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Dividing fractions word problems

5:23 AM

Muhammad Yusuf , Posted in Miscellaneous , 0 Comments

Example #1:

An Italian sausage is 8 inches long. How many pieces of sausage can be cut from the 8-inch piece of sausage if each piece is to be two-thirds of an inch ?

Solution

Since you are trying to find out how many two-thirds there are in 8, it is a division of fractions problem.

So, divide 8 by 2/3

8 ÷ 2/3 = 8/1 ÷ 2/3 = 8/1 × 3/2 = 24/2 = 12

Therefore, you can make 12 pieces having a length of 2/3 inches

Example #2

How many halves are there in six-fourth?

Again, since you are trying to find out how many halves there are in six-fourth, it is a division of fractions problem.

So, divide 6/4 by 1/2

6/4 ÷ 1/2 = 6/4 × 2/1 = 12/4 = 3

Therefore, there are 3 halves in six-fourth

Example #3:
An airplane covers 50 miles in 1/5 hours. How many miles can the airplane cover in 5 hours?

This problem is a combination of division and multiplication of fractions

First, find out how many fifths (1/5) are there in 5. This is a division of fractions problem

So, divide 5 by 1/5

5 ÷ 1/5 = 5/1 ÷ 1/5 = 5/1 × 5/1 = 25/1 = 25

Then, multiply 50 by 25 to get 1250

In 5 hours, the airplane will cover 1250 miles

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Junior MathBattle

9:17 PM

Muhammad Yusuf , Posted in Miscellaneous , 0 Comments

1. Could the altitude, bisector and median drawn from different vertices of a triangle create an equilateral triangle at intersection?

2. There are 300 trees in a park. If one marks any 201 trees, then there are always be an oak, a pine and a maple tree among the marked. How many oaks, pines and maple tree could be in the park (describe all possibilities)?

3. Basil places black, red and white rooks on a chessboard (the same number of each color) so that the rooks of different color do not attack each other. Find the maximal number of rooks that can be placed on the chessboard.

4. Each employee of the firm “Horns and Hoofs” is either a Lair or Truth Teller. They all are well acquainted with each other. Once, while they were sitting at a round table, every employee was asked two questions. On the first question ”‘How many Truth Tellers among your neighbors,” only one person answered ”one”, all the others said ”zero”. On the second question if there are Lairs among his/her neighbors all answered ”yes”. Given this information, is it possible to distinguish the Liars from Truth Tellers?

5. Alex thinks of a number (from 1 to 9), and Boris tries to guess it. If he is mistaken, Alex changes his number: he tries to divide his number by Boris’ number; however, if it does not divide evenly, he doubles it. Is there some strategy for Boris to guess Alex’ number?

6. 300 oranges are distributed into 10 boxes, so that no two boxes contain the same number of fruits. A packer picks up 9 oranges from the box with the largest number of fruits and distributes them into the other boxes, one orange to each box. The packer wants to get the same number of fruits at least in two boxes. Is there a guarantee that he can ever complete the task?

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MathBattle

9:14 PM

Muhammad Yusuf , Posted in Miscellaneous , 0 Comments

1. Each of 2004 boxes contains one pebble. Some of pebbles are white and the total number of them is even. It is allowed to chose any two boxes and ask questions whether there is a white ball (at least one) among of chosen two. Find the minimal number of questions that allow to determine two boxes which contain white pebbles.

Solution. Answer: 4005.

Let us numerate boxes (and pebbles in them) by the numbers from 1 to 2004. Let pair (i, j) denote a question. Show, that 4005 questions is enough. Really, consider the following sequence of the questions: (1, 2), (1, 3), . . . , (1, 2004), (2, 3), (2, 4), . . . , (2, 2004). If all the answers are positive then first and second pebbles are white. (If first pebble is, say, black, then pebbles from 2 to 2004 are white and the total number of white pebbles is odd. Contradiction). If one of the answers is negative (say on (1, k) question), then pebble number 1 is black. Therefore, pebble j is white if on questions (1, j) the answer was positive. Let us prove, that 4005 is the minimal number. Assume, that there is a procedure which allows to determine two white pebbles for the lesser number of questions. Let to answer positively on all the questions.

Then after 4004 questions (as maximum) one is able to determine two boxes, say m and n with white pebbles. Note, that one of the questions either (m,i) or (n,i) is unanswered (since the total number of such questions is 4005 ). Let it be (m, k) . Let us place into m and k boxes black pebbles. Then all the answers stay unchanged: however, box m contains black pebble.

2. A monkey has two coconuts. It is fooling around by throwing coconut down from the balconies of M-storey building. The monkey wants to know the lowest floor when coconut is broken. What is the minimal number of attempts needed to establish that fact, if

(a) M = 15.

