Direct Variation Question 3 with Solutions

Direct Variation Question 3 with Solutions

Last week, I went over direct variation, and I gave you three direct variation problems to try on your own. You can see that post here: Direct Variation

Today I would like to give a solution to the last of those three problems. You can see solutions to the first two problems here: Q1 Q2

**Example: **If *y* is directly proportional to *x*, which of the following could express *y* in terms of *x* ?

A) 7*x*

B) *x*^{7}

C) *x* + 7

D) 7/*x*

Try to solve the problem yourself before checking the solutions below.

**Solution: **Since is directly proportional to , we have that is a constant times . The only answer choice that satisfies this requirement is choice A.

**More Problems with Explanations**

If you are preparing for the SAT, ACT, or an SAT math subject test, you may want to take a look at the Get 800 collection of test prep books.

And if you liked this article, please share it with your Facebook friends:

Direct Variation Question 2 with Solutions

Direct Variation Question 2 with Solutions

Last week, I went over direct variation, and I gave you three direct variation problems to try on your own. You can see that post here: Direct Variation

Today I would like to give a solution to the second of those three problems. You can see solutions to the first problem here: Direct Variation Q1.

**Example: **The amount of revenue that an online magazine retailer makes in a month is directly proportional to the number of active subscribers to the magazine. In July, the magazine had a total of 1200 subscribers, and the retailer reported revenue of $7200. In August, the online magazine had a total of 1500 subscribers. How much revenue did the retailer make?

Try to solve the problem yourself before checking the solutions below.

**Solutions: **

**(1) **Since the revenue, *R*, is directly proportional to the number of subscribers, *x*, *R* = *kx* for some constant *k*. We are given that *R *= 7200 when *x *= 1200, so that 7200 = *k*(1200), or *k *= 7200/1200 = 6. Thus, *y* = 6*x*. When *x *= 1500, we have *y *= 6 ⋅ 1500 = **9000**.

**(2) **Since *R* is directly proportional to *x*, *R*/*x *is a constant. So we get the following ratio: 7200/1200 = *R*/1500. Cross multiplying gives 1200*R *= 7200 ⋅ 1500, or equivalently, *R* = (7200⋅1500)/1200 = **9000**.

**(3) **The graph of *R *= *f*(*x*) is a line passing through the points (0, 0) and (1200, 7200). The slope of this line is (7200 – 0)/(1200 – 0) = 6. Writing the equation of the line in slope-intercept form we have *y* = 6*x*. As in solution 1, when *x *= 1500, we have *y *= 6 ⋅ 1500 = **9000**.

**More Problems with Explanations**

If you are preparing for the SAT, ACT, or an SAT math subject test, you may want to take a look at the Get 800 collection of test prep books.

And if you liked this article, please share it with your Facebook friends:

Direct Variation Question 1 with Solutions

Direct Variation Question 1 with Solutions

Yesterday, I went over direct variation, and I gave you three direct variation problems to try on your own. You can see that post here: Direct Variation

Today I would like to give a solution to the first of those three problems. I will give solutions to the other three problems throughout this week.

**Example: **If *y* = *kx* and *y* = 7 when *x* = 11, then what is *y* when *x* = 33?

Try to solve the problem yourself before checking the solutions below.

**Solutions: **

**(1) **We are given that *y* = 7 when *x* = 11, so that 7 = *k*(11), or *k* = 7/11. Therefore *y* = 7*x*/11. When *x* = 33, we have *y* = 7(33)/11 = **21**.

**(2) **Since *y* varies directly as *x*, *y*/*x *is a constant. So, we get the following ratio: 7/11 = *y*/33. Cross multiplying gives 231 = 11*y*, so that *y* = **21**.

**(3) **The graph of *y* = *f*(*x*) is a line passing through the points (0,0) and (11, 7) The slope of this line is (7 – 0)/(11 – 0) = 7/11. Writing the equation of the line in slope-intercept form we have *y* = 7/11 *x*. As in solution 1, when *x* = 33, we have *y* = 7(33)/11 = **21**.

