Problem Solving for 780+ Aspirants.

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sreak1089
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PostThu Jun 11, 2009 7:43 am
I would like to change my answer..I will go with B (4/11) Smile
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muna_m
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PostThu Jun 11, 2009 7:56 am
IMO B. 4/11

Can u please post the OA?
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Musiq
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PostThu Jun 11, 2009 11:17 am
sureshbala wrote:
Musiq wrote:
sureshbala wrote:
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.
Hi,

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
Hi Suresh,

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.
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.

@yogami..thanks for the post...nice to hear that this thread helped you to brush up the concepts
Ok....that's ridiculous....who are you.....???

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)
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PostThu Jun 11, 2009 1:09 pm
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
IMO B.

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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PostThu Jun 11, 2009 9:26 pm
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
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bizwizashish
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PostSat Jun 13, 2009 12:58 pm
B

explanation as above
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joymukhi
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PostMon Jun 15, 2009 2:13 am
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
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PostMon Jun 22, 2009 1:12 pm
x = male, y = female, z = total

1.05x + 1.2y = 1.1z
x + y = z

a) 1/3

Reason: y/z = 1/3 -> z = 3y

x + y = 3y
x = 2y

1.05(2y) + 1.2y = 1.1(3y)
2.1y + 1.2y = 3.3y
3.3y = 3.3y

Therefore, 1/3 is correct.
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truplayer256
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PostMon Jun 22, 2009 1:29 pm
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.
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yogami
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PostTue Jun 23, 2009 2:36 pm
1/3 is the original fraction. You multiply that by 1.2/1.1 to get the new fraction which is 4/11 indeed
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PostWed Jun 24, 2009 2:47 am
[quote="sureshbala"]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.
[color=brown]
[b]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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PostWed Jun 24, 2009 12:49 pm
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
I know HCF * LCM = X*Y

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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PostWed Jun 24, 2009 12:59 pm
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.
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Musiq
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PostThu Jun 25, 2009 7:52 am
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.
I thought about this but at the 720+ level, I would'nt take chances with None of the Above.
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abhinav85
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PostThu Jun 25, 2009 8:29 am
Quote:
I thought about this but at the 720+ level, I would'nt take chances with None of the Above.
Well i 'll say gud for u man!!!!
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