Janice left her home at 11 a.m., traveling along Route 1 at 30 mph. At

Question

Janice left her home at 11 a.m., traveling along Route 1 at 30 mph. At 1 p.m., her brother Richard left home and started after her on the same road at 45 mph. At what time did Richard catch up to Janice?

(A) 2:45 p.m.
(B) 3:00 p.m.
(C) 3:30 p.m.
(D) 4:15 p.m.
(E) 5:00 p.m.

(A) 2:45 p.m.
(B) 3:00 p.m.
(C) 3:30 p.m.
(D) 4:15 p.m.
(E) 5:00 p.m.

IanStewart · Answer

In the first two hours, from 11am to 1pm, only Janice is traveling, at 30 mph, so she travels 60 miles.

After 1pm, both are traveling, and each hour Richard travels 45 miles and Janice travels 30 miles, so Richard is gaining 15 miles each hour. He needs to gain 60 miles to catch up, so it will take 4 hours, and the answer is 5pm.

100mitra · Answer

Correct Option E (5:00 pm)

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jaindevansh1 · Answer

Distance travelled by both will be the same when they meet 
Let T= Time travelled by Richard, then time travelled by Janice = T+2 
D1 = D2 
S1 * T1 = S2 * T2
30 (T+2) = 45 (T)
T = 4 hrs

Time Richard caught up is 4 hrs after he left i.e. 5:00 pm (E)

ShahadatHJ · Answer

In 2 hours SHE went 30*2 = 60 miles
Relative Speed 45-30 = 15 mph
HE catcher HER = 60/15 = 4 Hours @ 1:00+ 4 hours = 5:00 PM
Ans. E

Janice left her home at 11 a.m., traveling along Route 1 at 30 mph. At

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Janice left her home at 11 a.m., traveling along Route 1 at 30 mph. At

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