# سوالات کار

سرفصل: بخش ریاضی / سرفصل: مسائل کلمه / درس 10

### توضیح مختصر

• زمان مطالعه 9 دقیقه
• سطح سخت

### دانلود اپلیکیشن «زوم»

این درس را می‌توانید به بهترین شکل و با امکانات عالی در اپلیکیشن «زوم» بخوانید

## Work Questions

Work Problems. Some word problems concern machines or workers and how fast they can get certain jobs done. The fundamental equation for this is remarkably similar to the D equals RT equation that we saw in the motion problems. The work equation, I’ll write this as A equals RT.

A is the amount of work done, R is the work rate, the rate at which work is done, T is time. Some people remember this as art, any kind of work, it’s a sort of art that you’re doing that helps some people to remember the equation. The amount A can be products manufactured or houses painted or pizzas made etc. The work rate could be the number of products per minute or houses per day or pizzas per hour etc.

So different work rates would be expressed obviously in different amounts, in different amounts of time. As with the D = RT equation, it’s important to be able to solve this equation for all three variables. A = RT is already solved for the amount, if we solve for the rate that would be rate equals A over T or if we divide by R.

Then we’ll get the time equals the amount divided by the rate and just as it’s important to switch between those three forms, with the D equals RT, it’s also important to switch between these three forms for the A equals RT. There are two broad categories of work, work problems. One type involves using proportion, of course the work rate, like any rate, is a ratio, and can be used to set up a proportion.

For background, see the Rates and Ratio lesson, Ratio and Rates lesson, in the Percents and Ratio module. If you haven’t seen that lesson already, I highly recommend taking a look at that lesson first. The other in type involves multiple machines or workers working at different rates and figuring out how much work these get done together or apart something along those lines.

Here’s a practice problem with the first type, pause the video and do this without a calculator. A machine, working at a constant rate, manufactures 36 staplers in 28 minutes. How many staplers does it make in one hour and 45 minutes? So this looks like it could be challenging to do without a calculator and if you did this without a calculator you might have found yourself working with very big numbers.

Let’s talk about how to avoid running into such big numbers. So first of all I’m gonna change that time to a 105 minutes, so everything is in, in minutes now. So we’re gonna set up a ratio, staplers over time, so 36 over 28 that’s the initial ratio that we’re given and this is gonna be S, the number of staplers over a 105 minutes.

Well, first of all it would be a mistake to cross multiply right here, to cross multiply and do 36 times 105 then, 28 times S, and then trying to divide that gigantic number by 28 that would be a nightmare. Please do not make a mistake like that. So do not start by cross-multiplying all those big numbers together. Instead, simplify the fraction on the left that’s a very good place to start 36 and 28 are divisible by 4 cancel the 4 we get down to single digit numbers, 9 over 7.

Much better, but we’d still like to cancel more to make that 105 smaller before we cross multiply. Another huge mistake, we cannot cross-cancel between the 9 and the 105 that is an absolute nightmare, wrong thing to do and many people think that you are allowed to do that. Now if this is all brand new to you I would suggest go back and watch the video Operations on Proportions where we talk about this in great detail.

Instead we can cancel the sevens, a factor of 7 in the two denominators, so we’re cancelling across the denominators and again, if this, if this move’s, unfamiliar to you, go back and watch the Operation with Proportions that’s really an important video to understand, if what I’m saying here about proportions is not familiar. So we cancel the 7 and we get down to 9 over 1 equals S over 15.

Well now it’s ridiculously easy. Now we can cross multiply. We get S equals 9 times 15, so S equals 135, this is 135 staplers. So notice that once we did enough canceling it was very easy to do without a calculator. If you know the legitimate operations with proportions, and you remember to set up the ratios at the matching units on both sides, so stapler over time, stapler over time, then there aren’t many curveballs the test can throw you with these proportion questions, they’re relatively straightforward.

The other type with multiple work rates can be more confusing if you’re not familiar with it. So pause the video and think about this problem and then we’ll talk about it. Okay to detail a car means to clean it thoroughly, inside and out. When Amelia and Brad detail a car together, 1 car takes 3 hours to detail.

When Amelia details the car alone 1 car takes 4 hours. How long does it take Brad working alone to detail one car? If you don’t know the secret to handling this question type these questions can be confusing because most of the simple arithmetic operations one might try don’t lead to the correct answer. The secret to problems involving different machines or workers working at different work rates is this,.

The combined work rate of people or machines working together is the sum of the individual work rates. In other words, we add the rates, we can’t add or subtract the times, we can’t add or subtract the different amounts made in different times. Those are the numbers usually given in the problem, we have to construct the rates and add or subtract them very important.

So we’ll go back to the question and the rates that we’re gonna look at are cars over hours. How many, how many cars per hour? So the rate of A and B together is a rate of 1 over 3, one-third. They do 1 car in 3 hours and Amelia working alone has a rate of 1 over 4 that’s her rate, and so it must be that when we add the two individual rates, they add up to the combined rate.

And so that means that Brad’s rate alone, that must be the combined rate, minus Amelia’s rate alone and so that’s gonna be one-third minus one-quarter. Find a common denominator, subtract, it’s one-twelfth. So Brad works at a rate of one car in 12 hours. So in other words, it takes him 12 hours to detail a single car. So that is how we solve an equation like that.

Here’s another one along these lines. Pump X takes 28 hours to fill a pool. Pump Y takes 21 hours to fill the same pool. How long does it take them to fill the same pool if they are working simultaneously? So pause the video, think about this, and then we’ll talk about this.

Again, what we need to do is figure out the rates and then combine the rates. So here the rates are gonna be in pool per hour. So we’re talking about the same pool, that single pool in how many hours. The rate of X working alone is 1 over 28, 1 pool over 28 hours and Y working alone is 1, 21st, 1 over 1 pool in 21 hours.

We’re gonna get the combined rate by adding those two individual rates, so we’re gonna add them. In order to add them, we’re gonna find a common denominator. Notice that 28 is 4 times 7, 21 is 3 times 7, so we’ll multiply the first one by, by 3 over 3, and multiply the second one by 4 over 4 and that will get us to a common denominator of 84, that is the least common denominator, the least common multiple.

When we add in the numerator, we get 7 over 84 that cancels down to 12. So that is the combined rates, so that means that they’re working together, they do one pool in 12 hours, so it takes them 12 hours together to fill the pool. For work-related word problems, use A equals RT. A is the amount, R is the work rate, T is the time.

For proportion-related work problems, you need matching units on each side, and you need to understand the operations with proportions. For problems with multiple workers or machines, create rates for each one and then add the rates.

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