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INDEX Chapters Pages 1. Value Added Tax 4 2. Banking 7 3. Shares and Dividend 8 4. Linear Inequations 10 5. Quadratic Equation 15 6. Solving Problems (Based on Quadratic) 17 7. Ratio and Proportion 22 8. Remainder and Factor Theorem 24 9. Matrices 26 10. Arithmetic Progression 28 11. Geometric Progression 29 INDEX Chapters Pages 12. Reflection 31 13. Section and Midpoint Formula 32 14. Equation of a Line 35 15. Similarity 37 16. Loci 45 17. Circle, Tangents and Intersecting Chords 49 18. Constructions 72 19. Cylinder, Cone and Sphere 74 20. Trigonometric Ratios 81 23. Measures of Central Tendency 84 24. Probability 93 Ch. 1 Value Added Tax (Formulae) : r Sale price 100 1. Amount of Sales tax = 2. Total Amount = Sale Price + 3. Total Cost Price = Cost Price + Overheads 4. Profit/Loss% = 5. Discount% = r Sale price 100 Profit/Loss Total Cost Price Discount List Price 100 100 4 Ch. 1 Value Added Tax (Formulae) : 6. In case of single discount: Sale Price = 100 D% List Price 100 In case of double discount: Sale Price = 100 D1% 100 100 D2% List Price 100 5 Ch. 1 Value Added Tax (Formulae) : r S.P. 100 7. Tax Charged = 8. Tax Paid = 9. VAT = Tax Charged Tax Paid r r S.P. C.P. = 100 100 r C.P. 100 6 Ch. 2 Banking (Formulae) : 1. Amount deposited = P n 2. Interest = P 3. Maturity value = n(n + 1) 2 12 Maturity value = r 100 Amount deposited + P n + P Interest n(n + 1) 2 12 r 100 7 Ch. 3 Shares and Dividend (Formulae) : 1. M.V. = F.V. + Premium 2. M.V. = F.V. Discount 3. Dividend = No. of shares rate of dividend F.V. 4. Sum invested = No. of shares bought M.V. of 1 share. Income 5. Income /Return % = 6. Dividend = Income = Returns = Profit, but, Rate of Dividend Income% 7. Rate of dividend F.V. = Income % M.V. Investment 100 8 Ch. 3 Shares and Dividend (Formulae) : 8. Profit% = Profit Sum Invested 100 Profit = Amount received on selling shares Sum Invested 9 Ch. 4 Linear Inequations (Rules) Rule 1: On transferring a term from one side of an inequation to its other side, the sign of the term changes Example : Example : x-2 7 x+5 2 x 2 -5 x 7+2 Rule 2: If each term of an inequation be multiplied or divided by the same positive number , the sign of inequality remains the same. Example : Example : 5 > 3 5 2 > 3 2 10 > 6 12 > 8 312 > 8 2 4 4 1 1 > 3 2 10 Ch. 4 Linear Inequations (Rules) Rule 3: If each term of an inequation be multiplied or divided by the same negative number, the sign of terms is changed and inequality sign reverses. Example : Example : 12 > 8 5 > 3 312 > 8 2 10 < 4 1 1 4 5 2 > 3 2 3 6 < 2 Rule 4: If both the sides of an inequation are positive or both are negative, then on taking their reciprocals, the sign of inequality reverses. Example : 7 > 4 Example : 4 > 10 11 Ch. 4 Linear Inequations (Number Line) Graph on Number Line : 2 x < 5, x W 0, 1, 2, 3, 4 1 2 3 4 3 2 1 0 1 2 3 4 5 6 Graph on Number Line : 2 x < 5, x N 1, 2, 3, 4 1 2 3 4 3 2 1 0 1 2 3 4 5 6 Graph on Number Line : 2 x < 5, x I 2, 1, 0, 1, 2, 3, 4 1 2 3 4 3 2 1 0 1 2 3 4 5 6 12 Ch. 4 Linear Inequations (Number Line) Graph on Number Line : 2 x < 5, x R 1 2 3 4 3 2 1 0 1 2 3 4 5 6 OR 1 2 3 4 3 2 1 0 1 2 3 4 5 6 OR 1 2 3 4 3 2 1 0 1 2 3 4 5 6 13 Ch. 4 Linear Inequations (Operations on Sets) Let P = {2, 4, 6, 8, 10, 12} Q = {3, 6, 9, 12, 18} U = {2, 3, 4, 5, 6, 8, 9, 10, 12, 18} (i) P Q = {6, 12} (ii) P U Q = {2, 4, 6, 8, 10, 12 , 3, 9, 18} (iii) P Q = {2, 4, 8, 10} (iv) Q P = {3, 9, 18} (v) P' = U P = {3, 5, 9, 18} (vi) P' Q = {3, 9, 18} 14 Ch. 5 Quadratic Equations (Formula) ax2 + bx + c = 0 x = Decimal 2 significant figures 2 decimal places 3.233 3.2 3.23 0.762 0.76 0.76 0.329 0.33 0.33 9.359 9.4 9.36 2.756 2.8 2.76 15 Ch. 5 Quadratic Equations (Nature of Roots) 1. If then 2. If then 3. If then D = b2 4ac = 0 Roots are real and equal D = b2 4ac > 0 Roots are real and unequal D = b2 4ac < 0 Roots are not real 16 Ch. 6 Solving Problems based on Quadratic Q. Divide 20 into two parts such that three times the square of one part exceeds the other part by 10. PARTS OF 20 1ST 2ND 8 20 8 = 12 x 20 x 17 Ch. 6 Solving Problems based on Quadratic Q. Rs 480 is divided equally among x children. If the number of children were 20 more then each would have got Rs 12 less. Find x . Rs 480 1 80 2 80 3 80 4 80 5 80 6 80 Amount received by each child = Rs 480 6 = Rs 80 18 Ch. 6 Solving Problems based on Quadratic Q. Rs 480 is divided equally among x children. If the number of children were 20 more then each would have got Rs 12 less. Find x . Rs 480 1 2 3 4 5 x Amount received by each child = Rs 480 x 19 Ch. 6 Solving Problems based on Quadratic SUMS BASED ON WORK No of days required to complete work Work done in 1 day 8 1/8 6 1/6 x 1/x x + 10 1/(x + 10) 20 Ch. 6 Solving Problems based on Quadratic Q. The speed of a boat in still water is 15 km/hr. It can go 30 km upstream and return downstream to the original point in 4 hrs 30 mins . Find the speed of the stream. Soln: Let the speed of stream = x km/hr. Speed Upstream = Speed of the boat in still water Speed of the stream = (15 x) km/hr Speed Downstream = (15 + x) km/hr 21 Ch. 7 Ratio and Proportion TYPES OF RATIO 8 8 64 Duplicate ratio of = = 15 225 15 Sub-duplicate ratio of 64 = 64 = 8 225 15 225 2 2 8 Triplicate ratio of = = 3 27 3 Sub-triplicate ratio of 8 = 8 27 27 = 2 3 22 Ch. 7 Ratio and Proportion PROPERTIES OF PROPORTION a c = b d 3) By Componendo 1) By Invertendo b d = a c 2) By Alternendo a b or = c d d c = b a a+ b = b c + d d 4) By Dividendo a b = b c d d 5) By Componendo and Dividendo a+ b c + d = c d a b 23 Ch. 8 Remainder and factor theorem Steps to follow while using Remainder Theorem STEP 1 Put the divisor equal to zero and solve the equation obtained to get the value of its variable Substitute the value of the variable , obtained in step 1, in the given polynomial and STEP 2 simplify it to get the required remainder 24 Ch. 8 Remainder and factor theorem Factor Theorem If f(x) is a polynomial and f(a) =0 then (x a) is a factor of f(x) Q. Factorise completely: f(x) = 2x3 + x2 13x + 6 By Hit & Trial Method, Put x = 2, f(2) = 0 (x 2) is a factor Factors of 12 1 12 2 6 3 4 Try positive as well as negative factors 25 Ch. 9 Matrices (Solving Matrix Equation): Q. If A = 2 4 3 1 and B = , find matrix C such that AC = B. 2 5 AC = B B C= A 26 Ch. 9 Matrices (Solving Matrix Equation): Q. If A = 3 1 and B = , find matrix C such that AC = B. 2 5 2 4 Soln. A = 2 4 1 , 5 B= 2 2 3 2 , 2 1 Let order of C be m n C= a b 2 1 AC = B 2 4 1 5 A 2 2 2 a b = C 2 1 3 2 = B 2 1 1 27 Ch. 10 A.P. (Formulae): 1. tn = a + (n 1) d 2. Three consecutive terms of A.P. : a d, a, a + d Four consecutive terms of A.P. : a 3d, a d, a + d, a + 3d Five consecutive terms of A.P. : a 2d, a d, a, a + d, a + 2d 3. tn = Sn Sn 1 4. Sn = n [a + l] 2 5. Sn = n [2a + (n 1)d] 2 28 Ch. 11 G.P. (Formulae): 1. 