1 › Number, Measure & Accuracy Accuracy
∙06 › Decimal Places & Significant Figures
\(\require{cancel}\)
Approximations
When '2 ÷ 7' or '√200' or '259 ÷ 204' or '99999 ÷ 7' are worked out on the calculator, the decimal form of the number is tedious to write out in full:
2 ÷ 7 = 0.286714285…
√200 = 14.14213562…
259 ÷ 204 = 1.269607843…
99999 ÷ 7 = 14285.57143…
We may not need the answer to be given that accurately, so we may decide to approximate the number:
Decimal Places Approximation
To approximate a number to 3 decimal places (3 d.p.):
- Count to the 3rd digit after the decimal and then cover up all the digits after that one:
decimal
│
▼↓↓↓
e.g.1: 2 ÷ 7 = 0.285714285…
decimal
│
▼↓↓↓
e.g.2: √200 = 14.14213562…
decimal
│
▼↓↓↓
e.g.3: 259 ÷ 204 = 1.269607843…
decimal
│
▼↓↓↓
e.g.4: 99999 ÷ 7 = 14285.57143…
- Now take a peek at the next digit (i.e. the first of the digits you just covered up):
↓↓↓
e.g.1: 2 ÷ 7 = 0.285714285…
⭫
↓↓↓
e.g.2: √200 = 14.14213562…
⭫
↓↓↓
e.g.3: 259 ÷ 204 = 1.269607843…
⭫
↓↓↓
e.g.4: 99999 ÷ 7 = 14285.57143…
⭫
- If that digit is 'less than 5' then just ignore it; the number is as you wrote above in step 1.
But if the 4th digit is '5 or more' then increase the 3rd digit by '1':
┌────► this digit is 5-or-more ──────┐
│ ▼
e.g.1: 2 ÷ 7 = 0.285714285… so we
▲⭫ │
├────── increase this digit by 1 ◄────┘
▼
= 0.283
╘══╦══╛
╚══════════ this is now correctly approximated to 3 d.p.
┌────► this digit is less-than-5 ────┐
│ ▼
e.g.2: √200 = 14.14213562… so we
▲⭫ │
├────── leave this digit unchanged ◄──┘
▼
= 14.142
╘══╦═══╛
╚══════════ this is now correctly approximated to 3 d.p.
┌────► this digit is 5-or-more ──────┐
│ ▼
e.g.3: 259 ÷ 204 = 1.269607843… so we
▲⭫ │
├────── increase this digit by 1 ◄────┘
but since the digit we │
want to increase is a │
'9' we have to carry '1' │
▼
= 1.270
╘══╦══╛
╚══════════ this is now correctly approximated to 3 d.p.
┌────► this digit is less-than-5 ────┐
│ ▼
e.g.4: 99999 ÷ 7 = 14285.57143… so we
▲⭫ │
├────── leave this digit unchanged ◄──┘
▼
= 14285.571
╘══╦══════╛
╚══════════ this is now correctly approximated to 3 d.p.
- Finally, you MUST write '3 d.p.' next to the number to signify that it has been approximated to 3 decimal places:
↓↓↓
e.g.1: 2 ÷ 7 = 0.286 (3 d.p.)
↓↓↓
e.g.2: √200 = 14.142 (3 d.p.)
↓↓↓
e.g.3: 259 ÷ 204 = 1.270 (3 d.p.)
↓↓↓
e.g.4: 99999 ÷ 7 = 14285.571 (3 d.p.)
·
Significant Figures Approximation
The method is similar to decimal places approximation except that we start counting from the 'first non-zero digit' rather than from the decimal place:
↓↓↓ ↓↓↓
So for: 2013.596784… And for: 0.0075638…
▲ ▲ ▲▲▲
│ ignore these zeros ───┴─┴┘│
┌──────────────┴───────────────┐ ┌──────────────┴───────────────┐
│we start counting from the '2'│ │we start counting from the '7'│
└──────────────────────────────┘ └──────────────────────────────┘
To approximate a number to 3 significant figures (3 s.f.):
- Count out '3' digits of the number, starting from the 'first non-zero digit', then cover up the rest of the digits
↓↓↓
e.g.1: 2 ÷ 7 = 0.285714285…
▲
│
we start counting from the '2'
↓↓ ↓
e.g.2: √200 = 14.14213562…
▲
│
we start counting from the '1'
↓ ↓↓
e.g.3: 259 ÷ 204 = 1.269607843…
▲
│
we start counting from the '1'
↓↓↓
e.g.4: 99999 ÷ 7 = 14285.57143…
▲
│
we start counting from the '1'
- Now take a look at the next digit (i.e. the first of the digits you just covered up):
↓↓↓
e.g.1: 2 ÷ 7 = 0.285714285…
↓↓ ↓
e.g.2: √200 = 14.14213562…
↓ ↓↓
e.g.3: 259 ÷ 204 = 1.269607843…
↓↓↓
e.g.4: 99999 ÷ 7 = 14285.57143…
- If that digit is 'less than 5' then just ignore it; the number is as you wrote above in step 1.
