A cricket player scored \[180\] runs in the first match and \[257\] runs in the second match. Find the number of runs he should score in the third match so that the average of runs in the three matches be \[230\].
Answer
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Hint: Here we will be using the formula of calculating the average of any number which is also known as the arithmetic mean and it is calculated by adding a group of numbers given with the count of that particular number. For example, the average of numbers \[2\], \[3\], \[5\] and \[7\] will be equals to as below:
\[{\text{Average}} = \dfrac{{2 + 3 + 5 + 7}}{4}\]. By adding the terms in the numerator, we get:
\[ \Rightarrow {\text{Average}} = \dfrac{{17}}{4}\]
Complete step-by-step solution:
Step 1: Let the runs scored in the first match be denoted as
\[{x_1} = 180\]. The runs scored by the player in the second match are denoted as \[{x_2} = 257\] and the third match runs are denoted as \[{x_3}\] which we need to find.
Step 2: By using the formula of average we get:
\[{\text{Average}} = \dfrac{{{x_1} + {x_2} + {x_3}}}{3}\]
By substituting the value of \[{x_1} = 180\]and
\[{x_2} = 257\] in the above expression we get:
\[ \Rightarrow {\text{Average}} = \dfrac{{180 + 257 + {x_3}}}{3}\]
By substituting the value of average which is
\[230\] in the above expression we get:
\[ \Rightarrow 230 = \dfrac{{180 + 257 + {x_3}}}{3}\]
Step 3: By doing the addition to the RHS side of the above expression we get:
\[ \Rightarrow 230 = \dfrac{{437 + {x_3}}}{3}\]
Bringing \[3\] into the LHS side of the above expression and multiplying it with \[230\], we get:
\[ \Rightarrow 690 = 437 + {x_3}\]
By bringing \[437\] into the LHS side of the above expression and subtracting it from \[690\], we get:
\[ \Rightarrow {x_3} = 253\]
The runs scored by the player in the third match is \[253\].
Note: Students need to remember the basic formulas for calculating the average of the terms. There are some important properties of average as below:
When the difference between all the terms is the same then the average equal to the middle term.
If a particular variable \[x\] is added in all the terms then, the average will also be increased by \[x\]
If a particular variable \[x\] is subtracted in all the terms then, the average will also be decreased by \[x\]
\[{\text{Average}} = \dfrac{{2 + 3 + 5 + 7}}{4}\]. By adding the terms in the numerator, we get:
\[ \Rightarrow {\text{Average}} = \dfrac{{17}}{4}\]
Complete step-by-step solution:
Step 1: Let the runs scored in the first match be denoted as
\[{x_1} = 180\]. The runs scored by the player in the second match are denoted as \[{x_2} = 257\] and the third match runs are denoted as \[{x_3}\] which we need to find.
Step 2: By using the formula of average we get:
\[{\text{Average}} = \dfrac{{{x_1} + {x_2} + {x_3}}}{3}\]
By substituting the value of \[{x_1} = 180\]and
\[{x_2} = 257\] in the above expression we get:
\[ \Rightarrow {\text{Average}} = \dfrac{{180 + 257 + {x_3}}}{3}\]
By substituting the value of average which is
\[230\] in the above expression we get:
\[ \Rightarrow 230 = \dfrac{{180 + 257 + {x_3}}}{3}\]
Step 3: By doing the addition to the RHS side of the above expression we get:
\[ \Rightarrow 230 = \dfrac{{437 + {x_3}}}{3}\]
Bringing \[3\] into the LHS side of the above expression and multiplying it with \[230\], we get:
\[ \Rightarrow 690 = 437 + {x_3}\]
By bringing \[437\] into the LHS side of the above expression and subtracting it from \[690\], we get:
\[ \Rightarrow {x_3} = 253\]
The runs scored by the player in the third match is \[253\].
Note: Students need to remember the basic formulas for calculating the average of the terms. There are some important properties of average as below:
When the difference between all the terms is the same then the average equal to the middle term.
If a particular variable \[x\] is added in all the terms then, the average will also be increased by \[x\]
If a particular variable \[x\] is subtracted in all the terms then, the average will also be decreased by \[x\]
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