2. tn = arn 1 Three terms of G.P. : a , a , ar r Five terms of G.P. : a , r2 Four terms of G.P. : 3. a , a , ar , a r2 r a a , ar , ar3 , 3 r r tn = Sn Sn 1 29 Ch. 11 G.P. (Formulae): 4. Sn Case (i) : When r = 1 Sn of G.P. in terms of l : Sn = na Case (ii) : When r < 1 Sn = a (1 rn) 1 r Case (iii) :When r > 1 Sn = a (rn 1) r 1 When r < 1, a lr Sn = 1 r When r > 1, lr a Sn = r 1 30 Ch. 12 Reflection (Formulae): 1. Mx (a, b) = (a, b) 2. My (a, b) = ( a, b) 3. MO (a, b) = ( a, b) 4. Area of Trapezium = 5. Area of Rhombus = 1 (Sum of parallel sides) height 2 1 Product of diagonals 2 31 Ch. 13 Section and Midpoint Formula : 1. Section Formula: m1 : m2 A (x1, y1) (x, y) = P (x, y) B (x2, y2) m1x2 + m2x1 , m1y2 + m2y1 m1 + m2 m1 + m2 2. Midpoint Formula: A (x1, y1) (x, y) = P (x, y) 1 : 1 B (x2, y2) x2 + x1 , y2 + y1 2 2 32 Ch. 13 Section and Midpoint Formula : 3. Centroid Formula: C(x3, y3) G CENTROID B (x2, y2) A (x1, y1) x3 + x2 + x1 , y3 + y2 + y1 G= 3 3 33 Ch. 13 Section and Midpoint Formula : 4. Tri Section: 2 1 A (x1, y1) P 2 Q B (x2, y2) 1 34 Ch. 14 Equation of a line (Formulae) : y2 - y1 1. Slope (m) = tan = 2. If lines with slopes m1 and m2 are parallel, then m1 = m2 3. If lines with slopes m1 and m2 are perpendicular, Then m1 m2 = 1 4. Equation of a line in slope intercept form: y = mx + c 5. Equation of a line in slope point form: y y1 = m(x x1) x2 - x1 35 Ch. 14 Equation of a line (Formulae) : 6. Equation of line l : ax + by + c1 = 0 Equation of line perpendicular to l : bx ay + c2 = 0 7. Equation of line passing through origin: y = mx 8. If equation of line is given then slope is found by converting to y = mx + c 36 Ch. 15 Similarity (Properties): 1. Tests of similarity: (a) A A (b) S A S (c) S S S 37 Ch. 15 Similarity (Properties): A 2. Properties of similarity triangles: P B DABC ~ C DPQR AB BC AC = = PQ QR PR Perimeter of ABC = Perimeter of PQR Q Corresponding sides of similar triangles A = P B = Q C = R Corresponding angles of similar triangles 38 R Ch. 15 Similarity (Properties): A P Q B 3. R C AREAS OF SIMILAR TRIANGLES: DABC DPQR ( AB AB)2 ( BC BC)2 Ar. (DABC) = = = ( PQ PQ)2 ( QR QR)2 Ar. (DPQR) ( AC AC)2 ( PR PR)2 39 Ch. 15 Similarity (Properties): RESULTS RELATED TO THE RATIO OF THE AREAS OF TWO SIMILAR TRIANGLES The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding altitudes. A N P O E F B C M Ar. (DABC) Ar. (DPQR) DABC = A perpendicular drawn from the vertex of a triangle to its opposite side Q DPQR (AM AM )2 ( BN BN)2 = = ( QE QE)2 (PD PD )2 D R ( CO CQ)2 ( RF RF)2 40 Ch. 15 Similarity (Properties): RESULTS RELATED TO THE RATIO OF THE AREAS OF TWO SIMILAR TRIANGLES The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding medians. A P M N Y Z B L Ar. (DABC) Ar. (DPQR) = Segment joining the vertex of a triangle to the midpoint of its opposite side C Q DABC DPQR ( AL AL)2 ( PX AX)2 = ( BM BM)2 ( QY QY)2 = X R ( CN CN)2 ( RZ RZ)2 41 Ch. 15 Similarity (Properties): 4. Ratio of Areas of Similar triangles = Ratio of Squares of Corresponding Sides = Ratio of Squares of Corresponding Altitudes = Ratio of Squares of Corresponding Medians 42 Ch. 15 Similarity (Properties): 5. 6. For Maps/Models : 1] Length in map/model = k [Actual length] 2] Area in map/model = k2 [Actual Area] 3] Volume of model = k3 [Actual Volume] B.P.T D In DDEF, line XY || side EF X Y E DX XE = DY YF [By B.P.T] F 43 Ch. 15 Similarity (Properties): 7. Enlargement/Reduction If DABC is enlarged/reduced by scale factor k about centre O to form DA B C then: A' A OB OC A C OA A B B C = = k = = = = OB OC AC OA AB BC O B B' C C' A A' B' O B C' 44 C Ch. 16 Loci (Constructions) : 1. Locus of a point equidistant from a given point in the same plane We can have infinite such points O Let us locate points in the plane at a