Question

An amount of 65000 is inversed in three bonds at the rates of 6%,8%and 10% per annum respectively. The total annual income 4800. The income from the third bond is 600 more than that from the second bond. Determine the price of each bond.

Answer 1

6500(r upon 100)power n. 6500*1upon

Answer 2

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,

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

Now multiply equation 1 with 3 and then subtract from equation 2 we get

$(3x+4y+5z)-3(x+y+z)=240000-65000\times3\\\\\Rightarrow 3x+4y+5z-3x-3y-3z=240000-195000\\\\\Rightarrow y+2z=45000\text{....................equation-4}$

Now multiply equation 4 with 4 and then add with equation 3 we get

$4(y+2z)+(-4y+5z)=4\times45000+30000\\\\\Rightarrow 4y+8z-4y+5z=180000+30000\\\\\Rightarrow 13z=210000\\\\\Rightarrow z=16153.85$

Put the value of z in equation 4 we get

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

Put the value of y and z in equation 1 we get

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

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