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10 Jul 2025, 04:45
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Option A is the correct answer. IMO, easiest way to solve this is using, plugin method.
Let amount on Monday be 100 and m be 100%
Amount on Tuesday = 100 + 100% * 100 = 200
Amount on wendesday = 100/2 = 50 = 200 - n% * 200
so, n = 75%
Putting m = 100 , in the options, we get n = 75 from , option A.
Let amount on Monday be 100 and m be 100%
Amount on Tuesday = 100 + 100% * 100 = 200
Amount on wendesday = 100/2 = 50 = 200 - n% * 200
so, n = 75%
Putting m = 100 , in the options, we get n = 75 from , option A.
Bunuel
10 Jul 2025, 05:06
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money * (1+m/100) * (1-n/100) = 50*money/100
(m+100)/100 * (100-n)/100 = 1/2
(m+100) * (100-n) = 5000
100-n = 5000/(m+100)
n = 100 - 5000/(m+100) = (100m+10000-5000)/(m+100) = (100m+5000)/(m+100) = 100(m+50)/(m+100)
The right answer is A
(m+100)/100 * (100-n)/100 = 1/2
(m+100) * (100-n) = 5000
100-n = 5000/(m+100)
n = 100 - 5000/(m+100) = (100m+10000-5000)/(m+100) = (100m+5000)/(m+100) = 100(m+50)/(m+100)
The right answer is A
10 Jul 2025, 05:08
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Say Elena had $\(x\) in her account on Monday.
On Tuesday, the amount increased by \(m\)%.
Tuesday amount \(= (1+\frac{m}{100})x\)
On Wednesday, she withdrew \(n\)%
Wednesday amount \(= (1-\frac{n}{100})(1+\frac{m}{100})x\)
Wednesday amount is \(50\)% of monday amount.
\((1-\frac{n}{100})(1+\frac{m}{100})x=\frac{50}{100}x\)
\((\frac{100-n}{100})(\frac{100+m}{100})=1/2\)
\(n=\frac{100(m+50)}{100+m}\)
On Tuesday, the amount increased by \(m\)%.
Tuesday amount \(= (1+\frac{m}{100})x\)
On Wednesday, she withdrew \(n\)%
Wednesday amount \(= (1-\frac{n}{100})(1+\frac{m}{100})x\)
Wednesday amount is \(50\)% of monday amount.
\((1-\frac{n}{100})(1+\frac{m}{100})x=\frac{50}{100}x\)
\((\frac{100-n}{100})(\frac{100+m}{100})=1/2\)
\(n=\frac{100(m+50)}{100+m}\)