distance of 3 cm from O 3 cm The location (locus) of all theseGiven points will be a circle with centre O and plane radius 3 cm P Given point 45 Ch. 16 Loci (Constructions) : X 2. Locus of points equidistant from two points A and B Perpendicular bisector of AB A B Y 46 Ch. 16 Loci (Constructions) : 3. Locus of points equidistant from two lines AB and BC Angle bisector of B C X B A 47 Ch. 16 Loci (Constructions) : 4. Locus of a point equidistant from a given line. Q X 3 cm M M 3 cm 3 cm 3 cm P l Y 48 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 1. Chord Bisection Property: The perpendicular drawn from the centre of a circle to a chord bisects the chord Presentation: In the given figure, O is the centre of the circle and OP is perpendicular to chord AB AP = PB O A P B 49 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 1. Chord Bisection Property: The segment joining midpoint of a chord and the centre of a circle is perpendicular to the chord. Presentation: In the given figure, O is the centre of the circle and P is the midpoint of chord AB OP AB OPA = OPB = 90 O A lll P lll 50 B Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 2. Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres). C Presentation: In the given figure, O is the centre of the circle AB = CD, OM AB, ON CD OM = ON N O A M D B 51 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 2. Chords equidistant from the centre of a circle are equal in length. C N Presentation: In the given figure, O is the centre of the circle OM AB, ON CD and OM = ON AB = CD O A D B M 52 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 3. The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. C C C O O B B A O A B A Presentation: AOB = 2 ACB [Angle at the centre is twice the angle at the remaining circumference] 53 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 3. The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. O A B C Presentation: Reflex AOB = 2 ACB [Angle at the centre is twice the angle at the remaining circumference] 54 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 4. Angles in the same segment are equal. D Presentation: ACB = ADB [Angles in the same segment are equal] C O B A 55 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 5. Angle in a semi-circle is a right angle. C Presentation: ACB = 90 [Angle in a semi-circle is a right angle] A O B 56 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 6. Cyclic quadrilateral A D B C 57 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 7. Opposite angles of a cyclic quadrilateral are supplementary A D Presentation: A + B = 180 [Opposite angles of a C + D = 180 cyclic quadrilateral are supplementary] B C 58 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 8. Exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. B A S P D R ASP @ PQR BPQ @ QRS Q CQR @ PSR DRS @ QPS C 59 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 9. If in a circle two arcs are equal then they cut equal chords. Presentation: arc AB = arc CD Chord AB = Chord CD [Equal arcs cut equal chords] 60 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 9. If in a circle two chords are equal then they cut equal arcs. Presentation: Chord AB = Chord CD arc AB = arc CD [Equal chords cut equal arcs] 61 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 10. If in a circle two arcs are equal then they subtend equal angles at the centre Presentation: arc AB = arc CD AOB = COD [Equal arcs subtend equal angles at the centre] 62 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 