Crescent Girls’ School Mathematics Department

UPPER AND LOWER BOUND

Qn 4 from 2012 Sec 3 E Math Nov Hol HW

Fred used a ruler to measure the length and width of a rectangular postcard and they are 8 cm and 5 cm respectively, correct to the nearest cm. Calculate

(a) the smallest possible perimeter,

(b) the largest possible area.

Answer: (a) 24 cm (b) 46.75 cm2

GCE 2011 P1, Q 12

A glass block has a mass of 43 grams, correct to the nearest gram.

(a) Find the least possible mass of the glass block.

(b) The volume of the glass block is 15 cm3, correct to the nearest cubic centimeter.

Find the greatest possible mass of 1 cubic centimeter of the glass.

Cambridge Examiner’s Report:

A fully correct solution to this question was extremely rare.

(a) The correct answer was given by a significant number, but an answer of 40 was not uncommon.

(b) Very few candidates appreciated the need to divide the greatest mass by the least volume and indeed the majority of candidates did not realize that this part of the question had any

connection with part (a) and upper and lower bounds, and simply calculated 4315

. Of those

who did see some connection, most used the bound of the numerator only i.e. 43.415

or 43.515

.

Those who did see the need to work with the denominator as well often used 43.515.5

rather than

the correct 43.514.5

. Weaker candidates often started the question by calculating 3√15

Answers: (a) 42.5 g; (b) 3 g

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Crescent Girls’ School Mathematics Department

ABSOLUTE ERROR

Worked Example

Measured to the nearest 0.1 cm, the lengths of the sides of a rectangular carpet are 7.5 cm and 3.4 cm. Find the greatest and least possible values of the area of the rectangular carpet and the largest possible error in the area.

Solution

The largest possible absolute error = = ___________ cm

∴ 7.5 cm lies between _________________ cm. And 3.4 cm lies between __________________ cm.

The greatest area = _________ ______________ = _____________ cm2

The smallest area = ___________ ____________ = _______________ cm2

The largest possible error in the area = ______________

= ___________ cm2

Solution

The largest possible absolute error = 12×0.1 = 0.05 cm

∴ 7.5 cm lies between 7.5 ± 0.05 cm. And 3.4 cm lies between 3.4 ± 0.05 cm.

The greatest area = 7.55 3.45 = 26.0475 cm2

The smallest area = 7.45 3.35 = 24.9575 cm2

The largest possible error in the area = 26.0475 – 24.9575

= 1.09 cm2

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Present in this manner!