An anonymous reader came by seeking help with this question:

A motorist takes 1/2 h less than the usual time to travel from KTown to JTown when he increases his average speed by 10%. For a distance of 100km, he will take 1/3 h less than the usual time if his speed is 20% more than the average speed from KTown to JTown. Find the distance between KTown and JTown.

***************************

For those who want to take a stab at this, don't look further. Anyway, I'm not sure if the answer's right. Brian attempted this question and came up with the algebraic statements but did not get the right answer before he left for school; I think he stumbled when he used 30 min instead of 1/2 hour. I tried it myself after he left and I think I got the solution, for a speed question! Yay me!! :)

***************************

Let D = Distance and S = Speed; Time taken = D/S

D/S - D/(11/10)S = 1/2

D/S - 10D/11S = 1/2

D/11S = 1/2

2D = 11S --- (1)

When D is 100 and speed is 20% more

D/S - D/(12/10)S = 1/3

D/S - D/12S = 1/3

D/6S = 1/3

3D = 6S

300 = 6S

S = 50km/h --- (2)

Substitute (2) into (1)

2D = 11(50)

2D = 550

D = 275km

## 3 comments:

Thank you for solving the question.

Scenario A -> Speed (110%) & time is 1/2 h less. Scenario B -> Speed (120% for first 100 km & 100% for the rest of remaining distance) & time is 1/3 h less.Scenario C -> Speed (100%).

Ratio of speed

A : B : C

110 : 120 : 100

11 : 12 : 10

Common multiple of 11,12,10 is 660

Ratio of time

A : B : C

60 : 55 : 66

C-A -> 66-60 = 6 units

6 units-> 1/2 h

60 units-> 1/2 x 10 = 5 h (A)

Speed A : 11 units

Time A : 5 h

Distance = 11 x 5 = 55 units

For scenario B->

5+1/2-1/3 = 5 and 1/6 = 31/6 h

(55 units - 100 km)/10 + 100 km/12 = 31/6

(330 units - 600 km + 500 km)/60 = 31/6

1980 units - 600 km = 1860 units

120 units = 600 km

1 unit = 5 km

55 units = 5 x 55 = 275 km

Distance between the two towns is 275 km.

Another method using ratio plus algebra.

Ratio of speed

100 % : 110 %

1 : 1.1

Ratio of time

T-0.5 : T

where T is the time to reach the destination while travelling at normal speed.

Cross multiply them becomes :

T = 1.1(T-0.5)

T = 1.1T-0.55

0.1T = 0.55

T = 5.5 hours

Ratio of speed

100 % : 120 %

1 : 1.2

Ratio of time

T1-1/3 : T1

where T1 is the time to travel first 100 km of the journey.

Cross multiply them becomes :

T1 = 1.2(T1-1/3)

3T1 = 1.2(3T1-1)

3T1 = 3.6T1 - 1.2

0.6T1 = 1.2

T1 = 2 hours

2 hours-> 100 km

5.5 hours -> 100/2 x 5.5 = 275 km.

Distance between the two towns is 275 km.

Post a Comment