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+17 votes

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

 +   a   b   c 
 a b a c
 b a b c
 c a c b


 *   a   b   c 
a a b c
b b c a
c c c b

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 (59.4k points)
edited by | 955 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

4 Answers

+20 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

               (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

                 (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 

                  (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 solution is 2

answered by Veteran (54.5k points)
selected 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
+9 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 'a'

so (a∗x)+(a∗y)=c will results in the above 3 pair only.

so 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 Loyal (8.4k points)
+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 (4.2k points)

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