GMAT Problem Solving Discussions

MBA
3901
I can say if tht the eqn has a solution or not only by combining both the statements...So I selected option C.

-Deepak.



Statement 1 cannot be sufficient ,as the eq can be satisfied in more than one set of no.
{m+n=prime no.

2+3=5
2+5=7
}

Statement 1 is sufficient alone,mn=prime no.
{only one set,
1*2=2
}

Correct me if I'm wrong!!
3902
Statement 1 cannot be sufficient ,as the eq can be satisfied in more than one set of no.
{m+n=prime no.

2+3=5
2+5=7
}

Statement 1 is sufficient alone,mn=prime no.
{only one set,
1*2=2
}

Correct me if I'm wrong!!


I hope you meant statement 2 alone is suff...Bt still you are considering only 1*2 bt it can also be 1*3 or 1*7 etc.So either statement alone is insuff.
3903
either statements is not sufficient.. now combining them..

lets solve it using ur method, taking sum n product of roots.. let the roots be m and n.

equation is x^2 - bx + c = 0

(m+n)=b----->(1)
mn=c ---------->(2)

from (2) since c is prime, either m or n shd be 1(if both m & n are prime then c will become composite)

lets say m=1, then n=c

or from (1)
1+n = b => 1+c =b (where both b and c are primes)
this is possible only when c=2 and b=3.

so both statements r sufficient.. hence option C..


Yup. Looks "C" & OA is satisfying too :)

Regards,
3904

Could you please analyze, why it should be "E"?

