85% and 92% alcoholic solutions are mixed to get 35 litres of an 89% alcoholic solution. How many litres of each solution are there in the new mixture?
Aptitude
Percentage
Difficulty: Medium
Choose an option
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A10 of the first and 25 of the second
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B20 of the first and 15 of the second
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C15 of the first and 20 of the second
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D12 of the first and 23 of the second
Answer
Correct Answer: 15 of the first and 20 of the second
Explanation
### Concept & Strategy
When mixing two solutions of different concentrations to form a target concentration, the **Rule of Alligation** provides the exact ratio of the quantities required. It avoids complex linear equations by mapping the distances between the percentages.
$$ \frac{\text{Quantity of Cheaper}}{\text{Quantity of Dearer}} = \frac{\text{Dearer Price} - \text{Mean Price}}{\text{Mean Price} - \text{Cheaper Price}} $$
### Step-by-Step Solution
* **Given:**
* Solution 1 (Lower) = $85\%$
* Solution 2 (Higher) = $92\%$
* Target Mixture (Mean) = $89\%$
* Total Volume = $35$ litres
* **Calculation / Deduction:**
* Set up the Alligation cross:
* Place $85\%$ on the top left and $92\%$ on the top right.
* Place $89\%$ in the middle.
* Calculate the diagonal differences:
* Right diagonal (Difference between $92\%$ and $89\%$) $= 3$. This goes under the $85\%$ side.
* Left diagonal (Difference between $89\%$ and $85\%$) $= 4$. This goes under the $92\%$ side.
* The required ratio of Solution 1 to Solution 2 is $3 : 4$.
* The total number of ratio parts is $3 + 4 = 7$ parts.
* We know the total volume is $35$ litres. So, $7$ parts $= 35$ litres.
* $1$ part $= 5$ litres.
* Volume of Solution 1 ($3$ parts) $= 3 \times 5 = 15$ litres.
* Volume of Solution 2 ($4$ parts) $= 4 \times 5 = 20$ litres.
### Exam Strategy & Shortcut
The Alligation rule itself is the shortcut. However, to verify your answer even faster via Option Elimination:
The target mixture ($89\%$) is closer to $92\%$ than it is to $85\%$. Therefore, the mixture must contain a larger quantity of the $92\%$ solution. Looking at the options, only (a) and (c) have more of the second solution. A quick mental check of the $3:4$ ratio confirms (c) is correct.
### Common Pitfall
Students often place the calculated ratio components under the wrong initial solutions (swapping the $3$ and $4$). Remember: the difference calculated from the right side value dictates the proportion of the left side substance, and vice versa.
### Final Answer
**Therefore, the correct answer is 15 of the first and 20 of the second.**