IBPS SO Prelims Exam Model Paper 10

Let the mixture be of 4 liters and x litresof the mixture must be replaced.
Water =

3

4

×4=3 liters
Alcohol =

1

4

×4=1 liters
Amount of water removed in x liters of mixture =

3

4

×x
Amount of alcohol removed in xliters of mixture =

1

4

×x
Amount of alcohol added =x liters
∴ New amount of water =3−(

3

4

×x)
New amount of alcohol =1−(

1

4

×x)+x
According to the question,

New‌amount‌of‌water

New‌amount‌of‌alcohol

=

1

2

⇒

(3− 3 4 ×x)

(1− 1 4 ×x+x)

=

1

2

⇒6−3

x

2

=1−(

1

4

)×x+x
⇒5=3

x

2

−

x

4

+x
⇒5=9

x

4

⇒x=

20

9

⇒x=2.22
∴Required % =

2.22

4

×100=55.5 %