(b) M is arbitrary.

Solution.

Let us note that if the monkey has one coconut and the building is M = k storey high, then the number of attempts needed equals k. Also, in case of k attempts permitted, the maximal high of the building is k.

Consider the case of two coconuts. Let us fix number k, attempts permitted and find the maximal hight of the building when the answer on experiment is guaranteed. With two coconuts monkey starts from the k-th floor (if coconut is broken, it still has k − 1 attempts to investigate floors from the first to k − 1 to get the answer). If coconut is not broken, then monkey spends it second attempt, going up k − 1 floors (it is on (2k − 1)-th floor now) to continue its experiment. Again, if coconut is broken, then k−2 attempts is enough to answer the question, otherwise the monkey goes up k−2 floors. So, the maximal high of the building in case of k attempts permitted is k +(k −1)+(k −2)+· · ·+1 = k(k −1)/2. So, in order to evaluate the minimal numbers of attempts given the M is the high of the building, we should find the minimal k which satisfies inequality k(k − 1)/2 ≥ M

3. There are N trenches in a row. A soldier is hiding in one of them. A gunman shots at any particular trench, and kills the soldier if he is there. After each shot the soldier (if alive) runs to an adjacent trench. Is there a strategy for the gunman that ensures the kill after the number of shots?

(a) N = 2004.

(b) N = isarbitrary.

Answer: yes.

Solution. (i) Consider the case when N = 2k is even. Strategy: Gunman shots in sequence from number 1, 2, . . . , 2k, 2k − 1, 2k − 2, . . . , 1. Let us assume, that the soldier is hiding in a trench with odd number. Note, that after any odd number of shots the soldier be in even numbered trench, while after even number of shots he be in odd numbered trench. Let us verify, that the above strategy works. Gunman shots at trench numbered 1. Let us assume, that the soldier was not there, otherwise he is killed. After the first shot the soldier would be in an even numbered trench. The gunman shots at trench numbered 2. Assuming, that soldier was not there, after the shot he can occupy any odd numbered trench, except number 1. The gunman shots at trench numbered 3. Assuming, that soldier was not there, after the shot he can be in any even numbered trench, except 2. The gunman shots at trench numbered 4. Assuming that soldier was not there, after the shot he can be in any odd numbered trench, except number 3 and 1 (to run to number 3 he needs to be in trenches 4 or 2 prior to it, which is impossible). And so on. To a moment the gunman shots at 2k trench, he kills the soldier for sure unless our assumption was wrong and the soldier was hiding initially in an even numbered trench. So, after the gunman shot, we learn, that the soldier was not there, so prior to that he could occupy an any even numbered trench, except 2k. Next shot persuades us he is not in (2k − 1) trench, but in another trench odd numbered. The shot at (2k − 2) persuade us he was not in (2k − 3) prior to it. . . And so on.

4. 11 coins identical in appearance are arranged in a row. Among them there are 3 counterfeits, with the same weight, heavier than the real ones, and all three of them are placed next to each other. Using a simple balance find the minimal number of weighings that allow to determine all counterfeit coins. The simple balance shows whether both sides of the balance are at equilibrium or one side is heavier/lighter.

Solution.

Show that it could be done in two weighings.

Let us numerate the coins from 1 to 11. Let us start from weighing

1. (1,2,3) ? (9,10,11).

(a) In case of equality (1, 2, 3) = (9, 10, 11) counterfeits are among (4,5,6,7,8) with 6 be counterfeit for sure. Then 2. 5 ? 7. In case of 5=7 counterfeits are (5,6,7); if 5 > 7 then 7 is a real one and counterfeits are (4,5,6). If 5 < 7 then counterfeits are (6,7,8).
(b) If (1, 2, 3) > (9, 10, 11) then counterfeits are among (1,2,3,4,5) with 3 be counterfeit for sure. Then 2. 2 ? 4. In case of 2=4 counterfeits are (2,3,4); if 2 > 4 then 4 is a real one and counterfeits are (1,2,3). If 2 < 4 then counterfeits are (3,4,5).
(c) (1, 2, 3) < (9, 10, 11) is considered in the similar way.

One weighing in not enough to determine counterfeit coins.

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MathBattle : Problems offered for all Leagues

9:01 PM

Muhammad Yusuf , Posted in Miscellaneous , 0 Comments

1. Three cops are trying to catch a gangster on an infinite square grid. All four move along the horizontal and vertical lines of the grid at the same maximal speed. The gangster is caught if he is ever on the same line as a cop. At the start time the location of the gangster is unknown. Is there a strategy that ensures that the cops catch the gangster eventually?

