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First, let's establish what we know:
* At 60 RPM, efficiency = 36 units per liter
* For R > 60: Efficiency decreases by 1.2 units for every 8 RPM increase
* For R < 60: Efficiency increases by 0.8 units for every 4 RPM decrease
For Decreased Efficiency (R > 60):
The rate of decrease is 1.2 units per 8 RPM increase, which equals 0.15 units per 1 RPM increase.
If R > 60, then the RPM increase from 60 is (R - 60).
Total decrease in efficiency = 0.15 × (R - 60)
Efficiency at R RPM = 36 - 0.15(R - 60) = 36 - 0.15R + 9 = 45 - 0.15R
For Increased Efficiency (0 < R < 60):
The rate of increase is 0.8 units per 4 RPM decrease, which equals 0.2 units per 1 RPM decrease.
If R < 60, then the RPM decrease from 60 is (60 - R).
Total increase in efficiency = 0.2 × (60 - R)
Efficiency at R RPM = 36 + 0.2(60 - R) = 36 + 12 - 0.2R = 48 - 0.2R
Therefore, the correct formulas are:
Increased Efficiency (0 < R < 60): 48 - 0.2R
Decreased Efficiency (R> 60): 45 - 0.15R
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The prompt tells us that RPM and efficiency are negatively correlated.

2 cases can be formed:

  • When the efficiency increases
  • When the efficiency decreases

Firsty, When the efficiency increases:

Here the RPM decreaes and thus the efficiency increases -
Let's say the RPM is decreased by 4y units, this will increase the efficiency by 0.8y

RPM = 60 - 4y
Efficiency = 36 + 0.8y

We are given that for R, where 0<R<60, we need to find the efficiency

So RPM = R = 60-4y
solving for y, we get, y = \(\frac{(60-R)}{4}\)

Putting in this for efficiency, we get,

Efficiency = 36 + 0.8(\(\frac{(60-R)}{4}\))
Efficiency = 48-0.2R


Secondly, When the efficiency decreases:

Let's say the RPM is increased by 8x units, this will decrease the efficiency by 1.2x

RPM = 60 + 8x
Efficiency = 36-1.2x

Now, for R, where R>60
We can say, RPM = R = 60+8x

Solving for x, we get, x = \(\frac{(R-60)}{8}\)

Plugging it to the equation, we get,

Efficiency = 36 - 1.2(\(\frac{(R-60)}{8}\))
Efficiency = 45-0.15R

Answer:
Increased efficiency: 48 - 0.2R
Decreased efficiency: 45 - 0.15R
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Bunuel
 


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A machine operates at an efficiency of 36 units per liter of fuel when running at exactly 60 rotations per minute (RPM). Its efficiency decreases by 1.2 units for every 8 RPM increase above 60, and increases by 0.8 units for every 4 RPM decrease below 60.

Select for Increased Efficiency a formula for the machine’s efficiency when running at R RPM, where 0 < R < 60 and select for Decreased Efficiency a formula for the machine’s efficiency when running at R RPM, where R > 60. Make only two selections, one in each column.
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Quite straightforward.

Increase efficiency: 36 + ((60-R)/4) x 0.8 = 48 - 0.2R
Decreased Efficiency: 36 - ((R - 60)/8) x 1.2 = 45 - 0.15R
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given efficiency is 36 units per liter

RPM is 60

efficiency decreases by 1.2 units for every 8 RPM increase over 60

so efficiency becomes 34.8 at 68 RPM
R is 68 plug in value in options to get 34.8 ; 45 - 0.15R

efficiency increases by 0.8 units for every 4 RPM increase below 60
so efficiency becomes 36.8 at 56 RPM
R is 56 , plug in value in options to get 36.8 ; 48 - 0.2R

correct option

increased efficiency 48 - 0.2R ; decreased efficiency 45 - 0.15R
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A machine operates at an efficiency of 36 units per liter of fuel when running at exactly 60 rotations per minute (RPM). Its efficiency decreases by 1.2 units for every 8 RPM increase above 60, and increases by 0.8 units for every 4 RPM decrease below 60.

