# a boat goes 30 km upstream and 44 km downstream in 10 h. in 13 h, it can go 40 km upstream and 55 km downstream. determine the speed of the stream and that of the boat in still water.

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## The boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of stream and that of the boating till water.

Click here👆to get an answer to your question ✍️ The boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of stream and that of the boating till water.

Question

## The boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of stream and that of the boating till water.

**A**

## Speed of boat = 8 km/h & Speed of stream =3 km/hr

**B**

## Speed of boat = 8 km/h & Speed of stream =4 km/hr

**C**

## Speed of boat = 7 km/h & Speed of stream =2 km/hr

**D**

## Data insufficient

Hard Open in App Solution Verified by Toppr

Correct option is A)

Let the speed of boat in still water=x km\hr and The speed of stream=y km\hr

Speed of boat at downstream

⇒(x+y)km/hr

Speed of boat at upstream

⇒(x−y)km/hr ∵time= speed distance

Time taken to cover 30 km upstream ⇒

x−y 30

Time taken to cover 44 km downstream⇒

x+y 44

According to the first condition,

⇒ x−y 30 = x+y 44 =10

Time taken to cover 40 km upstream ⇒

x−y 40

Time taken to cover 55 km downstream ⇒

x+y 55

According to the second condition,

⇒ x−y 40 = x+y 55 =13 Let x−y 1 =uand x+y 1 =v ⇒30u+44v=10.....eq1 ⇒40u+55v=13.....eq2

Multiplying eq1 by 3 and eq2 by 5 and subtract both

⇒(150u+220v=50)−(160u+220v=52)

⇒−10u=−2⇒u= 5 1 put u= 5 1 in eq1 ⇒30× 5 1 +44v=10⇒44v=4⇒v= 4 1 ⇒u= x−y 1 = 5 1 ⇒x−y=5...eq3 ⇒v= x+y 1 = 11 1 ⇒x+y=11...eq4

Subtracting eq3 and eq4, we get

⇒x=8 Put x=8 in eq3 ⇒y=3

Hence, the speed of the boat in still water=8km\hr

The speed of stream=3km\hr

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## A boat goes 30 km upstream and 44 downstream in 10 hours. The same boat goes 40 km upstream and 55 km downstream in 13 hours. On this information some student guessed the speed of the boat in still water as 8.5 km / h and speed of the boat in still water as 8.5 km / h and speed of the stream as 3.8 km / h. Do you agree with their guess? Explain what do we learn from the incident?

A boat goes 30 km upstream and 44 downstream in 10 hours. The same boat goes 40 km upstream and 55 km downstream in 13 hours. On this information some student guessed the speed of the boat in still water as 8.5 km / h and speed of the boat in still water as 8.5 km / h and speed of the stream as 3.8 km / h. Do you agree with their guess? Explain what do we learn from the incident?

Byju's Answer Standard X Mathematics

Solving Simultaneous Linear Equation Using Method of Elimination

A boat goes 3... Question

A boat goes 30 km upstream and 44 downstream in 10 hours. The same boat goes 40 km upstream and 55 km downstream in 13 hours. On this information some student guessed the speed of the boat in still water as 8.5 km/h and speed of the boat in still water as 8.5 km/h and speed of the stream as 3.8 km/h. Do you agree with their guess?

Explain what do we learn from the incident?

Open in App Solution

Let the speed of boat = x km/hr.

Let the speed of stream = y km/hr.

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

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

Time taken to cover 30 km upstream

= 30 x − y hrs.

Time taken to cover 40 km downstream

= 44 x + y hrs.

According to question,

Total time taken = 10 hrs.

30 x − y + 44 x + y = 10 ...(i)

Now, Time taken to cover 55 km downstream

= 55 x + y h r s .

Time taken to cover 40 km upstream

= 40 x − y hrs.

Total time taken = 13 hrs.

40 x − y + 55 x + y = 13 ...(ii)

Solving eq. (i) and eq. (ii).

Let 1 x − y = u , 1 x + y = v 30 u + 44 v = 10 ...(iii) 40 u + 55 v = 13 ...(iv)

Multiplying eq. (iii) by 4 and eq. (iv) by 3, and subtracting we get

120 u + 176 v = 40 120 u + 165 v = 39 − − − 11 v = 1 v = 1 11

Putting the value of v in eq. (iii)

30 u + 44 v = 10 ⇒ 30 u + 44 × 1 11 = 10 ⇒ 30 u + 4 = 10 ⇒ 30 u = 6 ⇒ u = 6 30 or u = 1 5 Now, v = 1 11 ⇒ 1 x + y = 1 11 ⇒ x + y = 11 ...(v) And u = 1 5 ⇒ 1 x − y = 1 5 ⇒ x − y = 5 ...(vi)

On solving eq. (v) and (vi)

x + y = 11 x − y = 5 –––––––––––– 2 x = 16 o r x = 8

Put the value of x in eq. (v)

8 + y = 11 y = 11 - 8 y = 3

The speed of boat in still water = 8 km/hr.

