Question

Two cars leave an intersection at the same time. One car travels north and the other one east. When the travelling north had gone 16 km more than twice the distance travelled by the car heading east, the distance between the cars was 16 km less than three times the distance travelled by the car heading east. Find the distance between the cars at that time.

Explain with complete steps & justifications.

Points : 15 ☺

Answer 1

i hope this will usful to u

Answer 2

Let car A moves in East and car B moves in North ,
Let car A covered distance x km in East direction

and Car B covered distance (2x + 16) km in North direction .
because north and east are perpendicular
so, at that time right angle trainlge form and hypotenuse is distance between car at that time
distance between the cars = √(x)² + (2x +16)²

again ,

√{ x ² + (2x + 16)² } = 3x -16

take square both sides

x² + 4x² + 64x +256 = 9x² - 96x +256

4x² -96x -64x = 0

x = 40 km , 0 km

but x ≠ 0 km

so, x = 40 Km

hence distance travelled by car A = 40 km
and distance travelled by car B = 2{40} +16 = 96 km

so, distance between car = √(40² +96)² or 3(40) -16 = 120 -16 = 104 km

hence distance between car at that time = 104 km