Let p = the product of all the odd integers between 500 and

Question

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  \frac{1}{p} +  \frac{1}{q} ?

A.  \frac{1}{600q} 
B.  \frac{1}{359,999q} 
C.  \frac{1,200}{q} 
D.  \frac{360,000}{q} 
E.  359,999q

Bunuel · Accepted Answer

q=p*599*601=p(600-1)(600+1)=p*(360,000-1)=359,999p  -->  p=\frac{q}{359,999} .

\frac{1}{p} + \frac{1}{q}=\frac{359,999}{q}+\frac{1}{q}=\frac{3600,000}{q} .

Answer: D.

Zarrolou · Answer

p=501*503*...*597 
 q=501*503*...*597*599*601

\frac{1}{p}+\frac{1}{q}=\frac{p+q}{pq}=\frac{p(1+599*601)}{pq}

\frac{1+599*601}{q}=\frac{360000}{q}

BrentGMATPrepNow · Answer

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)   (p) (599)(601)] 
= (599)(601) /  (p) (599)(601) + 1 /  (p) (599)(601)    
=   /  (p) (599)(601)
= 360,000 /  (p) (599)(601)
= 360,000 / q   (p) (599)(601)]

Answer:  D

Virgilius · Answer

Let's formalise these expressions a bit :

p = the product of all the odd integers between 500 and 598

Meaning that \(p = 501*503*...*597\) (1)

Like wise for \(q\) being the product of all the odd integers between 500 and 602, we get, using (1) :

\(q = 501*503*...597*599*601 = p*599*601\) (2)

Since we are looking to express \(\frac{1}{p} + \frac{1}{q}\) in terms of \(q\), we get from (2) : \(p = \frac{q}{(599*601)}\)

So then :

\(\frac{1}{p} + \frac{1}{q}\) \(= \frac{1}{q/(599*601)} + \frac{1}{q} = \frac{1}{q} * (599*601 +1)\)

Since\(599 = 600 - 1\) then \(599*601 + 1 = (600 - 1)*601 + 1 = 360600 - 601 + 1 = 360599 + 1 = 360000\)

Which yields \(\frac{1}{p} + \frac{1}{q}\) = \(\frac{360000}{q}\)

Which is answer choice D.

Hope that helped :-D

ankur1901 · Answer

i think we can first simplify, expression 1/p +1/q to 1/q (q/p + 1). This way its easier to visualize that q/p will be only 599*601.

mvictor · Answer

oh wow..good question..requires some outside the box thinking...
p=q/599*601

1/p + 1/q = p+q/pq

first thing:
p+q
q/599*601 + q = q+q(599*601)/599*601

pq = q^2/599*601

now
 
  *

we can simplify by 599*601
we get q+q(599*601)/q^2
we can factor out q in the numerator = q(1+599*601)/q^2
divide both sides by q
1+599*601/q
599*601 = (600-1)(600+1) = 359,999
we add one and get 360,000
now...final step
360,000/q

answer is D

vmelgargalan · Answer

I am struggling a bit with this question. I do not understand how 359,999pq=p∗599∗601=p(600−1)(600+1)=p∗(360,000−1)=359,999p --> p=q359,999p=q359,999 .

This type of questions are just a big confusing in general. Any advice on where to revise this type of questions

Bunuel · Answer

We applied there (a-b)(a+b) = a^2 - b^2 , thus (600-1)(600+1)=600^2 - 1^2=(360,000-1) . Theory on Algebra: https://gmatclub.com/forum/algebra-101576.html Algebra - Tips and hints: https://gmatclub.com/forum/algebra-tips- ... 75003.html DS Algebra Questions to practice: https://gmatclub.com/forum/search.php?se ... &tag_id=29 PS Algebra Questions to practice: https://gmatclub.com/forum/search.php?se ... &tag_id=50 Hope it helps.

