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08 May 2018, 23:12
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D
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Jar 1 contains a solution which is 30% nitric acid and 70% water, while Jar 2 contains a solution which is 60% nitric acid and 40% water. How many ml of Jar 1 should be mixed with 100 ml of Jar 2 solution, in order to form an intended solution S?
(1) Solution S is intended to be 40% nitric acid and 60% water.
(2) If similar quantities of solution S and the solution in Jar 1 are taken, then solution S will contain 4/3 times the nitric acid than that present in Jar 1.
(1) Solution S is intended to be 40% nitric acid and 60% water.
(2) If similar quantities of solution S and the solution in Jar 1 are taken, then solution S will contain 4/3 times the nitric acid than that present in Jar 1.
08 May 2018, 23:39
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Choice D.
Statement 1: We know the resultant solution S has 40% nitric acid. Using rule of alligation, we can determine the ratio in which Jar 1 and Jar 2 have to be mixed as 2 : 1. So, we need to mix 200 ml of Jar 1 with 100 ml of Jar 2. The attachment has the working of the rule of alligation.
Alternatively, we can find the ratio by framing linear equations.
Let x ml of Jar 1 and y ml of Jar 2 be mixed to get solution s.
30% of x is nitric acid and 60% of y is nitric acid. So, resultant nitric acid after mixing the two jars will be (0.3x + 0.6y) ml .. (1)
Total volume of solution s is (x + y) ml and 40% of it is nitric acid. So, we can infer that 0.4(x + y) is the volume of nitric acid in solution s. ..(2)
Equating (1) and (2), 0.3x + 0.6y = 0.4x + 0.4y
Or 0.1x = 0.2y
or x/y = 2
2 parts of x and 1 part y.
Volume of y is 100 ml. So, volume of x is 200 ml.
So, statement 1 alone is sufficient.
Statement 2: We can deduce that 4/3 of 30% = 40%. The process to determine the volume of Jar 1 is the same as outlined for statement 2.
Statement 2 alone is also sufficient.
Choice D.
Statement 1: We know the resultant solution S has 40% nitric acid. Using rule of alligation, we can determine the ratio in which Jar 1 and Jar 2 have to be mixed as 2 : 1. So, we need to mix 200 ml of Jar 1 with 100 ml of Jar 2. The attachment has the working of the rule of alligation.
Alternatively, we can find the ratio by framing linear equations.
Let x ml of Jar 1 and y ml of Jar 2 be mixed to get solution s.
30% of x is nitric acid and 60% of y is nitric acid. So, resultant nitric acid after mixing the two jars will be (0.3x + 0.6y) ml .. (1)
Total volume of solution s is (x + y) ml and 40% of it is nitric acid. So, we can infer that 0.4(x + y) is the volume of nitric acid in solution s. ..(2)
Equating (1) and (2), 0.3x + 0.6y = 0.4x + 0.4y
Or 0.1x = 0.2y
or x/y = 2
2 parts of x and 1 part y.
Volume of y is 100 ml. So, volume of x is 200 ml.
So, statement 1 alone is sufficient.
Statement 2: We can deduce that 4/3 of 30% = 40%. The process to determine the volume of Jar 1 is the same as outlined for statement 2.
Statement 2 alone is also sufficient.
Choice D.
Attachments
File comment: How to set up rule of alligation?
Rule-of-alligation.png [ 70.42 KiB | Viewed 8049 times ]
03 Jun 2018, 03:54
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amanvermagmat
Given
Jar 1 - 30% Nitric acid & 70% water
Jar 2 - 60% Nitric acid & 40% water.
Quantity of Jar 2 = 100ml
Quantity of Jar 1 = X ml
No info in prompt about intended solution S.
Lets look at St.1 Alone
Intended Solution S = 40% Nitric acid & 60% water.
Quantity of Solution S = (100+X) ml
Hence Quantity of Nitric acid = Quantity of Nitric acid in Jar 1 + Quantity of Nitric acid in Jar 2
40%*(100+X) = 30% of X + 60% of 100
Equation in single unknown, solving which we can deduce Quantity of Jar 1 hence, statement 1 is Sufficient.
St. 2 Alone
When similar quantities of Solution S & Jar 1 are taken, the Nitric acid quantity in Solution 4/3 times of Nitric acid in Jar 1.
which means, if 100 ml of Solution S & Jar 1 are taken,
Quantity of Nitric acid in Jar 1 is 30% of 100 ml = 30ml
Therefore, Quantity of Nitric acid in Solution S = 4/3*(30) = 40 ml
Hence Solution S contains 40% Nitric acid & 60% water.
Which is similar to info provided in St.1, solving which we can deduce Quantity of Jar 1. Statement 2 is Sufficient.
St 1 & St 2 alone are Sufficient.
Answer D.
Thanks,
GyM
