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17 Aug 2021, 08:00
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Difficulty:
95%
(hard)
Question Stats:
36% (02:21) correct
64%
(02:06)
wrong
based on 383
sessions
History
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9/20th of water from Jar X is poured to jar Y. After this, does jar Y contain more water than jar X ?
(1) After pouring the water, the amount of water in Jar Y increased by more than 2/3 rds
(2) After pouring the water, the amount of water in Jar Y increased by less than 4/5 ths
M36-24
(1) After pouring the water, the amount of water in Jar Y increased by more than 2/3 rds
(2) After pouring the water, the amount of water in Jar Y increased by less than 4/5 ths
M36-24
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11 Oct 2021, 04:15
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Official Solution:
\(\frac{9}{20}^{th}\) of water from Jar X is poured to jar Y. After this, does jar Y contain more water than jar X ?
Say the amount of water in Jar X is \(x\) units and the amount of water in Jar Y is \(y\) units.
The question asks: is \(x - \frac{9}{20}*x < y + \frac{9}{20}*x\)?, which simplifies into is \(x < 10y\) ?.
(1) After pouring the water, the amount of water in Jar Y increased by more than \(\frac{2}{3}^{rds}\)
\(\frac{9}{20}*x > \frac{2}{3}*y\);
\(x > \frac{40}{27}*y\)
So, \(x\) can be less than \(10y\) as well as more than \(10y\). Not sufficient.
(2) After pouring the water, the amount of water in Jar Y increased by less than \(\frac{4}{5}^{ths}\)
\(\frac{9}{20}*x < \frac{4}{5}*y\);
\(x < \frac{16}{9}*y\)
\(x < \frac{16}{9}*y\) gives an YES answer to the question whether \(x < 10y\). Sufficient.
Answer: B
General Discussion
9/20th of water from Jar X is poured to jar Y. After this, does jar Y contain more water than jar X ?
(1) After pouring the water, the amount of water in Jar Y increased by more than 2/3 rds
(2) After pouring the water, the amount of water in Jar Y increased by less than 4/5 this
Lets solve this through plug in values
initial volume in jar X 20
initial volume in jar Y a
9 lt is transfer to Y
now volume in JAr Y a+9
Statement 1
volume in JAr Y increased by more than 2/3
9=2/3a
a(lhighest value )= 27/2=13.5
so volume in Jar Y will be less than 13.5+9=22.5 & volume in Jar X is 11
Not sufficient
State 2
volume in JAr Y increased by less than 4/5
9=4/5a
a(Lowest value)= 45/4= 11.2
so volume in Jar Y is 11.2+9=22.5 & volume in Jar X is 11
Sufficient
IMO-B
(1) After pouring the water, the amount of water in Jar Y increased by more than 2/3 rds
(2) After pouring the water, the amount of water in Jar Y increased by less than 4/5 this
Lets solve this through plug in values
initial volume in jar X 20
initial volume in jar Y a
9 lt is transfer to Y
now volume in JAr Y a+9
Statement 1
volume in JAr Y increased by more than 2/3
9=2/3a
a(lhighest value )= 27/2=13.5
so volume in Jar Y will be less than 13.5+9=22.5 & volume in Jar X is 11
Not sufficient
State 2
volume in JAr Y increased by less than 4/5
9=4/5a
a(Lowest value)= 45/4= 11.2
so volume in Jar Y is 11.2+9=22.5 & volume in Jar X is 11
Sufficient
IMO-B
