The Robert Smyth School
Mathematics Faculty
Topic 8
Approximations HW2
Innovation & excellence
Topic 8 (higher) – Homework on Approximations
1.
A notice board is a rectangle with a length of 80 cm and a width of 40 cm.
Both measurements are correct to the nearest centimetre.
80 cm
40 cm
(a)
Not drawn
accurately
What is the least possible length of the notice board?
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(1)
(b)
What is the greatest possible width of the notice board?
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(1)
(c)
What is the greatest possible area of the notice board?
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(3)
2.
A field is 50 m in width and 110 m in length. The width is given correct to the nearest 5 metres.
The length is given correct to the nearest 10 metres. Find the maximum area of the field.
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Answer ............................................
(4)
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Mathematics Faculty
3.
Innovation & excellence
Topic 8
Approximations HW2
A stop-watch records the time for the winner of a 100 metre race as 10.4 seconds, measured to
the nearest one tenth of a second.
(a)
What are the greatest and least possible times for the winner ?
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(2)
(b)
The length of the 100 metre track is correct to the nearest 10 cm.
What are the greatest and least possible lengths of the track ?
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(2)
(c)
What is the fastest possible average speed of the winner clocking 10.4 s in the 100 m
race?
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(2)
4.
p, q and r are three continuous measures.
p = 5.3 to an accuracy of two significant figures.
q = 0.64 and r = 0.64, each to an accuracy of two decimal places.
(a)
(i)
Calculate the lower bound of p – q.
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Answer .........................................................
(2)
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(ii)
Innovation & excellence
Calculate the upper bound of
Topic 8
Approximations HW2
pq
.
r
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Answer .........................................................
(2)
T is an integer where T = 50 to the nearest 5.
(b)
Calculate the largest possible value of Tp.
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Answer .........................................................
(2)
5.
A reservoir, in the shape of a cuboid, is 190 metres long and 80 metres wide.
On one day 3.7 inches of rain fell on the reservoir.
These measurements are all given correct to two significant figures.
What is the maximum possible volume of rain which could have fallen on the reservoir during
that day?
Give your answer in cubic metres correct to four significant figures.
Use 2.54 centimetres to be exactly 1 inch.
The volume of a cuboid is length x width x depth.
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Answer ........................................... cubic metres
(Total 6 marks)
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6.
Innovation & excellence
Topic 8
Approximations HW2
The measurements of a rectangular microchip are given as length 12 mm and width 7 mm,
correct to the nearest millimetre.
(a)
Between what limits must the length of the microchip lie?
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(2)
(b)
Between what limits must the area of the microchip lie?
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(3)
(c)
The area is given as (84  x) mm .
2
Suggest a suitable value for x.
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(2)
7.
The sides of a rectangle are measured as 7.00 m and 5.00 m.
(a)
Calculate the greatest possible value of the perimeter.
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(2)
(b)
Calculate the least possible value of the area.
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(2)
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The Robert Smyth School
Mathematics Faculty
8.
Innovation & excellence
Topic 8
Approximations HW2
The area of a rectangle is 54.4 square centimetres correct to 1 decimal place.
The length of this rectangle is 8.3 centimetres correct to 1 decimal place.
(a)
From this information, write down
(i)
the largest value that the area of the rectangle could have;
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(ii)
the smallest value that the length of the rectangle could have.
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(2)
(b)
Use your answers in (a) to calculate the largest possible width of the rectangle.
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(2)
9.
The volume of a cylinder is given as 880 ml correct to 2 significant figures.
John measures its height as 11.2 cm (to the nearest mm).
Between what limits must the radius of the cylinder lie?
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(7)
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Mathematics Faculty
10.
Innovation & excellence
Topic 8
Approximations HW2
A crane has a cable with a breaking strain of 5300 kg measured to 2 significant figures.
It is used to lift crates which weigh 100 kg measured to the nearest 10 kg.
What is the greatest number of crates that can be lifted at one time so that the cable does
not break?
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Answer .........................................................................
(Total 4 marks)
11.
In a series of experiments, all measurements are taken correct to 3 significant figures.
In one particular experiment, the results recorded are
v = 24.5, u = 19.2 and s = 115.
Using f 
v 2  u2
, what could have been the maximum value for f?
2s
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(4)
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12.
Innovation & excellence
Topic 8
Approximations HW2
A coffee machine dispenses 130 millilitres of black coffee into cups with a capacity of 175
millilitres.
These values are accurate to 3 significant figures.
Milk is supplied in small cartons which contain 21 millilitres, accurate to the nearest millilitre.
Beryl likes milky coffee and always puts 2 cartons of milk in her coffee.
Will Beryl’s cup ever overflow?
You must show all your working.
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(4)
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The Robert Smyth School
Mathematics Faculty
1.