**(4) **To get from *x* = 11 to *x* = 33 we multiply *x* by 3. So we have to also multiply *y* by 3. We get 3(7) = **21**.

**More Problems with Explanations**

If you are preparing for the SAT, ACT, or an SAT math subject test, you may want to take a look at the Get 800 collection of test prep books.

And if you liked this article, please share it with your Facebook friends:

Direct Variation

Direct Variation

Today I would like to talk about **direct variation** – a topic that shows up on standardized tests such as the ACT, SAT, and GRE.

The following are all equivalent ways of saying the same thing:

(1) *y* varies directly as *x. *

(2) *y* is directly proportional to *x.*

(3) *y *= *kx* for some constant *k.*

(4) *y*/*x *is constant.

(5) the graph of *y *= *f*(*x*) is a nonvertical line through the origin.

**Example:** In the equation *y* = 5*x*, *y* varies directly as *x*. Here is a partial table of values for this equation.

Note that we can tell that this table represents a direct relationship between *x* and *y* because 5/1 = 10/2 = 15/3 = 20/4. Here the **constant of variation** is 5.

Here is a graph of the equation.

Note that we can tell that this graph represents a direct relationship between *x* and *y* because it is a nonvertical line through the origin. The constant of variation is the slope of the line, in this case *m *= 5.

The various equivalent definitions of direct variation lead to several different ways to solve problems.

**Example: **If *y* = *kx* and *y* = 5 when *x* = 8, then what is *y* when *x* = 24 ?

Try to solve the problem yourself before checking the solutions below.

**Solutions: **

**(1) **We are given that *y* = 5 when *x* = 8, so that 5 = *k*(8), or *k* = 5/8. Therefore *y* = 5*x*/8. When *x* = 24, we have *y* = 5(24)/8 = **15**.

**(2) **Since *y* varies directly as *x*, *y*/*x *is a constant. So, we get the following ratio: 5/8 = *y*/24. Cross multiplying gives 120 = 8*y*, so that *y* = **15**.

**(3) **The graph of *y* = *f*(*x*) is a line passing through the points (0,0) and (8,5) The slope of this line is (5 – 0)/(8 – 0) = 5/8. Writing the equation of the line in slope-intercept form we have *y* = 5/8 *x*. As in solution 1, when *x* = 24, we have *y* = 5(24)/8 = **15**.

**(4) **To get from *x* = 8 to *x* = 24 we multiply *x* by 3. So we have to also multiply *y* by 3. We get 3(5) = **15**.

Here are a few more problems for you to try. I will provide solutions to these over the next few days.

1. If *y* = *kx* and *y* = 7 when *x* = 11, then what is *y* when *x* = 33?

2. The amount of revenue that an online magazine retailer makes in a month is directly proportional to the number of active subscribers to the magazine. In July, the magazine had a total of 1200 subscribers, and the retailer reported revenue of $7200. In August, the online magazine had a total of 1500 subscribers. How much revenue did the retailer make?

3. If *y* is directly proportional to *x*, which of the following could express *y* in terms of *x* ?

A) 7*x*

B) *x*^{7}

C) *x* + 7

D) 7/*x*

**More Problems with Explanations**

And if you liked this article, please share it with your Facebook friends:

**SAT Chemistry Problem with Solution –**

Product and Reactant Relationships

Product and Reactant Relationships

Today I would like to give a solution to yesterday’s SAT Chemistry problem*. * You can see the original question here: SAT Chemistry Problem – Product and Reactant Relationships

This problem is from the book *320 SAT Chemistry Problems arranged by Topic and Difficulty Level. *Here is the problem one more time, followed by the solution.

**Level 4 Product and Reactant Relationships**

How many moles of water can be produced by the reaction of 4 moles of hydrogen gas with 4 moles of oxygen gas?

A) 2

B) 4

C) 8

D) 10

E) 16

***Solution:** According to the stoichiometry, hydrogen is the limiting reactant. Only 2 moles of water will be produced, choice A.

