If v, w, x, y, and z are consecutive positive integers, z v, and

Question

If  v ,  w ,  x ,  y , and  z  are consecutive positive integers,  z  >  v , and the difference between  v^2  and  w^2  is 43, what is the difference between  y^2  and  z^2 ?

(A) 43

(B) 46

(C) 49

(D) 51

(E) 57

Source:  Marty Murray Coaching

sfasd · Answer

All of them are consecutive integers.
So we can write them as v, v+1, v+2, v+3, v+4.
Given: (v+1)^2 - v^2 = 43
So v = 21
(v+4)^2 - (v+3)^2 = 16 + 8v - 9 - 6v = 7 + 2v = 7 + 42 = 49.

Edskore · Answer

The key concept here is Difference of Squares — and once you see the pattern, this question becomes very fast.

Step 1: Set up the consecutive integers.
Since v, w, x, y, z are consecutive, let v = n. Then w = n+1, x = n+2, y = n+3, z = n+4.

Step 2: Use the Difference of Squares identity.
w2 - v2 = (w + v)(w - v)
Since consecutive integers differ by 1, (w - v) = 1, so:
w2 - v2 = w + v = 43

Step 3: Find the actual values.
w + v = 43, and w = v + 1, so:
(v + 1) + v = 43 → 2v + 1 = 43 → v = 21
Therefore: v = 21, w = 22, x = 23, y = 24, z = 25

Step 4: Apply the same identity to y2 and z2.
z2 - y2 = (z + y)(z - y) = (25 + 24)(1) = 49

Answer: (C) 49

The elegant shortcut: You don't need to solve for v at all. Notice the pattern:
- w2 - v2 = v + w = 43
- x2 - w2 = w + x = 45
- y2 - x2 = x + y = 47
- z2 - y2 = y + z = 49

Each consecutive pair's difference of squares is 2 more than the previous pair's. The question is asking 4 steps ahead of where we started, so the answer is 43 + 4(2) - 2 = ... actually just count: 43, 45, 47, 49. Done in 10 seconds.

Common trap: Many students try to solve for all five values first, then compute z2 - y2 directly by squaring. That works, but it's slower and more error-prone. The pattern shortcut is what separates a 700+ approach from a 600-level approach on this question type.

MartyMurray · Answer

If  v ,  w ,  x ,  y , and  z  are consecutive positive integers,  z  >  v , and the difference between  v^2  and  w^2  is 43, what is the difference between  y^2  and  z^2 ?

Five consecutive positive integers starting with  v  are the following:

v ,  v + 1 ,  v + 2 ,  v + 3 , and  v + 4

So,  w^2 - v^2 = 43  is the equivalent of  (v + 1)^2 - v^2 = 43 .

Also,  z^2 - y^2  is the equivalent of  (v + 4)^2 - (v + 3)^2 .

We can recognize that  (v + 1)^2 - v^2  is a difference of squares, which factors to  ((v + 1) + v)((v + 1) - v) .

So, we have the following:

((v + 1) + v)((v + 1) - v) = 43

(2v + 1)(1) = 43

(2v + 1) = 43

2v = 42

v = 21

So,  z^2 - y^2 = (v + 4)^2 - (v + 3)^2 = 25^2 - 24^2 = (25 + 24)(25 - 24) = (49)(1) = 49

(A) 43

(B) 46

(C) 49

(D) 51

(E) 57

Correct answer:   C

Fisayofalana · Answer

Everyone's technical method is spot on. However, I just plugged in values for v, w, x, y, z and found that z^2-y^2 is 6 greater than w^2-v^2. It must be the same for the original values. 6 greater than 43 is 49. The answer is 49.

zhangwma · Answer

I like your elegant shortcut! I had to think about it for a while before I understood what you meant by " Each consecutive pair's difference of squares is 2 more than the previous pair's," so in case this helps anyone else, I thought of it/broke it down as...

The difference between  each pair  of difference of squares is 2   because the integers v, w, x, y, and z are all consecutive   positive integers. Therefore, the difference between  w^2-v^2=(w+v)(w-v)  and  x^2-w^2=(x+w)(x-w)  is 2 because you've added 1 to v and 1 to x, while each expression is being multiplied by 1, and both expressions contain the variable w.

For example: If we ignore the fact that  w^2-v^2=(w+v)(w-v)=43 , and we sub in easier to work with, consecutive integers: v=1, w=2, and x=3, then we'd get (2+1)(2-1) = 3 & (3+2)(3-2) = 5, and 5 - 3 = 2.   
  This is also demonstrated/proven by x - v = 3 - 1 = 2

egmat · Answer

Step 1: Set up the variables

Since v, w, x, y, z are consecutive and z > v:
v, w, x, y, z = v, v+1, v+2, v+3, v+4

Step 2: Use the given information

We're told: w2 − v2 =  43

Apply the  difference of squares formula: a2 − b2 = (a+b)(a−b)

w2 − v2 = (w + v)(w − v)

Since w = v + 1:
= (v + 1 + v)(v + 1 − v)
= (2v + 1)(1)
=  2v + 1

So: 2v + 1 = 43
Therefore:  v = 21

Step 3: Find all five integers

v =  21 , w =  22 , x =  23 , y =  24 , z =  25

Step 4: Calculate z2 − y2

Using the same formula:
z2 − y2 = (z + y)(z − y) = (25 + 24)(25 − 24) =  49  ×  1  =  49

Answer: C

---

kevincan1 · Answer

W= v+1 
Y= v+3
Z= v+4

W2 =v2 + 2v + 1 thus 2v+1= 43

Y2 = v2 + 6v + 9
Z2 = v2 + 8v + 16

Z2 - y2 = 2v + 7 = (2v + 1) +6 = 43 + 6

If v, w, x, y, and z are consecutive positive integers, z > v, and

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