The digits of a two-digit number differ by 3. If the digits are interchanged and the
resulting number is added to the original number, we get 143. What can be the original number?
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Answer:-
Let the number be (10x + y).
Given:
Difference of the digits = 3
→ x - y = 3 -- equation (1) (or) y - x = 3
And,
The sum of the original number and the number formed by interchanging the digits is 143.
→ 10x + y + 10y + x = 143.
→ 11x + 11y = 143
→ 11(x + y) = 143
→ x + y = 143/11
→ x + y = 13 -- equation (2)
Adding equations (1) & (2) we get,
→ x - y + x + y = 13 + 3
→ 2x = 16
→ x = 8
Substitute x value in equation (1).
→ x - y = 3
→ 8 - y = 3
→ y = 8 - 3
→ y = 5
Hence, the number would be 85 or 58.
Question:
The digits of a two-digit number differ by 3. If the digits are interchanged and the
resulting number is added to the original number, we get 143. What can be the original number?
To Find:
[tex]{\boxed{The\:original\:Number}}[/tex]
Taken:
Let the two digit number be 10a + b.
Number obtained on reversing the digits is 10b + a.
Given:
➜ The Equation Formed is :
☞ [tex]a - b = 3[/tex] [Equation 1]
➜ The Equation Formed is :
☞ [tex]10a + b + (10b + a) = 143[/tex][Equation 2]
Solution:
Solving Equation 2 , we get:
[tex]10a + b + (10b + a) = 143[/tex]
[tex] \Rightarrow 10a + b + 10b + a = 143[/tex]
[tex] \Rightarrow 11a+ 11b= 143[/tex]
Taking the common;
[tex] \Rightarrow 11(a+ b)= 143[/tex]
[tex] \Rightarrow (a+ b)= 13[/tex] [Equation 3]
Putting the ex. 1 and ex. 3 , we get:
[tex]a - b = 3[/tex]
[tex]a+ b = 13[/tex]
_______[By Adding]
[tex]2a = 16[/tex]
[tex]a = 8[/tex]
[tex]\therefore a = 8[/tex]
Putting the value in the equation 1,
we get:
[tex]a - b = 3[/tex]
[tex]8 - b = 3[/tex]
[tex]- b = 3 - 8[/tex]
[tex]b = 5[/tex]
Answer:
[tex]\bold{Given}[/tex]
[tex]\text{The two digit number be 10a + b}[/tex]
[tex] 10(8) + 5 = 85[/tex]
[tex]\therefore the\:no.\:is\:85[/tex]
______________________________________
[tex]{\boxed{The\:number\:is\:85}}[/tex]