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31 Mar 2020, 02:56
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E
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[GMAT math practice question]
The function \(f(x)\) is defined as follows
\(f(x)\)={\(\frac{x}{5}\) , (\(n\) is a multiple of \(5\)) \(x+1\) , (\(n\) is not a multiple of \(5\))
Moreover, f(xn) = xn+1, and f(x1) = 1000. What is the value of f(x50)?
A. 1
B. 2
C. 3
D. 4
E. 5
The function \(f(x)\) is defined as follows
\(f(x)\)={\(\frac{x}{5}\) , (\(n\) is a multiple of \(5\)) \(x+1\) , (\(n\) is not a multiple of \(5\))
Moreover, f(xn) = xn+1, and f(x1) = 1000. What is the value of f(x50)?
A. 1
B. 2
C. 3
D. 4
E. 5
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31 Mar 2020, 20:38
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\(f(x_1) = x_2 = 1000\)
\(f(x_2) = x_3 = f (1000) = \frac{1000}{5} = 200\)
\(f(x_3) = x_4 = f (200) = \frac{200}{5} = 40\)
\(f(x_4) = x_5 = f (40) = \frac{40}{5} = 8\)
\(f(x_5) = x_6 = f (8) = 8+1 = 9\)
\(f(x_6) = x_7 = f (9) = 9+1 = 10\)
\(f(x_7) = x_8 = f (10) = \frac{10}{5} = 2\)
\(f(x_8) = x_9 = f (2) = 2+1 = 3\)
\(f(x_9) = x_{10} = f (3) = 3+1 = 4\)
\(f_(x_{10}) = x_{11} = f (4) = 4+1 = 5\)
Now the function becomes recursive with a cyclicity of 5.
\(f_(x_{10}) = f_(x_{15})= f_(x_{20})=.......= f_(x_{50}) = 5 \)
\(f(x_2) = x_3 = f (1000) = \frac{1000}{5} = 200\)
\(f(x_3) = x_4 = f (200) = \frac{200}{5} = 40\)
\(f(x_4) = x_5 = f (40) = \frac{40}{5} = 8\)
\(f(x_5) = x_6 = f (8) = 8+1 = 9\)
\(f(x_6) = x_7 = f (9) = 9+1 = 10\)
\(f(x_7) = x_8 = f (10) = \frac{10}{5} = 2\)
\(f(x_8) = x_9 = f (2) = 2+1 = 3\)
\(f(x_9) = x_{10} = f (3) = 3+1 = 4\)
\(f_(x_{10}) = x_{11} = f (4) = 4+1 = 5\)
Now the function becomes recursive with a cyclicity of 5.
\(f_(x_{10}) = f_(x_{15})= f_(x_{20})=.......= f_(x_{50}) = 5 \)
MathRevolution
02 Apr 2020, 01:23
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=>
f(x1) = x2 = 1000
f(x2) = x3 = 1000/5 = 200
f(x3) = x4 = 200/5 = 40
f(x4) = x5 = 40/5 = 8
f(x5) = x6 = 8 + 1 = 9 (because 8 is not a multiple of 5)
f(x6) = x7 = 9 + 1 = 10
f(x7) = x8 = 10/5 = 2
f(x8) = x9 = 2 + 1 = 3
f(x9) = x10 = 3 + 1 = 4
f(x10) = x11 = 4 + 1 = 5
f(x11) = x12 = 5/5 = 1
f(x12) = x13 = 1 + 1 = 2
f(x13) = x14 = 2 + 1 = 3
f(x14) = x10 = 3 + 1 = 4
f(x15) = x11 = 4 + 1 = 5
…
f(x5k) = 5
Therefore, E is the answer.
Answer: E
f(x1) = x2 = 1000
f(x2) = x3 = 1000/5 = 200
f(x3) = x4 = 200/5 = 40
f(x4) = x5 = 40/5 = 8
f(x5) = x6 = 8 + 1 = 9 (because 8 is not a multiple of 5)
f(x6) = x7 = 9 + 1 = 10
f(x7) = x8 = 10/5 = 2
f(x8) = x9 = 2 + 1 = 3
f(x9) = x10 = 3 + 1 = 4
f(x10) = x11 = 4 + 1 = 5
f(x11) = x12 = 5/5 = 1
f(x12) = x13 = 1 + 1 = 2
f(x13) = x14 = 2 + 1 = 3
f(x14) = x10 = 3 + 1 = 4
f(x15) = x11 = 4 + 1 = 5
…
f(x5k) = 5
Therefore, E is the answer.
Answer: E
