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
OA: D
Source: MGMAT
is there an easy approach to this? what would be the fastest way to tackle this problem?
lake loser
This topic has expert replies
- sanju09
- GMAT Instructor
- Posts: 3650
- Joined: Wed Jan 21, 2009 4:27 am
- Location: India
- Thanked: 267 times
- Followed by:80 members
- GMAT Score:760
If 2/7 evaporates in a periodic manner, then 5/7 is left in the same periodic manner, each year.rahul.s wrote: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
OA: D
Source: MGMAT
is there an easy approach to this? what would be the fastest way to tackle this problem?
On January 1, 2076 = x
On December 31, 2076 = (5/7) x
On December 31, 2077 = (5/7) (5/7) x
On December 31, 2078 = (5/7) (5/7) (5/7) x
On December 31, 2079 = (5/7) (5/7) (5/7) (5/7) x
WAIT...
Since, 5/7 is very close to 0.7, then (5/7) (5/7) is very close to 0.5, and hence (5/7) (5/7) (5/7) (5/7) is very close to 0.25 or ¼. See, this happened as on December 31, 2079. So from now and onwards, the water in the lake will be reduced to less than ¼ of the original x liters, and the applicable first year will be [spoiler]2080[/spoiler].
[spoiler]D[/spoiler]
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
sanju09 wrote:If 2/7 evaporates in a periodic manner, then 5/7 is left in the same periodic manner, each year.rahul.s wrote: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
OA: D
Source: MGMAT
is there an easy approach to this? what would be the fastest way to tackle this problem?
On January 1, 2076 = x
On December 31, 2076 = (5/7) x
On December 31, 2077 = (5/7) (5/7) x
On December 31, 2078 = (5/7) (5/7) (5/7) x
On December 31, 2079 = (5/7) (5/7) (5/7) (5/7) x
WAIT...
Since, 5/7 is very close to 0.7, then (5/7) (5/7) is very close to 0.5, and hence (5/7) (5/7) (5/7) (5/7) is very close to 0.25 or ¼. See, this happened as on December 31, 2079. So from now and onwards, the water in the lake will be reduced to less than ¼ of the original x liters, and the applicable first year will be [spoiler]2080[/spoiler].
[spoiler]D[/spoiler]
One more way as looks more complicated in calculations,
So, Take a LCM of 7 (from 2/7 as the reduced quantity for every year) and 4 ( need to check the conditions which is less than 1/4) .
LCM will come as 28 ( to make easy calculations)
During which year will the water in the lake be reduced to less than 1/4 of the original x liters?
so we need to check in which it will come as less than 7
If we calculate approximately will come as 5 years which include 2076. Answer would be 2080
Hope this save some time.
-
- Master | Next Rank: 500 Posts
- Posts: 222
- Joined: Mon Oct 13, 2008 4:04 pm
- Thanked: 3 times
- Followed by:2 members
I would do it this way
Let x= LCM(4,7) = 28
so let X = 2800 After how many years will it become 1/4(2800) = 700
So therefore n(700)= 2800 = 4
so after 4 years
2077+ 2078_2079+2080
answer 2080
D
Let x= LCM(4,7) = 28
so let X = 2800 After how many years will it become 1/4(2800) = 700
So therefore n(700)= 2800 = 4
so after 4 years
2077+ 2078_2079+2080
answer 2080
D