The sum of the digits of a two digit number is 15 the number obtained by interchanging

Let the number be 10x+ywith digits x,ytherefore x+y=15........(i)again,        10y+x-10x-y=9or,    9y-9x=9or,    x-y=-1..........(ii)solving equation (i) and (ii) we get x+y=15........(i)x-y=-1..........(ii)------------------------2x=14x=7andy=8therefore the number is 10x+y=10x7+8=78

Solution:

Let’s suppose that the tens and the units digits of the required number be $\mathrm{x}$ and $\mathrm{y}$, respectively. Required number $=(10 \mathrm{x}+\mathrm{y})$ $x+y=15\dots \dots(i)$ Number obtained on reversing its digits $=(10 \mathrm{y}+\mathrm{x})$ $\begin{array}{l} \therefore(10 \mathrm{y}+\mathrm{x})-(10 \mathrm{x}+\mathrm{y})=9 \\ \Rightarrow 10 \mathrm{y}+\mathrm{x}-10 \mathrm{x}-\mathrm{y}=9 \\ \Rightarrow 9 \mathrm{y}-9 \mathrm{x}=9\dots \dots(ii) \end{array}$ On adding equation(i) and equation(ii), we obtain: $\begin{array}{l} 2 y=16 \\ \Rightarrow y=8 \end{array}$ On substituting $\mathrm{y}=8$ in equation(i) we obtain $\begin{array}{l} x+8=15 \\ \Rightarrow x=(15-8)=7 \end{array}$ $\text { Number }=(10 x+y)=10 \times 7+8=70+8=78$

As a result, the required number is 78 .

Let the tens and the units digits of the required number be x and y, respectively.Required number = (10x + y)x + y = 15                  ……….(i)Number obtained on reversing its digits = (10y + x)∴ (10y + x) - (10x + y) = 9⇒10y + x – 10x – y = 9

⇒9y – 9x = 9

⇒y – x = 1                ……..(ii)On adding (i) and (ii), we get:2y = 16⇒y = 8On substituting y = 8 in (i) we getx + 8 = 15⇒ x = (15 - 8) = 7Number = (10x + y) = 10 × 7 + 8 = 70 + 8 = 78

Hence, the required number is 78.

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The sum of the digits of a two digit number is 15.The number obtained by interchanging the digits exceeds the given number by 9.The number is a 96 b 69 c 87 d78

Secondary School Mathematics X