Recreational Mathematics
Word Puzzles
Sam Loyd (1841-1911) has many such puzzles to his name (see Puzzles 1 to 4 ). Lewis Carroll (1832-1898) is another famous recreational mathematician. He's also known as Charles Lutwidge Dodgson. He's the creator of Alice in Wonderland. The following puzzles from 5 to 8 are due to him. Another great creator of puzzles is Henry Ernest Dudeney (1847-1930). He is England's greatest creator of puzzles. He and Sam Lyod are the contenders for the greatest puzzle creator in the world. See his creations in Puzzles 9 to 11.
(1) 3 Texas Drovers met on the highway and proceeded to dockers as follows.
Says Hank to Jim :"I'll give you 6 pigs for a horse, then you'll have twice as many animals as I'll have."
Says Duke to Hank :"I'll give you 14 sheep for a horse, then you'll have 3 times as many animals as I."
Says Jim to Duke :"I'll give you 4 cows for a horse, then you'll have 6 times as many animals as I"
From these facts, can you tell how many animals were on each docker?
(2) If an elastic ball is dropped from the Leaning Tower of Pisa at a height of 179 feet and on each rebound the ball rises 1/10 of its previous height, what distance will it travel before coming to rest?
(3) Mrs Wiggs explained to Lovey Mary that she has a larger square cabbage patch now than she had last year and will therefore raise 211 more cabbages. How many cabbages will Mrs Wiggs raise this year?
(4) A real estate speculator bought a piece of land for $243 which he divided into equal lots, then sold them at $18 per lot, cleaning up the whole transaction before his train arrived. His profit on the deal was exactly equal to what 6 lots originally cost. How many lots were in the piece of land?
(5) There are 5 sacks, of which Numbers 1 and 2 weigh a total of 12lbs; Numbers 2 and 3, 13.5lbs; Numbers 3 and 4, 11.5lbs; Numbers 4 and 5, 8lbs; Numbers 1, 3 and 5, 16lbs. What is the weight of each sack?
(6) An oblong garden, half a yard longer than wide consists entirely of gravel walk, spirally arranged, a yard wide and 3630 yards long. What are the dimensions of the garden?
(7) If 70% have lost an eye, 75% an ear, 80% an arm, 85% a leg, what % at least must have lost all 4?
(8) A captive queen and her son and daughter were shut up in the top of a tower. Outside the window was a pulley with a rope around it, and a basket of fastened to each end of the rope of equal weight. There was a weight. It would be dangerous for any of them to come down if they weighed 15lbs more than the contents in the other basket, and they also must not weigh less. The one basket coming down would naturally of course draw the other up. The queen weighed 194lbs, daughter 105, son 90 and the weight 75. How did they all escape safely?
(9) 'Between 2 and 3 o'clock yesterday,' said Colonel Crackham,'I looked at the clock and mistook the minute hand for the hour hand and consequently the time appeared to be 55 minutes earlier than actually was.' What was the correct time?
(10) 9 travellers each possessing a car meet on the eastern edge of a desert. They wish to explore the interior always going west. Each car can travel 40 miles on the contents of the engine tank, which holds a gallon of fuel and each can carry 9 extra gallons. Unopened cans can be transferred from car to car. What is the greatest distance at which they can enter the desert without making any depots of fuel for the return journey?
(11) One of Gubbins' 2 equal-length candles burns for 4 hrs, while the other 5 hrs. After burning for a while, one candle is exactly 4 times the length of the other. How long were the candles burning and what are their length after burning?
There are other puzzles which are more teasers than puzzles. Next Page.
Answers:
(1) The general method to solve this kind of problem is to form the simultaneous equations and solve them. The 3 equations are:
J + 15 = 2H ; 3D = H + 52 ; 6J = D + 21
Solving gives H = 11, J = 7, D = 21.
(2) This is a geometric series problem. The solution is 179 + 2(1/10)(179) + 2(1/10)2(179) + ... = 179 {1 + 2(1/10)[1 + 1/10 + (1/10)2 + ...]} = 218.777... feet.
(3) This is a problem in Number Theory, which involves solving the Diophantine Equation (Equation with integer solution(s)) x2 + 211 = y2. Suppose y = x + n. Then 2xn + n2 = 211. Since 211 is a prime, and n is a factor of 211, n must be 1. Therefore x = 105. Hence Mrs Wiggs will raise 1062 = 11236 cabbages this year. See Number Theory Puzzles for more problem on Number Theory.
(4) 18n - 243 = 6 (243/n) => n = 18
(5) 5.5, 6.5, 7, 4.5, 3.5. The sum of all the weighing, 61lbs, including sack 3 thrice and each of the others twice. Deducting twice the sum of the first and fourth weighings, i.e. 21lbs for thrice sack 3 - i.e. 7lbs for sack 3. The rest follows.
(6) 60 × 60.5 yards. Let one side be x, then the other is x+0.5. Then x(x+0.5)=3630, hence x=60.
(7) 10%. 30% are not blind, 25% not deaf, 20% not without an arm, 15% not without a leg, hence at least 100 - (30+25+20+15) = 10% have all 4.
(8)
a) weight down, basket up
b) son down, weight up
c) daughter down, son up
d) weight down, basket up
e) queen down, daughter and weight up
f) weight down, basket up
g) son down, weight up
h) daughter down, son up
i) weight down, basket up
j) son down, weight up
k) son gets out, weight falls to the ground
(9) This can only occur when the minute hand is between 1 and 2 o'clock. Let the actual time be 2 x/60, then the mistaken time is 2 x/60 - 55/60 = 1 (5+x)/60. Let the angle the actual hour hand makes with the vertical be A, then A = 60o + x/60 × 30o = 60 + x/2. Now consider the same hour hand but in the mistaken situation, the angle is A = (5+x)/60 × 360o = 30 + 60x. Equating and solving gives x = 5 5/11. Therefore the actual time was 5 5/11 minutes pass 2 o'clock.
(10) The 9 men A to I, all go 40 miles together on the 1 gallon, when A transfers 1 gallon to each of the other 8 and has 1 gallon left to return home. The 8 go another 40 miles, when B transfers 1 gallon to each of the 7 and has 2 gallons to take him home. The 7 go another 40miles, when C transfers 1 gallon to the 6 others and return home on the remaining 3 gallons. This goes on till finally, I goes 40 miles and has 9 gallons to take him home. Thus I has gone 360 miles out and home, the greatest distance.
(11) Solving this equation 1 - 1/5t = 4(1 - 1/4t), gives t = 3.75hrs. The lengths remaining are 1/16 of the length for one and 1/4 for the other.