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13+ · Mathematics

CE.M24 Number, Ratio and Algebra Depth

ISEB COMMON ENTRANCE ISEB CE 13+ Mathematics · Calculator allowed · about 105 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Simplify each expression, giving your answer as a single power.

(a)Simplify p⁶ * p^2.(1)

(b)Simplify q⁹ / q^4.(1)

(Total for Question 1 is 2 marks)

Simplify each expression.

(a)Simplify (r³)^4.(1)

(b)Simplify 5m² * 3m^5.(2)

(Total for Question 2 is 3 marks)

Evaluate each of the following. Use the laws of indices where this helps.

(a)Evaluate 6^0.(1)

(b)Evaluate 3^-2, giving your answer as a fraction.(2)

(c)Evaluate 2^-1 * 2⁴ by first writing it as a single power.(2)

(Total for Question 3 is 5 marks)

Simplify each expression, giving your answer in its simplest form.

(a)Simplify (4x³)^2.(2)

(b)Simplify (2m²ⁿ)^3.(2)

(Total for Question 4 is 4 marks)

Simplify (12a^7b^3)/(4a^3b), giving your answer in its simplest form.

(Total for Question 5 is 3 marks)

Write each number in standard form or as an ordinary number, as appropriate.

(a)Write 52,000,000 in standard form.(1)

(b)Write 0.000318 in standard form.(1)

(c)Write 9.04 x 10⁵ as an ordinary number.(1)

(Total for Question 6 is 3 marks)

A laboratory records the number of bacteria present in four samples:
Sample W: 3.6 x 10⁵
Sample X: 4.1 x 10⁴
Sample Y: 2.9 x 10⁶
Sample Z: 8.7 x 10⁵
Write the samples in order of increasing number of bacteria.

(Total for Question 7 is 2 marks)

A memory card can store 3.15 x 10⁸ bytes of data. Each photograph taken by a venue's security camera has a file size of 4.5 x 10⁶ bytes. Calculate how many such photographs the memory card can store.

(Total for Question 8 is 3 marks)

A charity quiz night raises £180, which is shared between venue hire and the prize fund in the ratio 2:7.

(a)Find the value of one share.(1)

(b)Work out how much money is used for the prize fund.(2)

(Total for Question 9 is 3 marks)

An allotment plot has an area of 108 m^2. It is divided between vegetables, flowers and a compost area in the ratio 7:3:2.

(a)Find the value of one share.(1)

(b)Work out the area used for vegetables, the area used for flowers, and the area used for the compost area.(3)

(Total for Question 10 is 4 marks)

In a school orchestra, the ratio of string players to wind players is 5:4. The ratio of wind players to percussion players is 2:3.

(a)Find the ratio of string players to wind players to percussion players, in its simplest form.(3)

(b)There are 24 wind players. Work out the number of string players.(2)

(Total for Question 11 is 5 marks)

Two cyclists, Leo and Nadia, share the cost of repairing a tandem bicycle in the ratio 3:5, because Nadia used it more often. Leo pays £18 less than Nadia.

(a)Using x for one share, write expressions for the amounts Leo and Nadia pay.(1)

(b)Form an equation using the given difference, and solve it to find the value of x.(2)

(c)Find the total cost of the repair.(2)

(Total for Question 12 is 5 marks)

6 metres of ribbon cost £4.50. Isla buys 10 metres of the same ribbon. Assuming the cost is directly proportional to the length, calculate the cost of Isla's ribbon.

(Total for Question 13 is 3 marks)

The cost, C pounds, of hiring a bouncy castle for a birthday party is directly proportional to the number of hours, h, it is hired for. Hiring the bouncy castle for 3 hours costs £45.

(a)Find the value of k in the formula C = kh.(2)

(b)Calculate the cost of hiring the bouncy castle for 7 hours.(2)

(Total for Question 14 is 4 marks)

The energy, E joules, stored in a stretched spring is directly proportional to the square of its extension, x cm. When the extension is 3 cm, the stored energy is 45 joules.

(a)Find the value of k in the formula E = kx^2.(2)

(b)Calculate the energy stored when the extension is 6 cm.(2)

(Total for Question 15 is 4 marks)

A groundskeeper is mowing a large field using several identical mowers, all working at the same constant rate. The time taken is inversely proportional to the number of mowers used. Using 3 mowers takes 8 hours to mow the whole field.

(a)Calculate the total number of mower-hours needed to mow the field.(2)

(b)Calculate how long it would take to mow the field using 5 mowers.(2)

(c)The field must now be mown in at most 3 hours. Calculate the minimum number of mowers needed.(2)

(Total for Question 16 is 6 marks)

y is inversely proportional to x. When x = 4, y = 15.

(a)Find the value of k in the formula y = k/x.(2)

(b)Find the value of x when y = 24.(2)

(Total for Question 17 is 4 marks)

Solve 4x - 9 = 23.

(Total for Question 18 is 2 marks)

Solve each equation, showing your working.

