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GMATPrep Questions

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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GMATPrep Questions

by prasant » Sun May 13, 2007 1:14 pm
Hey Guys,

I just finished taking a GMATPrep Practice Test and got the following quant questions wrong. Can someone please help me solve these problems? (I have fixed the mistake I made in #26)

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Thanks :)
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Re: GMATPrep Questions

by prasant » Sun May 13, 2007 1:16 pm
Thanks to a lot that I learnt on this forum, I am finally getting to where I wanna be:

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I have my test scheduled for May 19.
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by Stacey Koprince » Thu May 17, 2007 7:25 pm
Hi - you'll be more likely to get replies if you break your questions up into separate posts - most people who see so many questions in a row will be overwhelmed and move to another post. :)
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by atarafder » Fri May 18, 2007 2:26 pm
17.
If Don had travelled the whole of x-miles at 60 mph, his travel time would have been t= x/60

His actual travel time is = (x-5)/60 + 5/30
= x/60 + 1/12

This means that his travel time was increased by 1/12 hours.

=(1/12)*100/x/60 = 500/x %
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by atarafder » Fri May 18, 2007 2:32 pm
26.
1. Insufficient
2. for the tip t be $8 the bill must be >=40 and </49
15% of 49 <$7.5, so sufficient
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by atarafder » Fri May 18, 2007 2:41 pm
33.Let the # of cameras produced in 1993 = 100
# of cameras in 1994 (x% more of that produced in 1993) = 100*(100+x)/100 = 100 + x

# of cameras produced in 1995 (y% more than that produced in 1994) = (100+y) (100+x)/100 = 100 (1 + x+y + xy/100)

1. Insufficent

2. Sufficient
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by bww » Fri May 18, 2007 3:25 pm
For #26...

Isn't 1) sufficient? We know that Martin takes the tens digit of the amount of his bill x2 as the amount of his tip. Assuming his bill is $16, he would then leave a $2 tip. 15% of 16 if $1.80, so yes, the tip is greater than 15% of the amount of the bill. A 15% tip on a $49 bill would be $7.35, but Martin's rule of thumb has him leaving $8 as a tip. So, 1) is sufficient.

2) If he leaves an $8 tip, then his bill must have been between $40-49. 15% tip on $40 is $6, and $7.35 on a $49 tab. 2) alone must also be sufficient...

Answer is D?
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by prasant » Fri May 18, 2007 8:50 pm
Guys, Thanks for the responses. Very helpful.
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Correction

by Sadowski » Mon May 21, 2007 6:33 am
bww wrote:For #26...

Isn't 1) sufficient? We know that Martin takes the tens digit of the amount of his bill x2 as the amount of his tip. Assuming his bill is $16, he would then leave a $2 tip. 15% of 16 if $1.80, so yes, the tip is greater than 15% of the amount of the bill. A 15% tip on a $49 bill would be $7.35, but Martin's rule of thumb has him leaving $8 as a tip. So, 1) is sufficient.

2) If he leaves an $8 tip, then his bill must have been between $40-49. 15% tip on $40 is $6, and $7.35 on a $49 tab. 2) alone must also be sufficient...

Answer is D?
Actually, 15% of $16 is $2.40, so his $2 tip is not greater than 15%. B is the only sufficient answer.
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by mikeclarke44 » Sun Jun 10, 2007 1:33 pm
#27
I don't know why they posted it incorrect. I come up with a value for r when I use #1 and #2

1). r=3m
2). r/2 + m/2 =12 r + m = 24

3m + m = 24
4m = 24
m= 6, r=18

That answer should have been (C)
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by mikeclarke44 » Sun Jun 10, 2007 1:34 pm
#27
I don't know why they posted it incorrect. I come up with a value for r when I use #1 and #2

1). r=3m
2). r/2 + m/2 =12 r + m = 24

3m + m = 24
4m = 24
m= 6, r=18

That answer should have been (C)
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by vrider » Sun Jun 10, 2007 3:18 pm
From (1) r=3m , insufficient
From (2) r-12 = 12-m , insufficient

combining, r-12 = 12 - r/3
4/3*r = 24 or r = 18

So, C?

Hmm...I think I am missing something here...let me check.
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by vrider » Sun Jun 10, 2007 3:27 pm
Ok, I think I know why. Assumption that r=3m is wrong. If m is negative, distance from 0 to r can still be 3 times distance from m to 0 i.e. r=3m or r = -3m.

Hence, combining (1) & (2) is still insufficient to find r.
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by restero » Tue Mar 09, 2010 10:23 am
#27 is E
as r=3m or r=-3m

solving u get 6, 18 and -12, 36; So E!
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by harsh001 » Sat Mar 13, 2010 5:57 pm
#19

Let X = 1 and Y = 2 then equation = -1/3 < 1
Let X = 1 and Y = -2 then equation = -3/2 < 1
Let X =0 and Y = 1 then equation = 1 = 1
Let X = 0 and Y =-1 then equation = -1 < 1
Let X = -1 and Y = 2 then equation = -3 < 1
Let X =-1 and Y = -2 then equation = -1/3 < 1

therefore in all cases the equation in not greater than 1 if we take similar values of condition 2 we will see that each condition in its own answers the question therefore D is the answer!!
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