Latest Accenture Aptitude Question SOLUTION: A man travels three-fifths of distance AB at a speed of 3a, and the remaining at a speed of 2b, if he goes from A to B and back at a speed of 5c in the same time, then prove that 1

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A man travels three-fifths of distance AB at a speed of 3a, and the remaining at a speed of 2b, if he goes from A to B and back at a speed of 5c in the same time, then prove that 1/a + 1/b = 1/c

Asked In Accenture X22 (12 years ago)

Solved Jai Read Solution (10)

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Read Solution (10) :

sanjib chowdhury  (12 years ago)

Let the distance between A and B is X
Three-fifths of distance AB =3X/5
Remaining distance=2X/5
If he goes from B to A and back i.e X+X=2X

Now from the question we have

(3X/(5*3a))+(2x/(5*2b))=2X/5c
Or 1/a+1/b=2/c

Is this solution Helpfull? Yes (15) | No (4)  

Anil Purohit  (12 years ago)

this que shoud be like this...............

A man travels three fifths of a distance AB at a speed 3a, and the remaining at a speed 2b. If he goes from B to A and returns at a speed 5C in the same time then :
@opt a
1/a + 1/b =1/c

@opt b
a+b=c
@opt-ans
1/a + 1/b =2/c
@opt
None of these

Is this solution Helpfull? Yes (6) | No (4)  

Jai  (12 years ago)

(3/5 * d / 3a)+(2/5 * d / 2b) = d/5c
then

(1/a)+(1/b)=(1/c)

Is this solution Helpfull? Yes (3) | No (3)  

Anil Purohit  (12 years ago)

its ans will be 1/a +1/b =2/c

Is this solution Helpfull? Yes (2) | No (5)  

revathy.r  (12 years ago)

X be the distance AB
X/(3a+2b) = X/5c

5c = 3a + 2b

Is this solution Helpfull? Yes (1) | No (4)  

v s s sidharda  (10 years ago)

Let the distance between A and B is X
Three-fifths of distance AB =3X/5
time taken is (3X/5)/(3a)= 3X/15a
Remaining distance=2X/5
time taken is (2X/5)/(2a)= 3X/10a.
and on return journey X/5c.

so eq is

3X/15a + 2X/10b = X/5c

X/a + X/b = X/c

Is this solution Helpfull? Yes (1) | No   

srilatha  (12 years ago)

data in sufficient

Is this solution Helpfull? Yes | No (4)  

Swapnil Gupta  (11 years ago)

Let the distance between A and B is X
Three-fifths of distance AB =3X/5
Remaining distance=1-3x/5=2X/5
If he goes from A to B and back then total distance X+X=2X
and 2*[((3X/(5*3a))+(2x/(5*2b))]
Now acc. to question

2*[(3X/(5*3a))+(2x/(5*2b))]=2X/5c
solve it and you get 1/a+1/b=1/c

Is this solution Helpfull? Yes | No   

ABHISHEK  (7 years ago)

speed=dist./time
3a=(3/5AB)/t1
2b=(2/5AB)/t2
and also given
t1+t2=AB/5c
now, ((3/5AB)/3a)+((2/5AB)2b)=AB/5c
it becomes,3AB/15a + 2AB/5b=AB/5c
take 1/5AB common from both side and hence it is proved..

Is this solution Helpfull? Yes | No   

sonal agrawal  (7 years ago)

A....................C...............................B
AB = x => AC = 3x/5 and CB = 2x/5.
Distance = Speed * Time
Time taken to travel the distance between A and C = 3x/5 = 3a * T1 => T1 = x/5a
Time taken to travel the distance between C and B = 2x/5 = 2b * T2 => T2 = x/5b
Time taken to travel between A and B back and fourth = x= 5c * T => T = x/5c
Given that, T = T1 + T2
=> x/5a + x/5b = x/5c
=> 1/a + 1/b = 1/c.

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