**Remarks: **(1) The first step for this problem is to write out the chemicals involved and create a balanced equation. This will allow us to determine the relationship between the reactants. It is important to know the coefficients of the reaction because it is very likely that one reactant will be the **limiting reagent **or **limiting reactant**. The unbalanced equation will be H_{2} + O_{2} → H_{2}O. For a review of balancing equations, see problem 12 in *320 SAT Chemistry Problems*. The balanced reaction ends up being 2H_{2} + O_{2} → 2H_{2}O.

(2) Even though there is an equal amount of moles of hydrogen and oxygen given, the hydrogen will run out first. This is because the reaction uses twice as much hydrogen as oxygen. We can compare limiting reagent problems to building sandwiches. In this hypothetical, we will say that a sandwich is two slices of bread and three slices of cheese. The stoichiometry would be 2 Bread + 3 Cheese → 1 sandwich.

If you were given 10 slices of bread and 10 slices of cheese and told to make as many sandwiches as possible, you would see that the cheese runs out first. We can define this mathematically by dividing the total amount of each sandwich ‘reactant’ by its coefficient. Looking at the bread ‘reactant’, we get (10 slices of bread)/(2 slices per sandwich) = 5 sandwiches. Looking at the cheese ‘reactant’, we get (10 slices of cheese)/(3 slices per sandwich) = 3 . 3 sandwiches. We would run out of cheese and have leftover bread. This is what it means for a reactant to be limiting. It prevents the use of the other reactant because it runs out.

**More Test Prep Books from Get 800**

Check out the other SAT subject test books from the *Get 800* collection here: Get 800 SAT Subject Test Books

Thank you all for your continued support!

**SAT Chemistry Problem –**

Product and Reactant Relationships

Product and Reactant Relationships

Today I would like to give you an SAT Chemistry problem from the book* 320 SAT Chemistry Problems arranged by Topic and Difficulty Level. * I will post a full solution tomorrow.

**Level 4 Product and Reactant Relationships**

How many moles of water can be produced by the reaction of 4 moles of hydrogen gas with 4 moles of oxygen gas?

A) 2

B) 4

C) 8

D) 10

E) 16

**More Test Prep Books from Get 800**

Check out the other SAT subject test books from the *Get 800* collection here: Get 800 SAT Subject Test Books

If you think your friends might be interested in this article, please share it with them on Facebook:

Thank you all for your continued support!

**Draw Your Own Figure**

Question 4 with Solution

Question 4 with Solution

Last week, I went over a strategy for solving certain math problems on standardized tests by drawing a figure, and I gave you four problems to try on your own. You can see that post here: Math Strategy: Draw Your Own Figure

Today I would like to give a solution to the last of those four problems. You can see solutions for the first three problems here: Draw Your Own Figure: Q1 Q2 Q3.

**Example: **In rectangle *PQRS*, point *T *is the midpoint of side *PQ*. If the area of quadrilateral* QRST* is 0.8, what is the area of rectangle *PQRS* ?

Try to solve the problem yourself before checking the solution below.

**Solution: **Let’s begin by drawing a picture.

This picture alone really sheds some light on the situation. Lets now chop up our picture into 4 equal parts.

To get the area of one of those pieces simply divide 0.8 by 3. In our calculator we get

.266666666666667 or 4/15 if we change back to a fraction.

This is 1/4 of the area of the rectangle, so we simply multiply this result by 4 to get the answer **16/15**.

**More Problems with Explanations**

And if you liked this article, please share it with your Facebook friends:

**Draw Your Own Figure**

Question 3 with Solution

Question 3 with Solution

Last week, I went over a strategy for solving certain math problems on standardized tests by drawing a figure, and I gave you four problems to try on your own. You can see that post here: Math Strategy: Draw Your Own Figure

Today I would like to give a solution to the third of those four problems. You can see solutions for the first two problems here: Draw Your Own Figure: Q1 Q2.

**Example: **Point *A* is a vertex of a 6-sided polygon. The polygon has 6 sides of equal length and 6 angles of equal measure. When all possible diagonals are drawn from point *A* in the polygon, how many triangles are formed?

Try to solve the problem yourself before checking the solution below.

**Solution: **We draw a picture.

Observe that the number of triangles is **4**.

**More Problems with Explanations**

And if you liked this article, please share it with your Facebook friends: