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Let’s suppose $\bigoplus = a,\bigodot = b,\Delta = c,$ and $\bigotimes = d.$

Then, $a \div b =2;\: a \div c =3;\: b + c =5; \:c \times d =10$

Now, $a = 2b = 3c,$ and $b + c = 5 \implies c = 2,d = 5,a = 6$

$\therefore$ The value of $\left(\bigotimes \;– \;\bigoplus\right)^{2} = (d-a)^{2} = (5 – 6)^{2} = (-1)^{2} = 1.$

So, the correct answer is $(B).$

Then, $a \div b =2;\: a \div c =3;\: b + c =5; \:c \times d =10$

Now, $a = 2b = 3c,$ and $b + c = 5 \implies c = 2,d = 5,a = 6$

$\therefore$ The value of $\left(\bigotimes \;– \;\bigoplus\right)^{2} = (d-a)^{2} = (5 – 6)^{2} = (-1)^{2} = 1.$

So, the correct answer is $(B).$