Numbers and general problem solving
… The ANS button on your calculator remembers the last answer. This can be very useful when you are in mid-calculation and wish to keep the numbers as accurate as you can. For example:
2 ÷ 3 = 0.666666666.
ANS × 9 = 6
… If 1793 x 185 = 331705, then what is:
a) 1.793 x 185
b) 331705 ÷ 1.85
… Multiplying common bases means you can …?… the powers.
If you don’t have a common base, then it is often possible to make them match, for example:
4^5 x 2^-4
(2²)^5 x 2^-4
2^10 x 2^-4
= 2^6
where “^” means “to the power of”
Conversion that you will need in the AQA exams include:
… Miles to kilometres… What is 10 miles in km?
… Litres to gallons… What is 90 litres in gallons?
… Pounds to kilograms… what is 110 pounds in kilograms?
… Inches to centimetres… How many centimetres are there in 12 inches?
Here is a conversion table you could remember and use for converting between metric prefixes. For example, this one is for metres, but could also be used for grams, litres or other units.
Remember that Up is mUltiply and Down is Divide:
micro metre (μm) << think of micro as being the smallest and so floats to the top!
1000
milli metre (mm)
10
centi metre (cm)
100
metre
1000
kilo metre (km)
1000
mega metres (Mm)
So if you had 27cm, you could convert this to μm by going Up (multiply)
27 ×10 ×1000 = 27×10^4 μm ( = 270,000 μm )
If you needed to convert 627μm into cm, then move Down (Divide):
627μm ÷1000 ÷10 = 627 ×10^-4 cm ( = 0.0627 cm)
… When converting units of AREA, the multiplying or dividing factor is SQUARED.
So converting 23cm² to m² is going Down (Divide):
cm —- ÷ 100 —–> m (because there are 100cm in every 1m)
cm² —- ÷ 100² —–> m²
23 ÷ 100² = 23 ÷ 10000 = 23×10^-4 m² ( = 0.0023m² )
… When converting units of VOLUME, the multiplying or dividing factor is CUBED.
So 61cm³ in mm³ is going Up (mUltiply):
cm —- x 10 —–> mm
cm³ —- x 10³ —–> mm³
61 × 10³ = 61 × 1000 = 61×10^3 ( = 61000mm³ )
… How would you convert 500cm³ into m³?
… Do check the initial scale factor by trying an example and seeing if it feels right.
… Rounding to significant figures, e.g. 3.14159 rounded to 4 significant figures is 3.142 (count the first NON-ZERO figure as the first significant figure).
… The number 0.650 would have been rounded to 3 significant figures (because otherwise we wouldn’t need to put the ‘0’ on the end).
However, the number 650 is unclear… It could be rounded to 3 sig figs or 2 sig figs!
… What is £6789 rounded to 2 significant figures?
(The second significant figure is 7 so we are rounding to the nearest hundred)
… Estimating decimal calculations by rounding each term to 1 significant figure.
For example, estimate (42.4 × 3.81) / 0.51…
(40 × 4) / 0.5
160/0.5
320/1
= 320
The ‘approximately equal to’ sign is ≈
The ‘exactly equal to’ sign is =
… Rounding to decimal places is a little easier, for example:
What is 267.895 rounded to 1 decimal place?
Here, the first decimal place is just after the 8, so we ‘look over’ and see that the 9 is strong enough to round the 8 up to a 9, so we get:
267.895 ≈ 267.9 (1 d.p.)
where “≈” means “approximately equal to”.
… Take care with 9s that round up, for example:
What is 35.895 rounded to 2 decimal places?
Here, the 2nd decimal place is just after the 9 and we see that the 5 is ‘strong’ enough to round the 9 up to 10… but this ‘spills’ into the tenths column. So we get:
35.895 ≈ 35.90 (2 d.p.)
notice how we have kept the 0 at the end!
… Decimals that have been truncated have had numbers removed WITHOUT rounding up or down. We can use inequalities to show what the original number may have been. For example:
6.84 has been truncated to 2 decimal places.
The original number, x, could have been:
6.84 ≤ x < 6.85
For 6.84 we might have had “0000…” after the “4”, i.e. 6.840000… which truncates to 6.84
For 6.85 we might have had “99999…” after the “4”, i.e. 6.849999… which also truncates to 6.84
… Using ‘chunking’ to divide large numbers. E.g. Work out 2700 ÷ 25 using this method.
… Using a bus stop for small number divisors (the number doing the dividing). For example 57 ÷ 4. If you need to work out decimals, put in the decimal place and add zeros as required.
… Multiplying decimals. First get rid of the decimal place by multiplying the number by 10, 100 or 1000. You may then have to do a long multiplication. Remember to divide your answer by 10, 100 or 1000 at the end.
… Dividing decimals. Put the division into a fraction form. Multiply the top and bottom of the fraction by 10, 100 or some other easy number so that the decimal point ‘disappears’. Then divide using a bus stop, chunking or by cancelling down the fraction. (There is no need to divide by 10, 100 at the end because you have kept the fraction the same)
… When working on problem solving type questions involving many calculation stages, try to add brief descriptive labels after each calculation. This will help your working to be clearer and so you will make fewer mistakes. Putting in the correct units will also help.
