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each pair has a different sum

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each pair has a different sum

by sanju09 » Mon Mar 07, 2011 2:40 am
The four numbers p < q < r < s can be paired in six different ways. If each pair has a different sum, then what is the value of s?

(1) The four smallest sums are 1, 2, 3, and 4.

(2) The sum of q and r is 3.



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by Anurag@Gurome » Mon Mar 07, 2011 4:00 am
sanju09 wrote:The four numbers p < q < r < s can be paired in six different ways. If each pair has a different sum, then what is the value of s?

(1) The four smallest sums are 1, 2, 3, and 4.
(2) The sum of q and r is 3.
Statement 1: The four smallest sums are 1, 2, 3, and 4
The smallest sum = (p + q) = 1
2nd smallest sum = (p + r) = 2

Next two smallest sums are (p + s) and (q + r).
But we cannot compare between them.
Hence, we cannot assign any values to them and thus we cannot solve for s.

Not sufficient

Statement 2: The sum of q and r is 3
No information related to s.

Not sufficient

1 & 2 Together: Now we have (p + q) = 1, (p + r) = 2, and (q + r) = 3.
Hence (p + s) must be equal to 4.
Solving these equations we have s = 4

Sufficient

The correct answer is C.
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by dharan » Mon Mar 07, 2011 10:46 am
Let's take A alone: Plug in some number which gives different sum

p = 0, q = 1, r = 2, s = 4,

p + q = 1
p + r = 2
p + s = 4
q + r = 3
q + s = 5
r + s = 6

So four smallest sums are 1,2,3 and 4

Then the s value is 4

Let's take B Alone :

q + r is 3, then s can be 4, 5, 6, ....[ Multiple values ]
So B Alone is rejected.

So the answer is A.

Am I doing something wrong ?
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by sanju09 » Mon Mar 07, 2011 11:32 pm
dharan wrote:Let's take A alone: Plug in some number which gives different sum

p = 0, q = 1, r = 2, s = 4,

p + q = 1
p + r = 2
p + s = 4
q + r = 3
q + s = 5
r + s = 6

So four smallest sums are 1,2,3 and 4

Then the s value is 4

Let's take B Alone :

q + r is 3, then s can be 4, 5, 6, ....[ Multiple values ]
So B Alone is rejected.

So the answer is A.

Am I doing something wrong ?
By taking arbitrary numbers like those, you are making (p + s) and (q + r) comparable, which are definitely incomparable in the absence of further info. Anurag's take is excellent and [spoiler]C[/spoiler] is the right answer.
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by kareem.firoz » Tue Mar 08, 2011 2:01 am
@Anurag,
Can't it be this way?

Statement 1: The four smallest sums are 1, 2, 3, and 4
The smallest sum = (p + q) = 1
2nd smallest sum = (p + r) = 2

Next two smallest sums are (p + s) and (q + r). So, there are two possibilities, as given below.
(i) (p+s) = 3 & (q+r) =4 OR
(ii) (q+r) = 3 & (p+s)=4
Consider Possibility (i); We get p= -(1/2) on solving this which means p is not an integer. Hence (i) wrong. We go for 2.
On solving Possibility (ii); we get p=0 which leads to q=1, r=2 and s=4

Hence Statement 1 is SUFFICIENT. Answer is A.

Thanks,
FK
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by sanju09 » Tue Mar 08, 2011 5:30 am
kareem.firoz wrote:@Anurag,
Can't it be this way?

Statement 1: The four smallest sums are 1, 2, 3, and 4
The smallest sum = (p + q) = 1
2nd smallest sum = (p + r) = 2

Next two smallest sums are (p + s) and (q + r). So, there are two possibilities, as given below.
(i) (p+s) = 3 & (q+r) =4 OR
(ii) (q+r) = 3 & (p+s)=4
Consider Possibility (i); We get p= -(1/2) on solving this which means p is not an integer. Hence (i) wrong. We go for 2.
On solving Possibility (ii); we get p=0 which leads to q=1, r=2 and s=4

Hence Statement 1 is SUFFICIENT. Answer is A.

Thanks,
FK
Where did the question say that p must be an integer?
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by santosh_surathkal » Mon Mar 14, 2011 7:33 am
Dear Anurag,
This is in reference to the below math question

https://www.beatthegmat.com/each-pair-ha ... 77786.html
The smallest sum = (p + q) = 1
2nd smallest sum = (p + r) = 2
This means P is either (-)Ve ....
Next two smallest sums are (p + s) and (q + r).
In this case
if we assume P is -1 then
P = -1
Q = 2,
R = 3,
S = 4
we can get the sum as 1, 2, 3 but we do not get 4.
Now if we assume P = +1
Then Q = 2
R = 3
S = 4
We get all the sum mentioned in Option 1.
I do not see any other alternative value of P which satisfies option 1.
I feel A should be the answer.
Could you please help me?
Best Regards,
Santosh.
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