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

(a) The greatest lower bound is half of the unit below the number.

as the unit = 0.1m, half the unit = 0.05m

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.)

Note the keyword here ESTIMATE. In an estimation round all numbers to 1 significant figure.

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

380km expressed to 1 s.f. = 400km

3.  Show how you would estimate the answer to the following expression without using a calculator:

9.75 + 30.2
  0.2 x 48

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) 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)?

(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 volume of juice required using these numbers.

Volume = Area of Square Base x Height

= 6.75 x 6.75 x 16.35

= 744.95cm³ (to 2 d.p.)

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.

(a) What is the maximum area of the room?
(b) What is the minimum area of the room?

Here 10cm = 0.1m, make sure that you always do calculations in one unit (we will use metres here)

(a) Area = length x width

= 10.85 x 8.25 = 89.5m² (to 1 d.p.)

(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.

260m accurate to the nearest 5m can be anywhere between 257.5 m (260 - 2.5) and 262.5 m (260 + 2.5).

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