If $3 + 5 = 16; 7 + 9 = 64; 10 + 12 = 121$ , then $11 + 3 = ?$

56

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48

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49

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196

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Solution

The correct option is B $49$
The numbers are following one rule:
$A + B = {(\frac{A + B}{2})}^{2}$

$3 + 5 = {(\frac{3 + 5}{2})}^{2} = 16$

$7 + 9 = {(\frac{7 + 9}{2})}^{2} = 64$

$10 + 12 = {(\frac{10 + 12}{2})}^{2} = 121$
Similarly,
$11 + 3 = {(\frac{11 + 3}{2})}^{2} = 49$

Suggest Corrections

Similar questions

Fill in the blanks.

(i) $(\frac{- 3}{17}) + (\frac{- 12}{5}) = (\frac{- 12}{5}) + (. . . . .)$

(ii) $- 9 + \frac{- 21}{8} = (. . . . . .) + (- 9)$

(iii) $(\frac{- 8}{13} + \frac{3}{7}) + (\frac{- 13}{4}) = (. . . . . .) + [\frac{3}{7} + (\frac{- 13}{4})]$

(iv) $- 12 + (\frac{7}{12} + \frac{- 9}{11}) = (- 12 + \frac{7}{12}) + (. . . . .)$

(v) $\frac{19}{- 5} + (\frac{- 3}{11} + \frac{- 7}{8}) = {\frac{19}{- 5} + (. . . . .)} + \frac{- 7}{8}$

(vi) $\frac{- 16}{7} + . . . . . = . . . . . . + \frac{- 16}{7} = \frac{- 16}{7}$

(121 - 196), (144 - 225), (36 - 83), (16 - 49)

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\begin{matrix} 25 & 81 & (196) 16 & 49 & (121) 64 & 100 & (-) \end{matrix}

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