Sec 3 G1 Maths Indices and standard form worksheet with answers

A free Sec 3 G1 Maths worksheet on indices and standard form: 9 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.

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

Or practise it online with step-by-step help when you get one wrong.

Try a question: Indices and standard form

Free · no sign-up

Answer a real practice question. Get it wrong and MugCorner shows you where it went wrong, with hints and a step-by-step guide.

MugCornerT-Leng Tuition · mugcornersg.com

Sec 3 G1 Maths worksheet: Indices and standard form

Total: 30 marksSuggested time: 50 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

(a)Evaluate

10^2

Answer (a)[1]

(b)Evaluate

(-6)^5

Answer (b)[1]

(a)Evaluate

3^{0}

Answer (a)[1]

(b)Evaluate

2^{0}

Answer (b)[1]

(a)Write

9.08 \times 10^{-5}

as an ordinary number.

Answer (a)[1]

(b)Write

9.08 \times 10^{-4}

as an ordinary number.

Answer (b)[1]

Evaluate

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

Answer[2]

(a)Express

12\,000\,000

in standard form.

Answer (a)[1]

(b)Express

0.000\,000\,03

in standard form.

Answer (b)[1]

(a)Write

5 \times 5 \times 5 \times 5 \times 5

in index notation.

Answer (a)[1]

(b)Write

13 \times 13 \times 13

in index notation.

Answer (b)[1]

(a)Wei Ling invests $8000 at 2.5% per annum compound interest, compounded annually. Find the total amount after 2 years, correct to the nearest cent.

Answer (a)$[2]

(b)Priya invests $10 000 at 4% per annum compound interest, compounded annually. Find the interest earned after 4 years, correct to the nearest cent.

Answer (b)$[3]

Section B

(a)(i)Write

7 \times 7 \times 3 \times 7 \times 7 \times 3

in index notation.

Answer (a)(i)[1]

(a)(ii)Write

11 \times 11 \times 11 \times 11 \times 11 \times 11

in index notation.

Answer (a)(ii)[1]

(b)The value of a car bought for $40 000 depreciates at 15% per annum. Find its value after 5 years, correct to the nearest cent.

Answer (b)$[2]

(c)(i)Evaluate

\left(\dfrac{3}{2}\right)^{0}

Answer (c)(i)[1]

(c)(ii)Evaluate

3^{-1}

Answer (c)(ii)[1]

(a)The value of a car bought for $18 000 depreciates at 15% per annum. Find its value after 5 years, correct to the nearest cent.

Answer (a)$[2]

(b)Evaluate

\left(\dfrac{5}{4}\right)^{-3}

Answer (b)[2]

(c)(i)Calculate

(7.5 \times 10^{3}) \div (5 \times 10^{-4})

, giving your answer in standard form.

Answer (c)(i)[1]

(c)(ii)Calculate

(8.7 \times 10^{6}) + (2.9 \times 10^{4})

, giving your answer in standard form.

Answer (c)(ii)[2]

END OF PAPER

Answers and worked solutions

Question 1

(a)Answer:

= 100

[1]

10^2 = 10 \times 10

= 100

(b)Answer:

(-6)^5 = -7776

[1]

(-6)^5 = (-6) \times (-6) \times (-6) \times (-6) \times (-6)

There are 5 negative factors: an odd number, so the answer is negative.

(-6)^5 = -7776

Question 2

(a)Answer:

1

[1]

3^{0} = 1

1

(b)Answer:

1

[1]

2^{0} = 1

1

Question 3

(a)Answer:

9.08 \times 10^{-5} = 0.000\,090\,8

[1]

A negative power of 10 makes the number smaller: the point moves left.Number of places the point moves:

5

9.08 \times 10^{-5} = 0.000\,090\,8

(b)Answer:

9.08 \times 10^{-4} = 0.000\,908

[1]

A negative power of 10 makes the number smaller: the point moves left.Number of places the point moves:

4

9.08 \times 10^{-4} = 0.000\,908

Question 4

Answer:

\dfrac{9}{4}

[2]

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

\left(\dfrac{3}{2}\right)^{2} = \dfrac{9}{4}

\dfrac{9}{4}

Question 5

(a)Answer:

12\,000\,000 = 1.2 \times 10^{7}

[1]

Number of places the point moves:

7

12\,000\,000 = 1.2 \times 10^{7}

(b)Answer:

0.000\,000\,03 = 3 \times 10^{-8}

[1]

Number of places the point moves:

8

0.000\,000\,03 = 3 \times 10^{-8}

Question 6

(a)Answer:

5 \times 5 \times 5 \times 5 \times 5 = 5^5

[1]

5 \times 5 \times 5 \times 5 \times 5 = 5^5

(b)Answer:

13 \times 13 \times 13 = 13^3

[1]

13 \times 13 \times 13 = 13^3

Question 7

(a)Answer:

A = 8405.00

[2]

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

A = 8000\left(1 + \dfrac{2.5}{100}\right)^{2}

A = 8405.00

(b)Answer: $1698.59[3]

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

A = 10000\left(1 + \dfrac{4}{100}\right)^{4}

A = 11698.5856\ldots

Interest

= A - P

= 11698.5856\ldots - 10000

$1698.59

Question 8

(a)(i)Answer:

7 \times 7 \times 3 \times 7 \times 7 \times 3 = 3^2 \times 7^4

[1]

7 \times 7 \times 3 \times 7 \times 7 \times 3 = 3^2 \times 7^4

(a)(ii)Answer:

11 \times 11 \times 11 \times 11 \times 11 \times 11 = 11^6

[1]

11 \times 11 \times 11 \times 11 \times 11 \times 11 = 11^6

(b)Answer: $17748.21[2]

