Question

Jasmine needs Rs. 30 to buy a sweater. She uses the following plan to save money. The first day she sets aside Rs. 1. The second day Jasmine sets aside Rs. 1.50. The third day she sets aside Rs. 2, and on the fourth day she sets aside Rs. 2.50. If Jasmine continues this pattern, on which day will she have enough money to buy the sweater ?

• A. Day 30
• B. Day 35
• C. Day 10
• D. Day 60

Answer 1

Answer: (C)

Total money required $=Rs.30$

On first day money save $=Rs.1$

Second day $=Rs.1.5$

Third day $=Rs.2$

Fourth day $=Rs.2.5$

We observe the above pattern is AP

with $a=1$ $d=0.5$

Now $S\ge 30$ (30 is min. amount)

$\therefore S=\frac{n}{2}(2a+(n-1\left)d\right)$ formula where $n=$ no. of days

$\therefore \frac{n}{2}(2a+(n-1\left)d\right)\ge 30$

$\therefore \frac{n}{2}(2\times 1+(n-1\left)\frac{1}{2}\right)\ge 30$

$\therefore \frac{n}{2}(2+\frac{n}{2}-\frac{1}{2})\ge 30$

$\therefore \frac{n}{2}(\frac{n}{2}+\frac{3}{2})\ge 30$

$\therefore n(n+3)\ge 120$

$\therefore n^2+3n-120\ge 0$

$\therefore n\ge \frac{-3\pm \sqrt{9+480}}{2}$

$\therefore n\ge \frac{-3+\sqrt{489}}{2}$ Positive avalue

$\therefore n\ge \frac{19.11}{2};n\ge 9.55$

$\therefore n=10$ Nearest integer value

Hence on 10th day she will have just enough money.