The Manchester Grammar School 2017 Section A 11+ Maths Solutions

1

Write in figures the number three hundred and eight thousand, five hundred and twenty.

Answer

$308520$

2

Work out $825 - 573$ .

Borrowing not shown
$\rtqMathsColumnarArithmeticStyle \begin{array}{cccccccccccccc} & 8 & 2 & 5 \\ {}- & 5 & 7 & 3 \\ \hline & 2 & 5 & 2 \end{array}$

Borrowing shown
$\rtqMathsColumnarArithmeticStyle \begin{array}{cccccccccccccc} & {}^{\rtqMathsSubtractBorrow{7}} \cancel{8} & {}^{\rtqMathsSubtractBorrow{1}} 2 & 5 \\ {}- & 5 & 7 & 3 \\ \hline & 2 & 5 & 2 \end{array}$

Answer

$252$

3

Work out $2047 \div 23$ .

Bus-stop division
Bus-stop division of 2047 by 23. Quotient 89, remainder 0.

Long division
Full long division of 2047 by 23. Quotient 89, remainder 0.

Answer
$\begin{aligned} \text{Answer} &= 89 \end{aligned}$

Answer

$89$

4

Work out $5 \dfrac{1}{3} \times 3\dfrac{3}{4}$ , giving your answer in its simplest form.

$\begin{aligned} &= 5 \dfrac {1} {3} \times 3 \dfrac {3} {4} \\\\ &= \dfrac {16} {3} \times \dfrac {15} {4} \\\\ &= \dfrac {16 \times 15} {3 \times 4} \\\\ &= \dfrac {4 \times 4 \times 3 \times 5} {3 \times 4} \\\\ &= \dfrac {\cancel{4} \times 4 \times \cancel{3} \times 5} {\cancel{3} \times \cancel{4}} \\\\ &= 4 \times 5 \\\\ &= 20 \end{aligned}$

Answer

$20$

5

If the following numbers were put in order, which number would be in the middle?

$0.67, \quad 55 \%, \quad \dfrac{3}{5}, \quad \dfrac{7}{10}$ and $\dfrac{2}{3}$

Keep in mind
Compare all values in one common form.

Preferred order:

Decimals

Percentages

Same-denominator fractions
Compare
Convert all the values to decimals so they can be compared.
$\begin{aligned} & 0.67 \quad 55 \% \quad \dfrac{3}{5} \quad \dfrac{7}{10} \quad \dfrac{2}{3} \\\\ & 0.67 \quad 0.55 \quad 0.6 \quad 0.7 \quad 0.666 \end{aligned}$

Order
In ascending order:
$\begin{aligned} & 0.55 < 0.6 < 0.666 < 0.67 < 0.7 \\\\ & 55 \% < \dfrac{3}{5} < \dfrac{2}{3} < 0.67 < \dfrac{7}{10} \end{aligned}$

Answer
$\begin{aligned} \text{Number in the middle} &= \dfrac{2}{3} \end{aligned}$

Answer

$\dfrac {2} {3}$

6

Work out $0.03 \times 3200$ .

$\begin{aligned} &= 0.03 \times 3200 \\\\ &= \dfrac {3} {100} \times 3200 \\\\ &= \dfrac {3 \times 3200} {100} \\\\ &= \dfrac {3 \times 32\cancel{0}\cancel{0}} {1\cancel{0}\cancel{0}} \\\\ &= 3 \times 32 \\\\ &= 96 \end{aligned}$

Answer

$96$

7

Write $0.84$ as a fraction in its simplest form.

$\begin{aligned} &= 0.84 \\\\ &= \dfrac {84} {100} \\\\ &= \dfrac {4 \times 21} {4 \times 25} \\\\ &= \dfrac {\cancel{4} \times 21} {\cancel{4} \times 25} \\\\ &= \dfrac {21} {25} \end{aligned}$

Answer

$\dfrac {21} {25}$

8

Which sum has been set out correctly to give the answer $67$ ?

