Sec 3 E-Math Indices, compound interest and standard form worksheet with answers

A free Sec 3 E-Math worksheet on indices, compound interest and standard form: 8 questions, 30 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 3 E-Math worksheet: Indices, compound interest and standard form

Total: 30 marksSuggested time: 45 minutes

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

1 megabyte

= 10^{6}

bytes and 1 gigabyte

= 10^{9}

bytes. Express

16

megabytes as a percentage of

4

gigabytes, giving your answer in standard form.

Answer%[2]

Simplify

3a^{3} \times 3a^{3}

Answer[2]

The value of a car bought for $90 000 depreciates at 20% per annum. Find its value after 3 years, correct to the nearest cent.

Answer$[2]

(a)Evaluate

\left(\dfrac{16}{25}\right)^{-\frac{3}{2}}

Answer (a)[2]

(b)Given that

a = 5^{x}

, express

\left(\dfrac{1}{5}\right)^{x - 1}

in terms of

a

Answer (b)[2]

Section B

(a)Given that

2 = 2^k

, write down the value of

k

Answer (a)[1]

(b)Hence solve the equation

2^{x - 2} = 16

Answer (b)[2]

A collector buys a painting for $2000. Its value increases by 6% each year, compounded yearly.

(a)Show that the value of a painting after 2 years is $2247.20.

[2]

(b)Find the least number of complete years after which the value of a painting is more than $4000.

Answer (b)years[3]

(c)A collector sells a painting at the end of the year found in part (b). Find the profit, correct to the nearest dollar.

Answer (c)$[2]

(d)Instead, the buyer could have put the $2000 in a bank that pays 6% interest for the first year, then 3.5% per annum compound interest, compounded yearly, for each year after that. Find the amount in the bank after the same number of years as in part (b), correct to the nearest cent.

Answer (d)$[2]

(e)Which would have been worth more at that time?

Answer (e)[1]

Light travels at

3 \times 10^{8}

m/s. The distance from the Sun to the Earth is

1.5 \times 10^{11}

(a)Find the time, in seconds, that light takes to travel from the Sun to the Earth. Give your answer in standard form.

Answer (a)s[2]

(b)Convert your answer to minutes, correct to 3 significant figures.

Answer (b)min[1]

A new car costs $96 000. Its value depreciates by 15% each year.

(a)Find the value of the car after 3 years.

Answer (a)$[2]

(b)Find the percentage of the original value that the car has lost after 3 years, correct to 1 decimal place.

Answer (b)%[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

= 4 \times 10^{-1}\%

[2]

\dfrac{16 \times 10^{6}}{4 \times 10^{9}} \times 100\%

= 400 \times 10^{-3}\%

= 4 \times 10^{-1}\%

Question 2

Answer:

9a^{6}

[2]

Numbers:

3 \times 3 = 9

a

: add the indices,

3 + 3 = 6

9a^{6}

Question 3

Answer:

A = 46080.00

[2]

A = P\left(1 - \dfrac{r}{100}\right)^n

A = 90000\left(1 - \dfrac{20}{100}\right)^{3}

A = 46080.00

Question 4

(a)Answer:

\dfrac{125}{64}

[2]

\left(\dfrac{16}{25}\right)^{-\frac{3}{2}} = \left(\dfrac{25}{16}\right)^{\frac{3}{2}}

\sqrt{\dfrac{25}{16}} = \dfrac{5}{4}

\left(\dfrac{5}{4}\right)^{3} = \dfrac{125}{64}

\dfrac{125}{64}

(b)Answer:

= \dfrac{5}{a}

[2]

\dfrac{1}{5} = 5^{-1}

\left(\dfrac{1}{5}\right)^{x - 1} = 5^{-1(x - 1)}

= 5^{-1x} \times 5^{1}

= (5^{x})^{-1} \times 5

= \dfrac{5}{a}

Question 5

(a)Answer:

2^{1}

, so the index is

1

[1]

2 = 2^{1}

(b)Answer:

x = 6

[2]

2^{x - 2} = 2^{4}

x - 2 = 4

x = 6

Question 6

(a)[2]

\text{Value} = 2000\left(1 + \dfrac{6}{100}\right)^{2}

= 2000 \times 1.06^{2}

= 2247.2

(b)Answer:

12

years[3]

\text{After } 11 \text{ years: } 2000 \times 1.06^{11} = 3796.59\ldots

, less than

4000

\text{After } 12 \text{ years: } 2000 \times 1.06^{12} = 4024.39\ldots

, more than

4000

The least number of complete years is

12

(c)Answer: $2024[2]

\text{Profit} = 4024.39\ldots - 2000

= 2024.39\ldots

\approx 2024

(to the nearest dollar)

(d)Answer: $3095.14[2]

\text{Amount} = 2000\left(1 + \dfrac{6}{100}\right)\left(1 + \dfrac{3.5}{100}\right)^{11}

= 3095.13\ldots

\approx 3095.14

(to the nearest cent)

(e)Answer: a painting:

4024.39\ldots > 3095.14

[1]

4024.39\ldots > 3095.14

, so a painting would have been worth more.

