### All SAT Math Resources

## Example Questions

### Example Question #61 : Algebra

Phillip can paint square feet of wall per minute. What area of the wall can he paint in 2.5 hours?

**Possible Answers:**

**Correct answer:**

Every minute Phillip completes another * square feet of painting. To solve for the total area that he completes, we need to find the number of minutes that he works.*

There are 60 minutes in an hour, and he paints for 2.5 hours. Multiply to find the total number of minutes.

If he completes * square feet per minute, then we can multiply ** by the total minutes to find the final answer.*

### Example Question #1 : Direct And Inverse Variation

The value of varies directly with the square of and the cube of . If when and , then what is the value of * when and ?*

**Possible Answers:**

**Correct answer:**

Let's consider the general case when *y* varies directly with *x*. If *y* varies directly with *x*, then we can express their relationship to one another using the following formula:

*y* = *kx*, where *k* is a constant.

Therefore, if *y* varies directly as the square of *x* and the cube of *z*, we can write the following analagous equation:

*y* = *kx*^{2}*z*^{3}, where *k* is a constant.

The problem states that *y* = 24 when *x* = 1 and *z* = 2. We can use this information to solve for *k* by substituting the known values for *y*, *x*, and *z*.

24 = *k*(1)^{2}(2)^{3} = *k*(1)(8) = 8*k*

24 = 8*k*

Divide both sides by 8.

3 = *k*

*k* = 3

Now that we have *k*, we can find *y* if we know *x* and *z*. The problem asks us to find *y* when *x* = 3 and *z* = 1. We will use our formula for direct variation again, this time substitute values for *k*, *x*, and *z*.

*y* = *kx*^{2}*z*^{3}

*y* = 3(3)^{2}(1)^{3} = 3(9)(1) = 27

*y* = 27

The answer is 27.

### Example Question #3 : Variables

In a growth period, a population of flies triples every week. If the original population had 3 flies, how big is the population after 4 weeks?

**Possible Answers:**

**Correct answer:**

We know that the initial population is 3, and that every week the population will triple.

The equation to model this growth will be , where is the initial size, is the rate of growth, and is the time.

In this case, the equation will be .

Alternatively, you can evaluate for each consecutive week.

Week 1:

Week 2:

Week 3:

Week 4:

### Example Question #5 : Variables

and are the diameter and circumference, respectively, of the same circle.

Which of the following is a true statement? (Assume all quantities are positive)

**Possible Answers:**

varies inversely as the fourth power of .

varies directly as the fourth power of .

varies inversely as the fourth root of .

varies directly as the fourth root of .

varies directly as .

**Correct answer:**

varies directly as .

If and are the diameter and circumference, respectively, of the same circle, then

.

By substitution,

Taking the square root of both sides:

Taking as the constant of variation, we get

,

meaning that varies directly as .

### Example Question #1 : How To Use The Direct Variation Formula

is the radius of the base of a cone; is its height; is its volume.

; .

Which of the following is a true statement?

**Possible Answers:**

varies directly as the third power of .

varies directly as .

varies directly as the fifth power of .

varies directly as the fifth root of .

varies directly as the cube root of .

**Correct answer:**

varies directly as the fifth power of .

The volume of a cone can be calculated from the radius of its base , and the height , using the formula

, so .

, so .

, so by substitution,

Square both sides:

If we take as the constant of variation, then

,

and varies directly as the fifth power of .

### Example Question #1 : How To Use The Direct Variation Formula

and are the radius and volume, respectively, of a given sphere.

.

Which of the following is a true statement?

**Possible Answers:**

varies directly as the sixth power of .

varies inversely as the sixth root of .

varies directly as .

varies directly as the sixth root of .

varies inversely as the sixth power of .

**Correct answer:**

varies directly as the sixth power of .

The volume of a sphere can be calculated from its radius as follows:

Therefore, squaring both sides, we get

Substituting:

If we let the constant of variation be , we see that

,

and varies directly as , the sixth power of .

### Example Question #2 : Direct And Inverse Variation

The temperature at the surface of the ocean is . At meters below the surface, the ocean temperature is . By how much does the temperature decrease for every meters below the ocean's surface?

**Possible Answers:**

**Correct answer:**

This may seem confusing, but is pretty straightforward.

Thus, for every 125 meters below the surface, the temperature decreases by one degree.

To find how much it decreases with every 100 meters, we need to do the following:

Thus, the temperature decreases by every 100 meters.

### Example Question #1 : How To Use The Inverse Variation Formula

The square of varies inversely with the cube of . If when , then what is the value of when ?

**Possible Answers:**

**Correct answer:**

When two quantities vary inversely, their products are always equal to a constant, which we can call k. If the square of x and the cube of y vary inversely, this means that the product of the square of x and the cube of y will equal k. We can represent the square of x as x^{2} and the cube of y as y^{3}. Now, we can write the equation for inverse variation.

x^{2}y^{3} = k

We are told that when x = 8, y = 8. We can substitute these values into our equation for inverse variation and then solve for k.

8^{2}(8^{3}) = k

k = 8^{2}(8^{3})

Because this will probably be a large number, it might help just to keep it in exponent form. Let's apply the property of exponents which says that a^{b}a^{c} = a^{b+c}.

k = 8^{2}(8^{3}) = 8^{2+3} = 8^{5}.

Next, we must find the value of y when x = 1. Let's use our equation for inverse variation equation, substituting 8^{5} in for k.

x^{2}y^{3} = 8^{5}

(1)^{2}y^{3} = 8^{5}

y^{3} = 8^{5}

In order to solve this, we will have to take a cube root. Thus, it will help to rewrite 8 as the cube of 2, or 2^{3}.

y^{3} = (2^{3})^{5}

We can now apply the property of exponents that states that (a^{b})^{c} = a^{bc}.

y^{3} = 2^{3•5} = 2^{15}

In order to get y by itself, we will have the raise each side of the equation to the 1/3 power.

(y^{3})^{(1/3)} = (2^{15})^{(1/3)}

Once again, let's apply the property (a^{b})^{c} = a^{bc}.

y^{(3 • 1/3) }= 2^{(15 • 1/3)}

y = 2^{5} = 32

The answer is 32.

### Example Question #1 : How To Use The Inverse Variation Formula

varies directly as and inversely as .

and .

Which of the following is true about ?

**Possible Answers:**

varies inversely as the square root of and directly as the fourth root of .

varies directly as the square of and inversely as the fourth power of .

varies inversely as the square of and directly as the fourth power of .

varies directly as the square root of and inversely as the fourth root of .

varies directly as both the square root of and the fourth root of .

**Correct answer:**

varies inversely as the square of and directly as the fourth power of .

varies directly as and inversely as , so for some constant of variation ,

.

We can square both sides to obtain:

.

, so .

By substitution,

.

Using as the constant of variation, we see that varies inversely as the square of and directly as the fourth power of .

### Example Question #1 : How To Use The Inverse Variation Formula

The radius of the base of a cylinder is ; the height of the same cylinder is ; the cylinder has volume 1,000.

Which of the following is a true statement?

Assume all quantities are positive.

**Possible Answers:**

varies directly as the square of .

varies inversely as .

varies dorectly as the square root of .

varies inversely as the square of .

varies inversely as the square root of .

**Correct answer:**

varies inversely as .

The volume of a cylinder can be calculated from its height and the radius of its base using the formula:

, so ;

, so .

The volume is 1,000, and by substitution, using the other equations:

If we take as the constant of variation, we get

,

meaning that varies inversely as .