$8 + 4 \times 6 - 5$
$(8 + 4) \times (6 - 5)$
$(8 + 4) \times 6 - 5$

Keep in mind
Using BIDMAS
Option A
$\begin{aligned} \text{Option A} &= 8 + 4 \times 6 - 5 \\\\ &= 8 + 24 - 5 \\\\ &= 32 - 5 \\\\ &= 27 \neq 67 \end{aligned}$

Option B
$\begin{aligned} \text{Option B} &= (8 + 4) \times (6 - 5) \\\\ &= 12 \times 1 \\\\ &= 12 \neq 67 \end{aligned}$

Option C
$\begin{aligned} \text{Option C} &= (8 + 4) \times 6 - 5 \\\\ &= 12 \times 6 - 5 \\\\ &= 72 - 5 \\\\ &= \rtqMathsFilledValue{67} \end{aligned}$

Answer
$\begin{aligned} \text{Answer} &= \text{C} \end{aligned}$

Answer

$\text{C}$
$(8 + 4) \times 6 - 5$

9

Add together the three lengths $1.4 \ \text{m}$ , $33 \ \text{cm}$ and $580 \ \text{mm}$ giving your answer in cm.

Keep in mind
It is really important to use the right units in calculations and convert where required.
$\begin{aligned} &= 1.4 \ \text{m} + 33 \ \text{cm} + 580 \ \text{mm} \\\\ &= 1.4 \times 100 \ \text{cm} + 33 \ \text{cm} + \dfrac {580} {10} \ \text{cm} \\\\ &= 140 + 33 + 58 \\\\ &= 231 \ \text{cm} \end{aligned}$

Answer

$231 \ \text{cm}$

10

Work out $30 \%$ of $\pounds 16.80$ .

$\begin{aligned} &= 30 \% \ \text {of} \ \pounds 16.80 \\\\ &=\dfrac {30} {100} \times 16.80 \\\\ &=\dfrac {30 \times 16.80} {100} \\\\ &=\dfrac {504} {100} \\\\ &= \pounds 5.04 \end{aligned}$

Answer

$\pounds 5.04$

11

The average weight of five parcels is $3.7 \ \text{kg}$ . Four of the parcels are identical and each weigh $3.5 \ \text{kg}$ . What is the weight of the fifth parcel?

Formulas used
$\text{Mean} = \dfrac {\text{Total of items}} {\text{Number of items}}$
Define variables
Let the weight of the fifth parcel $= w \ \text{kg}$

Equation
From the question,
$\begin{aligned} w + 4 \times 3.5 &= 3.7 \times 5 \\\\ w + 14 &= 18.5 \\\\ w &= 18.5 - 14 \\\\ w &= 4.5 \ \text{kg} \end{aligned}$

Answer
$\begin{aligned} \text{Weight of the fifth parcel} &= 4.5 \ \text{kg} \end{aligned}$

Answer

$4.5 \ \text{kg}$

12

At a birthday party, $12$ children said they liked orange juice and $10$ children said they liked lemonade. If $7$ of these children said they liked both and $3$ others said they only liked water, how many children were at the party?

Formulas used
$(A \text{ or } B) = A + B - (A \text{ and } B)$
Given information
$\begin{aligned} \text{Orange juice} &= 12 \\\\ \text{Lemonade} &= 10 \\\\ \text{Both} &= 7 \\\\ \text{Only water} &= 3 \end{aligned}$

Answer
$\begin{aligned} \text{Orange juice or lemonade} &= 12 + 10 - 7 \\\\ &= 15 \\\\ \text{Children at the party} &= 15 + 3 \\\\ &= 18 \end{aligned}$

Answer

$18$

13

Aidan was told to divide a certain number by $17$ . Instead he divided the number by $7$ and got the answer $68$ . What was the answer to the division he was supposed to do?

Define variables
Let the original number $= a$

Equation
From the question,
$\begin{aligned} \dfrac {a} {7} &= 68 \\\\ a &= 68 \times 7 \\\\ a &= 476 \end{aligned}$

Answer
$\begin{aligned} \text{Actual answer} &= \dfrac {a} {17} \\\\ &= \dfrac {476} {17} \\\\ &= \dfrac {28 \times 17} {17} \\\\ &= \dfrac {28 \times \cancel {17}} {\cancel {17}} \\\\ &= 28 \end{aligned}$

Answer

$28$

14

A piece of string is $160 \ \text{cm}$ long. It is formed into a rectangle whose length is three times its width. Work out the area of the rectangle in $\text{square centimetres}$ $(\text{cm}^2)$ .

