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GMAT Club Olympics 2025 (Day 9): A machine operates at an efficiency

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10 Jul 2025, 10:16

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Increased Efficiency:
= 36 + ((60-R)/4)*0.8
= 36 + 12 - 0.2R
= 48 - 0.2R

Decreased Efficiency:
= 36 - ((R-60)/8)*1.2
= 36 + 9 - 0.15R
= 45 - 0.15R

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Bunuel

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For increased efficincy:
0.8 units for every 4RPM, So 0.2 units for 1 RPM, Here it is 60-R RPMs so 0.2(60-R) units.
Efficiency=36+0.2(60−R)=36+12−0.2R=48−0.2R.

Hence, Increased Efficiency: 48−0.2R.

Similarly,For Decreased Efficiency:
1.2 units for every 8RPM sp 0.15 for 1 RPM. So For R+60 RPMs => 0.15(R-60) units.
Efficiency=36−0.15(R−60) => 45 - 0.15R

Hence, IMO Decreased Efficiency: 45 - 0.15R and Increased Efficiency: 48−0.2R.

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For an RPM below 60, the efficiency increases by 0.2 units for every 1 RPM decrease, leading to the formula 36+0.2(60-R)=48-0.2R. For an RPM above 60, the efficiency decreases by 0.15 units for every 1 RPM increase, resulting in the formula 36−0.15(R-60)=45-0.15R.

Regards,
Lucas

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10 Jul 2025, 10:37

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for increased efficiency, we know that efficiency increases by 0.8 for every decrease of 4 RPMs,

So, it can be written as 36 + 0.8(60-R)/4 ===> 48 - 0.2R

Similarly for decreased efficiency,
36 - 1.2(R-60)/8 ===> 45 - 0.15R

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Base efficiency: 36 units per liter at 60 RPM.

Case 1: Increased Efficiency (when 0<R<60 RPM)

Efficiency increases by 0.8 units every 4 RPM decrease below 60.
rate of efficiency = 0.8/4 => 0.2 units/RPM
total increase in efficiency= 0.2 * (60-R)
E=36 + 0.2 * (60-R)
E=36 + 12-0.2R
E=48 - 0.2R

Case 2: Decreased Efficiency (when R>60 RPM)

Efficiency increases by 1.2 units every 8 RPM increase above 60.
rate of efficiency = 1.2/8 => 0.15 units/RPM
total decrease in efficiency= 0.15 * (R-60)
E=36 - 0.15 * (R-60)
E=36 - 0.15R+9
E=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.

Increased efficiency = 36 + .8(60 - R)/4 = 36 + .2(60-R) = 48 - .2R
Decreased efficiency = 36 - 1.2(R-60)/8 = 36 - .15(R-60) = 36 + 9 - .15R = 45 - .15R

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	Decreased Efficiency
48 - .2R	45 - .15R

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10 Jul 2025, 11:19

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E= 366 units / ltr of fuel at 60 RPM

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

equation for increased efficiency 0<R<60

we will use line slope formula

(y2-y1)/(x2-x1) = slope of line

(E -36)/(60-R) = 0.8/4

E -36 = (60-R)*0.2

E = 48 -0.2R

equation for decreased efficiency R>60

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

36 -E = 0.15*R - 9

E = 45 - 0.15R

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Given efficiency = 36 units/ litre for 60 RPM

For Increased efficiency:
where 0 < R < 60
Let k be the value by which the efficiency is increased.

R= 60 -4k
$k= \frac{60-R}{4}$

Efficiency: 36+ 0.8k
$= 36 + 0.8*\frac{60-R}{4}$
= 48 -0.2R

For decreased efficiency:

where R>60
Let k be the value by which the efficiency is increased.

R= 60 + 8k
$k= \frac{R-60}{8}$

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

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For Decreased efficiency when rpm > 60

for 8 rpm, 1.2 units decresed then for 1 rpm = 1.2/8 = 0.15
and increase in rpm is R-60

then equation = 36 - 0.15(R-60) = 45-0.15R

For Increase efficiency when rpm < 60

for 4 rpm, 0.8 units increased then for 1 rpm = 0.8/4 = 0.2
and decrese in rpm is 60-R

then equation = 36 + 0.2(60-R) = 48-0.2R

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Bunuel

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A machine operates at an efficiency of 36 units per litre of fuel, while running at exactly 60 rotations per minute (RPM).

If RPM is increased by 8 ( above 60 ), the efficiency decreases by 1.2 units.

If RPM is decreased by 4 ( below 60 ), the efficiency increases by 0.8 units.

If RPM is in the range 0 < R < 60

R = 60 - x

No of 4 rpm decrease = x/4

Efficiency increase = ( no of 4 rpm decrease )* 0.8 = (x/4)*0.8 = 0.2x

Efficiency = 36 + 0.2 x

= 36 + 0.2 * (60 - R)

= 36 + 12 - 0.2 R

= 48 - 0.2 R

Efficiency = 48 - 0.2R

If RPM is > 60,

R = 60+y , where y is the extra RPM beyond 60.

Number of 8 RPM increase = y/8

Efficiency = (y/8) *1.2 = 0.15 y

Efficiency = 36 - 0.15 y

= 36 - 0.15 * ( R - 60)

= 45 - 0.15 R

If 0 < R <60, Efficiency = 48 - 0.2R.

