If the sum and difference of two numbers are $20$ and $8$ respectively,then the difference of their squares is

  • A
    $12$
  • B
    $28$
  • C
    $160$
  • D
    $180$

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$A$ number consists of two digits such that the digit in the ten's place is less by $2$ than the digit in the unit's place. Three times the number added to $\frac{6}{7}$ times the number obtained by reversing the digits equals $108$. The sum of the digits in the number is:

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$\frac{(0.6^{3}-0.1^{3}-0.4^{3}-3 \times 0.6 \times 0.1 \times 0.4)}{(0.6^{2}+0.1^{2}+0.4^{2}+0.6 \times 0.1+0.6 \times 0.4-0.1 \times 0.4)}$

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