Question

t took Brian 6 hours to drive to a job interview. On the way home, he was able to increase his average speed by 19mph and make the return drive in only 4 hours. Find his average speed on the return drive.

Answer 1

$\huge \bf\mathfrak \red{ANSWER:}$

$\boxed {57 mph}$

$\boxed{Step-by-step verification:}$

LET, THE GOING SPEED IS x AND RETURNING SPEED IS x+19.

NOW,

DISTANCE FOR GOING(D¹)

$= v \times t \\ = x \times 6 \\ = 6x$

THEN,

DISTANCE FOR RETURNING(D²)

$= v \times t \\ = (x + 19) \times 4 \\ = 4x + 76$

WE KNOW,

D¹=D²

OR,

$6x = 4x + 76$

OR,

$2x = 76$

•°•

$x = 38mph$

AGAIN,

RETURNING SPEED

$=( x+19 )mph\\ = (38 + 19 )mph\\ =57mph$

SO, THE AVERAGE SPEED ON RETURNING IS 57 mph.

Answer 2

Given :- It took Brian 6 hours to drive to a job interview. On the way home, he was able to increase his average speed by 19mph and make the return drive in only 4 hours. Find his average speed on the return drive. ?

Solution :-

Let us assume that, Original speed of Brain while going for Job interview was x km/h .

so,

→ speed * Time = Distance

→ x * 6 = 6x km .

now,

→ New speed = (x + 19) km/h .

→ Time = 4 hours.

then,

→ Distance = 4(x + 19) km.

since, both distance is same ,

→ 4(x + 19) = 6x

→ 4x + 76 = 6x

→ 6x - 4x = 76

→ 2x = 76

→ x = 38 km/h .

therefore,

→ Speed while return drive = 38 + 19 = 57 km/h . (Ans.)

Hence, Brian average speed on the return drive is 57km/h .

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