rest... 20 mins for this..:lol: Thx sir

kevincan
20% = 1/5
20% of 26 is a little more 5 (26/5=52/10=5.2)
So Gwen would a 5 dollar tip on a $26 restaurant bill
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