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Figure for IPMAT Rohtak 2019 QA question 2 (Arithmetic)
Figure for IPMAT Rohtak 2019 QA question 2 (Arithmetic)
Figure for IPMAT Rohtak 2019 QA question 2 (Arithmetic)

If the time taken by boat to travel upstream on Monday is 271527 \frac{1}{5} hrs more than the time taken by it to travel downstream on the same day, then find the speed of boat in still water on Monday? (speed of boat in still water is the same in upstream as in downstream)

Solution

✅ Correct Option: 1

Let speed of the boat Upstream be - UU

Speed of the boat Downstream be - DD

Speed of boat in still water be - BB

Speed of the stream be - WW

U=B−WU = B - W

D=B+WD = B + W

On Monday distance covered Upstream =16% of 4800=768 km= 16\% \text{ of } 4800 = 768 \text{ km}

Distance covered Downstream =14% of 2400=336 km= 14\% \text{ of } 2400 = 336 \text{ km}

Let time taken to cover downstream be - t hrst \text{ hrs}

Then, time taken to cover upstream will be - t+27.2 hrst + 27.2 \text{ hrs}

On Monday, Speed of still water(stream) =W=5 kmph= W = 5 \text{ kmph}

U=B−5U = B - 5

D=B+5D = B + 5

D−U=B+5−(B−5)=10 kmphD-U = B + 5 - (B - 5) = 10 \text{ kmph}

Also, D=336tD = \frac{336}{t}

Also, U=768t+27.2U = \frac{768}{t+27.2}

D−U=10 kmphD - U = 10 \text{ kmph}

336t−768t+27.2=10\frac{336}{t} - \frac{768}{t+27.2} = 10

10t+704t−9139.2=010t + 704t - 9139.2 = 0

t=11.2 hourst = 11.2 \text{ hours}

So, D=33611.2=30 kmphD = \frac{336}{11.2} = 30 \text{ kmph}

Also, D=B+5D = B + 5

B+5=30B + 5 = 30

B=25 kmphB = 25 \text{ kmph}

Therefore, the speed of boat in still water on Monday is - 25 kmph25 \text{ kmph}

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