Re: 11-plus grammar
- From: the Omrud <usenet.omrud@xxxxxxxxxxxxxxxx>
- Date: Mon, 30 Jun 2008 19:50:29 GMT
Paul Wolff wrote:
the Omrud <usenet.omrud@xxxxxxxxxxxxxxxx> wroteArfur Million wrote:I did it differently. There's a base rectangle seven bricks long and nine bricks high, so put 7 x 9 bricks down."Leslie Danks" <leslie.danks@xxxxxx> wrote in message news:4868ab9c$0$2144$91cee783@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxIf you try the test, Question 4 will tell you whether you belong to theI didn't understand the scenario described (?) in question 2 at all - am I alone in this?
sheep or the rest:
<http://news.bbc.co.uk/today/hi/today/newsid_7478000/7478154.stm>
9 bricks in the first pile, 10 bricks in the second, etc. 7 piles, so the total number of bricks is 9 + 10 + 11 + 12 + 13 + 14 + 15.
As with many of this type of question, the total can be arrived at by the staggeringly useful formula for the area of a trapezium:
n/2(a + l)
where n is the number of steps, a is the first term and l is the final term:
7/2 (9 + 15)
3.5 x 24
84
Right.
Then there are 1, 2, 3, 4, 5 and 6 bricks on top, which are easily summed as 1+6, 2+5 and 3+4, which is three lots of seven.
Aha.
n/2 (a + 1)
6/2 (1 + 6)
3 x 7
Total 12 lots of seven, or 84. It saves working with fractions until they are really needed, in some hypothetical later question.
--
David
.
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