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Maths Short Cut Tricks For Board Exams

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Methods or  Tricks For Board Exams!

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    1. 2. 3. 4. 5 6. 7. Shortcuts in Quantitative Aptitude NUMBER SYSTEM Method to multiply 2-digit number. +20/35=12/41/35-1645 (ii) +C)/BC 74 +6) / 4 x 6 = 49 / 70 / 24 = 49 / 70/24 = 5624 (iii) = 35 x 4/ (3 + 5) x 4/5 x 4 Method to multiply 3-digit no. ABC x DEF = AD/ AE / AF + BE + CD / BF + CE / CF -8/ 12+ 10/ 16+ + 12/20+ 18/24 = 8 / 22 /43 / 38 / 24 = 106704 If in a series all number contains repeating 7. To find their sum, we start from the left multiply 7 by 1, 2, 3, 4, 5 & 6. Look at the example below. 777777 + 77777 + 7777 + 777 + 77 + 7 = ? 0.5555+0.555 + 0.55+0.5 = ? To find the sum of those number in which one number is repeated after decimal, then first write the number in either increasing or decreasing order. Then -find the sum by using the below method. 0.5555+0.555 +0.55+0.5 = 20/15 / 10/5=2.1605 Those numbers whose all digits are 3. (33)2 Those number. in which all digits are number is 3 two or more than 2 times repeated, to find the square of these number, we repeat 1 and 8 by (n — l) time. Where n —+ Number of times 3 repeated. (333)2 11108889 Those number whose all digits are 9. (99)2 = 9801 (999)2 = 998001 = 9999800001 Those number whose all digits are I. In this we count number of digits. We write l, 2, 3, ..... in their square the digit in the number, then write in decreasing order up to 1. 112 = 121 1112 = 12321
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    8. 9. 10. 11. 12. 13. 14. 101 Shortcuts in Quantitative Aptitude Some properties of square and square root: (i) Complete squareofano. is possible if its last digit is 0, 1, 4, 5, 6 & 9. If last digitofano. is 2, 3, 7, 8 then complete square root of this no. is not possible. (ii) If last digit of a no. is 1, then last digit of its complete square root is either 1 or 9. (iii) If last digit of a no. is 4, then last digit of its complete square root is either 2 or 8. (iv) If last digit of a no. is 5 or 0, then last digit of its complete square root is either 5 or 0. (v) If last digit of a no. is 6, then last digit of its complete square root is either 4 or 6. (vi) If last digit of a no. is 9, then last digit of its complete square root is either 3 or 7. Prime Number : (i) Find the approx square root of given no. Divide the given no. by the prime no. less than approx square root of no. If given no. is not divisible by any of these prime no. then the no. is prime otherwise not. For example : To check 359 is a prime number or not. Sol. Approx sq. root = 19 Primeno. < 19 are2, 3, 5, 7, 11, 13, 17 359 is not divisible by any of these prime nos. So 359 is a prime no. For example: Is 25001 + 1 is prime or not? 5001 2 Reminder = 0, 25001 + I is not prime. (ii) There are 15 prime no. from I to 50. (iii) There are 25 prime no. from 1 to 100. (iv) There are 168 prime no. from I to 1000. If a no. is in the form ofxn + an, then it is divisible by (x + a); if n is odd. Ifxn + (x — 1), then remainder is always 1. Ifxn+(x+ l) (i) (ii) (i) (ii) (iii) If n is even, then remainder is 1. If n is odd, then remainder is x. Value of Value of Value of 2 4P+1-1 2 (iv) Value of P P P PUF P [Where n —+ no. of times P repeated]. Note: If factors of P aren & (n + 1) type then value of p + p + p + Number of divisors : (i) If N is any no. and N = an x bill x cp x .... where a, b, c are prime no. No. of divisors of N = (n + l) (m + l) (p + 1) e.g. Find the no. of divisors of 90000. ..00 (n -k 1) and ..00 N = 90000 = 22 x 32 x 52 x 102 = 22 x 32 x 52 x (2 x = 24 x 32 x 54 So, the no. of divisors = (4 + l) (2 + 1) (4 + 1) = 75 (ii) N = an x bill x cP, where a, b, c are prime Then set of co-prime factors of N = [(n + 1) (m + 1) (p + 1) — I + nm + mp + pn + 3mnp] (an+l (iii) If N = anx bill x cp..., where a, b& c are prime no. Then sum of the divisors = (a — l) cp+l