The speed of stream = 3 km/hr.

We learn that the speed of boat is slow in upstream and fast in downstream

Suggest Corrections 69

SIMILAR QUESTIONS

**Q.**

The boat goes 30 k m upstream and 44 k m downstream in 10 h o u r s . In 13 h o u r s , it can go 40 k m upstream and 55 k m

downstream. Determine the speed of the stream and that of the boat in still water.

**Q.**A boat goes 30 km upstream and 44 km downstream in 10 hours. It can go 40 km upstream and 55 km downstream in 13 hours. Find the speed of the stream and that of the boat in still water.

**Q.**

A Boat goes 24 km upstream and 20 km downstream in 6 hours ago 30 kilometre upstream and 21 km downstream in 6 and half hours find speed of boat in still water and also speed of stream?

**Q.**The boat goes

30 km upstream and 44 km downstream in 10 hrs. In 13 hours, it can go 40 km upstream and 55

km downstream. Determine the product of speed of stream and that of the boat in still water.

**Q.**A boat goes 24 km upstream and 28 km downstream in 6 hrs. It goes 30 km upstream and 21 km downstream in

612hrs

. Find the speed of the boat in still water and also speed of the stream.

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Standard X Mathematics

## The boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of stream and that of the boating till water.

The boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of stream and that of the boating till water.. Ans: Hint – Here we will proceed by using the formulas Downs...

Last updated date: 12th Mar 2023

• Total views: 208.4k • Views today: 1.88k Answer Verified 208.4k+ views 3 likes

Hint – Here we will proceed by using the formulas Downstream speed :

a=u+v a=u+v , upstream speed: b=u−v b=u−v

. Then the speed of boat in still water

u= 1 2 (a+b) u=12(a+b)

, the speed of the boat in still water:

v= 1 2 (a−b) v=12(a−b) .

Complete step-by-step solution -

Let the speed of the boat in still water =

x x km per hour.

And, the speed of stream

=y =y km per hour.

Speed of boat at downstream

⇒(x+y) ⇒(x+y) km per hour.

Speed of boat at upstream

⇒(x−y) ⇒(x−y) km per hour. ∵ ∵ time= distance speed time=distancespeed

Time taken to cover 30 km upstream

⇒ 30 x−y ⇒30x−y

Time taken to cover 44 km downstream

⇒ 44 x+y ⇒44x+y

According to the first condition,

⇒ 30 x−y + 44 x+y =10 ⇒30x−y+44x+y=10

Time taken to cover 40 km upstream

⇒ 40 x−y ⇒40x−y

Time taken to cover 55 km downstream

⇒ 55 x+y ⇒55x+y

According to the second condition,

⇒ 40 x−y + 55 x+y =13 ⇒40x−y+55x+y=13 Let 1 x−y =u 1x−y=u and 1 x+y =v 1x+y=v ⇒30u+44v=10 ⇒30u+44v=10 …. (1) ⇒40u+55v=13 ⇒40u+55v=13 …. (2)

Multiply (1) by 5 and (2) by 4 and subtract both

⇒(150u+220v=50)−(160u+220v=52)

⇒(150u+220v=50)−(160u+220v=52)

⇒−10u=−2 ⇒u= 1 5 ⇒−10u=−2⇒u=15 Put u= 1 5 u=15 in equation (1) ⇒30× 1 5 +44v=10 ⇒44v=4 ⇒v= 1 11

⇒30×15+44v=10⇒44v=4⇒v=111

⇒u= 1 x−y = 1 5 ⇒u=1x−y=15 ⇒x−y=5 ⇒x−y=5 …. (3) ⇒v= 1 x+y = 1 11 ⇒v=1x+y=111 ⇒x+y=11 ⇒x+y=11 …. (4)

Adding equation (3) and equation (4), we get

⇒x−y=5 ⇒x−y=5 + ( x+y=11 x+y=11 ) ⇒x=8 ⇒x=8 Put x=8 x=8 in (3) x−y=5 ⇒8−5=y ⇒3=y x−y=5⇒8−5=y⇒3=y

Hence, the speed of the boat in still water = 8 km per hour and the speed of stream = 3 km per hour.

Note – Whenever we come up with this type of problem, one must know that when an object moves downstream in a river, the river current supports the object’s movement and hence the net speed of the object is the sum of its speed in still water and the speed of the river current. On the other hand, the current acts against the object’s movement if the object is travelling upstream, hence the net speed of the object is still water subtracted by the speed of the river current.

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