11 Plus Exams Forum

Hi All,

Any help with question 36 would be much appreciated.

I think its 44cm (guess work) but just wanted to be sure.

http://www.nlcs.org.uk/userfiles/nlcs/p ... 202014.pdf" onclick="window.open(this.href);return false;

Thanks
Bill

The answer is 48.

There is a way to do it using algebra, but for quickness in exams I often tell dd to put in numbers. So imagine the indent were 1cm, the total would be 28 for the horizontal lengths or if you put in 3, it would still be 28. The total vertical distance is 2*10.

Algebraically, calling the indent x.
Horizontal length of base = 10+4-x
Horizontal length of other 3 bits=10+x+4.
When you add these together the x cancels out giving 28.

You do overcomplicate things russet!

I recall doing this question when dd was appearing for her 11+. The question always remained in my memory because it tested the mind. The first thing we did was to mark out all the known and unknown distances.

All vertical parts add up to 20cm. As far as the horizontal parts are concerned, there are some unknowns. The base line is more then 10cm. by, say "a"cm. The other unknown is the small bit above the 4cm. horizontal line. Let's call this "b". Thus, the equation for the perimeter will read :-

20 (verticals) + 10 + b + 4 + (10+a) = 44 + a + b.

Now look at the diagram again and see if you can find what "a+b" is.

I hope this helps.
Tagore

Thanks.

Guest 55. Which bit were you finding difficult to follow?

Cheeky!

No need to use algebra ... the question is not designed for that.

Did anyone confirm the answer to this? I got to 40 +(4-x) + (4-x), or 48 - 2x, but have no idea how to work out what x is...?