Sevenoaks School 2018 Paper 1

IndependentCo-ed

20 questions

Original PDF

Questions

1. 1
  Sophie is putting sweets in party bags for her friends. She has:
  
  $80$ fizzy cola bottles
  
  $78$ milk bottles
  
  $68$ fried eggs
  
  $56$ jelly rings
  
  $44$ flying saucers
  
  $12$ strawberry boot laces
  1.a
  How many sweets does Sophie have in total?
  $\begin{aligned} \text{Total sweets} &= 80 + 78 + 68 + 56 + 44 + 12 \\\\ &= 338 \end{aligned}$
  Answer
  $338$
  1.b
  She shares the sweets equally amongst $13$ bags. How many sweets are in each bag?
  $\begin{aligned} \text{Sweets in each bag} &= \dfrac {338} {13} \\\\ &= \dfrac {13 \times 26} {13} \\\\ &= \dfrac {\cancel {13} \times 26} {\cancel {13}} \\\\ &= 26 \end{aligned}$
  Answer
  $26$
2. 2
  A concert hall has $128$ rows of $35$ seats.
  2.a
  How many seats are there in the concert hall?
  $\begin{aligned} \text{Total seats} &= 128 \times 35 \\\\ &= 4480 \end{aligned}$
  Answer
  $4480$
  2.b
  If the concert hall is $75\%$ full, how many seats are empty?
  If the concert hall is $75\%$ full, then the empty seats are $100\% - 75\% = 25\%$ of all the seats.
  
  Calculation
  $\begin{aligned} \text{Empty seats} &= 25\% \ \text{of} \ 4480 \\\\ &= \dfrac {25} {100} \times 4480 \\\\ &= \dfrac {25 \times 4480} {100} \\\\ &= \dfrac {25 \times 4 \times 1120} {25 \times 4} \\\\ &= \dfrac {\cancel {25} \times \cancel {4} \times 1120} {\cancel {25} \times \cancel {4}} \\\\ &= 1120 \end{aligned}$
  Answer
  $1120$
3. 3
  Calculate:
  3.a
  $-3 \times 2 - 9$
  Keep in mind
  Using BIDMAS
  $\begin{aligned} &= -3 \times 2 - 9 \\\\ &= -6 - 9 \\\\ &= -15 \end{aligned}$
  Answer
  $-15$
  3.b
  $4^3$
  $\begin{aligned} &= 4 ^ 3 \\\\ &= 4 \times 4 \times 4 \\\\ &= 16 \times 4 \\\\ &= 64 \end{aligned}$
  Answer
  $64$
  3.c
  $2 + -4 \times -3$
  Keep in mind
  Using BIDMAS
  $\begin{aligned} &= 2 + -4 \times -3 \\\\ &= 2 + 12 \\\\ &= 14 \end{aligned}$
  Answer
  $14$
  3.d
  $(5 - 11) \div 3$
  Keep in mind
  Using BIDMAS
  $\begin{aligned} &= (5 - 11) \div 3 \\\\ &= -6 \div 3 \\\\ &= -\dfrac {6} {3} \\\\ &= -2 \end{aligned}$
  Answer
  $-2$
4. 4
  Nicola has four number cards:
  
  $\Large{\boxed{1} \quad \boxed{2} \quad \boxed{8} \quad \boxed{9}}$
  
  She arranges them to form two $2$ digit numbers. She multiplies the numbers together. Calculate the largest possible answer.
  To make the largest product, use $9$ and $8$ as the tens digits. Then check the two ways to use $1$ and $2$ as the units digits.
  
