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The sum of the digits of a $2$-digit number is $11$. The number
nobtained by interchanging its digits exceeds the given number by
n$9$. To know the number :
n(i) form the linear equations representing the above situation.
n(ii) verify that the equations have a unique solution.
n(iii) solve the equations to get the given $2$-digit number.
nobtained by interchanging its digits exceeds the given number by
n$9$. To know the number :
n(i) form the linear equations representing the above situation.
n(ii) verify that the equations have a unique solution.
n(iii) solve the equations to get the given $2$-digit number.
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Let the digit at unit's place be $x$ and at ten's place be $y$.
nThe number is $10y + x$ (1/2 Mark)
(i) As per given statements
$x + y = 11$ --- (1) (1 Mark)
$10x + y = 10y + x + 9$
$\Rightarrow x - y = 1$ --- (2) (1 Mark)
(ii) Here, $\frac{a_1}{a_2} = \frac{1}{1}$, $\frac{b_1}{b_2} = \frac{1}{-1}$ or $-1$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ (1 Mark)
Therefore, system of equations have a unique solution.
(iii) Solving equations (1) & (2), we get
$x = 6$ and $y = 5$ (1/2 + 1/2 Mark)
Therefore, given number is $56$. (1/2 Mark)
nThe number is $10y + x$ (1/2 Mark)
(i) As per given statements
$x + y = 11$ --- (1) (1 Mark)
$10x + y = 10y + x + 9$
$\Rightarrow x - y = 1$ --- (2) (1 Mark)
(ii) Here, $\frac{a_1}{a_2} = \frac{1}{1}$, $\frac{b_1}{b_2} = \frac{1}{-1}$ or $-1$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ (1 Mark)
Therefore, system of equations have a unique solution.
(iii) Solving equations (1) & (2), we get
$x = 6$ and $y = 5$ (1/2 + 1/2 Mark)
Therefore, given number is $56$. (1/2 Mark)