Math, asked by Anonymous, 3 months ago

man covered a certain distance at some speed had he moved 3 km/hr faster he would have taken 40 minutes less If he had moved 2 km/hr slower he would have taken 40 minutes more The distance ( in km) 

.​

Answers

Answered by Anonymous
28

Answer:

man covered a certain distance at some speed had he moved 3 km/hr faster he would have taken 40 minutes less If he had moved 2 km/hr slower he would have taken 40 minutes more The distance ( in km) 

Answer:

The distance and speed are 40 km and 12 km/hr respectively.

Step-by-step explanation:

Let us assume that the certain distance and speed be x and y respectively.

Formula: Time = Distance/Speed

According to question:

He moved 3 km/hr faster he would have taken 40 minutes less,

x/y - x/(y + 3) = 40/60 [1 minute = 1/60 hour]

→ (xy + 3x - xy)/(y² + 3y) = 2/3

→ 3x/(y² + 3y) = 2/3

→ 9x = 2y² + 6y ___________(i)

If he had moved 2 km/hr slower he would have taken 40 minutes more,

x/(y - 2) - x/y = 40/60

→ (xy - xy + 2x)/(y² - 2y) = 2/3

→ 2x/(y² - 2y) = 2/3

→ 3x = y² - 2y ___________(ii)

On dividing equation (i) by equation (ii):

9x/3x = (2y² + 6y)/(y² - 2y)

→ 3 = y(2y + 6)/y(y - 2)

→ 3(y - 2) = 2y + 6

→ 3y - 6 = 2y + 6

→ y = 12

Substituting the value of y in equation (ii):

3x = (12)² - 2 × 12

Answered by Anonymous
5

Answer:

Let the distance = x km and usual rate = y km/hr Then

y

x

y+3

x

=

60

40

y(y+3)

x(y+3)−xy

=

3

2

y(y+3)

3x

=

3

2

⇒2y(y+3)=9x...............(i)

and

y−2

x

y

x

=

60

40

y(y−2)

xy−x(y−2)

=

3

2

y(y−2)

2x

=

3

2

⇒y(y−2)=3x...............(ii)

On dividing eqn (i) by eqn (ii) we get

y−2

2(y+3)

=3⇒2y+6=3y−6⇒y=12

∴ Putting the value of y in (i) we get 2 x 12 x 15 = 9x ⇒ x = 40 km/hr

Similar questions