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Monster Media 1996 #15
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Monster Media Number 15 (Monster Media)(July 1996).ISO
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1996-06-05
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Below are eight DISTANCE word problems.
Have your pencil and scratch paper ready.
Press the PAGE DOWN key to start...^15
*************************************************************
* *
* Sample DISTANCE problem #1 *
* *
*************************************************************
1. Bill walked into town at 4 mph. He rode back in a^14
friend's car along the same route at 20 mph. He^14
returned home in 3 hours. How far was it to town?^14
(a) 6 miles^14
(b) 8 miles^14
(c) 10 miles^14
(d) none of these^14
Press the PAGE DOWN key for solution...^15
SOLUTION.^15
1. Use distance formula^11
DISTANCE = RATE x TIME, so TIME = DISTANCE / RATE
2. Time is what is given, so solve for time^11
TIME TO THERE + TIME FROM THERE = 3 hours
DISTANCE#1 / RATE#1 + DISTANCE#2 / RATE#2 = 3 hours
3. Substitute numbers from problem into equation and solve^11
let X = distance to town^3
( X / 4 mph ) + ( X / 20 mph) = 3 hours
lcd=20, so^3
20(X/4) + 20(X/20) = 20(3)
5X + X = 60
6X = 60, so X = 10
Answer is (c), 10 miles^3
Press the PAGE DOWN key for next problem...^15
*************************************************************
* *
* Sample DISTANCE problem #2 *
* *
*************************************************************
2. Two cars traveling in opposite directions were 330 miles^14
apart after 3 hours. If one car travels 10 mph faster^14
than the other, find the rate of the faster car.^14
(a) 60 mph^14
(b) 55 mph^14
(c) 50 mph^14
(d) none of these^14
Press the PAGE DOWN key for solution...^15
SOLUTION.^15
1. Going in opposite directions means add the two distances^11
DISTANCE#1 + DISTANCE#2 = TOTAL DISTANCE
(RATE#1 x TIME#1) + (RATE#2 x TIME#2) = TOTAL DISTANCE
2. Total distance is 330 miles and time for each is 3 hours^11
(RATE#1 x 3 hrs) + (RATE#2 x 3 hrs) = 330 miles
3. Substitute numbers from problem into equation and solve^11
let X = speed of faster car^3
( (X) mph x 3 hrs) + ( (X-10) mph x 3 hrs) = 330 miles
3X + 3(X-10) = 330
3X + 3X - 30 = 330
6X - 30 = 330
6X = 360, so X = 60
Answer is (a), 60 mph^3
Press the PAGE DOWN key for next problem...^15
*************************************************************
* *
* Sample DISTANCE problem #3 *
* *
*************************************************************
3. Butch starts out at 8 AM traveling east at 5 mph;^14
John starts out at the same time and heads west at^14
