A) \[+\]
B) \[+\]
C) \[c\,\square \,b+a\]
D) \[a\,\phi \,b\,\square \,c\]
Correct Answer: D
Solution :
\[4+12=16\] is equivalent to \[16=16,\]. Hence between a and b, we have \[4\times 6-2=14\] or \[\times \text{to }\div ,2\text{and}4\]or \[\text{- to }\div ,2\text{and6}\] Further b = c implies that b and c are interchangeable. Hence (a), (b) and (c) are not possible. [Observe that (b) states \[\text{- to +},2\text{and6}\] which means \[\times \text{to +},4\text{and6}\] which is not possible. Similarly in (c) \[4\times 2+6=14\] which means \[8+6=14\] which contradicts the hypothesis.] (d) Is the correct answer which states that \[14=14\]and\[(6\div 2)\times 3=0\]. Both statements are possible.You need to login to perform this action.
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