My analysis:
~~~~~~~
Lets say male employee = M
female employee = F

(1) after recruiting 14 female the total female count would be (F+14)
Thus, M/(F+14) = 16/9.

This is insufficient.

(2) M-F = 105

This is also insufficient.

(1) and (2) together ---> M-F = 105 . Thus, M= F+105 .

Hence, M/(F+14) = 16/9
=> (F+105)/(F+14) = 16/9.

Hence, F can be derived to a finite value. Thus (C) should be the result.

What is the number of female employees in Company X ?

(1) If Company X were to hire 14 more people and all of these people were females, the ratio of the number of male employees to the number of female employees would then be 16 to 9.

(2) Company X has 105 more male employees than female employees.

A Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D EACH Statement ALONE is sufficient.
E Statements (1) and (2) TOGETHER are NOT sufficient.

Post 7
I think the answer should be E
3905

Hi!
I have my GMAT exam on 19th July.
Have completed OG & Kaplan 800 for quants.
I am looking for some good resource (tough maths problems) for further practise. I feel practising from CAT/XAT material in final 19-20 doesnt make sense.

Can somebody recommend me some good resource for quants?

3906

Hi guys...plz help me out with this one...

F(n) denotes the remainder when n is divided by k. Is k greater than 1?
1). f(k+32)=9
2). f(k+16) =6
3907

Hi guys...plz help me out with this one...

F(n) denotes the remainder when n is divided by k. Is k greater than 1?
1). f(k+32)=9
2). f(k+16) =6






can be solved directly using (divisor > remainder) funda.. D shd b da answer..
1. k(divisor) shd be greater than 9(remainder)--------->sufficient
2.k(divisor) shd b greater than 6(remainder)------------>sufficient
3908
Hi!
I have my GMAT exam on 19th July.
Have completed OG & Kaplan 800 for quants.
I am looking for some good resource (tough maths problems) for further practise. I feel practising from CAT/XAT material in final 19-20 doesnt make sense.

Can somebody recommend me some good resource for quants?


all da best dude..
well this question especially 4 u..

2 drunkards,Mission and Atrish,(no offense.. :)) play the following game with a fair coin. Mission( chooses (and announces) a triplet (HHH, HHT, HTH, HTT, THH, THT, TTH, or TTT) that might result from three successive tosses of the coin. Atrish then chooses a different triplet. The drunkards toss the coin until one of the two named triplets appears. The triplets may appear in any three consecutive tosses: (1st, 2nd, 3rd), (2nd, 3rd, 4th), and so on. The winner is the drunkard whose triplet appears first.

which of the two drunkards is more likely to win??
1. Mission.
2. Atrish.
3. Both Mission and Atrish have equal probability of winning.
4. None of them wins
5. Data insufficient
3909
can be solved directly using (divisor > remainder) funda.. D shd b da answer..
1. k(divisor) shd be greater than 9(remainder)--------->sufficient
2.k(divisor) shd b greater than 6(remainder)------------>sufficient


Hi,
Can you please elaborate the funda here?



Hi guys...plz help me out with this one...

F(n) denotes the remainder when n is divided by k. Is k greater than 1?
1). f(k+32)=9
2). f(k+16) =6





My take:

f(k+32) = k+32 mod k = 9
> 0 + 32 mod k = 9
> 32 mod k = 9

k= 23 is a possible value > 1
k = -41 also seems to be a possible value
32 mod -41 = -32 mod 41 = +9


Same is the case with B.

So, answer should be E.

What's the OA?
3910
Hi,
Can you please elaborate the funda here?

F(n) denotes the remainder when n is divided by k. Is k greater than 1?
1). f(k+32)=9
2). f(k+16) =6

look here F(n) denotes the remainder(given in question) so f(k+32) and f(k+16) are also remainders.

also given n = qk + F(n)---------->(main_equation)
dividend(D) = quotient x divisor + remainder

1. from (main_equation) and statement 1,
rem.= f(k+32)=9, divisor is k
=> k>9 (implies greater than 1)..... sufficient.
2. k>6.. sufficient

so option D..
3911
Hi,
Can you please elaborate the funda here?




My take:

f(k+32) = k+32 mod k = 9
> 0 + 32 mod k = 9
> 32 mod k = 9
k= 23 is a possible value > 1
k = -41 also seems to be a possible value
32 mod -41 = -32 mod 41 = +9


Same is the case with B.

So, answer should be E.

What's the OA?

bro u r complicating here(bold text)..
u can directly write
k+32 = qk + 9 (where q = quotient and k>9)
3912
F(n) denotes the remainder when n is divided by k. Is k greater than 1?
1). f(k+32)=9
2). f(k+16) =6

look here F(n) denotes the remainder(given in question) so f(k+32) and f(k+16) are also remainders.

also given n = qk + F(n)---------->(main_equation)
dividend(D) = quotient x divisor + remainder

1. from (main_equation) and statement 1,
rem.= f(k+32)=9, divisor is k
=> k>9 (implies greater than 1)..... sufficient.
2. k>6.. sufficient

so option D..


1) I think F(n) and f(n) should be the same here
2) f(k+32) means the remainder obtained when k+32 is divided by k'
Hence my derivation.

Btw, I'm getting a -ve value of k too using this approach. Why are you not getting a -ve value?
Where am i messing up, if so?
3913

Thanks a lot....:-)

3914
can be solved directly using (divisor > remainder) funda.. D shd b da answer..
1. k(divisor) shd be greater than 9(remainder)--------->sufficient
2.k(divisor) shd b greater than 6(remainder)------------>sufficient


thanks a lot guys...:-)
3915

1) For every positive even integer n, the function F(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of F(100)+1, then what is p ?

2) try to visualize this, cant draw figure:
In the semicircle with origin (0,0) in XY plane point p(-sqrt(3), 1) and Q(s,t) connects with origin at 90 angle. What is the value of s?

3)The rate of certain chemical reaction is directl proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of If the concentration of B is increased by 100%, which of the following is closest to percent change in the concentration of chemical A required to keep the reaction rate ? Answer is 40% increase but how?

plz give explanations.

3916

3)The rate of certain chemical reaction is directl proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of If the concentration of B is increased by 100%, which of the following is closest to percent change in the concentration of chemical A required to keep the reaction rate ? Answer is 40% increase but how?

plz give explanations.



Rate = k A^2 / B .

Assume that A = 100 and B = 100 at intial phase.Now, B = 200 ( since rate of increase is 100%).

Now we need increase A to match 200 so that rate of reaction remains the same. So the value of A which when squared becomes 200 is 14 - 15.

14 ^ 2 = 196 & 15^2 = 225.So A must be 140 to nullify the net rate.Now, the percent increase is from 100 to 140 which is 40%.

-Deepak.
3917
1) For every positive even integer n, the function F(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of F(100)+1, then what is p ?

2) try to visualize this, cant draw figure:
In the semicircle with origin (0,0) in XY plane point p(-sqrt(3), 1) and Q(s,t) connects with origin at 90 angle. What is the value of s?

3)The rate of certain chemical reaction is directl proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of If the concentration of B is increased by 100%, which of the following is closest to percent change in the concentration of chemical A required to keep the reaction rate ? Answer is 40% increase but how?

plz give explanations.

would be very helpful if u put options with the questions..
1.
F(n) = n!
now ((p-1)! + 1) is always divisible by p, where p=prime number (dont remember this theorem name)

F(100) + 1 = (101-1)! + 1 shd be divisible by 101(a prime number).

hence answer shd b 101.

2. i guess it is an incomplete question.. couldn't make out nething 4m it..

3. r = kA^2
r= l/B
r= mA^2/B--------->(1)
let new A be a, and new B be b
then
r=ma^2/b
or r=ma^2/2B (since b=2B given)-------->(2)
dividing (1) by (2)
a^2 = 2A^2 or
a= root(2)A=1.414A
hence increase is 41.4%
3918
would be very helpful if u put options with the questions..
1.
F(n) = n!
now ((p-1)! + 1) is always divisible by p, where p=prime number (dont remember this theorem name)

F(100) + 1 = (101-1)! + 1 shd be divisible by 101(a prime number).

hence answer shd b 101.

2. i guess it is an incomplete question.. couldn't make out nething 4m it..

3. r = kA^2
r= l/B
r= mA^2/B--------->(1)
let new A be a, and new B be b
then
r=ma^2/b
or r=ma^2/2B (since b=2B given)-------->(2)
dividing (1) by (2)
a^2 = 2A^2 or
a= root(2)A=1.414A
hence increase is 41.4%


first, my appologies for not putting up the choices. But i purposefully did that in order to get the explanation.
anyway, regarding first, u r missing the question. It says F(n) is product of all even integers between 2 and n inclusive.
so F(100) +1 = (2x4x6x8x10.....96x98x100)! + 1
= 2(50!) + 1
now p is a smallest prime factor of 2(50!) + 1. how do we find p now? 😞 too bad in number properties...
3919
first, my appologies for not putting up the choices. But i purposefully did that in order to get the explanation.
anyway, regarding first, u r missing the question. It says F(n) is product of all even integers between 2 and n inclusive.
so F(100) +1 = (2x4x6x8x10.....96x98x100)! + 1
= 2(50!) + 1 [ u r completely wrong here]
now p is a smallest prime factor of 2(50!) + 1. how do we find p now? 😞 too bad in number properties...


100! is not equal to 2(50!)..
3920
dare2 Says
100! is not equal to 2(50!)..


oops.. misread the question.. completely overlooked 'even integers' .

wait let me solve it for 2(50!)+1..

it wont b even 2(50!)+1 (damn!! i need some good sleep)..