Thread: Homework somone help me please?

1.On a quiz of 10 questions, every correct answer earns 5 points, but every rong answer deducts 2 points. Questions left blank earn 0 point. Tim got c questions correct, w questions wrong and left B questions blank. he earned a score of 31 points on the quiz. What is the order triple (c,w,b)?

2.Due to a bowling score of 204 in his last game, Remy raised his average from exactly 156 to 158. What score must eh bowl in the next game to raise his overall everage to exactly 159?


3. Of all four-digit positive integers contaning only digits from the set {2, 4, 6, 8} what fraction of them have at least one of their digits repeated? Exspress ur answer as a common fraction.

Originally Posted by Brent

1.On a quiz of 10 questions, every correct answer earns 5 points, but every rong answer deducts 2 points. Questions left blank earn 0 point. Tim got c questions correct, w questions wrong and left B questions blank. he earned a score of 31 points on the quiz. What is the order triple (c,w,b)?

Tim got 5c points for his correct answers, since they are worth 5 points each. similarly, he earned -2w points for his wrong answers, and 0b = 0 points for the questions he left blank. thus we can represent his total points by

5c - 2w = 31

moreover, since there were 10 questions in all,

c + w + b = 10

now finish up

2.Due to a bowling score of 204 in his last game, Remy raised his average from exactly 156 to 158. What score must eh bowl in the next game to raise his overall everage to exactly 159?

remember, the average score is the sum of all the scores divided by the number of scores.

Let the sum of his scores before the last game be x and the number of games he played at that point be n, then his average was



now, he played another game, which brings the number of his games to n + 1, moreover, the sum of his scores increased by 204, so that his new average of 158 is described by




with those two equations, you can find x and n, and then use that to answer the question

3. Of all four-digit positive integers contaning only digits from the set {2, 4, 6, 8} what fraction of them have at least one of their digits repeated? Exspress ur answer as a common fraction.

there are such integers in all (why?)

there are of those digits that have no repeats (why?)

now, how would you find those that have at least one repeat? then what would the answer be?

Tanks jhevon for the help. I understand how to finish out #1, but I still dont understand how to finish out the last 2. I picked this work book up at a used book fair, so there arent any resources in it to help me. I'm only in the 4th grade.

Originally Posted by Brent

Tanks jhevon for the help. I understand how to finish out #1, but I still dont understand how to finish out the last 2. I picked this work book up at a used book fair, so there arent any resources in it to help me. I'm only in the 4th grade.

be specific as to what you don't get. problem 2 is very similar to problem 1, it involves solving the simultaneous equations

note that the first equation says x = 156n and the second says x + 204 = 158n + 158

for the second, it is more of a counting argument, and you have to realize: the number of integers with repeats = the total number of integers - the integers without repeats. where are you stuck on this also?

Okay Jhevon after looking at #1 i think i may have set it up wrong.

5c-2w=31
5c=2w+31
c=5(2w+31)
c=10w+155

Am I on the right path so far?

Originally Posted by Brent

Okay Jhevon after looking at #1 i think i may have set it up wrong.

5c-2w=31
5c=2w+31
c=5(2w+31)
c=10w+155

Am I on the right path so far?

nope, that's wrong

to go from the second to the third line, you would have to divide by 5, won't you?

incidentally, for this problem, since we were only given 2 equations with 3 unknowns, some trial and error must be used. however, it should not be that hard, since all the numbers c, w, and b are less than or equal to 10. just think of two positive integers c and w such that 5c - 2w = 31. then you can find b when done

yea i know the answers to c=7 w=2 b=1 but i wont to make sure how i arrived at the answer.

Originally Posted by Brent

yea i know the answers to c=7 w=2 b=1 but i wont to make sure how i arrived at the answer.

in this case, saying that you found c and w "by inspection" or "by trial and error" is fine, since we do not have a third equation. then say you calculated b by the second equation

I want to thank you jhevon for your time, but I'm still confused how to set up #1 & #3 to come to the answers I have. But thank you for your time, I've gota get to bed b4 my mom kicks my butt.

Originally Posted by Brent

I want to thank you jhevon for your time, but I'm still confused how to set up #1 & #3 to come to the answers I have. But thank you for your time, I've gota get to bed b4 my mom kicks my butt.

i though we were done with 1?

for 3, note that to make the integers you have 4 spots to fill. you can choose any of the 4 integers for the first spot, and then any of the 4 for the second spot (since repeats are allowed), then any of the 4 for the third spot, and then any of the 4 for the fourth spot. so in all there are 4*4*4*4 = 4^4 integers that we can form in total.

the ones without repeats we can form in 4*3*2*1 = 4! ways. since we can choose any of the 4 for the first spot, once that has been chosen, there are only 3 choices for the second spot, then 2 for the third and one for the fourth.

and thus you get the numbers i gave you. now what can you do with them?