Solution (i) Let the length of the segments of the grid be 1 and the speed be 1. Let us assume that the three cops, (A, B, C), are gathered at point O, the origin of the coordinate system, and the gangster (G) is at a distance of no more than r (r is integer) from O.

Cops Strategy: A runs to the West and B runs to the East while C patrols along OY between (0,−2r) and (0, 2r) resting for 1 minute at each intersection.

Let us show that after r passes C, the x- coordinate of G is between those of A and B. If G even once went to outside of the y-interval [−2r, 2r], then he made at least r vertical moves and thus the x-coordinates of both A and B exceeded (or reached) the x-coordinate of G. On the other hand, if the y-coordinate of G never exceeded 2r, then during each pass of C he must at least once go vertically because if he stayed on the same horizontal line, he would be caught by C.

After we have ensured that the x-coordinate of G is been between those of A and B, let us change the strategy. Now A runs to the North, B run to the South, while C patrols in an East-West direction between the verticals of A and B, staying at each knot for one minute.

Note that G cannot cross these verticals. Further, after some time the y-coordinate of G is localized between those of A and B. Therefore, G is somewhere inside of the rectangle with A and B be as opposite vertices.

Further strategy of catching G is obvious (if he had not been caught so far). We denote this strategy by Σ(r).

(ii) Assume that G is at a distance of R from the origin (but the cops do not know R). They employ the strategy Σ(r).

If G is not caught, then our initial assumption is wrong. Then cops return to point O. Note that the time t(r) that the cops spent trying to catch G according to Σ(r) (and return to O) depends on r only. During this time G will be at distance R + t(r) from O.

Now, let us apply the strategy Σ(r₁) with r₁ = 2r + t(r); then strategy Σ(r₂) with r₂ =

3r + t(r) + t(r₁) and so on. G will be be caught eventually, since if he is not caught after Σ(r_k) it would mean that at the initial time he was at a distance of R ≥ kr from O.

2. A positive non-straight angle with vertex O and a point P inside of it are given. Construct a segment through P creating a triangle of minimal area.

Solution.

Point P must be the midpoint of the segment in question. To see this, let us consider any other segment through P and the triangle created by it. Its area is equal to the area of the triangle containing the segment in question plus and minus the areas of the two triangles created by both segments. One can prove that the triangle to be added has a larger area than the triangle to be subtracted.

Construction: Through point O’ symmetrical to O with respect to P draw two lines parallel to the sides of the angle. Then P is the centre of the created parallelogram.

3. Each inhabitant of the Three Parties Island is either a Truth- Teller or a Liar and a member of exactly one of the three political parties. One day each person was asked whether he belonged to

(a) the First Party

(b) the Second Party

(c) the Third Party

60%, 50%, and 40% of population answered affirmatively on questions (a), (b), and (c), respectively. Who (Truth-Tellers or Liars) constitutes a majority of the Second Party?

Solution

Each Truthteller answered affirmatively to exactly one question while each Liar did so on two questions. Note that the total number of positive answers is 150=40+50+60% (of the population). Therefore, the Liars constitute exactly 50% of the population. The second question was answered positively by either Truthtellers from The Second Party or by Liars who are not from The Second Party. Let Truthtellers from The Second Party constitute x% of the total population; then Liars who are not from The Second Party constitute (50 − x)% of the population. Also, the number of Liars from The Second Party constitute 50%−(50−x)% = x% of the population. Therefore, The Second Party is equally split betweenLiars and Truthtellers.

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A Sample of Mathematical Pseudoeponyms

3:31 PM

Muhammad Yusuf , Posted in Miscellaneous , 0 Comments

1. The Pythagorean theorem

The Pythagorean theorem states that in a right triangle, the sum of the squares of the legs equals the square of the hypotenuse. It would be difficult to overestimate the importance of this result. It is generally acknowledged that this remarkable theorem was known before the time of Pythagoras of Samos (ca. 582–500 B.C.), the Greek philosopher and mathematician (see, e.g., NCTM [1969], Swetz and Kao [1977], Ang [1978]). Van der Waerden (1983) hypothesized that since the Pythagorean theorem was known in four ancient civilizations—Babylonia, India, Greece, and China—it is probable that a common origin of the whole theory of right triangles exists. Using both written sources and archeological evidence, he constructed an interesting and compelling argument that led him to the following conclusion (van der Waerden 1983, 29): “I am convinced that the excellent neolithic mathematician who discovered the Theorem of Pythagoras had a proof of the theorem.” He also remarked that the best account of mathematical science in the Neolithic Age is to be found in Chinese texts.

