Find the Number | | DJS

Find the Number

One of the repeated problems I observe is an inability to create the maths with which to do a problem. This page will, I hope, assist the development of that skill. From the point of view of a Year Eight, decide what x is, write down some algebra with x in it, and solve that problem.

1    Find a number whose double is fourteen bigger than its quarter.

2    Find a number whose treble is four times bigger than one less than itself.

3    Find a number whose quintuple is four times bigger than two more than itself.

4    Find an number x such that a multiple of x is one more multiple of one less than x.

5    I’m thinking of a number that you’re going to find. I take my number and double it, add two, multiply what I've got by one less than my original number, add four, halve what I have, - and I’ve got 17. What was my number?

6    I have a straight line graph that goes through (3,7) and (6, 13). What is the equation of my line?

7     The stopping distance for cars is declared to be made of thinking distance plus braking distance. From 20 mph the stopping distance in feet is 20 (think) plus 20 (braking). At 30mph it is 30 plus 45 at 40mph it is 40 plus 80 and at 50 it is 50+125. Can you find the figures for 60mph?       and see essay 191. Also see Stopping.

Teachers: You can write these quite easily. 
Pupils: If you’d like some more, just write, they’re really not hard to create. The hard bit is writing the English, not making up the problem.

1    2x = x/4 +14  => x = 8

2    3x=4(x-1)       => x = 4

3    5x=4(x+2)      => x = 8

4    Ax=(A+1)(x-1) => x=1+A

5    ((2x+2)(x-1)+4)/4 = x²+1 = 17, so x = ±4 so you probably said “Four”.

6    y=ax+b; 7=3a+b AND 13=6a+b; the difference of these equations is 6=3a, so a=2 and then b=1, which works in both given points. Answer y=2x+1.

Alternatively, say “x goes up three when y goes up six, so y=2x plus B. 3x2+B=7 says B=1, so y=2x+1. Check in (6,13). Oh, I’m right. Next?”

7    Look at the thinking time series {20, 30, 40, 50} obviously the next number is 60. Look at braking series {20, 45, 80, 125} and then the differences {25, 35, 45}. Obviously the next of these differences is 55, so the next stopping distance is 125+55=180.  So the figures for S60=60+180=240. Done. In feet, which is weird, but it works.

Alternatively, let Sv = Tv + Bv. We can see that Tv = v ( the speed number but in different units). As v changes so the change in B is NOT straightforward; the differences are not constant. One answer that will fit is to use v². We now have B(20)=20, B(30)=45, B(40)=80, B(50)=125. This is not B=a+kv. The next simplest formula is B=a+bv+cv², but I picked nice numbers (lots of work hidden here) and B=cv², c= 1/20. So the formula for the whole stopping distance S=v+v²/20  and S(60)=60+3600/20 = 60+180=240.

Hidden answers.

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