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Hint: Here, we will proceed by forming the series obtained by the days on which Latika will deliver the bottled water to MTG’s bookstore and Sammy’s bookstore and then we will write the expression for any nth term for both these series and will equate these values with each other.

Complete step-by-step answer:

According to the problem statement, on 4th (Thursday) Latika delivers the bottled water to both MTG’s bookstore and Sammy’s bookstore. Then the days on which she will again the deliver bottled water to MTG’s bookstore will be 10th,16th,24th,… day and the days on which she will again deliver the bottled water to Sammy’s bookstore will be 13th,22th,31th,… day

As we know that nth term for any arithmetic progression with first term as a1 and common difference d is given by ${a_n} = {a_1} + \left( {n - 1} \right)d{\text{ }} \to {\text{(1)}}$

Clearly, the days after 4th when Latika will deliver the bottled water again to MTG’s bookstore are in an arithmetic progression with first term a1=10 and common difference d=6.

Using formula given by equation (1), we get

nth term of this AP, $

{a_n} = 10 + 6\left( {n - 1} \right) = 10 + 6n - 6 \\

\Rightarrow {a_n} = 6n + 4{\text{ }} \to {\text{(2)}} \\

$

Similarly, the days after 4th when Latika will deliver the bottled water again to Sammy’s bookstore are in an arithmetic progression with first term a1’=13 and common difference d’=9.

Using formula given by equation (1), we get

(n’)th term of this AP, $

{a_n}' = 13 + 9\left( {n' - 1} \right) = 13 + 9n' - 9 \\

\Rightarrow {a_n}' = 9n' + 4{\text{ }} \to {\text{(3)}} \\

$

In order to obtain the exact number of days after 4th when Latika will deliver the bottled water again to both MTG’s bookstore and Sammy’s bookstore, the value of the nth term given by equations (2) and (3) should be equal.

So, we will make the RHS of equations (2) and (3) equal.

i.e., $

6n + 4 = 9n' + 4 \\

\Rightarrow 6n = 9n' \\

\Rightarrow n = \dfrac{{3n'}}{2}{\text{ }} \to {\text{(4)}} \\

$

Since, in equation (4) both n and n’ should be natural numbers because the number of terms are always natural numbers.

Clearly, if we will put n’=2 (natural number), the value of n obtained according to equation (4) is n=3 which is also a natural number.

Now, put n=3 in equation (2) or put n’=2 in equation (3) we will get the value of nth term as

$ \Rightarrow {a_n} = \left( {6 \times 3} \right) + 4 = 18 + 4 = 22$ or $ \Rightarrow {a_n}' = \left( {9 \times 2} \right) + 4 = 18 + 4 = 22$

So, on the 22nd day Latika will deliver the bottled water to both the bookstores.

So, exactly after (22-4)=18 days Latika will deliver the bottled water to both the bookstores.

Hence, option B is correct.

Note: Here, we have made the nth term of the first arithmetic progression equal to the (n’)th term of the second arithmetic progression because for that day when Latika will deliver the bottled water to both the bookstores on the same day again that particular day should be same. Also, the values of n and n’ obtained are the minimum possible values in order to obtain the nearest day to 4th.

Complete step-by-step answer:

According to the problem statement, on 4th (Thursday) Latika delivers the bottled water to both MTG’s bookstore and Sammy’s bookstore. Then the days on which she will again the deliver bottled water to MTG’s bookstore will be 10th,16th,24th,… day and the days on which she will again deliver the bottled water to Sammy’s bookstore will be 13th,22th,31th,… day

As we know that nth term for any arithmetic progression with first term as a1 and common difference d is given by ${a_n} = {a_1} + \left( {n - 1} \right)d{\text{ }} \to {\text{(1)}}$

Clearly, the days after 4th when Latika will deliver the bottled water again to MTG’s bookstore are in an arithmetic progression with first term a1=10 and common difference d=6.

Using formula given by equation (1), we get

nth term of this AP, $

{a_n} = 10 + 6\left( {n - 1} \right) = 10 + 6n - 6 \\

\Rightarrow {a_n} = 6n + 4{\text{ }} \to {\text{(2)}} \\

$

Similarly, the days after 4th when Latika will deliver the bottled water again to Sammy’s bookstore are in an arithmetic progression with first term a1’=13 and common difference d’=9.

Using formula given by equation (1), we get

(n’)th term of this AP, $

{a_n}' = 13 + 9\left( {n' - 1} \right) = 13 + 9n' - 9 \\

\Rightarrow {a_n}' = 9n' + 4{\text{ }} \to {\text{(3)}} \\

$

In order to obtain the exact number of days after 4th when Latika will deliver the bottled water again to both MTG’s bookstore and Sammy’s bookstore, the value of the nth term given by equations (2) and (3) should be equal.

So, we will make the RHS of equations (2) and (3) equal.

i.e., $

6n + 4 = 9n' + 4 \\

\Rightarrow 6n = 9n' \\

\Rightarrow n = \dfrac{{3n'}}{2}{\text{ }} \to {\text{(4)}} \\

$

Since, in equation (4) both n and n’ should be natural numbers because the number of terms are always natural numbers.

Clearly, if we will put n’=2 (natural number), the value of n obtained according to equation (4) is n=3 which is also a natural number.

Now, put n=3 in equation (2) or put n’=2 in equation (3) we will get the value of nth term as

$ \Rightarrow {a_n} = \left( {6 \times 3} \right) + 4 = 18 + 4 = 22$ or $ \Rightarrow {a_n}' = \left( {9 \times 2} \right) + 4 = 18 + 4 = 22$

So, on the 22nd day Latika will deliver the bottled water to both the bookstores.

So, exactly after (22-4)=18 days Latika will deliver the bottled water to both the bookstores.

Hence, option B is correct.

Note: Here, we have made the nth term of the first arithmetic progression equal to the (n’)th term of the second arithmetic progression because for that day when Latika will deliver the bottled water to both the bookstores on the same day again that particular day should be same. Also, the values of n and n’ obtained are the minimum possible values in order to obtain the nearest day to 4th.

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