solved problems of maths

Solved Problems

There are many unsolved problems in mathematics. Several famous problems which have recently been solved include:

1. The Pólya conjecture (disproven by Haselgrove 1958, smallest counterexample found by Tanaka 1980).

2. The four-color theorem (Appel and Haken 1977ab and Appel et al. 1977 using a computer-assisted proof).

3. The Bieberbach conjecture (L. de Branges 1985).

4. Tait's flyping conjecture (Menasco and Thistlethwaite in 1991) and the other two of Tait's knot conjectures (by various authors 1987).

5. Fermat's last theorem (Wiles 1995, Taylor and Wiles 1995).

6. The Kepler conjecture (Hales 2002).

7. The Taniyama-Shimura conjecture (Breuil et al. in 1999).

8. The honeycomb conjecture (Hales 1999).

9. The Poincaré conjecture .

10. Catalan's conjecture .

11. The strong perfect graph theorem .

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Math Problem Answers

Math problem answers are solved here step-by-step to keep the explanation clear to the students. In Math-Only-Math you'll find abundant selection of all types of math questions for all the grades with the complete step-by-step solutions.

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Various types of Math Problem Answers are solved here.

1. Mrs. Rodger got a weekly raise of $145. If she gets paid every other week, write an integer describing how the raise will affect her paycheck.

Let the 1st paycheck be x (integer). Mrs. Rodger got a weekly raise of $ 145. So after completing the 1st week she will get $ (x+145). Similarly after completing the 2nd week she will get $ (x + 145) + $ 145. = $ (x + 145 + 145) = $ (x + 290) So in this way end of every week her salary will increase by $ 145.

3. Mr. Jones sold two pipes at $1.20 each. Based on the cost, his profit one was 20% and his loss on the other was 20%. On the sale of the pipes, he: (a) broke even, (b) lost 4 cents, (c) gained 4 cents, (d) lost 10 cents, (e) gained 10 cents Solution:

Selling price of the first pipe = $1.20

Profit = 20%

Let’s try to find the cost price of the first pipe

CP = Selling price - Profit

CP = 1.20 - 20% of CP

CP = 1.20 - 0.20CP

CP + 0.20CP = 1.20

1.20CP = 1.20

CP = \(\frac{1.20}{1.20}\)

Selling price of the Second pipe = $1.20

Let’s try to find the cost price of the second pipe

CP = Selling price + Loss

CP = 1.20 + 20% of CP

CP = 1.20 + 0.20CP

CP - 0.20CP = 1.20

0.80CP = 1.20

CP = \(\frac{1.20}{0.80}\)

Therefore, total cost price of the two pipes = $1.00 + $1.50 = $2.50

 And total selling price of the two pipes = $1.20 + $1.20 = $2.40

Loss = $2.50 – $2.40 = $0.10

Therefore, Mr. Jones loss 10 cents.

Answer:   (d) 

5. A man has $ 10,000 to invest. He invests $ 4000 at 5 % and $ 3500 at 4 %. In order to have a yearly income of $ 500, he must invest the remainder at: (a) 6 % , (b) 6.1 %, (c) 6.2 %, (d) 6.3 %, (e) 6.4 % Solution: Income from $ 4000 at 5 % in one year = $ 4000 of 5 %. = $ 4000 × 5/100. = $ 4000 × 0.05. = $ 200. Income from $ 3500 at 4 % in one year = $ 3500 of 4 %. = $ 3500 × 4/100. = $ 3500 × 0.04. = $ 140. Total income from 4000 at 5 % and 3500 at 4 % = $ 200 + $ 140 = $ 340. Remaining income amount in order to have a yearly income of $ 500 = $ 500 - $ 340. = $ 160. Total invested amount = $ 4000 + $ 3500 = $7500. Remaining invest amount = $ 10000 - $ 7500 = $ 2500. We know that, Interest = Principal × Rate × Time Interest = $ 160, Principal = $ 2500, Rate = r [we need to find the value of r], Time = 1 year. 160 = 2500 × r × 1. 160 = 2500r 160/2500 = 2500r/2500 [divide both sides by 2500] 0.064 = r r = 0.064 Change it to a percent by moving the decimal to the right two places r = 6.4 % Therefore, he invested the remaining amount $ 2500 at 6.4 % in order to get $ 500 income every year. Answer: (e) 6. Jones covered a distance of 50 miles on his first trip. On a later trip he traveled 300 miles while going three times as fast. His new time compared with the old time was: (a) three times as much, (b) twice as much, (c) the same, (d) half as much, (e) a third as much Solution: Let speed of the 1st trip x miles / hr. and speed of the 2nd trip 3x / hr. We know that Speed = Distance/Time. Or, Time = Distance/Speed. So, times taken to covered a distance of 50 miles on his first trip = 50/x hr. And times taken to covered a distance of 300 miles on his later trip = 300/3x hr. = 100/x hr. So we can clearly see that his new time compared with the old time was: twice as much. Answer: (b)