10. If two arcs of a circle subtend equal angles at the centre then the arcs are equal. Presentation: AOB = COD arc AB = arc CD [If two arcs of a circle subtend equal angles at the centre then the arcs are equal] 63 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 11. In the circle with centre O If AB is a side of regular polygon of n sides 3600 n Then AOB = 3600 n 64 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 11. In the circle with centre O If arc APB : arc CQD = m: n Then AOB : COD = m: n n m n m 65 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 12. The tangent at any point of a circle and the radius through this point are perpendicular to each other. Presentation: OAB = 90 [Angle between the radius and the tangent is 90 ] l O A B 66 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 13. The lengths of two tangent segments drawn from an external point to a circle are equal. A Presentation: PA = PB [Tangents to a circle from an exterior point are equal] P O B 67 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 13. Property of intersecting chords If two chords AB and CD of a circle intersect inside the circle at a point P then AP BP = CP DP D A P C B 68 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 14. Property of intersecting secants If two secants AB and CD of a circle intersect outside the circle at a point P then AP BP = CP DP A B P D C 69 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 15. Tangent secant property If PAB is a secant to a circle intersecting at points A and B and PT is a tangent then PA PB = PT . T P A B 70 Ch. 17 Circle, Tangents and Intersecting Chords (Properties): 15. Angles in alternate segments are equal. R C Angle in alternate segment R Angle in alternate segment B B O O D Angle between tgt and chord P A Q Presentation: BAQ = ACB [Angles in the alternate segments are equal] P A Q Angle between tgt and chord PAB = ADB [Angles in the alternate segments are equal] 71 Ch. 18 Constructions 1. To draw a tangent to a circle of given radius draw a perpendicular to the radius 2. Draw 2 tangents to a circle of radius 3 cm from a point P at a distance of 5 cm from the centre. A P O M B 72 Ch. 18 Constructions 3. Circumcircle In every triangle 3 perpendicular bisectors meet at one point called circumcentre. To draw circumscribing circle we need to draw perpendicular bisectors of any 2 sides. 4. Incircle In every triangle 3 angle bisectors meet at one point called incentre. To draw inscribing circle we need to draw angle bisectors of any 2 angles. 73 Ch. 19 Cylinder, Cone and Sphere 1. Cylinder CSA of cylinder = 2 rh Total Surface Area = 2 r (r + h) Volume = r2h h r 74 Ch. 19 Cylinder, Cone and Sphere 2. Hollow Cylinder 1. Thickness of its wall = R r 2. Area of cross section = R2 r2 = (R2 r2) 3. External curved surface area = 2 Rh R r 4. Internal curved surface area = 2 rh h 5. Total surface area = External CSA + Internal CSA + 2 (Area of cross section) = 2 Rh + 2 rh + 2 (R2 r2) 6. Volume = External volume Internal volume = R2h r2h = [ R2 r2 ] h 75 Ch. 19 Cylinder, Cone and Sphere 3. Cone Slant height (l ) = (r)2 + (h)2 CSA of cone = rl l h Total Surface Area = r (r + l) Volume = 1 r2h 3 r 76 Ch. 19 Cylinder, Cone and Sphere 4. Sphere Surface Area = 4 r Volume = 4 3 r3 r 77 Ch. 19 Cylinder, Cone and Sphere 5. Hollow Sphere Volume = Volume of Volume of external sphere internal sphere 4 r3 3 = 4 R3 3 = 4 (R3 r3) 3 R r 78 Ch. 19 Cylinder, Cone and Sphere 6. Hemi Sphere CSA = 2 r2 r TSA = 3 r2 Volume = 