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X, Y and Z are consecutive numbers (X>Y>Z). X +2Y +3Z = 5Y + 4. What

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X, Y and Z are consecutive numbers (X>Y>Z). X +2Y +3Z = 5Y + 4. What  [#permalink]

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New post 05 Jul 2016, 11:30
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A
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Difficulty:

  55% (hard)

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64% (01:41) correct 36% (02:08) wrong based on 153 sessions

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Re: X, Y and Z are consecutive numbers (X>Y>Z). X +2Y +3Z = 5Y + 4. What  [#permalink]

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New post 05 Jul 2016, 11:34
Let the 3 consecutive integers be k-1 (Z), k (Y) and k+1 (X). Given : X +2Y +3Z = 5Y + 4.
k+1 + 2k + 3k - 3 = 5k+ 4 => k =6.

Thus Z = k-1 = 5.
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Re: X, Y and Z are consecutive numbers (X>Y>Z). X +2Y +3Z = 5Y + 4. What  [#permalink]

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New post 05 Jul 2016, 11:43
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If X, Y and Z are consecutive numbers (X>Y>Z). X +2Y +3Z = 5Y + 4. What is the value of Z?

X+2Y+3Z=5Y+4
=> X+3Z=3Y+4
=> X+3Z=(3Z+3)+4 .. as Y=Z+1
=> X=7; Y=6; Z=5
Ans B.

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Re: X, Y and Z are consecutive numbers (X>Y>Z). X +2Y +3Z = 5Y + 4. What  [#permalink]

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New post 21 Feb 2018, 09:20
Let the three consecutive numbers be = n, n+1, n+2.

Therefore, n+2+2(n+1)+3n=5(n+1)+4.
Hence, n+2+2n+2+3n=5n+5+4. 6n+4=5n+9. n=5.

Smallest number=z=5.

Therefore, B.
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Re: X, Y and Z are consecutive numbers (X>Y>Z). X +2Y +3Z = 5Y + 4. What  [#permalink]

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New post 22 Feb 2018, 16:47
Bunuel wrote:
If X, Y and Z are consecutive numbers (X>Y>Z). X +2Y +3Z = 5Y + 4. What is the value of Z?

A. 6.
B. 5.
C. 4.
D. 3.
E. 2.


We can let Z = z, thus Y = z + 1 and X = z + 2. Thus, in terms of z, we can re-express the given equation as follows:

z + 2 + 2(z + 1) + 3z = 5(z + 1) + 4

4z + 2 + 2z + 2 = 5z + 5 + 4

6z + 4 = 5z + 9

z = 5

Answer: B
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Re: X, Y and Z are consecutive numbers (X>Y>Z). X +2Y +3Z = 5Y + 4. What  [#permalink]

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New post 23 Feb 2018, 08:51
Top Contributor
Bunuel wrote:
If X, Y and Z are consecutive numbers (X>Y>Z). X +2Y +3Z = 5Y + 4. What is the value of Z?

A. 6.
B. 5.
C. 4.
D. 3.
E. 2.


If , Y and Z are consecutive numbers AND Z < Y < X, then we know that....
Y = Z + 1
X = Z + 2

Now take the given equation (X +2Y + 3Z = 5Y + 4) and replace Y and X with Z+1 and Z+2
We get: (Z+2) +2(Z+1) + 3Z = 5(Z+1) + 4
Expand: Z+2 + 2Z + 2 + 3Z = 5Z + 5 + 4
Simplify both sides: 6Z + 4 = 5Z + 9
Solve: Z = 5

Answer: B

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Brent
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Re: X, Y and Z are consecutive numbers (X>Y>Z). X +2Y +3Z = 5Y + 4. What &nbs [#permalink] 23 Feb 2018, 08:51
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