Greg and brian are both at point a (above). Starting at the same time, greg drives to point b while brian drives to point c. Who arrives at his destination first? (1) greg's average speed is 2/3 that of brian's. (2) brian's average speed is 20 miles per hour greater than greg's.
Answers
Answer:
Step-by-step explanation:
Greg's speed: sgsg
Distance covered by Greg: DD
Time taken by Greg: tg=Dsgtg=Dsg
Brian's speed sbsb
Distance covered by Greg D2√D2 (Hypotenuse of right isosceles triangle = root(2) times base)
Time taken by Brian tb=D2√sbtb=D2sb
1.
sg=23sbsg=23sb
tg=Dsg=D23sb=3D2sbtg=Dsg=D23sb=3D2sb-------------1
tb=2√Dsbtb=2Dsb----------------2
Comparing 1 and 2:
32>2√32>2. Thus, Greg took relatively more time to reach his destination.
Sufficient.
2.
sb=sg+20sb=sg+20
tb=2√Dsb=2√Dsg+20tb=2Dsb=2Dsg+20
tg=Dsgtg=Dsg
Let's prove opposite of st1:
tb>tgtb>tg
2√Dsg+20>Dsg2Dsg+20>Dsg
2√sg+20>1sg2sg+20>1sg
sg2√>sg+20sg2>sg+20
0.4sg>200.4sg>20
sg>50sg>50
Thus, if greg's speed is approx more than 50, brian would take more time, otherwise greg would take more time.
Hope it helps you...