Mock revision (number, algebra, shape and space, data handling)

Question 1

How do you carry out long multiplication?

Answer 1

• put one number on top of the other
• multiply each number opposite each other, carrying if necessary
• add these numbers together, adding a 0 in the second row of addition

e.g. 6.3 x 2.1

63
21 x
63 1260 + 1323

! remember to add the . !
so the answer is 13.23

Question 2

How do you calculate the HCF and LCM?

Answer 2

1) draw a frequency tree for both numbers
2) write down all of the numbers circled multiplied together
3) simplify to include indexes if necessary

HCF is the numbers that appear in both (use lowest index)
LCM is all the numbers (use highest index)

e.g. 150 and 250

150 = 5² x 2 x 3
250 = 5³ x 2

HCF = 5² x2 = 50
LCM = 5³ x 2 x 3 = 750

Question 3

How do you write an error interval?

Answer 3

• write lower bound
• write ≤
• write the letter
• write >
• write upper bound

e.g. m = 7.6

UB = 7.65
LB = 7.55
so
7.55 ≤ m > 7.65

Question 4

How do you write an error interval when the number has been truncated?

Answer 4

• write the number
• write ≤
• write the letter
• write >
• write the number but with 1 decimal (lowest) added on

e.g. s = 6.57

6.57 ≤ s > 6.58

Question 5

How do you convert a recurring decimal to a fraction?

Answer 5

• write x = the number with the recurring digits written out
• multiply x by how many times it takes for the recurring decimals to be subtracted to equal nothing (so the recurring digits line up)
• subtract all of these numbers from each other (including the xs)
• put the remaining number over the number of xs left

e.g.

0.46 recurring

 x =     0.46464646 100x = 46.46464646 -   99x = 46

so 46/99

NOTE: if there is a non-recurring number in front of the recurring numbers, carry out the first 2 steps so that the non-recurring number is in front of the decimal, then multiply again so the recurring numbers line up again and continue

Question 6

How do you find how a total number splits between a ratio?

Answer 6

• add the numbers in the ration to find the total ratio
• divide the total number by the total ratio
• multiply this number by each individual number

e.g. if three people split £1440 between them in the ratio 3:5:7, how much money will each person get?

3+5+7 = 15
1440 / 15 = 96

96 x 3 = 288
96 x 5 = 480
96 x 7 = 672

(add these numbers together to check they add to the total amount)

Question 7

How do you find the total amount, given the amount of one number in the ratio?

Answer 7

• write the ratio
• write the amount underneath the number you know
• find the conversion factor (how many times the number from the ratio multiplies to give the amount)
• multiply the other numbers in the ratio by this conversion factor
• add all the amounts to find the total amount

e.g. if there are 16 sheep, how many animals are there altogether of sheep, cows and pigs in the ratio 8:1:3?

8 : 1 : 3
x2
16 2 6

16+2+6 = 24 animals

Question 8

How do you calculate percentage change?

(something increases, how much % did it increase by)

Answer 8

% change = actual change/ original amount x100

Question 9

How do you calculate percentage increase?

(something increases by x%, what is the new number)

Answer 9

• add the % increase to 100% (the original number)
• divide this number by 100
• multiple original number by this number

e.g. £1552 increase by 20%

100% + 20% = 120%
120 / 100 = 1.2

1552 x 1.2 = £1862.40

Question 10

How do you calculate percentage decrease?

(something decreases by x%, what is the new number)

Answer 10

• subtract the % decrease from 100% (the original number)
• divide this number by 100
• multiply original number by this number

e.g. £250 decreases by 15%

100% - 15% = 85%
85 / 100 = 0.85

250 x 0.85 = £212.50

Question 11

How do you calculate reverse percentages?

(finding the original amount after something has increased/decreased e.g. what was the cost before?)

Answer 11

• write out as an equation, with the original number as ?, multiplied by the percentage increase/decrease which equals the amount you are given
• rearrange the equation so that the ? equals the amount divided by the percentage increase/decrease
• solve

e.g. an item on sale for £22.05 reduced by 10%, what was the price before?

percentage decrease = 100% - 10% / 100 = 0.9

? x 0.9 = £22.05

? = 22.05/ 0.9
? = £24.50

Question 12

How do you calculate compound interest? (x amount in bank for x years, earns x% compound interest)

Answer 12

• total amount in bank = investment x % increase/decrease to the power of the number of years

e.g. £300 in bank for 3 years, earns 2% compound interest

percentage increase = 100% + 2% / 100 = 1.02

total = 300 x 1.02 to the power of 3
total = £3183.62

Question 13

What are the three rules of indicies?