But if the 4th digit is '5 or more' then increase the 3rd digit by '1'
At the same time, any digits in-front the decimal that are still covered up convert to zeros
┌────► this digit is 5-or-more ──────┐
│ ▼
e.g.1: 2 ÷ 7 = 0.285714285… so we
▲ │
├────── increase this digit by 1 ◄────┘
▼
= 0.286
╘══╦══╛
╚══════════ this is now correctly approximated to 3 s.f.
┌────► this digit is less-than-5 ────┐
│ ▼
e.g.2: √200 = 14.14213562… so we
▲ │
├────── leave this digit unchanged ◄──┘
▼
= 14.1
╘══╦═╛
╚══════════ this is now correctly approximated to 3 s.f.
┌────► this digit is 5-or-more ──────┐
│ ▼
e.g.3: 259 ÷ 204 = 1.269607843… so we
▲ │
├────── increase this digit by 1 ◄────┘
▼
= 1.27
╘══╦═╛
╚══════════ this is now correctly approximated to 3 s.f.
┌────► this digit is 5-or-more ──────┐
│ ▼
e.g.4: 99999 ÷ 7 = 14285.57143… so we
▲▲▲ remaining digits in-front │
├┼┼─── increase this digit by 1 ◄─────┘
▼▼▼ of the decimal become zeros
= 14300
╘══╦══╛
╚══════════ this is now correctly approximated to 3 s.f.
- Write '3 s.f.' next to the number to signify that it has been approximated to 3 significant figures:
↓↓↓
e.g.1: 1 ÷ 7 = 0.143 (3 s.f.)
↓↓ ↓
e.g.2: √200 = 14.1 (3 s.f.)
↓ ↓↓
e.g.3: 259 ÷ 204 = 1.27 (3 s.f.)
↓↓↓●●
e.g.4: 99999 ÷ 7 = 14300 (3 s.f.)
Which is Better?
For very large or very small numbers; significant figures is the best method of approximation to use:
Accurate Number │ 3 d.p. │ 3 s.f.
———————————————————┼—————————————————————┼—————————————
e.g. A very small number: 0.000235739… │ ╔► 0.000 │ 0.000236 ◄╗
---------------------------------------------│-║-------------------│--------------║
e.g. A very large number: 3,579,424.1 │ ║ 3,579,424.100 ◄╗ │ 3,580,000 ◄╣
│ ║ ║ │ ║
│ ╚══ both silly ═══╝ │both reasonable
For the 'very small number' the 3 decimal place approximation ends up destroying the number completely
Whereas, for the 'very large number', the 3 decimal place approximation doesn't approximate the number at all!
Approximating Numbers that are in Standard Form
Significant figures Approximation
·
Numbers written in standard form can be approximated using significant figures very easily:
·
e.g. Approximate 2.373687 × 10³ to 3 s.f.
- Ignore the × 10³ and approximate the number in the normal way:
2.373687 = 2.37 (3 s.f.)
- Put back in the ' × 10³ ':
2.373687 × 10³ = 2.37 × 10³ (3 s.f.)
·
Decimal Places Approximation
·
Numbers written in standard form must first be converted to normal numbers before making a decimal place approximation:
·
e.g. Approximate 2.373687 × 10³ to 1 d.p.
- Convert back to a normal number
2.373687 × 10³ = 2373.687
- Approximate as normal
2373.687 = 2373.7 (1 d.p.)
- Convert back to standard form
2373.7 (1 d.p.) = 2.3737 × 10³
·
Question 1, part (a): On the calculator: \(\pi =3.141592654\)...
