GCSE Mathematics Foundation Tier Non-Calculator Study Guide

Arithmetic Fundamentals and Powers

  • Calculation of square root:   49=7\sqrt{49} = 7

  • Calculation of cubing integers:   33=3×3×3=273^3 = 3 \times 3 \times 3 = 27

  • Expressing powers of 10:   10 000=10410\,000 = 10^4

  • Imperial unit conversion between pounds and ounces:

    • Standard conversion factor: 1 pound=16 ounces1\,\text{pound} = 16\,\text{ounces}
    • Calculation for 3 pounds3\,\text{pounds}:     3×16=48 ounces3 \times 16 = 48\,\text{ounces}

Fractions, Decimals, and Ordering

  • Mixed numbers and improper fractions:

    • Conversion of 32\frac{3}{2} to a mixed number:     32=112\frac{3}{2} = 1\frac{1}{2}
  • Addition of fractions with common denominators:   15+15=1+15=25\frac{1}{5} + \frac{1}{5} = \frac{1 + 1}{5} = \frac{2}{5}

  • Subtraction of mixed numbers and fractions:

    • Expression: 115−3101\frac{1}{5} - \frac{3}{10}
    • Step 1: Convert 1151\frac{1}{5} to an improper fraction:     115=651\frac{1}{5} = \frac{6}{5}
    • Step 2: Convert to a common denominator of 1010:     65=1210\frac{6}{5} = \frac{12}{10}
    • Step 3: Perform subtraction:     1210−310=910\frac{12}{10} - \frac{3}{10} = \frac{9}{10}
  • Ordering fractions, decimals, and percentages:

    • Values to order: 80%80\%, 0.70.7, 34\frac{3}{4}
    • Convert all values to decimals for comparison:     80%=0.8080\% = 0.800.7=0.700.7 = 0.7034=0.75\frac{3}{4} = 0.75
    • Order from smallest to largest:     0.7,34,80%0.7, \frac{3}{4}, 80\%

Number Properties, Factors, and Proof by Counterexample

  • Factors of an integer:

    • All positive factors of 2020: 1,2,4,5,10,201, 2, 4, 5, 10, 20
  • Counterexample in number theory:

    • Statement disproof: "When two multiples of 5 are added, the answer is always a multiple of 10."
    • Counterexample:     5+10=155 + 10 = 15     Both 55 and 1010 are multiples of 55, but their sum 1515 is not a multiple of 1010

Practical Financial Mathematics

  • Calculating unit costs from total purchases:

    • Scenario: Buying 2 hats2\,\text{hats} and 3 scarves3\,\text{scarves} for a total cost of £28.50\text{£}28.50
    • Cost of each hat: £4.50\text{£}4.50
    • Total cost of 2 hats2\,\text{hats}:     2×4.50=£9.002 \times 4.50 = \text{£}9.00
    • Remaining cost for 3 scarves3\,\text{scarves}:     28.50−9.00=£19.5028.50 - 9.00 = \text{£}19.50
    • Cost per scarf:     19.503=£6.50\frac{19.50}{3} = \text{£}6.50
  • Percentage discounts on items:

    • Scenario: A multipack costs 10%10\% less than 6 single bags6\,\text{single bags}
    • Discount multiplier:     1−0.10=0.901 - 0.10 = 0.90
    • Cost formula for the multipack:     Multipack Cost=0.90×(6×Cost of single bag)\text{Multipack Cost} = 0.90 \times (6 \times \text{Cost of single bag})

Data Representation and Frequency Trees

  • Frequency tree analysis:

    • Total visitors to a maze: 120 people120\,\text{people}
    • Age category breakdown: 80 children80\,\text{children}, 120−80=40 adults120 - 80 = 40\,\text{adults}
    • Directional decisions (left or right):
    • Total people turning left: 75 people75\,\text{people}
    • Children turning left: 45 children45\,\text{children}
    • Adults turning left: 75−45=30 adults75 - 45 = 30\,\text{adults}
    • Children turning right: 80−45=35 children80 - 45 = 35\,\text{children}
    • Adults turning right: 40−30=10 adults40 - 30 = 10\,\text{adults}
    • Calculating fraction of children turning left:     4580\frac{45}{80}
    • Simplifying the fraction by dividing numerator and denominator by 55:     45÷580÷5=916\frac{45 \div 5}{80 \div 5} = \frac{9}{16}
  • Critiquing statistical bar chart representations:

