
Find the two digit number if the digit at the tens place is 5 less than twice the digit at unit place. And the sum of the original number and the number obtained by interchanging the digits is 176.
Answer
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Hint: We will first let the digit at the place be $x$ and let the digit at one place be $y$. Then, form the pair of equations using the given conditions. Solve the equations using substitution or elimination to find the value of $x$ and $y$. The number will be of the form $10x + y$, written as xy.
Complete step-by-step answer:
Let the digit at the place be $x$ and let the digit at one place be $y$.
Now we are given that the digit at the tens place is 5 less than twice the digit at unit place.
That is, \[x = 2y - 5\] eqn. (1)
Label this as equation (1)
The second condition given is that the sum of the original number and the number obtained by interchanging the digits is 176.
The first number can be written as \[10x + y\] and the number after interchanging is written as \[10y + x\].
Then their sum is \[10x + y + 10y + x = 176\]
Which is
\[11x + 11y = 176\]
Taking 11 common from LHS, we will get,
\[11\left( {x + y} \right) = 176\]
Divide the equation throughout by 11
That is \[x + y = 16\] eqn. (2)
Label this as equation (2)
Now solve both the equations to get the value of $x$ and $y$
Subtract equations (1) and (2) to eliminate the value of $x$ and solve for the value of $y$.
\[
x - x - y = 16 - 2y + 5 \\
\Rightarrow - y + 2y = 21 \\
\Rightarrow 3y = 21 \\
\]
Divide the equation throughout by 3
$y = 7$
Substitute the value of $y$ in equation (2) to find the value of $x$.
\[
x + 7 = 16 \\
\Rightarrow x = 16 - 7 \\
\Rightarrow x = 9 \\
\]
Hence, the original number is 97.
Note: If a number is written in the form ab, then the value of the number is 10a+b. The value of the number of the sum of the product of the digits with the value of the corresponding place. Here, we have solved for the value of $x$ using elimination, but we can also solve it using other methods like substitution, cross-multiplication etc. also, the answers of $x$ and $y$ will be one digit numbers only.
Complete step-by-step answer:
Let the digit at the place be $x$ and let the digit at one place be $y$.
Now we are given that the digit at the tens place is 5 less than twice the digit at unit place.
That is, \[x = 2y - 5\] eqn. (1)
Label this as equation (1)
The second condition given is that the sum of the original number and the number obtained by interchanging the digits is 176.
The first number can be written as \[10x + y\] and the number after interchanging is written as \[10y + x\].
Then their sum is \[10x + y + 10y + x = 176\]
Which is
\[11x + 11y = 176\]
Taking 11 common from LHS, we will get,
\[11\left( {x + y} \right) = 176\]
Divide the equation throughout by 11
That is \[x + y = 16\] eqn. (2)
Label this as equation (2)
Now solve both the equations to get the value of $x$ and $y$
Subtract equations (1) and (2) to eliminate the value of $x$ and solve for the value of $y$.
\[
x - x - y = 16 - 2y + 5 \\
\Rightarrow - y + 2y = 21 \\
\Rightarrow 3y = 21 \\
\]
Divide the equation throughout by 3
$y = 7$
Substitute the value of $y$ in equation (2) to find the value of $x$.
\[
x + 7 = 16 \\
\Rightarrow x = 16 - 7 \\
\Rightarrow x = 9 \\
\]
Hence, the original number is 97.
Note: If a number is written in the form ab, then the value of the number is 10a+b. The value of the number of the sum of the product of the digits with the value of the corresponding place. Here, we have solved for the value of $x$ using elimination, but we can also solve it using other methods like substitution, cross-multiplication etc. also, the answers of $x$ and $y$ will be one digit numbers only.
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