Chikis

We were told that they could finish the job in 15 days. They did 6 days work remaining 9 days work, which A complete in 30 days? 9 days work took A 30 days. 6 days work will take A [math]\frac{6}{9}\times30[/math] = 20 days It will take A 20 + 30 = 50 days. I believe it will take B 50 days as well.

You mean I should divide the 1 whole by the days?

Let's the work is 1 whole. 1/9 is remaining after B fell ill because they had already done 1/6 of the work together. Am I right?

If this is a problem on ratio or propotion, it would have been more easier for me to handle. Am thinking this is a harder problem on rate. For example, I would find it more easier to handle this problem: 12 men complete a job in 9 days. How many men working at the same rate, would be required to complete the job in 6 days? Could you ask me to show you my work on this problem? Maybe you now show me how can apply the same method to solve the problem in op.

A and B can do a piece of work together in 15 days. They both started, but after 6 days B gets ill and then A takes 30 more days to finish it by himself. How long would each take, working alone? How do I tackle this? Am havina a hard time dealing with this problem. They can do the job together in 15 days. When they both started, they did the job for 6 days. But B could not complete the job due to ill health. This means that it remaining more 9 days to complete the job if both of them are working together. But in this case A spent 30 days doing the 9 days work. Ijust have had time dealing with this.

That is his expenses right?

What is overhead?

A shopkeeper made a net profit of N12,500 from his business in 1984. The cost price of all the goods he bought during amounted to N104000. His overheads amounted to N13 500. Calculate: (i) his average weekly sales during the year. (ii) his net profit as a percentage of his annual turnover. (Take 1 year = 52 weeks) What do the following terms mean as it is concerned with the solving of this problems: (i) turnover (ii) overheads (iii) net profit

It does. Thank you.

L = 9 W = 4.5 (9-2)(4.5-3) = 10.5 (4.5)9 = 40.5 40.5 - 10.5 = 30

Here is the work. Since L= 2w, the area of the triangle would be LW. Thus [math]LW= 2W(W) = 2W^2[/math] Based on the condition of the problem, we have [math](L-2)(w-3)=2w^2-30[/math] If expanded, we have [math]2w^2-6w-2w+6=2w^2-30[/math] [math]\to[/math] -8w = -36 [math]\therefore[/math] W = -36/-8 = 4.5cm L = 2(4.5) = 9cm

Am not in a hurry to solve the problem.I would like us to look at the way some problems on rectangle are phrased and study them. A rectangle is twice as long as it is wide. Mean L = 2W where L and W are lenght and width of the rectangle. It also means that the lenght is two times the width. Can the statement be written also as a lenght is two times as long as it wide? If the statement is, a rectangle is twice as wide as it is long, then the equation W = 2L. The same statement also mean that the width is twice it lenght. Hope my understanding concerning the following statement are very correct. [math](L-2)(W-3) = 2W^2-30[/math] W = 4.5cm L = 9cm

Am not in a hurry to solve the problem. I would like us to look at the way some problems on rectangle are phrased and study them. A rectangle is twice as long as it is wide. Mean L = 2W where L and W are lenght and width of the rectangle. It also means that the lenght is two times the width. Can the statement be written also as a lenght is two times as long as it wide? If the statement is, a rectangle is twice as wide as it is long, then the equation W = 2L. The same statement also mean that the width is twice it lenght. Hope my understanding concerning the following statement are very correct.

A rectangle is twice as long as it is wide. If the lengh and width are decreased by 2cm and 3cm respectively, the area is decreased by 30 cm^2. Find the original dimension of the rectangle. We have that 2L=w where L = lenght of rectangle and w = width of rectangle. (L-2)(w-3)-30 = area How do I form the equations?

I have solve the problem.

Could this problem be solved using only one variable.

Could this problem be solved in one variable?

I will soon go for a mini laptop. I believe that will enable me to draw a diagram. My mentality made wrote 3(x+y) as the total distance. I was thinking he did all the journey by foot as his car had broken down not knowing he had done some part of the journey with his car before it broke down. Or maybe, he woke up one morning and discovered that his car was not in good condition. He then decided to trek some distance and finally boarded/hired a bus and completed the journey. Which of the thinking is right?

I believe the extra distance walked is 2y. Why must we subtract 2y from x?

Ok, I know now, extra walking is 2y. Should I say that x could have been 2y the distance driven? Therefore x-2y = 0. That means there is no distance driven. What do I next?

So how do I get or know the extra walking? Ok, I know now, extra walking is 2y. Should I say that x could have been 2y the distance driven? Therefore x-2y = 0. That means there is no distance driven.

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 3(x+y)km = total time = 7/10 h I think I have problem making an equation for the second case. What do I do?

Thanks for helping me verify what I want to verify. Am now satisfied.

The path can be splited into 2 pairs of path. So what are the dimension of the two pairs of path? I want sum up all the areas of the pairs of path to get 36m^2.

What I mean is what is the lenght and width of the path such that the area of path = 36m^2 If x is the width of the path. What is the lenght of the path?