Veritas Test2 Q34

Hi Abhijit,

Good question!

p = 501 * ... * 597
q = 501 * .. * 597 * 599 * 601 = p * 599 * 601

p = q / (599*601)

We need to find 1/p + 1/q

(599*601)/q + 1/q

599 = 600 - 1 and 601 = 600 + 1

a^2 - b^2 = (a-b)(a+b)

Hence ( 600^2 - 1^2 + 1 )/q

360000/q

Answer D

Abhijit K wrote: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.1/359999q
C.1200/q
D.360000/q
E.359999*q

p = (501)(503)...(595)(597).
q = (501)(503)...(595)(597)(599)(601).
Notice the OVERLAP between p and q.
Implication:
q = (p)(599)(601).

Since the answer choices are IN TERMS OF A VARIABLE, we can PLUG IN any values for p and q such that q = (p)(599)(601).

Let p =1.
Then q = (1)(599)(601) = (600-1)(600+1) = 360000 - 1 = 359999.
Thus:
1/p + 1/q = 1/1 + 1/359999 = 359999/359999 + 1/359999 = 360000/359999. This is our target.

Now plug q = 359999 into the answers to see which yields our target of 360000/359999.
A quick scan of the answers reveals that only D works:
360000/q = 360000/359999.

The correct answer is D.

I had a slightly different approach here.

i concluded that p/q muse be equal to 1/(599)(601) which is equal to 1/(600-1)(600+1)

then p would be equal to Q/(600-1)(600+1)

1/p is the reciprocal of Q/(600-1)(600+1) which is (600-1)(600+1)/Q

Hence 1/p + 1/Q = (600-1)(600+1)/Q + 1/Q which equals to 600^2 / Q --> D

Brent@GMATPrepNow wrote:

Abhijit K wrote: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.1/359999q
C.1200/q
D.360000/q
E.359999*q

p = (501)(503)(505)...(597)
q = (501)(503)(505)...(597)(599)(601)
So, q = (p)(599)(601)

So, 1/p + 1/q = 1/p + 1/(p)(599)(601) [replaced q with (p)(599)(601)]
= (599)(601)/(p)(599)(601) + 1/(p)(599)(601) [found common denominator]
= [(599)(601) + 1]/(p)(599)(601)
= 360,000/(p)(599)(601)
= 360,000/q [since q = (p)(599)(601)]
= D

Cheers,
Brent

P= 501*503...*597
q= 501*503...597*599*601 or P(599*601)=p(359999)

Just plugin the values

1/P+1/q= 1/p+1/p(359999) =359999+1/p(359999)=360000/p(359999) now p(359999)=q answer is 360000/q

Abhijit K wrote: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.1/359999q
C.1200/q
D.360000/q
E.359999*q

We are given that p = the product of the odd integers from 500 to 598, i.e., from 501 to 597 inclusive. We are also given that q = the product of the odd integers from 500 to 602, i.e., 501 to 601 inclusive.

Thus:

q = p(599)(601)

Now we can evaluate 1/p + 1/q as:

1/p + 1/q = (599)(601)/[p(599)(601)] + 1/q = (599)(601)/q + 1/q = [(599)(601) + 1]/q

Notice that (599)(601) = (600 - 1)(600 + 1) = 600^2 - 1. Thus, the numerator (599)(601) + 1 becomes 600^2 - 1 + 1, or simply 600^2. Therefore:

1/p + 1/q = [(599)(601) + 1]/q = 600^2/q = 360,000/q

Answer: D