Question

SAT Math Help (MANY THANKS!)

A microbiologist observes that a certain species of thermophilic bacteria reproduces at its maximum rate at a temperature of 59 degrees Celsius, and that for every degree cooler than 59 degrees Celsius, the bacteria's reproduction rate is reduced by 20 percent. By what percent will the bacteria's reproduction rate be reduced from its maximum at a temperature of 56 degrees Celsius?"
***Please show me HOW you got the answer that you did. THANK YOU!!!!

Answer 1

48.8%

Explanation:

So let's start at the temperature
`59^\circ C` where maximum reproduction is achieved. Let's say the maximum reproduction rate number is
`x`.

The problem says the reproduction rate is reduced by 20 percent per decrease in degree of temperature in Celsius.

So a
`58^\circ C`, the reproduction number is
`x-.2(x)` since we are taking 20% of the number we began with.

Let's simplify
`x-.2(x)`.

These are like terms; 1-.2 is .8 so
`x-.2x=.8x`

Now let's go to
`57^\circ C`, the reproduction rate is decreased by 20% from the .8x. So we are taking 20% off of .8x:

`.8x-.2(.8x)`

`.8x-.16x`

`.64x`

Going to finally
`56^\circ C`, the reproduction rate is decreased by 20% from .64x. So we are taking 20% off of .64x:

`.64x-.2(.64x)`

`.64x-.128x`

`.512x`

So that is 51.2% of x.

So it was reduce by 100%-51.2% =48.8% .

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Another way if you are good with exponential functions:

`A=P_0(1-.2)^t`

The initial reproduction rate is
`P_0`.

Let 59 degree C represent what happens at time t=0.

So 58 degree C represents what happens at time t=1.

...

56 degree C represents what happens at time t=3.

Plug in t=3:

`A=P_0(1-.2)^3`

`A=P_0(.8)^3`

`A=P_0(.512)`

This says at time t=3 that you have 51.2% of your reproduction rate which means it was decreased by 100%-51.2%=48.8% from it's original reproduction rate at t=0 which was
`59^\circ C`.