Approximating this to 3 d.p.
Step 1: Identify the 3rd d.p:
we start counting from the decimal
│
▼↓↓↓
3.141592654…
Step 2: Round to 3 d.p:
┌────► this digit is 5-or-more ──────┐
│ ▼
3.141592564… so we
▲ │
├────── increase this digit by 1 ◄────┘
▼ and delete all digits after it
= 3.142
╘══╦═══╛
╚══════════ this is now correctly approximated to 3 d.p.
Approximating this to 3 s.f.
Step 1: Identify the 3rd s.f:
↓ ↓↓
3.141592654…
▲
│
we start counting from the '3'
Step 2: Round to 3 s.f:
┌────► this digit is less-than-5 ────┐
│ ▼
3.141592654… so we
▲ │
├────── leave this digit unchanged ◄──┘
▼ and delete all digits after it
= 3.14
╘══╦═╛
╚══════════ this is now correctly approximated to 3 s.f.
Question 1, part (b): Typing \(\sqrt{2}\) into the calculator, you get: 1.41421356237309…
Approximating this to 3 d.p.
Step 1: Identify the 3rd d.p:
we start counting from the decimal
│
▼↓↓↓
1.4142913562…
Step 2: Round to 3 d.p:
┌────► this digit is less-than-5 ────┐
│ ▼
1.4142913562… so we
▲ │
├────── leave this digit unchanged ◄──┘
▼ and delete all digits after it
= 1.41…
╘══╦═══╛
╚══════════ this is now correctly approximated to 3 d.p.
Approximating this to 3 s.f.
Step 1: Identify the 3rd s.f:
↓ ↓↓
1.4142913562…
▲
│
we start counting from the '1'
Step 2: Round to 3 s.f:
┌────► this digit is less-than-5 ────┐
│ ▼
1.4142913462… so we
▲ │
├────── leave this digit unchanged ◄──┘
▼ and delete all digits after it
= ….……
╘══╦═╛
╚══════════ this is now correctly approximated to 3 s.f.
Question 1, part (c): On the calculator: ³√1,400,000 = 111.8688942...
Approximating this to 3 d.p.
Step 1: Identify the 3rd d.p:
we start counting from the decimal
│
▼↓↓↓
111.86888942…
Step 2: Round to 3 d.p:
┌────► this digit is 5-or-more ──────┐
│ ▼
111.86888942… so we
▲ │
├────── increase this digit by 1 ◄────┘
▼ and delete all digits after it
= 111.86…
╘═══╦═══╛
╚══════════ this is now correctly approximated to 3 d.p.
Approximating this to 3 s.f.
Step 1: Identify the 3rd s.f:
↓↓↓
111.8688942…
▲
│
we start counting from the '1'
Step 2: Round to 3 s.f:
┌────► this digit is 5-or-more ──────┐
│ ▼
111.8688942… so we
▲ │
├────── increase this digit by 1 ◄─────┘
▼ and delete all digits after it
= 112
╘═╦═╛
╚══════════ this is now correctly approximated to 3 s.f.
Question 1, part (d): On the calculator: 20,000-1/3 = 0.03684031499…
Approximating this to 3 d.p:
Step 1: Identify the 3rd d.p:
we start counting from the decimal
│
▼↓↓↓
0.03684031499…
Step 2: Round to 3 d.p:
┌────► this digit is 5-or-more ──────┐
│ ▼
0.03684031499… so we
▲ │
├────── increase this digit by 1 ◄────┘
▼ and delete all digits after it
= 0.03…
╘══╦══╛
╚══════════ this is now correctly approximated to 3 d.p.
Approximating this to 3 s.f:
Step 1: Identify the 3rd s.f:
↓↓↓
0.03684031499…
▲
│
we start counting from the '3'
Step 2: Round to 3 s.f:
┌────► this digit is less-than-5 ────┐
│ ▼
0.03684031499… so we
▲ │
├────── leave this digit unchanged ◄──┘
▼ and delete all digits after it
= 0.036…
╘══╦═══╛
╚══════════ this is now correctly approximated to 3 s.f.
