Question

A mother, Ramkali would need to save Rs.\[1800\] For admission fee and books etc., for her daughter to start going to school from next year. She saved Rs.\[50\] In the first month of this year and increased her monthly savings by Rs. \[20\]. After a year how money will she save? Will she be able to dream of sending her daughter to school for studying?

Answer 1

Hint: In the given question, we have to compute how much money Ramkali will save for her daughter’s studies in a year, if she saves Rs.\[50\] In the first month and increases her savings by Rs.\[20\] For every consequent month. We will use a formula of arithmetic progression to solve the question. As stated in the question Ramkali needs a total of Rs.\[1800\] To send her daughter to school.

Complete step-by-step answer:
According to the question, we know that Ramkali needs a total of Rs.\[1800\] To send her daughter to school.
Given that, she saves Rs.\[50\] At first month and increases her savings by Rs.\[20\] For every consequent month.
Now, let’s say she starts savings Rs.\[50\] In January.
Therefore, her savings for February becomes,\[50 + 20 = 70\]
Also, saving for March will be\[50 + 20 + 20 = 90\]
Likewise, for all the subsequent months we will follow same operation,
Therefore, we can see that Ramkali’s savings are in an A.P.
\[ \Rightarrow 50 + 70 + 90 + .......\]
Where,\[50 = a\],\[20 = d\]and\[n = 12\].
We know, formula for sum an A.P. is,
\[ \Rightarrow {S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]\]
Putting the values of a, d and n in the formula we get,
\[ \Rightarrow {S_n} = \dfrac{{12}}{2}\left[ {2 \times 50 + (12 - 1)20} \right]\]
\[ \Rightarrow {S_n} = \dfrac{{12}}{2}\left[ {2 \times 50 + (11)20} \right]\]
\[ \Rightarrow {S_n} = 6\left[ {100 + 220} \right]\]
\[ \Rightarrow {S_n} = 6\left[ {320} \right]\]
\[ \Rightarrow {S_n} = 1920\] which is Ramkali’s total saving for one year.
According to the question, Ramkali only needed \[1800\] rupees to send her daughter to school, but she was able to save \[1920\] in total in a year. Hence, she will be able to fulfil her dream of sending her daughter to school.

Note: We know that an A.P. is an order of numbers where the difference between two consequents numbers is always the same. In the above question we have used the formula for computing sum of an A.P., likewise, we also have a formula for finding Nth term of an A.P. Which is, \[{a_n} = a(n - 1)d\] where, a is the first term of the A.P. and n, d are the number of terms and d is the common difference.