3 mph less than Butch. At noon, how far apart are they?^14
(a) 28 miles^14
(b) 30 miles^14
(c) 32 miles^14
(d) none of these^14
Press the PAGE DOWN key for solution...^15
SOLUTION.^15
1. Going in opposite directions means add the two distances^11
DISTANCE#1 + DISTANCE#2 = TOTAL DISTANCE
(RATE#1 x TIME#1) + (RATE#2 x TIME#2) = TOTAL DISTANCE
2. 8 AM to noon is 4 hours. Substitute into formula
(RATE#1 x 4 hrs) + (RATE#2 x 4 hrs) = TOTAL DISTANCE
3. Substitute rates into formula and solve
(5 mph x 4 hrs) + ( (5-3) x 4 hrs) = TOTAL DISTANCE
20 miles + 8 miles = TOTAL DISTANCE
28 miles = TOTAL DISTANCE
Answer is (a), 28 miles^3
Press the PAGE DOWN key for next problem...^15
*************************************************************
* *
* Sample DISTANCE problem #4 *
* *
*************************************************************
4. Bob starts walking east at 9 AM at 4 mph. Jack starts^14
from the same point at 10 AM, but walks west at 5 mph.^14
How far apart are they at noon?^14
(a) 18 miles^14
(b) 20 miles^14
(c) 22 miles^14
(d) none of these^14
Press the PAGE DOWN key for solution...^15
SOLUTION.^15
1. Going in opposite directions means add the two distances^11
DISTANCE#1 + DISTANCE#2 = TOTAL DISTANCE
(RATE#1 x TIME#1) + (RATE#2 x TIME#2) = TOTAL DISTANCE
2. Noon - 9 AM = 3 hrs, Noon - 10 AM = 2 hrs. Substitute.^11
(RATE#1 x 3 hrs) + (RATE#2 x 2 hrs) = TOTAL DISTANCE
3. Substitute rates into formula and solve
(4 mph x 3 hrs) + ( 5 mph x 2 hrs) = TOTAL DISTANCE
12 miles + 10 miles = TOTAL DISTANCE
22 miles = TOTAL DISTANCE
Answer is (c), 22 miles^3
Press the PAGE DOWN key for next problem...^15
*************************************************************
* *
* Sample DISTANCE problem #5 *
* *
*************************************************************
5. A girl walked into town at the rate of 4 mph. She rode^14
back with her boyfriend along the same route, at 20 mph^14
(they took their time). She spent one hour shopping in^14
town before she started back. If she arrived home^14
4 hours after she left, how far is it to town?^14
(a) 8 miles^14
(b) 10 miles^14
(c) 12 miles^14
(d) none of these^14
Press the PAGE DOWN key for solution...^15
SOLUTION.^15
1. Use distance formula. Solve for time^11
DISTANCE = RATE x TIME, so TIME = DISTANCE / RATE
TIME#1 + TIME#2 = TOTAL TIME
(DISTANCE#1 / RATE#1) + (DISTANCE#2 / RATE#2) = TOTAL TIME
2. TOTAL time spent is 3 hours. Add the two times together^11
distance to or from town is same, so let it = D^3
(D / RATE#1) + (D / RATE#2) = 3 hours
3. Substitute rates from problem and solve
(D / 4 mph) + (D / 20 mph) = 3 hours
(D/4) + (D/20) = 3
20(D/4) + 20(D/20) = 20(3)
