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

Solved it totally based on intuition, might not be the best representation, but I think this is faster to understand and solve

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Best ways to find the solution of the given equation less than a min is : we know that identity element of this algebraic structure ({a,b,c} , *) is "a" ..so a*x = x and a*y =y i mean we get x+y = c find how many satisfy x+y=c ==> only (a,c) ,(b,c),(c,b) satisfy x+y =c ..try to check these three pair for equation 2 ..after check we get only two will satisfy (b,c) and (c,b) ..So the number of solution = 2

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PS : 😊 I was try to give unique answer for this question ..but after giving answer and i was trying to read the comment ..i got to know @Sachin Mittal 1 sir and @Soumya29 answer is exactly same as me ... Sometime it happen 😊 ...That's why i believe on story of Leibniz and Newton can give same result from two different places.

Answer 1

Consider each pair

1. $(a,a):$ $(a*a) +(a*a) = a+a = b   \neq c.$ So, $(a,a)$ is not possible

2. $(a,b):$ $(a*a) + (a*b) = a +b = a \neq c.$  So, $(a,b)$ is not possible

3. $(a,c):$ $(a*a)+(a*c) = a+c = c$

$\quad \quad \quad \quad (b*a) +(c*c) = b + b =b \neq c.$ So, $(a,c)$ is not possible

4. $(b,a):$ $(a*b) +(a*a) = b +a = a \neq c.$ So, $(b,a)$ is not possible

5. $(b,b):$ $(a*b) + (a*b) = b+b =b \neq c.$ So, $(b,b)$ is not possible

6. $(b,c):$ $(a*b) + (a*c) = b + c = c$

$\quad \quad \quad \quad (b*b) + (c*c) = c+b = c.$ So, $(b,c)$ is a solution

7. $(c,a):$ $(a*c) + (a*a) = c+a = a \neq c.$ So, $(c,a)$ is not possible

8.$(c,b):$ $(a*c) + (a*b) = c+b = c$

$\quad \quad \quad \quad (b*c) +(c*b) = a+c = c.$ So, $(c,b)$ is a solution

9. $(c,c):$ $(a*c) +(a*c) =c+c= b \neq c.$ So, $(c,c)$ is not possible

So, no. of possible solutions is $2$.

Correct Answer: $C$

Comments

I have developed an approach let me know if it's correct

First things first what all pairs under $+$ make c?

(1)a+c (2)b+c (3)c+b

Now consider the first equation that is given

(I) $(a*x)+(a*y)=c$

If you look at the table of  $*$, a seems to be the identity element.

Now to satisfy an equation a+c=c, what pair do I need?

Simple take x=a and y=c and hence 

$(a*a)+(a*c)=a+c=c$ Seems good.

In a similar fashion if I follow,my possible pairs for equations (1),(2) and (3) are

(a)(a,c)

(b)(b,c)

(c)(c,b)

Now I only need to check whether these pairs satisfy the second equation that is

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

Now, for this let me draw two graphs which represent the operation $b*x$ and $c*y$

here an edge from b to c->a means b*c=a

Now using these two graphs, check whether any of the pairs (a,c),(b,c) and (c,b) make any one of the equations below

a+c

b+c

c+b

and only pair (a,c) fails to do that.

So we have only 2 solutions

(b,c) and (c,b)

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Wow! This should be the best answer.

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nice method ayush, even i was able to intuitively think like that, but the point here is in the 3 hr time how fast we can think likt this, most poepl would have left thse questions.

Answer 2

We want $operand_1 + operand_2=c$ ,

so the pairs can be:
$(a, c)$
$(b, c)$
$(c, b)$

Now, note that identity element of $({a,b,c},*)$ is $\color{RED}{a}.$
So $(a∗x)+(a∗y)=c$ will result in the above 3 pair only.

No need to check equation 1, just see if these pairs satisfy $(b∗x)+(c∗y)=c.$
$(b, c)$ and $(c, b)$ satisfy equation 2 also so there are 2 solutions.

$C$ is the answer.

Answer 3

Approaching the problem the following way might speed you up : observe that the final sum required is $c$ and also the fact that $a*x = x, x\in a,b,c$. To get $c$ as the sum the only ways are: $$\begin{align} a+c &=c\\ b+c&=c\\ c+b&=c \end{align}$$

Now, to construct $a+c$ using the second equation $(b * x) + (c * y) = c $:
It can be: $(b*\color{blue}{c}=a) \color{red}{\wedge} ((c*\color{blue}{a}=c) \color{red}{\vee} (c*\color{blue}{b}=c))$.
Check $(c,a)$ and $(c,b)$ with the first equation: $(a * x) + (a * y) = c$ of which only $\color{red}{(c,b)}$ satisfies.

Similarly for $b+c$ we can have $(a,a)$ and $(a,b)$ which satisfy second equation of which neither satisfies first equation.

For $c+b$  we only have $(b,c)$ satisfying second equation and also $(a*b)+(a*c)=b+c=c$. Therefore $\color{red}{(b,c)}$ also satisfies first equation.

$\text{Hence, we only have two solutions } \color{red}{(c,b) \text{ and } (b,c)}$.

Answer 4

ans (C).. 2 solutions (b, c) and (c, b)..

Answer 5

Shortest approach, without any use of other identities/properties.

Only 3 equations can satisfy the given set of function, and they are as follows:

$a+c=c\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-(i)$

$b+c=c\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-(ii)$

$c+b=c\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-(iii)$

Now $x$ can take only 3 values i.e. $a,b \: and \:  c$. Putting

$x=a$	a + __ b + __ can use only eqn $(i) \& (ii)$	But no value of $y$ will give us $c$ in both the equations.
$x=b$	b + __ c + __ can use only eqn $(ii) \& (iii)$	Putting y=c satisfies. Therefore x=b and y=c is 1st solution
$x=c$	c+ __ b + __ can use only eqn $(ii) \& (iii)$	putting y=b satisfies. Therefore x=c and y=b is 2nd solution

 

Therefore, option C.

Answer 6

This solution Credit Goes to @Deepak Poonia