Select for Increased Efficiency a formula for the machine’s efficiency when running at R RPM, where 0 < R < 60 and select for Decreased Efficiency a formula for the machine’s efficiency when running at R RPM, where R > 60. Make only two selections, one in each column.
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First, find how many units drop for every RPM.
So that's 1.2/8=0.15 units,
So efficiency is 36-0.15(R-60)(since R>60)
We get 45−0.15R

then
0.8/4=0.2units
So efficiency is 36+0.20(60-R)(sinceR<60)
We get 48−0.2R

Increased Efficiency: 48 - 0.2R
Decreased Efficiency: 45 - 0.15R

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A machine operates at an efficiency of 36 units per liter of fuel when running at exactly 60 rotations per minute (RPM). Its efficiency decreases by 1.2 units for every 8 RPM increase above 60, and increases by 0.8 units for every 4 RPM decrease below 60.

Select for Increased Efficiency a formula for the machine’s efficiency when running at R RPM, where 0 < R < 60 and select for Decreased Efficiency a formula for the machine’s efficiency when running at R RPM, where R > 60. Make only two selections, one in each column.
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Increased efficiency = 36 + (60-R)X0.8/4
= 36 + 12 - 0.2R
= 48 - 0.2R
Decreased efficiency = 36 - (R-60)X1.2/8
= 36 - 0.15R + 9
= 45 - 0.15R
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At 60 RPM, efficiency = 36 units/litre
RPM>60, efficiency decreases by 1.2 units for every 8 RPM increase = 1.2/8=0.15 units decrease per RPM
RPM<60, efficiency increases by 0.8 units for every 4 RPM decrease = 0.8/4=0.2 units increase per RPM

INCREASED EFFICIENCY-
Every 1 RPM decreases efficiency by 0.2
At 60 RPM, efficiency = 36 units/litre
Efficiency = 36+0.2(60-R) = 36+12-0.2R = 48-0.2R

DECREASED EFFICIENCY-
Every 1 RPM increase reduces efficiency by 0.15
Efficiency = 36-0.15(R-60)= 36-0.15R+9= 45- 0.15 R
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A machine operates at an efficiency of 36 units per liter of fuel when running at exactly 60 rotations per minute (RPM). Its efficiency decreases by 1.2 units for every 8 RPM increase above 60, and increases by 0.8 units for every 4 RPM decrease below 60.

Select for Increased Efficiency a formula for the machine’s efficiency when running at R RPM, where 0 < R < 60 and select for Decreased Efficiency a formula for the machine’s efficiency when running at R RPM, where R > 60. Make only two selections, one in each column.
. Clearly we can see that increase efficiency is 0.8 for every 4rpm IE 60/4=15*0.8=12
ie 36+12=48 but we have decreased efficiency of 1.2 at 8rpm therefore 1.2-0.8=0.4 and decrease efficiency is1.2*7.5=9 15-9=6 0.4/6=0.2 therefore 48+0.2R

Decreased efficiency is 36+9=
45-0.15R
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The machine oprates at an effiiency of 60rpm with effciently increase by 1.5
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Increase per rpm = 0.8/4=0.2
So, this is per rpm decrease
rpm decrease = 60-R
Decrease = 36 + 0.2(60-R) = 36+12-0.2R=48-0.2R

Decrease per rpm = 1.2/8=0.15
So, this is per rpm increase
rpm increase = R-60
Decrease = 36 - 0.15(R-60) = 36-0.15R+9=45-0.15R

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A machine operates at an efficiency of 36 units per liter of fuel when running at exactly 60 rotations per minute (RPM). Its efficiency decreases by 1.2 units for every 8 RPM increase above 60, and increases by 0.8 units for every 4 RPM decrease below 60.

Select for Increased Efficiency a formula for the machine’s efficiency when running at R RPM, where 0 < R < 60 and select for Decreased Efficiency a formula for the machine’s efficiency when running at R RPM, where R > 60. Make only two selections, one in each column.
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Bunuel
 


This question was provided by GMAT Club
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A machine operates at an efficiency of 36 units per liter of fuel when running at exactly 60 rotations per minute (RPM). Its efficiency decreases by 1.2 units for every 8 RPM increase above 60, and increases by 0.8 units for every 4 RPM decrease below 60.

Select for Increased Efficiency a formula for the machine’s efficiency when running at R RPM, where 0 < R < 60 and select for Decreased Efficiency a formula for the machine’s efficiency when running at R RPM, where R > 60. Make only two selections, one in each column.