Question 7

(a)Answer:

5 \times 10^{2}

s[2]

Time

= \dfrac{\text{distance}}{\text{speed}}

= \dfrac{1.5 \times 10^{11}}{3 \times 10^{8}}

= 0.5 \times 10^{3}

\phantom{} = 5 \times 10^{2}

(b)Answer:

8.33

minutes[1]

500 \div 60 = 8.333\ldots

\phantom{500 \div 60} = 8.33

minutes (3 s.f.)

Question 8

(a)Answer: $58 956[2]

96000 \times 0.85^{3}

= 58\,956

(b)Answer:

38.6\%

[2]

Loss

= 96\,000 - 58\,956

= 37\,044

\dfrac{37\,044}{96\,000} \times 100\% = 38.5875\%

\phantom{\dfrac{37\,044}{96\,000} \times 100\%} = 38.6\%

Practice questions by skill

Sec 3 E-Math Indices, compound interest and standard form worksheet with answers

In the print window, choose Save as PDF to download it.

Answers and worked solutions

Question 1

Answer:

= 4 \times 10^{-1}\%

[2]

\dfrac{16 \times 10^{6}}{4 \times 10^{9}} \times 100\%

= 400 \times 10^{-3}\%

= 4 \times 10^{-1}\%

Question 2

Answer:

9a^{6}

[2]

Numbers:

3 \times 3 = 9

a

: add the indices,

3 + 3 = 6

9a^{6}

Question 3

Answer:

A = 46080.00

[2]

A = P\left(1 - \dfrac{r}{100}\right)^n

A = 90000\left(1 - \dfrac{20}{100}\right)^{3}

A = 46080.00

Question 4

(a)Answer:

\dfrac{125}{64}

[2]

\left(\dfrac{16}{25}\right)^{-\frac{3}{2}} = \left(\dfrac{25}{16}\right)^{\frac{3}{2}}

\sqrt{\dfrac{25}{16}} = \dfrac{5}{4}

\left(\dfrac{5}{4}\right)^{3} = \dfrac{125}{64}

\dfrac{125}{64}

(b)Answer:

= \dfrac{5}{a}

[2]

\dfrac{1}{5} = 5^{-1}

\left(\dfrac{1}{5}\right)^{x - 1} = 5^{-1(x - 1)}

= 5^{-1x} \times 5^{1}

= (5^{x})^{-1} \times 5

= \dfrac{5}{a}

Question 5

(a)Answer:

2^{1}

, so the index is

1

[1]

2 = 2^{1}

(b)Answer:

x = 6

[2]

2^{x - 2} = 2^{4}

x - 2 = 4

x = 6

Question 6

(a)[2]

\text{Value} = 2000\left(1 + \dfrac{6}{100}\right)^{2}

= 2000 \times 1.06^{2}

= 2247.2

(b)Answer:

12

years[3]

\text{After } 11 \text{ years: } 2000 \times 1.06^{11} = 3796.59\ldots

, less than

4000

\text{After } 12 \text{ years: } 2000 \times 1.06^{12} = 4024.39\ldots

, more than

4000

The least number of complete years is

12

(c)Answer: $2024[2]

\text{Profit} = 4024.39\ldots - 2000

= 2024.39\ldots

\approx 2024

(to the nearest dollar)

(d)Answer: $3095.14[2]

\text{Amount} = 2000\left(1 + \dfrac{6}{100}\right)\left(1 + \dfrac{3.5}{100}\right)^{11}

= 3095.13\ldots

\approx 3095.14

(to the nearest cent)

(e)Answer: a painting:

4024.39\ldots > 3095.14

[1]

4024.39\ldots > 3095.14

, so a painting would have been worth more.

Question 7

(a)Answer:

5 \times 10^{2}

s[2]

Time

= \dfrac{\text{distance}}{\text{speed}}

= \dfrac{1.5 \times 10^{11}}{3 \times 10^{8}}

= 0.5 \times 10^{3}

\phantom{} = 5 \times 10^{2}

(b)Answer:

8.33

minutes[1]

500 \div 60 = 8.333\ldots

\phantom{500 \div 60} = 8.33

minutes (3 s.f.)

Question 8

(a)Answer: $58 956[2]

96000 \times 0.85^{3}

= 58\,956

(b)Answer:

38.6\%

[2]

Loss

= 96\,000 - 58\,956

= 37\,044

\dfrac{37\,044}{96\,000} \times 100\% = 38.5875\%

\phantom{\dfrac{37\,044}{96\,000} \times 100\%} = 38.6\%