    • Dataset: Library = 3030, Home = 6868, In class = 3737
    • Key potential structural errors in bar chart construction:
    • Incorrect or non-linear scale increments on the vertical frequency axis
    • Mislabeled or omitted horizontal/vertical axis categories
    • Inaccurate bar heights that do not correspond to dataset values
    • Non-uniform spacing or unequal bar widths across categories

Probability and Sample Spaces

  • Combined events and sample space grid construction:
    • Selection set 1 (first three positive odd numbers): {1,3,5}\{1, 3, 5\}
    • Selection set 2 (first four prime numbers): {2,3,5,7}\{2, 3, 5, 7\}
    • Score definition: Product of one number from set 1 and one number from set 2
    • Multiplication outcomes (3×4=12 total outcomes3 \times 4 = 12\,\text{total outcomes}):
    • For odd number 11: 1×2=21 \times 2 = 2, 1×3=31 \times 3 = 3, 1×5=51 \times 5 = 5, 1×7=71 \times 7 = 7
    • For odd number 33: 3×2=63 \times 2 = 6, 3×3=93 \times 3 = 9, 3×5=153 \times 5 = 15, 3×7=213 \times 7 = 21
    • For odd number 55: 5×2=105 \times 2 = 10, 5×3=155 \times 3 = 15, 5×5=255 \times 5 = 25, 5×7=355 \times 7 = 35
    • Square number outcomes in the product grid:
    • 9=329 = 3^2
    • 25=5225 = 5^2
    • Number of square number outcomes: 22
    • Probability of obtaining a square number score:     P(square)=212=16P(\text{square}) = \frac{2}{12} = \frac{1}{6}

Algebraic Simplification and Functions

  • Combining like terms:

    • Expression: 8m+4−2m+78m + 4 - 2m + 7
    • Grouping variables and constants:     (8m−2m)+(4+7)=6m+11(8m - 2m) + (4 + 7) = 6m + 11
  • Algebraic multiplication:

    • Expression: 12c×6d\frac{1}{2}c \times 6d
    • Multiplying coefficients and combining variables:     (12×6)×c×d=3cd\left(\frac{1}{2} \times 6\right) \times c \times d = 3cd
  • Truth evaluation of algebraic inequalities:

    • Given xx and yy are distinct positive numbers (x>0x > 0, y>0y > 0, x≠yx \neq y):
    • Statement x+yx + y is positive: Always true (sum of two positive quantities is strictly positive)
    • Statement x−yx - y is negative: Sometimes true (true if x<yx < y, false if x>yx > y)
  • Functional mapping machines:

    • Function y=4x+5y = 4x + 5: Input xx →\rightarrow Multiply by 44 →\rightarrow Add 55 →\rightarrow Output yy
    • Function y=3x−24y = 3x - 24: Input xx →\rightarrow Multiply by 33 →\rightarrow Subtract 2424 →\rightarrow Output yy
    • Identity function y=xy = x: Input xx →\rightarrow Multiply by 11 →\rightarrow Output yy

Ratios and Proportions

  • Simplifying ratios to n:1n : 1 form:

    • Ratio: 6:26 : 2
    • Divide both parts by 22:     62:22  ⟹  3:1\frac{6}{2} : \frac{2}{2} \implies 3 : 1
    • Value of nn: n=3n = 3
  • Division of quantities in ratios:

    • Total books: 3535
    • Ratio of adult books to children books: 6:16 : 1
    • Sum of ratio parts:     6+1=7 parts6 + 1 = 7\,\text{parts}
    • Books per part:     357=5 books\frac{35}{7} = 5\,\text{books}
    • Adult books quantity:     6×5=30 books6 \times 5 = 30\,\text{books}
    • Children books quantity:     1×5=5 books1 \times 5 = 5\,\text{books}
    • Difference between adult and children books:     30−5=25 more adult books30 - 5 = 25\,\text{more adult books}

Geometry, Transformations, and Speed

  • Congruence and geometric transformations:

    • Congruence: Two shapes are congruent if all corresponding side lengths and angles are identical
    • Fractional enlargement: Enlarging a shape by scale factor 13\frac{1}{3} scales every side length to 13\frac{1}{3} of its original length
  • Average speed calculations:

