The difference between the two digits of a numberis 2. If the digits are reversed and the number addedto original number then, the result is 132. Find theoriginal number. About the author Autumn
Given:- The difference between the two digits of a number is 2. If the digits are reversed and the number is added to the original number, then the result is 132. To Find:- The original number. Solution:- Let the digit in the ones place be y and the digit in the tens place be x. Hence, [tex]\dag{\boxed{\green{\rm{The\:original\:number = 10x + y}}}}[/tex] Now, On reversing the digits:- [tex]\dag{\boxed{\blue{\rm{The\:new\:number = 10y + x}}}}[/tex] Now, We are given that the difference between the two digits of the number is 2. Hence, [tex]\blue{\rm{10x – y = 2 \longrightarrow(i)}}[/tex] Also it is given that, the sum of original number and the new number is 132. Hence, [tex]\red{\rm{(10x + y) + (10y + x) = 132}}[/tex] [tex] = \red{\rm{10x + y + 10y + x = 132}}[/tex] [tex] = \red{\rm{11x + 11y = 132}}[/tex] [tex] :\implies \red{\rm{11(x + y) = 132}}[/tex] [tex] :\implies \red{\rm{x + y = \dfrac{132}{11}}}[/tex] [tex] = \red{\rm{x + y = 12 \longrightarrow(ii)}}[/tex] Now, On subtracting equation (i) from equation (ii) = (x + y) – (x – y) = 12 – 2 = x + y – x + y = 10 = 2y = 10 [tex] = \sf{y = \dfrac{10}{2}}[/tex] = y = 5 Putting the value of y in equation (i) = x – y = 2 = x – 5 = 2 = x = 2 + 5 = x = 7 Therefore the original number becomes:- [tex]\rm{\orange{10x + y = 10\times 7 + 5 = 75}}[/tex] [tex]\boxed{\underline{\red{\rm{\therefore\:The\:original\:number\:is\:75}}}}[/tex] ______________________________________ Reply
Given:-
To Find:-
Solution:-
Let the digit in the ones place be y and the digit in the tens place be x.
Hence,
[tex]\dag{\boxed{\green{\rm{The\:original\:number = 10x + y}}}}[/tex]
Now,
On reversing the digits:-
[tex]\dag{\boxed{\blue{\rm{The\:new\:number = 10y + x}}}}[/tex]
Now,
We are given that the difference between the two digits of the number is 2.
Hence,
[tex]\blue{\rm{10x – y = 2 \longrightarrow(i)}}[/tex]
Also it is given that, the sum of original number and the new number is 132.
Hence,
[tex]\red{\rm{(10x + y) + (10y + x) = 132}}[/tex]
[tex] = \red{\rm{10x + y + 10y + x = 132}}[/tex]
[tex] = \red{\rm{11x + 11y = 132}}[/tex]
[tex] :\implies \red{\rm{11(x + y) = 132}}[/tex]
[tex] :\implies \red{\rm{x + y = \dfrac{132}{11}}}[/tex]
[tex] = \red{\rm{x + y = 12 \longrightarrow(ii)}}[/tex]
Now,
On subtracting equation (i) from equation (ii)
= (x + y) – (x – y) = 12 – 2
= x + y – x + y = 10
= 2y = 10
[tex] = \sf{y = \dfrac{10}{2}}[/tex]
= y = 5
Putting the value of y in equation (i)
= x – y = 2
= x – 5 = 2
= x = 2 + 5
= x = 7
Therefore the original number becomes:-
[tex]\boxed{\underline{\red{\rm{\therefore\:The\:original\:number\:is\:75}}}}[/tex]
______________________________________