Average On Five Quizzes

[harpazo]

After taking 3 quizzes, your average is 72 out of 100. What must your average be on the 5 quizzes to increase your average to 77?

Solution:

Let A = my average on 5 quizzes to increase my average to 77.

What about setting up a proportion?

3/5 = (72/100)/(A/77)

5(72/100) = 3(A/77)

360/100 = 3A/77

3A(100)= = 77(360)

300A = 27,720

A = 27,720 ÷ 300

A = 92.4

Right?

 

[Dr.Peterson]

Did you check your answer? You should always do that, when possible, before asking someone else.

But I think there's a typo in your problem, which I only realized when I tried to do the check. Look carefully:

After taking 3 quizzes, your average is 72 out of 100. What must your average be on the 5 quizzes to increase your average to 77?​

What are "the 5 quizzes"? It certainly can't mean that there are a total of 5 quizzes, because then the answer would be obvious: 77! I think it's supposed to be "the next 5 quizzes". Is that what it actually said? I'll assume it is, so there are a total of 8 quizzes.

Let's suppose that the first three quizzes are all 72, to make it easy. And suppose the last five are all 92.4. Then your claim is that the overall average should be 77. Let's see:

(72+72+72+92.4+92.4+92.4+92.4+92.4)/8 = 678/8 = 84.75​

So, no, it turns out to be wrong. But before I look at your work, you'd better confirm what the problem said, and that your work is for that problem.

The next question will be, why should your proportion be valid? It's easy to write a proportion; it's not always so easy to know whether it is appropriate for the problem.

 

[Guillem_dlc]

After taking 3 quizzes, your average is 72 out of 100. What must your average be on the 5 quizzes to increase your average to 77?

Solution:

Let A = my average on 5 quizzes to increase my average to 77.

What about setting up a proportion?

3/5 = (72/100)/(A/77)

5(72/100) = 3(A/77)

360/100 = 3A/77

3A(100)= = 77(360)

300A = 27,720

A = 27,720 ÷ 300

A = 92.4

Right?

Let [MATH]x_k[/MATH] be the score in the [MATH]k[/MATH]-th test, for each [MATH]k[/MATH] natural such that [MATH]1\leq k\leq 9[/MATH].

According to the statement, [MATH]\sum_{k=1}^4 x_k=4\cdot 76=304[/MATH] and [MATH]\sum_{k=5}^9 x_k=5\cdot 85=425[/MATH], with which the average score in the [MATH]9[/MATH] tests was

[MATH]\dfrac{\sum_{k=1}^9x_k}{9}=\dfrac{\sum_{k=1}^4 x_k+\sum_{k=5}^9 x_k}{9}=\dfrac{304+425}{9}=81.[/MATH]

 

[harpazo]

Did you check your answer? You should always do that, when possible, before asking someone else.

But I think there's a typo in your problem, which I only realized when I tried to do the check. Look carefully:

After taking 3 quizzes, your average is 72 out of 100. What must your average be on the 5 quizzes to increase your average to 77?​

What are "the 5 quizzes"? It certainly can't mean that there are a total of 5 quizzes, because then the answer would be obvious: 77! I think it's supposed to be "the next 5 quizzes". Is that what it actually said? I'll assume it is, so there are a total of 8 quizzes.

Let's suppose that the first three quizzes are all 72, to make it easy. And suppose the last five are all 92.4. Then your claim is that the overall average should be 77. Let's see:

(72+72+72+92.4+92.4+92.4+92.4+92.4)/8 = 678/8 = 84.75​

So, no, it turns out to be wrong. But before I look at your work, you'd better confirm what the problem said, and that your work is for that problem.

The next question will be, why should your proportion be valid? It's easy to write a proportion; it's not always so easy to know whether it is appropriate for the problem.

I went back to the site. It turns out that the problem as posted here is correct.

After taking 3 quizzes, your average is 72 out of 100. What must your average be on the 5 quizzes to increase your average to 77?

 

[harpazo]

Let [MATH]x_k[/MATH] be the score in the [MATH]k[/MATH]-th test, for each [MATH]k[/MATH] natural such that [MATH]1\leq k\leq 9[/MATH].

