In a laboratory, three containers, A, B and C, have equal volumes of

Question

In a laboratory, three containers, A, B and C, have equal volumes of different mixtures of liquid oxygen and liquid nitrogen. The concentration (by volume) of liquid oxygen in the three mixtures is 20%, 40% and 80% respectively. First, one-fifth of the contents of A are poured into B and then two-fifths of the contents of B are poured into C. The final concentration (by volume) of liquid oxygen in C is approximately

(A) 61.7%
(B) 62.4%
(C) 65.9%
(D) 68.6%
(E) 78.9%

Are You Up For the Challenge: 700 Level Questions

(A) 61.7%
(B) 62.4%
(C) 65.9%
(D) 68.6%
(E) 78.9%

Are You Up For the Challenge: 700 Level Questions

D33T · Answer

Can anyone help? I can't wrap my head around how to do this question in a time-efficient manner... it took me >20 mins of trial and error to finally get to the solution but I have the feeling there might be an easier / more straightforward way to do this as the ratios of oxygen in A, B and C are 20, 40 and 80 (2^1, 2^2, 2^3-ish)???

1/5 = 0.2 = 20%, so adding 1/5th of A into B:
(1/5 * 1/5) + 2/5 = 11/25 (amount of oxygen in B')

Adding 2/5th of B' into C:
(2/5 * 11/25) + 4/5 = 122/125 (amt of oxygen in C')

Volume of C': 
If x = original volumes of A B and C, 
2/5 * 6/5x of B' + x of C = 37/25x

Concentration of oxygen in C':
(122/125) / (37/25) = 122/185 = 65.9%

Ans: C

Sajal2408 · Answer

Lets Suppose all the containers have volume = 10L each.

A has 10 L mixture and 2L of Liq. Oxygen
B has 10 L mixture and 4L of Liq. Oxygen
C has 10 L mixture and 8L of Liq. Oxygen

When one-fifth of the contents of A are poured into B then
B(new) total mixture quantity will be = 10L + 20% of 10L = 12 L
B(new) Liq. Oxygen quantity = 4L + 20% of 2L = 4.4L

and

When two-fifths of the contents of B are poured into C.
C(new) total mixture quantity will be = 10L + 40% of 12L = 10L + 4.8L = 14.8L
B(new) Liq. Oxygen quantity = 8L + 40% of 4.4L = 8L + 1.76L = 9.76L

So Final conc. of Liq. Oxygen in Container C will be =9.76/14.8 = 65.9%

Ans  C

KarishmaB · Answer

We can combine the average calculation in one step.

Say each A, B and C have 5 ml of mixes. 
1 ml of A is put in B so B now has 6 ml. Of this 1/6th is A and 5/6th is B. (as learnt in Mixtures)

2/5 of B i.e. (2/5)*6 ml = 12/5 ml of B is put in C. 
In this 12/5 ml, (1/6)*(12/5) = 2/5 ml will be A and rest 10/5 ml will be B. 
Also, there will be 5 ml of C.

Cavg = \frac{20*(2/5) + 40*2+ 80*5}{37/5} = 65.9%

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