Saved Bookmarks
| 1. |
the sum of two digit number and the number firmed by reversing the order of digit is 66 if two digit differ by 2 find the number how many such number are therw |
|
Answer» Answer: 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=6611x+11y=66x+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=6x−y=2−−−−−−−2x=8x=44+y=6y=2 Hence number is 42. it's reverse number is 24. Verification: 42+24=66 HOPE IT HELPS |
|