A cyclist rode the first 18 km of his trip at a constant rate, and for the remaining 6 km he reduced his rate by 6 km/hour. Find the cyclist's rate in the last 6 km, if the entire trip took 1.5 hours.

12 km/h

Explanation:

The formula for speed is distance divided by time

S=D/T

T=D/S

Let the rate for the first 18 km of his trip be x km/h

Time taken in this first phase of the trip = 18/ x hrs

Then the rate for the remaining 6 km of his trip will be x-6 km/h

Time taken in this second phase of the trip is : 6/ x-6 hrs

Equation for the entire trip is;

18/x + 6/x-6 = 1.5

multiply each term by the common divisor x(x-6)

x(x-6)*18/x + x(x-6)* 6/x-6= 1.5* x(x-6)

18(x-6) +6 x =1.5 (x²+6 x)

18 x -108 +6 x = 1.5 x² - 9 x

24 x -108 = 1.5 x² -9 x

1.5 x² -33 x + 108 = 0 divide each term by 1.5

x²-22 x + 72 = 0 solve the quadratic equation

x²-4 x -18 x + 72 =0

x(x-4) -18 (x-4) =0

(x-18) (x-4) =0

x-18 =0

x= 18 or

x-4 =0

x=4

The rate in the 6 km will be

x-6 = 18-6 = 12 km/h

the answer would be c

f |_| ck f |_| ck

explanation: