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Either paper 2 marks(a)
Find the value of \(10^2\)
(b)Find the value of \(9^3\)
Show answer
(a) \(100\)
(b) \(729\)\(10^2\) means 10 multiplied by itself.
\(10 \times 10 = 100\)
\(9^3\) means three lots of 9 multiplied together.
\(9 \times 9 \times 9 = 729\)
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Either paper 3 marks(a)
Write \(3 \times 3 \times 3 \times 3\) in index form.
(b)Find the value of \(2^3\)
Show answer
(a) \(3^4\)
(b) \(8\)3 is used as a factor four times, so the index is 4.
\(3 \times 3 \times 3 \times 3 = 3^4\)
\(2 \times 2 = 4\)
\(4 \times 2 = 8\)
- (b) M1\(2 \times 2 = 4\)
- (b) A1\(2^3 = 8\)
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Either paper 1 mark
The square root of \(x\) is \(3\).
Work out the value of \(x\).
Show answer
\(9\)Undo the square root by raising the number to the power 2.
\(x = 3^2\)
\(3^2 = 3 \times 3 = 9\)
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Either paper 1 mark
Circle the value of \(4^{ 3 } + 7^{ 2 }\)
\(113\) \(121\) \(599\) \(3136\)
Show answer
\(113\)Work out each power on its own first.
\(4^{ 3 } = 64\) and \(7^{ 2 } = 49\)
Now add them: \(64 + 49 = 113\)
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Either paper 3 marks(a)
Work out the value of \(\sqrt{ 196 }\)
(b)Work out the value of \(\sqrt[3]{ 216 }\)
(c)Work out the value of \(\sqrt{ 196 } \times \sqrt[3]{ 216 }\)
Show answer
(a) \(14\)
(b) \(6\)
(c) \(84\)\(14 \times 14 = 196\), so \(\sqrt{ 196 } = 14\)
\(6 \times 6 \times 6 = 216\), so \(\sqrt[3]{ 216 } = 6\)
Take each root first, then multiply.
\(14 \times 6 = 84\)
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Either paper 2 marks
Show that 661 is the sum of a power of 5 and a square number.
Show answer
\(661 = 625 + 36 = 5^{ 4 } + 6^2\)Subtract powers of 5 from 661 and look for a square number left over.
\(661 - 125 = 536\), which is not a square number, because \(23^2 = 529\) and \(24^2 = 576\).
\(661 - 625 = 36\), and \(36 = 6^2\).
So \(661 = 625 + 36 = 5^{ 4 } + 6^2\), the sum of a power of 5 and a square number.
- M1subtracts powers of 5 from 661 until a square remains: \(661 - 625 = 36\)
- A1states \(36 = 6^2\) so \(661 = 5^{ 4 } + 6^2\)
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Either paper 3 marks
Show that exactly sixty-eight of the 4-digit numbers are square numbers.
Show answer
\(31^2 = 961\) is too small and \(100^2 = 10000\) is too big, so the bases run from \(32\) to \(99\): \(68\) numbersFind the last square number with too few digits and the first one with too many.
\(31^2 = 961\), which is less than \(1000\), so it is too small.
\(32^2 = 1024\) has 4 digits, so it is the first one that fits.
\(99^2 = 9801\) has 4 digits, so it is the last one that fits.
\(100^2 = 10000\), which is not less than \(10000\), so it is too big.
The square numbers are \(32^2, \ldots, 99^2\). Counting the bases \(32\) to \(99\) gives \(99 - 32 + 1 = 68\).
So there are exactly sixty-eight 4-digit square numbers.
- M1below the range: \(31^2 = 961\)
- M1above the range: \(100^2 = 10000\)
- A1bases \(32\) to \(99\) give \(68\) numbers
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Either paper 3 marks
The diagrams show the first two cube numbers, 1 and 8.
Show that the sixth cube number is the sum of three different primes.
Show answer
e.g. \(2 + 17 + 197 = 216\), which is \(6^3\), and \(2\), \(17\) and \(197\) are all prime and all different (any other valid triple is also correct)The cubes in the diagrams have edges 1 and 2, so the cube numbers are \(1^3 = 1\) and \(2^3 = 8\).
The sixth cube number has edge 6: \(6^3 = 216\).
Three odd primes add to an odd number, so an even cube needs the only even prime, 2, as one of the three.
Try \(2 + 17 + 197\):
\(2 + 17 = 19\), and \(19 + 197 = 216\)
\(2\), \(17\) and \(197\) are all prime and no two are the same, so 216 is the sum of three different primes.
- B1the sixth cube number is \(6^3 = 216\)
- M1three different prime numbers, for example \(2\), \(17\), \(197\)
- A1sum shown to be \(216\), for example \(2 + 17 + 197 = 216\)
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Either paper 2 marks
Work out the value of \(\left(2\frac{1}{4}\right)^2\)
Give your answer as a mixed number.
Show answer
\(5\frac{1}{16}\)\(\left(2\frac{1}{4}\right)^2\) means \(2\frac{1}{4} \times 2\frac{1}{4}\)
Change the mixed number to an improper fraction first, then multiply.
\(\frac{9}{4} \times \frac{9}{4} = \frac{81}{16} = 5\frac{1}{16}\)
- M1repeated product \(2\frac{1}{4} \times 2\frac{1}{4}\)
- A1value \(5\frac{1}{16}\)
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Non-calculator 2 marks
Do not use a calculator.
(a) Here are four numbers.
Circle the number that is a square number.99 999 1,331 8,100 (b) Here are four more numbers.
Circle the number that is half of a square number.9 74 147 162 Show answer
(a) \(8100\)
(b) \(162\)A square number has a whole-number square root and a cube number has a whole-number cube root. Compare each number with the squares and cubes either side of it.
\(8100\): \(90^2 = 8100\) and \(91^2 = 8281\); \(8100\) is the first of these, so it is a square number. \(20^3 = 8000\) and \(21^3 = 9261\); \(8100\) lies between them, so it is not a cube number.
\(1331\): \(36^2 = 1296\) and \(37^2 = 1369\); \(1331\) lies between them, so it is not a square number. \(11^3 = 1331\) and \(12^3 = 1728\); \(1331\) is the first of these, so it is a cube number.
\(999\): \(31^2 = 961\) and \(32^2 = 1024\); \(999\) lies between them, so it is not a square number. \(9^3 = 729\) and \(10^3 = 1000\); \(999\) lies between them, so it is not a cube number.
\(99\): \(9^2 = 81\) and \(10^2 = 100\); \(99\) lies between them, so it is not a square number. \(4^3 = 64\) and \(5^3 = 125\); \(99\) lies between them, so it is not a cube number.
Only \(8100\) is a square number.
To test each number, double it. The four results are 18, 148, 294, 324.
The square numbers up to the size needed are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961.
Only \(324\) is a square number: \(18^{ 2 } = 324\).
So the number that is half of a square number is \(162\).