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+25 votes

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

  1. $0$
  2. $1$
  3. $2$
  4. $3$
asked in Set Theory & Algebra by Veteran (59.8k points)
edited by | 1.5k views
In current GO pdf. It seems this question formatting is not proper.
you are right. i got wrong answer by solving from pdf. :D

@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.

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

5 Answers

+23 votes
Best answer
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: $C$
answered by Veteran (55.9k points)
edited by
sir how can we say that it is asking for a pair which is true for both the set.....(b,c=satisfying both)

it can be a pair (a,c) satisfying first one not the second....

then the ans can be 3 solution

I think asu

Do for equation 1 then only You get very less pair then ony those pair check with equation 2.

this solution is too long, isn't there a better way to solve this in lesser time ?

We want $(a * x) + (a * y) = c$ and $(b * x) + (c * y) = c$.
Note that to make $\text{Operand 1 }+\text{ Operand 2}=c$, pairs can be like this

$ \left (\text{Operand 1},\text{ Operand 2}  \right )=$
$\begin{align} \left (\text{a},\text{ c}  \right )\\

\left (\text{b},\text{ c}  \right )\\

\left (\text{c},\text{ b}  \right )

for each pair above find the value of (x,y) for $(a * x) + (a * y) = c$ , and see if this pair also satisfies $(b * x) + (c * y) = c$.

You may ended up doing same computation as brute force, but it seems like we are doing smartly :P

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




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


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




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

So we have only 2 solutions

(b,c) and (c,b)

+15 votes

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.

answered by Boss (15.5k points)
edited by
+2 votes
ans (C).. 2 solutions (b, c) and (c, b)..
answered by Active (5k points)
+1 vote

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.

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

answered by Active (4.9k points)
0 votes
By Addition We can Find output C only three ways: a+c=c, b+c=c, c+b=c

Now we will check multiplication table for combination of x,y that will give above output.

For 1: eqaution x,y= (a,c), (b,c), (c,b)

but only two combination of  x,y satisfies 2nd equation: (a,c), (c,b)

then number of solution:2

So option C is correct
answered by (29 points)

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