Estimating

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Estimating Revision

Estimating

The way we estimate answers to calculations is simple – we round every number involved to 1 significant figure, unless stated otherwise, and then perform the calculation with those numbers instead.

Make sure you are happy with the following topics before continuing.

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Type 1: Simple Estimating

These types of questions are the easiest you will see.

Example: Estimate the answer to \dfrac{8.21}{3.97} \times 31.59.

Step 1: Round each number to 1 significant figure:

8.21 rounds to 8,

3.97 rounds to 4,

31.59 rounds to 30.

Step 2: Put the rounded numbers into the equation and calculate:

\dfrac{8.21}{3.97} \times 31.59 \approx \dfrac{8}{4} \times 30 = 2 \times 30 = 60.

Note: The \approx symbol means “approximately equal to”.

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Type 2: Estimating with Equations

Estimating with equations is a little bit more difficult, since we also have to interpret the question.

Example: The formula for the force, F on a moving object is F = ma, where m is the mass and a is the acceleration.

Estimate the force on an object which has mass 5.87 kg and acceleration 24.02 m/s^2.

Step 1: Round the numbers in the question to 1 significant figure:

5.87 rounds to 6,

24.02 rounds to 20.

Step 2: Put the rounded numbers into the equation and calculate:

\text{Force } = 5.87 \times 24.02 \approx 6 \times 20 = 120

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Type 3: Estimating Square Roots

Estimating square roots is the hardest type of estimating question you will see, and is only for HIGHER students.

Example: Find an estimate for \sqrt{40}.

The square root of 40 will be some number that we can square to make 40.

Step 1: Find 2 square numbers, one on each side of the number we are given:

We know that

6^2 = 36 and 7^2 = 49

So, the answer must fall somewhere between 6 and 7.

Step 2: Choose an estimate based on which square number it is closest to:

Since 40 is 4 away from 36 but 9 away from 49, we can conclude the answer will be somewhat closer to 6.

Therefore, 6.3 is a suitable estimate for \sqrt{40}.

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Estimating Example Questions

Round each number to 1 significant figure:

 

9.02 rounds to 9,

 6.65 rounds to 7,

0.042 rounds to 0.04,

11 rounds to 10.

 

Therefore we get,

 

\dfrac{9.02 + 6.65}{0.042 \times 11} \approx \dfrac{9 + 7}{0.04 \times 10} = \dfrac{16}{0.4}

 

To make this division easier, multiply the top and bottom of the fraction by ten, to find

 

\dfrac{16}{0.4} = \dfrac{160}{4} = 40

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Rounding each number to 1 significant figure:

 

57.33 rounds to 60
29.88 rounds to 30
8.66 rounds to 9
5.55 rounds to 6

Therefore, we get:

 

\dfrac{57.33-29.88}{8.66-5.55}\approx\dfrac{60-30}{9-6}=\dfrac{30}{3}=10

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Because the answer needs to be in pounds, we should turn the cost of the pencils into pounds first.

45p = \pounds0.45

Now we can start estimating.

1.89 rounds to 2
0.45 rounds to 0.5

 

And now we need to multiply these amounts by how many of each he wanted.

 

\textrm{(Pens) }\pounds2\times5=\pounds10
\textrm{(Pencils) }\pounds0.50\times3=\pounds1.50

 

And now all we need to do is add them together.

 

\pounds10+\pounds1.50=\pounds11.50

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Round each number to 1 significant figure:

 

32.60 rounds to 30,

 

17.50 rounds to 20,

 

Therefore, the approximate cost of the 3 child tickets is 3 \times 20 = \pounds 60.

 

The approximate cost of the 2 adult tickets is 2 \times 30 = \pounds 60.

 

Thus, the approximate total cost is 60 + 60 = \pounds 120.

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First, we need to find 2 square numbers either side of 98.

 

We know that

9^2 = 81 and 10^2 = 100

So the answer must be between 9 and 10.

Since 98 is only 2 away from 100, but 17 away from 81, we can conclude that the solution is going to be much closer to 10.

Therefore, the estimate is

\sqrt{98} \approx 9.9

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