Secondary 1 Maths sample questions with answers

Primes, HCF and LCM

1. 1List all the prime numbers between 30 and 50.

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   Answer: 31, 37, 41, 43, 47

   1. Test each number from 31 to 49 for factors other than 1 and itself.
   2. Only 31, 37, 41, 43 and 47 have exactly two factors.

2. 2Express 252 as a product of its prime factors, giving your answer in index notation.

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   Answer: 252 = 2^2 × 3^2 × 7

   1. 252 = 2 × 2 × 3 × 3 × 7
   2. In index notation, 252 = 2^2 × 3^2 × 7

3. 3Is 119 a prime number? Give a reason for your answer.

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   Answer: No. 119 = 7 × 17

   1. 119 = 7 × 17
   2. So 119 has factors other than 1 and itself (1, 7, 17 and 119).

4. 4By listing the factors, find the highest common factor (HCF) of 36 and 54.

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   Answer: 18

   1. Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
   2. Factors of 54: 1, 2, 3, 6, 9, 18, 27, 54
   3. The largest factor in both lists is 18.

5. 5Find the lowest common multiple (LCM) of 8 and 14.

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   Answer: 56

   1. 8 = 2^3 and 14 = 2 × 7
   2. LCM = 2^3 × 7 = 56

6. 6It is given that 1512 = 2^a × 3^b × 7. Find the value of a and the value of b.

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   Answer: a = 3, b = 3

   1. 1512 = 2 × 2 × 2 × 3 × 3 × 3 × 7
   2. 1512 = 2^3 × 3^3 × 7, so a = 3 and b = 3

7. 7Using prime factorisation, find the value of sqrt(576). Show your working.

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   Answer: 24

   1. 576 = 2^6 × 3^2
   2. sqrt(576) = 2^3 × 3 = 24

8. 8Using prime factorisation, find the value of the cube root of 1728. Show your working.

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   Answer: 12

   1. 1728 = 2^6 × 3^3
   2. Cube root of 1728 = 2^2 × 3 = 12

9. 9Using prime factorisation, (a) express 84 and 132 as products of their prime factors, (b) find the HCF of 84 and 132, (c) find the LCM of 84 and 132.

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   Answer: (a) 84 = 2^2 × 3 × 7 and 132 = 2^2 × 3 × 11 (b) 12 (c) 924

   1. (a) 84 = 2^2 × 3 × 7 and 132 = 2^2 × 3 × 11
   2. (b) HCF = 2^2 × 3 = 12
   3. (c) LCM = 2^2 × 3 × 7 × 11 = 924

10. 10Is 91 a prime number? Explain your answer.

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    Answer: No. 91 = 7 × 13, so 91 is a composite number.

    1. 91 is odd, so 2 is not a factor.
    2. Digit sum 9 + 1 = 10, which is not divisible by 3, so 3 is not a factor.
    3. 91 does not end in 0 or 5, so 5 is not a factor.
    4. Try 7: 91 ÷ 7 = 13 exactly. So 91 = 7 × 13.
    5. 91 has factors 1, 7, 13 and 91, which is more than two factors.

11. 11Express 2016 as a product of its prime factors in index notation.

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    Answer: 2016 = 2^5 × 3^2 × 7

    1. Divide by 2 five times: 2016, 1008, 504, 252, 126, 63
    2. 63 = 3 × 3 × 7
    3. 2016 = 2^5 × 3^2 × 7