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Category: Working with numbers

Worked GCSE Number Questions

Coming on in leaps and BOUNDS

For this post, you need to know your basics in rounding numbers. Here is a recap

  1. Pick out the digit in the last place that will survive the rounding process

    33962

  2. Look at the next digit… Is it a 0 1 2 3 or 4? If so take no action. Otherwise raise the underlined digit by 1. In this case a 9, so raise to 4
  3. In both cases, remove all following digits after the underlined one

    34

Now, for this subject of bounds we are going to look at this kinda back to front. We are told a number has been rounded to 2 decimal places. So what could that number be?

2.335 is the smallest number it could be. Any number smaller would be rounded to 2.33
2.34499999 is the largest number it could be. Any number larger would be rounded to 2.35.

2.335 and 2.3449999  are the upper and lower bounds to the number given the information we have.

Let’s look at something a bit more complicated


Now, if we know that I have ran 10.6m to 1 d.p  in  19.5 seconds to 1 d.p, then what is my highest possible speed and lowest possible speed, to 3 d.p ?

We are being asked for the ‘upper bound’ and ‘lower bound’ for the speed.

Remember  Speed = distance / time

 

Start by writing down the upper bound’ and ‘lower bound’ of my distance and time.

Upper Bound of distance is 10.6499m
Lower Bound of distance is 10.5500m
Upper Bound on Time is 19.5499 seconds
Lower Bound on Time is 19.4500 seconds
Now, this is where we have to think because to find the upper bound on the speed, we need the highest possible time BUT the lowest possible time. This is because as time goes up, speed comes down, for any given distance. I use the highest possible value for time in the lower bound calculation, for the same reason

So upper bound on Time = 10.6499/19.4500 = 0.547553

The lower bound on time = 10.5500/19.5499 = 0.539645

The bounds for the speed are [0.540, 0.548]  to 3 d p

Note that I used 4 d.p. in my calculations. I will always use one or more decimal places than required then round at the end. We risk losing accuracy otherwise.

Mixed fractions: Multiplying and adding

How do we multiply mixed fractions.

 

We could treat the parts separately.  I mean, its the multiplication of two mixed numbers and we’ve been asked to give the answer as a mixed number..  so we can use multiply each part of the first number with each part of the second.

1 x 3     +   1 x 1/3   + 3/4 x 3  +  3/4 x 1/3

= 3  + 1/3 + 9/4 + 1/4

Then we need to get the three fractions over the same number, 12

= 3 + (4 + 27 + 3)/12 = 3 + 34/12 = 3 + 17/6 = 3 + 2 5/6 = 5 5/6

But actually that’s actually quite a lot of work,  and changing these mixed numbers into ‘top-heavy’* fractions  and back again can save time….

In this pic you’ll see I’ve converted the two numbers into 7/4 and 10/3 – done just a little cancelling, and then multiplied

The answer is 35/6   which we can convert back into 6 5/6. This  the same answer as before,  which is always a relief!

We can also add these two mixed numbers. I think the first method is easier then.  Don’t convert into top heavy fractions. We don’t need to multiply all parts together.  Just add the whole numbers. Then add the fractions separately, which does mean looking for a common denominator

1 + 3 = 4
3/4 + 1/3 = 9/12 + 4/14 = 13/12 = 1 1/12

So the answer is 5 1/12

 

If you would like some more practice in working with mixed numbers, look at this document

 

*The proper word for top heavy fractions is ‘vulgar’ fractions and I’m sure you’ll here a teacher use this word..  Its just not a words I like for numbers..  all numbers are fun, none of them vulgar at all  IMHO!!

Working out Recurring Decimals

A clever trick that sometimes comes up in GCSE exams is to how to turn a repeating decimal back into a fraction.

At first glance the question looks difficult… but actually it’s not really.

 

Write the decimal 0.393939393939 (recurring)  as a decimal.

 

So how do we do this?

 

Well we start with a step familiar from algebra – We use a letter!

Let r = 0.393939393939

We need to get rid of that long train of decimals – se we use the same method as when we want to lose something in solving simultaneous equations – we do a subtract

But how will that work?

Well think what 100r is

r = 0.393939393939  so 100r = 39.3939393939

[Can you see where we are going with this now?]

Now we can do a subtraction and all the numbers in the recurring chain cancel out
39.3939393939    _
0.3939393939

= 39

What we have done here is 100r – r   so this is 99r

99r = 39

So r = 39/99…  A fraction!  We’ve done it!

