Solution

Solutions to the puzzles

1. The first brother is 70 inches tall, the second 72 the third 74 and the fourth brother 80 inches tall.

2. Twenty-six minutes.

3. Since the boys have as many brothers as sisters, there must be 1 boy more than the number of girls. If we try 2 and 1, 3 and 2, and 4 and 3, we will find that 4 boys and 3 girls is the solution to fulfill the requirement that each girl has twice as many brothers as sisters.


4. Naturally, the train travelling against the spin of the earth. This train will wear its wheels out more quickly, because the centrifugal force is less on this train.

5. No, the answer is not 32§ miles an hour, though this figure is the obvious answer! However, this represents the average of the 2 speeds and not the average speed for the whole trip.
If the time is equal to the distance divided by the average speed, then the time for the trip starting
from San Francisco equals s/40 and the time for the s return trip is is » which gives us a total time of 40 which equals "MO •
Therefore the average speed for the whole trip when the average speed equals the distance 1SS divided by the time is 2S divided by ~soo which, „_ . 200 , . , , 400S „„ 10 equals 2S times iss% which equals i3s~or 30— ~f3.

6. The lowest square number I can think of, contain-ing all the nine digits once and only once, is 139854276, the square of 11826, and the highest square number under the. same conditions is 932187456 the square of 30384.

7. Here is the formula that gives the minutes past twelve to which the hour hand points when the minute hand is exactly thirty minutes ahead.

30
Minutes past twelve Y=—[ (n—1) 2+11 where n is the next hour— Let's take the case of at what time between 4 and 5 will the hands be opposite each other? (n= 5).
30 270 . _, 6
.-.Y = -pp x 9 = "TP* 24tt. i.e. the hour hand will be 24 TT minutes past 4. The formula may be derived from the following: If X is distance moved by the minute hand Y is the distance moved by hour hand

then X—Y = 30 First time the hands move round X = 12 Y Second time the hands move round X = 12 Y—5 Third time the hands move round X = 12 Y—10 etc.

8. While striking 7 the clock strikes its first gong at 7 o'clock and it strikes 6 more at regular intervals. These 6 intervals take 7 seconds so that the intervals between gongs is 7/6 seconds. However to
strike 10 there are 9 intervals each taking (7/6)*9  seconds for a total of 10.5 seconds.

9. In order that the little girl should have disposed of the oranges she had remaining after her second sale, she; must have had at least one whole orange remaining so that she could deduct from it 'half of her oranges plus half an orange', for the third and the final sale. Therefore, if 1 orange represents half of the remaining after the second sale, then she must have sold two oranges in her second sale, leaving the 3 oranges after the first sale.
Lastly if three oranges only represent half the original number, plus half an orange, then she must have started with [ (3x2)+1] or 7 oranges.


10. The minimum number of weights required is five and these should weight 1, 3, 9, 27 and 81
pounds.