In a two digit number, the tens digit is 5 less than the units digit. If you reverse the number, the result is 7 greater than double the original number. Find the original number.
Let's assume the tens digit is $x$, and the units digit is $y$. We are told that $x = 5+y$. The original number can be expressed as $10x+y$, and the reversed number can be expressed as $10y+x$. The problem states that the reversed number is $7$ more than double the original number, so we can set up the equation $10y+x = 2(10x+y)+7$.
We can now substitute the value of $x$ from the first equation into the second equation: $10y+(5+y) = 2(10(5+y)+y)+7$.
Simplifying this equation, we get $10y+5+y = 2(50+11y)+7$.
Simplifying further, we have $11y+5 = 100+22y+7$.
Combining like terms, we get $11y-22y = 100+7-5$, which gives us $-11y = 102$.
Dividing both sides by $-11$, we find that $y = -\frac{102}{11}$. Since $y$ must be a positive whole number, we conclude that there is no solution to this problem.
No.
Apologies for the error. Let's try again.
Let the tens digit be $x$ and the units digit be $y$. We are given that $x = y - 5$.
The original number can be expressed as $10x + y$, and the reversed number can be expressed as $10y + x$.
We are given that the reversed number is 7 greater than double the original number. Therefore, we have the equation $10y + x = 2(10x + y) + 7$.
Substituting the value of $x$ from the first equation into the second equation, we get $10y + y - 5 = 2(10(y - 5) + y) + 7$.
Simplifying this equation, we have $11y - 5 = 2(11y - 50) + 7$.
Expanding the right side, we get $11y - 5 = 2(11y - 50) + 7 \implies 11y - 5 = 22y - 100 + 7$.
Combining like terms, we have $11y - 5 = 22y - 93$.
Moving variables to one side and constants to the other side, we get $22y - 11y = 93 - 5 \implies 11y = 88$.
Dividing both sides by 11, we find that $y = 8$.
Substituting this value back into the equation $x = y - 5$, we have $x = 8 - 5 = 3$.
Therefore, the original number is $10x + y = 10(3) + 8 = 30 + 8 = \boxed{38}$.
No
Apologies for the confusion. Let's approach the problem again.
Let the tens digit be $x$ and the units digit be $y$. We are given that $x = y - 5$.
The original number can be expressed as $10x + y$, and the reversed number can be expressed as $10y + x$.
We are given that the reversed number is 7 greater than double the original number. Therefore, we have the equation $10y + x = 2(10x + y) + 7$.
Substituting the value of $x$ from the first equation into the second equation, we get $10y + (y-5) = 2(10(y-5) + y) + 7$.
Simplifying this equation, we have $11y - 5 = 2(11y - 50) + 7$.
Expanding the right side, we get $11y - 5 = 22y - 100 + 7$.
Combining like terms, we have $11y - 5 = 22y - 93$.
Moving variables to one side and constants to the other side, we get $11y - 22y = -93 + 5 \implies -11y = -88$.
Dividing both sides by -11, we find that $y = 8$.
Substituting this value back into the equation $x = y - 5$, we have $x = 8 - 5 = 3$.
Therefore, the original number is $10x + y = 10(3) + 8 = 30 + 8 = \boxed{38}$.
The sum of five consecutive even numbers is 130. Find the consecutive even numbers.
Let's represent the first even number in the sequence as $x$. Since the numbers are consecutive even numbers, the next four consecutive even numbers can be represented as $x+2$, $x+4$, $x+6$, and $x+8$.
We are given that the sum of these five consecutive even numbers is 130. Therefore, we have the equation $x + (x+2) + (x+4) + (x+6) + (x+8) = 130$.
Simplifying this equation, we have $5x + 20 = 130$.
Subtracting 20 from both sides, we get $5x = 110$.
Dividing both sides by 5, we find that $x = 22$.
Therefore, the five consecutive even numbers are $22$, $24$, $26$, $28$, and $30$.
Hunter leaves his house to go on a bike ride. He rides at a speed of 15 km/hr. Hunter's brother decides to join Hunter and leaves the house 30 minutes after Hunter. He rides at a speed of 18 km/hr. How long will it take Hunter's brother to catch up with Hunter?
To solve this problem, we can set up a distance equation. Let $d$ represent the distance that Hunter travels (in km), and let $t$ represent the time it takes for Hunter's brother to catch up with him (in hours).
The distance that Hunter's brother travels in $t$ hours is $18t$.
Since Hunter's brother leaves 30 minutes (or 0.5 hours) after Hunter, Hunter has already been riding for 0.5 hours. Therefore, the distance that Hunter has traveled is $15 \times 0.5 = 7.5$ km.
In order for Hunter's brother to catch up with him, we must have $d = 18t$.
Since both Hunter and his brother are moving at a constant speed, the distances they travel must be equal. From this, we have the equation $d = 7.5 + 18t$.
Setting the two expressions for $d$ equal, we have $7.5 + 18t = 18t$.
Subtracting $18t$ from both sides of the equation, we get $7.5 = 0$.
This is clearly not possible, so there is no solution to the equation.
Therefore, Hunter's brother will never catch up with Hunter.