November 18, 2018 November 18, 2018 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score. November 18th, 7 AM PST November 20, 2018 November 20, 2018 09:00 AM PST 10:00 AM PST The reward for signing up with the registration form and attending the chat is: 6 free examPAL quizzes to practice your new skills after the chat.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 50623

If n and p are positive integers, what is the ratio of n to p?
[#permalink]
Show Tags
25 Jan 2018, 22:34
Question Stats:
52% (01:54) correct 48% (01:56) wrong based on 124 sessions
HideShow timer Statistics



Senior Manager
Joined: 17 Oct 2016
Posts: 321
Location: India
Concentration: Operations, Strategy
GPA: 3.73
WE: Design (Real Estate)

Re: If n and p are positive integers, what is the ratio of n to p?
[#permalink]
Show Tags
25 Jan 2018, 22:44
C Not much info from st1. From st2 we can say (p+n)(pn)=28=14*2=7*4. Not sufficient Combining we can say only 14*2 case fits as solving for p and n from st2 gives p=8 and p=6. Sufficient Sent from my iPhone using GMAT Club Forum
_________________
Help with kudos if u found the post useful. Thanks



Senior Manager
Joined: 15 Oct 2017
Posts: 311

Re: If n and p are positive integers, what is the ratio of n to p?
[#permalink]
Show Tags
25 Jan 2018, 22:49
IMO B
1) From 1) possible values of p & n=12*4, 24*2, 48*1, 6*8, 3*16 along with ve or +ve values or all possible combinations therefore Not Sufficient.
2) From 2) possible values of (p+n)(pn)=28 are p=/+8, n=/+6, therefore only value possible for the ratio p/n=8/6=4/3. Sufficient.



DS Forum Moderator
Joined: 21 Aug 2013
Posts: 1365
Location: India

Re: If n and p are positive integers, what is the ratio of n to p?
[#permalink]
Show Tags
25 Jan 2018, 23:42
urvashis09 wrote: IMO B
1) From 1) possible values of p & n=12*4, 24*2, 48*1, 6*8, 3*16 along with ve or +ve values or all possible combinations therefore Not Sufficient.
2) From 2) possible values of (p+n)(pn)=28 are p=/+8, n=/+6, therefore only value possible for the ratio p/n=8/6=4/3. Sufficient. Hi I also think the answer is B. Although we cannot even assume negative values because its given that both n and p are positive integers.



Senior Manager
Joined: 15 Oct 2017
Posts: 311

Re: If n and p are positive integers, what is the ratio of n to p?
[#permalink]
Show Tags
26 Jan 2018, 00:36
amanvermagmat wrote: urvashis09 wrote: IMO B
1) From 1) possible values of p & n=12*4, 24*2, 48*1, 6*8, 3*16 along with ve or +ve values or all possible combinations therefore Not Sufficient.
2) From 2) possible values of (p+n)(pn)=28 are p=/+8, n=/+6, therefore only value possible for the ratio p/n=8/6=4/3. Sufficient. Hi I also think the answer is B. Although we cannot even assume negative values because its given that both n and p are positive integers. Oh yes, I missed that part! Thank you for correcting!



Intern
Joined: 27 Sep 2017
Posts: 10

Re: If n and p are positive integers, what is the ratio of n to p?
[#permalink]
Show Tags
02 Feb 2018, 04:08
urvashis09 wrote: IMO B
1) From 1) possible values of p & n=12*4, 24*2, 48*1, 6*8, 3*16 along with ve or +ve values or all possible combinations therefore Not Sufficient.
2) From 2) possible values of (p+n)(pn)=28 are p=/+8, n=/+6, therefore only value possible for the ratio p/n=8/6=4/3. Sufficient. Hello urvashis09, How did you reach that; 2 possible values of (p+n)(pn)=28 are p=/+8, n=/+6. Is there any way faster than that of substituting real numbers?



