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## Quiz of the Month (November 2003)

### Hector C. Parr

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#### SOLUTION TO LAST MONTH'S QUIZ

```    1.  3.155 ins.
2.  3.633 ins.
3.  8.532 ins.```

Notes
 1. The balls must be in opposite corners of the box, so their centres must lie at opposite corners of an imaginary cube of side x ins, whose space-diagonal = 2 ins. We have x2 + x2 + x2 = 22. So x = sqr(4/3), and required solution = 2 + sqr(4/3) ins. 2. The centres of the balls lie at the vertices of a regular tetrahedron with sides 2 ins. Its perpendicular height is easily shown to be sqr(8/3). So total height = 2 + sqr(8/3) ins.
3. Each ball in an upper layer touches three balls in the next lower layer, as in Q.2. So the vertical distance between the plane of centres of successive layers is sqr(8/3)
So total height = 2 + 4*sqr(8/3)ins.

#### THIS MONTH'S QUIZ

1. A motorist sets off on a journey hoping to complete it at an average speed of 60 mph. But after travelling for half the expected journey time he finds he has been averaging only 50 mph. What must be his average speed for the remainder of the journey if he is to arrive on time?

2. Another motorist sets off on the same journey, also hoping to average 60 mph. But after travelling half the distance he finds he has averaged only 50 mph. What must he average for the remainder of the journey to arrive on time?

3. Yet another motorist sets off on the same journey with the same aspiration. He travels the first two-thirds of the distance at 50 mph. What must he average for the remainder of the journey to arrive on time?

Disclaimer. Readers attempting to check their answers by practical experiment do so at their own risk.
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(c) Hector C. Parr (2003)

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