Represent \[{A_4}\] = ( \[x\] : \[x\] is a two-digit number and sum of the digits of \[x\] is 7) in roster form.
Answer
362.1k+ views
Hint: In roster form, the elements of the set are listed within braces and separated by commas. So, with the given set-builder form, we find the elements and list them inside braces, separated by commas.
Complete step-by-step answer:
A set is a collection of well-defined objects. It can be represented in two forms:
(i). Roster form
(ii). Set-Builder form
The given question is set-builder form, where there is a variable representing any element of the set followed by a property satisfied by every member of the set.
We have to convert this into roster form, where the elements of the set are listed within braces and are separated by commas. The elements appear only once and the order does not matter.
Now, let us find the elements of the set \[{A_4}\] using the property given, that is, \[x\] is a two-digit number and the sum of the digits of \[x\] is 7.
Let \[a\] be the digit in the tenth’s place and \[b\] be the digit in the unit’s place.
Therefore, the two-digit number is represented as \[ab\] and we have:
\[a \ne 0\]
Hence, \[a\] can take values between 1 and 9 and \[b\] can take values between 0 and 9.
And also, we have:
\[a + b = 7\]
Hence, the possible values of \[a\] and \[b\] are as follows:
\[a = 1;b = 6\]
\[a = 2;b = 5\]
\[a = 3;b = 4\]
\[a = 4;b = 3\]
\[a = 5;b = 2\]
\[a = 6;b = 1\]
\[a = 7;b = 0\]
Hence, the two-digit numbers are:
\[\{ 16,{\text{ }}25,{\text{ }}34,{\text{ }}43,{\text{ }}52,{\text{ }}61,{\text{ }}70\} \]
Therefore, \[{A_4}\] is represented in roster form as follows:
\[{A_4} = \{ 16,{\text{ }}25,{\text{ }}34,{\text{ }}43,{\text{ }}52,{\text{ }}61,{\text{ }}70\} \]
Hence, the answer is \[{A_4} = \{ 16,{\text{ }}25,{\text{ }}34,{\text{ }}43,{\text{ }}52,{\text{ }}61,{\text{ }}70\} \] .
Note: \[{A_4}\] is a finite set, hence, you must find all the elements of this set. You might be tempted to include 07 as an element of the set, but 07 is nothing but 7 itself and is not a two-digit number, this is a common error.
Complete step-by-step answer:
A set is a collection of well-defined objects. It can be represented in two forms:
(i). Roster form
(ii). Set-Builder form
The given question is set-builder form, where there is a variable representing any element of the set followed by a property satisfied by every member of the set.
We have to convert this into roster form, where the elements of the set are listed within braces and are separated by commas. The elements appear only once and the order does not matter.
Now, let us find the elements of the set \[{A_4}\] using the property given, that is, \[x\] is a two-digit number and the sum of the digits of \[x\] is 7.
Let \[a\] be the digit in the tenth’s place and \[b\] be the digit in the unit’s place.
Therefore, the two-digit number is represented as \[ab\] and we have:
\[a \ne 0\]
Hence, \[a\] can take values between 1 and 9 and \[b\] can take values between 0 and 9.
And also, we have:
\[a + b = 7\]
Hence, the possible values of \[a\] and \[b\] are as follows:
\[a = 1;b = 6\]
\[a = 2;b = 5\]
\[a = 3;b = 4\]
\[a = 4;b = 3\]
\[a = 5;b = 2\]
\[a = 6;b = 1\]
\[a = 7;b = 0\]
Hence, the two-digit numbers are:
\[\{ 16,{\text{ }}25,{\text{ }}34,{\text{ }}43,{\text{ }}52,{\text{ }}61,{\text{ }}70\} \]
Therefore, \[{A_4}\] is represented in roster form as follows:
\[{A_4} = \{ 16,{\text{ }}25,{\text{ }}34,{\text{ }}43,{\text{ }}52,{\text{ }}61,{\text{ }}70\} \]
Hence, the answer is \[{A_4} = \{ 16,{\text{ }}25,{\text{ }}34,{\text{ }}43,{\text{ }}52,{\text{ }}61,{\text{ }}70\} \] .
Note: \[{A_4}\] is a finite set, hence, you must find all the elements of this set. You might be tempted to include 07 as an element of the set, but 07 is nothing but 7 itself and is not a two-digit number, this is a common error.
Last updated date: 23rd Sep 2023
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Total views: 362.1k
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