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The sum of a number, 1/6 of that number, 2 1/2 of that number, and 7 is 12 1/2. Find the number.

Tema [17]

Answer:

2 $\frac{1}{16}$

Step-by-step explanation:

The question state that; the sum of a number, 1/6 of that number, 2 1/2 of that number, and 7 is 12 1/2. Find the number.

First, we need to interpret and represent this statement mathematically

Let n be the number.

The sum means 'addition'

1/6 of that number is 1/6 of n = 1/6 × n = $\frac{n}{6}$

2 1/2 of that number is 2 1/2 of n = 2 1/2 × n = 5/2 × n = $\frac{5n}{2}$

So the question says that, when we add; $\frac{n}{6}$ , $\frac{5n}{2}$ and 7 all together, the value is 12 1/2

That is;

$\frac{n}{6}$ + $\frac{5n}{2}$ + 7 = 12 1/2

$\frac{n}{6}$ + $\frac{5n}{2}$ + 7 = $\frac{25}{2}$ ( changing 12 1/2 to improper fraction will give $\frac{25}{2}$ )

So we can now go ahead and solve for n

$\frac{n}{6}$ + $\frac{5n}{2}$ + 7 = $\frac{25}{2}$

subtract 7 from both-side of the equation

$\frac{n}{6}$ + $\frac{5n}{2}$ = $\frac{25}{2}$ - 7

$\frac{n + 15n}{6}$ = $\frac{25 -14}{2}$

$\frac{16n}{6}$ = $\frac{11}{2}$

$\frac{16n}{6}$ can be reduced to give us $\frac{8n}{3}$

$\frac{8n}{3}$ = $\frac{11}{2}$

Cross multiply

8n × 2 = 11× 3

16n = 33

Divide both-side of the equation by 16

$\frac{16n}{16}$ = $\frac{33}{16}$

(On the left-hand side of the equation, 16 will cancel-out 16 leaving us with just, while on the right-hand side of the equation 33 will be divided by 16)

n = $\frac{33}{16}$ = $2\frac{1}{16}$

Therefore the number is $2\frac{1}{16}$

5 0

4 years ago

The difference of two numbers is 16 the greater number is five less than four times a small number find the two numbers

Elis [28]

Answer:

The numbers are 23 and 7

Step-by-step explanation:

1. Set up the equations

Let x = the greater number

and y = the smaller number. Then

(1) x - y = 16

4y = 4 times the smaller number and

4y - 5 = 5 less than 4 times the smaller number. Then

(2) x = 4y - 5

You have a system of two equations:

$\begin{cases}(1) & x - y = 16\\(2) & x = 4y - 5\end{cases}$

2. Solve the equations

$\begin{array}{lrcll}(3) & 4y - 5 - y & = & 16 &\text{Substituted (2) into (1)}\\& 3y - 5 & = & 16 &\text{Simplified}\\ & 3y & = & 21 &\text{Added 5 to each side}\\(4) &y & = & \mathbf{7} &\text{Divided each side by 3} \\& x - 7 & = & 16 & \text{Substituted (4) into (1)}\\& x& = & \mathbf{23} & \text{Added 7 to each side}\\\end{array}$