101 Word Problems/Solutions
Problem #1: Antifreeze
A quart bottle contains a mixture that is 1/3 antifreeze, 2/3 water. A half gallon bottle contains a mixture that is 3/4 antifreeze, 1/4 water. The contents of the two bottles are poured into a gallon jug. What is the ratio (one integer to another, in the form a:b) of antifreeze to water in the jug?
Solution #1: Antifreeze
11:7
Problem #2: Consecutive Birthdays
A woman and her grandson have the same birthday. For six consecutive birthdays, she is an integral multiple of his age. How old is the grandmother at the sixth of these birthdays?
Solution #2: Consecutive Birthdays
66
Problem #3: Cube Numbers
Find all sets of four positive consecutive integers such that the sum of the cubes of the first three is the cube of the fourth.
Solution #3: Cube Numbers
3,4,5,6
Problem #4: Don't Spend It All At Once
If I start with $10 and spend all but $3, how much do I have left?
Solution #4: Don't Spend It All At Once
$3. If I spend all but $3, I must have $3 left.
Problem #5: Hands of a Clock
How many times do the two hands of a clock point in the same direction between 6:00 am and 6:00 pm of a single day?
Solution #5: Hands of a Clock
11. It is tempting to give the "obvious" answer 12, but the true answer is 11 as will be found by counting the individual occasions.

Between 6:30 am and 6:35 am.

Between 7:35 am and 7:40 am.

Between 8:40 am and 8:45 am.

Between 9:45 am and 9:50 am.

Between 10:50 am and 10:55 am.

At Noon.

Between 1:05 am and 1:10 am.

Between 2:10 am and 2:15 am.

Between 3:15 am and 3:20 am.

Between 4:20 am and 4:25 am.

