Ann and Bea leave Townville at the same time and travel

Hi

Distance travelled by Ann on her journey back - c.

Bea's speed - v

Ann's speed - 4v

(2k + c) / 4v = (2k - c) / v

5vc = 6kv

c = 6k/5

Distance travlled by Bea: 2k - 6k/5 = 4k/5

Answer E.

David has nicely demonstrated the input-output approach for this question.
Here's an algebraic solution:

Let B = the distance Bea traveled
Let R = Bea's speed.

NOTE: the total distance from Townville to Villageton and then BACK TO Townville = 4K.

So, 4K - B = the distance Ann traveled
And 4R = Ann's speed (since her speed is 4 times Bea's speed)

From here, let's create a WORD EQUATION that uses distance and speed.
How about: Ann's travel time = Bea's travel time

Time = distance/rate, so we get:
(4K - B)/4R = B/R
Cross multiply to get: (B)(4R) = (R)(4K - B)
Expand: 4BR = 4RK - BR
Add BR to both sides: 5BR = 4RK
Divide both sides by R to get: 5B = 4K
Divide both sides by 5 to get: B = 4K/5

Answer:

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when will they meet? (total distance/speeds added together)!

-> In total, they travel 4K (2K to go towards & 2K to go back)

-> Bea's Speed: B, Ann's Speed: 4B -> together: 5B

4K/(5B) = time which will have passed when they meet

Time travelled by Bea after 4K/5B: time*speed = (4K/5B)*B = 4K/5

when A reaches V, B has traveled 1/4*2K=K/2 kilometers
when A meets B, B has traveled an additional 1/5*3/4*2K=3K/10 kilometers
K/2+3K/10=4K/5 kilometers
E

Let K = 10
So total distance between two towns is 2*10 = 20
Let speed of Ann = 4 km/h
Then speed of Bea = 1 km/h (Ann's speed is 4 times of Bea)
Ann will take \(\frac{20}{4}\)= 5 hrs to reach town V
In that time Bea will travel 1*5 = 5 kms
Now both are travelling towards each other so total distance between them is 15 (total 20 minus 5 already traveled by Bea)
While traveling towards each other their relative speed = 5 km/hr (4 km/hr + 1 km/hr)
They will take 3 hrs to meet, \(\frac{15}{5}=3\)
In 3 hrs Bea will further cover 3 more kilometers.
So total distance traveled by Bea = 5 + 3 = 8
Now using K=10 in all option
Option E gives value 8
Thus option E

let 2k be 100
then Ann did 100 and Bea did 25 (4 times less)
Ann turned around and coming on Bea did 60 and Bea did 15 ( 4 times less
So total Bea did 40
if 2k - 100
then x - 40
80k/100 = 4k/5

easy-peasy

Given the distance is 2k.
Lets assume Beas' speed =v.
Ann's speed =4v

Distance travelled by Ann = 2k+V.
Distance travelled by Bea = 2k-V.
(2k+V)/4v= (2k-V)/v

8k-4V= 2k+V

V= 6k/5.

So Distance by Bea = 4/5 k.

Ans:E

Hello,
This is a problem based on the relationship between Speed and Distance when time is constant. When time is constant, Speed varies directly with Distance. Now, let us look at a time saving approach while attempting such problems with variables.

Since the distance between Townville and Villageton is being expressed as a variable i.e. 2K kilometres, the best thing to do in such cases is to assume simple but workable values for K. In our case, a good value to assume could be 100. So, the distance between Townville and Villageton becomes 200 kilometres.

Secondly, the actual speeds of Ann and Bea have not been given, but a relationship between their speeds is given in terms of a ratio. Therefore, it makes sense to assume simple values for these variables as well. It’s specified that Ann’s speed is 4 times Bea’s speed. So, we can assume Bea’s speed as 10 km per hour and consequently, Ann’s speed will become 40 km per hour.
Since they start off together from Townville, Ann will reach Villageton in 5 hours (200km/40kmph). In these 5 hours, Bea would have traveled 50 km or in other words, he would be 50 km away from Townsville.



At this point again, both of them start their respective journeys at the same time (albeit traveling towards each other). Hence, the total distance of 150 km between them will have to be divided in the ratio of their speeds, which is 4:1.
Therefore, distance travelled by Ann = (4/5) * 150 = 120 and
distance travelled by Bea = (1/5) * 150 = 30.



Hence, Bea has travelled a total of 80 km by the time they meet. 80, when compared to the total distance of 200 is 40% or (2/5)th. But we have assumed 200 as 2K.
Hence, the total distance travelled by (2/5) * 2K = 4K/5. Hence, the answer option is E.

An interesting aspect of this question is that, the second part of the question (where 150 km is to be covered together by Ann and Bea) can also be solved by using Relative Speed concepts. However, it is not a necessity, and it has already been shown above that, without knowing the relative speed concepts also, you will be able to arrive at the answer with a fair degree of ease.

Hope this helps!

Cheers,
CrackVerbal Academics Team
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Time is the same, so plug in 1:

Ann: 4R = 2k + x
Bea: R = 2k - x

Add equations: 5R = 4k, R = 4k/5, R x T = D, D = (4k/5) x 1 = 4k/5.

Answer: E

Together, they have travelled 4K kilometers.

A's speed is 4x so the ratio of distance travelled needs to be 4:1.

B's distance: total x 1/5 = 4K/5

E is my answer.

Total distance = 2k

Relative speed between A and B = 4:1

Distance traveled by B when A reaches Villageton = 2k/4 = k/2

Distance left between both = 3k/2

Total speed for both traveling in opposite direction towards each other = 5x

So, time taken to meet = 3k/10x

Distance traveled by B in that time = (3k/10x)*x = 3k/10

Hence, total distance traveled by B = 3k/10 + k/2 = 4k/5

Let the additional distance travelled by Ann be x. Since the time taken is the same when they meet;
Assuming Bea's speed to be s, we have the equation -

(2000 - x)/s = (2000+x)/4s

Solving this we get x to be 1200 and the distance covered by Bea is 2000 - x which is 800, which is option E. (K = 1000)

I wouldn't get into equations on this one, and just play with ratios.

In the first leg of the race, divide the distance (2K) in 4 parts -> so when A reaches V, B has covered 0.5K.
In the second leg when A turns back, they have 1.5K to cover, BUT now the ratios must be divided in 5 parts (1:4), since there speeds add up effectively.

Thus the remaining 1.5K gets split into 0.3K for B and 1.2K for A

So B travels 0.5K + 0.3K = 0.8K = 4K/5

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