  Check the possible products
  We can find the answer using trial and error.
  Numbers Product
  $91$ and $82$ $91 \times 82 = 7462$
  $92$ and $81$ $92 \times 81 = 7452$
  
  Answer
  $\begin{aligned} \text{Largest product} &= 7462 \end{aligned}$
  Answer
  $7462$
5. 5
  Round $325.48$ to the following degrees of accuracy:
  5.a
  Nearest $100$ .
  $\begin{aligned} &= 325.48 \\\\ & \approx 300 \end{aligned}$
  Answer
  $300$
  5.b
  Nearest $10$ .
  $\begin{aligned} &= 325.48 \\\\ & \approx 330 \end{aligned}$
  Answer
  $330$
  5.c
  Nearest whole number.
  $\begin{aligned} &= 325.48 \\\\ & \approx 325 \end{aligned}$
  Answer
  $325$
  5.d
  One decimal place.
  $\begin{aligned} &= 325.48 \\\\ & \approx 325.5 \end{aligned}$
  Answer
  $325.5$
6. 6
  Place the following numbers in ascending order:
  
  $\boxed{ 1.02 } \quad \boxed{ 1.2 } \quad \boxed{ 1.15 } \quad \boxed{ 1.008 } \quad \boxed{ 1.9 }$
  $1.008 \quad 1.02 \quad 1.15 \quad 1.2 \quad 1.9$
  Answer
  $1.008 \quad 1.02 \quad 1.15 \quad 1.2 \quad 1.9$
7. 7
  The sum of three different primes is $16$ . What are the three numbers?
  Keep in mind
  Prime numbers between $1$ and $20= 2, 3, 5, 7, 11, 13, 17, 19$
  Define variables
  The sum of the three prime numbers is $16$ , which is an even number.
  Since the sum of three odd numbers is always odd, one of the prime numbers must be even. The only even prime number is $2$ .
  Let the other two primes $= a, b$
  
  Equation
  From the question,
  $\begin{aligned} 2 + a + b &= 16 \\\\ a + b &= 16 - 2 \\\\ a + b &= 14 \\\\ b &= 14 - a \end{aligned}$
  
  Check the possible primes
  We can find the answer using trial and error.
  Test prime values of $a$ until a valid pair is found.
  $a$ $b = 14 - a$ $b$ Valid pair?
  $3$ $14 - 3$ $11$ Yes
  $5$ $14 - 5$ $9$ No
  $7$ $14 - 7$ $7$ No, the primes must be different
  
  Answer
  $\begin{aligned} \text{Prime numbers} &= 2,\ 3,\ 11 \end{aligned}$
  Answer
  $2$
  $3$
  $11$
8. 8
  Blackcurrant cordial is diluted in the ratio $1$ part cordial to $4$ parts water.
  8.a
  How much water should be mixed with $30 \ \text{ml}$ of cordial?
  Using ratios,
  $\begin{array}{cccccccccccccc} & \text{cordial} &:& \text{water} \\\\ = & 1 \ \text{ml} &:& 4 \ \text{ml} \\\\ = & 1 \times 30 &:& 4 \times 30 \\\\ = & 30 \ \text{ml} &:& 120 \ \text{ml} \end{array}$
  
  Answer
  $\begin{aligned} \text{Amount of water} &= 120 \ \text{ml} \end{aligned}$
  Method 2
  Define variables
  Let the amount of water mixed $= a \ \text{ml}$
  
  Solve the ratio
  Using equivalent ratio,
  $\begin{aligned} 1 \ \text{ml} : 4 \ \text{ml} &= 30 \ \text{ml} : a \ \text{ml} \\\\ \dfrac {1} {4} &= \dfrac {30} {a} \\\\ 1 \times a &= 30 \times 4 \\\\ a &= 120 \ \text{ml} \end{aligned}$
  
  Answer
  $\begin{aligned} \text{Amount of water} &= 120 \ \text{ml} \end{aligned}$
  Answer
  $120 \ \text{ml}$
  8.b
  How much cordial should be used to make $750 \ \text{ml}$ of the drink?
  Using whole to part ratio,
  $\begin{array}{cccccccccccccc} & \text{drink} &:& \text{cordial} \\\\ = & (1 + 4) \ \text{ml} &:& 1 \ \text{ml} \\\\ = & 5 &:& 1 \\\\ = & 5 \times 150 &:& 1 \times 150 \\\\ = & 750 \ \text{ml} &:& 150 \ \text{ml} \end{array}$
  
  Answer
  $\begin{aligned} \text{Cordial used} &= 150 \ \text{ml} \end{aligned}$
  Method 2
  Define variables
  Let the amount of cordial used $= a \ \text{ml}$
  Water used $= (750 - a) \ \text{ml}$
  
