standard form - gcse-maths

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STANDARD FORM
TIP: When doing these sort of problems, remember:
a) Standard form is simply a way of writing very large and
very small numbers without having to write loads of numbers.
b) Here is a number in standard form:
6.3 x 108
Note:
The part before the multiplication sign (6.3 in our example)
can only have one digit (number) before the decimal point.
The part after the multiplication sign (108 in our example) is
always 10 to the power a number.
c) To convert decimal numbers less than 1 to standard form:
e.g 0.0082 = 8.2/1000 = 8.2 x 1/1000 = 8.2 x 1/103 = 8.2 x 10-3
Do the same in reverse to convert a standard form number
back to a decimal one.
d) When converting numbers greater than one, use the same
technique but multiply, instead of divide.
Solve the following:
1. Write 0.000008 in standard form.
2. Write 453000000 in standard form
3. Change 5.2 x 108 to an ordinary
decimal number.
4. Change 1.65 x 10-5 to an ordinary decimal number.
5. What is
(2.34 x 103) x (5.0 x 106) ?
6. The diameter of the Earth is;
1.27 x 104 km
The diameter of Mars is:
6.79 x 103 km
What's the difference between the two diameters?
7. The mass of an electron is:
0.000 000 000 000 000 000 000 000 91 grams.
How could this mass be written in standard form?
What is the mass of 8,000,000 electrons?
8. The distance between the Earth and the star Sirius is:
81,900,000,000,000 km
If light travels 3 x 105 km per second, how many seconds does it take for light to travel from
Sirius to the Earth?
(give your answer to 2 significant figures)
1. Write 0.000008 in
standard form.
Quick way to do this - count the 0's between the decimal point
and the first non-zero digit and add 1 - this give us
5+1=6
Now, write the number as 8.0 x 10-6
(the -6 in the power is the 6 you've just worked out with a minus
sign in front of it)
2. Write 453000000 in
standard form
Count all the digits after the first number to get the power.
= 4.53 x 108
3. Change 5.2 x 108 to an
ordinary
decimal number.
520,000,000
4. Change 1.65 x 10-5 to an 0.0000165
ordinary decimal number.
5. What is
= (2.34 x 5) x (103 x 106)
3
6
(2.34 x 10 ) x (5.0 x 10 ) ?
Using the rules for indices:
= 11.7 x (103+6)
= 11.7 x 109
6. The diameter of the
Earth is;
1.27 x 104 km
Do this the easiest way as they're quite small numbers, convert
back to ordinary decimal numbers to do the subtraction.
12700 - 6790 = 5910
The diameter of Mars is:
6.79 x 103 km
= 5.91 x 103 km
(using the ordinary decimal numbers it's easy to see that the
What's the difference
between the two diameters? Earth's diameter is almost twice that of Mars - not so easy to see
with standard form)
7. The mass of an electron 9.1 x 10-25
is:
0.000 000 000 000 000 000 8,000,000 = 8 x 106
000 000 91 grams.
= (9.1 x 10-25) x (8 x 106)
How could this mass be
written in standard form?
= (9.1 x 8) x (10-25 x 106)
What is the mass of
8,000,000 electrons?
= 72.8 x (10-25+6)
= 72.8 x 10-19 grams
= 7.28 x 10-18 grams (we're only allowed 1 digit in front of the
decimal point in standard form)
8. The distance between the Distance = 8.19 x 1013 km
Earth and the star Sirius is:
81,900,000,000,000 km
Speed = Distance / Time, so a bit of algebraic rearrangement
gives us:
If light travels 3 x 105 km
per second, how many
Time = Distance / Speed
seconds does it take for
light to travel from Sirius to = (8.19 x 1013) / (3 x 105)
the Earth?
