Value mixture and uniform motion problems?

Question

I know how to do these type of problems but I can't figure out how to do these 2 particular ones.

1) A jogger runs a distance at a speed of 8 mph and returns the same distance at a running speed of 6 mph. Find the total distance that the jogger runs if the total time running is 1 hour and 45 minutes.

2) How many ounces of pure water must be added to 60 oz of an 8% salt solution to make a 3% salt solution?

Thanks!

Answer 1

1)
Time = distance / speed
Let x be the distance one way.
time to + time back = 1.75
x/8 + x/6 = 1.75
6x + 8x = 84
14x = 84
x = 6 miles

2)
Let x be the amount of salt in the solution.
x/60 = 8/100
x = 4.8 oz of salt

Let y be the amount of water altogether in the 3% solution.
4.8/y = 3/100
y = 160 oz

Hence you must add 160 - 60 = 100oz of water.

Answer 2

You need to know the equations you will work with.
Drawing pictures help
In the first one distance = speed x time.
The distance one way = distance the other way so we can set the two legs equal to each other.
So let the time for the first leg be x, the time for the other be 105-x (in minute).
Set and solve the equality for x. Then solve a speed x time eqtn for distance. Then double it.

In the second one, you conserve salt.
Salt = Volume * % solution.
You start with 4.8 ounce of salt. Now find the volume that provides that salt with 3% solution.
The volume difference is the answer.

Answer 3

ds, rs and ts are the area, fee and time that Sue rides dd, rd and td are the area, fee and time that Doreen rides To holiday one hundred and 5 miles, ds = one hundred and 5: eqn a million it takes Sue, utilising a moped,3 hours much less time than it takes Doreen to holiday sixty 4 miles utilising a bicycle. ts = td - 3: eqn 2 dd = sixty 4 : eqn 3 Sue travels 13 miles in line with hour speedier than Doreen. rs = rd + 13: eqn 4 distance = fee * time ds = rs ts or utilising eqn 2 & eqn 4 one hundred and 5 =(rd + 13) *(td -3) : eqn 5 dd = rd td : eqn 6 stumble on the situations and expenses of the two women individuals. increasing eqn 5 one hundred and 5 = rd td -3 rd + 13 td – 39 utilising eqn 6 and eqn 3 one hundred and 5 = sixty 4 -39 – 3 rd + 13 td or 13 td – 3 rd – 80 = 0 Multiply with the aid of td 13 td^2 – 80 td -3 rd td = 0 utilising eqn 6 and eqn 3 13 td^2 – 80 td – 3(sixty 4) = 0 13 td^2 – 80 td – 192 = 0 13 t² – 80 t – 192 =0 writing this in widely used form: 13 td² -80 td-192 = 0 A = 13, B = -80, C = -192 the roots are td = (-B +/- sqrt( B² -4 A C) )/(2A) -B/(2A) = 40/13 B² - 4 A*C = (-80)² -4 * (13)(-192) B² - 4 A*C = 16384 the discriminate is beneficial so there are 2 actual unequal roots td = 8 td = -24/13 The unfavourable does not make adventure so td = 8 hours utilising eqn 6 and 3 rd = sixty 4/8 = 8 miles utilising eqn 4 rs = 21 miles utilising eqn 4 and a million one hundred and 5 = 21 ts ts = 5 hours