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

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SaanjK26

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10 Jul 2025, 21:13

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At 60 RPM, the machine works at 36 units per liter.
If you speed up past 60, efficiency drops by 1.2 units for every 8 RPM, which breaks down to 0.15 units lost per RPM.
So, for speeds above 60, efficiency follows the formula
45-0.15R.

If you slow down below 60, efficiency improves by 0.8 units for every 4 RPM slower, or 0.2 units gained per RPM.
That means below 60 RPM, efficiency is 48- 0.2R, showing it gets better as you slow down.

Answer: increased efficiency is 48- 0.2R (R>0 and R<60)
Decreased efficiency is 45-0.15R. ( R>60).

BeachStudy

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10 Jul 2025, 21:47

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NOTES:
36 units per liter of fuel
60 rotations per minute RPM

Decreases by 1.2 units for every 8 RPM above 60

increase of .8 units for every 4 RPM below 60

SOLVE:
Formula for increase efficiency.

.8units for every 4 under 60. R is number RPM. so .8/4= .2 (36+[.2*60]))+.2R

Decrease:

1.2 units every 8 above 60. so 1.2/8 = 0.15R (36+[.15*60])-0.15R

INCREASE: 48+.2R
DECREASE: 45-0.15R

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Increase in efficiency would be 36+0.2(60-r) so 48-0.2r
Decrease in efficiency would be 36-0.15(r-60) so 45-0.5r

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.

Base efficiency=36 units/ltr

When rpm decrease below 60 by R revolutions

The new increased efficiency= 36+ (60-R)*0.8/4= 36+12-0.2R= 48-0.2R

When rpm increases above 60 by R revolutions

The new decreased efficiency=36-(R-60)*1.2/8=36-0.15R+9=45-0.15R

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10 Jul 2025, 23:02

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Increased Efficiency:

Decrease in RPM $=60-R$

Increase in Efficiency $=\frac{0.8}{4}(60-R)=0.2(60-R)=12-0.2R$

Final Efficiency $=36+(12-0.2R)=48-0.2R$

Decreased Efficiency:

Increase in RPM $=R-60$

Decrease in Efficiency $=\frac{1.2}{8}(R-60)=0.15(R-60)=0.15R-9$

Final Efficiency $=36-(0.15R-9)=45-0.15R$

Increased Efficiency: $48-0.2R$

Decreased Efficiency: $45-0.15R$

Manu1995

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The formula for Decreased Efficiency (R > 60)

The machine's efficiency decreases by 1.2 units for every 8 RPM increase above the base of 60 RPM.
==>The increase in RPM is represented by $(R - 60)$.
==>The number of 8 RPM increments is $(R - 60)/8$
==>The total efficiency decrease is $1.2(R - 60)/8$= $0.15*(R-60)$
==> Total efficiency= $36 - 0.15*(R-60)$ = $36- 0.15R + 9$= $45 - 0.15R$

The formula for Increased Efficiency (R < 60)

The machine's efficiency increases by 0.8 units for every 4 RPM decrease below the base of 60 RPM.
==>The decrease in RPM is represented by $(60 - R)$.
==>The number of 4 RPM decrements is $(60-R)/4$
==>The total efficiency increase = $0.8*(60-R)/4$= $0.2*(60-R)$

Total efficiency= $36+0.2(60-R)$
= $36+ 12- 0.2R$ = $48-0.2R$

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11 Jul 2025, 00:14

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Increased Efficiency:
For every 4RPM below 60, increase by 0.8 units
Increase factor = \frac{0.8*(60-R)}{4}

Increased efficiency = 36 + Increase factor
= 36 + \frac{0.8*(60-R)}{4}
= \frac{192-0.8R} {4}
= 48 - 0.2R

Decreased efficiency:
For every 8RPM below 60, decrease by 1.2 units
Decrease factor = \frac{1.2*(R-60)}{8}

Decreased efficiency = 36 - Decrease factor
= 36 - \frac{1.2*(R-60)}{8}
= \frac{360-1.2R} {8}
= 45 - 0.15R

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11 Jul 2025, 00:45

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Bunuel

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At 60 RPM
R = 36 u/l

Decreased Efficiency
At R > 60 RPM
$R_d$ = 36 - 1.2(R - 60)/8
=> 8$R_d$ = 288 + 72 - 1.2R
=> 8$R_d$ = 360 - 1.2R
=> $R_d$ = 45 - 0.15R

Increased Efficiency
At < R < 60 RPM
$R_i$ = 36 + 0.8(60 - R)/4
=> 4$R_i$ = 144 + 48 - 0.8R
=> 4$R_i$ = 192 - 0.8R
=> $R_i$ = 48 - 0.2R

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

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Bunuel

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Answers

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:

For every 4 RPM decrease below 60, the efficiency increases by 0.8 units.
For every 1 RPM decrease below 60, the efficiency will increase by 0.8/4 = 0.20;

Now, for R RPM lower than 60, the efficiency formula would be: 36 + 0.20(60-R) = 36 + 12 - 0.20R = 48-0.20R

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

For every 8 RPM increase over 60 the efficiency reduced by 1.2 units.
For every 1 RPM increase over 60 the efficiency will reduce by 1.2/8 = 0.15;

Now for R RPM greater than 60 the efficiency formula would be: 36 - 0.15(R-60) = 36 - 0.15R + 9 = 45-0.15R

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Bunuel

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Upto 60 RPM efficiency is 36 units.

