Maths-11-plus-LondonIndependent-2011

13,602

5,745

22,352

774

0,0,1,1

a) 5

¹/₇ * ³⁵/₁

Always cross cancel. So the 7 and 35 can be simplified further….
¹/₁ * ⁵/₁

b) 21

The best approach is given that the answer to a) is 7, then 3/7ths would be 3 * 7 = 21

36

3 x 4 is 12 so think of multiples of 12 between 30 and 40

0.2; 1/5; 0.027

1. With these types of questions, its sometimes a good idea to convert all the options to fractions
2. This would give the following:
3. 1/5 ; 1/5; 2/5; 3/8; 27/1000

If not sure, then an alternative approach is to convert all the numbers to the same denominator

20,306

15 & 25

1. The technique here is to start with 40 and divide by 2. So thats 20
2. Then to give a difference of 10 between the 2 numbers, you have to deduct 5 from one of the 20s and add 5 onto the other 20
3. So this gives 15 and 25

729, 927, 972, 792, 297

Different combinations = number of digits multiplied. So 3x2x1. So 6 combinations. 5 other ones listed above

£1.70

1. 10.30am to 2.15pm falls in between 3 and 4 hours

1. The easiest way to complete these questions is to think about the HCF multiple.
2. Look at how many cupcakes you have ingredients for. Its 8. You need to complete for 12.
3. HCF of 8 & 12 is 4. Therefore, if you work out the ingredients for 4 cupcakes, then you can multiply up by 3x to get to 12
4. So for 4, its half of what is required for 8
5. So butter would be 75g.
6. x3 would give you butter for 12 cakes etc

£10.50

1. Ratios. Josie:Holly = 3:1
2. Given its 4 shares and £42 in total, each part is worth £10.50
3. Therefore Holly has £10.50

a) £1.20

Just the 2 added together

b) 1 apple = 14p. 1 orange = 16p =30p

Use simultaneous equations approach (x = apple, y=orange):

3x + 1y = 58p

1x + 3y = 62p  multiply by -3 to solve for y

3x + 1y = 58p

-3x – 9y = -186p

-8y = -128p

y =  16p

Use this to solve for a

3x + 16p = 58p

3x = 58p – 16p

3x = 42p

x = 14p

c) 1 apple = 14p.

a) 24

1. Give that Cats = 48 = 120º
2. Hamsters = 60º. Therefore, they are half of cats = 24

The true way to look at this question is to say if 120º = 48, then what is the HCF of 60 & 120. This is 60. Therefore, take the 120 and say how many 60s go into 120. This is 2. Therefore, 60 = 24

b) 60

1. Cats 120º + Hamsters 60º + Other 30º = 210º
2. Remainder is 360 – 210 = 150º
3. Again, HCF principle. Given you know 60º = 24, what is 150º?
4. HCF = 30. So 30 is half of 60 so it equals 12
5. Then how many 30s into 150 = 5
6. Therefore 5 * 12 = 60

£2,636

Classic maths questions whereby its just converting all the words to numbers

a) 1 hour 47 mins

Ensure you look at the departure time from Sidcot

b) 9.42

1. 2 hours 32 mins Burnley to Wester
2. If I arrive in Wester at 12.14pm, I have to take away 2 hours 32 mins
3. Take away 2 hours to give 10.14am.
4. Then take away 32mins which gives 9.42am
5. To sense check, re add back 2 hours 32mins to 9.42am

a) £8.05

(6 bananas = £4) + (2 grapefruit = £1.80) + (1/2kg Cherries = £2.25)

b) £11.95

7 slices

1. This question is again one related to HCF and multiples.
2. So convert to 12ths to make the question easier
3. Matthew eats 1/4 = 3/12
4. Katie eats 1/6 = 2/12
5. So this leaves 7 slices

10, 31, 18

The trick with sequence questions is you really MUST put down the change number on number. So like the following…It will help immensely in answering the question. See this. It makes it obvious that the pattern change is 10,9,8,7,6

a) James

b) £8

Joe £8; Samuel £11; James £4; Olivia £12; Sarah £7 = £42. Therefore, £8 remaining to be saved

a) 4 hours

If she spent 2 hours on the train and this represents 1/2 of the remaining time, then 2 hours = 1/4 of the time

Therefore if 1/4 = 2hours, 1/2 = 4hours

b) 8 hours

a) 32cm²

Area of a rectangle = LxW

b) 4cm²

8 triangles altogether = 1/2 LxW

c) 8cm²

4 and 5 in the above

d) 12

6 small triangles from 1,2,3,6,7,8

4 medium triangles from (1,2) , (2,3), (6,7), (7,8)

2 larger triangles from (1,4,5,6) and (3,4,5,8)

Answer is the 3rd one

These take practice and the best way is to try out the permutations with paper

a) 13/25

In total, its a 5×5 box so 25 boxes

Of these, 13/25 are shaded

b) See below

2/5ths = 10/25ths so 10 boxes to shade

a) 9cm

Perimeter of 36cm . 4 sides. So 9cm per side (36/4)

b) 81cm²

9×9

a) -8ºC

1. Note we are in negatives
2. Note what each marker is worth which is 2C

b) +6ºC

1. Tuesday is at -2ºC
2. Change v Monday is +6ºC

6

A is 6 x 2. Given we have a 3 x 1, we can fit in 4 rectangles into this side

B is 3 x 2. Given we have a 3 x 1, we ca fit in 2 rectangles into this side

a) Point W which is (9,1)

b) Point X which is (5,7)

c) Point Y which is (4,8)

a) 14

9 in lower level, 4 in middle level, 1 at the top.

b) 30

1. Observe the sequence. This question is simply sequences in pictures
2. 1st tower is 1×1 on the base. 2nd is 2×2. 3rd is 3×3.
3. 4th will be (4 x 4)+ (3×3) + (2×2) + (1×1) = 16 + 9 + 4 + 1 = 30

c) 55

Add on a 5 x 5 to the answer for b)

d) 100

Noting the comments for b), this therefore is a 10×10

See below

The question is about visualisation and its one you get in Non Verbal Reasoning in 11+

a) 4kg

Reading charts like this is always horizontal and vertical. Just follow along the x axis for 8 weeks and then follow up vertically. Use a ruler if needed

b) 3.5 weeks

This time, the weight is the y axis so you need to start there. Go up to 3kg and again, use a ruler and read across

See below

Start with 7 and the permutations for 7 and work from there

So (1,6), (2,5), (3,4) etc

See below

Ensure BIDMAS rules are applied

a) 9

Work backwards. 47 – 2 = 45. 45/5 = 9

b) 48

Work backwards. 5 + 3 = 8. x 6 = 48

c) 12

(50-x)/2 = 19

50-x = 38

x=12

d) 8

100 – (x + x) = 36

x² = 100 – 36

x² = 64

x = 8

a) 12

1. 2A + 1D + 4P = 21
2. So in total 7 parts. Each one is worth 3
3. Preeya must be 12 (3×4)

b) 100cm

1. P + A + D = 300
2. P – D = D – A
3. P + A = 2D
4. 2D + D = 300
5. 3D = 300
6. D = 100

a) 38

(6 x 6) + (6 x 1/3)

b) 5

1. 3 * x = 24
2. (3 x 3) + (3 x ?) = 24
3. 9 + (3 x ?) = 24
4. (3 x ?) = 24 – 9
5. 3 = 15/?
6. ? = 5

c) 4

1. (y * y) + (y * y) = 32
2. y² + y² = 32
3. 2y² = 32
4. y² = 16
5. y =4

211

See Calculating the Lowest Common Denominator process on the technique. A process here has been written