Stealing Wine Thrice
There was a thief who went to a wine shop to steal wine. He went in at 10 pm, stole 15 L of wine and added 15 L of water to top up the barrel. He came back at 1 am, stole 15 L of the mixture from the same barrel and added 15 L of water to top it up. He came back at 4 am, stole 15 L of the mixture from the same barrel and added 15 L of water to top it up. In the morning, the ratio of wine to water in the barrel was 343:169. Find the initial amount of wine in the barrel.
The answer is 120.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
8 solutions
could you explain why after 3 thefts the ratio become ( x − 1 5 ) 3 / x 3 ?
Log in to reply
At last the ratio of wine is (343/512)% of initial amount of wine.
Let, Initial amount of wine is X.
After first action the wine is {(X-15)/X}% of Initial wine.
After second action the wine is {(X-15)/X}^2 % of Initial wine.
After third action the wine is {(X-15)/X}^3 % of Initial wine.
Log in to reply
I mean, how come you just raise the power of ( x − 1 5 ) / x to a certain value? rather than subtracting the amount of x on the numerator, like ( x − 1 5 ) / x become ( ( x − 1 5 ) − 1 5 ) / x ) after the 2nd theft
Log in to reply
@Muhammad Ridwan Apriansyah B. – Here's an alternative way of explaining it. Let V be the initial volume of the wine. Also let V n be the volume of the wine after the nth steal. One can use a recursive formula to determine the V n : V n = V n − 1 − 1 5 V V n − 1 with V n − 1 to denote the volume of wine just after the previous steal.
Factoring out V n − 1 results in a recursive formula similar to a geometric sequence: V n = V n − 1 ( 1 − V 1 5 )
Starting from n = 3 , we follow the formula until we arrive at the equation: V 3 = V 0 ( 1 − V 1 5 ) 3
But V 0 = V , thus the volume for wine after 3 steals is : V 3 = V ( 1 − V 1 5 ) 3 .
But V is also the total volume of water and wine after 3 steals, and the ratio of wine volume to total volume is V t o t a l V 3 = 1 6 9 + 3 4 3 3 4 3 = 5 1 2 3 4 3 .
Thus we end up with: ( 1 − V 1 5 ) 3 = 5 1 2 3 4 3
or we can write this as: ( V V − 1 5 ) 3 = ( 8 7 ) 3
Log in to reply
@Ray John Pamillo – wow, i get it now. thank you so much!
@Ray John Pamillo – a nice way of generalizing the equations.thank you.
@Ray John Pamillo – awesome! I could have finished this problem in much shorter time
@Ray John Pamillo – Still don get that with the power of three.
@Muhammad Ridwan Apriansyah B. – If you just subtract like how you have mentioned, its like subtracting 30 from X(Initial ) for the second theft. which is not true.!
Nice, I didn't see that 7 3 / 8 3 trick. I ended up expanding everything.
This is a great solution.
i didnt thought of taking ratio of wine from whole
awesome.......gr8 idea
This problem falls into a cool theorem that I found in a very old book that my dad used in his school days (Jadav Chakrabarti's Arithmetic).
Let there be X amount of wine in the barrel initially.
When the thief first removes 15 L of wine and adds the same amount of water, there is (X-15) L of wine left in the wine-water mixture.
Subsequently, when he removes 15 L again, it is from this wine-water mixture
Thus he will have removed 15*(X-15)/X of wine this time (from the unitary method)
Hence, the amount of wine left in the barrel is now X - 15*(X-15)/X or (X-15)^2/X
Similarly, after the next 15 L is removed there will be (X-15)^3/ X^2 amount of wine left.
Thus the amount of water in the mixture is X minus the quantity above.
Equating the ratio to 343/169 it can be easily found that X = 120!!
I do not get this. Formula after the first theft is somehow clear but I dont get why it gets to the power or 3 after 3 thefts.
The fraction of wine remaining in the bottle after ' n ' operations would be: ( 1 − P Q ) n .... where, ' P ' is the initial amount, ' Q ' is the replaced volume.
Here: Q = 1 5 ; n = 3 ; fraction remaining = 5 1 2 3 4 3 = ( 8 7 ) 3
equating, we have: 1 − P Q = 8 7
P = 1 2 0
well the formatting guide didn't work :(
There is a simple formula to it ::: [(P-x)/P]^n = (concentration of wine/Total mixture concentration) where, P = Original quantity of Wine, x = Quantity of mixture removed, n = no. of times the mixture is removed
Using it we need to find the value of P :- [(P- 15)/P]^3 = [343/(343+169)] It implies, P-15/P = 7/8
Solving it the value of P = 120
(7/8)^3=343/512=343/(343+169); (x-15)/x=7/8; then x=120.
At last the ratio of wine is (343/512)% of initial amount of wine.
Let, Initial amount of wine is X.
After first action the wine is {(X-15)/X}% of Initial wine.
After second action the wine is {(X-15)/X}^2 % of Initial wine.
After third action the wine is {(X-15)/X}^3 % of Initial wine.
So, {(X-15)/X}^3 = 343/512
Or, (X-15)/X = cube root of (343/512)
Or, (X-15)/X = 7/8
Or, 8X-120 = 7X Or, X = 120
the % symbol should be removed from all the steps, as it is only ratio, not %ge
After getting the required equations no one would like to solve this the genuine way ...
We can use the formula of mixture
X [(1- (Y/X))^n] where x is the volume of the wine at the first, y is the volume of wine that were replaced by water, n is the amount of the process that has been done.
It lead to a result (x-15)^3=343/512x^3 X-15=7/8x so x=120
Hope u understand
let initial amount be x.
purity of wine after 1 theft= (x-15)/x.
after 3 thefts = ( x − 1 5 ) 3 / x 3 .
final ratio of wine to water = 343 to 169.
thus ratio of wine to total = 343 to 343+169=343 to 512.
( x − 1 5 ) 3 / x 3 = 7 3 / 8 3 .
(x-15)/x = 7/8.
solving x = 120