For 0 < R < 60 total rotations will be 60-R
Efficiency increases by 0.8 units for every 4 RPM decrease below 60.
Efficiency increases by 0.2 units for every 1 RPM decrease below 60.

Increased Efficiency = 36 + 0.2 (60-R)= 36 + 12 -0.2R =48 - 0.2R

For R>60 total rotations will be 60 + (R - 60 )
Efficiency decreases by 1.2 units for every 8 RPM increase above 60
Efficiency decreases by 0.15 units for every 1 RPM increase above 60

Decreased Efficiency = 36 - 0.15 ( R-60) = 36 - 0.15R +9 = 45-0.15R

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11 Jul 2025, 02:09

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Bunuel

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11 Jul 2025, 03:47

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IE = 36 + 0.8*(60-R)/4 = 36 + 0.2*(60-R) = 36 + 12 - 0.2R = 48 - 0.2R
DE = 36 - 1.2*(R-60)/8 = 36 - 0.15(R-60) = 36 + 9 - 0.15R = 45 - 0.15R

Increased Efficiency = 48 - 0.2R and Decreased Efficiency = 45 - 0.15R

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The confusing thing about the question is the rate of change of efficiency. Is that supposed to be a step function where the efficiency increase only happens every 4 RPM below 60 and decrease only happens every 8 RPM above 60 or is it supposed to be continuous like if efficiency increase of 0.8 units happens for 4 RPM below 60, that means its an efficiency increase of 0.2 units every 1 RPM decrease below 60. I assumed it's continuous.

Formula for when efficiency has increased -> $36 + \frac{(60-R)}{4}(0.8) = 48-0.2R$

Formula for when efficiency has decreased -> $36 - \frac{(R-60)}{8}(1.2) = 45-0.15R$

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11 Jul 2025, 04:23

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The Answer is 48 - 0.2R for decreased efficiency for Increased efficiency and 45-0.15R for decreased efficiency

Since the stem states 36u/l for 60RPM
And Its efficiency decreases by 1.2 units for every 8 RPM increase above 60
so for 34.8u/l- 68RPM. If we use the above values 45-0.15R, by putting 68 in the equation where R is placed, it shows correct value for decreased efficiency.
And for increases by 0.8 units for every 4 RPM decrease below 60, so for 36.8u/l-56 RPM. If we use 48 - 0.2R, and input 56 in place of R, we get correct value.

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11 Jul 2025, 04:31

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E=36 at 60 rpm
E=34.8 AT 68 rpm and so on = unit wise 1.2/8=0.15
E=36.8 at 56 rpm and so on = unit wise 0.8/4=0.2
So for decreased efficiency,
36−0.15(R−60)=36−0.15R+9=45−0.15R
And for increased eff,
36+0.2(60−R)=36+12−0.2R=48−0.2R

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Given:
1. At 60 RPM, efficiency = 36 units/litre
2. For R > 60, efficiency decreases by 1.2 units per 8 RPM
Rate of decrease = 1.2 / 8 = 0.15 per RPM
3. For R < 60, efficiency increases by 0.8 units per 4 RPM
Rate of increase = 0.8 / 4 = 0.2 per RPM
To find:
Increased Efficiency a formula for the machine’s efficiency when running at R RPM, where 0 < R < 60
Decreased Efficiency a formula for the machine’s efficiency when running at R RPM, where R > 60.
Solution:
Decreased Efficiency when running at R RPM
= 36 – 0.15 (R – 60)
= 36 – 0.15R + 9
= 45 – 0.15R
Increased Efficiency when running at R RPM
= 36 + 0.2 (60 - R)
= 36 + 12 - 0.2R
= 48 – 0.2R

Ans:
1. Increased efficiency = 48 – 0.2R
2. Decreased Efficiency = 45 – 0.15R

Bunuel

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efficiency=36
>60 -> decreases by 1.2 units for every 8 RPM -> 1.2*(R-60)/8 = 0.15R - 0.15*60 = 0.15R - 9
<60 -> increases by 0.8 units for every 4 RPM -> 0.8*(60-R)/4 = 0.2*60 - 0.2R = 12 - 0.2R

Decreased Efficiency = 36 - (0.15R - 9) = 45 - 0.15R
Increased Efficiency = 36 + 12 - 0.2R = 48 - 0.2R

The right answers:
Increased Efficiency = 48 - 0.2R
Decreased Efficiency = 45 - 0.15R

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11 Jul 2025, 05:02

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For Decreased Efficiency (when R>60):
Rate of decrease = 1.2/8 = 0.15/RPM.
The formula (E) = 36 - 0.15*(R - 60) = 36 - 0.15R + 9 = 45 - 0.15R.

For Increased Efficiency (when 0<R<60):
Rate of increase = 0.8/4 = 0.2/RPM.
E = 36 + 0.2*(60 - R) = 36 + 12 - 0.2R = 48 - 0.2R.

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

Bunuel

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Let RPM be denoted by R.

For R > 60, decreased efficiency is given as:
E = 36 - {(R-60)/8}*1.2 units per liter
Upon simplification:
E = 45 - 0.15R

For R < 60
E = 36 + {(60 - R)/4}*0.8
Upon simplification:
E = 48 - 0.2R

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

Bunuel

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