Math, asked by yamunavision, 9 months ago

The sum of the digits of a two-digit number is 10 .The number obtained by interchanging its digit is 1 less than twice the original number. find the number.

Answers

Answered by Naihrik
7

Answer:

Let first digit be x and second digit be y

∴x+y = 10--------(1)

10y+x = 2(10x+y)-1

=> 10y+x = 20x+2y-1

=> 8y-19x = -1-----(2)

(1)×8 = 8x+8y = 80

- (2)×1 = 8y-19x = -1

-------------------------------------

27x = 81

=> x = 81/27

= 3

Putting x=3 in (1)

3+y = 10

=> y = 7

∴ The number is 37.

Answered by pranavirathod
7

Answer:

37 is the Answer

37 is the Answer Step-by-step explanation:

37 is the Answer Step-by-step explanation:Let x be the tens digit, and let y be the ones digit. It is easy to see that x + y = 10, and that the "original" number is 10x + y. The statement "If the digits are reversed, a new number is formed which is 1 less than twice the original number." is expressed by the equation 2(10x + y) - 1 = 10y + x. If you substitute 10 - x for y in this equation and solve, you get x = 3. Thus y = 7, so the original number is 37.

37 is the Answer Step-by-step explanation:Let x be the tens digit, and let y be the ones digit. It is easy to see that x + y = 10, and that the "original" number is 10x + y. The statement "If the digits are reversed, a new number is formed which is 1 less than twice the original number." is expressed by the equation 2(10x + y) - 1 = 10y + x. If you substitute 10 - x for y in this equation and solve, you get x = 3. Thus y = 7, so the original number is 37.Thank You For Attending

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