Question

A man saved $Rs.16,500$ in ten years. In each year after the first, he saved $Rs.100$ more than he did in the preceding year. How much did he save in the first place?
A) $Rs.1200$
B) $Rs.1300$
C) $Rs.1400$
D) $Rs.1400$

Answer 1

Hint: In this question, we have been told that a person saved $Rs.16,500$ in ten years. And every year, he saved $Rs.100$ more than he did in the preceding year. So, we have been asked how much did he save in the first year. If we read carefully, we will notice that it is forming an A.P. Start by assuming a certain amount in the first year. Then, add $Rs.100$ to that amount every year. You will get an A.P. Find the sum of the amounts using the formula of sum of an A.P and keep it equal to $Rs.16,500$. You will find your assumed money, simply by shifting.

Formula used: ${S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)$

Complete step-by-step solution:
Let the amount saved by the man in his first year be $Rs.x$. By reading the given statement, it can be stated that he saved $Rs.\left( {x + 100} \right)$ in the second year. And $Rs.\left( {x + 200} \right)$ in the third year. Following this rule, he would have saved $Rs.\left( {x + 900} \right)$ in the tenth year.
Therefore, we have the following A.P –
$x,x + 100,x + 200,x + 300,x + 400,.....,x + 900$
We will use the formula of A.P –
${S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)$, where $n = $ number of terms, $a = $ first term, ${a_n} = $ last term.
Putting all the values,
$ \Rightarrow 16500 = \dfrac{{10}}{2}\left( {x + x + 900} \right)$
Now, we will solve for x.
$ \Rightarrow 16500 = 5\left( {2x + 900} \right)$
Shifting and solving,
$ \Rightarrow \dfrac{{16500}}{5} = 2x + 900$
$ \Rightarrow 3300 = 2x + 900$
$ \Rightarrow 3300 - 900 = 2x$
Now, we have the value of x,
$ \Rightarrow \dfrac{{2400}}{2} = x$
$ \Rightarrow x = Rs.1200$
Therefore, that man saved $Rs.1200$ in the first year.

Option A is the correct answer.

Note: If you think it to be difficult to find the last term, you can also find it using the general formula of A.P.
$ \Rightarrow {a_n} = a + \left( {n - 1} \right)d$
Putting all the values,
$ \Rightarrow {a_n} = x + \left( {10 - 1} \right)100$
$ \Rightarrow {a_n} = x + 900$
Therefore, our last and 10th term is $x + 900$.