= (8.19 / 3) x ( 1013 - 105)
(give your answer to 2
significant figures)
= 2.73 x 1013-5
= 2.73 x 108 seconds
= 2.7 x 108 seconds (to 2 significant figures)
ESTIMATION AND APPROXIMATION
TIP: When doing these sort of problems, remember:
a) When you round a number to a number of significant figures
(e.g. 2 s.f., 3 s.f. etc.), you start counting the significant
figures as any which are not 0 from the left.
e.g. 0.000628314 to 2 s.f. is 0.00063
8793 to 3 s.f. is 8790
b) When you round a number to a number of decimal places (2
d.p. 3 d.p. etc.), you coutn the number of decimal places you
need and look at the next digit to the right. If this digit is less
than 5 round down, if its 5 or more round up.
e.g. 23.65789 to 2 .d.p. is 23.66
c) Things can be measured with inaccurate equipment, or a
degree of tolerance is allowed in measuring. If you measure
something which is correct to a given unit (where units could
be cm, mm, kg etc.) the true value could be anywhere in a
range between half a unit below (called the greatest lower
bound) and half a unit above (called the least upper bound).
e.g. a line measured as 156cm correct to the nearest cm
could really be anything between 155.5 cm and 156.5
cm.
d) To estimate answers round all numbers to 1 significant
figure.
e) Be very wary of exam questions - remember to re-read the
end of the question after working out the answer. The final
part of the question will often say things like "give your answer
to 3 s.f." etc. Normally a mark is given just for remembering
to do this!
Solve the following:
1. A rectangular carpet has a length of 6.4 m and a width of 3.5 m, where each measurement is
measured to the nearest 0.1m.
Calculate:
(a) The greatest lower bounds of the length and width
(b) The least upper bounds of the length and width
(c) The maximum area
(d) The minimum area
2. There are approximately 1.853 km in a nautical mile. Estimate how many kilometres there are
in 190 nautical miles giving your answer to 1 s.f.
3. Show how you would estimate the answer to the following expression without using a
calculator:
9.75 + 30.2
0.2 x 48
4. Perform the calculation 12.657 x 8.972
(a) Give your answer correct to 3 d.p.
(b) Give your answer correct to 2 d.p.
(c) Give your answer correct to 1 d.p.
(d) Give your answer correct to 3 s.f.
5. In a 1500 metre race a runner's time was calculated to be 4 minutes 12.43 seconds. If race times
are measured to the nearest 0.01 seconds, write down the range of times between which the
runner's exact time lies.
6. A carton of orange juice has a square base where each side is 6.7 cm and a height of 16.3 cm.
The measurements are correct to an accuracy of 1 decimal place.
(a) If a factory needs to make 3,000 of these cartons in a production run, how much juice must be
available to be sure of filling all the cartons?
(b) What is the maximum amount of cartons that could be made from the total amount of juice in
your answer to (a)?
7. A brochure in the estate agents gives the measurements of the sitting room in a house as 10.8m
by 8.2m. The measurements are taken to the nearest 10cm.
(a) What is the maximum area of the room?
(b) What is the minimum area of the room?
8. A rectangular field is 260m by 180m. If the measurements are taken to the nearest 5 metres,
what is the maximum perimeter of the field?
ESTIMATION AND APPROXIMATION - ANSWERS
1. A rectangular carpet has a length of (a) The greatest lower bound is half of the unit
6.4 m and a width of 3.5 m, where each below the number.
measurement is measured to the
nearest 0.1m.
as the unit = 0.1m, half the unit = 0.05m
Calculate:
(a) The greatest lower bounds of the
length and width
(b) The least upper bounds of the
length and width
(c) The maximum area
(d) The minimum area
So we have 6.4 - 0.05 = 6.35m
and 3.5 - 0.05 = 3.45m
(b) The least upper bound is half the unit above
the number.
giving: 6.4 + 0.05 = 6.45m
and 3.5 + 0.05 = 3.55m
(c) The area = length x width
maximum area is when the length and width are
at their maximum value (i.e. the answers to (b)):
giving:
Maximum Area = 6.45 x 3.55 = 22.90m2 (to 2
d.p.)
(d) Minimum Area = 6.35 x 3.45 = 21.91m2 (to
2 d.p.)
2. There are approximately 1.853 km Note the keyword here ESTIMATE. In an
in a nautical mile. Estimate how many estimation round all numbers to 1 significant
kilometres there are in 190 nautical
figure.
miles giving your answer to 1 s.f.
1.853 rounded to 1 s.f. = 2
So we can estimate that there are 2km to each
nautical mile.
