the sum of a two digit number and the number obtained by reversing the digits is 66 if the digits of the number differ by 2 find the number
Answers
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{gathered}10x + y + 10y + x = 66 \\ \\ 11x + 11y = 66 \\ \\ x + y = 6....eq1 \\ \\\end{gathered}
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{gathered}x - y = 2.....eq2 \\ \\\end{gathered}
x−y=2.....eq2
Now solve these two equations,here I am using Elimination method,add both equations
\begin{gathered}x + y = 6 \\ x - y = 2 \\ - - - - - - - \\ 2x = 8\\ \\ x = 4 \\ \\4+ y = 6 \\\\y=2\\\\\end{gathered}
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
Answer:
Step-by-step explanation:
Given:-
The sum of a two digit number and the number obtained by reversing the digits is 66 if the digits of the number differ by 2.
To find:-
find the number
Solution:-
Let the digits in a two digit number be "a" and "b"
"a" is in 10's place
The place value of a =10a
"b" is in 1's place
The place value of b =b
Now the two digit number =10a+b
The number obtained by reversing the digits =10b+a
Condition 1:-
the sum of a two digit number and the number obtained by reversing the digits is 66
=>(10a+b)+(10b+a)=66
=>10a+a+10b+b=66
=>11a+11b=66
=>11(a+b)=66
=>a+b=66/11
=>a+b=6.......(1)
Condition 2:-
the digits of the number differ by 2
=>a-b=2..........(2)
From (1)&(2)
a+b=6
a-b=2
(+)
______
2a+0=8
_______
=>2a=8
=>a=8/2
=>a=4
from(1)
=>4+b=6
=>b=6-4
=b=2
we have a=4 and b=2
number=10a+b=10(4)+2=40+2=42
Answer:-
The required number=42
Check:-
Condition 1:-
42+24=66
Condition 2:-
4-2=2