Latest Elitmus Aptitude Question SOLUTION: How many liters of 20% alcohol should be mixed with 60 % alcohol to get 32% alcohol (1) We get 70 liters of 32% alcohol (2) We used 21 lit of 60% alcohol, to prepare 32% A if the

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(#M40152815) ELITMUS Numerical Ability question Data sufficiency Keep an EYE

How many liters of 20% alcohol should be mixed with 60 % alcohol to get 32% alcohol
(1) We get 70 liters of 32% alcohol
(2) We used 21 lit of 60% alcohol, to prepare 32%
A if the question can be answered by using one of the statements alone, but cannot be answered by using the other statements alone
B if the question can be answered by using either statement alone
C if the question can be answered by using both statements together,but cannot be answered by using either statement alone
D if the question cannot be answered even by using both the statements together

Asked In Elitmus JUGAL KHAJURIYA (8 years ago)

Unsolved Read Solution (4)

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pranay  (8 years ago)

(B)
20 60
32
28 12
so the mixture will be in 28:12 ratio => 7:3 ratio
(1) if we have 70 liters of 32% alcohol that means we have 70*(7/(7+3)) liter = 49 liter of 20% alcohol
(2) if we have 21 liter of 60% alcohol that means 3x = 21(from the ratio)
=> x = 7
so we have 7 * 7(from ratio) liter of 20% of alcohol

so both statement alone is sufficient
that means Ans. B is correct

Is this solution Helpfull? Yes (28) | No (2)  

pranay  (8 years ago)

(B)
20 60
32
28 12
so the mixture will be in 28:12 ratio => 7:3 ratio
(1) if we have 70 liters of 32% alcohol that means we have 70*(7/(7+3)) liter = 49 liter of 20% alcohol
(2) if we have 21 liter of 60% alcohol that means 3x = 21(from the ratio)
=> x = 7
so we have 7 * 7(from ratio) liter of 32% of alcohol

so both statement alone is sufficient
that means Ans. B is correct

Is this solution Helpfull? Yes (5) | No (1)  

mushahid khan  (8 years ago)

option c..
the equation formed will be
0.2x+ 0.6y = .32z
where x is amout of 20% sol
y is amount of 60% sol
and z is amount of 32% sol
we can only find x by putting both the values of y and z from both the statements...

Is this solution Helpfull? Yes (1) | No (4)  

Sayantani Maiti  (6 years ago)

From statement 1 : The final output is 70 ltr 32% alcohal
so the final alcohal content is : 70 x .32 ltr = 22.4 ltr
From statement 2: 21 litre 60% alcohal was used.
so the initial alcohal content was: 21 x .6 = 12.6
The rest alcohal (22.4 - 12.6) = 9.8 ltr comes from 20% content.
The 20% content alcohal volume should be from (70 - 21) = 49 ltr
so its 20% alcohal will be .2 x 49 = 9.8 ltr hence satisfied
So both the statement are sufficient . [C]

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