2. Euler's polyhedral theorem

One of the most interesting formulas relating to simple polyhedra is F + V – E = 2, where F is the number of faces, V is the number of vertices, and E is the number of edges. The five simple polyhedra are tetrahedron (pyramid), hexahedron (cube), octahedron, dodecahedron, and icosahedron. For the cube, F = 6, V = 8 and E = 12.

Although this formula may have been known to Archimedes (ca. 225 B.C.), René Descartes, the French mathematician and philosopher, was the first to state this concept (ca. 1635). Leonard Euler independently discovered the theorem and announced his finding in Petrograd in 1752. Since Descartes's findings were not generally known until his unpublished mathematical works were made available in 1860, the polyhedral formula became known as Euler’s theorem rather than Descartes’s theorem (Smith 1958, 296).

To continues this text, you can see in [here]

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The magic of Primes

8:41 AM

Muhammad Yusuf , Posted in Miscellaneous , 0 Comments

We all know what a prime number is. It is a number that is divisible only by 1 and itself. But do you know how fascinating these numbers are?
You could call them the building block of all numbers.

Why do I say that?
We all know numbers that are not prime numbers are called composite numbers. But you may not know that all composite numbers can be built from prime numbers in a unique way by multiplication. In fact, this fact is called the “Fundamental Theorem of Mathematics” – that any number can be written as a factor of two or more primes in a unique way (by unique I mean there is only one way to write it as a factor of primes).

For example, take 96. If you factor it in primes, you will get 96 = 2 * 48 = 2 * 2 * 24 = 2 * 2 * 2 * 12 = 2 * 2 * 2 * 2 * 6 = 2 * 2 * 2 *2 * 2 * 3

So 96 can be represented as 2 * 2 * 2 * 2 * 2 * 3. There is no other way to factorize it using primes.
And this is true for all prime numbers. Isn’t it fascinating?

There are many fascinating things about prime numbers and their role as the building block for all numbers.
But let me tell you about one strange fact – there is no largest prime number. It is just not possible to have something called the “largest prime number”. A Greek Mathematician called Euclid proved it more than 2000 years ago.

There is an active search going on for finding larger and larger prime numbers. The largest one found so far is 243,112,609 − 1. What is that? It is 2 multiplied by itself 43 million, one lakh, twelve thousand, six hundred and nine times, and one subtracted from that number. That number is 12 million digits long.
How big is that? Let me tell you – If you could write 80 numbers in a line, and your notebook page had 40 lines, and your notebook had 50 pages, it would take you 80 notebooks to write down this number. Of course, the search for still larger prime numbers continues.

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Fake Problems

4:58 AM

Muhammad Yusuf , Posted in Miscellaneous , 0 Comments

One coin among N identical-looking coins is a fake and is slightly heavier than the others, which all have the same weight. To compare two groups of coins, you are allowed to use a set of scales with two pans which balance exactly when the weight in each pan is the same. Dan has to find the fake coin using at most two such comparisons.

(a) Explain how Dan can find the fake when N = 4.

(b) Explain how Dan can find the fake when N = 6.

(c) Explain how Dan can find the fake when N = 8.

(d) Can Dan find the fake when N = 9?

(e) Can Dan find the fake when N = 10?

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2011 in 4

5:03 AM

Muhammad Yusuf , Posted in Miscellaneous , 0 Comments

Write the number 2011 using only the digit 4 and any of the operations of addition, subtraction, multiplication, division, exponentiation, taking a square root and factorial. You can use any number composed of the digit 4, even if it's decimal, so 44 and 44.44 are both allowed. You're also allowed to use brackets.

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Cats Problems

4:23 AM

Muhammad Yusuf , Posted in Miscellaneous , 0 Comments

My neighbor, George, owns seven cats. The three oldest cats (Liberty, Lucky and Alexander) are 100% black and the only outdoor cats in the group. The sum of their weights equals the sum of the weights of the four indoor cats (Sassy, Moonlight, Bubba and Randy).Alexander weighs 2 oz. more than Lucky, who in turn weighs 6 oz. more than Liberty.

Randy, in turn, weighs 6 oz. more than Moonlight and 9 oz. more than Sassy. In addition Bubba weighs 28 oz. less than twice Randy’s weight.

Lucky and Randy are almost the same weight, but Lucky is the heavier of the two.

If their weights are all integers, what is the least that the cats could weigh?

Bonus: How many pounds do the 7 cats weigh in all?

Here is a variant on the story as given above. It has different answers…

My neighbor, George, owns seven kittens. The three oldest kittens are Liberty, Lucky and Alexander. The sum of their weights equals the sum of the weights of the four younger kittens (Sassy, Moonlight, Bubba and Randy).