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Unsolved Questions:

1. Fahrenheit temperature F is a linear function of Celsius temperature C. The ordered pair (0, 32) is an ordered pair of this function because 0°C is equivalent to 32°F, the freezing point of water. The ordered pair (100, 212) is also an ordered pair of this function because 100°C is equivalent to 212° F, the boiling point of water.

2. A sports field is 300 feet long. Write a formula that gives the length of x sports fields in feet. Then use this formula to determine the number of sports fields in 720 feet.

3. A recipe calls for 2 1/2 cups and I want to make 1 1/2 recipes. How many cups do I need?

4. Mario answered 30% of the questions correctly. The test contained a total of 80 questions. How many questions did Mario answer correctly?

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5 Simple Math Problems No One Can Solve

Easy to understand, supremely difficult to prove.

solved problems of maths

Mathematics can get pretty complicated. Fortunately, not all math problems need to be inscrutable. Here are five current problems in the field of mathematics that anyone can understand, but nobody has been able to solve.

Collatz Conjecture

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Pick any number. If that number is even, divide it by 2. If it's odd, multiply it by 3 and add 1. Now repeat the process with your new number. If you keep going, you'll eventually end up at 1. Every time.

Moving Sofa Problem

So you're moving into your new apartment, and you're trying to bring your sofa. The problem is, the hallway turns and you have to fit your sofa around a corner. If it's a small sofa, that might not be a problem, but a really big sofa is sure to get stuck. If you're a mathematician, you ask yourself: What's the largest sofa you could possibly fit around the corner? It doesn't have to be a rectangular sofa either, it can be any shape.

This is the essence of the moving sofa problem . Here are the specifics: the whole problem is in two dimensions, the corner is a 90-degree angle, and the width of the corridor is 1. What is the largest two-dimensional area that can fit around the corner?

The largest area that can fit around a corner is called—I kid you not—the sofa constant. Nobody knows for sure how big it is, but we have some pretty big sofas that do work, so we know it has to be at least as big as them. We also have some sofas that don't work, so it has to be smaller than those. All together, we know the sofa constant has to be between 2.2195 and 2.8284.

Perfect Cuboid Problem

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Remember the pythagorean theorem, A 2 + B 2 = C 2 ? The three letters correspond to the three sides of a right triangle. In a Pythagorean triangle, and all three sides are whole numbers. Let's extend this idea to three dimensions. In three dimensions, there are four numbers. In the image above, they are A, B, C, and G. The first three are the dimensions of a box, and G is the diagonal running from one of the top corners to the opposite bottom corner.

Just as there are some triangles where all three sides are whole numbers, there are also some boxes where the three sides and the spatial diagonal (A, B, C, and G) are whole numbers. But there are also three more diagonals on the three surfaces (D, E, and F) and that raises an interesting question: can there be a box where all seven of these lengths are integers?

The goal is to find a box where A 2 + B 2 + C 2 = G 2 , and where all seven numbers are integers. This is called a perfect cuboid. Mathematicians have tried many different possibilities and have yet to find a single one that works. But they also haven't been able to prove that such a box doesn't exist, so the hunt is on for a perfect cuboid.

Inscribed Square Problem

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Draw a closed loop. The loop doesn't have to be a circle, it can be any shape you want, but the beginning and the end have to meet and the loop can't cross itself. It should be possible to draw a square inside the loop so that all four corners of the square are touching the loop. According to the inscribed square hypothesis, every closed loop (specifically every plane simple closed curve) should have an inscribed square, a square where all four corners lie somewhere on the loop.

This has already been solved for a number of other shapes, such as triangles and rectangles. But squares are tricky, and so far a formal proof has eluded mathematicians.

