A man drives to a car park at an average speed of 40km/hr and then walks to his office at an average speed of 6km/h. The total journey takes him 25 minutes. One day his car broke down and he has to walk three times as far, so arriving at his office 17 minutes late. How far is it to his office? WHEN CAR WAS NORMAL Speed to cark park =40km/h Speed to office = 6km/h total time = 5/12 h distance to car park = xkm distance to office = ykm time = distance/speed x/40 + y/6 = 5/12 3x + 20y = 50 WHEN CAR BROKE DOWN He walked 3ykm = total time = 7/10 h Maybe the man had moved some distance before the car broke down. The total distance to be covered by foot and car is (x+y)km. The distance driven = [(x+y)-3y]km = (x-2y)km We have now \(\frac{x-2y}{40}+\frac{3y}{6}=7/10\) solving the equations in case (i) and case (ii) simultaneusly, we obtain y = 1km and x= 10km My problem now is how could this problem be solved in one variable?
As the total distance walked is 25 + 17 minutes and this distance is also 3 times the usually walked distance, the normal walking time (at 6 kph) is (25 + 17)/3 = 14 minutes.
The total time to walk the total distance is 25+17 minutes (42 minutes). The average walk speed is 6km/hr. Walking for 42 minutes at 6km/hr (100 meters per minute) means you walked a total distance of 4,200 meters in 42 minutes. So to answer the question in the OP of "How far is it to his office?", it's 4.2km to the office.
On the break down day he takes 17 mts extra and additionally walks twice the normal walking distance.... so, if the normal walk distance is y, then (2y/6) - (2y/40) = 17/60......This gives y = 1 km.
Using only one variable, things doesn't look all that write to me. ON THE DAY THINGS WERE NORMAL speed to car park = 40km/h speed to office = 6km/h total time = 25/60 = 5/12 hrs time to office = a hrs time to car park (5/12-a)hrs = \(\frac{(5-12a)}{12}\) distance to office = 6a km distance to car park = \(\frac{(5-12a)}{12}40\\=\frac{(5-12a)10}{3}\) km ON THE DAY THINGS WERE NOT TOO GOOD total time 42/60 = 7/10h distance to office = 3a(6) = 18a km distance to car park = (7/10 - a)40 = (7-10a)4 = (28-40a)km total distance travelled when car was normal = total distance travelled when car was bad \(28-40a=\frac{-102a+50}{3}\) thus 84-120a = 50-102a a = 34/18 =17/9 km distance to office = \(6\times\frac{17}{9}\) = 11.33km the distance to office gotten using two variable is not the same as that gotten using only one variable.
You are assuming that he walked the entire way, the question does not say that, I don't think. If tha is what it is saying that is a pretty easy question. Chikis generally asks very ambiguous questions.
Few slips in your solution above, corrected below in red... ON THE DAY THINGS WERE NOT TOO GOOD total time 42/60 = 7/10h distance to office = 3a(6) = 18a km distance to car park = (7/10 - 3a)40 = (7-30a)4 = (28-120a)km total distance travelled when car was normal = total distance travelled when car was bad 6a + (50-120a)/3 = 18 a + 28-120a Solve for a, you will get a = 1/6, thus distance for normal walk = 1 Km....Rest is simple. Why so many silly goof ups in your solution ??
It all depends on what the questions is. Did he drive some distance and then breakdown or did he walk the whole way. Apparently chikis does not know. I base that on the other 2 forums where he has asked the exact same question and his confused questions and answers. As always to figure out what the real question is that chikis is asking, he must go back to wherever he got the original question and CAREFULLY copy the question onto the forum.
