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# Some difficult problems (for me :)) anyone can help?

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Intern
Joined: 19 Dec 2009
Posts: 8
Some difficult problems (for me :)) anyone can help? [#permalink]

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02 Jan 2010, 00:10
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Hi guys, thank you very much for your quick replies for previous questions, appreciated..
There are some difficult problems for me to solve. Can anyone show how to solve them or give links if they are already solved..

1) A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first gigit cannot be 0. How many different identification numbers are possible?

a) 3024
b) 4536
c) 5040
d) 9000
e) 10000

OA is B

2) The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100%, which of the following is closet to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?

a) 100% dec.
b) 50% dec.
c) 40% dec.
d) 40% inc.
e) 50% inc.

3) All of the stocks on the over-the-counter market are designed by either a 4 letter of a 5 letter code, that is created by using the 26 letters of the alphabet. Which of the following gives the maximum number of different stocks that can be designated with these codes?

a) $$2(26^5)$$
b) $$26(26^4)$$
c) $$27(26^4)$$
d) $$26(26^5)$$
e) $$27(26^5)$$

OA is C
Intern
Joined: 31 Dec 2009
Posts: 13
Re: Some difficult problems (for me :)) anyone can help? [#permalink]

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02 Jan 2010, 02:41
Q1. I think the question tag must also include " no repetition allowed".. Only then the OA is obtained.

Four digits. Starting from the left most digit.. Given it cannot be '0'. hence 9 options (1,2,3,...9)
Second digit from left..Again 9 options(exclude the first digit but add 0. Hence 9)
Third digit from left...8 options(exclude first and second digit)
Last digit..7 options.
Total combinations = 9*9*8*7=4536..
However for the given question as is, I think 9000 is the correct answer..

Q2. Rate of chemical reaction R = k A^2/B
Now B becomes 2B(increased by 100%), for rate to remain constant numerator should be 2A^2. Hence A should become \sqrt{A} ~ 1.414 A.
Hence ~40% increase in A

Q3. Repetition of alphabets allowed.
Four digit ticker symbols = 26^4
Five digit ticker symbols = 26^5
Total possibilities = 26^4 + 26^5
==>26^4(1+26)
==>26^4(27)
Senior Manager
Joined: 22 Dec 2009
Posts: 359
Re: Some difficult problems (for me :)) anyone can help? [#permalink]

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02 Jan 2010, 06:08
acegre wrote:
Q1. I think the question tag must also include " no repetition allowed".. Only then the OA is obtained.

Four digits. Starting from the left most digit.. Given it cannot be '0'. hence 9 options (1,2,3,...9)
Second digit from left..Again 9 options(exclude the first digit but add 0. Hence 9)
Third digit from left...8 options(exclude first and second digit)
Last digit..7 options.
Total combinations = 9*9*8*7=4536..
However for the given question as is, I think 9000 is the correct answer..

Q2. Rate of chemical reaction R = k A^2/B
Now B becomes 2B(increased by 100%), for rate to remain constant numerator should be 2A^2. Hence A should become \sqrt{A} ~ 1.414 A.
Hence ~40% increase in A

Q3. Repetition of alphabets allowed.
Four digit ticker symbols = 26^4
Five digit ticker symbols = 26^5
Total possibilities = 26^4 + 26^5
==>26^4(1+26)
==>26^4(27)

For the 1st question, B is perfect. The question clearly states - different digits, which means no repetition allowed!

Q2 = 40% increase
Q3 $$= 27(26^4)$$

Cheers!
JT
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Intern
Joined: 19 Dec 2009
Posts: 8
Re: Some difficult problems (for me :)) anyone can help? [#permalink]

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02 Jan 2010, 20:24
acegre wrote:
Q1. I think the question tag must also include " no repetition allowed".. Only then the OA is obtained.

Four digits. Starting from the left most digit.. Given it cannot be '0'. hence 9 options (1,2,3,...9)
Second digit from left..Again 9 options(exclude the first digit but add 0. Hence 9)
Third digit from left...8 options(exclude first and second digit)
Last digit..7 options.
Total combinations = 9*9*8*7=4536..
However for the given question as is, I think 9000 is the correct answer..