Between 5:25 am and 5:30 am.
Problem #6: How Old Is My Daughter?
My daughter is twice as old as my son and half as old as I am. In twentytwo years my son will be half my age. How old is my daughter?
Solution #6: How Old Is My Daughter?
Let us assume my daughter is age x. We are told my daughter is twice as old as my son, so that my son must be age x/2. We are also told that I am twice as old as my daughter so my age is 2x. In 22 years time my son will be (x/2 + 22) and I will be (2x + 22). Since he will be half of my age at that time,
x/2 + 22 = 1/2 (2x+22)
Multiplying both sides by 2
x + 44 = 2x + 22
or
x = 22
My daughter is 22 years old.
Problem #7: The Striking Clock
I woke up one night and heard my clock strike "one." I was too tired to turn on the light to see what time it was. As I lay there pondering, it occurred to me to speculate how long I would have to lie awake in order to be sure what was the exact time. My clock strikes the hours and strikes "one" each half hour. I fell asleep before I solved the problem, but can you work out what is the longest time I would have to lie awake after hearing the strike "one," to be sure of the time?
Solution #7: The Striking Clock
The strike I heard was either one o'clock or some half hour. If, after another half hour I hear two more strikes, I know what hour it is, but if I again hear one strike, I may have heard successively 12:30 am and 1 o'clock or 1 o'clock and 1:30 am. After another half hour I either hear 1 strike (1:30 am) or two strikes (2 o'clock). I may, therefore, have to lie awake a whole hour before I can be sure of the exact time.
Problem #8: Relations
1. Is it legal for a man to marry his widow's sister?
2. Brothers and sisters I have none, but this man's father is my father's son. Who is "this man?"
Solution #8: Relations
1. Only dead men have widows. So... no, a dead man cannot legally marry anyone.
2. Since I have no brothers or sisters, "my father's son" must be referring to me; I am "this man's father." So, "this man" must be my son.
Problem #9: School Notes
If 6 boys fill 6 notebooks in 6 weeks and 4 girls fill 4 notebooks in 4 weeks, how many notebooks will a class of 12 boys and 12 girls fill in 12 weeks?
Solution #9: School Notes
If 6 boys fill 6 notebooks in 6 weeks,
then 12 boys fill 12 notebooks in 6 weeks,
and 12 boys fill 24 notebooks in 12 weeks;
If 4 girls fill 4 notebooks in 4 weeks,
then 12 girls fill 12 notebooks in 4 weeks,
and 12 girls fill 36 notebooks in 12 weeks;
Hence 12 girls and 12 boys fill 24+36=60 notebooks in 12 weeks.
Problem #10: Simple Algebra
Given x = 1 and y = 1, we have
x = y
Multiplying each side by x
x^{2} = xy
Subtracting y^{2} from each side
x^{2}  y^{2} = xy  y^{2}
Factoring each side
(x + y)(x  y) = y (x  y)
Dividing out the common term ((x  y) we have
x + y = y
Substituting the given values
1 + 1 = 1
Or
2 = 1
What is wrong with this proof?
Solution #10: Simple Algebra
There was nothing wrong up through line
(x + y)(x  y) = y (x  y)
If we substitute the values of x and y we have
(2)(0) = (1)(0)
or
0 = 0
However, when we divide both sides by (x  y), we break a fundamental rule of mathematics that we cannot divide out by a fraction which is equal to zero.
Problem #11: A Dinner Party
We were given a formal dinner party for ten (including ourselves) which is a number I always like because the host can sit at one end of the table and the hostess at the other, and still maintain the correct alternate male and female around the table. My wife was trying to work out the seating. "Tom and Jean have not been here to dinner before so they are the guests of honor. Tom must sit on my right and Jean on your right, but I don't know how I want to seat the others.: "Well," I said, "I would like Janet on my left. I have a soft spot for her." "You can have her," replied my wife, "but I will not have her husband Jack next to me; I think he should be next to Mary Ann."
Since we do not place husbands and wives next to each other, this determined the seating of everyone, including Howard's wife Lois, and Mary Ann's husband Bill. Can you work out the seating arrangement?
Solution #11: A Dinner Party
Numbering the ten places around the table 1 to 10 as shown in the figure below, if I take the head of the table (1), my wife will sit at (6), Jean will be at (10), Tom at (5), and Janet at (2).
Now my wife will not have Jack next to her so he must be at (9), because (3) would place him next to his wife. Since Mary Ann is next to Jack, she must be at (8); then Howard must be at (7), Bill at (3), and Lois at (4).
Problem #12: Dominoes on Board
A board, 9 inches by 7 inches, is marked out into 63 oneinch squares. To make the number of squares even, the middle square is blocked out by placing a chessman on it. John attempts to cover the board, other than the middle square, with dominoes which are 2 inches by 1 inch.
In the figure above he has 11 dominoes in place. Since 31 dominoes will be needed in all, more than one set will be required. Can you completely cover the board, other than the middle square, without having any domino stick out over the edge of the board, or can you prove it cannot be done?
Solution #12: Dominoes on Board
The use of a chessman to block out the middle square gives the clue to the solution of this problem. If the board is marked out as a chess board with alternate black and white squares, with the middle square black, we find that there are 32 white squares and 30 black squares (not counting the middle square). Now a domino however placed will cover one white square and one black square and hence having placed 30 dominoes we shall always be left with 2 white squares which cannot possibly be covered by a single domino. Hence, there is no possible solution to the problem.
Problem #13: An Unusual Series
Can you find the fifth term in the following series?
77, 49, 36, 18, ...
Solution #13: An Unusual Series
Each term consists of the first digit of the preceding term multiplied by the second digit of the same term. Thus,
49 = 7 x 7
36 = 4 x 9
18 = 3 x 6
Hence, the fifth term is 1 x 8 = 8.
Problem #14: Is the Rope Knotted?
A loop of rope is lying on the ground in the position show in the figure below: You are too far away to see which section of the rope is above or below at each of the crossovers at A, B and C.
If we assume that it is equally likely that either section is on top at each crossover, what is the probability that the rope is knotted?
Solution #14: Is the Rope Knotted?
We can assume without loss of completeness that at A the rope going from top left to bottom right is on top. (If it is the other way a mirror solution is produced which does not alter the probability.)
Then we have four equal possibilities. The segment BOC can be either on top or underneath at each of the points B and C.
Segment BOC
At B At C The rope is
On top On top Not knotted
On Top Under Not knotted
Under On top Knotted
Under Under Not knotted
Hence the chance that the rope is knotted is 1/4.
Problem #15: The Two Rugs
A woman has two rugs of the same material, one 10 feet by 10 feet and one a long runner 8 feet by 1 foot. How can she make a single continuous cut in the 10 foot by 10 foot rug, so as to form two pieces which, when combined with the 8 foot by 1 foot rug can be stitched into a rug 9 feet by 12 feet? The two rugs are of plain pattern and have a pile on one side only.
Solution #15: The Two Rugs
The 10 foot by 10 foot rug must be cut as shown below, each step being 2 feet long and 1 foot broad.
This gives two oddshaped pieces together, a gap will be left in the middle which can be exactly filled by the 8 foot by 1 foot strip, which is shaded below.
Problem #16: Mexican Table Mats
My son gave me a set of table mats he bought in Mexico. They were rectangular and were made of straw circles joined together in the form shown in the figure below:
All the circles on the edges of each mat were white and the inner circles red. I noted that there were 20 white circles and 15 red circles.
I wondered if it were possible to make a mat in this form where the number of white circles equaled the number of red circles. I found there were two possible solutions. Can you find them and show that there are only two such solutions?
Solution #16: Mexican Table Mats
Let the number of circles along the top edge of the mat be x and the number of circles along the side of the mat be y. Then the total number of circles in the mat is
xy
and the number of circles around the edge is
2x + 2y  4
Since this must be onehalf the total number of circles we have
xy = 4x + 4y  8
xy  4x  4y + 16 = 16  8
and (x  4) (y  4) = 8
Since x and y must be integers, so must (x  4) and (y  4) and these must be factors of 8. The only integer factors of 8 are (8 and 1) and (4 and 2), which give {x = 12, y = 5} and {x = 8, y = 6}.
Hence the mat must be 12 by 5 circles or 8 by 6 circles.
Problem #17: The Curious Sequence
What is the next letter in the following series?
O T T F F S S E
Solution #17: The Curious Sequence
This is on of those nasty catchy puzzles. The real solution is
O T T F F S S E N
being the initial letters of the counting numbers: One, Two, Three, Four, Five, Six, etc.
Problem #18: Colored Labels
Three intelligent women, Alice, Barb and Carol, sit down to try out a test in logical reasoning. They are so arranged that each can see the color of a label which is either red or blue, attached to the hats worn by the other two but no one of them can see the color of the label attached to her own hat. They are told that at least one of the labels is red. If any one of them can logically deduce the color of the label on her hat, she is to declare it. Carol decides to play this game with her eyes closed, knowing that the other two women have their eyes open. After a little time Carol, who has not seen the label on any of the hats declares her label is red. How can she deduce this?
Solution #18: Colored Labels
Carol argues as follows:

