the sum of two digit number and the number obtained by reversing the digits is a 56. if the digits of number differ by 2 2.find the number
Answers
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Answer:Number is 42.
Solution:
Let the two digits number is xy,this can be represented as 10x+y
on reversing the number the number become yx, now this can be represented as 10y+x
according to the question the sum of two digit number and the number reversing the digits is equal to 66
\begin{lgathered}10x + y + 10y + x = 66 \\ \\ 11x + 11y = 66 \\ \\ x + y = 6....eq1 \\ \\\end{lgathered}
10x+y+10y+x=66
11x+11y=66
x+y=6....eq1
the digit of the number differ by 2,hence it can be represented algebraically
\begin{lgathered}x - y = 2.....eq2 \\ \\\end{lgathered}
x−y=2.....eq2
Now solve these two equations,here I am using Elimination method,add both equations
\begin{lgathered}x + y = 6 \\ x - y = 2 \\ - - - - - - - \\ 2x = 8\\ \\ x = 4 \\ \\4+ y = 6 \\\\y=2\\\\\end{lgathered}
x+y=6
x−y=2
−−−−−−−
2x=8
x=4
4+y=6
y=2
Hence number is 42.
it's reverse number is 24.
Verification:
42+24=66
hope it helps you dear
Answer:
sorry I don't know this maths