Problem Solving for 780+ Aspirants.
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Musiq
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Ok....that's ridiculous....who are you.....???sureshbala wrote:That's nice to hear...My first boss is also Mr.JVM. I learnt a lot during my career as a mentor in TIME, Vizag. I worked there from 2003-2006.Musiq wrote:Hi Suresh,sureshbala wrote:Hi,axat wrote:Hi Suresh
The questions you post here are exciting and challenging, but are you sure GMAT(even at a 99%ile difficulty level) matches questions such as these?
I doubt that. Also, where have you had the experience of 6 years as a GMAT tutor?
In any case, thanks for the effort.
Probably the initial questions in this thread are quite tough and the scope for them to appear in the GMAT may be quite low but the questions after the first 3 or 4 are definitely much within the standard of the GMAT.
By the way, I worked for TIME and CL as a mentor/content developer.
Regards
I used to teach at TIME ( Andhra Pradesh). Mr. J.V. Murthy was my boss.
I really loved that experience.....it's nice to see TIME being mentioned.
@yogami..thanks for the post...nice to hear that this thread helped you to brush up the concepts
Dwarakanagar, 4th lane, man that was my hood.
I cant get a definite ID of you from your picture here.....John here (used to teach Verbal at TIME)
For love, not money.
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ssmiles08
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IMO B.sureshbala wrote:Here is the next question....
Participation in the local soccer league this year is 10% higher than last year.The number of males increased by 5% and the number of females increased by 20%. What fraction of the soccer league is now female?
A. 1/3
B. 4/11
C. 2/5
D. 4/9
E. 1/2
1.05M +1.2F = 1.1*(M+F)
2F = M
so the fraction is (1.2F)/[1.05(2F) +1.2F]
12F/33F = 4/11
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tohellandback
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IMO B
X- boys
Y-girls
x+y=z
x/z+y/z=1------------------------>1
21x/20+6y/5=11z/10
21x/20z+6y/5z=11/10-------------->2
solve 1 ans 2
6y/5z-21y/20z=11/10-21/20
3y/20z=1/20
y/z=1/3
x/z=2/3
z=1
females now=1/3+1/5*1/3=6/15
total now=1+10/100=11/10
fraction of females=6/15*10/11=4/11
X- boys
Y-girls
x+y=z
x/z+y/z=1------------------------>1
21x/20+6y/5=11z/10
21x/20z+6y/5z=11/10-------------->2
solve 1 ans 2
6y/5z-21y/20z=11/10-21/20
3y/20z=1/20
y/z=1/3
x/z=2/3
z=1
females now=1/3+1/5*1/3=6/15
total now=1+10/100=11/10
fraction of females=6/15*10/11=4/11
The powers of two are bloody impolite!!
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bizwizashish
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Hi Suresh ..what is the source of this question. A good one really. Cudnt find any book which explains this chord - angles concept. Where can I find detailed explanation on this concept please. Appreciate if you could let me know. Please post some more questions.
Regards,
Joy
Regards,
Joy
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truplayer256
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The correct answer should be 4/11.
1.10(M+F)=1.05M+1.20F
1.10M+1.10F=1.05M+1.20F
.05M=.10F
M/20=F/10
10M=20F
M=2F
We want to ratio of 1.2F/1.10(M+F)= 1.2(M/2)/(1.10(3M/2))=6/5*10/11*M/2*2/3M=2/5*10/11=20/55=4/11.
1.10(M+F)=1.05M+1.20F
1.10M+1.10F=1.05M+1.20F
.05M=.10F
M/20=F/10
10M=20F
M=2F
We want to ratio of 1.2F/1.10(M+F)= 1.2(M/2)/(1.10(3M/2))=6/5*10/11*M/2*2/3M=2/5*10/11=20/55=4/11.
sureshbala wrote:Here is the next one.....
Two positive numbers x and y are such that their LCM = 1188 and their HCF = 12. If x + y = 240, then find the smaller of the two numbers x and y.
A. 96
B. 108
C. 84
D. 144
E. None of these
Factor the LCM
LCM =2^2 x 3^3 x 11
GCF = 2^2 x 3
If LCM and GCF has 2^2 then both numbers Must have it. If GCF has 3 the one of them has only 3 and the other has 3^3. Since 11 is not in GCF only one of them can have 11. Assigning 11 to the factor with a single 3 meets the 240 condition.
X= 2^2 x 3^3
Y = 2^2 x 3 x 11
Smallest is 27 x 4 =108
Choose B.
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ssmiles08
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I know HCF * LCM = X*Ysureshbala wrote:Here is the next one.....
Two positive numbers x and y are such that their LCM = 1188 and their HCF = 12. If x + y = 240, then find the smaller of the two numbers x and y.
A. 96
B. 108
C. 84
D. 144
E. None of these
1188*12 = x*y
240 = x+y
I finally figured out that the answer was (B) by (240-x)(x) = 14256
But I find this algebra cumbersome, and I was wondering if anyone has any neat tips or tricks for factoring out big numbers such as these??
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abhinav85
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IMO B.
If LCM=1188 and HCF=12.
To get the product of x and y. multiply LCM with HCF,1188 x 12 = 14236.
Now we can see that 6 is in the unit's digit place.
Just go through the answer choices and look for combination
which ends with 6 in the end.........i e 108 x 132..........8 x 2 = 16.
Answer B.
If LCM=1188 and HCF=12.
To get the product of x and y. multiply LCM with HCF,1188 x 12 = 14236.
Now we can see that 6 is in the unit's digit place.
Just go through the answer choices and look for combination
which ends with 6 in the end.........i e 108 x 132..........8 x 2 = 16.
Answer B.
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Musiq
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I thought about this but at the 720+ level, I would'nt take chances with None of the Above.abhinav85 wrote:IMO B.
If LCM=1188 and HCF=12.
To get the product of x and y. multiply LCM with HCF,1188 x 12 = 14236.
Now we can see that 6 is in the unit's digit place.
Just go through the answer choices and look for combination
which ends with 6 in the end.........i e 108 x 132..........8 x 2 = 16.
Answer B.
For love, not money.
