GATE Mechanical 2021 Set 2 | GA | Question: 4

Question

If $\bigoplus \div \bigodot =2;\: \bigoplus \div\Delta =3;\:\bigodot +\Delta =5; \:\Delta \times \bigotimes =10$,

Then, the value of $\left ( \bigotimes - \bigoplus \right )^{2}$, is :

• A. $0$
• B. $1$
• C. $4$
• D. $16$

Comments

nice problem.

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$\oplus \div \odot = 2 \Rightarrow \oplus = 2 \odot \text{  or }\\\ \odot =  \frac{\oplus}{2}$

$\text{similarily}, \Delta =  \frac{\oplus}{3}$

$\text{Put in}, \\\ \Delta + \odot = 5$

$\frac{\oplus}{3} + \frac{\oplus}{2} = 5 \Rightarrow \oplus= 6$

$\text{By using $\oplus=2,$} \\\ \Delta = \frac{\oplus}{3} = \frac{6}{3} = 2 = \Delta$

$\Delta \times \otimes = 10 \Rightarrow \otimes = 5$

$(\otimes - \oplus)^2 = (5 - 6)^2 = 1$

Answer 1

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 $(B).$

Comments

detailed version of this answer:

given,

$\frac{\bigoplus}{\bigodot} = 2 \, \Rightarrow \, \bigoplus = 2\bigodot$       ...............................(1)

$\frac{\bigoplus}{\bigtriangleup} = 3  \, \Rightarrow \, \bigoplus = 3\bigtriangleup$           ..........................(2)

$2\bigodot = 3\bigtriangleup$            ...............................(3)

$ \bigodot \, + \, \bigtriangleup = 5$

$\Rightarrow \, \bigodot = 5 \, - \, \bigtriangleup$         .............................(4)

from eqn (3) and (4),

$2(5 \, - \, \bigtriangleup) = 3\bigtriangleup$

$\Rightarrow \, 10 \, - \, 2\bigtriangleup = 3\bigtriangleup$

$\Rightarrow \, 10 = 3\bigtriangleup \, + \, 2\bigtriangleup$

$\Rightarrow \, 10 = 5\bigtriangleup$

$\Rightarrow \, \bigtriangleup = 2$          ..............................(5)

again from eqn (4),

$\bigodot = 5 \, - \, 2$

$\Rightarrow \, \bigodot = 3$        ......................(6)

now, we have one more eqn,

$\bigtriangleup \, \times \, \bigotimes = 10$

$\Rightarrow \, 2 \, \times \, \bigotimes = 10$

$\Rightarrow \, \bigotimes = 5$            ........................(7)

from eqn (1) and (6) or (2) and (5),

$\bigoplus = 6$            ........................(8)

now,

$(\bigotimes \, - \, \bigoplus)^{2}$

$\Rightarrow \, (5 \, - \, 6)^{2}$

$\Rightarrow \, (-1)^{2}$

$\Rightarrow \, 1$

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easily solveable after assigning variables (a,b,c,d) to them respectively.

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why are you trying to solve in such a complex way, we can easily solve it by assigning them variable name a,b,c,d.

 

Answer 2

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 $(B).$