If Alice sees two blue labels she will know that her label is red and will declare this and since a little time has passed without this happening Betty and Carol cannot both have blue labels.

Similarly, if Betty sees two blue labels he will know that his label is red and will declare this, and hence, Alice and Carol cannot both have blue labels.

Carol can now deduce that if her label is blue, both Alice and Betty must have red labels. Now Carol knows that both Alice and Betty are intelligent. If Betty and seen a red label on Alice's hat and a blue label on Carol's hat, she would know her hat did not have a blue label, because in this case Alice would have declared her label to be red.

Since a little time has passed and neither Alice nor Betty has declared she could deduce the color of her label, Carol knows her label cannot be blue, and therefore must be red.
Problem #19: War in the Middle Ages
During a grim medieval battle, 85% of the warriors lost an ear, 80% lost an eye, 75% lost an arm, and 70% lost a leg. What is the smallest percentages possible of combatants who lost all of the above?
Solution #19: War in the Middle Ages
Since 85% of the warriors lost an ear and 80% an eye (a total of 165%), at least 65% of them must have lost both an ear and an eye.
Since at least 65% of the warriors lost both an ear an eye, and 75% lost an arm (a total of 140%), at least 40% of them must have lost an ear, an eye, and arm and a leg each.
Since at least 40% of the warriors lost an ear, an eye, and an arm and 70% lost a leg (a total of 110%), at least 10% of them must have lost an ear, an eye, and arm and a leg each.
(This problem was set by Lewis Carroll.)
Problem #20: How Many Are We?
How many are we?
You tell us the answer, knowing that the probability that at least two of us have birthdays on the same day is less than half, but that this would not be the case were we one more in number.
Solution #20: How Many Are We?
Probability that two people have different birthdays: 364/365
Probability that three people have different birthdays: (364/365) x (363/365), etc.
For n people
P = (364/365) x (363/365) x ... x ([653n+1]/365)
It can be checked that the product of these fractions becomes less than 1/2 when n goes from 22 to 23. We are therefore 22.
Problem #21: Jaws
A seaside resort along the Pacific Ocean is equipped with an electronic shark detection system which sounds an alarm on the average of 1 day out of 30. There are 10 times more false alarms than there are undetected sharks. We also know that only three out of four sharks are detected. Given the above information, what is the percentage of "peaceful" resort days if that term is defined as days unmarred by alarms or sharks?
Solution #21: Jaws
Let x be the percentage of a shark going undetected. We know that the probability of an alarm is 1/30. Hence the probability of having neither an undetected shark nor an alarm is
1  (1/30 + x) = 29/30  x
We have also been told that the probability of false alarm is 10x. Hence the probability of real alarm is 1/30 = 10x. But this is equal to 3x since 3 times as many sharks are detected as go undetected
1/30  10x = 3x
Hence x = 1/390
Thus the probability of a peaceful day is
29/30  1/390 = 0.964
Problem #22: Meteorology
It rains here one day out of three. Our local meteorologists, who are pessimistic by nature, are wrong in their daily forecasts one time out of two when they should have predicted good weather but wrong only one time out of five when they should have forecasted rain.
Each morning, Francine leaves home for the day. If she departs without an umbrella and it rains, she is twice as annoyed as if it had been fair and she had taken her umbrella with her. "Would I be wiser," she wonders, "to listen to the morning weather forecast and take my umbrella only if rain is predicted? Or should I consistently carry it? Or should I never take it? What would you advise Francine to do?
Solution #22: Meteorology
There are four possible situations:

Fair weather predicted and occurred: probability 1/2 * 2/3 = 1/3.

