---
title: "Mutual fund from growing salary, 15 years"
subject: "Engineering Economics"
chapter: 2
key: "semester_6/engineering_economics:mutual-fund-salary-growth-15yr"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/mutual-fund-salary-growth-15yr
site: BCT Engineering
---

# Mutual fund from growing salary, 15 years

Engineering Economics · Chapter 2: Interest and Time Value of Money

- Asked in: 2069 Bhadra, 4+4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/mutual-fund-salary-growth-15yr
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-2

## Question

Ramesh, a Civil Engineer, is planning to place a total of 20% of his salary, which is Rs. 250000 per year now, each year in mutual fund. He expects 7% salary increase each year for next 15 years. If the mutual fund will average 10% annual return, what will be the sum-amount at the end of 15 years? If salary increases by Rs. 25000 per year, what will be the amount?

## Short answer

With 7% annual salary growth, using the geometric series formula, the future amount is Rs 2,363,694. With an annual salary increase of Rs 25,000, using gradient and annuity factors, it is Rs 2,427,248.

## Answer

Assumption: the first deposit is at the end of year 1 and equals 20% of the present salary: A₁ = 0.20 × 2,50,000 = Rs 50,000. Later deposits grow with the salary. Interest i = 10%, n = 15 years.

#### (a) Salary grows 7% per year (geometric series)
```text
 deposits:  50,000   50,000(1.07)   50,000(1.07)²  ...  50,000(1.07)^14
 year:         1          2              3         ...        15
```
For a geometric series with growth g ≠ i:

F = A₁ [(1 + i)^n − (1 + g)^n] / (i − g)

- (1.10)^15 = 4.17725; (1.07)^15 = 2.75903
- F = 50,000 × (4.17725 − 2.75903) / (0.10 − 0.07)
- F = 50,000 × 47.2739

**Answer (a): F = Rs 2,363,694 (about Rs 23.64 lakh).**

#### (b) Salary increases by Rs 25,000 every year (gradient)
The deposit rises by 20% × 25,000 = Rs 5,000 each year: G = 5,000.

```text
 deposits: 50,000, 55,000, 60,000, ... , 50,000 + 14(5,000) = 1,20,000
```
- (F/A, 10%, 15) = 31.7725
- (A/G, 10%, 15) = 1/i − n/[(1 + i)^n − 1] = 5.2789
- Uniform equivalent A = 50,000 + 5,000 × 5.2789 = Rs 76,394.67
- F = A (F/A, 10%, 15) = 76,394.67 × 31.7725

**Answer (b): F = Rs 2,427,248 (about Rs 24.27 lakh).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the first deposit?
   A. Rs 250,000
   B. Rs 20,000
   C. Rs 267,500
   D. 20% of Rs 250,000 = Rs 50,000
   Answer: D. A1 = 0.20 x 250,000. (Key point: First deposit)

2. Future value with 7% growth, 10% interest?
   A. F = A1[(1+i)^n - (1+g)^n]/(i - g)
   B. F = A1 (F/A,i,n)
   C. F = A1 (1+g)^n
   D. F = G (F/G,i,n)
   Answer: A. Percentage growth uses the geometric series. (Key point: Geometric formula)

3. With Rs 25,000 yearly salary increase, what is G for deposits?
   A. Rs 25,000
   B. Rs 5,000 (20% of 25,000)
   C. Rs 50,000
   D. Rs 2,500
   Answer: B. Deposits rise by 20% of the increase. (Key point: Gradient)

4. How is the uniform equivalent found for part (b)?
   A. A = 50,000 + 5,000 x 15
   B. F = 50,000 x 15
   C. A = 50,000 + 5,000 (A/G,10%,15), then F = A (F/A,10%,15)
   D. A = 5,000 only
   Answer: C. A/G converts the gradient to uniform. (Key point: Uniform equivalent)

## Related in Engineering Economics

- [Deposit now to draw growing monthly amount](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Adeposit-now-growing-withdrawal) — Engineering Economics, chapter 2
- [Forecast CFL demand for next five years](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aforecast-cfl-demand-five-years) — Engineering Economics, chapter 1
- [Car loan interest rate and lump sum at year 3](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Acar-loan-installments-lump-sum-year3) — Engineering Economics, chapter 2
- [After-tax cash flow, sinking fund depreciation](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aaftertax-cash-flow-sinking-fund-25000) — Engineering Economics, chapter 7
- [Replace existing machine now; AEC and ESL](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Areplace-existing-machine-cash-flow-aec) — Engineering Economics, chapter 5
- [Replacement strategy for four-year need](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Areplace-machine-aec-four-years) — Engineering Economics, chapter 5

## Related in other subjects

- [Engineer opens confidential salary file](https://www.bctengineer.com/question.md?key=semester_8%2Fengineering_professional_practice%3Aconfidential-salary-file-conduct) — Engineering Professional Practice, chapter 7
- [EPS; total capital Rs 15,00,000](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Aeps-capital-1500000-pref-loan-a74) — Project Engineering, chapter 6
- [Total demand, village survey 2013, design 15 years](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ademand-c69) — Water Supply Engineering, chapter 3, asked 2×
- [Valuation of a cinema hall (land Rs 15 lakh)](https://www.bctengineer.com/question.md?key=semester_8%2Fconstruction_management%3Avalue-cinema-hall-2077) — Construction Management, chapter 13, asked 2×