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# Are You Good At Math? Take the Ultimate Math Quiz Part 2!?!?!?!?!?! 10 Points goes to the best explanations!?

6. Jane always works at a constant rate, and Frank always works at a constant rate. Working together, Jane & Frank can put out a fire in 5 minutes. If Jane were to work twice as fast as her normal constant rate, the 2 of them could put out the fire in 4 minutes. Find the number of minutes it would take Frank to put out the fire by himself. Express your answer as an improper fraction reduced to lowest terms.

7. The 4 members of the Frosh Math Team at Red Bud High School have first names of Al, Bonita, Carol, and Dan. If a member of the Frosh Math Team at Red Bud High School is selected at random, find the probability that the # of letters in that person's first name is more than 4. Express your answer as a common fraction reduced to lowest Terms.

8. Claire has a pet pig and a pet calf. One is male, and one is a female. Together, these 2 pets weigh 125 kilograms. The pig's weight is an odd #. The male weighs 4 times as much as the female. Find the number of kilograms in the calf.

### 4 Answers

- astazangastaLv 51 decade agoFavorite Answer
6. If Jane puts out j fire-units/min, and Frank f, then the number of fire-units put out in 5 minutes is: 5*j+5*f. This is equal to 4*2j+4*f, meaning 3j = f. There are thus 20/3*f fire-units to be put out, meaning it would take Frank 6 2/3 minutes.

7. Since only Bonita and Carol have more than 4 letters in their name, the condition can be true only if they are selected. Thus, the probability is 1/2.

8. The pig must be the female, since the male is a multiple of 4 and thus not odd. 4x+x = 125, x = 25, and so the (male) calf weighs 100 kilos.

- 1 decade ago
6. it is not possible.

Jane = x Frank = y

x + y =5

2x +y =4

x= -1 (no one can do less than nothing work)

7. There are two people with more than four letters in their first name. 2/4 = 1/2 or 50% of the people.

8. This is mainly a guess and check problem. Let the pig = p and the calf = c.

you know that p + c =125

then, we'll guess that the pig is the male, so p= 4c

plug 4c in for p and you get 5c = 125, c = 25. That would leave the pig to be 100 kg. That is an even number. However, if you said that the calf is the male, then c=4p. you would plug it in again and get 5p = 125, or p = 25. 25 is an odd number, so you knwo it is correct.

- 4 years ago
the answer to the first question: enable x be the type of one million's, so 10x is the type of two's, enable y be the type of 5's. So 21x + 5y = 221. (that is termed a Diophantine equation.) Rewrite as 20x + 5y - 220 = a million-x. because the left area is divisible with the help of 5, also x-a million should be. Say x=a million + 5k. Substituting in and simplifying we get, 21k + y = 40. Repeat the trick above with the help of writing 21k - 21 = 19 - y. So 19 - y is divisible with the help of 21. This leaves only 2 opportunities: y = 19, ok=a million, x = 6 or y = 40, ok = 0, x = a million. both a form of contain options to the equation. So the type of a possibility 2 dollar expenses is 60 or 10, which upload to 70.

- ironduke8159Lv 71 decade ago
J = Jane's rate

F = Frank's rate

Then F*J/(F+J) = 5 minutes <--- Eq 1

and 2*F*J/(F+2*J) = 4 minutes <--- Eq 2

Solve Eq 2 for J

FJ/(F+2J) =2

FJ = 2F + 4J

FJ -4J = 2F

J(F-4)= 2F

J = 2F/(F-4) Substitute into Eq 1

2F^2/(F-4)/(F + 2F/(F-4)) = 5

2F^2/(F-4)/(F^2-2F)/(F-4)) = 5

2F^2/(F^2-2F) = 5

2F/(F-4) = 5

2F= 5F- 20

3F + 20

F= 20/3

2/4 = 1/2

Let x = pig's weight

Then 4x = calf's weight.

4x+x = 125

5x = 125 so x = 25 = pig's weight (the pig is female)

4x = 100 = calf's weight {the calf is male)