If R > 60, Efficiency = 45 - 0.15 R

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10 Jul 2025, 12:34

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Efficiency at 60 RPM = 36units/L
For every 8 RPM increase, efficiency decrease by 1.2
Machine efficiency = 36 - 1.2(R-60)/8 = 36 - 0.15(R-60) = 36-0.15R+9 = 45-0.15R

For every 4 RPM decrease, efficiency increase by 0.8
Machine efficiency = 36 + 0.8(60-R)/4 = 36+0.2(60-R)=36+12-0.2R = 48-0.2R

Hence answer is:
Increased Efficiency: 48-0.2R
Decreased Efficiency: 45-0.15R

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Bunuel

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Increase Efficiency, where 0<R<60, every 4RPM decrease below 60 => increase efficiency by 0.8 units
=> 0.8/4=0.2
=> 36 + 0.2(60 - R) = 48 - 0.2R
Decrease Efficiency, where R>60, every 8RPM increase above 60 => decrease efficiency by 1.2 units.
=> 1.2/8=0.15
=> 36 - 0.15(R - 60) = 45 - 0.15R.

Answer: Increase Efficiency : 48 - 0.2R ; Decrease Efficiency: 45 - 0.15R

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1- Efficiency increases by 0.8 units for every 4 RPM decrease below 60

In this case, R<60, so the efficiency increases 0.8 for every $\frac{60-R}{4}$
So the new efficiency will be: $36+0.8* \frac{60-R}{4}$
If we simplify it, we'll have: $48-0.2R$

2- Efficiency decreases by 1.2 units for every 8 RPM increase above 60

In this case, R>60, so the efficiency decreases 1.2 for every $\frac{R-60}{8}$
So the new efficiency will be: $36-1.2* \frac{R-60}{8}$
If we simplify it, we'll have: $45-0.15R$

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10 Jul 2025, 14:14

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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.

For inceased efficiency we can deduce following formula, where R will be the new RPM lower than 60:

36 + 4/5 x ( (60 - R) /4 ) = 36 + 240/20 - 4R/20 = 36+12 - R/5 = 48 + 0.2R

For decreased efficiency we can deduce following formula, where R will be the new RPM higher than 60:

36 - 6/5 x ( (R - 60)/8 ) = 36 - 6R/40 + 360/40 = 36 - 3R/20 + 9 = 45 - 3R/20
= 45 - 0.15R

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10 Jul 2025, 14:16

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efficiency of machine be E=36 units/ltr. of fuel ---60RPM
1. E decreases by 1.2 units for every 8RPM increase above 60
2. E increase by 0.8 units for every 4RPM decrease below 60
if R is RPM
Increased E=36+(60-R)/4*0.8 where 0<R<60
=48-0.2R
decreased E=36-(R-60)/8*1.2 where R>60
=45-0.15R

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let E = efficiency , R = rotations per minute

Given data, base efficiency is 36 units per litre of fuel when running at exactly 60 rotations per minute (RPM)

Given conditions, 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.

we have to (a)solve for Increased Efficiency a formula for the machine’s efficiency when running at R RPM, where 0 < R < 60

decrease in RPM = 60−R
number of intervals = 60−R/4
then total increase = 0.8×( 60−R/4)= 0.2×(60−R)= 12−0.2R
Increased Efficiency (E) = Base Efficiency + Total Increase =36+(12−0.2R) =48−0.2R

(b) solve for Decreased Efficiency a formula for the machine’s efficiency when running at R RPM, where R > 60
increase in RPM = R−60
number of intervals = R−60/8
then total decrease = 1.2×( R−60/8)=0.15×(R−60)= 0.15R−9
Decreased Efficiency (E) = Base Efficiency − Total decrease = 36−(0.15R−9) = 45 − 0.15R

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10 Jul 2025, 16:14

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At 60 RPM, the machine's efficiency is 36 units.

when RPM goes below 60 =>for every 4 RPM decrease efficiency increases by 0.8 units so =>for every 1 RPM decrease = 0.8/4 = 0.2 units efficieny increase.

So here new effi = start effi + increase in effi
=>36+(60−R)*0.2
=>36+(60*0.2)−0.2R
=>36+12−0.2R
Increased Efficiency =>[b]48−0.2R[/b]

when RPM goes above 60 =>for every 8 RPM increase efficiency drops by 1.2 units so => for 1 RPM increase = 1.2/8 = 0.15 efficieny decrease.

So new effi = Start effi - Drop in effi
=>36−(R−60)*0.15
=>36−0.15R+(60*0.15)
=>36−0.15R+9
Decreased Efficiency=>45−0.15R

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0 < R < 60:
36+0.8(60-R)/4
48-0.2R

R > 60:
36-1.2(R-60)/8
45-0.15R

Bunuel

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10 Jul 2025, 19:25

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for Increased Efficiency =
it increases by 0.8 units for every 4 RPM dec. We can write the eq as
36 + [(60-R)/4]*0.8
36+ [15-0.25R]*0.8
36+ [12-02.R]
48-0.2R

For Decreased Efficiency ,
1.2 dec for every 8RPM increase

36-[(R-60)/8]*1.2
36-[R/8-7.5]*1.2
36-0.15R + 9
45-015R

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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).

Rate = 36units/1liter when RPM = 60

Its efficiency:

(i) decreases by 1.2 units for every 8 RPM increase above 60, and

(ii) 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
Increased = 36/1 + ( 60 - R )/4 * 0.8
Increased = 36 + ( 60 - R ) * 0.2
Increased = 36 + 12 - 0.2R
Increased = 48 - 0.2R

and select for Decreased Efficiency a formula for the machine’s efficiency when running at R RPM, where R > 60.
Decreased = 36/1 - (R - 60)/8 * 1.2
Decreased = 36 - (R - 60) * 12 / 80
Decreased = 36 - (R - 60) * 3 / 20
Decreased = 36 - 3R/20 + 180/20
Decreased = 36 - 3R/20 + 9
Decreased = 45 - 3R/20

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