Question

divide 42 into two parts such that 10 times one of these numbers may exceed 9 times the other by 1

Answer 1

$\textbf{Answer}$

Suppose one of the number is x

Since sum of both numbers = 42

=> Other number = 42 - x

$\textbf{According to the question,}$

$\textbf{10times first number exceeds}$ $\textbf{9 times the other number}$ by 1.

Suppose the 10 times the number x exceeds 9times the other number by 1.

=> 10x = 9(42-x) + 1

=> 10x = 378 - 9x + 1

=> 10x + 9x = 379

=> 19x = 379

=> x = 19.95

=> 42 - x = 42 - 19.05

$\textbf{Required numbers}$ are $\textbf{19.05 and 22.05}$

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Answer 2

Suppose one of the number is x

Since sum of both numbers = 42

=> Other number = 42 - x

\textbf{According to the question,}According to the question, 

\textbf{10times first number exceeds}10times first number exceeds \textbf{9 times the other number}9 times the other number by 1.

Suppose the 10 times the number x exceeds 9times the other number by 1.

=> 10x = 9(42-x) + 1

=> 10x = 378 - 9x + 1

=> 10x + 9x = 379

=> 19x = 379

=> x = 19.95

=> 42 - x = 42 - 19.05

\textbf{Required numbers}Required numbers are \textbf{19.05 and 22.05}19.05 and 22.05