(a)Solve 3(2x - 1) = 5x + 8.(2)

(b)Solve x/4 + 3 = 10.(2)

(Total for Question 19 is 4 marks)

Solve the equation (x+3)/4 - (x-1)/5 = 2. Show each step of your working clearly.

(Total for Question 20 is 4 marks)

A school fete sells raffle tickets, each costing the same price. Priya buys 5 tickets and a £2 programme, spending £14.50 in total. Jamal buys 8 tickets and no programme.

(a)Using t for the price of one ticket in pounds, form an equation from Priya's spending and solve it to find t.(3)

(b)Calculate how much Jamal spends on his 8 tickets.(2)

(c)Jamal has £5 left after buying his tickets. Calculate how much money he had at the start.(1)

(Total for Question 21 is 6 marks)

Here are the first five terms of a sequence: 6, 11, 16, 21, 26, ...

(a)Write down the term-to-term rule for this sequence.(1)

(b)Find a formula for the nth term of the sequence.(2)

(Total for Question 22 is 3 marks)

The nth term of a sequence is 6n - 4.

(a)Find the 12th term of the sequence.(2)

(b)Determine whether 100 is a term of this sequence. Show your working.(2)

(Total for Question 23 is 4 marks)

Here are the first five terms of a sequence: 4, 7, 12, 19, 28, ...

(a)Find the next term of the sequence.(1)

(b)Show that the nth term of the sequence is n² + 3.(3)

(c)Use your formula to find the 9th term of the sequence.(1)

(Total for Question 24 is 5 marks)

Solve the equation 3^{2x - 1} = 27, giving the value of x.

(Total for Question 25 is 3 marks)

Two friends, Zainab and Callum, invest money in a small business in the ratio 4:7. After one year, the profit is shared between them in the same ratio as their investment. Zainab receives £150 less than Callum.

(a)Using x for one share, write expressions for Zainab's and Callum's amounts of profit.(1)

(b)Form an equation using the given difference, and solve it to find the value of x.(2)

(c)Find the total profit shared between them.(2)

(d)The following year, the profit totals £880 and is shared in the same ratio, 4:7. Calculate Callum's share of this profit.(2)

(Total for Question 26 is 7 marks)

Mark scheme · CE.M24 Number, Ratio and Algebra Depth

Question 1

(a) B1 p⁸ cao
(a) Answer: p⁸
(b) B1 q⁵ cao
(b) Answer: q⁵

Question 2

(a) B1 r¹² cao
(a) Answer: r¹²
(b) M1 5 x 3 = 15 and m² * m⁵ = m⁷ seen (both elements identified)
(b) A1 15m⁷ oe cao
(b) Answer: 15m⁷

Question 3

(a) B1 1 cao
(a) Answer: 1
(b) M1 3² = 9 seen (writes the negative index as 1 over the positive power)
(b) A1 1/9 oe cao
(b) Answer: 1/9
(c) M1 2^-1+4 = 2³ seen (adds the indices)
(c) A1 8 cao
(c) Answer: 8

Question 4

(a) M1 4² = 16 seen, or x³ raised to the power 2 gives x⁶ seen
(a) A1 16x⁶ oe cao
(a) Answer: 16x⁶
(b) M1 2³ = 8 seen, or m² raised to the power 3 gives m⁶ seen
(b) A1 8m⁶ⁿ^3 oe cao
(b) Answer: 8m⁶ⁿ^3

Question 5

M1 coefficients divided: 12/4 = 3
M1 indices subtracted: a^7-3 = a⁴ and b^3-1 = b²
A1 3a^4b^2 oe cao
Answer: 3a^4b^2

Question 6

(a) B1 5.2 x 10⁷ cao
(a) Answer: 5.2 x 10⁷
(b) B1 3.18 x 10^-4 cao
(b) Answer: 3.18 x 10^-4
(c) B1 904,000 cao
(c) Answer: 904,000

Question 7

M1 compares the powers of 10 first, then the leading numbers within equal powers
A1 X, W, Z, Y cao
Answer: X, W, Z, Y

Question 8

M1 sets up 3.15 x 10⁸ / 4.5 x 10⁶
M1 (3.15/4.5) x 10^8-6 = 0.7 x 10² seen, dependent on previous method
A1 70 cao
Answer: 70 photographs

Question 9

(a) B1 £20 cao
(a) Answer: £20
(b) M1 7 x 20 oe, ft from part a
(b) A1 £140 cao
(b) Answer: £140

Question 10

(a) B1 9 m² cao
(a) Answer: 9 m²
(b) M1 7 x 9, 3 x 9 and 2 x 9 oe, ft from part a
(b) A1 at least two of 63 m², 27 m², 18 m² correct
(b) A1 all three of 63 m², 27 m², 18 m² correct cao
(b) Answer: Vegetables 63 m², flowers 27 m², compost area 18 m²

