You are pretty close; b = 100/120, that is not 4/5 right?
Other than that, good job!
Ah, dangit! Itâs funny - that same error got me when I was doing the handwritten part to work it all out, and here it is again! Argh!
Thanks for the catch Ms. Sparkle!
If b = 2 and t = 120, then m = 0+(2*100) = 240;
If m = 100 and t = 120, then b = (0-100)/(-120) = ~0,833
For Java Fans:
import java.util.Scanner;
public class Main {
static void calculate_m() {
float c = 0, b = 0, t = 0;
float result;
Scanner userInput = new Scanner(System.in);
System.out.print("\nPlease enter the value of c (Current Fuel): ");
try {
c = userInput.nextInt();
} catch(Exception e) {
}
System.out.print("\nPlease enter the value of b (Burn Rate): ");
try {
b = userInput.nextInt();
} catch(Exception e) {
}
System.out.print("\nPlease enter the value of t (Time): ");
try {
t = userInput.nextInt();
} catch(Exception e) {
}
result = c+b*t;
System.out.printf("Result: %f%n%n", result);
}
static void calculate_b() {
float c = 0, m = 0, t = 0;
float result;
Scanner userInput = new Scanner(System.in);
System.out.print("\nPlease enter the value of c (Current Fuel): ");
try {
c = userInput.nextInt();
} catch(Exception e) {
}
System.out.print("\nPlease enter the value of m (Maximum Fuel): ");
try {
m = userInput.nextInt();
} catch(Exception e) {
}
System.out.print("\nPlease enter the value of t (Time): ");
try {
t = userInput.nextInt();
} catch(Exception e) {
}
result = (m-c)/t;
System.out.printf("Result: %f%n%n", result);
}
static void calculate_c() {
float b = 0, m = 0, t = 0;
float result;
Scanner userInput = new Scanner(System.in);
System.out.print("\nPlease enter the value of b (Burning Rate): ");
try {
b = userInput.nextInt();
} catch(Exception e) {
}
System.out.print("\nPlease enter the value of m (Maximum Fuel): ");
try {
m = userInput.nextInt();
} catch(Exception e) {
}
System.out.print("\nPlease enter the value of t (Time): ");
try {
t = userInput.nextInt();
} catch(Exception e) {
}
result = m-(b*t);
System.out.printf("Result: %f%n%n", result);
}
static void calculate_t() {
float b = 0, m = 0, c = 0;
float result;
Scanner userInput = new Scanner(System.in);
System.out.print("\nPlease enter the value of b (Burning Rate): ");
try {
b = userInput.nextInt();
} catch(Exception e) {
}
System.out.print("\nPlease enter the value of m (Maximum Fuel): ");
try {
m = userInput.nextInt();
} catch(Exception e) {
}
System.out.print("\nPlease enter the value of c (Current Fuel): ");
try {
c = userInput.nextInt();
} catch(Exception e) {
}
result = (m-c)/ b;
System.out.printf("Result: %f%n%n", result);
}
public static void main(String[] args) {
String input;
Scanner userInput = new Scanner(System.in);
while(true) {
System.out.print("What do you want to calculate? (m, b, c, t)");
try {
input = userInput.next();
} catch(Exception e) {
System.err.println("Error! " + e);
continue;
}
if(input.equals("m") || input.equals("M")) {
calculate_m();
}
if(input.equals("b") || input.equals("B")) {
calculate_b();
}
if(input.equals("c") || input.equals("C")) {
calculate_c();
}
if(input.equals("t") || input.equals("T")) {
calculate_t();
}
}
}
}
Have a nice day 
1 Like
Youâre welcome, keep up the good work!
Enjoy your day!
Hereâs my solution
- Simple, but I couldnât resolve it for longer time
c = m - (b * t)
c + b * t = m
Check
0 + 2 * 120 = 240(m)
- Harder, but not that much
c = m - (b * t)
c - m = - bt â both sides * -1
-c + m = bt â both sides * t
(-c + m) / t = b
Check
(-0 + 240) / 120 = 240/120 = 2(c)
Look ok for me
1 Like
[spoiler] For b=5:
m=600
For m=100:
b=0,833âŚ
Is this correct? Had some issues but I am starting to get it now: [/spoiler]
1 Like
t = 120 c = 0
b = (m - c) / t
b = 0.8333333
m = c - (b * t);
m = 240
1 Like
I struggled with this one a bit so instead of wasting time trying to figure it out without a clue I looked at the answers others gave and spent time working backwards to figure out what the answer meant. I wasnât going to give up on it until I fully understood and I think I got there.
So for m = ? we have the numbers for burn rate at 2 and time at 120 so if we look at the figures once the m = 0 (and c = 0) we can fill out the equation m = c - (b * t) so 0 = 0 - (2 * 120) = 240.
For b = ? we can also go to the end of the process, knowing that the m = 100, c = 0 and t = 120 we can populate the equation b = (m - c) / t so (100 - 0) / 120 = 0.83.
I havenât explained it very well but it was important to me to understand the equations and hopefully understand algebra more moving forward.
In your explanation :
c = m - (b * t)
c - m = -bt â Subtract m from both sides
(c - m) / -t = b â Divide both sides by -t
b = (c - m) / -t â Swap the sides to tidy everything up
Why is b on the 3rd line not a negative?
I ended up with the following:
m = c + bt
For the time of 0 and burn rate of 2, this is 240 needed for max fuel.
And for burn rate:
b = (c - m) / -t
For a time of 120 seconds, this means a burn rate of 0.8333 per second.
It was a fun exercise. The last one took me some double-checking as I had mistakenly tried to equate it to burn rate of 2 from the previous examples⌠but b is what we are looking for!
burn rate 0.833âŚ
max fuel 240
Max fuel M=120
and burn rate b = 1
My Answers
t = (m - c) / b
120 = (120 - 0) / 1 or 120 = (240 - 0) / 2 and so forth
b = (m - c) / t
1 = (120 - 0) / 120 or 2 = (240 - 0) / 120 and so forth
I like how Martin_Castaneda wrote out his response.
Iâll have to try that in my notebook as well just for the extra practice!
1 Like