If w + z = 28, what is the value of wz?

Hi GMAThirst, let's examine these statements one at a time:

1. w and z are positive integers.
We know that both must be positive, so this eliminates 30 and -2, but it could be 1 and 27, 2 and 26, 3 and 25, etc. All of which have different products (1x27=27, 2x26=52, etc) A is insufficient on its own

2. w and z are consecutive odd integers
This would have to be 13 and 15. No other positive consecutive odd integers would add up to 28. Statement A makes us wonder about negative numbers, what about -13 and -15? Well they would add up to -28, which isn't the same as +28, so they're out as possibilities. What about other negative consecutive odd integers? Nothing is ever going to give you anything other than a negative number, so no negative numbers could work here. Statement 2 is thus sufficient on its own, with no need to bring in statement #1. (w*z must be either 13*15 or 15*13, either way giving the answer of 195 by commutativity)

Answer: B.

Hope this helps!
-Ron

I understand why statment 2 alone is sufficent in this example, but what if we were told that "w + z = \(|28|\)" would that mean the answer is C as statment 2 in itself is insufficent (w and z could be -15, -13 or 13, 15) but with statment 1 we can solve the stem (since only 13 and 15 are left).


Have I understood it correctly?

No. |28|=28, so w + z = |28| is the same as w + z = 28.

Guess, you meant: |w + z| = 28. Now, if the question were: "|w + z| = 28, what is the value of wz?", the correct answer would still be B. From |w + z| = 28 and w and z are consecutive odd integers, it follows that the integers can be 13 and 15 OR -13 and -15. In any case wz = 195.

Hope it's clear.

Hello...
Statement A : not suff. If w,z were positive is the only known condition, then there are numerous values for w and z such that w+ z =28 ; eg. (1,27); (3,25), (2,26) etc.
Statement B:
w and z are consecutive odd integers so,
lets take w=2k-1 and z=2k+1
Thus w+z= 4k= 28 and wz=4k^2 - 1
Two possible sets for w,z i.e (15,13) or (13,15) both have the same product ---> Sufficient

Thus B

Statement 1) w and z are +ve integers. w can be 1 and z can be 27 or w can be 2 and z can be 26. so in each case wz changes. Hence Not sufficient.

Statement 2) As w and z are consecutive odd integers, then w can be either 13 or 15 and z can be 15 or 13. In either case wz is same same.
Hence Sufficient.

Hence Option B)

Option 1 will lead to multiple answers for ex: W =27,z=1 then WZ=27 or W=15,Z=13 then WZ= 195. Answer will not Statement A

Option 2 :

Given W and Z are consecutive odd interger then W and Z can be written as W= 2n+1 and Z=2n+3 OR Z= 2n+1 and W=2n+3 (by using definition of Odd number)
Also from problem statement W+ Z = 28 => 2n+1 + 2n+3 = 28
Solving for n , we get n = 6
W= (2*6)+1=13
Z=(2*6)+3 =15


OR

Z= (2*6)+1=13
W=(2*6)+3 =15

(1) Option 1 can lead to many possibilities: ex., w=1 & z=27 or w=13 & z=15...
Insufficient;
(2) Since w and z are consecutive odd integers, we can find out quite easily that the 2 possible numbers should be -1 and +1 of the middle number(14).

Alternative approach, especially when the sum is big, is by assuming (2n+1) as the 1st consecutive odd integer and (2n+3) as the 2nd consecutive odd integer.

2n+1+2n+3=28;
4n=24;
=> n=6;

The 1st consecutive odd integer is 2(6)+1 = 13;
The 2nd consecutive odd integer is 2(6)+3 = 15;

Ans is (B).

I thought that you would need to know that both are not negative to ensure that they add up to 28.. thoughts?

There is no other solution exist for w + z = 28 and w and z are consecutive odd integers, other than 13 and 15.

Solution:

Question Stem Analysis:

We need to determine the value of wz, given that the sum of w and z is 28.

Statement One Alone:

Statement one alone is not sufficient. For example, if w = 1 and z = 27, then wz = 27. However, if w = 14 and z = 14, then wz = 196.

Statement Two Alone:

Statement two alone is sufficient since the only consecutive odd integers that add up to 28 are 13 and 15. Therefore, regardless of whether w = 13 and z = 15 OR w = 15 and z = 13, we have wz = 195.

Answer: B

w + z = 28 can be satisfied by infinite combinations of w and z.

From statement I alone, w and z are positive integers.

If w = 1 and z = 27, wz = 27.
If w = 2 and z = 26, wz = 52.
Statement I alone is insufficient. Answer options A and D can be eliminated.

From statement II alone, w and z are consecutive odd integers.

For the purists, if w = 2k + 1, then z = 2k + 3. Therefore, 2k + 1 + 2k + 3 = 28.
Simplifying, 4k = 24 or 2k = 12. Therefore, the numbers are 13 and 15.

Even otherwise, it's not hard to estimate that the only two consecutive odd numbers that add up to 28 are 13 and 15.

Note that multiplication is commutative and hence A * B = B * A. Therefore, regardless of the order, the product of w and z will always be 195.

Statement II alone is sufficient. Answer options C and E can be eliminated.

The correct answer option is B.

Hope that helps!
Aravind BT

Statement 1) w and z are +ve integers. w can be 1 and z can be 27 or w can be 2 and z can be 26. so in each case wz changes. Hence Not sufficient.

Statement 2) As w and z are consecutive odd integers, then w can be either 13 or 15 and z can be 15 or 13. In either case wz is same same.
Hence Sufficient.

Statement 1) w and z are +ve integers. w can be 1 and z can be 27 or w can be 2 and z can be 26. so in each case wz changes. Hence Not sufficient.

Statement 2) As w and z are consecutive odd integers, then w can be either 13 or 15 and z can be 15 or 13. In either case wz is same same.
Hence Sufficient.

Statement One Alone:

Statement one alone is not sufficient. For example, if w = 1 and z = 27, then wz = 27. However, if w = 14 and z = 14, then wz = 196.

Statement Two Alone:

Statement two alone is sufficient since the only consecutive odd integers that add up to 28 are 13 and 15. Therefore, regardless of whether w = 13 and z = 15 OR w = 15 and z = 13, we have wz = 195.

Answer: B

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