2 r3 3 79 Ch. 19 Cylinder, Cone and Sphere 7. Hollow Hemi Sphere Total = 2 (R2 + r2) + R2 r2 surface area Volume = R r 2 [ R3 r3 ] 3 80 Ch. 20 Trigonometry (Formulae) : 1. Trigonometric Ratios: sin = Opposite side Hypotenuse cosec = Hypotenuse Opposite side cos = Adjacent side Hypotenuse sec = Hypotenuse Adjacent side tan = Opposite side Adjacent side cot = Adjacent side Opposite side 81 Ch. 20 Trigonometry (Formulae) : 2. Reciprocal Relations: sin = 1 cosec cos = 1 sec and sec = 1 cos tan = 1 cot and cot = 1 tan tan = sin cos and cot = cos sin and cosec = 1 sin 82 Ch. 20 Trigonometry (Formulae) : 3. Complementary Angles: a) sin = cos (90 ) or cos = sin(90 ) b) tan = cot (90 ) or cot = tan(90 ) c) cosec = sec (90 ) or sec = cosec(90 ) 4. Identities: a) sin2 + cos2 = 1 b) 1 + tan2 = sec2 c) 1 + cot2 = cosec2 83 Ch. 23 Measures of Central Tendency 1. Mean of Raw Data Sum of all observations MEAN = ____________________________ Number of observations 84 Ch. 23 Measures of Central Tendency 2. Mean of Ungrouped Data Mean = fx f xi f 29 1 34 7 35 6 36 7 38 5 40 1 41 1 43 1 51 1 Total f = fx fx = 85 Ch. 23 Measures of Central Tendency 3. Mean of Grouped Data Direct Method class interval class mark ( x) frequency ( f) f = Total Mean ( ) = fx fx = fx f 86 Ch. 23 Measures of Central Tendency 3. Mean of Grouped Data Short Cut Method class interval class mark frequency d = x A ( x) ( f) fd A f = Total Mean ( ) = A + fd = fd f 87 Ch. 23 Measures of Central Tendency 3. Mean of Grouped Data Step Deviation Method Class width (i) = _____ class interval Assumed Mean (A) = ____ class mark t = x A ( x) i frequency ( f) ft A f = Total Mean ( ) = A + i ft = ft f 88 Ch. 23 Measures of Central Tendency 4. Median of Raw Data Steps to find Median: 1) Arrange the data in either ascending or descending order. 2) Find the total number of observations (n). 3) If n is odd then, Median = th n + 1 Observation 2 If n is even then, Median = Mean of th and + th Observation 89 Ch. 23 Measures of Central Tendency 5. Quartiles Quartiles When Total Frequency (n) is even n Lower Quartile (Q1) = 4 3n 4 Inter Quartile Range = Q3 - Q1 Upper Quartile (Q3) = When Total Frequency (n) is odd Lower Quartile (Q1) = n+1 4 3(n+1) Upper Quartile (Q3) = Inter Quartile Range 4 = Q3 - Q1 90 Y Cumulative frequency less than type Ch. 23 Measures of Central Tendency 6. Median of Grouped Data (From Ogive) 1. Median = 57 Marks 2. Q1 = 42 Marks 3. Q3 = 69 Marks 4. No. of students scoring more than 80 marks = 200 180 = 20 students 200 (90,194) (100,200) (80,180) 180 160 (70,151) 140 120 (60,111) 100 80 60 40 20 X Scale : On X axis: 1 cm = 10 marks On Y axis: 1 cm = 20 students 0 Y (50,73) (40,46) Q1 Median (30,26) Q3 (20,15) (10,5) 10 20 30 40 50 60 70 80 90 100 Upper class boundaries 91 X Y Scale : On X axis:2 cm = 10 marks 7. Mode of Grouped Data (From Histogram) On Y axis:1 cm = 2 students No. of students Ch. 23 Measures of Central Tendency X 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0 Y Mode = 37marks Mode 10 20 30 40 Marks 50 60 70 80 X 92 Ch. 24 Probability (Formulae): Total favourable outcomes Total possible outcomes 1. Probability = 2. P E P E 1 93 Ch. 24 Probability (Formulae): 3. Tossing 3 Coins: No. of Heads No. of Tails Possible Outcomes 3 0 HHH 2 1 HHT , HTH , THH 1 2 HTT , THT , TTH 0 3 TTT S = { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT } 94 Ch. 24 Probability (Formulae): 4. Throwing 2 Dice: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) 95 96
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