(a)
(b)
Innovation & excellence
Topic 8
Approximations HW2
79.5 (cm)
40.5 (cm)
B1
B1
Accept 40.499999...with indication of recurrence.
Truncated decimals get B0.
Sc: if “length” and “width” confused (a) will be 39.5 and (b)
will be 80.5 or 80.4999...
Allow B1, B1 for this
(c)
3260.25(cm2)
For greatest x value
M1
A1
For 80.5 × their value
A1 ft
For answer .ft 80.5 × their value (b). Accept 80.4999...
Must be exact. No rounding or truncating, even on ft notations
ending...9999...,are acceptable.
Sc: If “length” and “width” confused as above then M1 for
greatest × greatest
A1 for 40.5 × their value
A1 answer.ft 40.5 × their value (b). Accept 40.4999...
Must be exact. No rounding or truncating, even on fit notations
ending...9999..., are acceptable.
[5]
2.
Max width is 52.5 m
Max length is 115 m
Accept 52.499… and 114.99…
Max area is 52.5 × 115 m2
= 6037.5 m2
[SC 6040
+ unit mark unless 6037.5 seen]
6040m2
Accept 6037m2
B1
M1
A1
B1 (units)
M1 B1
B1 M1 B1
[4]
3.
(a)
greatest
least
10.45s
10.35s
B1
B1
(b)
greatest
least
10005cm
9995cm
B1
B1
(c)
10005/10.35 cm/s
=9.7m/s
M1
A1
[6]
4.
(a)
(i)
5.25 - 0.645
or 5.25 and 0.645 seen
M1
4.605
A1
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5.35  0.645 
(ii)
M1
0.635
or 5.35 + 0.645
9.440...
or 9.4... with method
(b)
Topic 8
Approximations HW2
A1
52 × 5.35
M1
or 52 and 5.35 seen
278.2
A1
or 278 with method
[6]
5.
Max lengths: 195, 80.5 and 3.75
B1 each
Accept 194,99 etc
B2 194.9, 80.49, 3.749
B1 for any two
3.75 inches =
B3
3.75  2.54
m
100
Max Volume = 195 × 80.5 ×
M1
3.75  2.54
100
M1
= 1495
A1 cao
1494 from 194.9 etc M2B2A1
[6]
6.
(a)
11.5 – 12.5
(b)
11.5 x 6.5 – 12.5 x 7.5
A1 A1
M1 M1
74.75 – 93.75
A1
(c)
[7]
7.
(a)
Perimeter = 2(7.005 + 5.005) = 24.02 (cm)
(b)
Area = 6.995 × 4.995 m2
= 34.940025 m2
M1 A1
M1
A1
[4]
8.
(a)
(b)
(i)
54.449 or 54.45
(ii)
8.25
54.45/8.25
6.6
Accept 54.449 or 54.4499 etc.
Reject 54.44
B1
B1
Follow through using candidate’s
answers to(a).
M1
A1
[4]
9.
Max vol 885 ml Min vol 875 ml
Min ht 11.15cm Max ht 11.25cm
Award B1 if 1 out of 2 correct in each case.
B1
B1
Accept 884.9, 884.999 ..., 11.249, 11.24999... as maximum values
Any two of four B1
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885
11.15π
rmax 
Topic 8
Approximations HW2
M1 M1 
correct pair
= 5.03 cm
rmin 
A1 f.t
875
= 4.98 cm
11.25π
Uses both their maxm values etc.
then both minm values
Uses their max. value and 11.2
then their min. value and 11.2
r=
880
112
. 
M1 A1
M1 M0 A0
M1 A1 f.t
M1 M0 A0
M1 A0
M1 A0 only
r = 5 by trial and error
M1 A0 only
[7]
10.
Sight of 5250 or 5350
B1
Sight of 95 or 105
B1
Their correct combination
Min strain 5250
ie.
Max crate 105
M1
= 50
A1
Accept 49, with explanation that 50 would be right on the limit,
hence 49 is the maximum
[4]
11.
f
max =
24.552  19.152
...c
 1030480349
.
2  114.5
24.55
19.15
11.45
1.08
M1
M1
M1
M1
[4]
12.
Upper limit coffee (130.5)
130 .49 , 130.49..........(implying recurrence)
Any truncated values e.g 130.4, 130.499 gets B0.
B1
Lower limit cup (174.5)
B1
Upper limits of both milks 2 × 21.5 = 43
42.9, 42.9......(implying recurrence)
Any truncated values e.g. 42.8, 42.9 gets B0.
B1
No with full justification
e.g 2 × 21.5 = 43 < 44
174.5 > 173.5
Follow through their limits providing fully justified.
Only award if attempt made to find correct limits of cup, coffee
and 2×milk.
DB1
[4]
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