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Each of the three tanks holds \(n\) liters of water. After some water is redistributed from one tank into the other two, the new ratio of water becomes 1:6:8. What amount of water, in terms of \(n\), was poured into the second tank?
A. \(\frac{n}{15}\)
B. \(\frac{2n}{15}\)
C. \(\frac{n}{5}\)
D. \(\frac{4n}{15}\)
E. \(\frac{n}{3}\)
A. \(\frac{n}{15}\)
B. \(\frac{2n}{15}\)
C. \(\frac{n}{5}\)
D. \(\frac{4n}{15}\)
E. \(\frac{n}{3}\)
17 May 2025, 12:25
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Official Solution:
Each of the three tanks holds \(n\) liters of water. After some water is redistributed from one tank into the other two, the new ratio of water becomes 1:6:8. What amount of water, in terms of \(n\), was poured into the second tank?
A. \(\frac{n}{15}\)
B. \(\frac{2n}{15}\)
C. \(\frac{n}{5}\)
D. \(\frac{4n}{15}\)
E. \(\frac{n}{3}\)
After the redistribution, the new ratio of water becomes \(x:6x:8x\), so the total amount of water is \(x + 6x + 8x = 15x\) liters. Assuming \(x = 1\), the total volume is 15 liters, which means each tank initially held 5 liters (\(n = 5\)). After redistribution, the second tank holds 6 liters, so it received 1 liter from tank 1. Substituting \(n = 5\) into the options, we find that option C (\(\frac{n}{5}\)) yields 1 liters.
Answer: C
Each of the three tanks holds \(n\) liters of water. After some water is redistributed from one tank into the other two, the new ratio of water becomes 1:6:8. What amount of water, in terms of \(n\), was poured into the second tank?
A. \(\frac{n}{15}\)
B. \(\frac{2n}{15}\)
C. \(\frac{n}{5}\)
D. \(\frac{4n}{15}\)
E. \(\frac{n}{3}\)
After the redistribution, the new ratio of water becomes \(x:6x:8x\), so the total amount of water is \(x + 6x + 8x = 15x\) liters. Assuming \(x = 1\), the total volume is 15 liters, which means each tank initially held 5 liters (\(n = 5\)). After redistribution, the second tank holds 6 liters, so it received 1 liter from tank 1. Substituting \(n = 5\) into the options, we find that option C (\(\frac{n}{5}\)) yields 1 liters.
Answer: C
30 Jan 2026, 02:32
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The answer provided by Bunuel was very helpful to build the kind of thinking that is required in a timed environment. Although I got it wrong because I approached the problem algebraically, and because of the time constraint, I had to mark i,t and that was a wrong guess.
After the test, I resolved this one with the method I previously used just to check whether I am arriving at the correct answer or not, and here is my approach:
Given Info before redistribution
T1:T2:T3=>n:n:n i.e 1:1:1
After redistribution n-x:n+a:n+b, where x=a+b=>a=x-b....(need to find)
so new T1:T2:T3=> n-x:n+a:n+b => 1:6:8
Solving the ratios
(n-x)/n+a=1/6
(n-x)/(n+x-b)=1/6
solving this we get x=(b+5n)/7....1
next, (n+a)/(n+b)=6/8
subsituting a=x-b
we get x=(7b-n)/4....2
equating 1 & 2, we get
b=3n/5
substituing b in eqn(1) we get
x=4n/5
since we are required to find the value of a, check above
so a=x-b=4n/5-3n/5=n/5
Hope this helps. and Thanks bunuel
After the test, I resolved this one with the method I previously used just to check whether I am arriving at the correct answer or not, and here is my approach:
Given Info before redistribution
T1:T2:T3=>n:n:n i.e 1:1:1
After redistribution n-x:n+a:n+b, where x=a+b=>a=x-b....(need to find)
so new T1:T2:T3=> n-x:n+a:n+b => 1:6:8
Solving the ratios
(n-x)/n+a=1/6
(n-x)/(n+x-b)=1/6
solving this we get x=(b+5n)/7....1
next, (n+a)/(n+b)=6/8
subsituting a=x-b
we get x=(7b-n)/4....2
equating 1 & 2, we get
b=3n/5
substituing b in eqn(1) we get
x=4n/5
since we are required to find the value of a, check above
so a=x-b=4n/5-3n/5=n/5
Hope this helps. and Thanks bunuel
Bunuel
