Could Anyone help me with worded problems please?

Question

I am making sure that my answers are correct but I am not sure with my answers: T_T Help

1.a messenger travel from A to B if he will leave A at 8am and travel 2kph he will arrive at B 3 minutes earlier than his expected time of arrival however if he will leave at 8:30 am and travel at 3 kph he will arrive 6 minutes late than the expected time of arrival what is the expected minutes time of arrival?

2. 2 airplane traveling in opposite direction, leave an airport at the same time if one plane quarages 480 mph & the other average 520 mph, how long will it take before they are 200 mi apart?

3. admission tickets to a motion picture theater were priced at $100 for adult and $75 for students, if 810 tickets were sold and the total receipt were $ 71,325.08 how many types of tickets were sold?

Please help me :( I wanna make sure I don't have any error both on solution and answer. Thanks a Lot :)

Answer 1

Hi,


1.a messenger travel from A to B if he will leave A at 8am and travel 2kph he will arrive at B 3 minutes earlier than his expected time of arrival however if he will leave at 8:30 am and travel at 3 kph he will arrive 6 minutes late than the expected time of arrival what is the expected minutes time of arrival?

Let x = expected minutes until arrival
Since rate x time = distance, the distance traveled will come from rate x time.

The 1st messenger goes 2mph as a rate and time is "x - 3", 3 minutes less than the expected time. So distance traveled is 2(x - 3)

The 2nd messenger goes 3 mph but traveling time is "x" minutes - 30 minutes for starting late plus 6 minutes late arriving. This distance is 3(x - 30 + 6)

Setting these distances equal gives

2(x - 3) = 3(x - 30 + 6)

Solving gives:

2x - 6 = 3x - 90 + 18
2x - 6 - 3x - 72
66 = x

Normal delivery time is 66 minutes. <== answer



2. 2 airplane traveling in opposite direction, leave an airport at the same time if one plane quarages 480 mph & the other average 520 mph, how long will it take before they are 200 mi apart?

Together they are going 480 + 520 or 1000 mph. 200/1000 or 200 miles out of 1000 hours reduces to 1/5 hour which is 12 minutes.



3. admission tickets to a motion picture theater were priced at $100 for adult and $75 for students, if 810 tickets were sold and the total receipt were $ 71,325.08 how many types of tickets were sold?

Let x = number of adults and y = number of students

Equations are:

x + y = 810 number of adults + number of students = 810

100x + 75y = 71325
$ from adults + $ from students =TOTAL MONEY

(Ignore the .08 - there was no change involved in prices.)

Solving by matrices on my calculator,

x = 423 and y = 387, so there were 423 adults and 387 students.



I hope that helps!! :-)

Answer 2

Time to travel= distance/speed
8 AM departure. Let x be time traveled by messenger in MINUTES. Let K be distance.
Then 60 * K/2 = x

8:30 AM departure.
60 * K/3 = (x-30+9)

So 60 K=2x and 60 K= 3(x-21) so 2x=3x-63
and x=63 minutes.

The expected time of arrival is 66 minutes.

2. D= v*t
Planes are moving apart at 1000 mph. So to separate 200 miles, they fly for 0.2 hours.

3. TR = SIGMA (price/ticket * tickets sold)
Let adult tickets be x
Student tickets are 810-x
Adult receipts are 100*x
Student receipts are 75(810-x)
TR = 25x + 810*75 = 71325
2815 - 2430 = x
x= 385
810-x= 425

Answer 3

3. A(dults) + S(tudents) = 810
A x $100 + S x $75 = $71,325
multiply 1st equation by 100
A x 100 + S x 100 = 81000
Sx25=9675
S=387
A=423

2. D(1) + D(2) = 200
D(1) =time x 480mph
D(2) = time x 520 mph
200 = time x (480 + 520) = time x 1000
time = 0.20 hours = 12 minutes

1. (time - 3 min) x (2/60) k/min =distance
(time -30 min + 6 min) x (3/60) k/min = distance
both distances are the same so-
(t-3) x (2/60) = (t -30+6)x(3/60)
(2/60)xt- 0.10 = (3/60)xt - 1.20
1.10 = (1/60)xt
t=66 minutes
66 minutes at 2kph expected travel time; arrive 9:06
66-30=36 minutes at 3kph expected travel time; arrive 9:06