ScottTargetTestPrep · Answer

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

Answer: D

Nunuboy1994 · Answer

We can solve this question using algebra:

p =  (501)(503)...(595)(597) . 
q =  (501)(503)...(595)(597) (599)(601). 
The overlap between P and Q implies that 
q = (p)(599)(601)

We could do this with another set of numbers (p is the odd integers between 2 and 8, q is the odd integers between 2 and 12)
p= 3 x 5 x 7
q=3 x 5 x 7 x 9 x 11

105= p (11)(9)

Anyways

The answer choices are in terms of a variable so are result must be in the form of P+q/pq

P =1. 
Q= (1)(599)(601) = (600-1)(600+1) = 360000 - 1 = 359999.

Therefore

1/p + 1/q = 1/1 + 1/359999 = 359999/359999 + 1/359999 = 360000/359999 = p + q/ q=

Plug in q = 359999 into the answers to see which equals 360000/359999.

360000/q = 360000/359999.

Thus D.

ydmuley · Answer

p = 501 * 503 * 505 * ............ * 597

q = 501 * 503 * 505 * ......................* 599 * 601

p = \frac{q}{599 * 601}

\frac{1}{p} + \frac{1}{q}

= \frac{1}{q/599 * 601} + \frac{1}{q}

= \frac{599 * 601}{q} + \frac{1}{q}

= \frac{599*601 + 1}{q}

= \frac{(600 - 1) (600 + 1) + 1}{q}

= \frac{((600)^2 + 600 - 600 - 1) + 1}{q}

= \frac{(600)^2}{q}

= \frac{360,000}{q}

Hence, Answer is D

Basshead · Answer

p = 501*503*505*..597

q = 501*503*505*..597  *599*601

q =  p*599*601

q = p(600-1)(600+1)

q = (360,000 - 1)p

q = 359,999p

p = \frac{q}{359,999}

\frac{1}{p} + \frac{1}{q} = \frac{359,999}{q} + \frac{1}{q} = \frac{360,000}{p}

rvgmat12 · Answer

OE:

Since p=(501)(503)...(595)(597) and q=(501)(503)...(595)(597)(599)(601), we can equally well express p as q/(599)(601)
 
So 1/p+1/q can be expressed as 1/q/(599)(601)+1/q, which cleans up to (599)(601)/q+1/q, which then further condenses to (599)(601)+1/q. Recognizing (599)(601) as the factored form of the difference of squares (600–1)(600+1) gives you a pass out of the long multiplication and enables you to shortcut this product to 600^2–1
So your numerator becomes 600^2–1+1, which is simply 600^2, or 360,000
The fraction, then, ultimately becomes 360,000/q

Ashu9876 · Answer

Sol: 
Given that, 
 p=501*503*…………………*597
q=501*503*…………………*597*599*601=p *599*601=p*359999
Hence, p= q/359999 and 1/p=359999/q
To find, 1/p+1/q= 359999/q+1/q= (359999+1 )/q=360000/q

luisdicampo · Answer

Deconstructing the Question  
Let  p  be the product of all odd integers from 501 to 597. 
Let  q  be the product of all odd integers from 501 to 601. 
So  q  includes everything in  p  plus the extra factors  599  and  601 :
 q=p\cdot 599\cdot 601 .

We want  \frac{1}{p}+\frac{1}{q}  in terms of  q .

Step-by-step  
From  q=p\cdot 599\cdot 601 :
 p=\frac{q}{599\cdot 601}

Invert:
 \frac{1}{p}=\frac{599\cdot 601}{q}

Add:
 \frac{1}{p}+\frac{1}{q}=\frac{599\cdot 601}{q}+\frac{1}{q}
=\frac{599\cdot 601+1}{q}

Compute:
 599\cdot 601=(600-1)(600+1)=600^2-1=360000-1=359999

So:
 599\cdot 601+1=360000

Therefore:
 \frac{1}{p}+\frac{1}{q}=\frac{360000}{q}

Answer: D

Let p = the product of all the odd integers between 500 and

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