… Inequalities on a number line. A filled-in dot means…? An ’empty’ dot means?
How would you represent the inequality x > -2 and x ≤ 6 on a number line?
… Plotting graphical inequalities and finding a region on the graph.
- First plot the lines of each inequality of the graph (treat them as equations).
- A dashed line represents an inequality which is < or >.
- A solid line represents ≤ or ≥ (more ink used!)
- Use each inequality to work out which side of the line you want.
- Finally, find the region which works for all the inequalities.
So, if y > 4x + 3, we would first draw the dashed line y = 4x + 3.
Then we would need the region ABOVE the line (because y > … )
LCM and HCF using a Venn diagram
… Find the “product of prime factors” of two numbers by using prime factor ‘trees’.
… Then put these numbers into a Venn Diagram to find the LCM and the HCF.
For example, if 68 = 17 x 2 x 2 and 100 = 5 x 5 x 2 x 2
- Two ‘2’s are common to both and so we write ‘2’ and ‘2’ in the intersection part of the Venn diagram.
- the ’17’ goes in the 68 circle
- the ‘5’ and ‘5’ go into the 100 circle.
The HCF is the product of the intersection part of the diagram = 2 x 2 = 4
The LCM is the product of ALL the numbers in the diagram = 17 x 2 x 2 x 5 x 5 = 1700
Recurring decimals
… How to write recurring decimals as fractions, for example:
What is 0.234343434… expressed as a fraction in its lowest form?
x = 0.234343434
100x = 23.4343434…
x = 0.234343434…
—————————— now subtracting:
99x = 23.2
x = 23.2/99 = 232/990 = 166/495
… If you need to write a fraction as a recurring decimal, use a bus stop division. For example,
Write 1/6 as a recurring decimal:
6 ⟌1.000
= 0.1666… (so this would be 0.16 with a dot over the 6)
… Terminating decimals (or ‘terminal decimals’) are numbers which are not recurring. For example, 3.45, or 1.
… Easy recurring decimals like 0.878787 can be converted to a fraction using the ‘9’ trick. Match the number of digits in the recurring pattern with 9, 99 or 999 in the denominator:
0.878787… = 87/99 = 29/33
… If the sum of a number’s digits are divisible by 3, then the number itself will be divisible by 3. For example:
12342
sum of digits = 12
So 12342 is divisible by 3
… Estimating square roots by finding the nearest square number.
For example, estimate the square root of 60:
Root(60) is nearly the same as root(64) = 8
How to work out square roots without a calculator (usually accurate to about 2dp). This is a trick not usually taught at GCSE, but very useful to know!
√X = √S + (X-S)/(2√S)
X = the number you want the square root of
S = the closest square number you know to X
Let’s try it on the square root of 75.
X=75 and so the nearest square number S = 81. This means that √S = 9.
Put the bits in the formula:
√75 = 9 + (75-81)/(2*9)
√75 = 9 + -6/18 = 9 – 0.333 = 8.667
Upper and Lower Bounds, Error Intervals, Truncation
… A mass m, of 67kg is correct to the nearest kilogram. What is the greatest and least mass this could be?
What would be the error interval for m?
… A time t, of 450 seconds was measured correct to 2 significant figures. What is the greatest and least possible values of this measurement?
What would be the error interval for t?
… If a length of 2.4cm is measured to the nearest 0.1cm, then the lower bound would be 2.35cm. The upper bound would be?
… A length of 35mm is measured to the nearest 5mm. What is the greatest possible value of the measurement?
… A height, h, is measured as being 168.4cm truncated to 1 decimal place. What is the error interval for h?
Remember that ‘truncated’ means that we DON’T round. So in this example, 168.49999… cm will be truncated to 168.4cm.
… Take care when finding (for example) the upper bound of a calculation – this can be sneaky!
For example, speed = distance / time
So to find an upper bound for calculating speed, you would need to use the …?… bound for distance, and the …?… bound for time.
… If you are asked to calculate a quantity to “a suitable degree of accuracy”, work out the upper and lower bounds for the quantity and then investigate how many decimal places you could quote your answer to (which could include the upper and lower bound values as part of ITS lower and upper bound!)
Significant figures and decimal places
… The first non-zero number is the ‘first significant figure’.
… If the question asks you (for example) to round the answer to 2 significant figures (or decimal places), then work to 3 significant figures or decimal places in your working out. This avoids making rounding errors during the calculations.
Standard form
… Always starts with one digit, then the decimal place (followed by any other numbers) x ten to the power of a positive or negative integer (whole number)
… What is 25300 written in standard form?
… What is 0.00876 written in standard form?
… When multiplying standard form numbers, rearrange the numbers (multiplication can be done in any order) so that you multiply the ‘normal numbers’ together and the powers of ten together. For example:
(1.2 × 10^3) × (3 × 10^6)
1.2 × 3 × 10^3 × 10^6
3.6 × 10^9
So the original brackets are actually a red herring to try and confuse you!
… A similar method can be used for when dividing standard form numbers. For example:
(8 × 10^12) / (2 × 10^4)
= 8/2 × 10^12 / 10^4
= 4 × 10^8
Please avoid writing out lots of zeroes!
Boost your GCSE physics at 