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Difficulty:
95%
(hard)
Question Stats:
46% (02:48) correct
54%
(02:49)
wrong
based on 1055
sessions
History
Date
Time
Result
Not Attempted Yet
On January 1, 2076, Lake Loser contains x liters of water. By Dec 31 of that same year, 2/7 of the x liters have evaporated. This pattern continues such that by the end of each subsequent year the lake has lost 2/7 of the water that it contained at the beginning of that year. During which year will the water in the lake be reduced to less than 1/4 of the original x liters?
A. 2077
B. 2078
C. 2079
D. 2080
E. 2081
A. 2077
B. 2078
C. 2079
D. 2080
E. 2081
22 Apr 2014, 09:54
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jlgdr
On January 1, 2076, Lake Loser contains x liters of water. By Dec 31 of that same year, 2/7 of the x liters have evaporated. This pattern continues such that by the end of each subsequent year the lake has lost 2/7 of the water that it contained at the beginning of that year. During which year will the water in the lake be reduced to less than 1/4 of the original x liters?
A. 2077
B. 2078
C. 2079
D. 2080
E. 2081
Each year 2/7 of the water evaporates, thus after each year 5/7 of the water remains compared to previous year. So:
By the end of 2076: \(\frac{5}{7}*x\) liters of water will be left;
By the end of 2077: \(\frac{5}{7}*\frac{5}{7}*x=\frac{25}{49}*x\approx{\frac{1}{2}*x}\) liters of water will be left;
By the end of 2078: \(\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x=\frac{125}{343}*x\approx{\frac{1}{3}*x}\) liters of water will be left;
By the end of 2079: \(\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x=\frac{625}{2401}*x\approx{\frac{1}{4}*x}\) liters of water will be left (still a bit more than 1/4);
By the end of 2080: \(\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*\frac{5}{7}*x\) liters of water will be left (this must be less than 1/4 since the previous one was just a bit more);
....
Basically after n evaporation the amount of water left is \((\frac{5}{7})^n*x\). Thus the question asks to find the smallest n for which \((\frac{5}{7})^n*x<\frac{x}{4}\).
Answer: D.
Similar questions to practice:
if-2-3-of-the-air-in-a-tank-is-removed-with-each-stroke-of-a-128432.html
if-3-4-of-the-mineral-deposits-in-a-reservoir-of-water-are-97300.html
if-1-2-of-the-air-in-a-tank-is-removed-with-each-stroke-of-a-114943.html
Hope it helps.
19 Aug 2009, 06:40
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Bookmarks
The answer is D
here is the long explanation
This fraction problem contains an "unspecified" total (the x liters of water in the lake). Pick an easy "smart" number to make this problem easier. Usually, the smart number is the lowest common denominator of all the fractions in the problem. However, if you pick 28, you will quickly see that this yields some unwieldy computation.
The easiest number to work with in this problem is the number 4. Let's say there are 4 liters of water originally in the lake. The question then becomes: During which year is the lake reduced to less than 1 liter of water?
At the end of 2076, there are 4 × (5/7) or 20/7 liters of water in the lake. This is not less than 1.
At the end of 2077, there are (20/7) × (5/7) or 100/49 liters of water in the lake. This is not less than 1.
At the end of 2078, there are (100/49) × (5/7) or 500/343 liters of water in the lake. This is not less than 1.
At the end of 2079, there are (500/343) × (5/7) or 2500/2401 liters of water in the lake. This is not less than 1.
At the end of 2080, there are (2500/2401) × (5/7) or 12500/16807 liters of water in the lake. This is less than 1.
Notice that picking the number 4 is essential to minimizing the computation involved, since it is very easy to see when a fraction falls below 1 (when the numerator becomes less than the denominator.) The only moderately difficult computation involved is multiplying the denominator by 7 for each new year.
The correct answer is D.
here is the long explanation
This fraction problem contains an "unspecified" total (the x liters of water in the lake). Pick an easy "smart" number to make this problem easier. Usually, the smart number is the lowest common denominator of all the fractions in the problem. However, if you pick 28, you will quickly see that this yields some unwieldy computation.
The easiest number to work with in this problem is the number 4. Let's say there are 4 liters of water originally in the lake. The question then becomes: During which year is the lake reduced to less than 1 liter of water?
At the end of 2076, there are 4 × (5/7) or 20/7 liters of water in the lake. This is not less than 1.
At the end of 2077, there are (20/7) × (5/7) or 100/49 liters of water in the lake. This is not less than 1.
At the end of 2078, there are (100/49) × (5/7) or 500/343 liters of water in the lake. This is not less than 1.
At the end of 2079, there are (500/343) × (5/7) or 2500/2401 liters of water in the lake. This is not less than 1.
At the end of 2080, there are (2500/2401) × (5/7) or 12500/16807 liters of water in the lake. This is less than 1.
Notice that picking the number 4 is essential to minimizing the computation involved, since it is very easy to see when a fraction falls below 1 (when the numerator becomes less than the denominator.) The only moderately difficult computation involved is multiplying the denominator by 7 for each new year.
The correct answer is D.