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

A = 40000\left(1 - \dfrac{15}{100}\right)^{5}

A = 17748.2125\ldots

$17748.21

(c)(i)Answer:

1

[1]

\left(\dfrac{3}{2}\right)^{0} = 1

1

(c)(ii)Answer:

\dfrac{1}{3}

[1]

3^{-1} = \left(\dfrac{1}{3}\right)^{1}

\dfrac{1}{3}

Question 9

(a)Answer: $7986.70[2]

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

A = 18000\left(1 - \dfrac{15}{100}\right)^{5}

A = 7986.6956\ldots

$7986.70

(b)Answer:

\dfrac{64}{125}

[2]

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

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

\dfrac{64}{125}

(c)(i)Answer:

= 1.5 \times 10^{7}

[1]

(7.5 \div 5) \times 10^{3 - (-4)}

= 1.5 \times 10^{7}

(c)(ii)Answer:

= 8.729 \times 10^{6}

[2]

2.9 \times 10^{4} = 0.029 \times 10^{6}

(8.7 + 0.029) \times 10^{6}

= 8.729 \times 10^{6}

Sec 3 G1 Maths Indices and standard form worksheet with answers

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

Or practise it online with step-by-step help when you get one wrong.

Section B

(a)(i)Write

7 \times 7 \times 3 \times 7 \times 7 \times 3

in index notation.

Answer (a)(i)[1]

(a)(ii)Write

11 \times 11 \times 11 \times 11 \times 11 \times 11

in index notation.

Answer (a)(ii)[1]

(b)The value of a car bought for $40 000 depreciates at 15% per annum. Find its value after 5 years, correct to the nearest cent.

Answer (b)$[2]

(c)(i)Evaluate

\left(\dfrac{3}{2}\right)^{0}

Answer (c)(i)[1]

(c)(ii)Evaluate

3^{-1}

Answer (c)(ii)[1]

Answers and worked solutions

Question 1

(a)Answer:

= 100

[1]

10^2 = 10 \times 10

= 100

(b)Answer:

(-6)^5 = -7776

[1]

(-6)^5 = (-6) \times (-6) \times (-6) \times (-6) \times (-6)

There are 5 negative factors: an odd number, so the answer is negative.

(-6)^5 = -7776

Question 2

(a)Answer:

1

[1]

3^{0} = 1

1

(b)Answer:

1

[1]

2^{0} = 1

1

Question 3

(a)Answer:

9.08 \times 10^{-5} = 0.000\,090\,8

[1]

A negative power of 10 makes the number smaller: the point moves left.Number of places the point moves:

5

9.08 \times 10^{-5} = 0.000\,090\,8

(b)Answer:

9.08 \times 10^{-4} = 0.000\,908

[1]

A negative power of 10 makes the number smaller: the point moves left.Number of places the point moves:

4

9.08 \times 10^{-4} = 0.000\,908

Question 4

Answer:

\dfrac{9}{4}

[2]

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

\left(\dfrac{3}{2}\right)^{2} = \dfrac{9}{4}

\dfrac{9}{4}

Question 5

(a)Answer:

12\,000\,000 = 1.2 \times 10^{7}

[1]

Number of places the point moves:

7

12\,000\,000 = 1.2 \times 10^{7}

(b)Answer:

0.000\,000\,03 = 3 \times 10^{-8}

[1]

Number of places the point moves:

8

0.000\,000\,03 = 3 \times 10^{-8}

Question 6

(a)Answer:

5 \times 5 \times 5 \times 5 \times 5 = 5^5

[1]

5 \times 5 \times 5 \times 5 \times 5 = 5^5

(b)Answer:

13 \times 13 \times 13 = 13^3

[1]

13 \times 13 \times 13 = 13^3

Question 7

(a)Answer:

A = 8405.00

[2]

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

A = 8000\left(1 + \dfrac{2.5}{100}\right)^{2}

A = 8405.00

(b)Answer: $1698.59[3]

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

A = 10000\left(1 + \dfrac{4}{100}\right)^{4}

A = 11698.5856\ldots

Interest

= A - P

= 11698.5856\ldots - 10000

$1698.59

Question 8

(a)(i)Answer:

7 \times 7 \times 3 \times 7 \times 7 \times 3 = 3^2 \times 7^4

[1]

7 \times 7 \times 3 \times 7 \times 7 \times 3 = 3^2 \times 7^4

(a)(ii)Answer:

11 \times 11 \times 11 \times 11 \times 11 \times 11 = 11^6

[1]

11 \times 11 \times 11 \times 11 \times 11 \times 11 = 11^6

(b)Answer: $17748.21[2]

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

A = 40000\left(1 - \dfrac{15}{100}\right)^{5}

A = 17748.2125\ldots

$17748.21

(c)(i)Answer:

1

[1]

\left(\dfrac{3}{2}\right)^{0} = 1

1

(c)(ii)Answer:

\dfrac{1}{3}

[1]

3^{-1} = \left(\dfrac{1}{3}\right)^{1}

\dfrac{1}{3}

Question 9

(a)Answer: $7986.70[2]

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

A = 18000\left(1 - \dfrac{15}{100}\right)^{5}

A = 7986.6956\ldots

$7986.70

(b)Answer:

\dfrac{64}{125}

[2]

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

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

\dfrac{64}{125}

(c)(i)Answer:

= 1.5 \times 10^{7}

[1]

(7.5 \div 5) \times 10^{3 - (-4)}

= 1.5 \times 10^{7}

(c)(ii)Answer:

= 8.729 \times 10^{6}

[2]

2.9 \times 10^{4} = 0.029 \times 10^{6}

(8.7 + 0.029) \times 10^{6}

= 8.729 \times 10^{6}