Hence in 190 nautical miles we get 190 x 2 =
380km
3. Show how you would estimate the
answer to the following expression
without using a calculator:
9.75 + 30.2
0.2 x 48
380km expressed to 1 s.f. = 400km
Note that keyword again...it's the same old thing
giving you:
10 + 30
0.2 x 50
which = 40/10 = 4
4. Perform the calculation 12.657 x
8.972
(a) Give your answer correct to 3 d.p.
(b) Give your answer correct to 2 d.p.
(c) Give your answer correct to 1 d.p.
(d) Give your answer correct to 3 s.f.
The answer to this calculation done on a
calculator is:
111.280344.
(a) Look at the next number after the 3 decimal
places - this is 3. As this number is less than 5
we round down, giving:
111.280
(b) Same again with 2 decimal places, we round
down again, giving:
111.28
(c) Same again with 1 decimal place - though we
need to round up here (because .28 is closer to
0.3 than it is to 0.2), so we get:
111.3
(d) Count the first three non-zero numbers from
the left, giving:
111
5. In a 1500 metre race a runner's time
was calculated to be 4 minutes 12.43
seconds. If race times are measured to
the nearest 0.01 seconds, write down
the range of times between which the
runner's exact time lies.
Here the unit is 0.01 seconds so half a unit is
0.005 seconds.
The maximum possible time (or least upper
bound) is: 12.43 + 0.005 = 12.435
The minimum possible time (or greatest lower
bound) is:
12.43 - 0.005 = 12.425
So, the range is between 12.425 and 12.435.
6. A carton of orange juice has a
square base where each side is 6.7 cm
and a height of 16.3 cm. The
measurements are correct to an
accuracy of 1 decimal place.
(a) The maximum side of the base = 6.75cm
The maximum height = 16.35cm
If it turns out that the accurate measurement of
the carton is the maximum (given above), then to
fill all the cartons we need to work out the
(a) If a factory needs to make 3,000 of volume of juice required using these numbers.
these cartons in a production run, how
much juice must be available to be sure Volume = Area of Square Base x Height
of filling all the cartons?
= 6.75 x 6.75 x 16.35
(b) What is the maximum amount of
cartons that could be made from the
= 744.95cm3 (to 2 d.p.)
total amount of juice in your answer to
(a)?
But, we need 3,000 times this volume to fill
3,000 cartons:
= 3,000 x 744.95 = 2, 234, 840.63 cm3 (to 2
d.p.)
(b) To fill the most cartons from 2,234,840.63,
the cartons will need to have the minimum side
of base = 6.65 and height = 16.25
Each carton made with the minimum size has a
volume of
6.65 x 6.65 x 16.25 = 718.62 cm3 (to 2 d.p.)
Now divide the total volume of juice by the
volume of each carton = 2,234,840.63 / 718.62
= 3,110 cartons (we can't round up in this
question as we have a limited amount of juice)
NOTE: Although each step has been written
down to 2 d.p. to make the numbers clearer, each
step was worked out with the accurate number
calculated on the calculator. Only the final
answer was rounded in the calculation.
7. A brochure in the estate agents
gives the measurements of the sitting
room in a house as 10.8m by 8.2m.
The measurements are taken to the
nearest 10cm.
Here 10cm = 0.1m, make sure that you always
do calculations in one unit (we will use metres
here)
(a) What is the maximum area of the
room?
(b) What is the minimum area of the
room?
= 10.85 x 8.25 = 89.5m2 (to 1 d.p.)
(a) Area = length x width
(b) Area = 10.75 x 8.15 = 87.6m2 ( to 1 d.p.)
Why do we give the answer to 1 d.p.? - we
weren't told to do this.
If you are not told how to provide your answer it
is best to give your answer to the same accuracy
as the other numbers in the problem.
8. A rectangular field is 260m by
260m accurate to the nearest 5m can be
180m. If the measurements are taken to anywhere between 257.5 m (260 - 2.5) and 262.5
the nearest 5 metres, what is the
m (260 + 2.5).
maximum perimeter of the field?
180m lies between 177.5m and 182.5m
The perimeter is the distance all the way round
the rectangle which is twice the length + twice
the width
The maximum perimeter = (2 x 262.5) + (2 x
182.5)
= 525 + 365 = 890m
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