Q2. Rate of chemical reaction R = k A^2/B
Now B becomes 2B(increased by 100%), for rate to remain constant numerator should be 2A^2. Hence A should become \sqrt{A} ~ 1.414 A.
Hence ~40% increase in A

Q3. Repetition of alphabets allowed.
Four digit ticker symbols = 26^4
Five digit ticker symbols = 26^5
Total possibilities = 26^4 + 26^5
==>26^4(1+26)
==>26^4(27)

Could you clarify this "Hence A should become \sqrt{A} ~ 1.414 A."?
Intern
Joined: 22 Dec 2009
Posts: 13
Re: Some difficult problems (for me :)) anyone can help? [#permalink]

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02 Jan 2010, 22:06
acegre wrote:
Total possibilities = 26^4 + 26^5
==>26^4(1+26)

For me, this step is the tricky part. Having done the math, I see that the answer is the same in either case, but I don't understand why 1+26 was chosen, or why multiplying by it should give the same answer as adding 26^5. I've hunted around for this rule for awhile and haven't found it; I would really appreciate an explanation or link. Thanks!
Senior Manager
Joined: 22 Dec 2009
Posts: 359
Re: Some difficult problems (for me :)) anyone can help? [#permalink]

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03 Jan 2010, 03:00
1
KUDOS
bekbek wrote:
acegre wrote:

Q2. Rate of chemical reaction R = k A^2/B
Now B becomes 2B(increased by 100%), for rate to remain constant numerator should be 2A^2. Hence A should become \sqrt{A} ~ 1.414 A.
Hence ~40% increase in A

Could you clarify this "Hence A should become \sqrt{A} ~ 1.414 A."?

As given $$R= \frac{k*A^2}{B}$$

Now B is increased to 2B... Therefore new R is:
$$R_n = \frac{k*A^2}{2B} = \frac{R}{2}$$ which tells us that the previous R is reduced to half because of increasing B to 100% more. This implies that $$A^2$$ should be increased by a factor 2 to counter the effect of increase in B...

Therefore new $$A^2$$ should be:
$$A_n^2 = 2 * A^2$$
$$A_n = \sqrt{2} * A$$

Therefore the increase in quantity of A is $$A_n - A = \sqrt{2} * A - A = (\sqrt{2} - 1)* A = 0.414 A \approx 40 %$$

Hope this is clear...

Cheers!
JT
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Senior Manager
Joined: 22 Dec 2009
Posts: 359
Re: Some difficult problems (for me :)) anyone can help? [#permalink]

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03 Jan 2010, 03:05
1
KUDOS
handsomebrute wrote:
acegre wrote:
Total possibilities = 26^4 + 26^5
==>26^4(1+26)

For me, this step is the tricky part. Having done the math, I see that the answer is the same in either case, but I don't understand why 1+26 was chosen, or why multiplying by it should give the same answer as adding 26^5. I've hunted around for this rule for awhile and haven't found it; I would really appreciate an explanation or link. Thanks!

Not very sure as to what you mean...

Guess the maths is pretty simple for this...

Four digit ticker symbols $$= 26^4$$
Five digit ticker symbols $$= 26^5$$
Total possibilities $$= 26^4 + 26^5$$
Now if you take $$26^4$$ common above (distributive property of multiplication), you get $$26^4 (1+26) = 26^4 (27)$$..

Is that clear... or there is something else which is unclear... do let me know.... !

Cheers!
JT
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Intern
Joined: 22 Dec 2009
Posts: 13
Re: Some difficult problems (for me :)) anyone can help? [#permalink]

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03 Jan 2010, 13:19
jeeteshsingh, I did not understand how to use the distributive property of multiplication in this case. Now I'm doing OK. Thanks for the help!
Re: Some difficult problems (for me :)) anyone can help?   [#permalink] 03 Jan 2010, 13:19
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# Some difficult problems (for me :)) anyone can help?

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