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If v^2 – t^2 = 24, then w^2 – u^2 =

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If v^2 – t^2 = 24, then w^2 – u^2 =  [#permalink]

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12 Jul 2017, 23:38
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The letters represent consecutive integers

If $$v^2 – t^2 = 24$$, then $$w^2 – u^2 =$$

(A) 28
(B) 24
(C) 20
(D) 18
(E) 16

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If v^2 – t^2 = 24, then w^2 – u^2 =  [#permalink]

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13 Jul 2017, 00:35
$$v^2 – t^2 = 24$$
So, (v+t)(v-t) = 24

One way to get the product of 24 is 2*12
We can use the equations
v+t = 12
v-t = 2
Solving these two equations, we get v=7 and t=5
Since the letters represent consecutive integers, we can say that w = 8 and u = 6

Hence, the value of the expression becomes $$w^2 – u^2 = 8^2 - 6^2 = 64-36 = 28$$(Option B)
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If v^2 – t^2 = 24, then w^2 – u^2 =  [#permalink]

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Updated on: 13 Jul 2017, 04:15
Bunuel wrote:

The letters represent consecutive integers

If $$v^2 – t^2 = 24$$, then $$w^2 – u^2 =$$

(A) 28
(B) 24
(C) 20
(D) 18
(E) 16

$$v^2 – t^2 = 24$$
(v-t)(v+t) = 24
2(v+t) = 24
(v+t) = 12
t+t+2 = 12
t = 5

so, t = 5 , u = 6, v = 7 , w = 8
$$w^2 - u^2 = (w-u)(w+u ) = 2 (6+8) = 28$$

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Originally posted by shashankism on 13 Jul 2017, 00:53.
Last edited by shashankism on 13 Jul 2017, 04:15, edited 1 time in total.
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If v^2 – t^2 = 24, then w^2 – u^2 =  [#permalink]

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13 Jul 2017, 01:45

Anyway this is my approach: v^2-t^2=24 ----> (v-t)(v+t)=24. As there is one number between v and t (u) the difference between v and t is necessarily even, whether v and t are odd or even numbers (odd-odd=even, even-even=even). So 24 is build from a multiplication of two even numbers. The options for the multiplications are 2 and 12 as well as 4 and 6. As we checking the options we revealed that 4 and 6 is impossible because if (v-t)=4 and (v+t)=6 ----> v=5. It means that t=3 and it doesn't satisfy the equation (5^2-3^2 equal 16 not 24). So the only option is 2 and 12. When we add the equations (v-t=2 and v+2=12) we get that v=7 and t=5. It satisfies the equation in the question (7^2-5^2=24). Now, as we deal with consecutive numbers, we can also calculate to value of u and w (6 and 8) and solve the question: 8^2-6^2 ----> 64-36 ----> 28.

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Re: If v^2 – t^2 = 24, then w^2 – u^2 =  [#permalink]

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13 Jul 2017, 02:20
oryahalom wrote:

Anyway this is my approach: v^2-t^2=24 ----> (v-t)(v+t)=24. As there is one number between v and t (u) the difference between v and t is necessarily even, whether v and t are odd or even numbers (odd-odd=even, even-even=even). So 24 is build from a multiplication of two even numbers. The options for the multiplications are 2 and 12 as well as 4 and 6. As we checking the options we revealed that 4 and 6 is impossible because if (v-t)=4 and (v+t)=6 ----> v=5. It means that t=3 and it doesn't satisfy the equation (5^2-3^2 equal 16 not 24). So the only option is 2 and 12. When we add the equations (v-t=2 and v+2=12) we get that v=7 and t=5. It satisfies the equation in the question (7^2-5^2=24). Now, as we deal with consecutive numbers, we can also calculate to value of u and w (6 and 8) and solve the question: 8^2-6^2 ----> 64-36 ----> 28.

Since t, u, v .... represent consecutive integers... the difference between them is 1 and hence difference between v and t is 2.
Simply v-t = 2
u = t+1
v = t+2
w = t+3
x = t+4
y = t+5
z = t+6

I donno why are u even trying to decode from the factors of 24,... This step is not at all required...
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Re: If v^2 – t^2 = 24, then w^2 – u^2 =  [#permalink]

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13 Jul 2017, 03:38
Thanks! Your approach is much easier. You mentioned mistakenly that t+t+2=24 instead of t+t+2=12

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Re: If v^2 – t^2 = 24, then w^2 – u^2 =  [#permalink]

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13 Jul 2017, 04:17
oryahalom wrote:
Thanks! Your approach is much easier. You mentioned mistakenly that t+t+2=24 instead of t+t+2=12

Sent from my iPhone using GMAT Club Forum

Yes t+t+2 = 12... I have done a mistake there.. Edited in the original solution..
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Re: If v^2 – t^2 = 24, then w^2 – u^2 =  [#permalink]

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13 Jul 2017, 04:48
Bunuel wrote:

The letters represent consecutive integers

If $$v^2 – t^2 = 24$$, then $$w^2 – u^2 =$$

(A) 28
(B) 24
(C) 20
(D) 18
(E) 16

assume t = x -----1
as these are consecutive integer ..
v = x + 2
t = x

v^2 - t ^2 = 24-------2
replacing the values in equation 2
(x+2-x)(x+2+x) = 24
x+1 = 6
x = 5
t= 5-----------------------3

using 3 we will get

w = 8
u=6

8^2 - 6^2 = 28 ...
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If v^2 – t^2 = 24, then w^2 – u^2 =  [#permalink]

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13 Jul 2017, 11:40
shashankism wrote:
Bunuel wrote:

The letters represent consecutive integers

If $$v^2 – t^2 = 24$$, then $$w^2 – u^2 =$$

(A) 28
(B) 24
(C) 20
(D) 18
(E) 16

$$v^2 – t^2 = 24$$
(v-t)(v+t) = 24
2(v+t) = 24

(v+t) = 12
t+t+2 = 12
t = 5

so, t = 5 , u = 6, v = 7 , w = 8
$$w^2 - u^2 = (w-u)(w+u ) = 2 (6+8) = 28$$

shashankism, maybe I need more coffee, but I don't follow how you got from:

(v-t)(v+t) = 24, to

2(v+t) = 24

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Re: If v^2 – t^2 = 24, then w^2 – u^2 =  [#permalink]

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13 Jul 2017, 11:48
1
genxer123

v = t+2(since v and t are consecutive numbers)
So, in essence, v-t =(t+2)-t = 2

Hence, (v-t)(v+t) becomes 2(v+t)

Hope that helps!
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If v^2 – t^2 = 24, then w^2 – u^2 =  [#permalink]

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13 Jul 2017, 11:51
pushpitkc wrote:
genxer123

v = t+2(since v and t are consecutive numbers)
So, in essence, v-t =(t+2)-t = 2

Hence, (v-t)(v+t) becomes 2(v+t)

Hope that helps!

It does. Beautifully explained. Thanks!
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If v^2 – t^2 = 24, then w^2 – u^2 =   [#permalink] 13 Jul 2017, 11:51
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