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    15. 16. 17. 18. 19. 20. 21. 101 Shortcuts in Quantitative Aptitude (........5) To find the last digit or digit at the unit's place of an. (i) If the last digit or digit at the unit's place of ais 1, 5 or 6, whatever bethe valueofn, it will have the same digit at unit's place, i.e., .6) (ii) n 4X4-I 4X4-2 4X4-3 If the last digit or digit at the units place of a is 2, 3, 5, 7 or 8, then the last digit of an depends upon the value of n and follows a repeating pattern in terms of 4 as given below : last digit of ...2P 2 4 8 6 last digit 3 9 7 1 last digit 7 9 3 1 last digit 8 4 2 6 (iii) If the last digitor digit at the unit's place of ais either 4 or 9, then the last digit of an depends upon the value of n and follows repeating pattern in terms of 2 as given below. n last digit Of 6 4 last digit 1 9 (i) (ii) (iii) (i) (ii) (iii) (i) (ii) (iii) (i) (ii) Sum of n natural number 2 Sum of n even number = (n) (n + 1) 2 Sum of n odd number = n Sum of sq. of first n natural no. = 6 n 4n Sum of sq. of first n odd natural no. = Sum of sq. of first n even natural no. = Sum of cube of first n natural no. — Sum of cube of first n even natural no Sum of cube of first n odd natural no xn — yn is divisible by (x + y) When n is even xn — yn is divisible by (x — y) When n is either odd or even. 3 4 3 2 2 2 . = 2n2 (n + 1) 2 . = n2 (2n2 — 1) 13 — n is divisible by 13. Some articles related to Divisibility : (i) A no. of3-digits which is formed by repeating a digit 3-times, then this no. is divisible by 3 and 37. e.g., 111, 222, 333 (ii) A no. of6-digit which is formed by repeating a digit 6-times then this no. is divisible by 3, 7, 11, 13 and 37.
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    22. 23. 24. 25. 26. 27. 28. 101 Shortcuts in Quantitative Aptitude Divisible by 7 : We use osculator (— 2) for divisibility test. 99995 : 9999 _ 2 x 5 = 9989 Now 98 is divisible by 7, so 99995 is also divisible by 7. Divisible by 11 : In a number, if difference of sum of digit at even places and sum of digit at odd places is either 0 or multiple of 11, then no. is divisible by 11. For example, 12342 + I I Sum of even place digit = 2 + 4 = 6 Sum ofodd place digit = I + 3 + 2 = 6 Difference = 6 — 6 = 0 . 12342 is divisible by 11. Divisible by 13 : We use (+ 4) as osculator. e.g., 876538+ 13 8 x 4 + 2 + 6 = 40 I x 8 = 13 13 is divisible by 13. 876538 is also divisible by 13. Divisible by 17 : We use (— 5) as osculator. e.g., 294678: 29467-5 = 29427 27427: 2942-5 x 7 = 2907 2907: 290-5 x 7 = 255 255: 25-5 x 294678 is completely divisible by 17. Divisible by 19 : We use (+ 2) as osculator. 149264: 4 x 14 1 X 2+14-9= 12 9 19 19 is divisible by 19 149264 is divisible by 19. HCF (Highest Common factor) There are two methods to find the HCF— (a) (i) (ii) (iii) (iv) (v) Factor method (b) Division method For two no. a and b if a < b, then HCF of a and b is always less than or equal to a . The greatest number by which x, y and z completely divisible is the HCF of x, y and z. The greatest number by which x, y, z divisible and gives the remainder a, b and c is the HCF of (x —a), (y—b) and (z—c). The greatest number by which x, y and z divisible and gives same remainder in each case, that number is HCF of (x—y), (y—z) and (z—x). e H.C.M. of (a, c, e) H.C.F. of — and b'd f L.C.M. of (b, d, f) LCM (Least Common Multiple) There are two methods to find the LCM— (a) (i) (ii) Factor method (b) Division method For two numbers a and b if a < b, then L.C.M. of a and b is more than or equal to b. If ratio between two numbers is a : b and their H.C.F. is x, then their L.C.M. = abx.
