How about time, rate and distance?
José and Chuchie took part in their barangay’s walkathon which started at 7:30 in the afternoon. José’s average speed was 20 meters per minute faster than Chuchie’s. When José completed the walkathon in 30 minutes, Chuchie had only walked 5/6 of the distance.
(a) At what time was it that Chuchie completed the walkathon?
(b) Find José’s average speed for the walkathon in meters per minute.
Distance, d, in terms of Chuchie’s speed, s:
d/(s + 20) = 30
d = 30s + 600
5/6d = 30s
d = 36s
Speed of Chuchie—s:
30s + 600 = 36s
5s + 100 = 6s
s = 100
Jose’s average:
= 100 + 20
= 120
Time Chuchie completed the walkathon:
= (30[100] + 600)/100 + 07:30
= (3,000 + 600)/100 + 07:30
= 3,600/100 + 07:30
= 00:36 + 07:30
= 07:66 or 08:06
Answer: 08:06 PM was the time Chuchie completed the walkathon; 120 meters per minute is the speed of Jose.
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Let’s assume.
Total distance=X meter
Chuchie’s avg. speed=Y meter/min
SO jose’s average speed= Y+20 meter/min
SO jose will cover= 30(Y+20)= X= total distance
X= 30(Y+20)= 30Y+600 ——-Eq.1
in 30 min chuchie run= 5/6 of X=5X/6
=> 30Y=5X/6—————Eq.2
here X=36Y—————Eq.3
So put Eq.3 in Eq.1
36Y=30Y+600
Y=100 meter/ min= Chuchie’s avg. speed
So Jose’s avg. speed= Y+20=120 meter/min
X=36Y=3600meter.
Time taken bu chuchie to finish the distance=3600/100=36 min.
So chuchie will finish the distance at= 7:30+36=8:06 PM.
Ans:
(a) 8:06PM
(b)120 meter/Min
References :
Distance, d, in terms of Chuchie’s speed, s:
d/(s + 20) = 30
d = 30s + 600
5/6d = 30s
d = 36s
Speed of Chuchie—s:
30s + 600 = 36s
5s + 100 = 6s
s = 100
Jose’s average:
= 100 + 20
= 120
Time Chuchie completed the walkathon:
= (30[100] + 600)/100 + 07:30
= (3,000 + 600)/100 + 07:30
= 3,600/100 + 07:30
= 00:36 + 07:30
= 07:66 or 08:06
Answer: 08:06 PM was the time Chuchie completed the walkathon; 120 meters per minute is the speed of Jose.
References :