Answer 13

when x, + the indicies
when /, - the indicies
when there are (), multiply indicies
e.g. (2³)³ = 2 to the power of 9

Question 14

What is the rule of indicies when it is to the power of 0?

Answer 14

it equals 1

Question 15

What is the rule of indicies when it is to a negative power?

Answer 15

it becomes 1/x to the power of x
e.g. 2-³ = 1/2³

Question 16

What is the rule of indicies when the power is a fraction? (fractional indicies)

Answer 16

a number to the power of a fraction is the square root of that number
the denominator is the root
the numerator is the power

e.g. 27 to the 2/3 = (the cube root of 27)² = 3² = 9

Question 17

What is the rule of indicies when there is a fraction to the power of -1?

Answer 17

you just flip the fraction

e.g. (2/3) to the -1 = 3/2

Question 18

What is the rule of indicies when there is a fraction to the power of x?

Answer 18

you add the power to both numbers in the fraction

e.g. (2/5)² = 2²/5² = 4/25

remember to simplify if possible!!

Question 19

What is the rule for negative fractional indicies? (a number to the power of a negative fraction)

Answer 19

the same as fractional indicies but you just add 1/x

e.g. 27 to the -1/3 = 1/ the cube root of 27 = 1/3

Question 20

How do you simplify surds?

Answer 20

find two factors of the number - one must be a square number

e.g. √75 = √25√3 = 5√3

Question 21

How do you add or subtract surds?

Answer 21

the surds must be the same
it’s like collecting ‘like terms’

e.g. 3√2 + 5√2 = 8√2

2√27 - 3√3 = 6√3 - 3√3 = 3√3
simplify = 2 x√9√3 = 2x3√3 = 6√3

if the surds are not the same, simplify them so that they have a surd in common and a square number, quick can be square rooted and multiplied with the normal number

e.g. 5√7 + 3√28

5√7 + 3√7 x √4
5√7 + 3√7 x 2
5√7 + 6√7
11√7

Question 22

What is the rule for multiplying surds that are the same?

Answer 22

they equal the normal number

e.g. √3 x √3 = √9 = √

√5 x √5 = 5
√2 x √2 = 2 etc

Question 23

How do you expand double brackets with surds?

Answer 23

expand brackets like normal except :

multiplying the same surd equals the normal number
if you are multiplying two different surds you just multiply the two numbers and add the √
e.g. √3 x √3 = 3

√3 x √2 = √6

you then collect the like terms

e.g. (3 + √2) (5 - √2) = 15 - 3√2 + 5√2 - 2 = 13 + 2√2

(3 - √5) (4 - √2) = 12 - 3√2 - 4√5 + √10

Question 24

How do you rationalise surds?

Answer 24

it is just multiplying out the surd in the denominator - so multiple the numerator and denominator by the denominator

e.g. 7/√2 = 7/√2 x √2/√2 = 7√2/2

remember to simplify fractions if possible!

Question 25

How do you solve a simultaneous equation (by elimination)? (4 steps)

Answer 25

1) multiply 1 or both equation so they have the same number of xs or ys (regardless of sign) 2) + or - resulting equations to cancel out the unknowns that are the now the same size 3) solve the resulting equation 4) substitute this into either equation (depending on which looks easiest) and solve resulting equation eg. 1) 3x + 4y = 29 2) 5x - 2y = 5 2) x2 --> 10x - 4y = 10 1) -------> 3x + 4y =29 + ------------------------------------ 13x = 39 ( / 13) x = 3 sub x into 1) 3x + 4y = 29 3(3) + 4y = 29 9 + 4y = 29 (-9) 4y = 20 ( / 4) y = 5

Question 26

How do you expand single brackets (1 step)

Answer 26

- multiple term on outside of brackets by each term on the inside of the brackets eg. 2(a + 5) 2 x a = 2a 2 x 5 = 10 2a + 10

Question 27

How do you expand double brackets? (2 steps)

Answer 27

1) multiple each term in the left bracket by each term in the right bracket 2) collect the terms eg. (q + 4) (p + 3) q x p = pq q x 3 = 3q 4 x p = 4p 4 x 3 = 12 pq + 3q + 4p + 12

