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3 candidates A,B and C contest an election. A gets at least 40% of all the votes. B gets at least 20% of the number of votes that A gets and cannot get more than 80% of number of votes that c gets. What is the maximum percentage of votes that a can get?

I am looking at a generic way how to solve these questions on finding maximum values of variables where the values are inter-dependent. Thanks.

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You don't need to sign your posts: see Can I use a signature or tagline? from the FAQ. –  Ben Millwood Jul 21 '12 at 15:13
    
@BenMillwood Thanks. Removed the signature. –  Manty Jul 21 '12 at 15:30

3 Answers 3

The technique to be used depends heavily on context. The problem is a standard linear programming problem. But since you are using the tag "quantitative aptitude," we will first try to solve the problem by reasoning.

The number of votes $A$ gets is largest if (i) $B$ gets exactly $20\%$ of the votes $A$ gets, and (ii) $B$ gets $80\%$ of the votes $C$ gets. Let $z$ be $C$'s vote percentage. Then $B$'s is $(0.8)z$, and $A$'s is $(5)(0.8z)$.

But the vote percentages add up to $100$. This gives the equation $(5)(0.8z)+(0.8z)+z=100$. So $5.8z=100$, and therefore $z=\frac{100}{5.8}$. It follows that $A$ gets the percentage $\frac{400}{5.8}$, a bit under $69\%$ of the vote.

Another way: Another approach is graphical. We stop using percentages, and use fractions instead. Let $x$, $y$, and $z$ respectively be the fractions of the vote obtained by $A$, $B$, and $C$. We have $x+y+z=1$, and therefore $z=1-x-y$. So effectively we have a two variable problem.

Now you will need to use paper and ruler, and draw carefully the lines described below. We are told that $x \ge \frac{2}{5}$. Draw the line $x=\frac{2}{5}$. We must be to the right of it.

We are told that $y \ge \frac{1}{5}x$. Draw the line $y=\frac{x}{5}$ or equivalently $5y=x$. We must be above it.

We are told that $y \le \frac{4}{5}z$. So $y \le \frac{4}{5}(1-x-y)$, which simplifies to $\frac{9y}{5} \le \frac{4}{5} -\frac{4}{5}x$. Draw the line $\frac{9y}{5}=\frac{4}{5}-\frac{4x}{5}$, that is, $9y=4-4x$. We must be below this line.

If you consider the three geometrical constraints, and have drawn the diagram carefully, you will see that we must be in a certain triangle. And it is obvious that $x$ is biggest at the rightmost corner of that triangle, which is where the lines $5y=x$ and $9y=4-4x$ meet. Solve to find $x$.

Remark: In the real world, problems with a similar structure occur, often with hundreds (or more) variables and hundreds (or more) linear constraints. There are efficient procedures, such as the simplex method, for solving such problems, of course by computer. The field is called Linear Programming. It is of great practical importance.

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The first step is to interpret the information you are given into equations. Let $a,b,c$ be the percentages gathered by $A, B, C$. The first sentence tells you $a+b+c=100$. The second says $a \ge 40$. The third says $b \ge 0.2a$ and also that $b \le 0.8c$.

In this case, if we want to maximize the number of votes $a$ gets, we should minimize the number $b$ and $c$ get. So let $b=0.2a$, which is the minimum for $b$. Then $c \ge 1.25b = 0.25a$ and to minimize that, we have equality as well. Now plug those into the first equation, getting $a + 0.2a + 0.25a=100$ or $a=\frac {100}{1.45}$. It is a bit surprising this didn't come out cleanly. Then you can verify that in fact $a \ge 40$ holds.

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Since you are asking on the question where the variables are inter-dependent, the BEST way to set up a proper solution is to assign each task with a variable and consider the relations between them. For instance, if we let the percentage of candidate A is x percent, analogously, we also assign y and z for candidate B and C respectively. The obvious implication would be $x+y+z=100$ and other restrictions are also taken into account. Hope this gives you some hint on solving the problem.

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