  Solve the ratio
  Using equivalent ratio,
  $\begin{aligned} 1 \ \text{ml} : 4 \ \text{ml} &= a \ \text{ml} : (750 - a) \ \text{ml} \\\\ \dfrac {1} {4} &= \dfrac {a} {750 - a} \\\\ 750 - a &= 4a \\\\ 4a + a &= 750 \\\\ 5a &= 750 \\\\ a &= \dfrac {750} {5} \\\\ a &= \dfrac {5 \times 150} {5} \\\\ a &= \dfrac {\cancel 5 \times 150} {\cancel 5} \\\\ a &= 150 \ \text{ml} \end{aligned}$
  
  Answer
  $\begin{aligned} \text{Cordial used} &= 150 \ \text{ml} \end{aligned}$
  Answer
  $150 \ \text{ml}$
9. 9
  Add or subtract these fractions. Give your answers in their simplest terms.
  9.a
  $\dfrac {3} {5} + \dfrac {2} {10}$
  $\begin{aligned} &= \dfrac {3} {5} + \dfrac {2} {10} \\\\ &= \dfrac {3} {5} + \dfrac {2} {2 \times 5} \\\\ &= \dfrac {3} {5} + \dfrac {\cancel {2}} {\cancel {2} \times 5} \\\\ &= \dfrac {3} {5} + \dfrac {1} {5} \\\\ &= \dfrac {3 + 1} {5} \\\\ &= \dfrac {4} {5} \end{aligned}$
  Answer
  $\dfrac {4} {5}$
  9.b
  $\dfrac {24} {28} - \dfrac {3} {7}$
  $\begin{aligned} &= \dfrac {24} {28} - \dfrac {3} {7} \\\\ &= \dfrac {6 \times 4} {7 \times 4} - \dfrac {3} {7} \\\\ &= \dfrac {6 \times \cancel {4}} {7 \times \cancel {4}} - \dfrac {3} {7} \\\\ &= \dfrac {6} {7} - \dfrac {3} {7} \\\\ &= \dfrac {6 - 3} {7} \\\\ &= \dfrac {3} {7} \end{aligned}$
  Answer
  $\dfrac {3} {7}$
10. 10
  Insert an inequality sign $<$ or $>$ to make the following statements correct:
  10.a
  $\dfrac {6} {10} \quad \boxed {\phantom {\dfrac {a} {b}}} \quad \dfrac {3} {4}$
  $\begin{aligned} &= \dfrac {6} {10} \quad \rtqMathsBoxedEmptyValue \quad \dfrac {3} {4} \\\\ &= 0.6 \quad \rtqMathsBoxedEmptyValue \quad 0.75 \\\\ &= 0.6 \quad \rtqMathsBoxedFilledValue{<} \quad 0.75 \\\\ &= \dfrac {6} {10} \quad \rtqMathsBoxedFilledValue{<} \quad \dfrac {3} {4} \end{aligned}$
  Answer
  $\dfrac {6} {10} \quad < \quad \dfrac {3} {4}$
  10.b
  $\dfrac {8} {7} \quad \boxed {\phantom {\dfrac {a} {b}}} \quad \dfrac {11} {13}$
  $\begin{aligned} &= \dfrac {8} {7} \quad \rtqMathsBoxedEmptyValue \quad \dfrac {11} {13} \\\\ &\approx 1.14 \quad \rtqMathsBoxedEmptyValue \quad 0.85 \\\\ &\approx 1.14 \quad \rtqMathsBoxedFilledValue{>} \quad 0.85 \\\\ &= \dfrac {8} {7} \quad \rtqMathsBoxedFilledValue{>} \quad \dfrac {11} {13} \end{aligned}$
  Answer
  $\dfrac {8} {7} \quad > \quad \dfrac {11} {13}$
11. 11
  Find:
  11.a
  $\dfrac {1} {3}$ of $36$
  $\begin{aligned} &= \dfrac {1} {3} \ \text{of} \ 36 \\\\ &= \dfrac {1} {3} \times 36 \\\\ &= \dfrac {36} {3} \\\\ &= \dfrac {12 \times 3} {3} \\\\ &= \dfrac {12 \times \cancel{3}} {\cancel{3}} \\\\ &= 12 \end{aligned}$
  Answer
  $12$
  11.b
  $\dfrac {4} {5}$ of $80$
  $\begin{aligned} &= \dfrac {4} {5} \ \text{of} \ 80 \\\\ &= \dfrac {4} {5} \times 80 \\\\ &= \dfrac {4 \times 80} {5} \\\\ &= \dfrac {4 \times 5 \times 16} {5} \\\\ &= \dfrac {4 \times \cancel{5} \times 16} {\cancel{5}} \\\\ &= 4 \times 16 \\\\ &= 64 \end{aligned}$
  Answer
  $64$
12. 12
  Archie thinks of a number. Five sixths of the number is $60$ . What is three quarters of the number?
  Define variables
  Let the number $= a$
  
