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Veritas Prep Challenge Question

by Ashley@VeritasPrep » Thu Jul 07, 2011 7:10 pm
As advertised on the main page of Beat the GMAT, Brian, David and I will be posting some original questions today and tomorrow with prizes for the first five correct responders who show their work!

Good Luck!


Let p = the product of all the odd integers between 500 and 598, and let q = the product of all the odd integers between 500 and 602. In terms of q, what is the value of 1/p + 1/q ?

(A) 1/(600q)
(B) 1/(359,999q)
(C) 1,200/q
(D) 360,000/q
(E) 359,999q
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by swetha2 » Thu Jul 07, 2011 7:19 pm
Let p = the product of all the odd integers between 500 and 598, and let q = the product of all the odd integers between 500 and 602. In terms of q, what is the value of 1/p + 1/q ?

(A) 1/(600q)
(B) 1/(359,999q)
(C) 1,200/q
(D) 360,000/q
(E) 359,999q

Answer: D 360,000/q

q=p*599*601= 359999p
therefore, p=q/359,999

1/p + 1/q = 1/(q/359,999) + 1/q

= 359,999/q +1/q = 360,000/q
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by krishnasty » Thu Jul 07, 2011 7:30 pm
p = 501*503*..........597
q = 501*503*..........597*599*601
hence,
q = p * 599*601
p = q/(599*601)

1/p + 1/p = (599*601)/q + 1/q =
360000/q

Ans: D
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by anandc19 » Thu Jul 07, 2011 7:32 pm
p=501 X 503 ....X 597
q=501 X 503 ....X 597 X 599 X 601

1/P + 1/q =
{(599 X 601) +1 }/ Q= 360000/Q

So option D
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by soumava » Thu Jul 07, 2011 7:43 pm
p = 501*503*..........597
q = 501*503*..........597*599*601
hence,
q = p * 599*601
p = q/(599*601)

1/p + 1/p = (599*601)/q + 1/q =
360000/q

Hence the answer is D
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by nav!n » Thu Jul 07, 2011 7:53 pm
p= 501x503...x597
q=501x503....x599x601

p =q/(599x601)


1/p +1/q = [(599x601)+1]/q = 360000/q

D is correct.
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by srini1988 » Thu Jul 07, 2011 10:09 pm
p=501*...*597

q=p*599*601 => p=q/(599*601)

therefore (599*601)/q+1/q = 360000/q

D is the answer..

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by Nemesis09 » Fri Jul 08, 2011 12:12 am
Answer: (D)

Let p = the product of all the odd integers between 500 and 598, and let q = the product of all the odd integers between 500 and 602. In terms of q, what is the value of 1/p + 1/q ?

(A) 1/(600q)
(B) 1/(359,999q)
(C) 1,200/q
(D) 360,000/q
(E) 359,999q

Solution:
p=501*502*503*..... ...*597
q=501*502*503*..... ...*597*599*601

--> q=p*599*601
--> p=q/(599*601)

(1/p) + (1/q)= (599*601)/q + 1/q
= (359999 + 1)/q
= 360,000/q -- Answer(D)
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by eubschoolasp » Fri Jul 08, 2011 2:51 am
Solution:
Since, p=501*503*505*....*597 (product of all odd integers between 500 and 598)
and q=501*503*505*.....*597*599*601 (product of all odd integers between 500 and 602),

We may re-write q= p*599*601
or,q= p*359999
or,p= q/359999

Hence, 1/p + 1/q= 359999/q + 1/q
= (359999+1)/q
= 360,000/q
Choose (D).
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