Management Aptitude Test Sep 2020 Paper

Question : 107 of 200

Marks: +1, -0

Suppose that a maximum of

25 {g}

of salt dissolves in

100 {g}

of water. Any more salt, if added, remains undissolved and a sediment falls at the bottom. Now, water is evaporated from

1 {kg}

of a

4 \%

solution at the rate of

28 {g}

per hour. After how long will it start sedimenting, approximately?

Solution:

Maximum

25 {g}

of salt dissolves in

100 {g}

water, after that sedimentation start.
Amount of salt in given solution

=1000 × \;{4}/{100}=40 {g}

∴

Minimum amount of water required after which sedimentation starts

=\;{40}/{25} × 100=160 {g}

∴ \;\;

Minimum

840 {g}

needs to evaporated. Time required to evaporated

840 {g}

of water

=\;{840}/{28}=30 {h}

∴

After

31 {h}

, the given solution starts sedimenting.

Go to Question:

Prev Question

Next Question