Question

Sum of the digits of a two-digit number is 9. When we interchange the digits, it is
found that the resulting new number is greater than the original number by 27. What
is the two-digit number?​

Answer 1

Answer:

92 or 29

Step-by-step explanation:

Sum of the 2digit n.o = 9

new n.o=greater than o.no by 27

if the n.o is 10x +1y= 9.........eq 1

(1y +10x)=27....eq 2 interchange

now EQ 2-EQ 1,

(1y + 10x)=27

(10x+1y)=9

- 9y+9x=18

xy=18

18 can be written as 9×2 or 2×9

we can take the n.o as 92 or 29

Answer 2

AnswEr :

$\:\bullet\:\sf\ Let \: the \: one's \: place \: digit \: be \: \bf\ y$

$\:\bullet\:\sf\ Let \: the \: ten's \: place \: digit \: be \: \bf\ x$

$\rule{100}1$

$\normalsize\star{\underline{\boxed{\sf{Original \: number = (10 \times\ ten's \: place \: digit + one's \: place \: digit) }}}}$

$\normalsize\sf\quad\ Hence, \: Original \: number = 10x + y$

$\normalsize\star{\underline{\boxed{\sf{Reversing \: number = (10 \times\ one's \: place \: digit + ten's \: place \: digit) }}}}$

$\normalsize\sf\quad\ Hence, \: Reversing \: number = 10y + x$

$\underline{\dag\:\textsf{According \: to \: question \: now:}}$

$\normalsize\twoheadrightarrow\sf\ Sum \: of \: two \: digit \: number = 9$

$\normalsize\twoheadrightarrow\sf\ (x + y) = 9 \: \quad\ ---(eq.1)$

$\normalsize\twoheadrightarrow\sf\ Reversing number = (Original number + 27)$

$\normalsize\twoheadrightarrow\sf\ 10y + x = (10x + y + 27)$

$\normalsize\twoheadrightarrow\sf\ 10y + x = 10x + y + 27$

$\normalsize\twoheadrightarrow\sf\ 9x - 9y = -27$

$\scriptsize\sf{\: \: \: \: \: \: \: \: \: \dag\ Take \: 9 \: common \: from \: L.H.S}$

$\normalsize\twoheadrightarrow\sf\ 9(x - y) = -27$

$\normalsize\twoheadrightarrow\sf\ (x - y) =\frac{\cancel{-27}}{\cancel{9}}$

$\normalsize\twoheadrightarrow\sf\ (x - y) = -3 \: \quad\ ---(eq.2)$

$\underline{\dag\:\textsf{From \: equation \: 1 \: and \: 2 :}}$

$\normalsize\twoheadrightarrow\sf\ x + \cancel{y} = 9$

$\normalsize\twoheadrightarrow\sf\ x - \cancel{y} = -3$

$\rule{200}1$

$\normalsize\twoheadrightarrow\sf\ 2x = 6$

$\normalsize\twoheadrightarrow\sf\ x = \frac{6}{3}$

$\normalsize\twoheadrightarrow\sf\red{x = 3}$

$\normalsize\sf\ Putting \: the \: value \: of \: x \: in \: eq.1$

$\normalsize\twoheadrightarrow\sf\ x + y = 9$

$\normalsize\twoheadrightarrow\sf\ 3 + y = 9$

$\normalsize\twoheadrightarrow\sf\red{y = 6}$

$\rule{100}1$

$\underline{\dag\:\textsf{Calculation \: of \: Original \: number :}}$

$\normalsize\dashrightarrow\sf\ Original_{no} = 10x + y$

$\normalsize\dashrightarrow\sf\ Original_{no} = 10 \times\ 3 + 6$

$\normalsize\dashrightarrow\sf\ Original_{no} = 30 + 6$

$\normalsize\dashrightarrow\sf\ Original_{no} = 36$

$\normalsize\maltese\: \: {\underline{\boxed{\sf \green{ Original \: number = 36}}}}$