5D + D = 60
6D = 60, so D = 10
Answer is (b), 10 miles^3
Press the PAGE DOWN key for next problem...^15
*************************************************************
* *
* Sample DISTANCE problem #6 *
* *
*************************************************************
6. Two airplanes with lines of flight at right angles to^14
each other pass each other (slightly) at noon. One is^14
flying at 140 mph and the other is flying at 160 mph.^14
How far apart are they at 12:30 pm?^14
(a) 100 miles^14
(b) 106 miles^14
(c) 110 miles^14
(d) none of these^14
Press the PAGE DOWN key for solution...^15
SOLUTION - page 1.^15
1. Use distance formula and pythagorean formula
DISTANCE = RATE x TIME A^^2 + B^^2 = C^^2
LEG#1^^2 + LEG#2^^2 = DISTANCE^^2
2. Pythagorean formula is used when you have right triangles.
** **
^^ |** **
| | ** ** = DISTANCE between planes
| | ** **
one | ** **
plane | ** **
+-----** **
other plane -->
Press PAGE DOWN for rest of solution...^15
SOLUTION - page 2.^15
3. Substitute distances of the legs into pythagorean formula^11
LEG#1^^2 + LEG#2^^2 = DISTANCE^^2
(RATE#1 x TIME#1)^^2 + (RATE#2 x TIME#2)^^2 = DISTANCE^^2
remember that 30 minutes = .5 hour^3
(140 mph x .5 hr)^^2 + (160 mph x .5 hr)^^2 = DISTANCE^^2
4. Solve equation for total DISTANCE^11
( 140(.5) )^^2 + ( 160(.5) )^^2 = DISTANCE^^2
70^^2 + 80^^2 = DISTANCE^^2
4900 + 6400 = DISTANCE^^2
11300 = DISTANCE^^2
106 = DISTANCE
Answer is (b), 106 miles^3
Press PAGE UP for previous page, PAGE DOWN for next problem^15
*************************************************************
* *
* Sample DISTANCE problem #7 *
* *
*************************************************************
7. A plane flies 300 miles with a tailwind of 10 mph and^14
returns against a wind of 20 mph. What is the speed^14
of the plane in still air if the total flying time is^14
4.5 hours?^14
(a) 140 mph^14
(b) 120 mph^14
(c) 100 mph^14
(d) none of these^14
Press the PAGE DOWN key for solution...^15
SOLUTION - page 1.^15
1. Use the distance formula and solve for TIME^11
DISTANCE = RATE x TIME, so TIME = DISTANCE / RATE
2. The two times add up to 4.5 hours^11
TIME#1 + TIME#2 = TOTAL TIME
(DISTANCE#1 / RATE#1) + (DISTANCE#2 / RATE#2) = 4.5 hours
3. Substitute rate of plane with and against wind^11
let rate in still wind = X^3
(DISTANCE#1 / (X+10)) + (DISTANCE#2 / (X-20)) = 4.5 hours
Press PAGE DOWN for rest of solution...^15
SOLUTION - page 2.^15
4. Distance is 300 miles. Substitute and solve
((300 miles) / (X+10)) + ((300 miles) / (X-20)) = 4.5 hours
300 / (X+10) + 300 / (X-20) = 4.5
lcd is (X+10)(X-20)(2); multiply each term by lcd^3
600(X-20) + 600(X+10) = 9(X-20)(X+10)
600X - 12000 + 600X + 6000 = 9X^^2 - 90X - 1800
1200X - 6000 = 9X^^2 - 90X - 1800
0 = 9X^^2 - 1290X + 4200
0 = 3X^^2 - 430X + 1400
(3X - 10) (X - 140) = 0
X = 10/3 or X = 140
10/3 is an extraneous solution because the tailwind
would cause the plane to go backwards!
Answer is (a), 140 mph^3
Press PAGE UP for previous page, PAGE DOWN for next problem^15
*************************************************************
* *
* Sample DISTANCE problem #8 *
* *
*************************************************************
8. Two ships leave port at the same time. One ship sails^14
north at 30 mph and the other sails east at 40 mph.^14
How far apart are they after 30 minutes?^14
(a) 15 miles^14
(b) 20 miles^14
(c) 25 miles^14
(d) none of these^14
Press the PAGE DOWN key for solution...^15
SOLUTION - page 1.^15
1. Use distance formula and pythagorean formula
DISTANCE = RATE x TIME A^^2 + B^^2 = C^^2
LEG#1^^2 + LEG#2^^2 = DISTANCE^^2
2. Pythagorean formula is used when you have right triangles.
** **
^^ |** **
| | ** ** = DISTANCE between ships
| | ** **
one | ** **
ship | ** **
(north) +-----** **
other ship --->
(east)
Press PAGE DOWN for rest of solution...^15
SOLUTION - page 2.^15
3. Substitute distances of the legs into pythagorean formula^11
LEG#1^^2 + LEG#2^^2 = DISTANCE^^2
(RATE#1 x TIME#1)^^2 + (RATE#2 x TIME#2)^^2 = DISTANCE^^2
remember that 30 minutes = .5 hour^3
(30 mph x .5 hr)^^2 + (40 mph x .5 hr)^^2 = DISTANCE^^2
4. Solve equation for total DISTANCE^11
( 30(.5) )^^2 + ( 40(.5) )^^2 = DISTANCE^^2
15^^2 + 20^^2 = DISTANCE^^2
225 + 400 = DISTANCE^^2
625 = DISTANCE^^2
25 = DISTANCE
Answer is (c), 25 miles^3
Press PAGE UP for previous page, ESC to end^15