Rahman left town A and drove to town C. After travelling 1/3 of the journey at an average speed of 72km/h for 45 minutes,he stopped at townB to have his lunch. He took 1/2 hour for lunch. Then he carried on with journey at an average speed of 90km/h. He reached Town C at 1.15pm. Qns-(a) What was the distance between TownB and TownC?(b) What was the time when Rahman left TownA?
2006-09-03 03:57:34 · 13 answers · asked by lousydotcom r 1 in Science & Mathematics ➔ Mathematics
Distance from A to B : Speed X Time = 75km/h X 45 mins
= (75/60) X 45
= 54
Since the distance from A to B is 1/3 of the journey, the distance from B to C must be 2/3 of the journey.
1/3 of journey = 54 km
2/3 of journey = 108km (ANS A)
Total distance = 54 X 3
= 162
Time taken from B to C: Distance/Speed = 108/ 90
= 1.2 hr
So 1.15pm - 1.2hr - 1/2 hr - 45 mins = 10.48AM (ANS B)
2006-09-03 04:08:49 · answer #1 · answered by lee-wlcj 2 · 0⤊ 0⤋
1st ... he was traveling for 45 minutes and passed 1/3 of the way. 45 minutes is 3/4 of an hour, so, when you multiply 3/4 w 72 km/h you get 54 km.
since 54 is 1/3 of the way, from b to c is 2/3 left, so distance between b and c is 108 km.
ok ... if he travelled 108 km at speed of 90 km/h, he was in town c in 72 minutes (108/90=1,2h=72minutes since 0.2=2/10=1/5 of a hour is 12 minutes) ... so, spent 45 + 72 minutes driving, and 30 more relaxing and eating, that's 147 minutes ...
since he arrived at 1.15 pm, he started at 10.48am
2006-09-03 11:14:13 · answer #2 · answered by cybrdng 2 · 0⤊ 0⤋
So far from Town A to Town B it takes 45 minutes, Which is 54km. Then he spends 1/2 hour there. Making his journey 1 hour and 15minutes thus far. He departs Town B Doing 90/km per hour. Which is 1.5Km per minute. He arrives at 1:15pm in Town C. There is not enough information to calucate departure of Town A and distance to Town C
2006-09-03 11:11:47 · answer #3 · answered by crissyll22 4 · 0⤊ 0⤋
Let the total distance = x km
Therefore 1/3 distance from A to B= x/3 km
Speed = 72 km/hr and time =45mt=3/4 hours
Time = distance/speed=(x/3)/(3/4)
3/4 = (x/3)/72=x/216
x = 216*3/4 = 162 km............................i
Therefore distance from B to C =2x/3=2*162/3=108 km.....ii
Time from B to C = 108/90=1.2 hrs=1 hour 12minutes......iii
Total time taken from A to C=45 minutes+30minutes+72 minutes
=147 minutes
=2hrs27 minutes
Time of starting from A =13hrs15mintes- 2hours 27 minutes
=10hrs 48 minutes
2006-09-03 11:39:46 · answer #4 · answered by Amar Soni 7 · 0⤊ 0⤋
Let's use hours instead of minutes across the board.
Also, let's use this with military time. So the 45 minutes
is .75 h, and 1:15PM is 13.25
The distance from A to B is .75(72)=54K.
This is 1/3 the distance from A to C, so distance from
B to C is 2/3 or 2 times as much as from A to B, so
it's 2X54=108k. The time it took is 108K/90k/h=1.2h
The time he left A is (13.25)h - 1.2h -.5h -.75h=10.8.
10.8h is 10:48 AM.
a) 108K b) 10:48 AM
Do you like the clean methodology and report form.
2006-09-03 11:45:47 · answer #5 · answered by albert 5 · 0⤊ 0⤋
(a) Distance from Town B to Town C is 108km
(b) Rahman left Town A at 10:48am
2006-09-03 11:11:41 · answer #6 · answered by DougieWang 1 · 0⤊ 0⤋
1/3 of the journey fron town c to b. i would say around 50 somthin km
2006-09-03 11:08:33 · answer #7 · answered by james 2 · 0⤊ 0⤋
a)108 b) 10:48am
2006-09-03 12:08:51 · answer #8 · answered by lenz 2 · 0⤊ 0⤋
hmmm...thanks for the 2 points!
2006-09-03 11:02:51 · answer #9 · answered by me. 2 · 0⤊ 0⤋
i hate these questions. they think they need math for everything. all u need is a watch. no normal person ever uses this stuff.
2006-09-03 15:30:09 · answer #10 · answered by R.B. 1 · 0⤊ 0⤋