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

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

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47