To find m:
m= c + bt = 0 + 2*120 = 240
To find b:
b = (c-m)/-t = (0-100)/-120 = 100/120 = 5/6
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b=(m+c)/t and m=(bt)+c
b=(100+0)/120=5/6
m=(2*120)+0=240
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Equation for working m: m = c + bt
Equations for working b: b = (-c + m)/t
Working on the idea that b = 2, c = 0, and t = 120:
m = 0 + (2 x 120)
m=240
Working on the idea that m = 100, c = 0, and t=120:
b = (0 + 100)/120
b = 100/120 = 5/6 = 0.833…
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For modified MaxFuel, I used:
m = c + bt; where b = 2, c = 0, t = 120
m = 240
For modified BurnRate, I used:
b = (-c+m)/t; where c = 0, m = 100, t = 120
b = 5/6
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Wasn’t 100% sure I’d understood the question, but thankfully arrived at similar answers to others!
I arrived at a general rearranged equation of 12b = m
So for b = 2:
m = 120 x 2 = 240
For m = 100:
120b = 100
b = 5/6 or 0.83333333
I solved for a general case first:
0 = m - ( b x 120 )
0 = m - 120b
- m = -120b
m / b = 120
Then I solved two possible cases, based on the original m and b values:
If we keep m as 100:
100 = 120 b
100 / 120 = b
b = 5 / 6
if we keep b = 2:
m = 120 x 2
m = 240
To think of it in concrete terms, as long as the max amount of fuel is 120 times bigger than the burn rate it will last for 120 seconds.
Ok here is how I worked it out
c = m -(t b)
c -m =-tb
m -c = tb
t = (m-c)/b
since c == 0 I drop the c here and get
t = m/b
so
t = 120
m = 240 b = 2 120 = 240/2 m = bt
or
m = 120 b=1 b = m/t
t = (m-c)/b
120 = (100-0) / b
b = 100/120 = .83
t = (m-c)/b
120 = (m-0)/2
m = 240
Time = 120
CurrentFuel = 0
Calculate max fuel: (If burn rate = 2)
c = m - bt
c - m = m - bt (Cancel out the m on the right side)
c - m = -bt - c (Cancel out the c on the left side)
-m = -bt - c * -1 (Negate the value)
(b * t) + c = m
Answer:
(2 * 120) + 0 = 240
Calculate burn rate: (If max fuel = 100)
m = bt + c
m - c = bt + c (Cancel out the c on the right side)
m - c = bt
(m - c) / t = bt (Cancel out the t on the right side)
(m - c) = b
Answer:
(100 - 0) / 120 = 0.833
Solved for m. Just input any burn rate b and get any max possible fuel m.
Solved for b. Just input any max possible fuel m and get any posible burn rate b.
Cheers smiley:
With a Max fuel of 120 we need a BurnRate of 1 ;then we would have 0=120-(120x1) that is 120=120 ,so the equation is satisfied.
Good morning everyone!
Here is my submission to the challenge!
-
Finding (m)
if c = m - (bt)
c - m = -(bt)
-1 (c - m) = -1 ( -bt)
-c +m = + bt
m = +c +bt
[ m= c+ bt ] -
Finding (b)
if m= c + bt
m - c = bt
(m - c)/ t = b
[b = (m - c) / t] -
Finding (m) for c = 0 ; t = 120; b = 2
m = 0 + (2 * 120)
= 0 + 240
[ m = 240 ] -
Finding (b) for m = 100; t = 120; c = 0
b = (100 - 0)/ 120
= 100 / 120
[ b = 0.834 ]
m = b*t + c
b = (c - m) / (-t)
Max fuel:
c = m - (b * t)
c - m = - (b * t)
-m = - (b * t) - c
m = (b * t) + c
m = (2 *120) + 0
m = 240
Burn rate:
c = m - (b *t)
c - m = -bt
(c - m) / -t = b
b = (c - m) / -t
b = (0 - 100) / -120
b = 0.833
solve for burn rate (b)
0 = 100 - (b * 120)
-100 = -120b
-100 / -120 = b
0.83 = b
solve for max fuel (m)
0 = m - (2 * 120)
0 = m - 240
0 + 240 = m - 240 + 240
240 = m
Hey Gary!
One quick question for you. On line seven of your Strikethrough example, you have the formula written as…
-1 * (-m) = -1 * (bt - c)
Should it not be…
-1 * (-m) = -1 * (-bt - c)
I know this is pedantic but I want to ensure I’m understanding the formula correctly.
Thanks in advance!
I have no clue if I am doing this correctly. So, here’s my shot. 
Burnrate Adjustment:
(c - m) / (-t) = b
(0 - 100) / (-120) = 0.83 (Burn Rate)
Maxfuel Adjustment:
c + (bxt) = m
0 + (2 x 120) = 240 (Max Fuel)
Variables
MaxFuel(m) = 100
BurnRate(b) = 2
CurrFuel(c) = m - (b*t)
Time(t) = (c-m) / -b #Seconds spent thrusting engines
Max fuel equation
m = c+(b*t)
m = 0+(2*120)
m = 240
To fly for 120 seconds we'd need 240 max fuel.
Burn rate equation
b = (c-m)/-t
b = (100-100)/-120
b = -100/-120
b = -5/-6
b = 0.83333(Repeating)
To fly for 120 seconds we'd need a burn rate of 0.83333(Repeating)
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It was a little bit messy getting there, but here’s my answers for the challenge:
- Given a desired flight time of 120 seconds, solve for m with a burn rate of 2/second and solve for b with a maximum fuel of 100
-
For a flight time of 120 seconds and with a burn time of 2/second, the maximum fuel will need to be 240:
2a. Using the formula t = (m - c) / b
2b. 120 = (m - 0) / 2
2c. 120 = m / 2
2d. 240 = m -
For a maximum fuel of 100 and a flight time of 120 seconds, the burn rate will need to be 5/6 (or about 8.333…)
3a. Using the formula b = (m - c) / t
3b. b = (100 - 0) / 120
3c. b = 100 / 120
3d. b = 5/6 or about 8.333…
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