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.
No..
Apologies for the confusion. Let's try again.
We know that distance equals speed multiplied by time.
Let $t$ represent the time it takes for Hunter's brother to catch up with Hunter (in hours). Since Hunter's brother leaves 30 minutes (or 0.5 hours) later than Hunter, the time Hunter has been riding is $t + 0.5$ hours.
The distance Hunter travels is given by $d = 15(t + 0.5)$ (distance equals speed multiplied by time).
The distance Hunter's brother travels is given by $d = 18t$.
Since both Hunter and his brother are traveling the same distance when his brother catches up, we have $15(t + 0.5) = 18t$.
Expanding the left side, we get $15t + 7.5 = 18t$.
Subtracting $15t$ from both sides, we have $7.5 = 3t$.
Dividing both sides by 3, we find that $t = 2.5$.
Therefore, it will take Hunter's brother $2.5$ hours to catch up with Hunter.
No...
Apologies for the confusion. Let's try again.
Let's start by converting the 30 minutes that Hunter's brother leaves after Hunter into hours. Since 1 hour is equal to 60 minutes, 30 minutes is equal to $\frac{30}{60} = \frac{1}{2}$ hour.
Let $t$ represent the time it takes for Hunter's brother to catch up with Hunter (in hours).
In the time it takes for Hunter's brother to catch up, Hunter has been riding for $t + \frac{1}{2}$ hours.
Since the speed of Hunter is 15 km/hr, the distance Hunter has traveled is $15(t + \frac{1}{2})$ km.
Hunter's brother rides at a speed of 18 km/hr, so the distance Hunter's brother travels is $18t$ km.
When Hunter's brother catches up, they would have traveled the same distance. Therefore, we have the equation $15(t + \frac{1}{2}) = 18t$.
Expanding the left side, we get $15t + \frac{15}{2} = 18t$.
Subtracting $15t$ from both sides, we have $\frac{15}{2} = 3t$.
Dividing both sides by 3, we find that $t = \frac{15}{6} = \frac{5}{2} = 2.5$.
Therefore, it will take Hunter's brother 2.5 hours to catch up with Hunter.
There are nickels, dimes and quarters in a piggy bank. THe total value of the coins is $3.65. The number of dimes is three times greater than the number of nickels, and the number of quarters is on greater than double the number of nickels. How many quarters, dimes and nickels are there?
Let's represent the number of nickels as $n$, the number of dimes as $d$, and the number of quarters as $q$.
The total value of the coins is $3.65. We know that a nickel is worth 5 cents, a dime is worth 10 cents, and a quarter is worth 25 cents. Therefore, we can set up the equation $5n + 10d + 25q = 365$ (in cents).
We are also given that the number of dimes is three times greater than the number of nickels. Therefore, we have the equation $d = 3n$.
Additionally, we are given that the number of quarters is one greater than double the number of nickels. Therefore, we have the equation $q = 2n + 1$.
To solve the system of equations, let's substitute the values of $d$ and $q$ from the second and third equations into the first equation:
$5n + 10(3n) + 25(2n + 1) = 365$
Simplifying this equation, we get $5n + 30n + 50n + 25 = 365$.
Combining like terms, we have $85n + 25 = 365$.
Subtracting 25 from both sides, we get $85n = 340$.
Dividing both sides by 85, we find that $n = 4$.
Using the value of $n$, we can substitute it back into the second equation to find $d$: $d = 3(4) = 12$.
Similarly, substituting $n$ into the third equation, we find $q = 2(4) + 1 = 9$.
Therefore, the piggy bank contains 4 nickels, 12 dimes, and 9 quarters.