3 brothers A, B, and C started simultaneously from a common

Ans: B

In 2 hours C walks 20 km(2x 10 miles/h) and B walks 80 km (2x 40 miles/h).
Let a is the distance between the point x and y. therefore 2a+ 20=80 i.e. a=30. therefore distance of x from starting point is 30 + 20 = 50 miles

so, distance covered by walk by A=60-50= 10 MILES

Ans B

A, B and C started at the same time.
Total time B had travelled till he picks up C = 2 hours and total distance travelled by B till he picks up C = 2*40miles/hr = 80miles

therefore, total distance travelled by C till the time he picks up C i.e two hours from start = 10miles/hour*2 Hours = 20 miles
Distance between point X and Point Y (let's Say Q km) is traveled [b]two times by B ( forward and backward)
hence, 20miles + Q miles + Q miles = 80 Km , so Q = 30miles

now, 20 miles( start to Point Y)+ 30miles (Point Y to Point X) + (remaining distance which A needs to walk) = 60 miles ( total distance)
Distance travelled by A by walking = 10 miles

Let T1 be the amount of time A & B travel together on a bike before A is dropped off at X point
Let T2 be the amount of time B travel by himself before meeting up with C at Y point

Given that A was dropped off at point X, which takes T1 hours and A & B travel at 40 Miles/hour with a bicycle:
1. X = 40*T1

Given that B and C meet up at point Y, which takes (T1+T2) hours, B is travelling backward from point X at 40 Miles/hour and C travels at 10 Miles/hour:
2. 10*[T1+T2] + 40*T2 = X

Given that it takes 2 hours for B & C to meet up at point Y,
3. T1 + T2 = 2

Rearrange Equation 3: T2 in terms of T1

3a. T2 = 2 - T1

Using three equations 1, 2 and 3a:

10*[2] + 40*(2 - T1) = 40*T1

Now solve for T1:
1 + 2*(2 - T1) = 2*T1
T1 = 5/4

Given that the total distance is 60 Miles, the distance A has to travel on foot is:
60 - X = 60 - 40*T1 = 60 - 40*(5/4) = 60 - 50 = 10 Miles

Answer is (b) 10 Miles

Correct Answer is B.

Little correction in the table : time is 2 hours till point x for B
time is 2 hours till point y for C

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(1) B travelled for 2 hours before it picked up C at point X => C travelled @10mph for 2 hours i.e. 20 miles (distance of point Y from origin) before it was picked up by B.

(2) Hence C covered remaining 40 miles @40mph in 1 hour and travelled for 3 hours in total. [Using (1)]

(3) All three reached at same time => All three travelled for 3 hours each [Using (2)]

(4) B travelled at a constant speed of 40mph for 3 hours => B covered 120 miles, which equals 60 miles it had to cover plus 30 miles (from point X to Y) to pick up C and 30 miles from Y to X.

(5) We already know that Y is 20 miles from origin using (1) and distance between X and Y is 30 miles using (4). Hence distance between origin and X is 20 + 30 = 50 miles => A travelled from X to end (i.e. 60 - 50 = 10 miles) on foot. Answer is choice B.

ronilbhan Congrats on nailing this. please pm me to get your GMAT Club tests!

Let’s call the starting point S and the ending point E. We are given that SE (the distance from S to E) = 60 miles and we need to determine the distance A had to walk, which is XE (the distance from point X to E).

We are given that B had ridden the bike from S to X and then from X to Y (where Y is some point between S and X) and he had ridden the bike for 2 hours. Since his rate is 40 mph, he had travelled 2 x 40 = 80 miles. In other words, SX + XY = 80.

Since B had traveled for 2 hours by time time he picked up C at point Y, C also had traveled for 2 hours by walking. Since C’s rate is 10 mph, he had traveled 2 x 10 = 20 miles. In other words, SY = 20.

Notice that since Y is some point between S and X, we can break SX into two parts: SY and YX. Thus, we have:

SX + XY = 80 → SY + YX + XY = 80

Since SY = 20 and XY = YX, we have:

20 + YX + YX = 80

2(YX) = 60

YX = 30

Notice that Y and X are some points between S and E and we know that SE = 60, SY = 20 and YX = 30 and we need to determine XE. So we have:

SY + YX + XE = SE

20 + 30 + XE = 60

XE = 10

Alternate Solution:

Let the distance of the point where B drops A to the starting point be d. Since B rides his bike for two hours at 40 mph, he travels a total of 2 x 40 = 80 miles. Since C walks at 10 mph, he travels a total of 2 x 10 = 20 miles. We see that B first travels a distance of d miles to drop off A and then turns around and travels another d - 20 miles to pick up C (since C walked 20 miles in that time); thus traveling d + d - 20 = 2d - 20 miles in total. Since we know the total distance traveled by B is 80, we can create the equation:

2d - 20 = 80

2d = 100

d = 50

So, B dropped A off at a distance of 50 miles to the starting point; which means A has to walk another 60 - 50 = 10 miles.

Answer: B

Hi souvik101990,

How can I get access to GMAT tests, as i have been trying to reach you
through pm but there has been no reply.

souvik101990, am I missing something?
I think this problem is inconsistent:

If "10 miles" is the answer, A and B traveled together for 50 miles at 40 miles per hour; therefore, they rode together for 5/4 hrs. Then A continued walking for 10 miles at 10 miles per hour; therefore, A took an additional hour to get to the fair. So, A took 5/4 + 1 hrs = 2.25 hrs to get to the fair.
However, as everybody calculated, B and C took 3 hrs to get to the fair: 2 hrs until B met C and another hour to ride the remaining 40 miles (30+10).
In conclusion they can't arrive at the same time, but the problem says that they arrive at the same time.