Question

A sum of 13000 is divided into three parts such that the simple interests accrued on them for two three and four year's respectively may be equal find the amount diposited for 4 Years?

Answer 1

Answer:

Let the amounts be x, y, z in ascending order of value. As the interest rate and interest accrued are same for 4 years 3 years and 2 years i.e. 4x = 3y = 2z = k.

L.C.M. of 4,3,2 = 12 So x:y:z: = 3000 : 4000 :6000

The amount deposited for 4 years = 3000

Answer 2

A sum of 13000 is divided into three parts such that the simple interests accrued on them for two, three, and four years respectively may be equal. The amount deposited for 4 Years is 3000.

Step-by-step explanation:

Given:

• The total Amount 13000 is divided into three parts.
• Time periods = 2, 3, and 4 years.
• Simple interest is equal on all three deposits.

To be found: The amount deposited for 4 years.

Formula used: Simple Interest = $\frac{Principal \times Rate of interest \times Time period}{100}$

Solution:

Let the first amount be $x$, the second amount be $y$ , and the third amount will become $13000-(x+y)$.

Let the Rate of interest be $r$ for all deposits.

Now using the formula of simple interest we will get:

• $S_{1}= \frac{x\times r \times 4}{100}$
• $S_{2}= \frac{y\times r \times 3}{100}$
• $S_{3}= \frac{(13000-(x+y))\times r \times 2}{100}$

Since all the simple interests are equal,

$S_{1}=S_{2}=S_{3}\\ \frac{x\times r \times 4}{100} = \frac{y\times r \times 3}{100} =\frac{(13000-(x+y))\times r \times 2}{100}$

Now eliminating the common terms we will have,

$x\times4=y\times 3= (13000-(x+y))\times 2\\ 4x=3y= 26000-(2x+2y)$

taking last two terms,

$3y=26000-2x-2y\\ 3y+2y=26000-2x\\ 5y=26000-2x\\ y= \frac{26000}5-\frac{2x}5\\ \therefore y=5200- \frac{2x}5$

Now putting the value of $y$ in,

$4x=3y\\ 4x=3\times(5200-\frac{2x}5)\\4x=15600-\frac{6x}5\\4x=\frac{78000-6x}5\\4x\times5=78000-6x\\20x=78000-6x\\20x+6x=78000\\26x=78000\\x=\frac{78000}{26}\\\therefore x= 3000$

So, the amount deposited for 4 years is 3000.