 # A child puts one five-rupee coin of her savings in the piggy bank on the first day. Question:

A child puts one five-rupee coin of her savings in the piggy bank on the first day. She increases her saving 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 can contribute to put the five-rupee coins into it and find the total money she saved.

Solution:

Saving of the child on the first day = Rs 5
Saving on the second day = Rs 5 + Rs 5 = Rs 10
Saving on the third day = Rs 5 + 2 × Rs 5 = Rs 15 and so on
The saving of the child on different days are Rs 5, Rs 10, Rs 15, ....
Since the savings of the child for each succeeding day is Rs 5 more than for the preceeding day, therefore the savings for different days forms an AP with first term a = Rs 5 and common difference d = Rs 5.
Suppose the number of days she continued to put the five-rupees coin in the piggy bank be n.
It is given that the total number of five-rupees coins in the piggy bank is 190.
So, the total sum of money saved by the child in n days = 190 × 5 = Rs 950

$\therefore S_{n}=190 \times 5=950$

$\Rightarrow \frac{n}{2}[2 \times 5+(n-1) \times 5]=950$

$\Rightarrow \frac{n}{2}(n+1)=190$

$\Rightarrow n^{2}+n=380$

$\Rightarrow n^{2}+n-380=0$

$\Rightarrow n^{2}+20 n-19 n-380=0$

$\Rightarrow n(n+20)-19(n+20)=0$

$\Rightarrow(n+20)(n-19)=0$

$\Rightarrow n+20=0$ or $n-19=0$

$\Rightarrow n=-20$ or $n=19$

Since the number of days cannot be negative, so n = 19.
So, the number of days she continued to put the five-rupees coin in the piggy bank is 19.
Also,
Total sum of money saved by her = Rs 950