Word problem help

[Guest]

Using a 5 minute and an 8 minute hourglass timer, how can you measure 1 minute?

Together a baseball and a football weigh 1.25 pounds, the baseball and a soccer ball weigh 1.35 pounds, and the football and soccer ball weigh 1.9 pounds. How much does each of the balls weigh?

 

[tkhunny]

Using a 5 minute and an 8 minute hourglass timer, how can you measure 1 minute?

Hint: You'll have to wait 15 minutes before your measured minute starts.

 

[Mrspi]

Together a baseball and a football weigh 1.25 pounds, the baseball and a soccer ball weigh 1.35 pounds, and the football and soccer ball weigh 1.9 pounds. How much does each of the balls weigh?

Let b = weight of baseball
Let f = weight of football
Let s = weight of soccer ball

From the statements in the problem, we know this:
b + f = 1.25
b + s = 1.35
f + s = 1.9

Solve the first equation for b in terms of f:
b = 1.25 - f

Substitute (1.25 - f) for b in the second equation:

(1.25 - f) + s = 1.35

Now, you have two equations in two variables:
(1.25 - f) + s = 1.35
f + s = 1.9

I hope you can take it from here......

 

[TchrWill]

Using a 5 minute and an 8 minute hourglass timer, how can you measure 1 minute?

1--Turn both over.
2--After the 5 min. one runs out, turn it back over there being 3 left in the 8.
3--When the 8 runs out there is 2 in the 5.
4--Turn the 8 over again.
5--When the 2 runs out of the 5, there will be 6 in the 8.
6-Turn the 5 over again.
7--When the 5 runs out there is 1 left in the 6.
8--When the remaining 1 in the 8 runs out, you have timed your 1 minute.

 

[soroban]

Hello, bware!

Together a baseball and a football weigh 1.25 pounds,
a baseball and a soccer ball weigh 1.35 pounds,
and a football and a soccer ball weigh 1.9 pounds.
How much does each of the balls weigh?

Let \(\displaystyle B\) = weight of a baseball.
Let \(\displaystyle F\) = weight of a football.
Let \(\displaystyle S\) = weight of a soccer ball.

We have: \(\displaystyle \,\begin{array}{ccc}\;[1]\;B\,+\,F\;=\;1.25\\ \:[2]\;B\,+\,S\;=\;1.35 \\[3]\;F\,+\,S\;=\;1.9\end{array}\)

Subtract [2] from [3]:\(\displaystyle \,F\,-\,B\:=\:0.55\)
. . . . . . . . . . Add [1]: \(\displaystyle F\,+\,B\;=\;1.25\)

And we have: \(\displaystyle \.2F\,=\,1.8\;\;\Rightarrow\;\;\)F = 0.9

Substitute into [1]: \(\displaystyle \,B\,+\,0.9\:=\:1.25\;\;\Rightarrow\;\;\)B = 0.35

Substitute into [3]: \(\displaystyle \,0.9\,+\,S\:=\:1.9\;\;\Rightarrow\;\;\)S = 1.0