Increased Efficiency

\(36 + \frac{(60 - R)}{4} * 0.8\)

\(36 + \frac{(60 - R)}{4} * \frac{8}{10}\)

\(36 + \frac{(60 - R)}{5}\)

\(36 + 12 - 0.2R\)

\(48 - 0.2R\)

Decreased Efficiency

\(36 - \frac{(R-60)}{8} * 1.2\)

\(36 - \frac{(R-60)}{8} * \frac{12}{10}\)

\(36 - (R-60) * \frac{3}{20}\)

\(36 - 0.15R + 9\)

\(45 - 0.15R\)

Decreased Efficiency: \(45 - 0.15R\)

Increased Efficiency: \(48 - 0.2R\)
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1. Increased efficiency: 0<R<60
36+ (0.8*(60-R)/4)= 48-0.2R

2. Decreased efficiency: R>60
36- (1.2*(R-60)/8)=45-0.15R
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Given:
At 60 RPM, efficiency = 36 units per liter
For every 8 RPM above 60, efficiency decreases by 1.2
For every 4 RPM below 60, efficiency increases by 0.8

We are to find formulas for:
When R < 60 --> Increased efficiency
When R > 60 --> Decreased efficiency

Case 1: Increased Efficiency (R < 60)
For every 4 RPM decrease, efficiency increases by 0.8
That means for each 1 RPM decrease, increase = 0.8 / 4 = 0.2
Let the decrease from 60 be: 60 - R
Then increase in efficiency = 0.2 × (60 - R)

So, Efficiency = 36 + 0.2(60 - R)
= 36 + 12 - 0.2R
= 48 - 0.2R
Increased Efficiency formula: 48 - 0.2R

Case 2: Decreased Efficiency (R > 60)
For every 8 RPM increase, efficiency decreases by 1.2
That means for each 1 RPM increase, decrease = 1.2 / 8 = 0.15
Let the increase from 60 be: R - 60
Then decrease in efficiency = 0.15 × (R - 60)
So, Efficiency = 36 - 0.15(R - 60)
= 36 - 0.15R + 9
= 45 - 0.15R
Decreased Efficiency formula: 45 - 0.15R
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Bunuel
 


This question was provided by GMAT Club
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A machine operates at an efficiency of 36 units per liter of fuel when running at exactly 60 rotations per minute (RPM). Its efficiency decreases by 1.2 units for every 8 RPM increase above 60, and increases by 0.8 units for every 4 RPM decrease below 60.

Select for Increased Efficiency a formula for the machine’s efficiency when running at R RPM, where 0 < R < 60 and select for Decreased Efficiency a formula for the machine’s efficiency when running at R RPM, where R > 60. Make only two selections, one in each column.
Given:
Case I: The efficiency decreases with each 8 unit increase in RPM over 60 which can be written as (R-60)/8
and the decrease is of 1.2 then the formula becomes 36- units*1.2 i.e. ....=> 36-((R-60)/8)*1.2. => 36 - ((R-60)*3)/20 => 45 - 0.15R Hence the answer.

Similarly, efficiency decreases with each 4 unit decrease in RPM below 60 which can be written as (60-r)/4
and the decrease is of 0.8 then the formula becomes 36- units*0.8 i.e. ....=> 36-((60-R)/4)*0.8 => 36-(60-R)*0.2 => 48 - 0.2R Hence the answer
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At 60 RPM, efficiency is 36 units.

Scenario A:
For every 8 RPM increase above 60 RPM, efficiency drops by 1.2 units.
This means for every 1 RPM increase, efficiency decreases by 0.15 units.

Scenario B:
For every 4 RPM decrease below 60 RPM, efficiency increases by 0.8 units.

So, we can set up the equations:

For RPM > 60:
Efficiency = 36 - 0.15(R - 60) = 45 - 0.15R

For RPM < 60:
Efficiency = 36 + 0.2(60 - R) = 48 - 0.2R
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Given, efficiency of 36 units per liter (@ 60RPM)
Efficiency decreases by 1.2 units for every 8 RPM increase above 60
Efficiency increases by 0.8 units for every 4 RPM decrease below 60.

Increased Efficiency a formula for the machine’s efficiency when running at R RPM, where 0 < R < 60

= 36+(0.8(60-R)/4)
=48-0.2R


Decreased Efficiency a formula for the machine’s efficiency when running at R RPM, where R > 60.

= 36-(1.2(R-60)/8)
=45-0.15R
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