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Miscellaneous Exercise 2(B) II · Q101

Q.An amount of Rs. 5000 is invested in three types of investments, at interest rates 6.7, 7.7, 8% per annum respectively. The total annual income from these investments is Rs. 350/-. If the total annual income from first two investments is Rs. 70 more than the income from the third, find the amount of each investment using matrix method.

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Step 1: Let Rs. x,y,zx,y,z be invested at 6.7%,7.7%,8%6.7\%,7.7\%,8\% respectively. "Amount invested is Rs. 5000": x+y+z=5000x+y+z=5000. "Total annual income is Rs. 350": 0.067x+0.077y+0.08z=3500.067x+0.077y+0.08z=350. "Income from first two investments is Rs. 70 more than the third": 0.067x+0.077y=0.08z+700.067x+0.077y=0.08z+70, i.e. 0.067x+0.077y−0.08z=700.067x+0.077y-0.08z=70.

Step 2: Multiplying the last two equations by 10001000 for cleaner coefficients: 67x+77y+80z=35000067x+77y+80z=350000 and 67x+77y−80z=7000067x+77y-80z=70000.

Step 3: Subtracting: (67x+77y+80z)−(67x+77y−80z)=350000−70000⇒160z=280000⇒z=1750(67x+77y+80z)-(67x+77y-80z)=350000-70000\Rightarrow160z=280000\Rightarrow z=1750.

Step 4: Adding instead: (67x+77y+80z)+(67x+77y−80z)=350000+70000⇒2(67x+77y)=420000⇒67x+77y=210000(67x+77y+80z)+(67x+77y-80z)=350000+70000\Rightarrow2(67x+77y)=420000\Rightarrow67x+77y=210000.

Step 5: From x+y+z=5000x+y+z=5000: x+y=5000−1750=3250⇒y=3250−xx+y=5000-1750=3250\Rightarrow y=3250-x. Substituting: 67x+77(3250−x)=210000⇒67x+250250−77x=210000⇒−10x=−40250⇒x=402567x+77(3250-x)=210000\Rightarrow67x+250250-77x=210000\Rightarrow-10x=-40250\Rightarrow x=4025.

Step 6: y=3250−4025=−775y=3250-4025=-775. …

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