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Bunuel
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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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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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Bunuel
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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Bunuel
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.

z<y<x
difference between consecutive num is 1
z-y and y-x are always 1

x+2y+3z=5y+4
x-3y+3z=4
x-3(y-z)=4
x-3(1)=4
x=7, y=6, z=5

Ans (B)
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The prompt should read as follows:

Bunuel
If X, Y and Z are consecutive integers (X>Y>Z). . What is the value of Z?

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

Since Y and Z are consecutive integers such that Y>Z, we get:
Y-Z = 1

Given equation:
X + 2Y + 3Z = 5Y + 4
X = 3Y - 3Z + 4
X = 3(Y-Z) + 4
X = 3(1) + 4
X = 7

Thus:
X=7, Y=6, Z=5

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