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The number of times 99 is subtracted from 1111 so that the remainder is less than 99 is
1) 10
2) 11
3) 12
4) 13

Answer
VerifiedVerified
582.9k+ views
Hint: We will first let the number be \[x\] and form the equation as we are given in the question that the number of times 99 is subtracted from 1111 will be equal to the remainder which should be less than 99. After this, we will substitute the value of \[x\] and the value which will satisfy the equation is the value of \[x\].

Complete step-by-step answer:
First, consider the value of the number be \[x\].
We need to find the value of \[x\] using the given information.
The number of times 99 implies that the number is multiplied by 99 which is further subtracted from 1111.
Thus, we get that,
\[ \Rightarrow 1111 - 99x\]
Next, we are given that the obtained expression is equal to the remainder which is represented by \[R\].
Thus, we get,
\[ \Rightarrow 1111 - 99x = R\]
Since, it is given that the remainder should be less than 99 which reduces the equation as,
\[
   \Rightarrow 1111 - 99 < 99x \\
   \Rightarrow 1012 < 99x \\
\]
Next, substitute the value of \[x\] and check which value will satisfy the equation,
Hence, first we will substitute \[x = 10\],
\[
   \Rightarrow 1012 < 99\left( {10} \right) \\
   \Rightarrow 1012 < 990 \\
\]
Hence, \[x = 10\] does not satisfy the equation.
Further, we will substitute \[x = 11\]
\[
   \Rightarrow 1012 < 99\left( {11} \right) \\
   \Rightarrow 1012 < 1089 \\
\]
Thus, \[x = 11\] satisfies the equation and hence we get the number as \[x = 11\].
Thus, option B is correct.

Note: As the equation is obtained in inequality form, thus to find the value of \[x\] we have to use the hit and trial method. We can form the equation by using division algorithm also as \[{\text{Divisor}} = p\left( {{\text{quotient}}} \right) + {\text{remainder}}\] where we are given \[{\text{Divisor}} = 1111\], \[{\text{Quotient}} = 99\] and remainder should be less than 99 to find the value of \[p\]. Thus, we will get, \[1111 = n\left( {99} \right) + R\].


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