In a two-digit number, the digit at the units place is thrice the digit in the tens place. The number exceeds the sum of its digits by $27$ . Find the number.

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Solution

Number $= y \times 10 + x$
$x = 3 y . . . (1)$
$(y \times 10 + x) = x + y + 27 . . . (2)$
$10 y + 3 y = 3 y + y + 27$
$9 y = 27 \Rightarrow y = 3$
$x = 9$
Number $= 39$

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In a two-digit number, the digit at the units place is thrice the digit in the tens place. The number exceeds the sum of its digits by 27. Find the number.

Number =y×10+x x=3y...(1)(y×10+x)=x+y+27...(2) 10y+3y=3y+y+279y=27⇒y=3 x=9Number =39

In a two-digit number, the digit at the units place is thrice the digit in the tens place. The number exceeds the sum of its digits by $27$ . Find the number.

Number $= y \times 10 + x$
$x = 3 y . . . (1)$
$(y \times 10 + x) = x + y + 27 . . . (2)$
$10 y + 3 y = 3 y + y + 27$
$9 y = 27 \Rightarrow y = 3$
$x = 9$
Number $= 39$