Question 1068612: In a two-place number the sum of the digits is 12.If 6 is subtracted from the tens' digit and 5 is added to the units' digit, the result is 38. Find the number. Ans is 93, I am getting it as 93/11 or x=3, y=9.
Found 3 solutions by josgarithmetic, ikleyn, MathTherapy:
Answer by josgarithmetic(39626)   (Show Source):
Answer by ikleyn(52858)  (Show Source):
You can put this solution on YOUR website! .
In a two-place number the sum of the digits is 12. If 6 is subtracted from the tens' digit and 5 is added to the units' digit,
the result is 38. Find the number. Ans is 93, I am getting it as 93/11 or x=3, y=9.
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The "units" digit of 38 is 8. Subtract 5 from 8 and get 3. It is your "units" digit.
The "tens" digit of 38 is 3. Add 6 to 3 to get 9. It is your "tens" digit.
Thus your number is 93.
WHAT CAN BE EASIER ???
The first phrase of the condition "In a two-place number the sum of the digits is 12." is excessive, not necessary and is not used in the solution.
You can use it as an additional checkpoint, but in reality it is excessive.
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The solution by "josgarithmeic" is ABSURD and is not relevant and is not adequate to the problem.
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Answer by MathTherapy(10556)  (Show Source):
You can put this solution on YOUR website!
In a two-place number the sum of the digits is 12.If 6 is subtracted from the tens' digit and 5 is added to the units' digit, the result is 38. Find the number. Ans is 93, I am getting it as 93/11 or x=3, y=9.
Let the tens and units digits be T and U, respectively
Then number is: 10T + U
2nd clue: 10(T - 6) + U + 5 = 38
10T - 60 + U + 5 = 38
10T + U = 38 + 55
10T + U = 93
Therefore, the number is 93.
As seen, the sum of the number's digits was not necessary! That's how it is with these problems, sometimes!
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