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HIGHER

3.8H Percentage Change: Higher Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 70 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A jacket costs 85 pounds. It is reduced by 15% in a sale. Work out the sale price of the jacket to the nearest penny.

(Total for Question 1 is 2 marks)

A TV is sold with 20% discount and then a further 10% off the reduced price. The original price was 450 pounds.
Work out the final selling price. Give your answer to the nearest penny.

(Total for Question 2 is 3 marks)

A smartphone price increases from 320 pounds to 400 pounds.
Work out the percentage increase. Give your answer correct to 1 decimal place.

(Total for Question 3 is 3 marks)

A car's value is depreciated by 12% each year. Its current value is 16,200 pounds.
Work out the value of the car after 2 years. Give your answer to the nearest pound.

(Total for Question 4 is 4 marks)

A shop shows an item with price 48 pounds including 20% VAT. Work out the price of the item before VAT. Give your answer to the nearest penny.

(Total for Question 5 is 5 marks)

A population increases by 8% each year for 3 years and then decreases by 5% in the fourth year.
Starting population = 12 500.
Work out the population after the 4 years. Give your answer to the nearest whole number.

(Total for Question 6 is 6 marks)

The price of a commodity first increases by p% then decreases by q% and ends up at the same price as it started.
Show that to first order (neglecting the small product term pq/100), q approximately equals p.
You must show the algebraic steps leading to the approximation.

(Total for Question 7 is 7 marks)

A fund grows by 6% per year compounded annually. Show that after n years the amount is A = P(1.06)ⁿ.
Then find how many whole years are needed for the fund to at least double if P = 8000. Give your answer as an integer number of years.

(a)Show that after n years the amount is A = P(1.06)ⁿ.(3)

(b)Find the smallest whole number n such that A ≥ 2P when P = 8000.(5)

(Total for Question 8 is 8 marks)

A machine is sold after 4 years. Its value follows V = 60 000 x (0.74)^t, where t is the age in years.
(a) Work out the value after 4 years. (3 marks)
(b) The owner expects the value to be at least 20 000 pounds. Use the formula to show whether this expectation is met after 6 years. Give a brief justification with your answer. (4 marks)
(c) Find the age t when the value will first fall below 25 000 pounds. Give your answer to 1 decimal place. (5 marks)

(a)Work out the value after 4 years.(3)

(b)Use the formula to show whether value after 6 years is at least 20 000 pounds. Give a brief justification.(4)

(c)Find t when V first falls below 25 000 pounds. Give answer to 1 decimal place.(5)

(Total for Question 9 is 12 marks)

Mark scheme · 3.8H Percentage Change: Higher Tier Practice

Question 1

M1 calculate 15% of 85 or apply 85 x 0.85 method
A1 72.25 pounds cao
Answer: 72.25

Question 2

M1 apply first reduction: 450 x 0.80 = 360 or show method
M1 apply second reduction: 360 x 0.90 = 324 or equivalent
A1 324.00 pounds cao
Answer: 324.00

Question 3

M1 calculate change 400 - 320 = 80 and form 80/320
M1 evaluate 80/320 = 0.25 and convert to percent
A1 25.0% cao
Answer: 25.0

Question 4

M1 use factor 0.88 for one year and show one step
M1 apply factor twice: 16200 x 0.88 x 0.88 or 16200 x 0.88²
A1 12545.28 to 2 d.p. (intermediate) ft
A1 12545 pounds (nearest pound) cao
Answer: 12545

Question 5

M1 identify that 48 is 120% of the pre-VAT price or use reverse percentage
M1 set up 48 = 1.20 x original and divide: original = 48 / 1.20
M1 calculate 48 / 1.20 = 40.0
A1 40.00 pounds cao
B1 state check: 40 x 1.20 = 48 oe
Answer: 40.00

Question 6

M1 use growth factor 1.08 for each of first 3 years
M1 use decrease factor 0.95 for year 4
M1 combine factors: 12500 x 1.08³ x 0.95
M1 evaluate 1.08³ = 1.259712 (or equivalent) and multiply through
A1 14959.08 (intermediate) ft
A1 14959 people (nearest whole number) cao
Answer: 14959

Question 7

M1 state final price = initial x (1 + p/100) x (1 - q/100)
M1 set (1 + p/100)(1 - q/100) = 1 since final = initial
M1 expand: 1 + p/100 - q/100 - (pq)/10000 = 1
M1 cancel 1 and rearrange to p/100 - q/100 - (pq)/10000 = 0
M1 rearrange to q/100 = p/100 - (pq)/10000
M1 neglect (pq)/10000 as small to get q/100 approx= p/100
A1 conclude q approx= p with brief justification cao
Answer: q approx= p

Question 8

(a) M1 state amount after 1 year = P(1 + 0.06), after 2 years = P(1.06)²
(a) M1 state pattern and generalise for n years
(a) A1 A = P(1.06)ⁿ cao
(a) Answer: A = P(1.06)ⁿ
(b) M1 set up (1.06)ⁿ ≥ 2
(b) M1 take logs or use growth table / calculator to solve n ≥ ln2 / ln1.06
(b) M1 calculate ln2 / ln1.06 = approx 11.897... (or use log10 equivalent)
(b) A1 state n must be integer ≥ 11.897
(b) A1 n = 12 years cao
(b) Answer: 12

Question 9

(a) M1 substitute t = 4: 60000 x 0.74⁴ shown
(a) M1 evaluate 0.74⁴ = 0.299865... or equivalent
(a) A1 17991.95 to 2 d.p. (nearest penny) cao
(a) Answer: 17991.95
(b) M1 substitute t = 6: 60000 x 0.74⁶ shown
(b) M1 evaluate 0.74⁶ approx= 0.1642 (or equivalent) and multiply
(b) M1 calculate value approx= 9852.4 (or approx 9852) and compare with 20000
(b) A1 conclude value < 20000 and expectation not met cao
(b) Answer: No; value approx= 9852 (< 20000)
(c) M1 set up inequality 60000 x 0.74^t < 25000
(c) M1 rearrange: 0.74^t < 25000/60000 = 5/12
(c) M1 take logs: t ln0.74 < ln(5/12) and solve for t
(c) A1 compute t > ln(5/12)/ln(0.74) approx= 2.91... and choose first t where value falls below
(c) A1 t = 2.9 years (1 d.p.) cao
(c) Answer: 2.9

Percentage Change: Higher Tier Practice - Worksheets, Questions and Revision