Junior Cycle Mathematics Revision Guide

Revision Test Overview

  • Subject: Junior Cycle Maths Revision Test.

  • Time Allotted: 90minutes90\,minutes.

  • Instructions:

    • Answer all questions.

    • Show all workings clearly.

    • Calculators are permitted unless instructions specify otherwise.

Section A: Algebra

  • Evaluating Algebraic Expressions:

    • Process: Substitute given numerical values into the variables of the expression and solve.

    • Example (Question 1): Evaluate 3x22y3x^2 - 2y where x=4x = 4 and y=5y = 5.

      • Substitute: 3(4)22(5)3(4)^2 - 2(5)

      • Square the value: 3(16)103(16) - 10

      • Multiply: 481048 - 10

      • Final Result: 3838

  • Expanding and Simplifying:

    • Process: Use the distributive law to multiply terms outside parentheses by terms inside, then combine like terms.

    • Example (Question 2): Expand and simplify 2(3x5)+4(x+2)2(3x - 5) + 4(x + 2).

      • Distribute: (6x10)+(4x+8)(6x - 10) + (4x + 8)

      • Group like terms: 6x+4x10+86x + 4x - 10 + 8

      • Simplify: 10x210x - 2

  • Factorisation:

    • Highest Common Factor (HCF): Identify the largest factor that divides into all terms.

    • Example (Question 3): Factorise completely 6x2+9x6x^2 + 9x.

      • Identify HCF: The HCF of 66 and 99 is 33; the HCF of x2x^2 and xx is xx. Common factor is 3x3x.

      • Divide terms by HCF: 3x(2x+3)3x(2x + 3)

  • Solving Linear Equations:

    • Process: Isolate the variable using inverse operations.

    • Example (Question 4): Solve 5x7=185x - 7 = 18.

      • Add 77 to both sides: 5x=18+75x = 18 + 7

      • Simplify: 5x=255x = 25

      • Divide by 55: x=5x = 5

  • Solving Simultaneous Equations:

    • Method (Elimination): Align equations to eliminate one variable by addition or subtraction.

    • Example (Question 5): Solve x+y=10x + y = 10 and 2xy=52x - y = 5.

      • Add the equations to eliminate yy: (x+2x)+(yy)=10+5(x + 2x) + (y - y) = 10 + 5

      • Solve for xx: 3x=15x=53x = 15 \rightarrow x = 5

      • Substitute xx back into the first equation: 5+y=105 + y = 10

      • Solve for yy: y=5y = 5

      • Solution: (5,5)(5, 5)

  • Solving Quadratic Equations:

    • Method (Factorisation): Find two numbers that multiply to the constant term and add to the coefficient of the middle term.

    • Example (Question 6): Solve x27x+10=0x^2 - 7x + 10 = 0.

      • Factorise: (x2)(x5)=0(x - 2)(x - 5) = 0

      • Set each factor to zero: x2=0x - 2 = 0 or x5=0x - 5 = 0

      • Solutions: x=2,x=5x = 2, x = 5

  • Rearranging Formulas (Change of Subject):

    • Process: Use algebraic operations to isolate a specific variable.

    • Example (Question 7): Rearrange the formula V=πr2hV = \pi r^2 h to make hh the subject.

      • Divide both sides by πr2\pi r^2: Vπr2=h\frac{V}{\pi r^2} = h

      • Final Formula: h=Vπr2h = \frac{V}{\pi r^2}

Section B: Coordinate Geometry

  • Distance Formula:

    • The distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by: d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

    • Example (Question 8): Distance between A(2,3)A(2, 3) and B(8,11)B(8, 11).

      • Substitute: d=(82)2+(113)2d = \sqrt{(8 - 2)^2 + (11 - 3)^2}

      • Calculate: d=62+82=36+64=100d = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100}

      • Result: 10units10\,units

  • Midpoint Formula:

    • The midpoint is given by: M=(x1+x22,y1+y22)M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

    • Example (Question 9): Midpoint of (4,6)(-4, 6) and (2,10)(2, 10).

      • Substitute: M=(4+22,6+102)M = \left(\frac{-4 + 2}{2}, \frac{6 + 10}{2}\right)

      • Calculate: M=(22,162)M = \left(\frac{-2}{2}, \frac{16}{2}\right)

      • Result: (1,8)(-1, 8)

  • Slope (Gradient) Formula:

    • The slope mm is calculated by: m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}

    • Example (Question 10): Slope of the line joining (1,2)(1, 2) and (5,14)(5, 14).

      • Substitute: m=14251=124m = \frac{14 - 2}{5 - 1} = \frac{12}{4}

      • Result: m=3m = 3

  • Equation of a Line:

    • Formula: yy1=m(xx1)y - y_1 = m(x - x_1)

    • Example (Question 11): Find the equation of the line with slope 33 passing through (2,1)(2, -1).

      • Substitute: y(1)=3(x2)y - (-1) = 3(x - 2)

      • Simplify: y+1=3x6y + 1 = 3x - 6

      • Standard form: 3xy7=03x - y - 7 = 0 or y=3x7y = 3x - 7

  • Parallel and Perpendicular Lines:

    • Parallel Lines: Have identical slopes (m1=m2m_1 = m_2).

    • Perpendicular Lines: Slopes are negative reciprocals (m1×m2=1m_1 \times m_2 = -1).

    • Example (Question 12): Compare y=2x+1y = 2x + 1 and y=2x5y = 2x - 5.

      • Both lines have a slope of m=2m = 2.

      • Conclusion: The lines are parallel.

