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

Question

If n and p are positive integers, what is the ratio of n to p?
 
(1) np = 48 
(2) p^2 – n^2 = 28

Sasindran · Answer

C

Not much info from st1.

From st2 we can say (p+n)(p-n)=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

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US09 · Answer

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)(p-n)=28 are p=-/+8, n=-/+6, therefore only value possible for the ratio p/n=8/6=4/3. Sufficient.

amanvermagmat · Answer

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.

US09 · Answer

Oh yes, I missed that part! Thank you for correcting!

ahmed.abumera · Answer

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?

US09 · Answer

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).

amanvermagmat · Answer

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 .

mau5 · Answer

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

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