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The principal amount is P, rate of interest is R and time of loan is T
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A sum of money gets doubled in 5 years at X% simple interest. If the interest was Y%, the sum of money would have become ten-fold in thirty years. What is Y-X
X = (200-100)/5 = 20% Y = (1000-100)/30 = 30% So, Y-X = 10%
Rs. 5000 was invested at 8%p.a simple interest for a period of N years. How much more money should be invested for N years at a simple interest of 14% p.a for the overall interest to be 10% p.a for N years
5000*8*N + X*14*N =(5000+X)*10*N
Or X = Rs. 2,500
At the start of 2010, Ajay invested Rs.10000 in compound interest at the rate of X%p.a (X>0). At the end of every year, he withdrew Y% of the total amount accumulated. If at the start of 2014, Ajay had exactly Rs.10000 in his account, which of the following is necessarily true?
If the amount left at start of 2014 is 10000 then Amount withdrawn in 2013=Amount accumulated over 2013 = 100X.
Hence, Amount withdrawn every year = 100X. So, Y = $$\frac{100X}{10000+100X}$$ = $$\frac{X}{100+X}$$ < X
Alternate approach:
At the beginning of 2010, Ajay has 10,000. Let us take this as '10k' for ease of calculation.
Let y = (Y/100) where Y is the %value withdrawn every time.
At the end of 2010, Ajay has $$10k\left(1+\frac{x}{100}\right)$$
At the beginning of 2011, he has $$10k\left(1+\frac{x}{100}\right)-y\left(10k\left(1+\frac{x}{100}\right)\right)$$
This can be simplified as $$10k\left(1+\frac{x}{100}\right)\left(1-y\right)$$
At the end of 2011, he will have $$10k\left(1+\frac{x}{100}\right)^2\left(1-y\right)$$
Similarly, at the start of 2012, he will take $$10k\left(1+\frac{x}{100}\right)^2\left(1-y\right)^2$$ for his investment
and at the end, he will have $$10k\left(1+\frac{x}{100}\right)^3\left(1-y\right)^2$$
In the same way, at the start of 2013, he will take (1-y) times his previous value which means that by the end of 2013 $$10k\left(1+\frac{x}{100}\right)^4\left(1-y\right)^3$$
This means that at the start of 2014 he will have (1-y) times the previous value= $$10k\left(1+\frac{x}{100}\right)^4\left(1-y\right)^4$$
It is mentioned in the question that this value is equal to 10k
On equating this, we get:
$$10k\left(1+\frac{x}{100}\right)^4\left(1-y\right)^4=10k$$
We can further simplify this to: $$\left(1+\frac{x}{100}\right)\left(1-y\right)=1$$
$$1=\frac{1}{\left(1-y\right)}-\frac{x}{100}$$
On rearranging the terms, we get: $$y=\frac{x}{x+100}$$
This value of y is definitely lower than x.
Hence we can deduce that x>y.
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