Question

$\huge \circ + \circ = 10 \\ \huge\circ \times \square + \square = 12 \\ \huge \circ \times \square - \triangle \times \circ = \circ \\ \huge \triangle = ?$



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Answer 1

Answer:

1

Step-by-step explanation:

As on solving above eqns we get Circle as 5,Square as 2 and Triangle as 1

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\: \circ + \circ = 10$

$\rm :\longmapsto\: 2\circ = 10$

$\rm :\longmapsto\: \circ = \dfrac{10}{2}$

$\bf\implies \:\boxed{\tt{ \circ = 5 \: }}$

Again, given that

$\rm :\longmapsto\:\circ \times \square + \square = 12$

$\rm :\longmapsto\:5\times \square + \square = 12$

$\rm :\longmapsto\:6\times \square = 12$

$\rm :\longmapsto\:\square = \dfrac{12}{6}$

$\bf\implies \:\boxed{\tt{ \:\square = 2 \: }}$

Again, Given that

$\rm :\longmapsto\:\circ \times \square - \triangle \times \circ = \circ$

can be rewritten as

$\rm :\longmapsto\:5 \times 2 - \triangle \times5 = 5$

$\rm :\longmapsto\:10 - \triangle \times5 = 5$

$\rm :\longmapsto\: - \triangle \times5 = 5 - 10$

$\rm :\longmapsto\: - \triangle \times5 = - 5$

$\rm :\longmapsto\: \triangle =\dfrac{ - 5}{ - 5}$

$\sf\implies\: \boxed{\tt{ \triangle =1 }}$