 

Just two more things to note

  1. We should always ‘cancel down’ fractions is we can to use the lowest numbers at top an bottom.
    So I should have finished this question by writing as  13/33.
  2. How did I know to multiply by 100? Would this work for every example?  Well, no, a careful study of what we have done shows that the key is to ‘cancel out’ all the recurring decimals. To do that, we need to ‘line up’ the same numbers in the subtraction. To do this we need to count how many numbers in the repeating pattern.. we then need that number of 0s in the ‘multiplying factor’

So if we have 0.123123123123123…   we have 3 numbers in the pattern

r = 0.123123123123123

1000r = 123.123123123123..

999r = 123

R = 123/999  = 41/333

 

Here are three more to work on,,,

 

  1. Write 0.777777777777… as a decimal
  2. Write 0.787878787878.. as a decimal

    And an advanced one…

  3. Write 0.0200200200200..  as a decimal  (That’s a bit tougher)

    Here are the answers
    0.7777 = 7/9
    0.78787878 = 78/99 = 26/33
    0.0200200200 = 20/999

Car Parking

Quick problem of the day  – Can you tell what number space the parked car is in?  Answer added tomorrow!

 

 

 

 

 

 

(Spoiler alert- I you page down)

 

 

 

 

 

 

 

 

 

OK  – No so much an answer today, as a clue…   – From which direction are we looking at these numbers?

 

Yes  . Upside down – The answer is 87

Adding up – Quickly!


Do you ever find yourself adding up a long list of numbers – checking a receipt maybe – and when you get to the end, you are not sure you included that 42 near the top? Or maybe you are halfway through and your dog jumps up at you, or you get a text you need to check?

Were you adding on each number at a time?
23 plus 145 , Hmm  168..  now add on 283..  that makes… Lets look at a quicker way.

 

 

We would normally add up one column at a time. But we can go further than that….

Look for sets of numbers that add up to 10. I’ve made a start here with the 7 and the 3.

 

 

Then I have collected together the two 5’s and the 8 and the 2

Sometimes its not two numbers that add up to 10. Sometimes it can be 3 numbers, like the 1, 4 and 5

Of course not all numbers can be grouped in 10s. The last three numbers here are the 3, 3 and 1 and we can add them together to make 7 with a low risk of being interrupted.

We do need to count up how many sets of 10 we had. I counted 4. That is what that small 4 is doing at the foot of the second column…. and to be included in the ‘groupings’ of course .

The second column can be added up in the same way

And here we have more left over numbers, but still easier than adding them all up!

Where is my missing £1

This is a puzzle I’ve known since I was 10, and then yesterday a friend tagged me on it on Facebook. The solution has now been added to the post.  Try answering it first before looking at the spoilers!

 

 

 

 

 

 

 

 

 

 

OK  –  Here is the answer! 

The skill in solving problems like this is noting down the information you have and trying to ignore any misdirection.  And boy, does the setter of this problem go in for misdirection!

And if you are thinking, that means bringing in skills learnt in an English comprehension class, you’d be right. School subjects may be split between Maths, English, Science, History – but in life things are never so simple. Actually I like to persuade students to write their answers in full English sentences, especially in problem solving cases.

In this situation, we have a situation involving some money, and there is money coming in (the money that each guest pays at the start) and money coming out (to the guests, the hotel and the bell boy).

Each Guest pays £10. That’s £30 ‘in the system’

But who has this money at the end?  The hotel has £25. The bell boy keeps £2. Each of the guests has £1 return to them by the bell boy. That is a total of £30 (£25 + £2 + £1 + £1 + £1)

Or if you prefer lets look and gains and losses. The guests have spent £9 each, making £27.   £25 goes to the hotel, £2 to the bell boy. A total of £27. Gains = Losses. Just as we’d expect.

 

 

 

 

 

So lets remind ourselves of the question. This asked us to add the £9s that the guests paid in and £2 the bell boy kept.  But these numbers are from different parts of the question! (As highlighted in grey and blue)   There is really no reason why they should be added. The fact that the total came close to £30 was a coincidence (or not, since the questioner chose the numbers to confuse!)

So in summary, ignore the last paragraph until you have come to your own understanding about the ins and outs. And then you won’t fall into the trap of adding an ‘in’ amount to an ‘out’ amount.