Maths Puzzles

Here was the puzzle .  Now let us say the restaurant is a at a distance of x miles from start point and y miles from end point.  From the diagram and Mr. Smith's answers, it is clear that,  x = distance from start to question 1 point + distance from question 1 point to restaurant But distance from start to q1 is 1/2 of distance from q1 to restaurant d1/2+d1 = x 3d1/2 = x d1 = 2/3 x If y is the distance from restaurant to end point, then d2  - distance from restaurant to  question2 point is 2/3rd of y So d2 = 2/3 y But it is given that d1+d2 = 200 2/3 x + 2/3 y = 200 2/3 (x+y) = 200 x+y = 200 * 3/2 = 300 miles Total distance from start to end points is 300 miles.

  At ten o'clock one morning Mr. Smith and his wife left their house in Connecticut to drive to the home of Mrs. Smith's parents in Penn- sylvania. They planned to stop once along the way for lunch at Patri- cia Murphy's Candlelight Restaurant in Westchester. The prospective visit with his in-laws, combined with business worries, put Mr. Smith in a sullen, uncommunicative mood. It was not until eleven o'clock that Mrs. Smith ventured to ask: "How far have we gone, dear?" Mr. Smith glanced at the mileage meter. "Half as far as the distance from here to Patricia Murphy's," he snapped. They arrived at the restaurant at noon, enjoyed a leisurely lunch, then continued on their way. Not until five o'clock, when they were 200 miles from the place where Mrs. Smith had asked her first question, did she ask a second one. "How much farther do we have to go, dear?" "Half as far," he grunted, "as the distance from here to Patricia

The question was  Two missiles speed directly toward each other, one at 9,000 miles per hour and the other at 21,000 miles per hour. They start 1,317 miles apart. Without using pencil and paper, calculate how far apart they are one minute before they collide.  If two objects are moving in opposite direction in speeds x and y , their relative speeds are x+y. (If they are moving in the same direction, their relative speeds are x-y). In this question, they are moving towards each other at a speed of 21000+9000 = 30000 miles/hour = 30000/60 = 500 miles/minute. Which means, just one minute before their collision, they were 500 miles apart.

 Two missiles speed directly toward each other, one at 9,000 miles per hour and the other at 21,000 miles per hour. They start 1,317 miles apart. Without using pencil and paper, calculate how far apart they are one minute before they collide. Find the solution here .

The question was  What is the pair of numbers whose product is equal to the product of its least common multiple (LCM) and greatest common divisor (GCD)? The solution is any pair of positive integers. Product of LCM and GCD is equal to product of the given numbers for any two positive integers.

Solution: Draw a diagonal AB from the same corner.   Area of triangle ABD is 1/2 of area of trisection. i.e. it is 1/6 area of rectangle. Area of triangle DBC is 1/3 area of rectangle. As these two triangles have a common altitude BC, their bases must be in the ratio 1:2 Ratio of AD:DC = 1:2 So point D trisects the line AC. Similar thing is true for  the the other line.  

 From one corner of a square, draw two straight lines such that the lines trisect area of the triangle exactly. In what ratio, do these line cut the two sides of the square?

 Start both the timers - 11 minutes and 7 minutes.  When 7 minutes timer is empty, invert it - turn it over.  After 11 minute timer is empty, turn the 7 minute timer over again. (It will now act as 4 minute timer.) When this 7 minute timer is empty, your 15 minutes have elapsed. 

 With an egg timer of 11 minutes and another timer of 7 minutes, what is the quickest way to boil an egg for 15 minutes?

 What is the pair of numbers whose product is equal to the product of its least common multiple (LCM) and greatest common divisor (GCD)? You can find the solution here . Hint: The problem is much much easier than you think.  The GCD is the largest positive integer that divides each of two or more integers.   The LCM is the smallest positive integer that is divisible by both integers

Let us say the side of equilateral triangle is x.  Then its perimeter is 3x. If the side of hexagon is y, its perimeter is 6y. 3x=6y  ---------since perimeters are same y = 1/2x Area of triangle is 2 2 = (√3/4)x² Area of hexagon= (3√3 y 2 )/2   Replacing y with x/2, we get Area of hexagon =  (3√3 (x/2) 2 )/2 = ( 3√3 x²/4)/2 = 3/2 *  ( √3 x²/4) = 3/2 * 2 = 3 units

An equilateral triangle and a regular hexagon have perimeters of the same length. If the triangle has an area of two square units, what is the area of the hexagon?