Question 28

How do you expand triple brackets? (3 steps)

Answer 28

1) multiple the first two brackets as double brackets 2) simplify the resulting expression 3) multiply these terms by the 3rd bracket and simplify eg. (2x + 1) (x - 3) (3x + 2) 2x x x = 2x² 2x x 3 = -6x 1 x x = x 1 x -3 = -3 (2x² - 6x + x - 3) (3x + 2) (2x² -5x -3) (3x + 2) 2x² x 3x = 6x³ 2x² x 2 = 4x² -5x x 3x = -15x -5x x 2 = -10x -3 x 3x = -9x -3 x 2 = -6 6x³ + 4x² - 15x - 10x - 9x - 6 6x³ + -11x² - 19x - 6

Question 29

How do you factorise quadratics? (single x²)

Answer 29

- the reverse of expanding a double bracket 1) there will be an x at the front of each bracket 2) we need to find 2 numbers that multiply to give the number at the end and add to give the number of xs (middle number) eg. x² + 5x + 6 (x + ) (x + ) 3 x 2 = 6 3 + 2 = 5 so (x + 3) (x + 2)

Question 30

How do you complete the square?

Answer 30

- we need to rearrange the expression into the form (x + a)² + b 1) half the number of xs to get a 2) subtract a² and combine it with any other number in the original expression eg. x² + 2x + 6 (x + )² - + 6 2x / 2 = 1 so (x + 1)² - + 6 1² = 1 so (x +1)² - 1 + 6

Question 31

How do you solve a quadratic equation by factorising?

Answer 31

1) factorise equation eg. x² + 5x = 0 x(x + 5) = 0 2) separate term outside of bracket from terms inside bracket so x = 0 or x + 5 = 0 3) simplify expression of terms that were inside of bracket x + 5 = 0 (-5) x = -5 so x = 0 or x = -5

Question 32

How do you solve equations by using the quadratic formula?

Answer 32

- quadratic formula is x = -b ± √b² - 4ac / 2a - a is the number of x² - b is the number of xs - c is the number eg. x² - 5x + 3 = 0 a = 1 b = -5 c = 3 so x = -(-5) ± √(-5)² - 4(1)(3) / 2(1) (solve on calculator) x = 5 + 3.6055 / 2 or x = 5 - 3.6055 / 2 x = 4.18 or x = 0.70

Question 33

What is the rule for angles in a semi-circle?

Answer 33

they equal 90°

Question 34

What is the rule for angles in the same segment?

Answer 34

they are equal

Question 35

What is the rule for angles at a centre?

Answer 35

they are twice the angle on the edge

Question 36

What is a cyclic quadrilateral? What is the rule?

Answer 36

all four corners are on the edge of the circle diagonally opposite angles add to 180°

Question 37

What is the rule for alternate segments?

Answer 37

diagonally opposite angles are equal

Question 38

What is the rule for the radius and a tangent?

Answer 38

the radius and tangent make a right angle

Question 39

What is the rule for a triangle formed by two radii?

Answer 39

it is isosceles

Question 40

What is the equation for speed?

Answer 40

speed = distance/time

Question 41

What is the equation for density?

Answer 41

density = mass/volume

Question 42

How do you find the volume of a cone?

Answer 42

volume = 1/3πr² x h

Question 43

How do you find the nth term of a linear sequence?

Answer 43

- find the difference between each number (should be the same), this is the number of ns - plus or minus the amount it takes to get from the start of the sequence to n e.g. 1, 3, 5, 7 2 2 2 so 2n 1+1 = 2 so 2n + 1

Question 44

How do you find the nth term of a quadratic sequence?

Answer 44

- find the difference between each number (should be different) - find the (second) difference between these numbers (should be the same) - divide this number by 2 to find the amount of n²s - write out the original sequence, and minus the square numbers from the sequence (if it was 2n², it would be the square numbers multiplied by 2) - find the nth term of the sequence you are left with (should be linear) - add the nth term of the linear sequence to the n²s e.g. 1, 5, 11, 19, 29 4 6 8 10 2 2 2 2/2 = 1n² -1 5 11 19 29 1 4 9 16 25 -------------------- 0 1 2 3 4 1 1 1 1 so nth term is n + 1 final nth term would be n² + n + 1