A boat goes up-stream 12 km in the same time in which it goes down-str

1.	A boat goes up-stream 12 km in the same time in which it goes down-stream 28 km. If the current flows at 6 km/h, find the rate of rowing on still water and also the rate at which the boat goes down the stream
Answer» Answer :- $\: \: \boxed{\boxed{\rm{\mapsto \: \: \: Firstly \: let's \: understand \: the \: concept \: used}}}$ Here the concept of Linear Equations in Two Variables has been used. According to this, if we make the value of one variable depend on other we can find the value of both. Here, are going to take the distance between the places and speed of boat as unknown quantities. Let's do it !! _____________________________________________ ★ Question :- A boat goes up-stream 12 km in the same time in which it goes down-stream 28 km. If the current flows at 6 km/h, find the rate of rowing on still water and also the rate at which the boat goes down the stream. _____________________________________________ ★ Solution :- Given, » Distance COVERED in Upstream = 12 Km » Distance covered in Downstream = 28 Km » Speed of the stream = 6 Km/hr • Let the speed of the boat be 'x' Km/hr • Let the time taken by the boat in both journeys be 'y' hr. So, » Speed of boat in upstream = (x - 6) Km/hr » Speed of boat in downstream = (x + 6) Km/hr => Distance = Speed × Time Then, according to the question :- ~ Case I (for upstream) :- ✒ (x - 6) × y = 12 ✒ xy - 6y = 12 ... (i) ~ Case II (for downstream) :- ✒ (x + 6) × y = 28 ✒ xy + 6y = 28 ... (ii) From equation (i) and (ii), we get ✒ xy - 6y - (xy + 6y) = 12 - 28 ✒ xy - 6y - xy - 6y = -16 ✒ -12y = -16 ✒ 12y = 16 $\: \: \large{\bf{\longmapsto \: \: y \: = \: \dfrac{16}{12} \: = \: \dfrac{4}{3}}}$ $\bf{\large{\boxed{y\: = \: \dfrac{4}{3}}}}$ From the value of y and equation (i), we get, ✒ (x - 6) × (4/3) = 12 ✒ (x - 6) = 3 × 3 ✒ x - 6 = 9 ✒ x = 9 + 6 $\bf{\large{\qquad \qquad \qquad{\boxed{x \: = \: 15 \: Kmhr^{-1}}}}}$ Then, • Speed of boat in still water = x = 15 Km/hr • Speed of boat while going downstream = (x + 6) Km/hr = 15 + 6 Km/hr = 21 Km/hr $\: \: \boxed{\sf{\leadsto \: \: Thus, \: the \: speed \: of \: boat \: in \: still \: water \: is \: \boxed{\underline{15 \: Kmhr^{-1}}} \: and \: that \: in \: downstream \: is \: \boxed{\underline{21 \: Kmhr^{-1}}}}}$ _____________________________________________ $\: \: \: \: \underbrace{\large{\underline{\sf{Confused? \; Don't \: worry \: let's \: verify \: it \: :-}}}}$ For verification, we need to simply APPLY the values we GOT into our equations. Then, ~ Case I :- => xy - 6y = 12 => 15(4/3) - 6(4/3) = 12 => (5×4) - (2×4) = 12 => 20 - 8 = 12 => 12 = 12 Clearly, LHS = RHS ~ Case II :- => xy + 6y = 28 => 15(4/3) + 6(4/3) = 28 => 20 + 8 = 28 => 28 = 28 Clearly, LHS = RHS Here both the conditions satisfy, so our answer is CORRECT. Hence, Verified. _____________________________________________ $\: \: \: \huge{\boxed{\tt{\large{More \: to \: know \: :-}}}}$ • Linear Equations are the equations formed using CONSTANT and variable terms but of single degree. • Polynomials are the equations formed using constant and variable terms and can have many different degrees.

A boat goes up-stream 12 km in the same time in which it goes down-stream 28 km. If the current flows at 6 km/h, find the rate of rowing on still water and also the rate at which the boat goes down the stream

Answer»

Answer :-

$\: \: \boxed{\boxed{\rm{\mapsto \: \: \: Firstly \: let's \: understand \: the \: concept \: used}}}$

Here the concept of Linear Equations in Two Variables has been used. According to this, if we make the value of one variable depend on other we can find the value of both. Here, are going to take the distance between the places and speed of boat as unknown quantities. Let's do it !!

_____________________________________________

★ Question :-

A boat goes up-stream 12 km in the same time in which it goes down-stream 28 km. If the current flows at 6 km/h, find the rate of rowing on still water and also the rate at which the boat goes down the stream.

_____________________________________________

★ Solution :-

Given,

» Distance COVERED in Upstream = 12 Km

» Distance covered in Downstream = 28 Km

» Speed of the stream = 6 Km/hr

• Let the speed of the boat be 'x' Km/hr

• Let the time taken by the boat in both journeys be 'y' hr.

So,

» Speed of boat in upstream = (x - 6) Km/hr

» Speed of boat in downstream = (x + 6) Km/hr

=> Distance = Speed × Time

Then, according to the question :-

~ Case I (for upstream) :-

✒ (x - 6) × y = 12

✒ xy - 6y = 12 ... (i)

~ Case II (for downstream) :-

✒ (x + 6) × y = 28

✒ xy + 6y = 28 ... (ii)

From equation (i) and (ii), we get

✒ xy - 6y - (xy + 6y) = 12 - 28

✒ xy - 6y - xy - 6y = -16

✒ -12y = -16

✒ 12y = 16

$\: \: \large{\bf{\longmapsto \: \: y \: = \: \dfrac{16}{12} \: = \: \dfrac{4}{3}}}$

$\bf{\large{\boxed{y\: = \: \dfrac{4}{3}}}}$

From the value of y and equation (i), we get,

✒ (x - 6) × (4/3) = 12

✒ (x - 6) = 3 × 3

✒ x - 6 = 9

✒ x = 9 + 6

$\bf{\large{\qquad \qquad \qquad{\boxed{x \: = \: 15 \: Kmhr^{-1}}}}}$

Then,

• Speed of boat in still water = x

= 15 Km/hr

• Speed of boat while going downstream

= (x + 6) Km/hr

= 15 + 6 Km/hr

= 21 Km/hr

$\: \: \boxed{\sf{\leadsto \: \: Thus, \: the \: speed \: of \: boat \: in \: still \: water \: is \: \boxed{\underline{15 \: Kmhr^{-1}}} \: and \: that \: in \: downstream \: is \: \boxed{\underline{21 \: Kmhr^{-1}}}}}$

_____________________________________________

$\: \: \: \: \underbrace{\large{\underline{\sf{Confused? \; Don't \: worry \: let's \: verify \: it \: :-}}}}$

For verification, we need to simply APPLY the values we GOT into our equations. Then,

~ Case I :-

=> xy - 6y = 12

=> 15(4/3) - 6(4/3) = 12

=> (5×4) - (2×4) = 12

=> 20 - 8 = 12

=> 12 = 12

Clearly, LHS = RHS

~ Case II :-

=> xy + 6y = 28

=> 15(4/3) + 6(4/3) = 28

=> 20 + 8 = 28

=> 28 = 28

Clearly, LHS = RHS

Here both the conditions satisfy, so our answer is CORRECT.

Hence, Verified.

_____________________________________________

$\: \: \: \huge{\boxed{\tt{\large{More \: to \: know \: :-}}}}$

• Linear Equations are the equations formed using CONSTANT and variable terms but of single degree.

• Polynomials are the equations formed using constant and variable terms and can have many different degrees.