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QUESTION: John drives along a straight motorway at a constant speed for 120 kilometres. The next day he completes the same journey travelling at 10 km/h faster. He completes this journey in 10 minutes less that the first journey. Find the two speeds at which he travelled.

ANSWER: I put the problem in Excel and let it do the calculations for me.

The main problem to watch out for is the speed is in kilometers per hour

and the difference it time needs to be 10 minutes.

This means we need to remember to adjust the difference (which is in hours)

by multiplying by 60 minutes per hour.

At 60 k/h, it takes 2 hours. Now 60 + 10 = 70, and at 70 k/h, it takes 1.71 hours.

The difference between 1.71 hours and 2 hours is roughly 17 minutes, so this is to slow.

At 70 k/h, it takes 1.71 hours. Now 70 + 10 = 80, and at 80 k/h, it takes 1.5 hours.

The difference between 1.5 hours and 1.71 hours is roughly 13 minutes, so this is to slow.

At 80 k/h, it takes 1 1/2 hours. Now 80 + 10 = 90, and at 90 k/h, it takes 1 1/3 hours.

The difference between 1 1/2 hours and 1 1/3 hours is 1/6 of an hour, which is exactly 10 minutes. That means 80 k/h and 90 k/h are the two speeds that answer the question.

---------- FOLLOW-UP ----------

QUESTION: I was looking for an actual equation and not just trial and error and I am not allowed to use Excel as this is preparation for a maths exam?

OK, lets redo the question with equations.

Take t1 and v1 are the 1st time and velocity.

Take t2 and v2 are the 2nd time and velocity.

It can be seen that since time * speed = distance,

(A) t1*v1 = 120 and (B) t2*v2 = 120

with velocity in km / hour and time in hours.

The question gives us

(C) v2 = v1 + 10 and (D) t2 = t1 - 10

when time is in minutes.

Note that t*v = 120 is true no matter what units are used,

so both equations can be taken as equation in minutes.

Also, note that velocity are both in hourly units,

so the difference needs to be divided by 60.

The units are all now in minutes. This makes the equations be

(A) t1*v1 = 120;

(B) t2*v2 = 120;

(C) v2 = v1 + 1/6; and

(D) t2 = t1 - 10.

Using (A), it can be seen that it is (A) t1 = 120 / v1.

Using (B), it can be seen that (B) t2 = 120 / v2.

Putting (A) and (B) into (D) gives us (D) 120/v2 = 120/v1 - 10.

We can then take (C) and put it into (D), giving (D) 120/(v1 + 1/6) = 120/v1 - 10.

Multiplying all three terms by v1(v1 + 1/6) gives (D) 120*v1 = 120(v1 + 1/6) - 10*v1(v1 + 1/6).

Multiplying out each term gives (D) 120*v1 = 120*v1 + 20 - 10*v1² - 10*v1/6.

Subtract 120*v1 from both sides and note that 10/6 is the same as 5/3.

This gives 0 = 20 - 10*v1² -5*v1/3.

Multiplying by -3/5 and reordering the terms gives 0 = 6*v1² + v1 - 12.

Note that the right side can be written as (3v-4)(2v+3), so our equation is really 0 = (3v-4)(2v+3).

That gives us two choices. It says either that v = 4/3 or v = -3/2.

Since we are only interested in positive time, throw out the 2nd answer.

If v = 80 km/h minutes, t = 120/80 = 3/2 of an hour. In minutes, that is t1 = 90 minutes.

Since v2 is v1 + 10, and v1 is 80, v2 is 90.

Since t2 is t1 - 10, and t1 is 90, v2 is 80.

As can be seen, both 80 km/h in and hour and a half and 90 km/h in an hour and a third

give a total of 120 km, so the answer checks out.

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