Dear J, I came across a question which can be solved using weighted avg but I want able to. Kindly have a look.
In a laboratory , three containers A, B, C have equal volumes of different mixtures of liquid oxygen and liquid nitrogen. The concentration(by vol) of liquid Oxygen in different mixtures is 20% , 40% and 80% resp. First 1/5th of the contents of A are poured into B, then 2/5th of the contents of B are poured into C. The final concentration of liquid Oxygen in C is approximately:
a) 62.4 %
b) 61.7 %
c) 68.6 %
d) 65.9 %
I did it this way:
20% (wt : 1/5 ) ————- 40% (wt: 1)
5 : 1 . so avg conc = 40 – (40-20)*1/6 = 40 – 10/3 = 110/3
then:
10/3% (wt : (2/5)*(1+1/5) ) ————- 80% (wt: 1)
(1+ 1/5) because the volume of the resultant mixture has increased.
so ratio becomes: 25 : 12
and the final conc of C comes out to be 80 – 20.8 = 59.6 BUT THE ANSWER IS 65.9
Even if I take the Wt of 10/3% mixture as only 2/5 ,
then also the answer is coming out to be: 67.6
Please help me out.
The mistake is towards the end – you have taken
“so ratio becomes: 25 : 12
and the final conc of C comes out to be 80 – 20.8 = 59.6 ”
The 20.8 is wrong….you have taken 12/25 of the difference but you should have taken 12/(12+25) = 12/37 of the difference….12/37 * (80 – 110/3) = 14.1x which when subtracted from 80 leaves 65.8x ~65.9%.
I did not understand the last two example of weighted average 1 .as how do we get 15 and 9 on dividing 24 into the ratio of 5;3,and so on in last question . plz help sir as soon as possible.
5x + 3x = 24. So 8x = 24. So x = 3. So 5x and 3x are 15 and 9. Knowledge of ratios is a basic prerequisite for applying this logic! So get comfortable with them first.
CAT-holics 
Dear J, I came across a question which can be solved using weighted avg but I want able to. Kindly have a look.
In a laboratory , three containers A, B, C have equal volumes of different mixtures of liquid oxygen and liquid nitrogen. The concentration(by vol) of liquid Oxygen in different mixtures is 20% , 40% and 80% resp. First 1/5th of the contents of A are poured into B, then 2/5th of the contents of B are poured into C. The final concentration of liquid Oxygen in C is approximately:
a) 62.4 %
b) 61.7 %
c) 68.6 %
d) 65.9 %
I did it this way:
20% (wt : 1/5 ) ————- 40% (wt: 1)
5 : 1 . so avg conc = 40 – (40-20)*1/6 = 40 – 10/3 = 110/3
then:
10/3% (wt : (2/5)*(1+1/5) ) ————- 80% (wt: 1)
(1+ 1/5) because the volume of the resultant mixture has increased.
so ratio becomes: 25 : 12
and the final conc of C comes out to be 80 – 20.8 = 59.6 BUT THE ANSWER IS 65.9
Even if I take the Wt of 10/3% mixture as only 2/5 ,
then also the answer is coming out to be: 67.6
Please help me out.
The mistake is towards the end – you have taken
“so ratio becomes: 25 : 12
and the final conc of C comes out to be 80 – 20.8 = 59.6 ”
The 20.8 is wrong….you have taken 12/25 of the difference but you should have taken 12/(12+25) = 12/37 of the difference….12/37 * (80 – 110/3) = 14.1x which when subtracted from 80 leaves 65.8x ~65.9%.
regards
J
Can you please help me understand second part of the sum?
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I did not understand the last two example of weighted average 1 .as how do we get 15 and 9 on dividing 24 into the ratio of 5;3,and so on in last question . plz help sir as soon as possible.
5x + 3x = 24. So 8x = 24. So x = 3. So 5x and 3x are 15 and 9. Knowledge of ratios is a basic prerequisite for applying this logic! So get comfortable with them first.
regards
J
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12/25 1
————————
36.67% x 80%
25 12
25+12= 37 parts are being divided into 43.33%
Hence 1 part will be 43.33/37
and 25 parts will be 43.33*25/37
Thus resultant concentration will be 36.67+29.27=65.94%