You will be able to find at least 11 genuine coins. To do this, start anywhere and number the coins clockwise around the circle. Put coin 1 on the left pan of the beam balance and coin 11 on its right pan. If the pans remain balanced, both coins must be genuine because there are only 10 fake coins, which are all in a row. And because there are only 20 genuine coins, and you’ve already found two, the 19 coins in the clockwise segment from coin 12 to coin 30 cannot all be genuine—the streak of 10 fake ones must be within that range. So all 11 coins from coin 1 to coin 11 are genuine. If the left pan goes down, coin 1 is genuine and coin 11 is fake. The fake coins can only be in the range from coin 2 to coin 20. Therefore, the 11 coins from coin 21 to coin 1 are genuine. And finally, if the right pan sinks, coin 11 is genuine, and coin 1 is fake. The fake coins can only be in the range from coin 22 to coin 10. Therefore, the 11 coins from coin 11 to coin 21 are genuine.

  Thirty €1 coins are arranged in a circle on a table. Ten of these coins are fake and a little lighter than the 20 real coins. You don’t know where in the circle the fake coins are, but you have been told they are all directly next to one another in the circle. Your task is to find as many real coins as possible. To do this, you have at your disposal a beam balance, a device that can be used to compare the masses of two objects or sets of objects. Your beam balance does not have weights, and it can be used only once. Which coins should you weigh, and how many real coins can you definitely find? You can find the solution here .

If Mr. Cog is the king, then the statements of both Mr. Cog and Mrs. Cog will be false. We know that only one in the couple tells lies. Mr. Cog is not the king.  If Mr.Ark is the king, then he is lying and also Mr. Bog is lying. Which means Mrs.Ark and Mrs. Bog both must be telling truth. Since their two statements are contradictory, they are lying. Which means Mr. Ark is not the king.  So Mr. Bog is the king.

There is a place called the Island of Knights and Knaves, where knights always tell the truth and knaves always lie and every inhabitant is either a knight or a knave. On this particular island, each woman is either a constant liar or a constant truth teller. The men are as usual—knights and knaves. In this island, I was introduced to three married couples—the Arks, the Bogs, and the Cogs. One of the three couples is the king and queen of the island. I was reliably informed that in none of the couples are both of them liars. They all made the following statements:  Mr. Ark: I am not the king.   Mrs. Ark: The king was born in Italy.    Mr. Bog: Mr. Ark is not the king. Mrs. Bog: The king was really born in Spain.    Mr. Cog: I am not the king. Mrs. Cog:  Mr. Bog is the king. Who is really the king? You can find the solution here.

 The question was Carla was enjoying her new calculator and discovering all the features and functions. She accidentally multiplied a number by 5 when she should have divided by 5. The incorrect answer displayed was 75. What should have been the correct answer? Well, the puzzle is obviously ancient as most of us nowadays have forgotten the word calculator.  Solution: Let us say the number was x. x*5 = 75 x = 75/5 = 15 But she should have divided the number by 5. So the answer is 15/5 = 3

 This was the question.  Two friends were playing a ring toss game where you throw 10 rings over the tops of cylinders 15 feet away. For each ring that goes over a cylinder, you receive 5 points. For each ring that misses, you lose 3 points. One of the friends scored 26 and the other scored 18. How many rings did each have that were successful tosses? Solution : Let us say first friend had x successful ring tosses and y failed ones. 5x - 3y = 26 -----(i)   x+y = 10 --------(ii) ( because each person can throw only 10 tosses. ) Multiplying the equation (ii) by 3, we get 3x+3y = 30 ----(iii) Adding (i) and  (iii) we get 8x = 56, x =56/8 = 7 y = 10-x = 10-7 =3 First person threw 7 rings over the cylinder and 3 rings which missed the cylinder.  For the second person, the equations are  5x-3y=18 ----(i)   x+y=10   -----(ii) Multiplying (ii) by 3 and adding (i) and (ii) we get,  5x-3y=18 3x+3y=30 8x=48 x=48/8 = 6 y=10-x=10-6=4 Second person threw 6 rings over the cylinder and 4 rings wh

 Here is the solution to the scrambled equation.

