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Home » A boat goes upstream and downstream in . It goes upstream and downstream in hours. Find the speed of the boat in still water and also speed of the stream.
A boat goes 24km upstream and 28km downstream in 6hrs. It goes 30km upstream and 21km downstream in 6.5 hours. Find the speed of the boat in still water and also speed of the stream.
CBSE | Class 10 | Exercise 3.10 | Maths | Pair of Linear Equations In Two Variables | RD Sharma
A boat goes 24km upstream and 28km downstream in 6hrs. It goes 30km upstream and 21km downstream in 6.5 hours. Find the speed of the boat in still water and also speed of the stream.

Let’s assume,

The speed of the boat in still water as Ckm/hr

And,

The speed of the stream as Dkm/hr

We know that,

Speed of the boat in upstream =(C-D)km/hr

Speed of the boat in downstream =(C+D)km/hr

So, time taken to cover 28 km downstream =28/(C+D)hr [∵ time = distance/ speed]

Time taken to cover 24 km upstream =24/(C-D)hr [∵ time = distance/ speed]

It’s given that the total time of journey is 6 hours. So, this can expressed as

24/(C-D)+28/(C+D)=6…… (i)

Similarly,

Time taken to cover 30 km upstream =30/(C−D) [∵ time = distance/ speed]

Time taken to cover 21km downstream =21/(C+D) [∵ time = distance/ speed]

And for this case the total time of the journey is given as 6.5 i.e 13/2 hours.

Hence, we can write

30/(C-D)+21/(C+D)=13/2 ….. (ii)

Hence, CD solving (i) and (ii) we get the required solution

Taking, 1/(C-D)=u and 1/(C+D)=v in equations (i) and (ii) we have (after rearranging)

24u+28v-6=0 …… (iii)

30u+21v-13/2=0 ……. (iv)

Solving these equations CD cross multiplication we get,

\frac{u}{28x-6.5-21x-6}=\frac{-v}{24x-6.5-30x-6}=\frac{1}{24\times 21-30\times 28}

u=1/6 and v=1/14

Now,

u=1/(C−D)=1/6

C-D=6 …. (v)

v=1/(C+D)=1/14

C+D=14……. (vi)

On Solving (v) and (vi)

Adding (v) and (vi), we get

2C=20

⇒ C=10

Using C=10 in (v), we find D

10+D=14

⇒ D=4

Therefore,

Speed of the stream =4km/hr.

Speed of boat =10km/hr.

i

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