Problem solving: distance/rate/time, investments, mixtures

[Mustang]

1. Alex and Judy are 32 miles apart on a calm lake paddling toward each other. Alex paddles at 5 mph, while Judy paddles at 8 mph. How lomng will it take them to meet?

2. Mellissa invested a sum of money at 3% simple interest. She invested three times that sum at 5% annual simple interest. If her total yearly interst from both investments was $7,200, how much was invested at 3%?

3. A chemist needs 140 ml of a 64% solution but has only 20% and 76% solution available. Find how many ml of each that should be mixed to get the desired solutiuon?

 

[Deleted member 4993]

Please show us what you have tried - so that we can help you properly.

 

[Deleted member 4993]

Re: Help problem sovling

1. Alex and Judy are 32 miles apart on a calm lake paddling toward each other. Alex paddles at 5 mph, while Judy paddles at 8 mph. How lomng will it take them to meet?

Hint:

Assume they meet after 't' hours

During that time Judy has travelled = 8*t miles

During that time Alex has travelled = 5*t miles

You know the total distance they need to travel (=32 miles)

Now solve for 't'

2. Mellissa invested a sum of money at 3% simple interest. She invested three times that sum at 5% annual simple interest. If her total yearly interst from both investments was $7,200, how much was invested at 3%?

Use simple interest formula and start forming equations and solve those

3. A chemist needs 140 ml of a 64% solution but has only 20% and 76% solution available. Find how many ml of each that should be mixed to get the desired solutiuon?

Assume the scientist requres 'x' mL of 20% solution

Then he needs (140-x) mL of 76% solution.

Applying mass balance

20% of x + 76% of (140-x) = 64% of 140

Solve for 'x' from above

 

[Kristy]

Re: Help problem sovling

3. A chemist needs 140 ml of a 64% solution but has only 20% and 76% solution available. Find how many ml of each that should be mixed to get the desired solutiuon?

Assume the scientist requres 'x' mL of 20% solution

Then he needs (140-x) mL of 76% solution.

Applying mass balance

20% of x + 76% of (140-x) = 64% of 140

Solve for 'x' from above

\

.2x + .75(140-x) = .64 * 140
.2x + .75*140 - .75x = .64*140
.2x - .75x + 75*140 = .64*140
get out a calculator or pen & paper:
-0.55x + 105 = 89.6
-0.55x + 105 -105 = 89.6 -105
-0.55x = -15.4
divide both sides by -0.55x


\(\displaystyle \L\\\frac{-0.55x}{-0.55}=\frac{-15.4}{-0.55}\)

x = \(\displaystyle \L\\\frac{-15.4}{-0.55}\)=28

So the scientist needs 28 mls of the 20% solution,
And 140-x = 140-28 = 112

 

[Deleted member 4993]

Re: Help problem sovling

.2x + .75(140-x) = .64 * 140... small mistake the solutions are 20% and 76% conc.
.2x + .75*140 - .75x = .64*140
.2x - .75x + 75*140 = .64*140
get out a calculator or pen & paper:
-0.55x + 105 = 89.6
-0.55x + 105 -105 = 89.6 -105
-0.55x = -15.4
divide both sides by -0.55x


\(\displaystyle \L\\\frac{-0.55x}{-0.55}=\frac{-15.4}{-0.55}\)

x = \(\displaystyle \L\\\frac{-15.4}{-0.55}\)=28

So the scientist needs 28 mls of the 20% solution,
And 140-x = 140-28 = 112

The answer would be

30 mL of 20% solution and
110 mL of 76% solution