Two cars A and B start from the same point. \'A\' travels at a constant speed of 10 km/hr but \'B\'

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26 Feb 2013 speed

Two cars A and B start from the same point. 'A' travels at a constant speed of 10 km/hr but 'B' starts with 5 km/h and every hr its speed increases by 0.25 km/hr. After how many hours will B overtake A?

OPtion

1) 21 hr
2) 26 hr
3) 31.5 hr
4) 41 hr
5) 41.5 hr
6) 46 hr
7) 59.5 hr
8) 52 hr
9) 57 hr
10) None of these

Solution

Let the time taken by B to overtake A be t hours
So distance travelled by B = Distance Travelled by A (i)
distance Travelled by A in t hr = speed * time
= 10t
and Distance travelled by B =
5, 5.25, 5.50, 5.75, ........... t
Total Distance = n/2(2a+(n-1)d) (by using AP)
therefor 10t = t/2(2*5+(t-1).25) ........by (i) equation
so t = 41 (Solve)
option no 4

Correct Option: 4 Best Solution (12)

sumit jindal   11 years AGO

let x be the hours to overtake
distance traveled by A=10x
distance traveled by B=10x
=>5+(5+0.25)+(5+(.25*2))+...+(5+(x-1)*0.25)=10x
=>5x+(.25*(1+2+3+...+(x-1)))=10x
=>5x+x*(x-1)/8=10x
=>x^2-41x=0
=>x=41
so 4)41 hr.

Like? Yes (14) | No   

phani   11 years AGO

b is in AP ....

n/2(2a+(n-1)d)=10n====>a=5,d=.25
n=41

Like? Yes (7) | No (3)  

Sneha Agarwal   11 years AGO

it will take 20 hours to B to match the speed of A and then again 20 hours to cover the remaining distance.
hence total time = 20+1+20 = 41

Like? Yes (5) | No (1)  

Vinay Sipani    11 years AGO

Distance travelled by A and B = d
Time taken in hours = n

A => d = 10n
B => d = d1 + d2 + d3 + ... = (5*1) + (5.25*1) + (5.5*1) + ...

=> (n/2) (10 + (n-1)0.25 ) = 10n
=> n(9.75 + 0.25n) = 20n
=> n = 41

Like? Yes (3) | No   

Amar kumar singh   11 years AGO

its simple we have to find out the time t=s.d
both train run in A.p series look.
10.n=n/2(2.5+(n-1).25) find n i.e 41

Like? Yes (3) | No   

Rats   11 years AGO

considering 41hour
A's distance=10*41=410km
B's distance=(41/2)*(10+39*.25)=410km
So after 41hour B overtake A.

Like? Yes (3) | No (1)  

Anurag Arora   11 years AGO

initial speed is 5 km/hr
5, 5.25, 5.5 ----- 10
apply ap formula => means it will take 21 hours to reach at average speed of 10 km/hr
now, we have to increase 20 times to overtake another person whose avg speed is 10 km/hr
total hours = 21 + 20 = 41 hours

Like? Yes (2) | No   

Sahil   11 years AGO

After 41 hrs
distance travelled by B= 5+5.25+5.5+...... 41 terms = (41/2)*(10+40*0.25) = 20*41/2 = 10*41 = distance travelled by A in 41 hrs

Like? Yes (2) | No   

sc mandota   11 years AGO

solution is 41

Like? Yes (2) | No (1)  

Mahendra Singh   11 years AGO

solution is 41

Like? Yes (2) | No (1)  

kathija    11 years AGO

let n be the no.of hrs B overtake A
10n = n/2(2a+(n-1)d)
10n= n/2(10+(n-1)1/4)
on solving
80n=40n+n^2-n
n^2=41n
n=41

Like? Yes (1) | No   

AVINASH KUMAR   11 years AGO

let B overtakes A in x hours
then distance traveled by B and A in x hours is equal
hence
10x=(5+5.25+5.50+............+(5+x*2.5))x
10=5+5.25+5.50+5.75+........
10=5+(n-1)*.25
5=(n-1)*.25
20=n-1
hence
n=20+1=21 hours

Like? Yes (4) | No (5)