Question

an amount of Rs 65000 is invested in three bonds at rates of 6%,8%,10% per annum respectively. The total annual income is Rs.4800 . The income from the third bond is Rs.600 more than that of the second bond. determine the price each bond (use gaussian method of elemention methods )â

Answer 1

Answer:

Step-by-step explanation:

Consider

Amount invested in bond A is = Rs. x

Amount invested in bond B is = Rs. y

Amount invested in bond C is = Rs. z

Total amount invested,

$(x+y+z)=65000\text{..........................equation-1}$

The income from the bond A (rates of 6%)=$\frac{6x}{100}$

The income from the bond B (rates of 8%)=$\frac{8y}{100}$

The income from the bond C (rates of 10%)=$\frac{10z}{100}$

Total return

$\frac{6x}{100}+\frac{8y}{100}+\frac{10z}{100}=4800\\\\\Rightarrow 6x+8y+10z=480000\\\\\Rightarrow 3x+4y+5z=240000\text{.................................equation-2}$

The income from the third bond is Rs.600 more than that of the second bond

Therefore,

$\frac{10z}{100}=\frac{8y}{100}+600\\\\\Rightarrow \frac{10z}{100}-\frac{8y}{100}=600\\\\\Rightarrow 10z-8y=60000\\\\\Rightarrow -4y+5z=30000\text{.....................equation-3}$

system of equation in matrix form

$\left[\begin{array}{ccc}1&1&1\\3&4&5\\0&-4&5\end{array}\right]\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}65000\\240000\\30000\end{array}\right]$

Gaussian elimination is a method of solving a linear system by bringing the augmented matrix .

Therefore, its augmented matrix is

$C=\left[\begin{array}{ccc}A\end{array}\right]:\left[\begin{array}{ccc}B\end{array}\right]=\left[\begin{array}{cccc}1&1&1&65000\\3&4&5&240000\\0&-4&5&300000\end{array}\right]$

Applying operation $R_{2}=R_{2}-3R_{1}$ we get

$C\sim\left[\begin{array}{cccc}1&1&1&650000\\0&1&2&45000\\0&-4&5&30000\end{array}\right]$

Applying operation $R_{3}=R_{3}+4R_{2}$ we get

$C\sim\left[\begin{array}{cccc}1&1&1&650000\\0&1&2&45000\\0&0&13&210000\end{array}\right]$

We can write the above matrix as

$\left[\begin{array}{ccc}1&1&1\\0&1&2\\0&0&13\end{array}\right]\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}65000\\45000\\210000\end{array}\right]$

Therefore from this matrix we get,

$13z=210000\\\\\Rightarrow z=\frac{210000}{13}= 16153.85$

$y+2z=45000\\\\\Rightarrow y=45000-2z\\\\\Rightarrow y=45000-2\times16153.85\\\\\Rightarrow y=12692.30$

$x+y+z=65000\\\\\Rightarrow x=65000-(y+z)\\\\\Rightarrow x=65000-(16153.85+12692.30)\\\\\Rightarrow x=36153.85$

Therefore the price of the bond A is Rs 36153.85

The price of the bond B is Rs 12692.30

The price of the bond C is Rs 16153.85

Answer 2

Answer: Price of 8% bond is $126.92\times 100=12692$

price of 10% bond is $80(126.92)+6000=16153.85$

Price of 6% bond is $70000-\dfrac{800}{3}\times 126.92=36154.67$

Step-by-step explanation:

Since we have given that

Amount invested in three bonds = Rs. 65000

Rate of interest = 6%, 8% and 10%

Let the amount on 8% be 100x.

So, the interest on 8% = 8x

Interest on 10%=8x+600

Amount=80x+6000

Interest on 6%=4800-8x-8x-600

=4200-16x

Amount=$70000-(\dfrac{800}{3})x$

According to question,

$100x+80x+6000+70000-\dfrac{800}{3}x=65000\\180x-\dfrac{800}{3}x=65000-76000\\\\\dfrac{540-800}{3}x=-11000\\\dfrac{260}{3}x=11000\\x=\dfrac{11000\times 3}{260}=126.92$

So, Price of 8% bond is $126.92\times 100=12692$

price of 10% bond is $80(126.92)+6000=16153.85$

Price of 6% bond is $70000-\dfrac{800}{3}\times 126.92=36154.67$