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For the function g, the value of \(\mathrm{g(x)}\) decreases by 45% for every increase in the value of x by...

GMAT Advanced Math : (Adv_Math) Questions

Source: Practice Test
Advanced Math
Nonlinear functions
HARD
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Notes
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For the function \(\mathrm{g}\), the value of \(\mathrm{g(x)}\) decreases by \(\mathrm{45\%}\) for every increase in the value of \(\mathrm{x}\) by 1. If \(\mathrm{g(0) = 14}\), which equation defines \(\mathrm{g}\)?

A

\(\mathrm{g(x) = 0.55(14)^x}\)

B

\(\mathrm{g(x) = 1.45(14)^x}\)

C

\(\mathrm{g(x) = 14(0.55)^x}\)

D

\(\mathrm{g(x) = 14(1.45)^x}\)

Solution

1. TRANSLATE the problem information

  • Given information:
    • \(\mathrm{g(x)}\) decreases by 45% for every increase in x by 1
    • \(\mathrm{g(0) = 14}\)
  • What this tells us: We need a function where the output consistently shrinks by the same percentage

2. INFER the function type

  • When a quantity changes by a fixed percentage repeatedly, this describes an exponential function
  • Since the value decreases, this is exponential decay
  • The general form is \(\mathrm{g(x) = a(r)^x}\) where a is the starting value and r is the multiplier

3. TRANSLATE the percentage change

  • "Decreases by 45%" means the new value is 45% smaller than the previous value
  • This means the new value retains \(\mathrm{100\% - 45\% = 55\%}\) of the previous value
  • As a decimal: \(\mathrm{55\% = 0.55}\)

4. INFER the specific equation

  • From \(\mathrm{g(0) = 14}\), we know \(\mathrm{a = 14}\)
  • From the 45% decrease, we know \(\mathrm{r = 0.55}\)
  • Therefore: \(\mathrm{g(x) = 14(0.55)^x}\)

Answer: C. \(\mathrm{g(x) = 14(0.55)^x}\)




Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students misinterpret "decreases by 45%" and think this means the multiplier should be 1.45 instead of 0.55.

They reason: "The problem mentions 45%, so I need 1.45 in my equation." This backward thinking focuses on the percentage mentioned rather than what mathematical relationship it represents.

This may lead them to select Choice D (\(\mathrm{g(x) = 14(1.45)^x}\)) - which would actually represent a 45% increase each time.

Second Most Common Error:

Poor INFER reasoning: Students recognize they need the 0.55 multiplier but put it in the wrong position, thinking the base should be related to the percentage rather than the initial value.

They create \(\mathrm{g(x) = 0.55(14)^x}\), mixing up which number serves as the base versus the coefficient.

This may lead them to select Choice A (\(\mathrm{g(x) = 0.55(14)^x}\)).

The Bottom Line:

Success requires carefully translating percentage language into mathematical relationships - "decreases by 45%" becomes "multiplied by 0.55" - and then recognizing exponential structure.

Answer Choices Explained
A

\(\mathrm{g(x) = 0.55(14)^x}\)

B

\(\mathrm{g(x) = 1.45(14)^x}\)

C

\(\mathrm{g(x) = 14(0.55)^x}\)

D

\(\mathrm{g(x) = 14(1.45)^x}\)

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