  Equation
  From the question,
  $\begin{aligned} \dfrac {5} {6} \times a &= 60 \\\\ \dfrac {5 \times a} {6} &= 60 \\\\ a &= \dfrac {60 \times 6} {5} \\\\ a &= \dfrac {5 \times 12 \times 6} {5} \\\\ a &= \dfrac {\cancel {5} \times 12 \times 6} {\cancel {5}} \\\\ a &= 72 \end{aligned}$
  
  Answer
  $\begin{aligned} \text{Three quarters} &= \dfrac {3} {4} \times 72 \\\\ &= \dfrac {3 \times 72} {4} \\\\ &= \dfrac {3 \times 4 \times 18} {4} \\\\ &= \dfrac {3 \times \cancel {4} \times 18} {\cancel {4}} \\\\ &= 3 \times 18 \\\\ &= 54 \end{aligned}$
  Answer
  $54$
13. 13
  Complete the table:
  
  Fraction Decimal Percentage
  $2\%$
  $0.68$
  $1\dfrac {1} {4}$
  First number
  $\begin{aligned} \text{First number} &= 2\% \\\\ \text{Fraction} &= \dfrac {2} {100} \\\\ &= \dfrac {2} {2 \times 50} \\\\ &= \dfrac {\cancel {2}} {\cancel {2} \times 50} \\\\ &= \dfrac {1} {50} \\\\ \text{Decimal} &= \dfrac {2} {100} \\\\ &= 0.02 \end{aligned}$
  
  Second number
  $\begin{aligned} \text{Second number} &= 0.68 \\\\ \text{Fraction} &= \dfrac {68} {100} \\\\ &= \dfrac {4 \times 17} {4 \times 25} \\\\ &= \dfrac {\cancel {4} \times 17} {\cancel {4} \times 25} \\\\ &= \dfrac {17} {25} \\\\ \text{Percentage} &= \dfrac {68} {100} \\\\ &= 68\% \end{aligned}$
  
  Third number
  $\begin{aligned} \text{Third number} &= 1\dfrac {1} {4} \\\\ \text{Decimal} &= 1 + \dfrac {1} {4} \\\\ &= 1 + 0.25 \\\\ &= 1.25 \\\\ \text{Percentage} &= 1.25 \times 100 \\\\ &= 125\% \end{aligned}$
  
  Completed table
  Fraction Decimal Percentage
  $\dfrac {1} {50}$ $0.02$ $2\%$
  $\dfrac {17} {25}$ $0.68$ $68\%$
  $1\dfrac {1} {4}$ $1.25$ $125\%$
14. 14
  The picture shows a tipper truck.
  
  `
  14.a
  Find the angles marked $a$ and $b$ .
  Formulas used
  $\text{Sum of angles in a triangle} = 180 \degree$
  $\text{Angles on a straight line add up to} = 180 \degree$
  Find a
  In the diagram,
  $\begin{aligned} a + 80 + 70 &= 180 \\\\ a + 150 &= 180 \\\\ a &= 180 - 150 \\\\ a &= 30 \degree \end{aligned}$
  
