Question

Yasmeen saves Rs. 32 during the first month, Rs. 36 in the second month and Rs. 40 in the third month. If she continues to save in this manner, in how many months she will save Rs. 2000?

Answer 1

Hint: Here we will notice that the pattern in which Yasmeen saves the money is an A.P.
Hence since here the sum of all terms is given hence we will apply the formula for sum of n terms and then find the value of n.
The sum of n terms of an A.P is given by:-
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]

Complete step-by-step answer:
Here since the pattern in which Yasmeen saves the money is an A.P.
Hence, the A.P. so formed is given by:-
\[Rs.32,Rs.36,Rs.40...............\]
Now here the first term (a) is \[Rs.32\]
And the common difference (d) which the difference of two consecutive terms of an A.P is given by:
\[d = Rs.36 - Rs.32\]
Solving it we get:-
\[d = Rs.4\]
Now the sum of the money saved is given as Rs. 2000
Hence we will apply the formula of sum of n terms of an A.P.
The sum of n terms of an A.P is given by:-
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
Putting in the respective values and finding the value of n we get:-
\[2000 = \dfrac{n}{2}\left[ {2\left( {32} \right) + \left( {n - 1} \right)\left( 4 \right)} \right]\]
Simplifying it we get:-
\[2000 = \dfrac{n}{2}\left[ {64 + 4n - 4} \right]\]
\[ \Rightarrow 2000 = \dfrac{n}{2}\left[ {4n + 60} \right]\]
Solving it further we get:-
\[2000 = n\left( {2n + 30} \right)\]
Now multiplying n inside the bracket we get:-
\[2000 = 2{n^2} + 30n\]
Simplifying it further we get:-
\[2{n^2} + 30n - 2000 = 0\]
Dividing the equation by 2 we get:-
\[{n^2} + 15n - 1000 = 0\]
Now solving for value of n using middle term split we get:-
\[{n^2} + 40n - 25n - 1000 = 0\]
Solving it further we get:-
\[n\left( {n + 40} \right) - 25\left( {n + 40} \right) = 0\]
\[ \Rightarrow \left( {n - 25} \right)\left( {n + 40} \right) = 0\]
Putting each of the terms equal to zero we get:-
\[ \Rightarrow n - 25 = 0;n + 40 = 0\]
\[ \Rightarrow n = 25;n = - 40\]
Now since number of months cannot be negative
Therefore, \[n = - 40\] is rejected.
Therefore, \[n = 25\]

Hence, it will take 25 months to save Rs. 2000.

Note: This question can also be solved using the quadratic equation by using quadratic formula.
For a standard equation \[a{x^2} + bx + c = 0\] the quadratic formula is given by:-
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Hence applying this formula for the quadratic equation \[{n^2} + 15n - 1000 = 0\] we get:-
\[n = \dfrac{{ - 15 \pm \sqrt {{{\left( {15} \right)}^2} - 4\left( 1 \right)\left( { - 1000} \right)} }}{{2\left( 1 \right)}}\]
Simplifying it further we get:-
\[n = \dfrac{{ - 15 \pm \sqrt {225 + 4000} }}{2}\]
\[ \Rightarrow n = \dfrac{{ - 15 \pm \sqrt {4225} }}{2}\]
Solving it further we get:-
\[ \Rightarrow n = \dfrac{{ - 15 \pm 65}}{2}\]
Hence,
\[n = \dfrac{{ - 15 + 65}}{2};n = \dfrac{{ - 15 - 65}}{2}\]
\[ \Rightarrow n = \dfrac{{50}}{2};n = \dfrac{{ - 80}}{2}\]
\[ \Rightarrow n = 25;n = - 40\]
Now the number of months cannot be negative therefore,
\[n = 25\]