an amount of Rs 5000 is put into three/investments at the rate of interest of 6%, 7%, 8% per annum respectively; the total annual income is Rs 35800. If the combined income from first two investments is Rs 70 more than the income from the third. Find amount of each installment by matrix method.
If Muntaha were younger by 5 years than what she really is then the square of her age would have been 11 more than five times her actual age. What is her age now?
Let's define the variables for the three investments:
Let x represent the amount invested at 6%.
Let y represent the amount invested at 7%.
Let z represent the amount invested at 8%.
We can set up the following equations based on the given information:
Equation 1: x + y + z = 5000 (since the total investment amount is Rs 5000)
Equation 2: 0.06x + 0.07y + 0.08z = 35800 (since the annual income is Rs 35800)
Equation 3: (0.06x + 0.07y) - (0.08z) = 70 (since the combined income from the first two investments is Rs 70 more than the income from the third)
Now, let's represent these equations in matrix form:
⎡ 1 1 1 ⎤ ⎡ x ⎤ ⎡ 5000 ⎤
⎢ 0.06 0.07 0.08 ⎥ ⎢ y ⎥ = ⎢ 35800 ⎥
⎣ 0.06 0.07 -0.08 ⎦ ⎣ z ⎦ ⎣ 70 ⎦
We can solve this matrix equation to find the values of x, y, and z.
To find the amount invested in each investment, we can represent the problem using a matrix. Let's denote the amount invested in the first, second, and third investments as x, y, and z respectively.
Based on the given information, we can form the following equations:
Equation 1: x + y + z = 5000 (since the total amount invested is Rs 5000)
Equation 2: (x * 6/100) + (y * 7/100) + (z * 8/100) = 35800 (since the total annual income is Rs 35800)
Equation 3: (x * 6/100) + (y * 7/100) = (z * 8/100) + 70 (since the combined income from the first two investments is Rs 70 more than the income from the third)
Now, let's represent these equations in matrix form:
[1, 1, 1] * [x, y, z] = [5000]
[0.06, 0.07, 0] * [x, y, z] = [35800]
[0.06, 0.07, -0.08] * [x, y, z] = [70]
To solve this system of equations using matrix methods, we can invert the coefficient matrix and multiply it with the constant matrix to find the values of x, y, and z.
Let's represent the coefficient matrix as A and the constant matrix as B:
A = [1, 1, 1; 0.06, 0.07, 0; 0.06, 0.07, -0.08]
B = [5000; 35800; 70]
To find the values of x, y, and z, we need to calculate the inverse of matrix A and multiply it with matrix B:
A^-1 * B = [x, y, z]
Once we obtain the values of x, y, and z, we will have the amounts invested in each investment.
If the amounts are x,y,z then
x+y+z = 5000
.06x + .07y + .08z = 3580
.06x + .07y = .08z + 70
I assume the interest is 358.00, not 35800!
(x,y,z) = (1000,2200,1800)