Questions & Answers
Question
AnswersWrite the following numbers in generalised form:
\[39,52,106,359,628,3458,9502,7000\]
Answer
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Hint: We have given two-digit, three-digit and four-digit numbers. It can be represented in a generalised form; if it is expressed as the sum of the product of its digits with their respective place values. So, a two digit number having \[a\] and \[b\] as its digits at tens and the ones places respectively is written in the form as \[10a + b\], where \[a\] can be any of the digits from \[1\] to \[9\] and \[b\] can be any of the digits from \[0\] to \[9\].
Complete step-by-step answer:
We can solve this question with three simple steps, which are given below:
Step 1: \[39\] is a two-digit number where digits \[9\] on unit place and \[3\] on tens place. So, it can be written in generalised form i.e., \[39 = 10 \times 3 + 9\].
Similarly, \[52\] is two-digit number, so it can be written in generalised form i.e. \[52 = 10 \times 5 + 2\]
Step 2: \[106\] is a three-digit number. So, it can be written in generalised form i.e.\[106 = 100 \times 1 + 10 \times 0 + 6\].
\[359\] is a three-digit number. So, it can be written in generalised form i.e., \[359 = 100 \times 3 + 10 \times 5 + 9\]
\[628\] is a three-digit number. So, it can be written in generalised form i.e., \[628 = 100 \times 6 + 10 \times 2 + 8\]
Step 3: \[3458\] is a four-digit number. So, \[3458 = 1000 \times 3 + 100 \times 4 + 10 \times 5 + 8\] is a generalised form.
Similarly, \[9502 = 1000 \times 9 + 100 \times 5 + 10 \times 0 + 2\]
and \[7000 = 1000 \times 7 + 100 \times 0 + 10 \times 0 + 0\] is a generalised form.
Note: Playing with numbers using any of the digits from \[0\] to \[9\] in generalised form of number. In four-digit number, we can write as \[1000a + 100b + 10c + d\], where \[a\] be any number from \[1\] to \[9\] while \[b,c\] and \[d\] be any number from \[0\] to \[9\].
Similarly, a three-digit number can be written as \[100a + 10b + c\] in generalised form where \[a\] be any number from \[1\] to \[9\] while \[b\] and \[c\] be any number from \[0\] to \[9\].
Complete step-by-step answer:
We can solve this question with three simple steps, which are given below:
Step 1: \[39\] is a two-digit number where digits \[9\] on unit place and \[3\] on tens place. So, it can be written in generalised form i.e., \[39 = 10 \times 3 + 9\].
Similarly, \[52\] is two-digit number, so it can be written in generalised form i.e. \[52 = 10 \times 5 + 2\]
Step 2: \[106\] is a three-digit number. So, it can be written in generalised form i.e.\[106 = 100 \times 1 + 10 \times 0 + 6\].
\[359\] is a three-digit number. So, it can be written in generalised form i.e., \[359 = 100 \times 3 + 10 \times 5 + 9\]
\[628\] is a three-digit number. So, it can be written in generalised form i.e., \[628 = 100 \times 6 + 10 \times 2 + 8\]
Step 3: \[3458\] is a four-digit number. So, \[3458 = 1000 \times 3 + 100 \times 4 + 10 \times 5 + 8\] is a generalised form.
Similarly, \[9502 = 1000 \times 9 + 100 \times 5 + 10 \times 0 + 2\]
and \[7000 = 1000 \times 7 + 100 \times 0 + 10 \times 0 + 0\] is a generalised form.
Note: Playing with numbers using any of the digits from \[0\] to \[9\] in generalised form of number. In four-digit number, we can write as \[1000a + 100b + 10c + d\], where \[a\] be any number from \[1\] to \[9\] while \[b,c\] and \[d\] be any number from \[0\] to \[9\].
Similarly, a three-digit number can be written as \[100a + 10b + c\] in generalised form where \[a\] be any number from \[1\] to \[9\] while \[b\] and \[c\] be any number from \[0\] to \[9\].