Ratio Practice Problems - Master Ratios and Proportions

To solve this problem, we'll determine the number of orange balls by calculating the fraction of the total number of balls:

• Step 1: Identify the total number of balls, $28$.
• Step 2: Note the fraction representing the orange balls, $\frac{1}{4}$.
• Step 3: Apply the formula to find the number of orange balls:
  Number of orange balls $= 28 \times \frac{1}{4}$

Now, let's perform the calculation:
$28 \times \frac{1}{4} = 28 \div 4 = 7$

Therefore, the number of orange balls in the box is $7$.

To solve this problem, we will determine the number of white balls in the box using a fraction of the total number of balls.

We are given the total number of balls in the box as 18, and we know that $\frac{2}{3}$ of these balls are white. To find the number of white balls, we follow these steps:

• Step 1: Identify the total quantity, which is 18 balls.
• Step 2: Use the given fraction $\frac{2}{3}$ to find the number of white balls.
• Step 3: Multiply the total number of balls by the fraction of white balls: $18 \times \frac{2}{3}$.

Perform the calculation:

$18 \times \frac{2}{3} = 18 \times 0.6667 = 12$

Alternatively, calculate directly using fractions:

$18 \times \frac{2}{3} = \frac{18 \times 2}{3} = \frac{36}{3} = 12$

Thus, the total number of white balls in the box is 12.

Therefore, the correct answer is choice 12.

The area of a circle is calculated using the following formula:

where r represents the radius.

Using the formula, we calculate the areas of the circles:

Circle 1:

π*4² =

π16

Circle 2:

π*10² =

π100

To calculate how much larger one circle is than the other (in other words - what is the ratio between them)

All we need to do is divide one area by the other.

100/16 =

6.25

Therefore the answer is 6 and a quarter!

Let's find side BC

Based on what we're given:

$\frac{AB}{BC}=\frac{x}{BC}=\sqrt{\frac{x}{2}}$

$\frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}$

$\sqrt{2}x=\sqrt{x}BC$

Let's divide by square root x:

$\frac{\sqrt{2}\times x}{\sqrt{x}}=BC$

$\frac{\sqrt{2}\times\sqrt{x}\times\sqrt{x}}{\sqrt{x}}=BC$

Let's reduce the numerator and denominator by square root x:

$\sqrt{2}\sqrt{x}=BC$

We'll use the Pythagorean theorem to calculate the area of triangle ABC:

$AB^2+BC^2=AC^2$

Let's substitute what we're given:

$x^2+(\sqrt{2}\sqrt{x})^2=m^2$

$x^2+2x=m^2$

We know that:

$\frac{AB}{BC}=\sqrt{\frac{x}{2}}$

We also know that AB equals X.

First, we will substitute the given data into the formula accordingly:

$\frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}$

$x\sqrt{2}=BC\sqrt{x}$

$\frac{x\sqrt{2}}{\sqrt{x}}=BC$

$\frac{\sqrt{x}\times\sqrt{x}\times\sqrt{2}}{\sqrt{x}}=BC$

$\sqrt{x}\times\sqrt{2}=BC$

Now let's look at triangle ABC and use the Pythagorean theorem:

$AB^2+BC^2=AC^2$

We substitute in our known values:

$x^2+(\sqrt{x}\times\sqrt{2})^2=m^2$

$x^2+x\times2=m^2$

Finally, we will add 1 to both sides:

$x^2+2x+1=m^2+1$

$(x+1)^2=m^2+1$