The tens digit of a two-digit positive integer is 2 more than three times the ones digit. If the digits are interchanged, the new number 13 less than half the given number. Find the given integer. (Hint: Let x = tens-place digit and y = ones-place digit, then 10x + y is the number.)
The tens digit of a two-digit positive integer is 2 more than three times the ones digit. If the digits are interchanged, the new number 13 less than half the given number. Find the given integer. (Hint: Let x = tens-place digit and y = ones-place digit, then 10x + y is the number.)
i think 82 is the answer.
original:
unit digit: x
tens digit: y
the actual number: 10y + x
number reversed: 10x + y
"the new number 13 less than half the given number" ---> 10x+y < (1/2)(10y+x) by 13
so , take 13 away from the larger part to make them equal.
10x + y = (1/2)(10y+x) - 13
20x + 2y = 10y + x - 26
19x -8y = -26
also y = 3x + 2
use substitution:
19x - 8(3x+2) = -26
19x - 24x - 16 = -26
-5x = -10
x = 2 , then y = 3(2)+2 = 8
the original number is 82
check: number reversed = 28
half the given number = 41
is half the given number (41) less than the number reversed (28) by 13 ??
yes!
To solve this problem, let's follow the hint. Let x be the tens-place digit and y be the ones-place digit. Therefore, the given number is 10x + y.
According to the problem, the tens digit is 2 more than three times the ones digit. So we can write the equation:
x = 3y + 2
Next, we are given that if the digits are interchanged, the new number is 13 less than half the given number. If we interchange the tens and ones digits, the new number becomes 10y + x.
So we can write another equation:
10y + x = (1/2)(10x + y) - 13
Now we have a system of two equations. Let's solve them simultaneously to find the values of x and y.
Substituting the value of x from the first equation into the second equation, we get:
10y + (3y + 2) = (1/2)(10(3y + 2) + y) - 13
Simplifying this equation, we have:
10y + 3y + 2 = (1/2)(30y + 20 + y) - 13
Combining like terms:
13y + 2 = (1/2)(31y + 20) - 13
Multiplying both sides by 2 to remove the fraction:
26y + 4 = 31y + 20 - 26
Combine like terms again:
26y + 4 = 31y - 6
Now, subtracting 26y from both sides:
4 = 5y - 6
Adding 6 to both sides:
10 = 5y
Dividing both sides by 5:
y = 2
Now that we have the value for y, we can substitute it back into the first equation to find the value of x:
x = 3(2) + 2
x = 6 + 2
x = 8
Therefore, the tens-place digit (x) is 8, and the ones-place digit (y) is 2.
The given integer is 10x + y, so it is 10(8) + 2 = 80 + 2 = 82. So the given integer is 82.