According to the statement, [MATH]\sum_{k=1}^4 x_k=4\cdot 76=304[/MATH] and [MATH]\sum_{k=5}^9 x_k=5\cdot 85=425[/MATH], with which the average score in the [MATH]9[/MATH] tests was

[MATH]\dfrac{\sum_{k=1}^9x_k}{9}=\dfrac{\sum_{k=1}^4 x_k+\sum_{k=5}^9 x_k}{9}=\dfrac{304+425}{9}=81.[/MATH]

The answer is 81. Yes? Is there an easier way to solve this problem?

 

[Dr.Peterson]

I went back to the site. It turns out that the problem as posted here is correct.

After taking 3 quizzes, your average is 72 out of 100. What must your average be on the 5 quizzes to increase your average to 77?

Odd. When we know what the correct answer is, we'll have to see whether it makes sense of the question. I think it's a typo on their part, unless you can think of a sensible interpretation of the wording as given.

The answer is 81. Yes? Is there an easier way to solve this problem?

You may have noticed that he was answering a different question from another thread, which is of a slightly different type. (I hope you did, or you haven't learned the lesson that you should check every answer, including what a book or a stranger tells you.)

So we don't yet have a solution for this thread's problem; but the other exercise gives practice in the same underlying idea.

What is the sum of the first three grades? What is the sum of all 8 grades (assuming I'm interpreting it right)? What must be the sum of the last 5?

You can also do it with algebra, letting x = the average of the last 5, and solving for x.

The answer turns out to be a nice number, so I think I've got it right. And this problem can be solved by a proportion, though quite different from yours, and I would feel more confident doing it this way.

 

[harpazo]

Odd. When we know what the correct answer is, we'll have to see whether it makes sense of the question. I think it's a typo on their part, unless you can think of a sensible interpretation of the wording as given.


You may have noticed that he was answering a different question from another thread, which is of a slightly different type. (I hope you did, or you haven't learned the lesson that you should check every answer, including what a book or a stranger tells you.)

So we don't yet have a solution for this thread's problem; but the other exercise gives practice in the same underlying idea.

What is the sum of the first three grades? What is the sum of all 8 grades (assuming I'm interpreting it right)? What must be the sum of the last 5?

You can also do it with algebra, letting x = the average of the last 5, and solving for x.

The answer turns out to be a nice number, so I think I've got it right. And this problem can be solved by a proportion, though quite different from yours, and I would feel more confident doing it this way.

Dr. Peterson,

Here is the link with the solution:

Algebra: Average Word Problems (solutions, videos, examples)

Algebra Average Problems: Word Problems Involving Averages and arithmetic mean, how to calculate a weighted mean or weighted average, Average (Arithmetic Mean), Weighted Average and Average Speed word problems with examples and step by step solutions

www.onlinemathlearning.com

I just now noticed that the solution to all three questions is posted at the above link. Check it out.

 

[Dr.Peterson]

Did you notice that in the video, they do have the word "next" in the problem? Careless of them to leave it out on their page, isn't it?

Also, he uses the algebraic method I suggested as an alternative; my questions amount to doing exactly the same things you do to solve his equation, more informally. (I hope you answered my questions, and then noticed that all the numbers he uses are the same.)

Finally, you might look at the answer and notice something that is in the ratio 5:3 that you could have used in a proportion; this is a well-known alternative technique to solve this specific kind of problem. (Algebra is a general-purpose tool that can be applied to a wide range of problems without needing to learn a special method for each.)

 

[harpazo]

Did you notice that in the video, they do have the word "next" in the problem? Careless of them to leave it out on their page, isn't it?

Also, he uses the algebraic method I suggested as an alternative; my questions amount to doing exactly the same things you do to solve his equation, more informally. (I hope you answered my questions, and then noticed that all the numbers he uses are the same.)

Finally, you might look at the answer and notice something that is in the ratio 5:3 that you could have used in a proportion; this is a well-known alternative technique to solve this specific kind of problem. (Algebra is a general-purpose tool that can be applied to a wide range of problems without needing to learn a special method for each.)

I saw the similarity. I think there is nothing more important to me as an online student here than to increase my word problem solving skills. If I can learn to form equations to solve algebra problems at least at the high school level, mission accomplished.