If the speed of boat in still water on Saturday was 27 km/hr and the speed of boat in still water on Wednesday was 66 2/3% more than that of Saturday and time taken to travel upstream on Wednesday is 16/13 times than time taken by it to travel downstream on Saturday, then find the speed of stream (in kmph) on Saturday? Option 1: 2 Option 2: 4 Option 3: 9 Option 4: 8

Answer: 9. Solution: Let speed of boat Upstream be - U Let speed of boat Downstream be - D Let speed of boat in still water be - B Let speed of stream be - W U = B - W D = B + W Given on Saturday: B_sat = 27 kmph On Wednesday: B_wed = B_sat + 662/3% x B_sat B_wed = 27 + 200/3 x 27/100 = 45 kmph Let time taken to travel downstream on Saturday be - t Time taken to travel upstream on Wednesday = 16/13t On Wednesday: W_wed = 6 kmph U_wed = B_wed - W_wed = 45 - 6 = 39 kmph Distance travelled upstream on Wednesday = 12% of 4800 = 576 km Time taken upstream on Wednesday = 576/39 = 192/13 hours 16/13t = 192/13 t = 12 hours Time taken downstream on Saturday = 12 hours Distance covered downstream on Saturday = 432 km Speed downstream on Saturday = 432/12 = 36 kmph D_sat = B_sat + W_sat 36 = 27 + W_sat W_sat = 9 kmph Therefore, speed of stream on Saturday is 9 kmph

Arithmetic

Time, Speed & Distance

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Figure for IPMAT Rohtak 2019 QA question 3 (Arithmetic)

If the speed of boat in still water on Saturday was 27 km/hr and the speed of boat in still water on Wednesday was $66 \frac{2}{3}\%$ more than that of Saturday and time taken to travel upstream on Wednesday is $\frac{16}{13}$ times than time taken by it to travel downstream on Saturday, then find the speed of stream (in kmph) on Saturday?

✅ Correct Option: 3

Let speed of boat Upstream be - $U$

Let speed of boat Downstream be - $D$

Let speed of boat in still water be - $B$

Let speed of stream be - $W$

$U = B - W$

$D = B + W$

Given on Saturday:

$B_{sat} = 27 \text{ kmph}$

On Wednesday:

$B_{wed} = B_{sat} + 66\frac{2}{3}\% \times B_{sat}$

$B_{wed} = 27 + \frac{200}{3} \times \frac{27}{100} = 45 \text{ kmph}$

Let time taken to travel downstream on Saturday be - $t$

Time taken to travel upstream on Wednesday $= \frac{16}{13}t$

On Wednesday:

$W_{wed} = 6 \text{ kmph}$

$U_{wed} = B_{wed} - W_{wed} = 45 - 6 = 39 \text{ kmph}$

Distance travelled upstream on Wednesday $= 12\% \text{ of } 4800 = 576 \text{ km}$

Time taken upstream on Wednesday $= \frac{576}{39} = \frac{192}{13} \text{ hours}$

$\frac{16}{13}t = \frac{192}{13}$

$t = 12 \text{ hours}$

Time taken downstream on Saturday $= 12 \text{ hours}$

Distance covered downstream on Saturday $= 432 \text{ km}$

Speed downstream on Saturday $= \frac{432}{12} = 36 \text{ kmph}$

$D_{sat} = B_{sat} + W_{sat}$

$36 = 27 + W_{sat}$

$W_{sat} = 9 \text{ kmph}$

Therefore, speed of stream on Saturday is $9 \text{ kmph}$

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