ACCA PM · Chapter 11
Quantitative analysis in budgeting MCQs with Answers
10 multiple-choice questions on Quantitative analysis in budgeting for ACCA PM Performance Management. Try each one before revealing the answer and explanation.
Practise this chapter interactivelyQuestion 1
The first unit of a new product took 100 hours to make. An 80% learning curve applies. What is the cumulative average time per unit when 8 units have been produced?
- A) 64.0 hours
- B) 409.6 hours
- C) 40.0 hours
- D) 51.2 hours
Show answer & explanation
Answer: D) 51.2 hours
Each time cumulative output doubles, the cumulative average time falls to 80% of its previous value. 1 unit: 100 hours; 2 units: 80; 4 units: 64; 8 units: 51.2 hours. Total time for 8 units = 8 x 51.2 = 409.6 hours.
Question 2
The first unit of a product took 50 hours. An 85% learning curve applies, and the learning index is b = log 0.85 / log 2 = -0.2345 (to four decimal places). How long will the 20th unit take to produce (to one decimal place)?
- A) 24.8 hours
- B) 495.4 hours
- C) 26.1 hours
- D) 19.1 hours
Show answer & explanation
Answer: D) 19.1 hours
Using y = ax^b: cumulative average for 20 units = 50 x 20^-0.2345 = 24.767 hours, so total time = 20 x 24.767 = 495.35 hours. Cumulative average for 19 units = 50 x 19^-0.2345 = 25.067 hours, so total = 19 x 25.067 = 476.27 hours. Time for the 20th unit = 495.35 - 476.27 = 19.08, which is 19.1 hours to one decimal place (the unrounded index gives 19.08 as well). 24.8 hours is the cumulative average, 495.4 hours the cumulative total, and 26.1 hours the 16-unit average (50 x 0.85^4).
Question 3
The first batch of a new product took 200 hours. The first four batches took a total of 512 hours. What learning rate is being experienced?
- A) 64%
- B) 80%
- C) 86%
- D) 90%
Show answer & explanation
Answer: B) 80%
Cumulative average time for 4 batches = 512 / 4 = 128 hours. Four batches is two doublings from one, so 200 x r^2 = 128, giving r^2 = 0.64 and r = 0.80, an 80% learning rate.
Question 4
In which of the following situations is a learning curve most likely to apply?
- A) A highly automated process where machines set the pace of work
- B) A labour-intensive assembly process for a new, complex product, with a stable workforce and continuous production
- C) A long-established product whose workers have made it for many years
- D) A process where there are long breaks between production runs and high staff turnover
Show answer & explanation
Answer: B) A labour-intensive assembly process for a new, complex product, with a stable workforce and continuous production
Learning effects arise when a new task is repeated by people who retain what they have learned. They are strongest in labour-intensive, complex, early-stage production with continuity. Machine-paced work, mature products and frequent breaks or staff turnover limit or eliminate the learning effect.
Question 5
Total production costs were $74,000 at an activity level of 12,000 units and $102,000 at 20,000 units. Using the high-low method, what is the expected total cost at 17,000 units?
- A) $104,833
- B) $86,700
- C) $59,500
- D) $91,500
Show answer & explanation
Answer: D) $91,500
Variable cost per unit = ($102,000 - $74,000) / (20,000 - 12,000) = $3.50. Fixed cost = $74,000 - (12,000 x $3.50) = $32,000. Cost at 17,000 units = $32,000 + (17,000 x $3.50) = $91,500.
Question 6
Total costs were $61,000 at 10,000 units and $93,000 at 18,000 units. Fixed costs increase by $8,000 when activity exceeds 15,000 units; the $93,000 includes this step. Using the high-low method, what is the expected total cost at 16,000 units?
- A) $85,000
- B) $79,000
- C) $87,000
- D) $95,000
Show answer & explanation
Answer: C) $87,000
Remove the step before applying high-low: $93,000 - $8,000 = $85,000. Variable cost = ($85,000 - $61,000) / (18,000 - 10,000) = $3.00 per unit. Fixed cost below 15,000 units = $61,000 - (10,000 x $3.00) = $31,000. Cost at 16,000 units = $31,000 + $8,000 + (16,000 x $3.00) = $87,000.
Question 7
A company uses linear regression (y = a + bx) to estimate overhead costs (y, $000) from machine hours (x, 000 hours). From five observations: n = 5, sum of x = 50, sum of y = 500, sum of xy = 5,240 and sum of x squared = 540. Using b = (n sum xy - sum x sum y) / (n sum x^2 - (sum x)^2) and a = (sum y / n) - b(sum x / n), what is the forecast overhead cost when 15,000 machine hours are worked?
- A) $90,000
- B) $130,000
- C) $150,000
- D) $49,000
Show answer & explanation
Answer: B) $130,000
b = (5 x 5,240 - 50 x 500) / (5 x 540 - 50^2) = (26,200 - 25,000) / (2,700 - 2,500) = 1,200 / 200 = 6. a = (500 / 5) - 6 x (50 / 5) = 100 - 60 = 40. Forecast y = 40 + (6 x 15) = 130, so overheads are $130,000.
Question 8
The correlation coefficient between advertising spend and sales volume has been calculated as 0.9. What proportion of the variation in sales volume is explained by variation in advertising spend?
- A) 90%
- B) 81%
- C) 95%
- D) 10%
Show answer & explanation
Answer: B) 81%
The coefficient of determination equals r squared = 0.9 x 0.9 = 0.81. Therefore 81% of the variation in sales volume is explained by the variation in advertising spend; the remaining 19% is due to other factors.
Question 9
In the learning curve formula y = ax^b, b = log r / log 2. What is the value of b for a 90% learning curve (to three decimal places)?
- A) -0.105
- B) -0.100
- C) -0.322
- D) -0.152
Show answer & explanation
Answer: D) -0.152
b = log 0.9 / log 2 = -0.0458 / 0.3010 = -0.152. The same result is obtained using natural logs, because the ratio of logs does not depend on the base.
Question 10
Which of the following is a limitation of the high-low method of cost estimation?
- A) It requires complex statistical software to calculate
- B) It uses only the highest and lowest activity observations, which may not be representative of normal conditions
- C) It cannot separate a cost into fixed and variable elements
- D) It considers every observation in the data set equally
Show answer & explanation
Answer: B) It uses only the highest and lowest activity observations, which may not be representative of normal conditions
The high-low method is simple and separates fixed and variable elements, but it relies on just two observations. If either is unusual, the estimate will be distorted. Regression analysis uses all data points and is therefore usually more reliable.