Happy Ending Problem

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The happy ending problem is so named because it led to the marriage of two mathematicians who worked on it, George Szekeres and Esther Klein. Essentially, the problem works like this:

Make five dots at random places on a piece of paper. Assuming the dots aren't deliberately arranged—say, in a line—you should always be able to connect four of them to create a convex quadrilateral, which is a shape with four sides where all of the corners are less than 180 degrees. The gist of this theorem is that you'll always be able to create a convex quadrilateral with five random dots, regardless of where those dots are positioned.

So that's how it works for four sides. But for a pentagon, a five-sided shape, it turns out you need nine dots. For a hexagon, it's 17 dots. But beyond that, we don't know. It's a mystery how many dots is required to create a heptagon or any larger shapes. More importantly, there should be a formula to tell us how many dots are required for any shape. Mathematicians suspect the equation is M=1+2 N-2 , where M is the number of dots and N is the number of sides in the shape. But as yet, they've only been able to prove that the answer is at least as big as the answer you get that way.

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Solving Problems With Math

Learning outcomes.

In this section we will bring together the mathematical tools we’ve reviewed, and use them to approach more complex problems. In many problems, it is tempting to take the given information, plug it into whatever formulas you have handy, and hope that the result is what you were supposed to find. Chances are, this approach has served you well in other math classes.

This approach does not work well with real life problems, however. Read on to learn how to use a generalized problem solving approach to solve a wide variety of quantitative problems, including how taxes are calculated.

Problem Solving and Estimating

Problem solving is best approached by first starting at the end: identifying exactly what you are looking for. From there, you then work backwards, asking “what information and procedures will I need to find this?” Very few interesting questions can be answered in one mathematical step; often times you will need to chain together a solution pathway , a series of steps that will allow you to answer the question.

Problem Solving Process

In most problems we work, we will be approximating a solution, because we will not have perfect information. We will begin with a few examples where we will be able to approximate the solution using basic knowledge from our lives.

In the first example, we will need to think about time scales, we are asked to find how many times a heart beats in a year, but usually we measure heart rate in beats per minute.

How many times does your heart beat in a year?

This question is asking for the rate of heart beats per year. Since a year is a long time to measure heart beats for, if we knew the rate of heart beats per minute, we could scale that quantity up to a year. So the information we need to answer this question is heart beats per minute. This is something you can easily measure by counting your pulse while watching a clock for a minute.

Suppose you count 80 beats in a minute. To convert this to beats per year:

\displaystyle\frac{80\text{ beats}}{1\text{ minute}}\cdot\frac{60\text{ minutes}}{1\text{ hour}}\cdot\frac{24\text{ hours}}{1\text{ day}}\cdot\frac{365\text{ days}}{1\text{ year}}=42,048,000\text{ beats per year}

The technique that helped us solve the last problem was to get the number of heartbeats in a minute translated into the number of heartbeats in a year. Converting units from one to another, like minutes to years is a common tool for solving problems.

In the next example, we show how to infer the thickness of something too small to measure with every-day tools. Before precision instruments were widely available, scientists and engineers had to get creative with ways to measure either very small or very large things. Imagine how early astronomers inferred the distance to stars, or the circumference of the earth.

How thick is a single sheet of paper? How much does it weigh?

Solution: While you might have a sheet of paper handy, trying to measure it would be tricky. Instead we might imagine a stack of paper, and then scale the thickness and weight to a single sheet. If you’ve ever bought paper for a printer or copier, you probably bought a ream, which contains 500 sheets. We could estimate that a ream of paper is about 2 inches thick and weighs about 5 pounds. Scaling these down,

\displaystyle\frac{2\text{ inches}}{\text{ream}}\cdot\frac{1\text{ ream}}{500\text{ pages}}=0.004\text{ inches per sheet}

The first two example questions in this set are examined in more detail here.

We can infer a measurement by using scaling.  If 500 sheets of paper is two inches thick, then we could use proportional reasoning to infer the thickness of one sheet of paper.

In the next example, we use proportional reasoning to determine how many calories are in a mini muffin when you are given the amount of calories for a regular sized muffin.

A recipe for zucchini muffins states that it yields 12 muffins, with 250 calories per muffin. You instead decide to make mini-muffins, and the recipe yields 20 muffins. If you eat 4, how many calories will you consume?

