Question

(1) The speed of a boat in stiIl water is 8 kms/hr. If the boat can go 15 kms. down stream and 22 kms. up stream in 5 hours, then let us write by calculating, the speed of the stream
(2) A superfast train runs having the speed 15 kms/hr more than that of an express train. Leaving same station the superfast train reached at a station of 180 kms. distance 1 hour before than the express train. Let us determine the speed of the superfast train in km/hr.​

Answer 1

Q1 Ans: Let x be the speed of the stream.

⇒  Speed of the boat in still water is 8km/hr

⇒  The speed of the boat in upstream is 8−xkm/hr

⇒  The speed of the boat in downstream is 8+xkm/hr

⇒  The time taken by the boat to cover 15km= 15/8 - x hr

The time taken by the boat to cover 22km = 22/8 + x hr

15/8-x + 22/8 + x = 5

15(8+x)+22(8−x)=5(8−x)(8+x)

⇒  120+15x+176−22x=5(64−x^2)

⇒  296−7x=320−5x^2

⇒  5x^2 −7x−24=0

⇒  5x^2 −15x+8x−24=0

⇒  5x(x−3)+8(x−3)=0

⇒  (x−3)(5x+8)=0

⇒  x−3=0 and 5x+8=0

⇒  x=3 and x= −8/5

Speed cannot be negative.

∴ The speed of the stream is 3km/hr.

Q2 Ans: Let the speed of the superfast train was x km/hr.

∴ the speed of the express train was (x-5)km/hr.

∴ to travel 180 km, time taken by the superfast train is 180x hr and by express train is 180x−15 hr As per questions, 180x − 15 − 180x = 1

or, 15x^2 − 15x = 1180

or, x^2 − 15x = 2700

or, x^2 − 15x − 2700=0

=120/2 and x= −90/2

= 60 and x= -45

But the speed of a train cannot be negative, ∴x = 60.

Hence the speed of the superfast train was 60 km/hr.

Hence, Q1 answer = 3km/hr and Q2 answer = 60 km/hr.

​

Answer 2

${ \sf {\huge{ \underline{ Answer \: 1}}}}$

Given:

• The speed of a boat in stiIl water is 8 kms/hr.

• The boat goes 15 kms. down stream and 22 kms. up stream in 5 hours.

To Find:

• The speed of the stream.

Solution:

Suppose the speed of the stream be x km/hr

The speed of the boat in upstream = 8 - y km/hr

The speed of the boat in downstream = 8 + y km/hr

We know that,

$\sf{Time = \frac{Distance}{Speed} }$

$\small\sf{Time \: taken \: by \: the \: boat \: to \: cover \: 15m = \frac{15}{8x}hr }$

$\small\sf{Time \: taken \: by \: the \: boat \: to \: cover \: 22km = \frac{22}{8 + x}hr } \\ </p><p>$

A.T.Q,

${ \sf {\Rightarrow \: { \frac{15}{8 - x} + \frac{22}{ 8 + x} = 5 }}} \\ </u></strong></p><p><strong><u>[tex]{ \sf {\Rightarrow \: { \frac{15}{8 - x} + \frac{22}{ 8 + x} = 5 }}}$

$\small \sf\Rightarrow \: {15(8 + x) + 22(8 - x) = 5(8 - x)(8 + x)}$

$\small \sf\Rightarrow \: {120 + 15x + 176 - 22x =5 (64 - {x}^{2})}$

$\small \sf\Rightarrow \: {296 - 7x = 320 - 5 {x}^{2} }$

$\small \sf\Rightarrow \: {5 {x}^{2} - 7x - 24 = 0 }</p><p>$

$\small \sf\Rightarrow \: {5 {x}^{2} - 15x + 8x - 24 = 0 }</p><p>$

$\small \sf\Rightarrow \: {5x(x - 3) + 8(x - 3) = 0}$

$\sf{}\Rightarrow \: {(5x + 8)(x - 3) = 0}$

$\sf{}\Rightarrow \: {5x + 8 = 0} \: \: \: or \: \: x - 3 = 0$

$\sf{}\Rightarrow \: {x = \frac{ - 8}{5} = 0 } \: \: or \: \: x = 3$

$\sf{}\Rightarrow \: {x = 3}$

Since speed cannot be negative.

Thus the speed of the stream is 3km/hr

${ \sf {\huge{ \underline{ Answer \:2}}}}$

Given:

• A superfast train runs having the speed 15 kms/hr more than that of an express train.

• Leaving same station the superfast train reached at a station of 180 kms. distance 1 hour before than the express train.

To Find:

• The speed of the super fast train in km/hr

Solution:

Suppose the speed of the superfast was y km/hr

The speed of the express train was(x - 5) km/hr

$\sf \small{Time \: taken \: by \: the \: superfast \: train \: = \frac{180}{x}hrs }$

$\sf \small{Time \: taken \: by \: the \: express\: train \: = \frac{180}{x - 15}hrs }$

A.T.Q,

$\sf \Rightarrow \: { \frac{180}{x - 15} - \frac{180}{x} = 1 }$

$\sf \Rightarrow \: {180( \frac{1}{x - 15} - \frac{1}{x}) = 1 }$

$\sf \Rightarrow \: { \frac{x - x + 15}{(x - 15)x} = \frac{1}{180} }$

$\sf \Rightarrow \: { \frac{15}{ {x}^{2} - 15x } = \frac{1}{180} }$

$\sf \Rightarrow \: { {x}^{2} - 15x = 2700 }$

$\sf \Rightarrow \: { {x}^{2} - 15x - 2700 = 0 }$

$\sf \Rightarrow \: {(x - 60)(x + 45) = 0}$

$\sf \Rightarrow \: {x - 60 = 0} \: \: or \: \: x + 45 = 0$

$\sf \Rightarrow \: {x = 60} \: , \: - 45$

Since speed cannot be negative.

Thus, the speed of the superfast train was 60km/hr.