Courier A departs from Warehouse X heading toward Warehouse Y at 8:00 AM traveling at a co… — GRE General Test

Courier A departs from Warehouse X heading toward Warehouse Y at 8:00 AM traveling at a constant speed of $40$ miles per hour. Courier B departs from Warehouse Y heading toward Warehouse X along the same straight route at 9:00 AM traveling at a constant speed of $60$ miles per hour. If the total distance between Warehouse X and Warehouse Y is $190$ miles, at what time will the two couriers meet?

A
9:54 AM
B
10:00 AM
10:30 AMAnswer
D
11:00 AM
E
11:15 AM

Answer

10:30 AM

The correct answer is 10:30 AM. Between 8:00 AM and 9:00 AM, Courier A travels 40 miles alone. At 9:00 AM, the distance remaining between them is 150 miles. Because they travel toward each other, their speeds combine to 100 mph (40 + 60). Dividing 150 miles by 100 mph gives 1.5 hours (1 hour and 30 minutes). Adding 1 hour and 30 minutes to 9:00 AM gives a meeting time of 10:30 AM.

Step-by-Step Solution

Calculate the distance traveled by Courier A before Courier B starts moving.

From 8:00 AM to 9:00 AM (1 hour), Courier A travels

40 \text{ mph} \times 1 \text{ hour} = 40 \text{ miles}

Courier A has a 1-hour head start.

Determine the remaining distance to be covered between the two couriers at 9:00 AM.

Remaining distance

= 190 - 40 = 150 \text{ miles}

Subtract Courier A's distance from the total distance of 190 miles.

Calculate the combined rate of both couriers and solve for elapsed time after 9:00 AM.

Combined rate

= 40 + 60 = 100 \text{ mph}

. Elapsed time

t = \frac{150}{100} = 1.5 \text{ hours} = 1 \text{ hour } 30 \text{ minutes}

Since they move toward each other, their speeds add up to close the gap.

Add the elapsed combined time to 9:00 AM to find the meeting time.

9:00 AM

+ 1 \text{ hour } 30 \text{ minutes} = 10:30 \text{ AM}

The combined movement began at 9:00 AM when Courier B started traveling.

Key Concept

Distance-rate-time relationship with staggered start times

Estimated Time:1m 30s

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