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Numerical Ability
Averages
Two cars start from the same point at same time.the first car travels at a constant speed of 12 km/h while the second car travels at an initial speed of 6 km/h,which increases by 1 km/h every hour.what will be the diatance traveled by the second car by the time it meets the first car?
Read Solution (Total 5)
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- 156 kms
- 10 years agoHelpfull: Yes(2) No(6)
- after 11 hrs 1st car travels 11x12=132km
after 11 hrs 2 nd car travels 6x11+1+2+3.....+11=132km(car increases its speed by 1km for 1hr)
ans:132km - 10 years agoHelpfull: Yes(1) No(8)
- 156 kms
13×12=156
12×6+13+12+11......+1=156
- 10 years agoHelpfull: Yes(1) No(9)
- to meet the distance 156km at 18hours
- 10 years agoHelpfull: Yes(0) No(5)
- Use sum of 1st n terms of AP
Sn= n/2[2a+(n-1)d]
For 1st car a = 12KM/h d = 0
Sn= n/2[2(12)+(n-1)0]
= 12n............> Eq
For 2nd car a = 6KM/h d = 1KM/h
Sn= n/2[2(6)+(n-1)1]
=1/2[11n+n^2] ...............> Eq2
Now equate Eq 1 and Eq 2
12n = 1/2[11n+n^2]
24n = 11n+n^2
13n = n^2
n = 13
Substitute n = 13 in any equation
We get the distance as 156 KM - 7 years agoHelpfull: Yes(0) No(0)
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