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)
Thanks
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- Stacey Koprince
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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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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 %
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 %
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
# 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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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?
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?
- Sadowski
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Actually, 15% of $16 is $2.40, so his $2 tip is not greater than 15%. B is the only sufficient answer.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?
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#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)
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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- Junior | Next Rank: 30 Posts
- Posts: 13
- Joined: Sat Jun 09, 2007 11:57 am
#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)
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)
#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!!
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!!