Question

A child puts one five rupee coin as her savings in the piggy bank on the first day. She increases her savings by one five rupee-coin daily. If the piggy bank can hold 190 coins of five rupees in all, find the number of days she continues to put the five rupee coins into it and find the total money she saved.

Answer 1

Hint: In the above given question, the given data follows AP, so with the help of the given values, find the number of days the child continues to put the coins in the piggy bank and then substitute this number so obtained in the formula for obtaining the total money saved by her.

Complete step-by-step answer:
It is given that the child increases her saving by one five rupee coin daily.
So, child's daily input is 5, 10, 15, …
Thus, the number of coins on each day can be given as
1, 2, 3, …
Here, it can be seen that the AP is formed with
$a = 1$ … (1)
$d = 1$ … (2)
It is also given that the piggy bank can hold 190 coins of five rupees in all, this implies that
${S_n} \leqslant 190$ … (3)
Where sum ${S_n} = \dfrac{n}{2}(2a + (n - 1)d)$
After substituting the equations (1) and (2) in (3), we get,
$\Rightarrow \dfrac{n}{2}(2 + (n - 1)1) \leqslant 190$
$\Rightarrow \dfrac{n}{2}(2 + n - 1) \leqslant 190$
$\Rightarrow n + \dfrac{{{n^2}}}{2} - \dfrac{n}{2} \leqslant 190$
$\Rightarrow \dfrac{{{n^2}}}{2} + \dfrac{{2n - n}}{2} \leqslant 190$
$\Rightarrow \dfrac{{{n^2} + n}}{2} \leqslant 190$
$\Rightarrow {n^2} + n \leqslant 380$
$\Rightarrow {n^2} + n - 380 \leqslant 0$
$\Rightarrow {n^2} + 20n - 19n - 380 \leqslant 0$
$\Rightarrow n(n + 20) - 19(n + 20) \leqslant 0$
$\Rightarrow (n - 19)(n + 20) \leqslant 0$
$n \in [ - 20,19]$
Therefore, she can put coins for 19 days.
For amount 5, 10, 15, …
AP = 5(1,2,3, ...)
So, here we have
a=5, d=5, n=19
${S_n} = \dfrac{n}{2}(2a + (n - 1)d)$
Sum of 19 terms$= \dfrac{{19}}{2}(2(5) + (19 - 1)5)$
$= \dfrac{{19}}{2}(10 + (18)5)$
$= \dfrac{{19}}{2}(10 + 90)$
$= \dfrac{{19}}{2}(100)$

=Rs 950
Hence, the total money she saved is Rs 950.

Note: Carefully identify the series which is being followed. Here, AP is followed, which is a sequence in which the difference between all the consecutive terms is constant. Then by substituting the required values in the formula of sum of n terms in AP, the solution can be obtained.