Formulas used
$\text{Area of rectangle} = \text{length} \times \text{width}$
$\text{Perimeter of rectangle} = 2 \times (\text{length} + \text{width})$
Define variables
Let the width of the rectangle $= w$
Length of the rectangle $= 3w$

Equation
From the question,
$\begin{aligned} 2 \times (3w + w) &= 160 \\\\ 2 \times 4w &= 160 \\\\ 8w &= 160 \\\\ w &= \dfrac {160} {8} \\\\ w &= \dfrac {8 \times 20} {8} \\\\ w &= \dfrac {\cancel 8 \times 20} {\cancel 8} \\\\ w &= 20 \ \text{cm} \end{aligned}$

Answer
$\begin{aligned} \text{Length} &= 3w \\\\ &= 3 \times 20 \\\\ &= 60 \ \text{cm} \\\\ \text{Area} &= 60 \times 20 \\\\ &= 1200 \ \text{cm}^{2} \end{aligned}$

Answer

$1200 \ \text{cm}^2$

15

In the following diagram the two shaded rectangles have equal areas. What is the total shaded area?

Formulas used
$\text{Area of rectangle} = \text{length} \times \text{width}$
Define variables
Let the length of the bottom shaded rectangle $= a$
Length of the top shaded rectangle $= 14 - a$

Equation
In the diagram,
$\begin{aligned} 3 \times a &= 4 \times (14 - a) \\\\ 3a &= 56 - 4a \\\\ 3a + 4a &= 56 \\\\ 7a &= 56 \\\\ a &= \dfrac {56} {7} \\\\ a &= \dfrac {7 \times 8} {7} \\\\ a &= \dfrac {\cancel {7} \times 8} {\cancel{7}} \\\\ a &= 8 \ \text{cm} \end{aligned}$

Answer
$\begin{aligned} \text{Total shaded area} &= 2 \times (3 \times 8) \\\\ &= 2 \times 24 \\\\ &= 48 \ \text{cm}^2 \end{aligned}$

Answer

$48 \ \text{cm}^2$

16

Nancy bought five oranges and two apples and they cost $\pounds 3.40$ . At the same shop, David bought three oranges and one apple and paid $\pounds 2.00$ .

What would Steve have paid if he bought four oranges and two apples from this shop?

Keep in mind
It is really important to use the right units in calculations and convert where required.
Define variables
Let the cost of an apple $= a$
Let the cost of an orange $= b$

Equations
From the question,
$\begin{aligned} 5b + 2a &= 3.40 & \rtqMathsEquationNumber{1} \\\\ 3b + a &= 2.00 & \rtqMathsEquationNumber{2} \\\\ a &= 2.00 - 3b & \rtqMathsEquationNumber{3} \end{aligned}$

Price of orange
Using equation $\rtqMathsEquationNumber{3}$ , substitute $(2.00 - 3b)$ for $a$ in equation $\rtqMathsEquationNumber{1}$ .
$\begin{aligned} 5b + 2a &= 3.40 \\\\ 5b + 2 \times (2.00 - 3b) &= 3.40 \\\\ 5b + 4.00 - 6b &= 3.40 \\\\ 4.00 - b &= 3.40 \\\\ b &= 4.00 - 3.40 \\\\ b &= \pounds 0.60 \end{aligned}$

Price of apple
Substitute $b$ into equation $\rtqMathsEquationNumber{3}$ .
$\begin{aligned} a &= 2.00 - 3b \\\\ &= 2.00 - 3 \times 0.60 \\\\ &= 2.00 - 1.80 \\\\ &= \pounds 0.20 \end{aligned}$

Answer
$\begin{aligned} \text{Steve paid} &= 4b + 2a \\\\ &= 4 \times 0.60 + 2 \times 0.20 \\\\ &= 2.40 + 0.40 \\\\ &= \pounds 2.80 \end{aligned}$

Answer

$\pounds 2.80$

17

St Hugh's School hall is in the shape of a large box or cuboid.
The area of the floor is $200 \ \text{m}^2$ , the area of one of the side walls is $150 \ \text{m}^2$ and the area of an end wall is $48 \ \text{m}^2$ .

What is the height, in metres, of the hall?

Formulas used
$\text{Area} = \text{length} \times \text{width}$
Define variables
Let the length of the school hall $= l$
Let the width of the school hall $= w$
Let the height of the school hall $= h$

Equations
From the question,
$\begin{aligned} l \times h &= 150 & \rtqMathsEquationNumber{1} \\\\ w \times h &= 48 & \rtqMathsEquationNumber{2} \\\\ l \times w &= 200 & \rtqMathsEquationNumber{3} \end{aligned}$

Calculation
Multiply equations $\rtqMathsEquationNumber{1}$ and $\rtqMathsEquationNumber{2}$ and divide by equation $\rtqMathsEquationNumber{3}$ .
$\begin{aligned} \dfrac {l \times h \times w \times h} {l \times w} &= \dfrac {150 \times 48}{200} \\\\ \dfrac {\cancel{l} \times \cancel{w} \times h^2} {\cancel{l} \times \cancel{w}} &= \dfrac {150 \times 48} {200} \\\\ h^2 &= \dfrac {150 \times 48} {200} \\\\ h^2 &= \dfrac {50 \times 3 \times 12 \times 4} {50 \times 4} \\\\ h^2 &= \dfrac {\cancel{50} \times 3 \times 12 \times \cancel{4}} {\cancel{50} \times \cancel{4}} \\\\ h^2 &= 3 \times 12 \\\\ h^2 &= 36 \\\\ h &= \sqrt {36} \\\\ h &= 6 \ \text{m} \end{aligned}$

Answer
$\begin{aligned} \text{Height of the school hall} &= 6 \ \text{m} \end{aligned}$

Answer

$6 \ \text{m}$

18

A number is pro-added if I multiply the number by itself and then add the original number to that answer.

So, for example, pro-add $(8) = 8 \times 8 + 8 = 72$ .

What number gives the answer $210$ when it is pro-added?

Define variables

Let the number $= a$

Equation

From the question,

$\begin{aligned} a \times a + a &= 210 \\\\ a (a + 1) &= 210 \end{aligned}$

Conditions

We can find the answer using trial and error.

The two numbers $a$ and $a + 1$ must satisfy both of these conditions:

Their product is $210$
Their difference is $1$

Check for possible values

Factor pairs of $210 = (1, 210), (2, 105), (3, 70), (5, 42), (6, 35), (7, 30), (10, 21), (14, 15)$

The table below determines which factor pair meets the required conditions.

Factor pair	Difference	$= 1$ ?
$(1, 210)$	$210 - 1 = 209$	No
$(2, 105)$	$105 - 2 = 103$	No
$(3, 70)$	$70 - 3 = 67$	No
$(5, 42)$	$42 - 5 = 37$	No
$(6, 35)$	$35 - 6 = 29$	No
$(7, 30)$	$30 - 7 = 23$	No
$(10, 21)$	$21 - 10 = 11$	No
$(14, 15)$	$15 - 14 = 1$	Yes

Answer

The correct pair $(a, a + 1)$ is $(14, 15)$ , so $a = 14$ .

$\begin{aligned} \text{Number} &= 14 \end{aligned}$

Answer

$14$

19

How many whole numbers less than $100$ cannot be divided exactly by $4$ or by $5$ ?

Formulas used
$(A \text{ or } B) = A + B - (A \text{ and } B)$
$(\text{Neither } A \text{ nor } B) = \text{Total} - (A \text{ or } B)$
Given information
There are $100$ whole numbers from $0$ to $99$ . To find how many are divisible by neither $4$ nor $5$ , first count the numbers divisible by $4$ or $5$ . Then subtract this count from $100$ .
When we add the numbers divisible by $4$ and the numbers divisible by $5$ , any number divisible by both is counted twice.
The numbers counted twice are $0$ , $20$ , $40$ , $60$ , and $80$ and are multiples of $4 \times 5 = 20$ .
Subtract these numbers once to remove the extra count.

Count divisible numbers
$\begin{aligned} \text{Multiples of } 4 &= 0, 4, 8, \ldots, 20 \ldots, 96 \\\\ &= \dfrac{96}{4} + 1 \\\\ &= 25 \text{ numbers} \\\\ \text{Multiples of } 5 &= 0, 5, \ldots, 20 \ldots, 95 \\\\ &= \dfrac{95}{5} + 1 \\\\ &= 20 \text{ numbers} \\\\ \text{Multiples of } 20 &= 0, 20, 40, 60, 80 \\\\ &= \dfrac{80}{20} + 1 \\\\ &= 5 \text{ numbers} \\\\ \text{Divisible by } 4 \text{ or } 5 &= 25 + 20 - 5 \\\\ &= 40 \text{ numbers} \end{aligned}$

Answer
$\begin{aligned} \text{Not divisible by } 4 \text{ or } 5 &= 100 - 40 \\\\ &= 60 \text{ numbers} \end{aligned}$

Answer

$60$

20

Two crosses can be put in three squares in three different ways, as follows.

$\large{\text{X}}$	$\large{\text{X}}$	$\large{\text{O}}$
$\large{\text{X}}$	$\large{\text{O}}$	$\large{\text{X}}$
$\large{\text{O}}$	$\large{\text{X}}$	$\large{\text{X}}$

In how many different ways can you put three crosses in five squares?

Arrangements

The possible ways are:

$\text{Way}$	$1$	$2$	$3$	$4$	$5$
$1$	$\large{\text{X}}$	$\large{\text{X}}$	$\large{\text{X}}$	$\large{\text{O}}$	$\large{\text{O}}$
$2$	$\large{\text{X}}$	$\large{\text{X}}$	$\large{\text{O}}$	$\large{\text{X}}$	$\large{\text{O}}$
$3$	$\large{\text{X}}$	$\large{\text{X}}$	$\large{\text{O}}$	$\large{\text{O}}$	$\large{\text{X}}$
$4$	$\large{\text{X}}$	$\large{\text{O}}$	$\large{\text{X}}$	$\large{\text{X}}$	$\large{\text{O}}$
$5$	$\large{\text{X}}$	$\large{\text{O}}$	$\large{\text{X}}$	$\large{\text{O}}$	$\large{\text{X}}$
$6$	$\large{\text{X}}$	$\large{\text{O}}$	$\large{\text{O}}$	$\large{\text{X}}$	$\large{\text{X}}$
$7$	$\large{\text{O}}$	$\large{\text{X}}$	$\large{\text{X}}$	$\large{\text{X}}$	$\large{\text{O}}$
$8$	$\large{\text{O}}$	$\large{\text{X}}$	$\large{\text{X}}$	$\large{\text{O}}$	$\large{\text{X}}$
$9$	$\large{\text{O}}$	$\large{\text{X}}$	$\large{\text{O}}$	$\large{\text{X}}$	$\large{\text{X}}$
$10$	$\large{\text{O}}$	$\large{\text{O}}$	$\large{\text{X}}$	$\large{\text{X}}$	$\large{\text{X}}$

Answer

$\begin{aligned} \text{Number of different ways} &= 10 \end{aligned}$

Answer

$10$

The Manchester Grammar School 2017 · Section A

Questions

Borrowing not shown

Borrowing shown

Bus-stop division

Long division

Answer

Compare

Order

Answer

Option A

Option B

Option C

Answer

Define variables

Equation

Answer

Given information

Answer

Define variables

Equation

Answer

Define variables

Equation

Answer

Define variables

Equation

Answer

Define variables

Equations

Price of orange

Price of apple

Answer

Define variables

Equations

Calculation

Answer

Define variables

Equation

Conditions

Check for possible values

Answer

Given information

Count divisible numbers

Answer

Arrangements

Answer

The Manchester Grammar School 2017 Section A