Answer:
60 miles
Step-by-step explanation:
Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him?
Statement 1. Brian’s walking speed is twice the difference between Ashok’s walking speed and his own
Statement 2. If Ashok’s walking speed were five times as great, it would be three times the sum of his and Brian’s actual walking speeds
Solution
A. Brian’s walking speed is twice the difference between Ashok’s walking speed and his own.
Let Brian speed=b
Ashok speed=a
Brian's walking speed=2(a-b)
b=2(a-b)
Divide both sides by 2
b/2=a-b
Ashok catches up in (time)= distance /( relative rate
=30/(a-b)
=30/(b/2)
=30÷b/2
=30*2/b
=60/b.
By that time Brian will cover a distance of
distance=rate*time
=b*60/b
=2(a-b)*60/2(a-b)
=60 miles
(2) If Ashok’s walking speed were five times as great, it would be three times the sum of his and Brian’s actual walking speeds.
5a=3(a+b)
5a=3a+3b
5a-3a=3b
2a=3b
Answer: 0.25 as a decimal
The 3rd quadrant bc they are both negative points
Answer:
D. 56%
Step-by-step explanation:
For the fractions we will do:
It depends on if he puts the first name back in, but I am going to asume Mr. Rosart does not, <em>so we will subtract one from the boys for the second pull</em>.
[Wanted over possible] 
[Dividing] 14 / 25 = 0.56
[Making a percent] .56 * 100 = 56%
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Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)
- Heather
Well you would have them work together which can possibly cut the work time in half. So if you have them both working then it can speed up the process to B. I hope this helped :D