Define: x = distance to office, in km (What we want to solve for) y = normal distance walked, in km (what we don't care about) \(\frac{x - y}{40} + \frac{y}{6} = \frac{25}{60} \\ \frac{x - 3y}{40} + \frac{3y}{6} = \frac{25}{60} + \frac{17}{60}\) Term by term, take 3 times the first equation minus the second equation: \(\frac{3x - 3y}{40} - \frac{x - 3y}{40} + \frac{3y}{6} - \frac{3y}{6} = \frac{75}{60} - \frac{25}{60} - \frac{17}{60}\) and expand terms: \(\frac{3x}{40} - \frac{3y}{40} - \frac{x}{40} + \frac{3y}{40} + \frac{3y}{6} - \frac{3y}{6} = \frac{75}{60} - \frac{25}{60} - \frac{17}{60}\) Simplify: \(\frac{3x}{40} - \color{red}{\frac{3y}{40}} - \frac{x}{40} + \color{red}{\frac{3y}{40}} + \color{orange}{\frac{3y}{6}} - \color{orange}{\frac{3y}{6}} = \frac{75}{60} - \frac{25}{60} - \frac{17}{60} \\ \frac{2x}{40} = \frac{50-17}{60} \\ \frac{x}{20} = \frac{33}{60}\) Solve this equation in one variable: x = 11 (km) // Edit: And with matrices: Same thing, more formally: \(\begin{pmatrix} \frac{1}{40} & \frac{1}{6} - \frac{1}{40} \\ \frac{1}{40} & \frac{3}{6} - \frac{3}{40} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{25}{60} \\ \frac{25}{60} + \frac{17}{60} \end{pmatrix} \) Since \(\begin{pmatrix} \frac{1}{40} & \frac{1}{6} - \frac{1}{40} \\ \frac{1}{40} & \frac{3}{6} - \frac{3}{40} \end{pmatrix} = \begin{pmatrix} \frac{1}{40} & \frac{17}{120} \\ \frac{1}{40} & \frac{17}{40} \end{pmatrix} \) and \( \begin{pmatrix} 60 & -20 \\ -\frac{60}{17} & \frac{60}{17} \end{pmatrix} \begin{pmatrix} \frac{1}{40} & \frac{17}{120} \\ \frac{1}{40} & \frac{17}{40} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) we have \( \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 60 & -20 \\ -\frac{60}{17} & \frac{60}{17} \end{pmatrix} \begin{pmatrix} \frac{25}{60} \\ \frac{25}{60} + \frac{17}{60} \end{pmatrix} = \begin{pmatrix} 25 - 14 \\ -\frac{25}{17} + \frac{42}{17} \end{pmatrix} = \begin{pmatrix} 11 \\ 1 \end{pmatrix} \) And since \(\begin{pmatrix} 60 & -20 \\ -\frac{60}{17} & \frac{60}{17} \end{pmatrix} = \begin{pmatrix} 20 & 0 \\ 0 & \frac{60}{17} \end{pmatrix} \begin{pmatrix} 3 & -1 \\ -1 & 1 \end{pmatrix} \) we get the prescription from the top line of the second matrix to "take 3 times the first equation minus the second equation" if we wish to solve just for x.
That's the question in the OP, and my question to you is; What's your answer to the question in the OP? If this was a written test question you have the option to answer or leave blank, your choice. If you leave blank you have no room to talk! BTW, Did I ever tell you I completed a test construction course after my instructor training course?
I agree with chikis answer, not yours. I like rpenner's single variable equation.Please Register or Log in to view the hidden image! Not that I recall
So 11.33km? So you like rpenner's 11km answer vs chikis's 11.33km? Which one do you like the most? So much I forgot to tell you.
It doesn't meet the initial conditions - it doesn't take him 25 minutes to get to work if he walks 1.4 kilometers of it and drives the rest at 40kph.
If it's 4.2 k total, 2/3 of that or 2.8 klicks worth, .o7 hours, or a little over 4 minutes. Then 1.4 k at 6 per, a little under 9 minutes. Around 13 minutes to get to work. We need 25.
The OP solved this for originally 10 km of driving and 1 km of walking. Since on the day the care broke down walking was 3 km, then there must have been 8 km of driving. 10 km of driving at 40 km/hour takes 15 minutes. 1 km of walking at 6 km/hour takes 10 minutes. 15 + 10 = 25 minutes. 8 km of driving at 40 km/hour takes 12 minutes. 3 km of walking at 6 km/hour takes 30 minutes. 12 + 30 = 25 + 17. So he was 17 minutes late on the day his car broke down.