Question 1, part (f): Typing \(0.2^{\frac{3}{5}}\) into the calculator, you get: 0.3807307877040014318…
Approximating this to 3 d.p:
Step 1: Identify the 3rd d.p:
we start counting from the decimal
│
▼↓↓↓
0.38073078770…
Step 2: Round to 3 d.p:
┌────► this digit is 5-or-more ──────┐
│ ▼
0.38073078770… so we
▲ │
├────── increase this digit by 1 ◄────┘
▼ and delete all digits after it
= 0.……
╘══╦══╛
╚══════════ this is now correctly approximated to 3 d.p.
Approximating this to 3 s.f:
Step 1: Identify the 3rd s.f:
↓↓↓
0.38073078770…
▲
│
we start counting from the '3'
Step 2: Round to 3 s.f:
┌────► this digit is 5-or-more ──────┐
│ ▼
0.3807308770… so we
▲ │
├──────────────────◄────────────────┘
▼
= 0.………
╘══╦═══╛
╚══════════ this is now correctly approximated to 3 s.f.
Question 1, part (i): Approximating '36,000' to 3 d.p.
decimal
│
▼↓↓↓
=> 36,000.000000 (counting 3 digits from the decimal - I had to put in lots of zeros)
=> 36,000.000000 (looking at the next digit)
=> 36,000.000000 (leaving the previous digit unchanged since pink red digit is 'less than 5')
=> 36,000.000000 (3 d.p.)
Approximating '36,000' to 3 s.f.
↓↓ ↓
=> 36,000.0 (counting 3 digit from the '3')
=> 36,000.0 (looking at the next digit)
=> 36,000.0 (leaving the previous digit unchanged since the pink digit is 'less than 5')
and the remaining digits in-front of the decimal become zeros)
=> 36,000 (3 s.f.)
This question highlights that, for very large or very small numbers, decimal place approximation can produce silly answers
Question 1, part (j): Approximating '0.0042' to 3 d.p:
decimal
│
▼↓↓↓
=> 0.0042 (counting 3 digits from the decimal)
=> 0.0042 (looking at the next digit)
=> 0.004 (leaving the previous digit unchanged since the pink digit is 'less than 5')
=> 0.004 (3 d.p.)
Approximating '0.0042' to 3 s.f.:
↓↓↓
=> 0.0042000 (counting from the '4' - I had to include the extra zeros)
=> 0.0042000 (looking at the next digit)
=> 0.0042000 (leaving the previous digit unchanged since the pink digit is 'less than 5')
=> 0.0042000 (3 s.f.)
Question 2, part (ai): This number is written in Standard Index Form (S.I.F.)
It makes no difference to us if a number is written in S.I.F. when we are doing a SIGNIFICANT FIGURE approximation...
So, ignoring the × 10-4 and approximating the 5.28741 to 4 s.f:
5.28741 = 5.287 (4 s.f.)
An then putting back the × 10-4:
5.28741 × 10-4 = 5.278 × 10-4 (4 s.f.)
Question 2, part (aii): But, when doing a DECIMAL PLACE approximation, we must first take the number out of STANDARD FORM:
5.2 8 7 4 1 × 10-4
╘════════╤═════════╛
is the same as
╒══════════════════╧═════════╕
0 0 0 0 0 5.2 8 7 4 1 × 10-4
0 0.0.0.0.5.2 8 7 4 1 ——————————► 0.0 0 0 5 2 8 7 4 1
▲ ▲ ▲ ▲ │
└—└—└—└—┘
moving the decimal
left by 4 places
Only once it is written as an ordinary number, can we then make the 4 d.p. approximation:
Step 1: Identify the 4th d.p:
decimal
│
▼ ↓↓↓↓
0.0000528741…
Step 2: Round to 4 d.p:
┌────► this digit is 5-or-more ──────┐
│ ▼
0.0000528741… so we
▲ │
├───── increase this digit by 1 ◄─────┘
▼ and delete all digits after it
= 0.000…
╘══╦═══╛
╚══════════ this is now correctly approximated to 4 d.p.
Finally, we need to convert this back to S.I.F...
Question 2, part (bi): So, we can do the 4 s.f. approxmation while it remains in Standard Form
Question 2, part (bii): But we can't do the 4 d.p. approximation unless we first take it out of standard form
Question 3: This is really about being able to use your calculator properly and, where necessary, putting in brackets.
If your answer doesn't match the one given, then try re-typing the expression into your calculator, using brackets where you think they might help...
To type this in, press these buttons on your calculator:
Question 4, part (a): We need to find an approximate answer to:
л √ 4.15² + 2.92² .
