Question 1 OR (a)

Splitting ₹60,000 among savings, bonds and business (matrix inverse)

6 marks

Time to write: About 12 minutes
Topic: Matrices and determinants

Question

Question 1 OR (a) · 6 marks

Mr. Y invested ₹60,000 in three investments. A part is placed in a savings account with an annual return of 5%, part in 8% annual yield bonds and the rest in business. In 2022, when he lost 3% of the money invested in the business, his net income from all three investments is ₹1,520. If the sum invested in the business exceeds that in the savings account by ₹12,000, how much was invested in each. Use matrix algebra.

Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.

Before you write

Read off the question and its marks

12 minfor 6 marks

Read from the question

how muchFormula first, then the working, then the answer with its unit (₹, %, units).

What the paper leaves open

The year is a date swap only; the amounts and rates are the question's own.

Write it in this order

Read the data; note exactly what is asked 1 min
1 Equations in matrix form 1½ marks 3 min
2 Determinant, adjoint, inverse 2½ marks 4 min
3 Amounts with check 2 marks 3 min
Check: arithmetic, units (₹, %), final answer underlined 1 min

123

Try it first. About 12 minutes for 6 marks. Write it out, then open the solution.

Solution

Hover or tap a marked figure, like 72,000, to see how it is made and where else it appears.

Let ₹x be placed in the savings account (5%), ₹y in bonds (8%) and ₹z in business (a loss of 3%).

Equations

x + y + z = 60,000 (total invested) 0.05x + 0.08y − 0.03z = 1,520 (net income), i.e. 5x + 8y − 3z = 1,52,000 −x + z = 12,000 (business exceeds savings by ₹12,000)

In matrix form AX = B with A = [[1, 1, 1], [5, 8, −3], [−1, 0, 1]], B = [60000, 152000, 12000]ᵀ.

Inverse. |A| = 1(8 − 0) − 1(5 − 3) + 1(0 + 8) = 8 − 2 + 8 = 14 ≠ 0.

Cofactors: A₁₁ = 8, A₁₂ = −2, A₁₃ = 8; A₂₁ = −1, A₂₂ = 2, A₂₃ = −1; A₃₁ = −11, A₃₂ = 8, A₃₃ = 3.

adj A = [[8, −1, −11], [−2, 2, 8], [8, −1, 3]], A⁻¹ = (1/14) adj A.

Solution X = A⁻¹B:

x = (8 × 60,000 − 1 × 1,52,000 − 11 × 12,000)/14 = (4,80,000 − 1,52,000 − 1,32,000)/14 = 1,96,000/14 = ₹14,000 y = (−2 × 60,000 + 2 × 1,52,000 + 8 × 12,000)/14 = (−1,20,000 + 3,04,000 + 96,000)/14 = 2,80,000/14 = ₹20,000 z = (8 × 60,000 − 1 × 1,52,000 + 3 × 12,000)/14 = (4,80,000 − 1,52,000 + 36,000)/14 = 3,64,000/14 = ₹26,000

Investment	Amount (₹)	Return (₹)
Savings account at 5%	14,000	700
Bonds at 8%	20,000	1,600
Business (loss 3%)	26,000	−780
Total	60,000	1,520

Check: the amounts total ₹60,000; net income 700 + 1,600 − 780 = ₹1,520; business − savings = 26,000 − 14,000 = ₹12,000.

Where the 6 marks sit

#	Component	Marks
1	Equations in matrix form	1.5
2	Determinant, adjoint, inverse	2.5
3	Amounts with check	2
	Total	6

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