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The following pattern was given to me by one of my friends:
If $$\quad 5 \star 5 = 10$$ $$\quad 6 \star 6=18$$ $$\quad 7 \star 7=36$$
then $$\quad 7 \star 8 = ?$$ a) $54 \quad$ b) $51 \quad$ c) $30$

My Thoughts:
$$5 \times 5 = 25 \to 2\times 5 =10$$ similarly, $$6 \times 6 = 36 \to 3\times 6 =18$$ similarly, $$7 \times 7 = 49 \to 4\times 9 =36$$

$$\implies 7 \times 8 = 56 \to 5\times 6 =30 \implies \text{option } \mathbf{(c)}$$

But my friend has told its answer as $\mathbf{a)\,54}$;
so, can anyone explain the relevant logic ...

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  • $\begingroup$ There is insufficient information provided to yield a unique solution in this case. $\endgroup$ – TheSimpliFire Apr 13 at 6:17
  • $\begingroup$ @TheSimpliFire Are you getting option b as its answer? $\endgroup$ – Suresh Apr 13 at 6:44
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    $\begingroup$ No, as 51 is odd. But a) and c) are equally reasonable. $\endgroup$ – TheSimpliFire Apr 13 at 7:06
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If you break it down as the following

5x5=10
6x6 = 10+8
7x7 = 18+18=10+8+10+8


if you look at the number of groups of

(10+8)

You'll see that in relation to the second number

5+0 = 5 , 5+1=6 , 5+2=7

so

5x5=10 ( zero groups of '8+10' )
6x6 = 10+8 ( one group of '8+10')
7x7 = 18+18=10+8+10+8 (two groups of '8+10')

if you follow the same logic

7x 8 , second number is 8 and 5+3 = 8
so
7*8 = (8+10)+(8+10)+(8+10) = 54

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    $\begingroup$ I see that this answer has been accepted, but it doesn't make any sense at all to me. It's essentially saying that you ignore the first number and take (second number - 3) "groups of 10+8", i.e., that the answer is 3*(second number-3). But that doesn't work at all for the first example, and you only make it "work" here by arbitrarily using 10 instead of 0 when you have "zero groups of 8+10". $\endgroup$ – Gareth McCaughan Apr 13 at 11:39
  • $\begingroup$ @Gareth McCaughan yes you are correct; But until anyone post its another answer(with a better logic), this answer may be considered as a solution for 54 ,that's why accepted this answer $\endgroup$ – Suresh Apr 14 at 10:15
  • $\begingroup$ I feel that this answer would be more suited for 8⋆8 than 7⋆8 $\endgroup$ – ielyamani Apr 14 at 23:38

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