I need help with a few word problems, please?

Question

1) The excursion boat on a river takes two and a half hours to make the trip to a point 12 miles upstream and to return. If the rate at which the boat travels in still water is 5 times the rate of the river current, what is the rate of the current?

2) The capacities of two trucks are 3 tons and 4 tons. If the larger truck makes 3 more trips than the smaller, its total haul is 20 more tons than the smaller one. How many trips does each truck make?

3) Rick can run the 440-yard dash in 55 seconds, and Jack can run it in 88 seconds. How great of a handicap must Rick give Jack for the boys to finish the race at the same time?

I'm just not sure how to set up the equations for each question. Any help is greatly appreciated.

calculator, abacus, pen & paper, anything can be used.

Answer 1

3)

handicap = 440 - 440 * (55/88) = 165 yards
-

Answer 2

2) The larger truck makes 11 trips and the smaller truck makes 8 trips. 11*4=44, 8*3=24, difference being 20.

Here's what I did:

Take the 3 extra trips times the amount per trip (4). Result is 12. Since the difference in load size is only 1 you can simply subtract the 12 from 20. You get 8 which is the amount of trips the small truck makes. Add the 3 extra and you get the amount of the larger truck.

3) Jack needs a 264 yard head start.

Here's why:

Rick does 440/55 sec.
Jack does 440/88 sec.
Jack needs to run "x" amount less. (440-x)
So set 440/55 = (440-x)/88
Cross multiply and you get: 38720 = 24200-55x
Solve for x and you get 264 yards.

Answer 3

1) This is a fairly straightforward word problem. You just need to stick it into equations.

X is the speed of the ship
Y is the speed of the current
A and B are the time to do each trip, upstream and downstream respectively.

upstream trip (X-Y)A=12
downstream trip (X+Y)B=12
The current is 1/5th of the speed of the ship so Y=1/5X

So we get
upstream 4/5XA=12
downstream 6/5XB=12
so 4/5XA=6/5XB
A=3/2B

as A+B=2.5 we get 3/2B+B =2.5 or 2.5B=2.5

B must be 1 and A must be 1.5

sticking it back into the earlier equation 6/5XB=12
we get 6/5X=12
so the speed of the ship (X) is 10
as the current is 1/5th of the speed it is 2

2) This one is simpler

X is the number of trips they both take
Y is the number of trips the big truck takes alone
A is the capacity of the big truck
B is the capacity of the small truck

(A-B)X + AY = 20

as we know Y=3, A =4 and B=3 plug them in and solve for X

(4-3)X +4*3=20
X+12=20
X=8

So the small truck took 8 trips and the big truck took 11

3)Ignoring variables like acceleration or possible fatigue at the end this can go one of two ways.

i) If you assume by handicap they mean time, it's simply a matter of the difference in their 440 yard times. 33 seconds. That's probably not what you are lloking for.

ii) If you mean distance, you need to figure out how many yards Jack can run in a second

X is Jack's speed in yards per second

X*88=440

so X = 5yards per second

now multiply that by the difference between their total times

5*33=165yards

Jack needs a 165yard headstart.

Answer 4

The trick with word problems is always finding the equations.

1) After reading it through, let's use X to denote the speed of the current. The speed of the boat is 5X (from the last statement).

Since Distance = Rate*Time, Time = Distance/Rate.

We know from the first sentence

12/(5X-X) + 12/(5X+X) = 5/2 (simplify the denominators)
12/(4X) + 12/(6X) = 5/2 (Simplify the fractions)
3/X + 2/X = 5/2 (Multiply both sides by X)
3 + 2 = (5/2) X (Multiply both sides by 2/5)
2/5 (5) = X = 2 Miles/Hr

2) Let X be the number of trips in the Small Truck, The larger truck then makes X+3 trips. The Tonnage would be 3X for the small truck and 4(X+3) for the large truck.

4(X+3) = 3X + 20 (from the second statement.
4X+12 = 3X + 20 (Subtract 3X and 12 from both sides to get)
4X-3X = 20 - 12
X = 8. The small truck makes 8 trips, the large truck makes 11.

3) Let X be the number of second head start. To give Jack a head start, means he get's to run X more seconds than Rick.

Recall from the first problem Distance = Rate*Time. See if you can figure it out from here.

Answer 5

Pay attention, now:
12/(vb-vc) +12/(vb+vc)=2.5, where vb is speed of boat and vc is speed of current. The current helps one direction and hinders in the other.

vb=5vc is given, then substituting and adding/subtracting in parens gives:

12/4vc + 12/6vc=2.5
multiplying by 12vc gives:
36+24=30vc
adding and dividing, gives vc=2 mph

Next:
4(S+3) - 3S = 20 where s is the numbe of trips of the small truck

4s +12 -3s =20
s=8, s=3=11

the small truck makes 8 trips for 24 tons, the large truck makes 11 trips for 44 tons

next

ricks speed is 440/55 and jacks is 440/88
t=distance/speed

rick is going to run 440 yards, while jack is going to run 440-x where x is the headstart.
the times must be equal, thus:
440/(440/55)=(440-x)/(440/88)
55(440)=(440-x)88 I do some stuff in my head to get to:
440(55-88)=-88x
440(-33)/(-88)=x
165 yards = x

Answer 6

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Answer 7

My guess for #1 is:
time=2.5
distance=24
boat rate=5*current

So using r=d/t...5*current = distance / time
5x=24/2.5
x=1.92 mph

Answer 8

are you allowed to use a calculator? because there is no way i can do that in my head