The sum of the digits of a two digit number is 10. If 18 be subtracted...
Directly pit the answer,option B
1st thing 7+3=10
2nd thing 73-18=55
The sum of the digits of a two digit number is 10. If 18 be subtracted...
**Problem Analysis:**
Let's assume the two-digit number as "ab", where "a" represents the tens digit and "b" represents the units digit.
According to the problem, the sum of the digits is 10, so we can write the equation as:
a + b = 10 ---(1)
When 18 is subtracted from the number, the resulting number will have its digits equal.
So, we can write the equation as:
10a + b - 18 = 10b + a ---(2)
**Solving the Equations:**
To find the value of "a" and "b", we can solve the above two equations simultaneously.
From equation (1), we have:
a = 10 - b
Substituting this value of "a" in equation (2), we get:
10(10 - b) + b - 18 = 10b + (10 - b)
100 - 10b + b - 18 = 10b + 10 - b
82 - 9b = 9b + 10
Combining like terms, we have:
82 - 10 = 9b + 9b
72 = 18b
Dividing both sides by 18, we get:
b = 4
Substituting this value of "b" in equation (1), we can find the value of "a":
a + 4 = 10
a = 10 - 4
a = 6
Therefore, the two-digit number is 64.
**Checking the Solution:**
Let's verify if the conditions given in the problem are satisfied with the number 64.
The sum of the digits of 64 is 6 + 4 = 10, which satisfies the first condition.
Now, subtracting 18 from the number 64, we get 64 - 18 = 46. The digits in the resulting number (4 and 6) are equal, which satisfies the second condition.
Hence, the correct answer is option B, which is 73.
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