Fair weather predicted but rain occurred: probability 1/5 * 1/3 = 1/15.

Rain predicted and occurred: probability 4/5 * 1/3 = 4/15.

Rain predicted but fair weather occurred: probability 1/2 * 2/3 = 1/3.
Let i be the measure of the inconvenience for Francine to carry her umbrella around for one whole fine day and 2i the measure of inconvenience of not having an umbrella when it rains. If she systematically carries her umbrella every day, the inconvenience will amount to i on two days out of three, hence an average inconvenience of 2i/3.
If she never takes her umbrella, the inconvenience will amount to 2i one day out of three, for an average inconvenience again of 2i/3.
If she decides to follow the advice of the weather forecasters and takes her umbrella only when rain is predicted why will suffer an inconvenience of i on 1 day out of 3, and ii on 1 day out of 15, which amounts to and average penalty of 7/i/15 (less than the two previous average penalties). We recommend, therefore, that Francine follow the advice of the weather forecasters.
Problem #23: Amateur Doctor
Aspirin does wonders for my headache and helps my rheumatic knee but it gives me nausea and upsets my stomach. Herbal medicine cures my nausea and stomach upset but gives me hip pain. Antibiotics sooth my headache and nausea but irritate my stomach and knee and give me a stiff neck. Cortisone helps my stiff neck and rheumatic knee but aggravates my hip condition. Hot compresses work wonders for my upset stomach and stiff neck. I woke up today with a pounding headache which prevents me from figuring out how best to doctor myself. What do you think I should do?
Solution #23: Amateur Doctor
I advise you to take aspirin and antibiotics and to apply hot compresses to your neck. Aspirin relieves the headache but gives you nausea and upsets your stomach; the antibiotics relieve the nausea but add pains to your knee and give you a stiff neck. The knee is taken care of my the aspirin. As far as the upset stomach and the stiff neck are concerned, the hot compresses will take care of them.
Problem #24: Bar Flies
At the same instant that Pete was leaving Harry's Bar to go to the Shamrock Tavern, Johnny was leaving the Shamrock Tavern to go to Harry's Bar. They walked at a constant speed. When they met, Pete announced that he had covered 200 meters more than Johnny had. The latter, his mind befuddled by alcohol, took this as a personal insult and started to beat up Pete, who returned the punches. When the fighting stopped, the men embraced in tears, then each continued his original journey, but at half his original speed, as both were slightly hurt. Pete thus arrived at the Shamrock Tavern in 8 minutes while Johnny took 18 minutes to get to Harry's Bar. How far apart are the two bars.
Solution #24: Bar Flies
Let x be the distance between the two bars.
Let d be the distance covered by Pete when he met Johnny.
Let V be Pete's speed and v be Johnny's speed before the fight.
The sum of the distances covered at the time they met is x.
d + (d200) = x (1)
Hence x = 2d  200
When they met, each had walked for the same length of time:
d/V = (d  200)v (2)
After the fight, Pete walked for 8 minutes:
(d  200)(V/2) = 8
Hence V = (d  200)/4 (3)
and Johnny for 18 minutes:
d(v/2) = 18
Hence v = d/9 (4)
By substituting the values for V and v into equation (2), we have
4d/(d  200) = 9(d  200)/d
Hence
5d^{2}  3600d + 360,000 = 0
Solving this quadratic in the classic way yields
d = 600 or 120 and x = 2d  200 = 1000 or 40
However, d must be smaller than x, so only one solution is possible:
d = 600 and x = 1000
The distance between the two bars is exactly 1000 meters.
Problem #25: Subway Escalator
I am in the habit of walking up the subway escalator while it is running. I climb 20 steps at my normal pace and it takes me 60 seconds to reach the top, whereas my wife climbs only 16 steps and it takes her 72 seconds to reach the top. If the escalator broke down tomorrow, how many steps would I have to climb?
Solution #25: Subway Escalator
Let x be the unknown number of steps. When I ride the escalator, I cover x  20 steps in 60 seconds. When my wife uses the escalator, she covers x  16 steps in 72 seconds. The escalator therefore moves at a rate of 4 steps every 12 seconds or 20 steps in 60 seconds. Its total height is therefore the sum of these 20 steps and of the additional 20 steps I climb: a total of 40 steps.