Senior Manager
Joined: 15 Oct 2017
Posts: 311

Re: If n and p are positive integers, what is the ratio of n to p?
[#permalink]
Show Tags
02 Feb 2018, 11:42
ahmed.abumera wrote: urvashis09 wrote: IMO B
1) From 1) possible values of p & n=12*4, 24*2, 48*1, 6*8, 3*16 along with ve or +ve values or all possible combinations therefore Not Sufficient.
2) From 2) possible values of (p+n)(pn)=28 are p=/+8, n=/+6, therefore only value possible for the ratio p/n=8/6=4/3. Sufficient. Hello urvashis09, How did you reach that; 2 possible values of (p+n)(pn)=28 are p=/+8, n=/+6. Is there any way faster than that of substituting real numbers? Hi, I used the substitution method only to reach the possible solutions of 8+6 & 86. Also, I could not find any other possible values fitting here anyway since 28 can be factorised into 7*2*2 and to fit the equation here of (p+q)(pq) the only possible combinations can be 28*1 (not possible to be deduced to our required form) and 14*2 (fits our required form).



DS Forum Moderator
Joined: 21 Aug 2013
Posts: 1365
Location: India

Re: If n and p are positive integers, what is the ratio of n to p?
[#permalink]
Show Tags
03 Feb 2018, 00:45
ahmed.abumera wrote: urvashis09 wrote: IMO B
1) From 1) possible values of p & n=12*4, 24*2, 48*1, 6*8, 3*16 along with ve or +ve values or all possible combinations therefore Not Sufficient.
2) From 2) possible values of (p+n)(pn)=28 are p=/+8, n=/+6, therefore only value possible for the ratio p/n=8/6=4/3. Sufficient. Hello urvashis09, How did you reach that; 2 possible values of (p+n)(pn)=28 are p=/+8, n=/+6. Is there any way faster than that of substituting real numbers? Hi Urvashi has already explained. But I will also try to explain. n&p are positive integers, so (p+n) and (pn) will also be integers. Now p+n will be always positive, and so (pn) also must be positive in order for the product of (p+n) and (pn) to be positive, which is 28. (p+n)(pn) = 28 Now 28 can be product of 28*1 or 14*2 or 7*4. lets take each case one by one. p+n = 28 pn = 1 Here if we add the two equations, we get 2p = 29 or p = 14.5, which is not possible since p should be an integer. Case rejected. p+n = 14 pn = 2 Add the two equations, we get 2p = 16 or p = 8. This gives n = 6. Both are integers, so this case is possible. p+n = 7 pn = 4 Add the two equations, we get 2p = 11 or p = 5.5, which is not possible since p should be an integer. Case rejected. So the only case possible is where p=8 and n=6. This statement is thus sufficient.



Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 611

Re: If n and p are positive integers, what is the ratio of n to p?
[#permalink]
Show Tags
03 Feb 2018, 05:12
Bunuel wrote: If n and p are positive integers, what is the ratio of n to p? (1) np = 48 (2) p^2 – n^2 = 28 We need to find out the value of \(\frac{n}{p}\), given \(n>0, p>0\) Fact Statement 1np = 48. For n=1,p=48, the ratio of \(\frac{n}{p} = \frac{1}{48}\) Again, for n=48,p=1, the ratio of \(\frac{n}{p} = \frac{48}{1}\) As we get different values for the fraction, this statement is InsufficientFact Statement 2 \(p^2  n^2 = 28\) Now, as p and n are both positive integers, we can infer that p>n. On simplifying the given expression, we get (p+n)*(pn) = 28 If we were to treat (p+n) and (pn) as two distinct integers, and factorize 28 as a product of two unique integers too, we would get the following pairs (28,1) , (1,28) , (14,2) , (2,14) , (7,4) , (4,7) Thus, (p+n)*(pn) = (28)*(1) OR (p+n)*(pn) = (14)*(2) and so on. Notice that (p+n)*(pn) = (28)*(1) means i can also say (p+n) = 12 AND (pn) = 1 > Solving for p > 2p = 13. However, this is against what was told about p (p is a positive integer). Based on this logic, its easy to see that all pairs above, all get eliminated except (14,2) or (2,14)*** The value of p comes out to be 2p = 16 > p =8, n = 6. We can clearly get the ratio of \(\frac{n}{p}\). Sufficient. ***Note that its easy to ascertain that the pair (2,14) is also not possible, and (14,2) is the only combination that stands.B.
_________________
All that is equal and notDeep Dive Inequality
Hit and Trial for Integral Solutions




Re: If n and p are positive integers, what is the ratio of n to p? &nbs
[#permalink]
03 Feb 2018, 05:12