  Find b
  In the diagram,
  $\begin{aligned} b + 70 &= 180 \\\\ b &= 180 - 70 \\\\ b &= 110 \degree \end{aligned}$
  Answer
  $a$
  $=$
  $30 \degree$
  $b$
  $=$
  $110 \degree$
  14.b
  The $80 \degree$ angle decreases to $75 \degree$ as the tipper tips further. What happens to angle $c$ ?
  Formulas used
  $\text{Angles on a straight line add up to} = 180 \degree$
  Find c
  From the question,
  $\begin{aligned} c + 75 &= 180 \\\\ c &= 180 - 75 \\\\ c &= 105 \degree \end{aligned}$
  
  Answer
  Angle $c$ increases to $105 \degree$ .
  Answer
  It increases to $105 \degree$ .
15. 15
  Calculate the perimeter of this rectangle.
  
  Formulas used
  $\text{Perimeter of rectangle} = 2 \times (\text{length} + \text{width})$
  $\begin{aligned} \text{Perimeter} &= 2 \times (3.6 + 6.1) \\\\ &= 2 \times 9.7 \\\\ &= 19.4 \ \text{cm} \end{aligned}$
  Answer
  $19.4 \ \text{cm}$
16. 16
  Calculate the area of this right-angled triangle.
  
  Formulas used
  $\text{Area of triangle} = \dfrac{1}{2} \times \text{base} \times \text{height}$
  $\begin{aligned} \text{Area} &= \dfrac {8 \times 4.5} {2} \\\\ &= \dfrac {2 \times 4 \times 4.5} {2} \\\\ &= \dfrac {\cancel {2} \times 4 \times 4.5} {\cancel {2}} \\\\ &= 4 \times 4.5 \\\\ &= 18 \ \text{cm} ^ 2 \end{aligned}$
  Answer
  $18 \ \text{cm} ^ 2$
17. 17
  A piece of wire $42 \ \text{cm}$ long is bent into the shape of a rectangle. If the rectangle is twice as long as it is wide, what is its area?
  Formulas used
  $\text{Perimeter of rectangle} = 2 \times (\text{length} + \text{width})$
  $\text{Area of rectangle} = \text{length} \times \text{width}$
  Define variables
  Let the width of the rectangle $= w$
  Length of the rectangle $= 2w$
  
  Equation
  From the question,
  $\begin{aligned} 2 \times (2w + w) &= 42 \\\\ 2 \times 3w &= 42 \\\\ 6w &= 42 \\\\ w &= \dfrac {42} {6} \\\\ w &= \dfrac {7 \times 6} {6} \\\\ w &= \dfrac {7 \times \cancel {6}} {\cancel {6}} \\\\ w &= 7 \ \text{cm} \end{aligned}$
  
  Answer
  $\begin{aligned} \text{Width} &= 7 \ \text{cm} \\\\ \text{Length} &= 2 \times 7 \\\\ &= 14 \ \text{cm} \\\\ \text{Area} &= 14 \times 7 \\\\ &= 98 \ \text{cm} ^ 2 \end{aligned}$
  Answer
  $98 \ \text{cm} ^ 2$
18. 18
  In how many ways can you join the two shapes shown below to make a figure with one line of symmetry?
  
  Symmetric joins
  There are $3$ places where the extra square can be joined to make a figure with one line of symmetry.
  
  Diagram
  The three successful joins are shown on the diagram.
  Answer
  $3$
19. 19
  The numbers $1$ to $12$ are to be placed such that the sum of the four numbers in each of the six rows in the diagram below is the same. Which letter will be replaced with the number $7$ ?
  
  $\begin{aligned} 3 + A + B + 1 \\\\ 3 + 8 + 6 + 9 \\\\ 4 + A + 8 + C \\\\ 4 + B + 5 + E \\\\ C + 6 + D + E \\\\ 1 + 5 + D + 9 \\\\ \end{aligned}$
  The complete row gives the common row sum $3 + 8 + 6 + 9 = 26$ .
  The numbers still to place are $2, 7, 10, 11,$ and $12$ .
  
