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 from the choices.

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19 days, Rs.900

22 days, Rs.850

19 days, Rs.950

22 days, Rs.800

answer is A.

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Detailed Solution

It is given that, the total number of five rupees coins in the piggy bank is n = 190.
The savings on the first day, a = 5.
The savings on the second day,

a 2 = 5 + 5 = 10

On proceeding in the same way we get,
5,10,15,.....
Above is in the form of AP where,
a=5,
The common difference,

d = 10 − 5 d = 5

n = 190

The total sum of money saved by the child in n days is given by,

S n = 190 × 5 S n = 950

We know that sum of n-terms in AP is,

S n = n 2 2 a + n − 1 d

Substitute the known values in above formula,

n 2 [2 × 5 + (n − 1) × 5] = 950

n 2 (10 + 5 n − 5) = 950 n 2 (5 n + 5) = 950

n 2 + n − 380 = 0

Solve the above equation by factorization,

n 2 + 20 n − 19 n − 380 = 0

n (n + 20) − 19 (n + 20) = 0

(n − 19) (n + 20) = 0

When (n-19)=0,

n = 19

When n + 20 = 0,

n = − 20

which is a negative number.
So it is not possible.
The number of days she continued to put the five-rupees coin in the piggy bank is 19 and the total sum of money saved by her is Rs.950.
Hence, option 1) is correct.

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