Question

An amount of ₹ 5,000/- is to be deposited in three different bonds bearing 6%, 7% and 8% per year respectively. Total annual income is ₹ 358/-. If the income from the first two investments is ₹ 70/- more than the income from the third, then find the amount of investment in each bond by the rank method

Answer 1

Let the three different bonds be x, y, and z

x + y + z = 5000  ........(1)

`x(6/00) +  y(7/00) +  z(8/100)` = 358

`(6x)/100 + (7y)/100 + (8z)/100` = 358

6x + 7y + 8x = 35800   .........(2)

`x(6/100) + y(7/100) = z(8/100) + 70`

`(6x)/100 + (7y)/100 = (8z)/100 + 70`

6x + 7y = 8z + 7000

6x + 7y – 8z = 7000  ........(3)

The matrix equation corresponding to the given system is

`[(1, 1, 1),(6, 7, 8),(6, 7, -8)] [(x),(y),(z)] = [(5000),(35800),(7000)]`
        A              X     =         B

Augmented Matrix [A, B]	Elementary Transformation
`[(1, 1, 1, 5000),(6, 7, 8, 358000),(6, 7, -8, 7000)]`
`∼[(1, 1, 1, 5000),(6, 7, 8, 35800),(0, 0, -16, -28800)]`	`{:"R"_3 -> "R"_3 - "R"_2:}`
`∼[(1, 1, 1, 5000),(0, 1, 2, 5800),(0, 0, -16, -28800)]`	`{:"R"_2 -> "R"_2 - 6"R"_1:}`
p(A) = 3; P(A,B) = 3

∴ The given system is equivalent to the matrix equation

`[(1, 1, 1),(0, 1, 2),(0, 0, -16)][(x),(y),(z)] = [(5000),(5800),(-28800)]`

x + y + z = 5000  ........(1)

y + 2z = 5800  .........(2)

– 16z = – 28800  .........(3)

Equation (3) ⇒ z = `(-28800)/(-16)`

∴ z = 1800

Equation (2) ⇒ y + 2(1800) = 5800

y + 3600 = 5800

y = 5800 – 3600

∴ y = 2200

Equation (1) ⇒ x + 2200 + 1800 = 5000

x + 4000 = 5000

x = 5000 – 4000

∴ x = 1000

The amount of investment in each bond is ₹ 1000, ₹ 2200 and ₹ 1800.