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25 Dec 2024, 07:38
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Let the maximum capacity be 100
Main wallet holds m% and backup wallet holds b%
To solve this we need to keep in mind the following points:
1. b > m
2. Main wallet is filled to 100% capacity
3. Half of b value when added with m should give 100 (i.e, if b=80, then b/2=40, m should be 60, since 60+40 = 100)
So, lets go along the given values -> half of the below value when added with the given option should give 100
30/2 = 15, => 100-15=85, Option not available
45/2 = 22.5 => 100-22.5=77.5, Option not available
50/2 = 25 => 100-25 =75, Option not available
55/2 = 27.5 => 100-27.5=72.5, Option not available
60/2 = 30 => 100-30=70, Option not available
90/2 = 45 => 100-45 = 55, Option is available
So, b=90 and m=55
Main wallet holds m% and backup wallet holds b%
To solve this we need to keep in mind the following points:
1. b > m
2. Main wallet is filled to 100% capacity
3. Half of b value when added with m should give 100 (i.e, if b=80, then b/2=40, m should be 60, since 60+40 = 100)
So, lets go along the given values -> half of the below value when added with the given option should give 100
30/2 = 15, => 100-15=85, Option not available
45/2 = 22.5 => 100-22.5=77.5, Option not available
50/2 = 25 => 100-25 =75, Option not available
55/2 = 27.5 => 100-27.5=72.5, Option not available
60/2 = 30 => 100-30=70, Option not available
90/2 = 45 => 100-45 = 55, Option is available
So, b=90 and m=55
25 Dec 2024, 08:42
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b% > m%
New balance backup: 1/2*b/100*C = b*C/200
New balance main account: C = 100C/100 = C
b = 200 -2*m
Testing Values:
New balance backup: 1/2*b/100*C = b*C/200
New balance main account: C = 100C/100 = C
b = 200 -2*m
Testing Values:
- [color=#000000][size=100]m=55[/size][/color]
- [color=#000000][size=100]b=90[/size][/color]
25 Dec 2024, 08:43
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Let's say that maximum capacity is 100. Then: \(m=\frac{m}{100}\) and \(b=\frac{b}{100}\)
After the transfer of x, we get \(\frac{b}{2}=b-\frac{x}{100} \) and \(m+x=100\) (as the main wallet was completely full)
Remembering that \(b=\frac{b}{100}, \frac{b}{2}=\frac{b}{200} \)
=> then \(b-\frac{x}{100}=\frac{b}{200}, 2(b-x)=b, 2b-2x=b \) and \(x=\frac{b}{2}\)
Inserting b into the equation for main wallet: \(100=m+x=m+\frac{b}{2}\)
And multiplying it by 2, we get \(2m+b=200 \) and \(b=200-2m\)
As we know that \(b>m\), and trying a couple of values for both, we're seeing that the values suitable are \(90=200-2*55\)
Therefore, m=55 and b=90
After the transfer of x, we get \(\frac{b}{2}=b-\frac{x}{100} \) and \(m+x=100\) (as the main wallet was completely full)
Remembering that \(b=\frac{b}{100}, \frac{b}{2}=\frac{b}{200} \)
=> then \(b-\frac{x}{100}=\frac{b}{200}, 2(b-x)=b, 2b-2x=b \) and \(x=\frac{b}{2}\)
Inserting b into the equation for main wallet: \(100=m+x=m+\frac{b}{2}\)
And multiplying it by 2, we get \(2m+b=200 \) and \(b=200-2m\)
As we know that \(b>m\), and trying a couple of values for both, we're seeing that the values suitable are \(90=200-2*55\)
Therefore, m=55 and b=90
