The sum of two numbers is 48. If one third of one number is 5 greater than one sixth of another number, which of the following is the smaller number?
To solve this problem, let's use algebra. Let's assume the first number is x and the second number is y.
We are given two pieces of information:
1. The sum of two numbers is 48, so we can write the equation: x + y = 48.
2. One third of one number is 5 greater than one sixth of another number, which can be written as: (1/3)x = (1/6)y + 5.
To solve this system of equations, we can use substitution or elimination. Let's use substitution:
From the first equation, we can solve for x: x = 48 - y.
Now substitute the value of x in the second equation: (1/3)(48 - y) = (1/6)y + 5.
Multiply both sides of the equation by 6 to get rid of the fractions: 2(48 - y) = y + 30.
Distribute the 2: 96 - 2y = y + 30.
Combine like terms: 96 - 30 = y + 2y.
Simplify: 66 = 3y.
Divide both sides by 3: y = 22.
Now, since we've found the value of y, we can substitute it back into the equation x + y = 48 to find x: x + 22 = 48. Subtract 22 from both sides: x = 26.
So, the two numbers are x = 26 and y = 22.
Now let's compare the numbers to find the smaller one. Since 22 is less than 26, the smaller number is 22.
Therefore, the smaller number is 22.
x + y = 48
x / 3 = (y / 6) + 5
... 2x = y + 30 ... 2x - y = 30
add equations to eliminate y
substitute back to find x