  Equations
  From the question,
  $\begin{aligned} 3 + A + B + 1 &= 26 \\\\ A + B &= 26 - 3 - 1 \\\\ A + B &= 22 & \rtqMathsEquationNumber{1} \\\\ 4 + A + 8 + C &= 26 \\\\ A + C &= 26 -4 - 8 \\\\ A + C &= 14 & \rtqMathsEquationNumber{2} \\\\ 4 + B + 5 + E &= 26 \\\\ B + E &= 26 - 4 - 5 \\\\ B + E &= 17 & \rtqMathsEquationNumber{3} \\\\ C + 6 + D + E &= 26 \\\\ C + D + E &= 26 - 6 \\\\ C + D + E &= 20 & \rtqMathsEquationNumber{4} \\\\ 1 + 5 + D + 9 &= 26 \\\\ D &= 26 - 1 - 5 - 9 \\\\ D &= 11 & \rtqMathsEquationNumber{5} \end{aligned}$
  
  The numbers still to place are $2, 7, 10,$ and $12$ .
  
  Find C and E
  Substitute $D$ into equation $\rtqMathsEquationNumber{4}$ .
  $\begin{aligned} C + D + E &= 20 \\\\ C + 11 + E &= 20 \\\\ C + E &= 20 - 11 \\\\ C + E &= 9 & \rtqMathsEquationNumber{6} \end{aligned}$
  From the remaining numbers $2, 7, 10,$ and $12$ , the only pair with sum $9$ is $2$ and $7$ .
  Use equation $\rtqMathsEquationNumber{3}$ .
  Check whether $E = 2$ .
  $\begin{aligned} B + E &= 17 \\\\ B + 2 &= 17 \\\\ B &= 15 \end{aligned}$
  But $15$ is not one of the numbers to place, so $E \neq 2$ .
  So $C$ is $2$ and $E$ is $7$
  
  Find A and B
  $\begin{aligned} B + E &= 17 \\\\ B + 7 &= 17 \\\\ B &= 10 \\\\ A + B &= 22 \\\\ A + 10 &= 22 \\\\ A &= 12 \end{aligned}$
  
  Answer
  The number $7$ replaces letter $E$ .
  Answer
  $E$
20. 20
  A twenty-four hour digital watch shows only hours and minutes. How many times each day does the watch show at least two fours simultaneously?
  Hours 04 and 14
  When the hour display is $04$ or $14$ , the hour display already has one $4$ .
  So the minute display needs at least one $4$ .
  The possible minute displays are: $04,$ $14,$ $24,$ $34,$ $40,$ $41,$ $42,$ $43,$ $44,$ $45,$ $46,$ $47,$ $48,$ $49,$ $54$ .
  There are $15$ possible minute displays.
  There are $2$ possible hour displays, so this gives $2 \times 15 = 30$ valid displays.
  
  Other hours
  When the hour display is not $04$ or $14$ , the hour display does not already have a $4$ .
  So the minute display needs two $4$ s.
  The only possible minute display is $44$ .
  There are $22$ possible hour displays, so this gives $22 \times 1 = 22$ valid displays.
  
  Total
  $\begin{aligned} \text{Total} &= 30 + 22 \\\\ &= 52 \end{aligned}$
  Answer
  $52$

Numbers	Product
$91$ and $82$	$91 \times 82 = 7462$
$92$ and $81$	$92 \times 81 = 7452$

$a$	$b = 14 - a$	$b$	Valid pair?
$3$	$14 - 3$	$11$	Yes
$5$	$14 - 5$	$9$	No
$7$	$14 - 7$	$7$	No, the primes must be different

Fraction	Decimal	Percentage
$\dfrac {1} {50}$	$0.02$	$2\%$
$\dfrac {17} {25}$	$0.68$	$68\%$
$1\dfrac {1} {4}$	$1.25$	$125\%$

Sevenoaks School 2018 · Paper 1

Questions

Calculation

Check the possible products

Answer

Define variables

Equation

Check the possible primes

Answer

Answer

Define variables

Solve the ratio

Answer

Answer

Define variables

Solve the ratio

Answer

Define variables

Equation

Answer

First number

Second number

Third number

Completed table

Find a

Find b

Find c

Answer

Define variables

Equation

Answer

Symmetric joins

Diagram

Equations

Find C and E

Find A and B

Answer

Hours 04 and 14

Other hours

Total

Sevenoaks School 2018 Paper 1