There are several possible solution pathways to answer this question. We will explore one.

To answer the question of how many calories 4 mini-muffins will contain, we would want to know the number of calories in each mini-muffin. To find the calories in each mini-muffin, we could first find the total calories for the entire recipe, then divide it by the number of mini-muffins produced. To find the total calories for the recipe, we could multiply the calories per standard muffin by the number per muffin. Notice that this produces a multi-step solution pathway. It is often easier to solve a problem in small steps, rather than trying to find a way to jump directly from the given information to the solution.

We can now execute our plan:

\displaystyle{12}\text{ muffins}\cdot\frac{250\text{ calories}}{\text{muffin}}=3000\text{ calories for the whole recipe}

View the following video for more about the zucchini muffin problem.

We have found that ratios are very helpful when we know some information but it is not in the right units, or parts to answer our question. Making comparisons mathematically often involves using ratios and proportions. For the last

You need to replace the boards on your deck. About how much will the materials cost?

There are two approaches we could take to this problem: 1) estimate the number of boards we will need and find the cost per board, or 2) estimate the area of the deck and find the approximate cost per square foot for deck boards. We will take the latter approach.

For this solution pathway, we will be able to answer the question if we know the cost per square foot for decking boards and the square footage of the deck. To find the cost per square foot for decking boards, we could compute the area of a single board, and divide it into the cost for that board. We can compute the square footage of the deck using geometric formulas. So first we need information: the dimensions of the deck, and the cost and dimensions of a single deck board.

Suppose that measuring the deck, it is rectangular, measuring 16 ft by 24 ft, for a total area of 384 ft 2 .

From a visit to the local home store, you find that an 8 foot by 4 inch cedar deck board costs about $7.50. The area of this board, doing the necessary conversion from inches to feet, is:

\displaystyle{8}\text{ feet}\cdot4\text{ inches}\cdot\frac{1\text{ foot}}{12\text{ inches}}=2.667\text{ft}^2{.}

This will allow us to estimate the material cost for the whole 384 ft 2 deck

\displaystyle\$384\text{ft}^2\cdot\frac{\$2.8125}{\text{ft}^2}=\$1080\text{ total cost.}

Of course, this cost estimate assumes that there is no waste, which is rarely the case. It is common to add at least 10% to the cost estimate to account for waste.

This example is worked through in the following video.

Is it worth buying a Hyundai Sonata hybrid instead the regular Hyundai Sonata?

To make this decision, we must first decide what our basis for comparison will be. For the purposes of this example, we’ll focus on fuel and purchase costs, but environmental impacts and maintenance costs are other factors a buyer might consider.

It might be interesting to compare the cost of gas to run both cars for a year. To determine this, we will need to know the miles per gallon both cars get, as well as the number of miles we expect to drive in a year. From that information, we can find the number of gallons required from a year. Using the price of gas per gallon, we can find the running cost.

From Hyundai’s website, the 2013 Sonata will get 24 miles per gallon (mpg) in the city, and 35 mpg on the highway. The hybrid will get 35 mpg in the city, and 40 mpg on the highway.

An average driver drives about 12,000 miles a year. Suppose that you expect to drive about 75% of that in the city, so 9,000 city miles a year, and 3,000 highway miles a year.

We can then find the number of gallons each car would require for the year.

\displaystyle{9000}\text{ city miles}\cdot\frac{1\text{ gallon}}{24\text{ city miles}}+3000\text{ highway miles}\cdot\frac{1\text{ gallon}}{35\text{ highway miles}}=460.7\text{ gallons}

If gas in your area averages about $3.50 per gallon, we can use that to find the running cost:

\displaystyle{460.7}\text{ gallons}\cdot\frac{\$3.50}{\text{gallon}}=\$1612.45

While both the absolute and relative comparisons are useful here, they still make it hard to answer the original question, since “is it worth it” implies there is some tradeoff for the gas savings. Indeed, the hybrid Sonata costs about $25,850, compared to the base model for the regular Sonata, at $20,895.

To better answer the “is it worth it” question, we might explore how long it will take the gas savings to make up for the additional initial cost. The hybrid costs $4965 more. With gas savings of $451.10 a year, it will take about 11 years for the gas savings to make up for the higher initial costs.