So, if we start by approximating each individual number to 1 s.f:
л √ 4.15² + 2.92² .
└┬┘ └─┬─┘ └─┬──┘
л=3.14159…= 3 (1 s.f.)──┘ │ └───────── 2.92 = 3 (1 s.f.)
│
4.15 = 4 (1 s.f.)
So, the sum can be approximated as:
3 √ 4² + 3² .
Which is pretty easy to figure out, even without a calculator
Note: This is the same expression that you CALCULATED in question 3, part (a) - so comparing the two answers will show you just how close your approximation was...
Question 4, part (b): In this case, it is fairly each to add the 62 + 59 (exactly) and the 6.7 + 8.6 (exactly), before we start approximating. And, if there is ever a part of the sum that you can (easily) do without having to approximate, then that is always going to result in a better approximation.
·
So, the sum can be re-written as: \(\sqrt[3]{\frac{121}{15.3}}\)
·
Now, I think it is a good idea to approximate each number in the fraction, but instead of using 1 s.f, I think is would be quite easy to divide these numbers once they approximated to 2 s.f. (which is thus more accurate):
\[\sqrt[3]{\frac{\color{#62be53}{121}}{\color{#2b83c3}{15.3}}}\begin{array}{c}
\color{#62be53}{\longleftarrow \longleftarrow 121=120 \left( 2 s.f. \right)}\\
\color{#2b83c3}{\longleftarrow \longleftarrow 15.3=15 \left( 2 s.f. \right)}\\
\end{array}\]
Which leaves us with an easy cube-root!
Again, comparing with the answer to question 3, part (b) will tell you how reasonable your approximation is...
Question 4, part (c): This is really easy, because we remember that when multiplying numbers in S.I.F:
╔═════════════╦═══╗
▼ ▼ ║
7.22 × 105 × 6.8 × 10-2 add
╘═╦═╛ ╘═╦═╛ ╔══╝
╚══multiply══╦═╝ ▼
.... × 103
└─┬──┘
└────── To find this we can just
use the approximates:
7.22 = 7 (1 s.f.)
6.8 = 7 (1 s.f.)
The only thing left to do is to ensure our number IS in SIF (I think in this case, we need to move the decimal left by 1-place, which means we must also --- 1 to the power)
Again, we can check against the (almost) exact value the calculator gave us...
Question 4, part (d): The sum we want is: \(\frac{86.05 \times \,\,0.07825}{5.95}\)
But, if we first approximate the numbers: 86.05 = 90 (1 s.f.), 0.07824 = 0.08 (1 s.f.) and 5.95 = 6 (1 s.f.)
Then, it is probably easier to work out 90 ÷ 6 first, then multiply that by 0.08... (its the same, since they are all multiplied and divided, we can swap the order!)
Question 4, part (j): Approximating 17.26 to 1 s.f.
Step 1:Identify the 1st s.f:
↓
17.22
▲
│
we start counting from the '1'
Step 2: Round to 1 s.f:
┌────► this digit is 5-or-more ──────┐
│ ▼
17.26 so we
▲▲ remaining digit in-front of │
├┼──────increase this digit by 1 ◄────┘
▼▼ the decimal becomes a zero
= ……
╘╦═╛
╚══════════ this is now correctly approximated to 1 s.f.
Next, approximating 9.64 to 1 s.f.
Step 1:Identify the 1st s.f:
↓
9.64
▲
│
we start counting from the '9'
Step 2: Round to 1 s.f:
┌────► this digit is 5-or-more ──────┐
│ ▼
9.64 so we
▲ │
├────── increase this digit by 1 ◄─────┘
▼
= ……
╘═╦╛
╚══════════ this is now correctly approximated to 1 s.f.
So now, the calculation becomes: \(\sqrt{\frac{20^2-10^2}{2\times 20^2+4\times 10^2}}\)
Question 5: Approximate the numbers first, but only as much as you need to...
Question 6, part (a): You can factorise this expression as:
┌────────────────────────┐
│ COMPARE WITH THIS │
│x² + 0.5x = x(x + 0.5)│ 2.5² + 0.5×2.5 = 2.5(2.5 + 0.5)
└────────────────────────┘ └────┬────┘
┌┴┐
= 2.5( 3 )
└──────┬──────┘
┌─┴─┐
= 7.5