Two friends Ana and Boris are approaching towards each other, each one

Speed of A = 1 kmph
Speed of B = 1 kmph
Speed of Dog = 9 kmph

Time taken by dog to reach from A to B = Distance / Relative Speed = 10 / (1+9) = 1 Hour

Distance traveled by A in this duration = Speed * Time = 1*1 = 1 Km
Distance traveled by Dog in this duration = Speed * Time = 9*1 = 9 Km

Distance between Dog and A when Dog starts coming back to A = 9 - 1 = 8 km

i.e. Time taken by dog to come back to A = Distance / Relative speed = 8 / (1+9) = 8/10 hours = 48 minutes

Distance travelled by A in 48 minutes = 1*(4/5) = 0.8 km

Total Distance travelled by Ana = 1 Km (When dog ran from A to B) + 0.8 km = 1.8 km

Answer: Option D

First hour,
A traveled 1km toward B,
B traveled 1km toward A,
Dog traveled 9km toward B and met B.
Distance between A and dog was 8km by that time.

Additional 0.8 hours later,
A traveled additional 0.8km
Dog traveled additional 7.2km toward A and met A (dog speed is 9x A's speed).

The distance travelled by Ana between the time the dog leaves her and comes back to her =
1+ 0.8 = 1.8km

FINAL ANSWER IS (D)

Posted from my mobile device

Two friends Ana and Boris are approaching towards each other, each one at 1 kmph. Ana is walking with a dog, which can run at speed of 9 kmph. The dog leaves Ana and runs towards Boris when Ana and Boris are 10 km apart. After reaching Boris the dog immediately runs back to Ana. What is the distance travelled by Ana between the time the dog leaves her and comes back to her?

A. 1.2 km
B. 1.4 km
C. 1.6 km
D. 1.8 km
E. 2 km

When Dog leaves Ana,
let time taken be t
Therefore t+9t = 10 [Distance while walking toward each other will be added and the time remains same]
Therfore t = 1hr (Dog will reach Boris in 1 hr)
Distance covered by Ana within 1 hr = 1 km --- i)

Now the distance between Dog and Ana will remain 8 kms which is same as the distance between Anna and Boris (since both traveled 1 km in that 1 hr)
Thus For Dog and Ana to meet again
t+9t = 8 => \(t=\frac{8}{10}\)
Distance covered by Ana = \(1*\frac{8}{10}\) --- ii)

Total Distance covered by Ana while Dog leaves her and meets again = i) + ii) = 1.8 Kms - Answer D

ar=1
br=1
dr=9

time d to reach b: 10km/(9+1)kmph=1hr
ana traveled: 1hr*1kmph=1km
d traveled to b: 1hr*9=9km
a distance from d: 9-1=8km
time d to reach a: 8/10=0.8hr
ana traveled: 0.8hr*1=0.8km
total ana traveled: 1.8km

Ans (D)

for 1st 1 hour ; relative speed of dog & Boris ; 10 kmph and it takes 10/10 ; 1 hr for dog to reach Boris
in the same 1 hour Ana would have covered 1 km and so would have Boris ; now the net distance b/w them is 8 km
which now Dog who is running towards Ana would cover in 8/10; .8 hr and in this period Ana would cover 1*.8 ; .8 km
so total distance of Ana in these 2 hrs ; 1+.8 ; 1.8km
IMO D

Two friends Ana and Boris are approaching towards each other, each one at 1 kmph. Ana is walking with a dog, which can run at speed of 9 kmph. The dog leaves Ana and runs towards Boris when Ana and Boris are 10 km apart. After reaching Boris the dog immediately runs back to Ana. What is the distance travelled by Ana between the time the dog leaves her and comes back to her?

A. 1.2 km
B. 1.4 km
C. 1.6 km
D. 1.8 km
E. 2 km

Ana and Boris are travelling towards each other
Dogs speed = 9kmph
hence dog+boris would have met in\( \frac{10}{9+1}\) hrs = 1 hr
In that 1 hr Ana would have traveled 1km ------------(a)
now dog distance between ana and dog is (8) km
(1 km boris would have moved and dog and boris met 1km away from where boris started)

Dog and ana would meet in \(\frac{8}{9+1 }\)= 0.8 hrs
in 0.8 hrs ana would have traveled 0.8*1 = 0.8 km-------------(b)
hence total would have traveled a +b = 1+0.8km = 1.8 km
_________________

Ana(1km) ----->Dog(9km)----->Boris(1km)
----------------10 km-----------------

The dog met Boris in an hour in 9th kilometer. \((9+1)t=10\) --> \(t=1\) hour--> \(9*1=9\)
Dog---------><-Boris
-------------9km---

Dog comes back to Ana--> Now the distance between them is 8 km (Ana walked 1 km in one hour)
--> \((9+1)*t_1= 8\)
\(t_1= 0.8\) hour

The total distance Ana walked -- \(Speed(t+t_1)= 1(1+0.8)= 1.8 km\)

The answer is D.

Two friends Ana and Boris are approaching towards each other, each one at 1 kmph. Ana is walking with a dog, which can run at speed of 9 kmph. The dog leaves Ana and runs towards Boris when Ana and Boris are 10 km apart. After reaching Boris the dog immediately runs back to Ana. What is the distance travelled by Ana between the time the dog leaves her and comes back to her?

A. 1.2 km
B. 1.4 km
C. 1.6 km
D. 1.8 km
E. 2 km

Time taken by dog to meet Boris = \(\frac{10}{(9+1)}\) = 1 hr
distance traveled by Ana in 1 hr = 1km
distance traveled by Boris in 1 hr = 1 km
Now, the distance between dog and Ana = 10 - 1 - 1 = 8km
time taken by by them meet = \(\frac{8}{(9+1)}\) = \(\frac{4}{5}\)hr
Total distance covered by Ana in (\(1 + \frac{4}{5}\)) hr = \(1 + \frac{4}{5}\) = 1.8km

Answer D.
_________________

Distance between dog and Boris is 10km


Time taken by dog to reach Boris is 10/(9+1)=1hr


In that 1hr Ana travelled 1km

Now dog is returning from Boris to ana....now the distance between them is 8km

Time taken by dog to reach Ana is 8/(9+1)=0.8hr

In that 0.8hr Ana travelled 0.8km

Total distance travelled is 1.8km

OA:D

Posted from my mobile device

Imagine a number line. Ana and Doggy are at 0 and Boris is at 10.
After one hour Ana is at 1 and Boris is at 9. So the Doggy can reach Boris at the end of the first hour, since its speed is 9km/hr.
In the second hour Ana travels some more distance and Doggy reaches to Ana.

Total distance traveled by Ana is D_1 (which is greater than 1 km)
and total distance traveled by Doggy is 18-D_1

Since Distance = Speed x Time,
(Distance traveled by Ana/Speed of Ana) = (Distance traveled by Doggy/Speed of Doggy)

(D_1/1) = ((18-D_1)/9)
9 * D_1 = 18 - D_1
10 * D_1 = 18
D_1 = 1.8