We can conclude that if you expect to own the car 11 years, the hybrid is indeed worth it. If you plan to own the car for less than 11 years, it may still be worth it, since the resale value of the hybrid may be higher, or for other non-monetary reasons. This is a case where math can help guide your decision, but it can’t make it for you.

This question pulls together all the skills discussed previously on this page, as the video demonstration illustrates.

#1. Click here to try this problem.

#2. Click here to try this problem.

#3. Click here to try this problem.

Governments collect taxes to pay for the services they provide. In the United States, federal income taxes help fund the military, the environmental protection agency, and thousands of other programs. Property taxes help fund schools. Gasoline taxes help pay for road improvements. While very few people enjoy paying taxes, they are necessary to pay for the services we all depend upon.

Taxes can be computed in a variety of ways, but are typically computed as a percentage of a sale, of one’s income, or of one’s assets.

Example: Sales Tax

The sales tax rate in a city is 9.3%. How much sales tax will you pay on a $140 purchase? Solution:

The sales tax will be 9.3% of $140. To compute this, we multiply $140 by the percent written as a decimal: $140(0.093) = $13.02.

Click here to try another version of this problem:

When taxes are not given as a fixed percentage rate, sometimes it is necessary to calculate the effective tax rate :   the equivalent percent rate of the tax paid out of the dollar amount the tax is based on.

Example: Property Tax

Jaquim paid $3,200 in property taxes on his house valued at $215,000 last year. What is the effective tax rate? Solution:

We can compute the equivalent percentage: 3200/215000 = 0.01488, or about 1.49% effective rate.

Taxes are often referred to as progressive, regressive, or flat.

Example: Federal Income Tax

The United States federal income tax on earned wages is an example of a progressive tax. People with a higher wage income pay a higher percent tax on their income.

For a single person in 2011, adjusted gross income (income after deductions) under $8,500 was taxed at 10%. Income over $8,500 but under $34,500 was taxed at 15%.

Earning $10,000

Stephen earned $10,000 in 2011. He would pay 10% on the portion of his income under $8,500, and 15% on the income over $8,500.

8500(0.10) = 850     10% of $8500 1500(0.15) = 225      15% of the remaining $1500 of income Total tax:   = $1075

What was Stephen’s effective tax rate?

The effective tax rate paid is 1075/10000 = 10.75%

Example: Gasoline Tax

A gasoline tax is a flat tax when considered in terms of consumption. A tax of, say, $0.30 per gallon is proportional to the amount of gasoline purchased. Someone buying 10 gallons of gas at $4 a gallon would pay $3 in tax, which is $3/$40 = 7.5%. Someone buying 30 gallons of gas at $4 a gallon would pay $9 in tax, which is $9/$120 = 7.5%, the same effective rate.

However, in terms of income, a gasoline tax is often considered a regressive tax. It is likely that someone earning $30,000 a year and someone earning $60,000 a year will drive about the same amount. If both pay $60 in gasoline taxes over a year, the person earning $30,000 has paid 0.2% of their income, while the person earning $60,000 has paid 0.1% of their income in gas taxes.

A sales tax is a fixed percentage tax on a person’s purchases. Is this a flat, progressive, or regressive tax?

While sales tax is a flat percentage rate, it is often considered a regressive tax for the same reasons as the gasoline tax.

Click here to try other tax problems .


This chapter contains material taken from Math in Society (on OpenTextBookStore ) by David Lippman, and is used under a CC Attribution-Share Alike 3.0 United States  (CC BY-SA 3.0 US) license.

This chapter contains material taken from of Math for the Liberal Arts (on Lumen Learning ) by Lumen Learning, and is used under a CC BY: Attribution license.

Solving Problems With Math by Gail Poitrast is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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Famous Math Problems

Throughout history, there have been many famous math problems posed that could not be solved at the time. Some conjectures lasted for hundreds of years before being proven or disproven, and some remain unsolved. Wolfram|Alpha has knowledge of many of these famous math problems, including Hilbert's 23 problems and the Millennium Prize problems.

Open Conjectures

Find information about an unsolved math problem.

Get information about a mathematical conjecture:

Learn about one of the smale problems:, ask a question about a conjecture:, solved problems.

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Learn about a mathematical paradox:, get historical information about a theorem:, related examples.


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