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Quantitative Aptitude > Learn Quantitative Aptitude > A car travels from P to Q at a constant speed. If its...
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A car travels from P to Q at a constant speed. If its speed were increased by 10 km/hr, it would have been taken one hour lesser to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/hr. The distance between the two cities is ?

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The distance between The two Cities P and Q is 420 kms
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Maths and Quantitative Aptitude Trainer with 10 years of experience

A car travels from P to Q at a constant speed. If its speed were increased by 10 km/hr, it would have taken one hour lesser to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/hr. What is the distance between the two cities? d/s - d/(s...
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A car travels from P to Q at a constant speed. If its speed were increased by 10 km/hr, it would have taken one hour lesser to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/hr. What is the distance between the two cities? d/s - d/(s + 10) = 1 d/s - d/(s + 20) = 1.75 d(s + 10) - ds = s(s + 10) d(s + 20) - ds = 1.75s(s + 20) 10d = s^2 + 10s 20d = 1.75s^2 + 35s 10d = 0.75s^2 + 25s s^2 + 10s = 0.75s^2 + 25s 0.25s^2 - 15s = 0 s = 60 10d = 3600 + 600 d = 420 The distance between the two cities is 420 km. read less
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Let X be the distance and Y be the initial speed. So initial time require is X/Y hrs. Now form two simultaneous equations. X/(Y+10 = X/Y - 1 and X/(Y+20) = X/Y - 1.75 Solving these two equations we get X = Distance = 420. So answer is 420
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dist = x, speed = s, time=t x=st, (s+10)(t-1)=x, (s+20)(t-7/4)=x solving (s+10)(t-1)=st 10t-s=10...(1) solving (s+20)(t-7/4)=st 20t-7/4s=35 80t-7s=140....(2) (2) - (1)*4= t=7 so s=60 therefore, x=420 Ans: distance between P and Q is 420km
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420
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Solutions: Let d= distance between P and Q t= Time taken for normal v= velocity d = v t -------(1) If speed is increased by 10km/h then time taken reduced by 1hr, then d= (v+10) (t -1) ......(2) If speed is further increased by another 10km/h, time taken decreases by another 45 min. =0.75hrs...
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Solutions: Let d= distance between P and Q t= Time taken for normal v= velocity d = v t -------(1) If speed is increased by 10km/h then time taken reduced by 1hr, then d= (v+10) (t -1) ......(2) If speed is further increased by another 10km/h, time taken decreases by another 45 min. =0.75hrs then d= (v+20) (t - 1.75)......(3) solving equations (1), (2) and (3), we get t = 7hrs, v= 60 km/h and the distance d= 450km read less
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d/s - d/(s + 10) = 1 --->d(s + 10) - ds=s(s + 10) --->10d = s^2 + 10s--->20d=2s^2 + 20s d/s - d/(s + 20) = 1.75 ---->d(s + 20) - ds=1.75s(s + 20)---->20d = 1.75s^2 + 35s Therefore, 2s^2 + 20s=1.75s^2 + 35s--->0.25s^2=15s---->s = 60 10d = 3600 + 600 =4200 d = 420 Ans.
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Let X be the distance and Y be the initial speed. So initial time require is X/Y hrs. Now form two simultaneous equations. X/(Y+10 = X/Y - 1 and X/(Y+20) = X/Y - 1.75 Solving these two equations we get X = Distance = 420. So answer is 420
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420
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