Question

Help please :)

A boat covers 32km upstream and 36 km downstream in 7 hours. Also it covers 40 km upstream and 48 km downstream in 9 hours . Find the speed of boat in still water and that of stream.

Class 10 CBSE
Pair of linear equations in two variables.
-Cahira :)​

Answer 1

Given :-

Seed upstream = 32 km/h

Speed downstream = 36 km/h

Time $\sf _1$ = 7 hours

Distance Upstream = 40 km

Distance downstream = 48 km/h

Time $\sf _2$ = 9 hrs

To Find :-

The speed of boat in still water and that of stream.

Solution :-

We know that,

• t = Time
• d = Distance
• s = Speed

Given that,

Distance traveled ($\sf d_1$) = 36 km

Distance traveled ($\sf d_2$) = 32 km

Time $\sf _1$ = 7 hours

Time $\sf _2$ = 9 hours

According to the question,

Let us consider the speed of the boat in the water to be x km/hr

Speed of the stream = y km/hr

Case I:

Speed of the boat in downstream = (x + y) km/hr

Given that, distance traveled, $\sf d_1$ = 36 km

$\underline{\boxed{\sf Speed= \dfrac{Distance}{Time} }}$

Substituting them,

$\sf \therefore \: Time \: (t_1)=\dfrac{36}{(x+y)}$

Speed of the boat in upstream = (x - y) km/hr

Distance traveled, $\sf d_2$ = 32

$\sf \therefore \: Time \: (t_2)=\dfrac{32}{(x+y)}$

Total time = 7 hours

$\sf \therefore \: t_1+t_2=7$

$\sf \dfrac{36}{(x+y)}+\dfrac{32}{(x-y)} =7 \qquad ...(1)$

$\sf Let \ \dfrac{1}{(x+y)} =a \ and \ \dfrac{40}{(x-y)} =b$

Equation (1) and (2) becomes,

$\sf 36a+32b=7 \qquad ...(3)$

$\sf 48a+40b=9 \qquad ...(4)$

Multiply equation (4) with 4 and (3) with 5,

$\sf 192a+160b=36 \qquad ...(5)$

$\sf 180a+160b=35 \qquad ...(6)$

Subtracting equation (6) from (5),

$\sf a=\dfrac{1}{12}$

Put the value of a in equation (3), we get

$\sf b=\dfrac{1}{8}$

$\sf \dfrac{1}{(x+y)}=a=\dfrac{1}{12} \longrightarrow (x+y)=12 \qquad ...(7)$

$\sf \dfrac{1}{(x-y)} =b=\dfrac{1}{8} \longrightarrow (x-y)=8 \qquad ...(8)$

After adding equation (7) and (8),

$\longrightarrow \sf x= 10$

$\longrightarrow \sf y=2$

Therefore,

Speed of the boat still in water = 10 km/hr

Speed of the stream = 2 km/hr

Case II:

Speed of the boat downstream = (x + y) km/hr

Distance traveled, $\sf d_3$ = 48

$\sf \therefore \: t_3=\dfrac{48}{(x+y)}$

Speed of the boat in upstream = (x + y) km/hr

Distance traveled, $\sf d_4$ = 40

$\sf \therefore \: t_4=\dfrac{40}{(x-y)}$

Total time = 9 hours

$\sf \therefore \: t_1+t_2=9$

$\sf \dfrac{48}{(x+y)}+\dfrac{40}{(x-y)} =9 \qquad ...(2)$

Answer 2

Answer:

Heya,

Let the speed of the boat

in still water = x kmph

Speed of the stream = y kmph

i ) relative speed of the  

boat in downstream

= ( x + y ) kmph

Distance travelled = d1 = 36

Time = t1 hr

t1 = d1 / s1

t1 = 36/ ( x + y )  

ii) relative speed of the boat  

in upstream = ( x - y ) kmph

Distancw = d2 = 32 km

Time = t2  

t2 = 32/ ( x - y )

Therefore ,

Total time = 7 hr

t1 + t2 = 7hr

36 / ( x + y ) + 32/ ( x - y ) = 7 ----( 1 )

iii) second time ,

Relativespeed of the boat in

downstream = ( x + y ) kmph

d3 = 48 km

Time = t3

t3 = 48/ ( x + y )

iv ) in upstream

Relative speed of the boat = ( x - y )  

kmph

time = t4 hr

d4 = 40km

t4 = 40/ ( x - y )

Total time = 9 hr

48 / ( x + y ) + 40/ ( x - y ) = 9 ---( 2 )

Let 1 / ( x + y ) = a ,

1 / ( x - y ) = b

Then rewrite ( 1 ) and ( 2 ) we get  

36 a + 32 b = 7 -----( 3 )

48a + 40b = 9 ------( 4 )

Multiply ( 4 ) with 4 and equation ( 3 ) with 5 and  

192a + 160b = 36 ---( 5 )

180a + 160b = 35 -----( 6 )

Subtract ( 6 ) from ( 5 )

we get  

a = 1/ 12  

put a = 1/ 12 in ( 3 )

we get ,

b = 1/ 8

Now 1/ ( x + y ) = 1/ 12  

1/ ( x - y ) = 1/ 8

Therefore ,

x + y = 12 ----( 7 )

x - y = 8 ----- ( 8 )

add ( 7 ) and ( 8 )

2x = 20

x = 10

put x = 10 in ( 7 ) we get  

y = 2

Speed of the boat in  

still water = x = 10 kmph

speed of the stream

= y = 2kmph

I hope this helps you.

Step-by-step explanation: