There is a 120 liter mixture of alcohol and water. The ratio of alcoho

Question

There is a 120 liter mixture of alcohol and water. The ratio of alcohol to water is 7 : 5. A shopkeeper mixes a certain amount of water in order to make the ratio of alcohol to water as 5 : 6. Find the new quantity of water is what percentage of original quantity of water in the mixture?

(A) 160%
(B) 168%
(C) 172%
(D) 175%
(E) 178%

(A) 160%
(B) 168%
(C) 172%
(D) 175%
(E) 178%

yashikaaggarwal · Accepted Answer

There is a 120-litre mixture of alcohol and water. The ratio of alcohol to water is 7:5.

So, there are 70 litres of alcohol and 50 litres of water in the solution.

A shopkeeper mixes a certain amount of water to make the ratio of alcohol to water as 5:6.

Here, 70 litres of alcohol is 5/11 part of the new solution
Then the quantity of new solution => 70*11/5 => 154 litres, which has 84 litres of water.

To calculate, the new quantity of water percentage of the original quantity of water in the new mixture
Calculate 84 litres of water is what percentage of 50 litres of the original solution.

84 = 50*x/100
X = 168

Answer is B.

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binzdelabinz · Answer

(7/5) (x) = 5/6
(7/5)(6/5) = x 
x = 42/25=168/100 (B)

onlyPlanA · Answer

Initial amount of alcohol = (7/12) * 120 = 70lt
Initial amount of water = (5/12) * 120 = 50lt

\frac{7}{12} * 120 = \frac{5}{11} * (120 + x)  
x = 34

So, new percentage of water =  \frac{6}{11} * (120 + 34)  = 84

Hence, we want the value of =  \frac{84}{50} * 100  = 168%

Riyaagoell · Answer

Can you please explain from alligation?

KarishmaB · Answer

When dealing with volume of parts, direct algebra is typically more straight forward. Still, here is the weighted averages approach: In the original solution, alcohol concentration is 7/12. In water, the alcohol concentration is 0. When they are mixed, alcohol concentration is 5/11. \frac{w1}{w2 }= \frac{(0 - 5/11)}{(5/11 - 7/12)} = \frac{60}{17} Hence for every 60 liter of original solution, 17 liters of water was added. The 60 liter original solution would have 5/12th of water which means that the 60 liter solution will have 25 liters of water. So new quantity of water is 17 + 25 = 42 which is 42/25 * 100 =168% of original solution. Answer (B) Weighted Averages Approach: https://anaprep.com/arithmetic-weighted-averages/ https://anaprep.com/arithmetic-mixtures/

egmat · Answer

Hi Riyaagoell,

Every solution in the thread got to  84 L  of water and  168% , just through direct ratios. You want the same result from  alligation , so let me set that up. The key idea: alligation compares  concentrations , so first pick one component to track and treat this as mixing two things - the  original mixture  and the  pure water  you pour in.

Track  alcohol  as a fraction of the whole:

- Original mixture: alcohol =  7/12 
- Water being added: alcohol =  0  (it's pure water)
- Final mixture: alcohol =  5/11

The alligation cross

Put the two ingredients on top, the target in the middle, and take the differences on the diagonals:

- Arm for the  original mixture  = (target - water) =  5/11 - 0 = 5/11 
- Arm for the  added water  = (mixture - target) =  7/12 - 5/11 = 17/132

So the ratio of  mixture : added water  is:

5/11 : 17/132 = 60/132 : 17/132 = 60 : 17

Turn parts into liters

The original mixture is  120 L , and that is the  60  part. So one part =  120 ÷ 60 = 2 L .

Added water =  17 parts = 17 × 2 = 34 L .

Now finish exactly as the thread did:

- Original water =  50 L 
- New water = 50 + 34 =  84 L 
-  84 ÷ 50 = 1.68 = 168%  -  B

The one thing to remember with alligation here: because you're adding a  pure  component, its concentration is simply  0  - that's what makes the second arm work. Everything else is the same cross you'd use for any two-ingredient mix.

Answer: B

sset92 · Answer

I find using ratios as a simple method when one quantity remains unchanged.

In this case 120 liter mixture is in the ration A:W as 7:5. 
Adding ratios gives us 12k so clearly the mixture is 70 liter A and 50 liter W

New ratio is 5:6. which can also be written as 50:60. 
Since Alcohol value has not changed lets find LCM and get both values of A same = 350

Therefore we have initial ratio as 350:250 and new ratio as 350:420.

New Quantity of water / Old quantity of water = 420/250 which gives us 168%

There is a 120 liter mixture of alcohol and water. The ratio of alcoho

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There is a 120 liter mixture of alcohol and water. The ratio of alcoho

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