The difference between a two digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?
A.3
B.4
C.9
D.Cannot be determined
E.None of these
Answer
670.8k+ views
Hint: To solve the question given, we will assume that the two digit number be A. we will consider the number on unit place be equal to x and the number at the tense place be equal to y. Then we will write A in terms of x and y. Now after reversing, the new number be B. we will write B in terms of x and y. Then we will subtract A from B and we will equate it to 36.
Complete step-by-step answer:
Now, we will assume that the two digit number be A. If A is a two digit number then it will have a unit place and a tens place, let the one digit number at unit place be x and one digit number at tens place by y.
Thus, we can write A as follows:
$A=10y+x$ …………………. (1)
Now, we are given that after interchanging the digits, the new number obtained is assumed to be B. In B the one digit number present at unit place will be y and tens digit will be x, Thus, we can write B as follows:
$B=10x+y$ ………………………… (2)
Now, we need to find the value of $\left( y-x \right)$. For this, we will subtract equation (2) from equation (1). After doing this, we will get:
$\begin{align}
& A-B=\left( 10y+x \right)-\left( 10x+y \right) \\
& \Rightarrow A-B=10y+x-10x-y \\
\end{align}$
$\Rightarrow A-B=9y-9x$…………………………….. (3)
Now, it is given in question that the difference between A and B is 36. Thus, we have:
$A-B=36$ ………………………. (4)
From (3) and (4) we have:
$\begin{align}
& \Rightarrow 9y-9x=36 \\
& \Rightarrow 9\left( y-x \right)=36 \\
& \Rightarrow \left( y-x \right)=\dfrac{36}{9} \\
& \Rightarrow \left( y-x \right)=4 \\
\end{align}$
Hence, option (b) is correct.
Note: Likewise can also solve the above question with the help of hit and trial method. In this method, we will assume a two digit number and then we will add 36 to it. If the new number obtained is reverse of the original then that will be our required number. For, consider $A=15$ .Now $A+36=51$ . Here we can see that difference of digits $=5-1=4$ .
Complete step-by-step answer:
Now, we will assume that the two digit number be A. If A is a two digit number then it will have a unit place and a tens place, let the one digit number at unit place be x and one digit number at tens place by y.
Thus, we can write A as follows:
$A=10y+x$ …………………. (1)
Now, we are given that after interchanging the digits, the new number obtained is assumed to be B. In B the one digit number present at unit place will be y and tens digit will be x, Thus, we can write B as follows:
$B=10x+y$ ………………………… (2)
Now, we need to find the value of $\left( y-x \right)$. For this, we will subtract equation (2) from equation (1). After doing this, we will get:
$\begin{align}
& A-B=\left( 10y+x \right)-\left( 10x+y \right) \\
& \Rightarrow A-B=10y+x-10x-y \\
\end{align}$
$\Rightarrow A-B=9y-9x$…………………………….. (3)
Now, it is given in question that the difference between A and B is 36. Thus, we have:
$A-B=36$ ………………………. (4)
From (3) and (4) we have:
$\begin{align}
& \Rightarrow 9y-9x=36 \\
& \Rightarrow 9\left( y-x \right)=36 \\
& \Rightarrow \left( y-x \right)=\dfrac{36}{9} \\
& \Rightarrow \left( y-x \right)=4 \\
\end{align}$
Hence, option (b) is correct.
Note: Likewise can also solve the above question with the help of hit and trial method. In this method, we will assume a two digit number and then we will add 36 to it. If the new number obtained is reverse of the original then that will be our required number. For, consider $A=15$ .Now $A+36=51$ . Here we can see that difference of digits $=5-1=4$ .
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