Carla was enjoying her new calculator and discovering all the features and functions. She accidentally multiplied a number by 5 when she should have divided by 5. The incorrect answer displayed was 75. What should have been the correct answer? Find the solution here . 

 Two friends were playing a ring toss game where you throw 10 rings over the tops of cylinders 15 feet away. For each ring that goes over a cylinder, you receive 5 points. For each ring that misses, you lose 3 points. One of the friends scored 26 and the other scored 18. How many rings did each have that were successful tosses?   Find the solution here . 

 In the equation given below, the numbers and symbols and operators are scrambled.   Can you re-arrange the numbers and symbols so that the equation on left hand side will have the value given on the right hand side.  You should use all the numbers and all the symbols given.  Find the solution here .

 Puzzle 97 was to solve for A and B AB+AB+AB = 19B Solution: Now if we look at unit's position, B+B+B = 10+B  or it can be B+B+B = 20+B But second part is not possible because if 3B=20+B, 2B=20, which makes B as 10. As A and B are single digit numbers, B can't be 10. So now we have in unit's place.   3B = 10+B 2B = 10 B = 5   And after we carry 1 to ten's place, we get  3A+1=19  3A = 18 A = 6 So A is 6 and B is 5

 Cryptarithmetic is mathematical game is a type of mathematical game consisting of a mathematical equation among unknown numbers, whose digits are represented by letters of the alphabet. Most commonly used example is    SEND +MORE ----------- MONEY (The solution to this puzzle is O = 0, M = 1, Y = 2, E = 5, N = 6, D = 7, R = 8, and S = 9.)   Now let us look at much simpler example.        AB    AB +  AB -------   19B  What are the values of A and B, if B is not zero? Hint: B can take only one value other than zero.    You can find the solution here . 

 The answer is 6. If you look closely at the numbers, you will see that the numbers opposite to each other add up to 21. So the missing number which is diagonally opposite to 15 is 21-15 = 6.

 Find the missing number in the pie given below.  Hint: It all adds up. You will find the solution here .

The question was - draw a triangle of numbers - 3 on each side. Fill the numbers 1 to 6 in them so that, each side adds to the sum of 9.  Here is the solution. My method was trial and error. A lay woman's method.

 The numbers 1 to 6 are placed in the triangle below so that sum of each side is 12. Can you use the same set of numbers, arrange them in a triangle so that each side totals to 9? You will find the solution here .

 We have to solve all the unknown numbers in the diagram. Let us give names to these unknowns. Now from the diagram, we can immediately find z.  35 = z+19 z = 35-19 = 16 If z is 16, then 13+y = 16 y = 3 19 = x+y ,    which means x = 19-3 = 16 Also t+35 = 78 t= 78 - 35 = 43 As t = z+p    43 = 16+p p = 27 As p = q+13 27 = q+13 ------->   q = 14   27 = q+r = 14+r r = 13   s = 27 + p = 27 + 27 = 54   u = s + t = 54 + 43 = 97   v = u + 78 = 97+78 = 175   Here is how to puzzle solved looks like    

Here is a pyramid of numbers where each number is obtained by adding two numbers below it. Fill in the unknown numbers You can find the solution here . 

 The question was In the addition problem below, the digits B and C represent a number different from any of the other numbers shown (that is, not 1, 4, 5, 6, or 9).      B5   +C9 -------  164 What are the only two possible values for B and C? Solution: When we add digits on units place, we get 5+9 = 14 and 1 is carried to tens place.  Ten's place addition will be B+C+1 = 16 B+C = 16-1 = 15 If B and C are unique, the possible solutions are 9,6 or 8,7. As the questions says, 9 and 6 etc. are to be excluded, the possible values for B and C are 8 and 7.  

  In the addition problem below, the digits B and C represent a number different from any of the other numbers shown (that is, not 1, 4, 5, 6, or 9).      B5   +C9 -------  164 What are the only two possible values for B and C?  Find the solution here .