Albert was driving on a straight highway, from Edenborough to Megiddo,

Question

Albert was driving on a straight highway, from Edenborough to Megiddo, at 60 mph. Beatrice was driving faster than Albert, on the same highway, in the same direction. She started behind Albert. She passed Albert at 1:20 pm. At 3:20 pm, Beatrice arrived at Megiddo, and Albert arrived 50 minutes later, at 4:10 pm. Assume both drivers kept up constant speeds without stops. What was Beatrice's speed, in mph?

A. 80
B. 85
C. 90
D. 95
E. 100

Are You Up For the Challenge: 700 Level Questions

A. 80
B. 85
C. 90
D. 95
E. 100

Are You Up For the Challenge: 700 Level Questions

firas92 · Answer

Lets say Beatrice met Albert at  d  miles from Megiddo

Beatrice traveling at  x  mph covers these  d  miles in  2  hours while Albert traveling at  60mph  covers the same  d  miles in  2  hours and  50  minutes or  2\frac{5}{6}=\frac{17}{6} hours

60*\frac{17}{6}=d 
 x*2=d

Therefore,  60*\frac{17}{6}=2x

x=85mph

Answer is  (B)

exc4libur · Answer

B to X took 15:20 - 13:20 = 2 hrs
A to X took 16:10 - 13:20 = 2 hours + 50 mins = 2+5/6 = 17/6 hrs

distance = rate * time
b's rate * 2hrs = d
a's rate * 17/6hrs = d
b's rate * 2hrs = a's rate * 17/6hrs, 2b = 60*17/6, b=85

Ans (B)

Archit3110 · Answer

first using info of Albert find d= 
total time taken by Albert to reach Megiddo = 2 Hrs : 50 mins or say ; 2+ 50/60 ; 17/6 hrs

so distance = 17/6 * 60 = 170 miles
time for Beatrice = 2 hrs
her speed = 170/2 ; 85 mph
IMO B

uchihaitachi · Answer

Distance = speed * time.

Distance is constant so speed is inversely proportional to time. 
Time taken by Beatrice = 2 hours = 120 minutes
Time taken by Albert = 2 hours 50 min = 170 minutes

Speed of Albert = 60 mph
Let speed of Beatrice = S

Now,  \frac{S}{60} = \frac{170}{120} 
From here, we get S = 85.

OA - B

satya2029 · Answer

When Betrice(2) passed albert(1), remaining distance for both of them will be same 
Speed2*t2=S1*t1
S2*120=60*170
S2=85mph
B:)

ScottTargetTestPrep · Answer

We see that Beatrice drove for 2 hours after catching up with Albert and passing him. Meanwhile, for the same distance, Albert had to drive 2 hours and 50 minutes at 60 mph. Therefore, the distance from the point where Beatrice caught up with Albert and passed him to Megiddo is (2 + 50/60) x 60 = 120 + 50 = 170 miles. Therefore, Beatrice’s speed is 170/2 = 85 mph.

Answer: B

luisdicampo · Answer

Deconstructing the Question  
This problem describes two drivers covering the  same distance  (from the passing point to the destination), but in different amounts of time.
When distance is constant, Speed and Time are  inversely proportional .

Passing Point Time:  1:20 pm  
 Albert's Speed ( R_A ):  60 mph  
 Goal: Find Beatrice's Speed ( R_B ).

Step 1: Calculate the Time taken for the remaining trip  
We only care about the segment from the passing point (1:20 pm) to Megiddo.

Beatrice's Time ( T_B ): 
Starts at 1:20 pm, arrives at 3:20 pm.
 T_B = 2 	ext{ hours} = 120 	ext{ minutes}

Albert's Time ( T_A ): 
Starts at 1:20 pm (the moment he is passed), arrives at 4:10 pm.
 T_A = 2 	ext{ hours } 50 	ext{ minutes} = 120 + 50 = 170 	ext{ minutes}

Step 2: Use the Inverse Ratio Formula  
Since Distance is constant:
 \frac{R_B}{R_A} = \frac{T_A}{T_B}

Substitute the known values:
 \frac{R_B}{60} = \frac{170}{120}

Step 3: Solve for Beatrice's Speed  
Simplify the fraction:
 \frac{R_B}{60} = \frac{17}{12}

Multiply by 60:
 R_B = 60 	imes \frac{17}{12} 
 R_B = 5 	imes 17 
 R_B = 85 	ext{ mph}

Answer:  B

Fisayofalana · Answer

Albert speed = 60
Albert time = 2 hours + 50 minutes which is 170 mins or 17/6 hours
Albert total distance = 60*17/6 = 170 miles

Beatrice speed = 60+ x
Beatrice time = 2 hours
Distance = (60+x) *2 = 170 
x = 25

Plug x back into Beatrice's speed. 60+25 =85

Albert was driving on a straight highway, from Edenborough to Megiddo,

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GMAT Problem Solving (PS) Questions

Albert was driving on a straight highway, from Edenborough to Megiddo,

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