Question

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?
Help me please...

Answer 1

Welcome to openstudy!

Answer 2

Are there any answer choices given?

Answer 3

Do you need to write a system of equations and then solve it?

Answer 4

Use algebra. Label your variable
Let x = value of one number then,
48-x is the value of the other number. This is true because their sum is 48 and was given.

Now write their equivalent relationships into an algebraic equation.
x/3 + 5 = (48 - x)/6 Now solve for x ......it is that simple.

Answer 5

Now I rewrite the equation:
x/3 + 5 = 8 - x/6 all I did here was the the quantity in the parentheses by the 6
I now subtract 5 from both sides of the equal sign, getting:
x/3 = 8 -5 - x/6, I now add x/6 to both sides. Getting:
x/3 + x/6 = 8 - 5 Now simplify by doing the required arithmetic operations on the equation:
1/2 x = 3 Multiplying both sides by 2 to get 1x
x = 6 one of the values and the smallest
the other value would be 48 - x or 48 - 6 = 42
Good luck with your studies.

Answer 6

@radar Your second equation is incorrect.
Check your numbers (6 and 42) with the original statements of the problem, and you'll see that they don't work.
This is your mistake:
In general, if a is 5 greater than b, then a = b + 5.
You did a + 5 = b.

If 1/3 of x is 5 greater than 1/6 of y, then
1/3 x = 1/6 y + 5
You have
1/3 x + 5 = 1/6 y

Answer 7

You can do this also with a system of equations.
Let x = one number
and let y = the other number

You can get an equation out of this statement.
The sum of two numbers is 48.
\(x + y = 48\)

You get the second equation out of this statement.
If one third of one number is 5 greater than one sixth of another number
\(\dfrac{1}{3}x = \dfrac{1}{6}y + 5\)

Now we have a system of equations:

\(x + y = 48\)
\(\dfrac{1}{3}x = \dfrac{1}{6}y + 5\)

We multiply both sides of the second equation by 6 to get rid of the denominators.
\(x + y = 48\)
\(6 \times \dfrac{1}{3}x = 6 \times \dfrac{1}{6}y + 6 \times 5\)

\(x + y = 48\)
\(2x = y + 30\)

Subtract y from both sides of the second equation.
\(x + y = 48\)
\(2x -y = 30\)

Add the equations to eliminate y.
3x = 78
x = 26

Substitute x = 26 in the first original equation and solve for y.
x + y = 48
26 + y = 48
y = 22

We have one number is 26 and the other number is 22.

Now let's check to see if our solution makes sense.

"The sum of two numbers is 48."
\(22 + 26 = 48\)

Yes, the sum of the numbers is 48.
That makes the first statement true.

Now let's check the second statement.
"one third of one number is 5 greater than one sixth of another number"
\(\dfrac{26}{3} = \dfrac{22}{6} + 5\)

\(\dfrac{26}{3} = \dfrac{11}{3} + \dfrac{15}{3} \)

\(\dfrac{26}{3} = \dfrac{26}{3} \)

This makes the second statement true.

The numbers are 26 and 22.

Answer 8

@mathstudent55 Thanks, you know I thought I was off on that, but went ahead any way. I did check my answer and with my misinterpretation of which was greater verified what I did, the only problem is....I did the wrong thing lol.

Answer 9

@mathstudent By the way, your explanation with the step by step was great

Answer 10

@sawyer345 Sorry if I erred (which I did), I hope you saw mathstudent55 solution and now understand.

Answer 11

@radar Thanks for your kind words.