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The ratio of water and ammonia in solution A is 1 : 4; the ratio of

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The ratio of water and ammonia in solution A is 1 : 4; the ratio of [#permalink]

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The ratio of water and ammonia in solution A is 1 : 4; the ratio of the same two substances in solution B is 3 : 1. If solutions A and B are mixed, what will be the ratio of water and ammonia in the resultant solution?

(1) Solutions A and B are mixed in ratio 2 : 3.
(2) 20 liters of solution A is used
[Reveal] Spoiler: OA

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Re: The ratio of water and ammonia in solution A is 1 : 4; the ratio of [#permalink]

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nkmungila wrote:
The ratio of water and ammonia in solution A is 1 : 4; the ratio of the same two substances in solution B is 3 : 1. If solutions A and B are mixed, what will be the ratio of water and ammonia in the resultant solution?

(1) Solutions A and B are mixed in ratio 2 : 3.
(2) 20 liters of solution A is used



Hi...
we have ratio of two items in two different solutions and they are mixed..

we are to find the ratio of items in RESULTANT solution..
here we are not finding the EXACT quantity so even the ratio of the solutions in MIX will give us our answer..


Statement I gives us the ratio of the two solutions so Sufficient..
Statement II is to make you fall in C trap

A
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Re: The ratio of water and ammonia in solution A is 1 : 4; the ratio of [#permalink]

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New post 25 Oct 2017, 11:22
nkmungila wrote:
The ratio of water and ammonia in solution A is 1 : 4; the ratio of the same two substances in solution B is 3 : 1. If solutions A and B are mixed, what will be the ratio of water and ammonia in the resultant solution?

(1) Solutions A and B are mixed in ratio 2 : 3.
(2) 20 liters of solution A is used



Let the amount of A used be X and Amount of B is Y.

water in A = 1/5* X and Ammo in A= 4/5*Y

Similarly Water in B= 3/4 * Y and Ammo+=1/4*Y

when we mix A & B :

Total amount of water = (1/5*X + 3/4*Y)
Total amount of Ammo = (4/5*X + 1/4*Y)

After Dividing both the equation by "Y" ; we need only value of X/Yin order to find the ratio.

So statement 1 sufficient

So answer is A

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Re: The ratio of water and ammonia in solution A is 1 : 4; the ratio of [#permalink]

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New post 28 Oct 2017, 09:28
nkmungila wrote:
The ratio of water and ammonia in solution A is 1 : 4; the ratio of the same two substances in solution B is 3 : 1. If solutions A and B are mixed, what will be the ratio of water and ammonia in the resultant solution?

(1) Solutions A and B are mixed in ratio 2 : 3.
(2) 20 liters of solution A is used


Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

When a question asks a ratio, if one condition is about a ratio and the other condition is just a number, then the condition about a ratio would be sufficient. Thus A could be the answer most likely for this quesiton.

The first step of VA(Variable Approach) method is modifying the original condition and the question, and rechecking the number of variables and the number of equations.
We can modify the original condition and question as follows.

Assume \(Aw\) and \(Aa\) are amounts of water and ammonia in Solution A and \(Bw\) and \(Ba\) are assumed in the similar way.

\(Aw : Aa = 1:4\) => \(Aw = x\) and \(Aa = 4\)x
\(Bw : Ba = 3:1\) => \(Bw = 3y\) and \(Ba = y\)

Assume \(a\) and \(b\) are amounts of solutions A and B, respectively.
\(a = 5x\) and \(b = 4y\)

The question asks \(( a*Aw + b*Bw ) / (a*Aa + b*Ba) = (ax + by) / ( 3ax + 3by ) = ( 5x^2 + 4y^2 ) / ( 15x^2 + 12y^2 )\).

Condition 1)
\(a : b = 2 : 3\)
\(5x : 4y = 2 : 3\)
\(15x = 8y\)
\(x/y = 8/15\)

\(( 5x^2 + 4y^2 ) / ( 15x^2 + 12y^2 )\)
\(= ( 5(x/y)^2 + 4 ) / ( 15(x/y)^2 + 12 )\)
\(= ( 5(8/15)^2 + 4 ) / ( 15(8/15)^2 + 12\)).
This is sufficient.

Condition 2)
\(a = 20\)
\(x = 4\)
Since we don't know b or y, this is not sufficient.

Then answer is A
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Re: The ratio of water and ammonia in solution A is 1 : 4; the ratio of   [#permalink] 28 Oct 2017, 09:28
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