Section C: Speed, Distance & Time

  • Average Speed Formula:

    • Average Speed=Total DistanceTotal Time\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}

    • Example (Question 13): A car travels 180km180\,km in 3hours3\,hours.

      • Calculation: 1803=60km/h\frac{180}{3} = 60\,km/h

  • Unit Conversion (Speed):

    • km/h to m/s: Since 1km=1000m1\,km = 1000\,m and 1hour=3600seconds1\,hour = 3600\,seconds, multiply by 10003600\frac{1000}{3600} (or divide by 3.63.6).

    • Example (Question 14): Convert 72km/h72\,km/h to m/sm/s.

      • Calculation: 72×10003600=20m/s72 \times \frac{1000}{3600} = 20\,m/s

  • Time Conversion for Speed:

    • When time is given in minutes, convert to hours to find km/hkm/h.

    • Example (Question 15): A cyclist travels 15km15\,km in 45minutes45\,minutes.

      • Convert time: 45minutes=0.75hours45\,minutes = 0.75\,hours

      • Calculation: 150.75=20km/h\frac{15}{0.75} = 20\,km/h

  • Multi-stage Journeys:

    • Example (Question 16):

      • Stage 1: 120km120\,km at 80km/h80\,km/h. Time = 12080=1.5hours\frac{120}{80} = 1.5\,hours.

      • Stage 2: 90km90\,km at 60km/h60\,km/h. Time = 9060=1.5hours\frac{90}{60} = 1.5\,hours.

      • Total Journey Time: 1.5+1.5=3hours1.5 + 1.5 = 3\,hours.

      • Average Speed: Total DistanceTotal Time=120+903=2103=70km/h\frac{\text{Total Distance}}{\text{Total Time}} = \frac{120 + 90}{3} = \frac{210}{3} = 70\,km/h.

Section D: Financial Maths

  • Value Added Tax (VAT):

    • Example (Question 17): Jacket costs 80€80 before 23%23\% VAT.

      • VAT Amount: 80×0.23=18.4080 \times 0.23 = €18.40

      • Total Price: 80+18.40=98.40€80 + €18.40 = €98.40

  • Profit Percentage:

    • Formula: Profit %=ProfitCost Price×100\text{Profit \%} = \frac{\text{Profit}}{\text{Cost Price}} \times 100

    • Example (Question 18): Phone bought for 250€250, sold for 320€320.

      • Profit: 320250=70320 - 250 = €70

      • Calculation: 70250×100=28%\frac{70}{250} \times 100 = 28\%

  • Discounts (Sale Price):

    • Example (Question 19): Runners cost 150€150, reduced by 20%20\%.

      • Discount: 150×0.20=30150 \times 0.20 = €30

      • Sale Price: 15030=120150 - 30 = €120

  • Compound Interest:

    • Formula: A=P(1+i)tA = P(1 + i)^t, where PP is principal, ii is interest rate (decimal), and tt is time.

    • Example (Question 20): 2,000€2,000 invested at 5%5\% for 3years3\,years.

      • Calculation: 2000(1+0.05)3=2000(1.157625)2000(1 + 0.05)^3 = 2000(1.157625)

      • Final Amount: 2,315.25€2,315.25

Section E: Sets

  • Set Notations and Operations:

    • Let A={1,2,3,4}A = \{1, 2, 3, 4\} and B={3,4,5,6}B = \{3, 4, 5, 6\}.

    • (a) Union (A ∡ B): All elements in either set.

      • Answer: {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}

    • (b) Intersection (A ∩ B): Elements found in both sets.

      • Answer: {3,4}\{3, 4\}

    • (c) Difference (A \ B): Elements in set A that are not in set B.

      • Answer: {1,2}\{1, 2\}

  • Venn Diagrams and Surveys:

    • Problem (Question 22): Total playing football (FF) = 1818, basketball (BB) = 1212, playing both (FBF \cap B) = 55.

    • Drawing Steps:

      • Place the overlap (55) in the intersection of circles F and B.

      • Determine "Only Football": 185=1318 - 5 = 13.

      • Determine "Only Basketball": 125=712 - 5 = 7.

    • Solution: 1313 students play only football.

Section F: Probability and Counting

  • Basic Probability:

    • Formula: P(E)=Number of Successful OutcomesTotal Number of Possible OutcomesP(E) = \frac{\text{Number of Successful Outcomes}}{\text{Total Number of Possible Outcomes}}

    • Question 23 (Fair Die):

      • (a) Probability of getting a 6: 16\frac{1}{6}

      • (b) Probability of an even number (2,4,62, 4, 6): 36=12\frac{3}{6} = \frac{1}{2}

    • Question 24 (Counters): Bag contains 4red4\,red, 3blue3\,blue, 5green5\,green (Total = 1212).

      • Probability of picking green: 512\frac{5}{12}

  • Sample Space and Multi-stage Events:

    • Example (Question 25): Two coins are tossed.

      • Sample Space: {HH,HT,TH,TT}\{HH, HT, TH, TT\}

      • Probability of exactly one head: Outcomes are HTHT and THTH.

      • Result: 24=12\frac{2}{4} = \frac{1}{2}

  • Fundamental Principle of Counting:

    • Multiply the number of choices for each stage of the event.

    • Example (Question 26): Password with 2letters2\,letters followed by 3digits3\,digits.

      • Assuming standard English alphabet (26letters26\,letters) and digits 00-99 (10digits10\,digits):

      • Calculation: 26×26×10×10×1026 \times 26 \times 10 \times 10 \times 10

      • Result: 676,000passwords676,000\,passwords