w, x, y, and z are consecutive odd integers such that w

Question

w, x, y, and z are consecutive odd integers such that w < x < y < z. Which of the following statements must be true?

I. wxyz is odd
II. w + x + y + z is odd
III. w + z = x + y

(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) I, II, and III

GMATinsight · Answer

consecutive odd integers such that w < x < y < z

let, w = 1, x = 3, y = 5, z = 7

I. wxyz is odd TRUE because all numbers ar eodd

II. w + x + y + z is odd FALSE Sum of 4 odd integers = even

III. w + z = x + y, i.e. 1+7 = 3+5 TRUE

Answer: Option D

Kinshook · Answer

Given : w, x, y, and z are consecutive odd integers such that w < x < y < z. 
Asked: Which of the following statements must be true?

z = y + 2 = x + 4 = w + 6

I. wxyz is odd
w (w + 2) (w + 4) (w + 6) -> odd * odd * odd *odd = odd : YES
II. w + x + y + z is odd
odd + odd + odd + odd = even; NO
III. w + z = x + y
w + w + 6 = w + 2 + w + 4 = 2w + 6; YES

IMO D

SakshiSingh02 · Answer

1. Multiplication of odd numbers is odd. So it is true.
2. Addition of odd numbers is even. So it is false.
3.Let w=1, x=3, y=5, z=7
So, w+z = x+y
 1+7 = 3+5
 8 = 8
Sim, let w=11, x= 13, y=15, z=17
So, 11+17 = 13+15
 28 = 28
So, it is true.
D is the answer.

Posted from my mobile device

SakshiSingh02 · Answer

1. Multiplication of odd numbers is odd. So it is true.
2. Addition of odd numbers is even. So it is false.
3.Let w=1, x=3, y=5, z=7
So, w+z = x+y
 1+7 = 3+5
 8 = 8
Sim, let w=11, x= 13, y=15, z=17
So, 11+17 = 13+15
 28 = 28
So, it is true.
D is the answer.

Posted from my mobile device

rajatchopra1994 · Answer

Explanation:
Given w,x,y,z are consecutive odd & w < x < y < z

I. wxyz is odd
odd multiply with gives always odd
Right
II. w + x + y + z is odd
No, Sum of 4 odd number will be even
III. w + z = x + y
yes, as in consecutive numbers; 1st+4th = 2nd + 3rd

IMO-D

Neha2404 · Answer

IMO-D

Given w,x,y,z are consecutive odd number
W<x<y<z

1wxyz is odd

Odd *odd=odd
Correct

2-w+x+y+z = odd
No, sum of even no of odd integer is even

3-w+z =x+y
If w=n
x=n+2
y=n+4
z=N+6

Then. 
w+z =x+y
 2n+6=2n+6
Correct

Answer is D

Posted from my mobile device

vipulshahi · Answer

Given: w,x,y,z = consec. odd int
w<x<y<z

(1) w*x*y*z = odd
all are odd then multiplication is also odd. True

(2) w+x+y+z = odd
sum of two odd is always even. False

(3) w+z = x+y
sum of two odd is always even. True

(1) & (3) satisfy the condition.

Answer : D

ScottTargetTestPrep · Answer

Solution:
 
Since w, x, y, and z are all odd, the product of four odd integers is odd. So I is true. However, the sum of four odd integers is even. So II is not true.

Since they are also consecutive odd integers, x = w + 2, y = w + 4, and z = w + 6. Therefore, in terms of w, we see that w + z = w + w + 6 = 2w + 6 and x + y = w + 2 + w + 4 = 2w + 6. Thus, w + z = x + y. So III is also true.

Answer: D

bumpbot · Answer

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w, x, y, and z are consecutive odd integers such that w < x < y < z. W

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