Last visit was: 23 May 2024, 12:40 It is currently 23 May 2024, 12:40
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
avatar
Director
Director
Joined: 29 Nov 2012
Posts: 579
Own Kudos [?]: 6090 [121]
Given Kudos: 543
Send PM
Most Helpful Reply
Math Expert
Joined: 02 Sep 2009
Posts: 93435
Own Kudos [?]: 626117 [47]
Given Kudos: 81940
Send PM
User avatar
Director
Director
Joined: 02 Sep 2012
Status:Far, far away!
Posts: 859
Own Kudos [?]: 4900 [34]
Given Kudos: 219
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Send PM
GMAT Club Legend
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6817
Own Kudos [?]: 30308 [14]
Given Kudos: 799
Location: Canada
Send PM
Let p = the product of all the odd integers between 500 and [#permalink]
8
Kudos
6
Bookmarks
Expert Reply
Top Contributor
fozzzy 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 \(\frac{1}{p}\)+ \(\frac{1}{q}\)?

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


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)]

Answer:

Originally posted by BrentGMATPrepNow on 30 Aug 2016, 17:03.
Last edited by BrentGMATPrepNow on 11 Jan 2021, 11:28, edited 1 time in total.
General Discussion
User avatar
Intern
Intern
Joined: 15 Feb 2013
Status:Currently Preparing the GMAT
Posts: 25
Own Kudos [?]: 58 [6]
Given Kudos: 11
Location: United States
GMAT 1: 550 Q47 V23
GPA: 3.7
WE:Analyst (Consulting)
Send PM
Re: Let p = the product of all the odd integers between 500 and [#permalink]
5
Kudos
1
Bookmarks
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
avatar
Manager
Manager
Joined: 23 May 2013
Posts: 76
Own Kudos [?]: 255 [0]
Given Kudos: 109
Send PM
Re: Let p = the product of all the odd integers between 500 and [#permalink]
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.
Board of Directors
Joined: 17 Jul 2014
Posts: 2160
Own Kudos [?]: 1180 [0]
Given Kudos: 236
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE:General Management (Transportation)
Send PM
Re: Let p = the product of all the odd integers between 500 and [#permalink]
fozzzy 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 \(\frac{1}{p}\)+ \(\frac{1}{q}\)?

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


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

[q+q(599*601)/599*601] * [599*601/q^2]

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
Manager
Manager
Joined: 16 Jan 2017
Posts: 55
Own Kudos [?]: 16 [0]
Given Kudos: 2
GMAT 1: 620 Q46 V29
Send PM
Re: Let p = the product of all the odd integers between 500 and [#permalink]
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
Math Expert
Joined: 02 Sep 2009
Posts: 93435
Own Kudos [?]: 626117 [1]
Given Kudos: 81940
Send PM
Re: Let p = the product of all the odd integers between 500 and [#permalink]
1
Bookmarks
Expert Reply
vmelgargalan wrote:
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


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.
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 18899
Own Kudos [?]: 22304 [2]
Given Kudos: 285
Location: United States (CA)
Send PM
Re: Let p = the product of all the odd integers between 500 and [#permalink]
2
Bookmarks
Expert Reply
fozzzy 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 \(\frac{1}{p}\)+ \(\frac{1}{q}\)?

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



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
Director
Director
Joined: 12 Nov 2016
Posts: 569
Own Kudos [?]: 118 [0]
Given Kudos: 167
Location: United States
Schools: Yale '18
GMAT 1: 650 Q43 V37
GRE 1: Q157 V158
GPA: 2.66
Send PM
Re: Let p = the product of all the odd integers between 500 and [#permalink]
fozzzy 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 \(\frac{1}{p}\)+ \(\frac{1}{q}\)?

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


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.
Retired Moderator
Joined: 19 Mar 2014
Posts: 817
Own Kudos [?]: 970 [2]
Given Kudos: 199
Location: India
Concentration: Finance, Entrepreneurship
GPA: 3.5
Send PM
Re: Let p = the product of all the odd integers between 500 and [#permalink]
1
Kudos
1
Bookmarks
fozzzy 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 \(\frac{1}{p}\)+ \(\frac{1}{q}\)?

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



\(p = 501 * 503 * 505 * ............ * 597\)

\(q = 501 * 503 * 505 * ......................* 599 * 601\)

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

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

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

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

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

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

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

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

\(= \frac{360,000}{q}\)

Hence, Answer is D
Director
Director
Joined: 09 Jan 2020
Posts: 960
Own Kudos [?]: 228 [0]
Given Kudos: 434
Location: United States
Send PM
Let p = the product of all the odd integers between 500 and [#permalink]
fozzzy 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 \(\frac{1}{p}\)+ \(\frac{1}{q}\)?

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


p = 501*503*505*..597

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

q = \(p*599*601\)

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

\(q = (360,000 - 1)p\)

\(q = 359,999p\)

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

\(\frac{1}{p} + \frac{1}{q} = \frac{359,999}{q} + \frac{1}{q} = \frac{360,000}{p}\)
Senior Manager
Senior Manager
Joined: 19 Oct 2014
Posts: 390
Own Kudos [?]: 328 [0]
Given Kudos: 188
Location: United Arab Emirates
Send PM
Let p = the product of all the odd integers between 500 and [#permalink]
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
Intern
Intern
Joined: 05 Sep 2022
Posts: 12
Own Kudos [?]: 7 [0]
Given Kudos: 1
Location: India
Schools: Wharton '25
Send PM
Re: Let p = the product of all the odd integers between 500 and [#permalink]
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
User avatar
Non-Human User
Joined: 09 Sep 2013
Posts: 33151
Own Kudos [?]: 829 [0]
Given Kudos: 0
Send PM
Re: Let p = the product of all the odd integers between 500 and [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
Re: Let p = the product of all the odd integers between 500 and [#permalink]
Moderator:
Math Expert
93435 posts