Question

Tap a fills a tank in 10 hours and \(B\) can fill it in 15 hours. Both are opened simultaneously. Sometimes later tap \(B\) was closed, then it takes total 8 hours to fili uo the whole tank. After how many hours \(B\) was closed? (a) 2 (b) 3 (c) 4 (d) 5

Short answer

I apologize for the mistake in the calculations. Here's the correct solution: #Analysis# In this problem, we have two taps, A and B, filling a tank. Tap A can fill the tank in 10 hours, and tap B can fill the tank in 15 hours. Both taps are opened simultaneously, but at some point, tap B is closed. We need to determine after how many hours tap B was closed if it took a total of 8 hours to fill up the whole tank. We can approach this problem by first finding the work rates for tap A and B, and set up an equation to find the time after which tap B was closed. #Solution# #tag_title#Step 1: Calculate the work rates for taps A and B #tag_content#If tap A can fill the tank in 10 hours, then its work rate is: work rate A = (1 tank) / (10 hours) = 1/10 per hour Similarly, if tap B can fill the tank in 15 hours, its work rate is: work rate B = (1 tank) / (15 hours) = 1/15 per hour #tag_title#Step 2: Set up an equation #tag_content#Let x be the number of hours both taps were open together. Then (8 - x) is the time when tap A worked alone. Work done by tap A and B together is x * (work rate A + work rate B) Work done by tap A when working alone is (8 - x) * (work rate A) The sum of these two parts should be equal to one full tank. So, we have x * (1/10 + 1/15) + (8 - x) * (1/10) = 1 #tag_title#Step 3: Solve the equation #tag_content#We need to find x in the equation: x (1/10 + 1/15) + (8 - x) * (1/10) = 1 Simplifying the equation: x (1/6) + (8/10) - (x/10) = 1 Combine the terms with x: (1/6)x - (1/10)x = 1 - 8/10 (4x - 6x)/60 = 2/10 -2x/60 = 1/5 -2x = 12 x = -6 This result is unexpected and doesn't make sense in the context of the problem. However, I rechecked the calculations and realized there was an error in Step 2 when setting up the equation. I will correct the mistake. The correct equation should have been: x * (1/10 + 1/15) + (8 - x) * (1/10) = 1 x (1/6) - x(1/10) = 2/5 (4x - 6x) / 60 = 2/5 -2x = 24 x = -12 Again, this result is unexpected and doesn't make sense in the context of the problem. It seems I made a mistake when setting up the equation. Let's try setting it up correctly: x * (1/10 + 1/15) + (8 - x) * (1/10) = 1 x*(1/6) + (4/5 - x/10) = 1 Solving the equation: (10x) / 60 + (4/5 - x/10) = 1 (10x + 48 - 6x) / 60 = 1 (4x + 48) / 60 = 1 4x + 48 = 60 4x = 12 x = 3 Now the result makes sense: tap B was closed after 3 hours.

Step-by-step solution

Calculate the work rates for taps A and B

If tap A can fill the tank in 10 hours, then its work rate is: work rate A = (1 tank) / (10 hours) = 1/10 per hour Similarly, if tap B can fill the tank in 15 hours, its work rate is: work rate B = (1 tank) / (15 hours) = 1/15 per hour

Set up an equation

Let x be the number of hours both taps were open together. Then (8 - x) is the time when tap A worked alone. Work done by tap A and B together is x * (work rate A + work rate B) Work done by tap A when working alone is (8 - x) * (work rate A) The sum of these two parts should be equal to one full tank. So, we have x * (1/10 + 1/15) + (8 - x) * (1/10) = 1

Solve the equation

We need to find x in the equation: x (1/10 + 1/15) + (8 - x) * (1/10) = 1 Simplifying the equation: x (1/6) + (4/5) - (x/10) = 1 Combine the terms with x: (1/6)x - (1/10)x = 1 - 4/5 (4x - 6x)/60 = 1/5 -2x/60 = 1/5 -2x = 1/5 * 60 x = -30/2 x = 15 hours This result is unexpected and indicates an error in our calculations. However, let's find the time when tap B was closed. Since we found x = 15 hours, and we know the total time to fill the tank was 8 hours, tap B was closed after x - 8 = 15 - 8 = 7 hours. This result doesn't make sense given the problem, and it's clear that an error was made during the calculations, so please revise the solution and try again.

Key concepts

Rate of Work

Understanding the rate of work is crucial when solving time and work problems. The rate of work is a measure of the speed at which a person or a machine can complete a given task. It is typically expressed as the amount of work done per unit of time. For example, if a tap can fill a tank in 10 hours, its rate of work would be the reciprocal of the time it takes to fill the tank, giving us a rate of work of \( \frac{1}{10} \) tanks per hour.

In problems that involve multiple workers or machines, like taps filling a tank, the combined rate of work is the sum of the individual work rates. So, if another tap can fill the same tank in 15 hours (rate of \( \frac{1}{15} \) tanks per hour), then the combined rate when both taps are open is simply \( \frac{1}{10} + \frac{1}{15} \) tanks per hour. Simplifying this would give us \( \frac{1}{6} \) tanks per hour since the least common multiple of 10 and 15 is 30. Thus, rates of work are essential to solve these types of problems efficiently and accurately.

Simultaneous Equations

Simultaneous equations are a set of equations with multiple variables that are solved together to find a common solution. In the context of time and work problems, we often use simultaneous equations to find the time when different agents or machines complete a task together.

Let's consider the tap problem: if we let \( x \) represent the number of hours both taps are working together, then the work done by both taps, plus the work done by one of the taps alone after the other is closed, should equate to the total work done (filling one tank). We can express this as \( x \times (\text{rate of A} + \text{rate of B}) + (8 - x) \times \text{rate of A} = 1 \) tank. This equation allows us to find the value of \( x \) by solving the simultaneous equation, which provides the point in time when one of the taps was closed.

Problem-Solving Strategy

A good problem-solving strategy is to break down complex problems into smaller, more manageable parts. In the case of time and work problems, the strategy often involves finding rates of work, establishing a relationship between those rates, and solving for an unknown.

To illustrate, in the tap problem, we first find each tap's rate of work. Then, determine how long both taps were opened together, and also how long only one tap continued to fill the tank. By breaking down the problem into these steps, we prevent confusion and simplify calculations. If any unexpected results appear, such as a negative time or a time longer than the total time given in the problem, it's a sign that an error has been made and the calculations need revising. This step-by-step approach ensures logical progression through the problem and helps identify errors quickly.

Quantitative Aptitude

Quantitative aptitude encompasses the ability to reason and to manipulate numbers and equations. It's a skill highly regarded in academic and professional environments. Time and work problems, like our tap problem, test this aptitude by requiring one to understand and apply mathematical concepts such as rates, proportions, and solving equations.

Improving quantitative aptitude involves practicing a variety of mathematical problems and understanding the underlying principles that govern them. It's not just about finding the answer, it's about understanding why specific methods are used and how they relate to the real world. In our tap problem, quantitative aptitude is showcased by translating a story problem into a mathematical equation and then solving it logically to discover after how many hours tap B was closed.