# 8.2 Estimation by clustering

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This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses how to estimate by clustering. By the end of the module students should understand the concept of clustering and be able to estimate the result of adding more than two numbers when clustering occurs using the clustering technique.

## Section overview

• Estimation by Clustering

## Cluster

When more than two numbers are to be added, the sum may be estimated using the clustering technique. The rounding technique could also be used, but if several of the numbers are seen to cluster (are seen to be close to) one particular number, the clustering technique provides a quicker estimate. Consider a sum such as

$\text{32}+\text{68}+\text{29}+\text{73}$

Notice two things:

1. There are more than two numbers to be added.
2. Clustering occurs.
1. Both 68 and 73 cluster around 70, so $\text{68}+\text{73}$ is close to $\text{80}+\text{70}=2\left(\text{70}\right)=\text{140}$ .
2. Both 32 and 29 cluster around 30, so $\text{32}+\text{29}$ is close to $\text{30}+\text{30}=2\left(\text{30}\right)=\text{60}$ .

The sum may be estimated by

$\begin{array}{ccc}\hfill \left(2\cdot 30\right)+\left(2\cdot \text{70}\right)& =& 6+140\hfill \\ & =& \text{200}\hfill \end{array}$

In fact, $\text{32}+\text{68}+\text{29}+\text{73}=\text{202}$ .

## Sample set a

Estimate each sum. Results may vary.

$\text{27}+\text{48}+\text{31}+\text{52}$ .

27 and 31 cluster near 30. Their sum is about $\text{2}\cdot \text{30}=\text{60}$ .

48 and 52 cluster near 50. Their sum is about $\text{2}\cdot \text{50}=\text{100}$ .

Thus, $\text{27}+\text{48}+\text{31}+5\text{2}$ is about $\begin{array}{ccc}\left(2\cdot 30\right)+\left(2\cdot 50\right)& =& 60+100\hfill \\ & =& 160\hfill \end{array}$

In fact, $\text{27}+\text{48}+\text{31}+\text{52}=\text{158}$ .

$\text{88}+\text{21}+\text{19}+\text{91}$ .

88 and 91 cluster near 90. Their sum is about $\text{2}\cdot \text{90}=\text{180}$ .

21 and 19 cluster near 20. Their sum is about $\text{2}\cdot \text{20}=\text{40}$ .

Thus, $\text{88}+\text{21}+\text{19}+\text{91}$ is about $\begin{array}{ccc}\left(2\cdot 90\right)+\left(2\cdot 20\right)& =& 180+40\hfill \\ & =& 220\hfill \end{array}$

In fact, $\text{88}+\text{21}+\text{19}+\text{91}=\text{219}$ .

$\text{17}+\text{21}+\text{48}+\text{18}$ .

17, 21, and 18 cluster near 20. Their sum is about $\text{3}\cdot \text{20}=\text{60}$ .

Thus, $\text{17}+\text{21}+\text{48}+\text{18}$ is about $\begin{array}{ccc}\left(3\cdot 20\right)+50& =& 60+50\hfill \\ & =& 110\hfill \end{array}$

In fact, $\text{17}+\text{21}+\text{48}+\text{18}=\text{104}$ .

$\text{61}+\text{48}+\text{49}+\text{57}+\text{52}$ .

61 and 57 cluster near 60. Their sum is about $\text{2}\cdot \text{60}=\text{120}$ .

48, 49, and 52 cluster near 50. Their sum is about $3\cdot \text{50}=\text{150}$ .

Thus, $\text{61}+\text{48}+\text{49}+\text{57}+\text{52}$ is about $\begin{array}{ccc}\left(2\cdot 60\right)+\left(3\cdot 50\right)& =& 120+150\hfill \\ & =& 270\hfill \end{array}$

In fact, $\text{61}+\text{48}+\text{49}+\text{57}+\text{52}=\text{267}$ .

$\text{706}+\text{321}+\text{293}+\text{684}$ .

706 and 684 cluster near 700. Their sum is about $\text{2}\cdot \text{700}=\text{1,400}$ .

321 and 293 cluster near 300. Their sum is about $\text{2}\cdot \text{300}=\text{600}$ .

Thus, $\text{706}+\text{321}+\text{293}+\text{684}$ is about $\begin{array}{ccc}\left(2\cdot 700\right)+\left(2\cdot 300\right)& =& 1,400+600\hfill \\ & =& 2,000\hfill \end{array}$

In fact, $\text{706}+\text{321}+\text{293}+\text{684}=\text{2,004}$ .

## Practice set a

Use the clustering method to estimate each sum.

$\text{28}+\text{51}+\text{31}+\text{47}$

$\left(2\cdot 30\right)+\left(2\cdot 50\right)=60+100=160$

$\text{42}+\text{39}+\text{68}+\text{41}$

$\left(3\cdot 40\right)+70=120+70=190$

$\text{37}+\text{39}+\text{83}+\text{42}+\text{79}$

$\left(3\cdot 40\right)+\left(2\cdot 80\right)=120+160=280$

$\text{612}+\text{585}+\text{830}+\text{794}$

$\left(2\cdot 600\right)+\left(2\cdot 800\right)=1,200+1,600=2,800$

## Exercises

Use the clustering method to estimate each sum. Results may vary.

$\text{28}+\text{51}+\text{31}+\text{47}$

$\text{42}+\text{19}+\text{39}+\text{23}$

$\text{88}+\text{62}+\text{59}+\text{90}$

$\text{76}+\text{29}+\text{33}+\text{82}$

$\text{19}+\text{23}+\text{87}+\text{21}$

$\text{41}+\text{28}+\text{42}+\text{37}$

$\text{89}+\text{32}+\text{89}+\text{93}$

$\text{73}+\text{72}+\text{27}+\text{71}$

$\text{43}+\text{62}+\text{61}+\text{55}$

$\text{31}+\text{77}+\text{31}+\text{27}$

$\text{57}+\text{34}+\text{28}+\text{61}+\text{62}$

$\text{94}+\text{18}+\text{23}+\text{91}+\text{19}$

$\text{103}+\text{72}+\text{66}+\text{97}+\text{99}$

$\text{42}+\text{121}+\text{119}+\text{124}+\text{41}$

$\text{19}+\text{24}+\text{87}+\text{23}+\text{91}+\text{93}$

$\text{108}+\text{61}+\text{63}+\text{96}+\text{57}+\text{99}$

$\text{518}+\text{721}+\text{493}+\text{689}$

$\text{981}+\text{1208}+\text{1214}+\text{1006}$

$\text{23}+\text{81}+\text{77}+\text{79}+\text{19}+\text{81}$

$\text{94}+\text{68}+\text{66}+\text{101}+\text{106}+\text{71}+\text{110}$

## Exercises for review

( [link] ) Specify all the digits greater than 6.

7, 8, 9

( [link] ) Find the product: $\frac{2}{3}\cdot \frac{9}{\text{14}}\cdot \frac{7}{\text{12}}$ .

( [link] ) Convert 0.06 to a fraction.

$\frac{3}{\text{50}}$

( [link] ) Write the proportion in fractional form: "5 is to 8 as 25 is to 40."

( [link] ) Estimate the sum using the method of rounding: $\text{4,882}+\text{2,704}$ .

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