Question 11

(a) M1 finds a common value for wind players in both ratios, e.g. LCM of 4 and 2 is 4
(a) M1 scales the ratios so the wind term is 4: string:wind = 5:4 (unchanged), wind:percussion = 4:6, dependent on previous method
(a) A1 5:4:6 cao
(a) Answer: 5:4:6
(b) M1 scale factor = 24/4 = 6, ft from part a
(b) A1 30 cao
(b) Answer: 30

Question 12

(a) B1 Leo = 3x and Nadia = 5x, both required, oe
(a) Answer: Leo = 3x, Nadia = 5x
(b) M1 5x - 3x = 18 oe (correct equation formed)
(b) A1 x = 9 cao
(b) Answer: x = 9
(c) M1 total = 3x + 5x = 8x oe, ft from part b
(c) A1 £72 cao
(c) Answer: £72

Question 13

M1 cost per metre = 4.50/6 = 0.75 oe (unitary method)
M1 10 x 0.75 oe, dependent on previous method
A1 £7.50 cao
Answer: £7.50

Question 14

(a) M1 45/3 oe
(a) A1 k = 15 cao
(a) Answer: k = 15
(b) M1 substitutes h = 7 into C = 15h, ft from part a
(b) A1 £105 cao
(b) Answer: £105

Question 15

(a) M1 45/3² = 45/9 oe
(a) A1 k = 5 cao
(a) Answer: k = 5
(b) M1 substitutes x = 6 into E = 5x², ft from part a
(b) A1 180 J cao
(b) Answer: 180 J

Question 16

(a) M1 3 x 8 oe
(a) A1 24 mower-hours cao
(a) Answer: 24 mower-hours
(b) M1 24/5, ft from part a
(b) A1 4.8 hours cao
(b) Answer: 4.8 hours
(c) M1 24/3, ft from part a
(c) A1 8 mowers cao
(c) Answer: 8 mowers

Question 17

(a) M1 4 x 15 = 60 oe
(a) A1 k = 60 cao
(a) Answer: k = 60
(b) M1 60/24, ft from part a
(b) A1 x = 2.5 cao
(b) Answer: x = 2.5

Question 18

M1 4x = 32 seen
A1 x = 8 cao
Answer: x = 8

Question 19

(a) M1 6x - 3 = 5x + 8 (bracket expanded)
(a) A1 x = 11 cao
(a) Answer: x = 11
(b) M1 x/4 = 7 seen
(b) A1 x = 28 cao
(b) Answer: x = 28

Question 20

M1 multiplies every term by 20, the LCM of 4 and 5: 5(x+3) - 4(x-1) = 40
M1 expands both brackets: 5x + 15 - 4x + 4 = 40
dM1 simplifies to x + 19 = 40 (dependent on previous method)
A1 x = 21 cao
Answer: x = 21

Question 21

(a) M1 5t + 2 = 14.50 formed
(a) M1 5t = 12.50 seen
(a) A1 t = £2.50 cao
(a) Answer: t = £2.50
(b) M1 8 x 2.50, ft from part a
(b) A1 £20 cao
(b) Answer: £20
(c) B1 £25 ft from their answer to part b
(c) Answer: £25

Question 22

(a) B1 add 5 (oe: +5)
(a) Answer: Add 5
(b) M1 common difference = 5, so 5n seen
(b) A1 5n + 1 oe cao
(b) Answer: 5n + 1

Question 23

(a) M1 substitutes n = 12 into 6n - 4
(a) A1 68 cao
(a) Answer: 68
(b) M1 sets 6n - 4 = 100 and solves to n = 17.33 (or 17 1/3, oe)
(b) A1 concludes 100 is not a term, since n is not a whole number
(b) Answer: No, 100 is not a term (n = 17.33, which is not a whole number)

Question 24

(a) B1 39 cao
(a) Answer: 39
(b) M1 second differences found to be constant and equal to 2, giving the n² coefficient equal to 1 (i.e. n²)
(b) M1 sequence minus n² calculated for at least two terms, giving 3, 3, 3, ... i.e. a constant of 3
(b) A1 nth term = n² + 3 stated, confirmed against given terms (cso, answer given)
(b) Answer: n² + 3
(c) B1 84 ft from their formula in part b
(c) Answer: 84

Question 25

M1 writes 27 as a power of 3: 27 = 3³
M1 sets 2x - 1 = 3 (equates the indices)
A1 x = 2 cao
Answer: x = 2

Question 26

(a) B1 Zainab = 4x and Callum = 7x, both required, oe
(a) Answer: Zainab = 4x, Callum = 7x
(b) M1 7x - 4x = 150 oe (correct equation formed)
(b) A1 x = 50 cao
(b) Answer: x = 50
(c) M1 total = 4x + 7x = 11x oe, ft from part b
(c) A1 £550 cao
(c) Answer: £550
(d) M1 one share = 880/11 = 80 oe
(d) A1 £560 cao
(d) Answer: £560

Number, Ratio and Algebra Depth - Worksheets, Questions and Revision