Simple Interest Problems

Question

Divide rs 2379 into 3 parts so that their amount after 2,3 and 4 years respectively may be equal, the rate of interest being 5% per annum at simple Interest, what will be first Part

**I have Tried: amount will be equal for all the three parts for n=2,n=3 and n=4 Amount = Simple Interest + Principal 2379/3= will give the value for each part it constitutes 793, we cannot directly divide that three parts May be another three different types of Number

Let the first part Number part I will be take $x$, surely the second part will be $2379-x$ and how will take the value for third part? Please anyone share the Answer and Logic**

Please anyone do it for Compound interest also

Answer 1

As you wrote in your comment you have $x+y+z=2379$.

You also have $1.1x=1.15y=1.2z$. From this you can express all variables using one of them. For example, you have \begin{align*} y&=\frac{1.1}{1.15}x=\frac{22}{23}x\\ z&=\frac{1.1}{1.2}x=\frac{11}{12}x \end{align*}

Plugging this into the equation $x+y+z=2379$ you get \begin{align*} x+\frac{22}{23}x+\frac{11}{12}x&=2379\\ \frac{276+264+253}{276}x&=2379\\ \frac{793}{276}x&=2379\\ x&=\frac{276\cdot2379}{793}\\ x&=276\cdot3\\ x&=828 \end{align*}

Now you can check whether the conditions of the original problem are fulfilled.

Answer 2

We know that:

• $x+y+z=2379$
• $1.05^2x=1.05^3y=1.05^4z$

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We can easily replace $x$ and $y$ in the equation $x+y+z=2379$, because:

• $1.05^2x=1.05^4z \implies x=1.05^2z$
• $1.05^3y=1.05^4z \implies y=1.05^1z$

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Therefore:

• $x+y+z=2379 \implies 1.05^2z+1.05^1z+z=2379 \implies z=\frac{2379}{3.1525}$

Therefore:

• $y=1.05^1z \implies y=1.05^1\cdot\frac{2379}{3.1525}=\frac{2497.95}{3.1525}$
• $x=1.05^2z \implies x=1.05^2\cdot\frac{2379}{3.1525}=\frac{2622.8475}{3.1525}$

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Summary:

• $x=\frac{2622.8475}{3.1525}\approx832$
• $y=\frac{2497.95}{3.1525}\approx792$
• $z=\frac{2379}{3.1525}\approx755$