If n and p are positive integers, what is the ratio of n to p?

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!

Hello urvashis09,

How did you reach that; 2 possible values of (p+n)(p-n)=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 & 8-6. 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)(p-q) the only possible combinations can be 28*1 (not possible to be deduced to our required form) and 14*2 (fits our required form).

Hi

Urvashi has already explained. But I will also try to explain.
n&p are positive integers, so (p+n) and (p-n) will also be integers. Now p+n will be always positive, and so (p-n) also must be positive in order for the product of (p+n) and (p-n) to be positive, which is 28.

(p+n)(p-n) = 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
p-n = 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
p-n = 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
p-n = 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.

We need to find out the value of \(\frac{n}{p}\), given \(n>0, p>0\)

Fact Statement 1

np = 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 Insufficient


Fact 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)*(p-n) = 28

If we were to treat (p+n) and (p-n) 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)*(p-n) = (28)*(1)

OR

(p+n)*(p-n) = (14)*(2) and so on.

Notice that (p+n)*(p-n) = (28)*(1) means i can also say

(p+n) = 12 AND (p-n) = 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.

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