GATE CSE 2003 | Question: 38

Question

Consider the set \(\{a, b, c\}\) with binary operators \(+\) and \(*\) defined as follows:

$$\begin{array}{|c|c|c|c|} \hline \textbf{+} & \textbf{a}& \textbf{b} &\textbf{c}\\\hline \textbf{a} & \text{b}& \text{a} & \text{c} \\\hline \textbf{b} & \text{a}& \text{b} & \text{c}\\\hline \textbf{c} & \text{a}& \text{c} & \text{b}\\\hline \end{array}\qquad \begin{array}{|c|c|c|c|} \hline \textbf{*} & \textbf{a}& \textbf{b} &\textbf{c}\\\hline \textbf{a} & \text{a}& \text{b} & \text{c} \\\hline \textbf{b} & \text{b}& \text{c} & \text{a}\\\hline \textbf{c} & \text{c}& \text{c} & \text{b}\\\hline \end{array}$$

For example, \(a + c = c, c + a = a, c * b = c\) and \(b * c = a\).

Given the following set of equations:

• $(a * x) + (a * y) = c $
• $(b * x) + (c * y) = c$

The number of solution(s) (i.e., pair(s) $(x, y)$ that satisfy the equations) is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Comments

@Arjun and all other admins Errata/Correction for this Question( in GO-PDF 2018 version)

Slight error is there in printing. [Delete this comment after correction or if already done]

 

 

There is this unnecessary c present in 2nd eqn.

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Thanks. That is not extra 'c' but two equations coming in same line. "new line" getting ignored in Math mode. Changed now.

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Detailed Video Solution

Answer 1

Given Equation

$\cdot (a*x) + (a*y) = c  ............ eq(1)$

$\cdot (b*x) + (c*y) = c ............ eq(2)$

Now adding both the eqn we get

$(a+b)*x + (a+c)*y =2c$

Now the solution pair(s) should also satisfy this eqn as well.

Simplifying the above eqn using table

$(a+b)*x + (a+c)*y =2c$

$a*x + c*y =2c$ [using ‘+’ Table a+b=a and a+c=c] ……………..eq(3)

Now we have to find (x,y) such that [ $a*x=c$ and $c*y=c$ in order to get 2c in RHS in eq 3]

Using ‘*’ Table we found that

(i) $a*x=c$ only if x=c

(ii) $c*y=c$ only if y={a,b}

Therefore the Solution pair [ {c,a}, {c,b} ]

Hence 2 solution.

Answer 2

An alternative way of solving this question is as follows:

The system of equations that have been given to us is:

$$(a*x)+(a*y)=c \tag1$$
$$(b*x)+(c*y)=c\tag2$$

Now we consult the first to get those entries whose value is $c$. So we have

$$a+c=c\tag3$$
$$b+c=c\tag4$$
$$c+b=c\tag5$$

Now we need to choose $x,y$ in such a manner that the LHS of the equations in $(1)$ and $(2)$ matches with the LHS of the equations given in $(3),(4),(5)$.

Let us make 3 tables from the right table:

 

Table 1

$$\begin{array}{|c|c|} \hline a*a&a\\  
   \hline
b*c&a\\  
   \hline
\end{array}$$

Table 2

$$\begin{array}{|c|c|} \hline c*c&b\\  
   \hline
 a*b&b\\  
   \hline b*a&b\\  
   \hline \end{array}$$

Table 3

$$\begin{array}{|c|c|} \hline c*b&c\\  
   \hline b*b&c\\  
   \hline  a*c&c\\  
   \hline c*a&c\\ \hline\end{array}$$

Trying match $(1)$ of the form $(3)$ we see from table 1 and 3, $x=a, y=c$ but this when put into $(2)$ we get $b+b=c$ which is not present in $(3)$ to $(5)$. So $x=a,y=c$ is not possible.

So we shall continue in this manner, first trying to make equation $(1)$ like the identities $(3)$ to $(5)$ and then checking the corresponding values of $x,y$ in equation $(2)$.

Then we shall continue, trying to make equation $(2)$ like the identities $(3)$ to $(5)$ and then checking the corresponding values of $x,y$ in equation $(1)$.