---
title: "Find A and G with A = 3G"
subject: "Engineering Economics"
chapter: 2
key: "semester_6/engineering_economics:find-a-and-g-a-equals"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/find-a-and-g-a-equals
site: BCT Engineering
---

# Find A and G with A = 3G

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

- Asked in: 2067 Mangsir, 8 marks
- Video: [Arithmetic Gradient - Engineering Economics Lightboard](https://www.youtube.com/watch?v=ykQ7M9UNuok) — Engineering Economics Guy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/find-a-and-g-a-equals
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-2

## Question

Find the value of A and G if i = 10%. A = 3G.
[Figure: outflow of 50000 at year 0; receipts at years 1 to 10 shown on the diagram: a uniform amount A each year, plus a gradient series that begins after year 3 and rises steadily (2G marked at year 5, 8G marked at year 8), with no gradient beyond year 8 (only A in years 9 and 10)]

## Short answer

Set present worth of receipts equal to 50,000: A(P/A,10%,10) + 2G(P/G,10%,5)(P/F,10%,3) = 50,000. With A=3G, G = Rs 1,739.47 and A = Rs 5,218.40.

## Answer

#### Reading the diagram (assumption)
Outflow of Rs 50,000 at year 0. Receipts: uniform A in years 1 to 10, plus a gradient that is zero at year 4, 2G at year 5, 4G at year 6, 6G at year 7 and 8G at year 8 (increase of 2G per year), nothing after year 8. So the gradient series has gradient amount 2G, five terms (years 4 to 8), and its base is at year 3. The project is taken as just equivalent, PW of receipts = PW of outflow.

```text
 receipts:  A  A  A  A  A+2G  A+4G  A+6G  A+8G   A   A
 year:      1  2  3  4    5     6     7     8     9   10
 outflow:  50,000 at year 0
```

#### Factors (i = 10%)
- (P/A, 10%, 10) = 6.1446
- (P/G, 10%, 5) = 6.8618  [(A/G, 10%, 5) = 1.8101, (P/A, 10%, 5) = 3.7908]
- (P/F, 10%, 3) = 0.7513

#### Equation
PW of receipts = PW of outflow

A (P/A, 10%, 10) + 2G (P/G, 10%, 5)(P/F, 10%, 3) = 50,000

Put A = 3G:

- 3G × 6.1446 + 2G × 6.8618 × 0.7513 = 50,000
- 18.4337 G + 10.3107 G = 50,000
- 28.7444 G = 50,000

**Answer: G = Rs 1,739.47 and A = 3G = Rs 5,218.40.**

## Check yourself

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

1. Gradient 0 at year 4, 2G at 5, ... 8G at 8: where is the base?
   A. Year 0
   B. Year 4
   C. Year 8
   D. Year 3 with 2G increments
   Answer: D. P/G value lies at year 3, so multiply by (P/F,10%,3). (Key point: Gradient base)

2. Equation used?
   A. A(P/A,10%,10) + 2G(P/G,10%,5)(P/F,10%,3) = 50,000
   B. A(P/A,10%,10) + G(P/G,10%,5) = 50,000
   C. A(F/A) + G(F/G) = 50,000
   D. A(P/A,10%,8) = 50,000
   Answer: A. Gradient increment is 2G. (Key point: Equation)

3. After substitution, what gives G?
   A. G x 6.1446 = 50,000
   B. G(18.4337 + 10.3107) = 50,000
   C. G(3+2) = 50,000
   D. G = 50,000 x 28.7444
   Answer: B. Combined coefficient 28.7444. (Key point: Substituting A=3G)

4. G and A?
   A. G = 5,218, A = 1,739
   B. G = 50,000, A = 150,000
   C. G = Rs 1,739.47, A = Rs 5,218.40
   D. G = 28.74, A = 86
   Answer: C. A = 3G. (Key point: Result)

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