    • Car travels 4 miles4\,\text{miles} in 5 minutes5\,\text{minutes}
    • Time conversion factor to 1 hour1\,\text{hour} (60 minutes60\,\text{minutes}):     605=12\frac{60}{5} = 12
    • Average speed calculation:     Speed=4×12=48 mph\text{Speed} = 4 \times 12 = 48\,\text{mph}
  • Circle area and concentric shapes:

    • Large circle radius R=12 cmR = 12\,\text{cm}
    • Ratio of large circle radius to small circle radius: 4:14 : 1
    • Radius of small circle rr:     r=124=3 cmr = \frac{12}{4} = 3\,\text{cm}
    • Area of large circle:     Arealarge=πR2=π×122=144π cm2\text{Area}_{\text{large}} = \pi R^2 = \pi \times 12^2 = 144\pi\,\text{cm}^2
    • Area of small circle:     Areasmall=πr2=π×32=9π cm2\text{Area}_{\text{small}} = \pi r^2 = \pi \times 3^2 = 9\pi\,\text{cm}^2
    • Shaded area between circles:     Shaded Area=144π−9π=135π cm2\text{Shaded Area} = 144\pi - 9\pi = 135\pi\,\text{cm}^2

Coordinate Geometry and Linear Vectors

  • Line segment vector steps:
    • Points J(0,12)J(0, 12) and K(5,10)K(5, 10) lie on straight line JKLMJKLM
    • Segment equality: JK=KL=LMJK = KL = LM
    • Step displacement vector from JJ to KK:     Δx=5−0=5\Delta x = 5 - 0 = 5Δy=10−12=−2\Delta y = 10 - 12 = -2
    • Coordinates of point LL:     (5+5,10−2)=(10,8)(5 + 5, 10 - 2) = (10, 8)
    • Coordinates of point MM:     (10+5,8−2)=(15,6)(10 + 5, 8 - 2) = (15, 6)

Statistical Data Shift Properties

  • Effect of increasing every dataset value by 1010:
    • Mode: True (increases by 1010)
    • Median: True (increases by 1010)
    • Range: False (remains unchanged since (Max+10)−(Min+10)=Max−Min(\text{Max} + 10) - (\text{Min} + 10) = \text{Max} - \text{Min})

Sequences and Progressions

  • Geometric progression:

    • Sequence: 1,4,16,…,2561, 4, 16, \dots, 256
    • Common ratio rr:     r=41=4r = \frac{4}{1} = 4
    • Missing term:     16×4=6416 \times 4 = 64
    • Check term:     64×4=25664 \times 4 = 256
  • Fibonacci-type sequence:

    • Term generation rule: Tn=Tn−1+Tn−2T_n = T_{n-1} + T_{n-2}
    • Sequence terms given: …,5,−9,…\dots, 5, -9, \dots
    • Next term (T4T_4):     5+(−9)=−45 + (-9) = -4
    • Subsequent term (T5T_5):     −9+(−4)=−13-9 + (-4) = -13

Solid Geometry, Trigonometry, and Proportion

  • Geometric properties and cross-sectional area of prisms:

    • Total faces on an nn-gonal prism: n+2 facesn + 2\,\text{faces}
    • Prism volume relationship:     Volume=Cross-Sectional Area×Length\text{Volume} = \text{Cross-Sectional Area} \times \text{Length}
    • Given: Volume=3500 cm3\text{Volume} = 3500\,\text{cm}^3, Length=20 cm\text{Length} = 20\,\text{cm}
    • Cross-sectional area calculation:     Cross-Sectional Area=350020=175 cm2\text{Cross-Sectional Area} = \frac{3500}{20} = 175\,\text{cm}^2
  • Exact trigonometric values:   sin⁡(90∘)=1\sin(90^\circ) = 1

  • Inverse proportion and varied work rates:

    • Uniform worker rate scenario: 10 people10\,\text{people} complete a job in 9 hours9\,\text{hours}
    • Total workload:     10×9=90 worker-hours10 \times 9 = 90\,\text{worker-hours}
    • Time required for 15 people15\,\text{people} at standard rate:     9015=6 hours\frac{90}{15} = 6\,\text{hours}
    • Non-uniform worker rate analysis (6 slower6\,\text{slower}, 9 faster9\,\text{faster}):
    • Without specific quantitative performance ratios for the faster and slower workers, the net average rate cannot be determined
    • Logical conclusion: It is not possible to say