
Sum of the digits of a two-digit number is 9. When we interchange the digits, it’s found that the resulting new number is greater than the original number by 27. What is the two-digit number?
Answer
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Hint: We need to first assume the tenth and units digits of the two-digit number as a and b respectively. From the condition of sum being 9 we get our first linear equation. Then from interchanging the digits we get the new number. We form it to get the second linear equation. We solve both to find the solution.
Complete step by step answer:
It’s given that the sum of the digits of a two-digit number is 9. Let’s assume that the digits are a and b. Here a is the tenth place digit and b is the unit placed digit.
Forming the mathematical equation according to the condition we find $a+b=9......(i)$.
The digit formed with a being the tenth place digit and b being the unit placed digit is $10\times a+1\times b=10a+b$.
It’s also given that if we interchange the digits then the resulting new number is greater than the original number by 27.
After the interchange of numbers, we find a to be the unit placed digit and b be the tenth place digit.
The digit formed with b being the tenth place digit and a being the unit placed digit is $10\times b+1\times a=10b+a$.
The difference is 27. So, \[\left( 10b+a \right)-\left( 10a+b \right)=27\].
Simplifying the equation, we get
\[\begin{align}
& \left( 10b+a \right)-\left( 10a+b \right)=27 \\
& \Rightarrow 9b-9a=27 \\
& \Rightarrow b-a=\dfrac{27}{9}=3 \\
& \Rightarrow b=a+3....(ii) \\
\end{align}\]
Now we have two equations of linear form of a and b. We solve them to find the values of a and b.
From equation (ii) we have value of b and we place that in equation (i).
$\begin{align}
& a+b=9 \\
& \Rightarrow a+a+3=9 \\
& \Rightarrow 2a=9-3=6 \\
& \Rightarrow a=\dfrac{6}{2}=3 \\
\end{align}$
Putting the value of a in equation (ii) we get $b=a+3=3+3=6$.
So, the original number is 36.
Note: Writing a number in the form $ab$ instead of $10a+b$ is wrong. $ab$ defines the multiplication of two numbers. We can write that only when we have exact numbers, not for variable form. So, in case of variables we have to form it using a metric system.
Complete step by step answer:
It’s given that the sum of the digits of a two-digit number is 9. Let’s assume that the digits are a and b. Here a is the tenth place digit and b is the unit placed digit.
Forming the mathematical equation according to the condition we find $a+b=9......(i)$.
The digit formed with a being the tenth place digit and b being the unit placed digit is $10\times a+1\times b=10a+b$.
It’s also given that if we interchange the digits then the resulting new number is greater than the original number by 27.
After the interchange of numbers, we find a to be the unit placed digit and b be the tenth place digit.
The digit formed with b being the tenth place digit and a being the unit placed digit is $10\times b+1\times a=10b+a$.
The difference is 27. So, \[\left( 10b+a \right)-\left( 10a+b \right)=27\].
Simplifying the equation, we get
\[\begin{align}
& \left( 10b+a \right)-\left( 10a+b \right)=27 \\
& \Rightarrow 9b-9a=27 \\
& \Rightarrow b-a=\dfrac{27}{9}=3 \\
& \Rightarrow b=a+3....(ii) \\
\end{align}\]
Now we have two equations of linear form of a and b. We solve them to find the values of a and b.
From equation (ii) we have value of b and we place that in equation (i).
$\begin{align}
& a+b=9 \\
& \Rightarrow a+a+3=9 \\
& \Rightarrow 2a=9-3=6 \\
& \Rightarrow a=\dfrac{6}{2}=3 \\
\end{align}$
Putting the value of a in equation (ii) we get $b=a+3=3+3=6$.
So, the original number is 36.
Note: Writing a number in the form $ab$ instead of $10a+b$ is wrong. $ab$ defines the multiplication of two numbers. We can write that only when we have exact numbers, not for variable form. So, in case of variables we have to form it using a metric system.
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