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by sureshbala » Tue May 05, 2009 9:36 pm
sureshbala wrote:Here is the next one.....

If x and y are positive integers such that 5 * x^3 = 6 * y^4, find the minimum possible value of x+y.

A. 1240

B. 1260

C. 1060

D. 1080

E. None of these
Folks, different solutions are given by our friends which are really quick and lucid.

I just thought to put forward another lucid way ....


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by sureshbala » Tue May 05, 2009 9:46 pm
Folks, here is the next question..


A factor of 840 is chosen at random. What is the probability that it is divisible by 15?


A. 1/4

B. 7/32

C. 1/2

D. 1/56

E. None of these
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by dtweah » Wed May 06, 2009 5:29 am
sureshbala wrote:Folks, here is the next question..


A factor of 840 is chosen at random. What is the probability that it is divisible by 15?


A. 1/4

B. 7/32

C. 1/2

D. 1/56
840=2^3 x 3 x 5 x 7
15= 3 x 5

The only factors of 840 that will be factors of 15 must have a 3 and a 5:

This reduces to in how many ways can (3,5) be paired with 2 2 2 7
By Fundamental Counting Principle. This is 2 x 4=8 ways. So only 8 factors will be divisible by 15.

There are 6 prime factors if we count the 2’s individually, which we must in this case. How many different factors can we get by combining these 6?

Well 6C1+6C2+6C3+6C4+6C5+6C6

6+15+20+15+6+1=63

Prob = 8/63.

I choose E.


E. None of these
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by Musiq » Wed May 06, 2009 11:16 am
Hi Suresh,

I have 1/4 as the probability. Please advise if this is correct, so I can post my response.
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by Uri » Thu May 07, 2009 4:28 am
hi suresh, missed you in the forum for few days. great that you are back!
for the above problem, i am also getting 1/4. is it the correct answer?
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by sureshbala » Thu May 07, 2009 12:26 pm
Uri wrote:hi suresh, missed you in the forum for few days. great that you are back!
for the above problem, i am also getting 1/4. is it the correct answer?
Hi, yes I really missed BTG. Anyway we will have a continuous season from now..

Hey by the way 1/4 is the correct answer. Please post your solution.

@musiq, please post your solution as well....
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by Uri » Thu May 07, 2009 10:10 pm
sureshbala wrote:A factor of 840 is chosen at random. What is the probability that it is divisible by 15?
A. 1/4
B. 7/32
C. 1/2
D. 1/56
E. None of these
840=2^3 * 3 * 5 * 7

Total number of factors of 840 = (3+1) * (1+1) * (1+1) * (1+1)= 32

Now, all factors that are multiples of 15, will have 3*5 in them. So, if we find the total number of factors that we can have with 2^3 and 7, then that will give us the total number of factors divisible by 15.

Total number of factors with 2^3 and 7= (3+1)*(1+1) = 8

So, the probability of choosing any random factor of 840 that will be divisible by 15 is 8/32 = 1/4

Ans. (A)
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by Musiq » Fri May 08, 2009 8:20 am
Uri wrote:
sureshbala wrote:A factor of 840 is chosen at random. What is the probability that it is divisible by 15?
A. 1/4
B. 7/32
C. 1/2
D. 1/56
E. None of these
840=2^3 * 3 * 5 * 7

Total number of factors of 840 = (3+1) * (1+1) * (1+1) * (1+1)= 32

Now, all factors that are multiples of 15, will have 3*5 in them. So, if we find the total number of factors that we can have with 2^3 and 7, then that will give us the total number of factors divisible by 15.

Total number of factors with 2^3 and 7= (3+1)*(1+1) = 8

So, the probability of choosing any random factor of 840 that will be divisible by 15 is 8/32 = 1/4

Ans. (A)
Word for word...thats exactly what I would have typed.

Great minds think alike ... Uri? :D
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by Uri » Fri May 08, 2009 9:12 am
Musiq wrote:Word for word...thats exactly what I would have typed.

Great minds think alike ... Uri? :D
:)

please wish me luck, dear! now-a-days i am quite tensed. only 6 more days to go. :oops:
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by Musiq » Fri May 08, 2009 12:36 pm
Uri wrote:
Musiq wrote:Word for word...thats exactly what I would have typed.

Great minds think alike ... Uri? :D
:)

please wish me luck, dear! now-a-days i am quite tensed. only 6 more days to go. :oops:
Wow, I didnt know that your test was approaching.

Here are my words of advice and well-wishes for you. For other readers, please note that I am tailoring my advice specifically for Uri ( who I know to be very good at Math). That's not an insult to anyone else...it just is.

1) MAKE SURE YOU FINISH THE TEST. You probably know this already, but the best of test-takers somehow manage to screw this one thing up. Yours truly can attest to that.

Quick Tip: When on your last question....before you even read the question, fill a random bubble up with your mouse. This records your response as "Answered" . So, even if you run out of time while reading the question you will not be penalized for NOT answering the question.

2) For Reading Comprehension, if you are stuck between 2 choices ALWAYS guess on the option that is more in alignment with the Author's convictions and/ or opinions.

3) For Verbal: Practice as many GMAC based questions as you can.

4) For Quant I have 2 pieces of advise for you:

a) Do NOT judge the diffuculty of the exam while taking the CAT. There can be no greater torture than this.

If you feel the questions are easy:
-----It could be because you are scaling down on the adaptive platform of the CAT algorithm
OR
----- Maybe you really are THAT GOOD!

Leave second guessing to crazy folks like me ....you want to crack the GMAT wide open.....you should only focus on answering a question and soldiering on.

b) Because your Math is really good, there is a tendency to linger on a question UNTIL you have solved it. This is dangerous....especially when you consider that anywher from 9 to 11 questions are Experimental anyway.

You should not be on the same screen for more than 210 seconds on any question.

Good luck!
Last edited by Musiq on Sat May 09, 2009 6:02 am, edited 1 time in total.
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by Uri » Fri May 08, 2009 2:02 pm
thanks, musiq, for your great advice. i shall keep these in mind and follow them.
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by francopiccolo » Sat May 16, 2009 4:08 pm
I think I am missing sth here. How do I know that:

Total number of factors of 840 = (3+1) * (1+1) * (1+1) * (1+1)= 32

Please explain the 3+1, 1+1, etc.
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by nitingupta5 » Mon May 18, 2009 6:23 am
i believe if the triangles ACB and ACP can be proved similar than the solution becomes easy but i don't know if i am in the right direction.
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by Musiq » Mon May 18, 2009 1:01 pm
francopiccolo wrote:I think I am missing sth here. How do I know that:

Total number of factors of 840 = (3+1) * (1+1) * (1+1) * (1+1)= 32

Please explain the 3+1, 1+1, etc.
To find the number of factors of a number:

Step 1: Prime Factorize the number
Step 2: Express the prime factorization in the exponent form, making sure to show an exponent 1 where necesary.
Step 3: Number of factors = ( Exp +1) ( Exp +1) ( Exp +1)...

Example: To find the number of factors of 24

Prime Facotrize 24: (3) (2)(2)(2)
Express in Exponent form:{ (2) ^ 3 } { (3)^1}

Number of factors = (3 +1) ( 1 +1) = 8.

Please let me know if you have any further questions regarding this technique.
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by francopiccolo » Mon May 18, 2009 1:53 pm
Thank you musik, I spared some thought on it and I understood why is that technique used. I thanked you with the button thanks, but I don't know if that is noticed by the thanked, hehe.[/quote]
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