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    101 Shortcuts in Quantitative Aptitude (iii) (iv) (v) (vi) (vii) (viii) Ox) x Ifratio between two numbers is a : b and their L.C.M. is x, then their H.C.F. ab The smallest number which is divisible by x, y and z is L.C.M. of x, y and z. The smallest number which is divided by x, y and z give remainder a, b and c, but (x — a) = (y — b) = (z — c) = k, then number is (L.C.M. of (x, y and z) — k). The smallest number which is divided by x, y and z give remainder k in each case, then number is (L.C.M. of x, y and z) + k. e L.C.M. of (a, c, e) L.C.M. of — and — b'd f H.C.F. of (b, d, f) For two numbers a and b — LCM x HCF b If a is the H.C.F. of each pair from n numbers and L is L.C.M., then product of n numbers = ALGEBRA an-I.L 29. 30. 31. 32. Algebra Identities: (iii) (v) (vii) Ox) (x) (ii) (iv) (vi) (viii) 4 ab b j— a i 1 If a + b + c = abc, then 2a 2b 2 I-b2 2 I—a 3 3a — a 2 1-3b2 1-3a 1 — (a — = 4ab a3 — b3 = (a — b) (a2 + ab + b2) = 0, then a3 + b3 + c3=3abc 2 2 2 and 3 2 then 3 h 3b b3h [3c c3h If al x + blY = and a2X + b2Y = c2, al (i) If — , one solution. b2 al (ii) al If a2 b 2 , Infinite many solutions. — — , NO solution (iii) If b2 If u and are roots of ax2 + bx + c = 0, then If u and are roots of ax2 + bx + c = 0, then and — are roots ofcx2 + bx + a (i) (ii) (iii) (iv) (v) One root is zero if c = 0. Both roots zero if b = 0 and c = 0. Roots are reciprocal to each other, if c = a. If both roots u and are positive, then sign of a and b are opposite and sign Ofc and a are same. If both roots u and are negative, then sign of a, b and c are same. 4 4 b , 01B — , then a a u + 4uß 2 2 2u2ß2 2 01 -k
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    101 Shortcuts in Quantitative Aptitude 33. 34. 35. 36. 37. Arithmetic Progression: (i) If a, a + d, a + 2d, ..... are in A.P., then, nth term ofA.P. an = a + (n — Sum of n terms of this A.P. — Sn n 2 A.M. = Arithmetic mean] = —la +1] whereC— 2 — last term a = first term d = common difference (ii) A.M. = 2 Geometric Progression: 2 (i) G.P. a, ar, ar Then, nth term of G.P. a — arn I s n (r (l 1) rn) r) [where r = (ii) G.M. 111 If a, b, c are in H.P., are in A.P. n term of H.M. — 2ab H.M. a+b 1 nth term of A.P. Note : Relation between A.M., G.M. and H.M. (i) A.M. x H.M. =G.M.2 A.M. —+ Arithmetic Mean G.M. —+ Geometric Mean H.M. Harmonic Mean (i) Average of first n natural no. = 2 (ii) Average of first n even no. = (n + l) (iii) Average of first n odd no. = n (i) Average of sum of square of first n natural no. = 6 • , = first term] common rat10 a AVERAGE (ii) Average of sum of square of first n even no. = 3 1 (iii) Average of sum of square of first odd no. = 411 3
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    101 Shortcuts in Quantitative Aptitude 38. 39. 40. 41. 42. 43. 44. 45. (i) Average of cube of first n natural no. = (ii) Average of cube of first n even natural no (iii) Average of cube of first n odd natural no. = n(2n2 — 1) Average of first n multiple of m = 2 4 . = 2n(n + 1) 2 (i) (ii) (iii) (iv) (v) 2 If average of some observations is x and a is added in each observations, then new average is (x + a). If average of some observations is x and a is subtracted in each observations, then new average is (x — a). If average of some observations is x and each observations multiply by a, then new average is ax. x If average of some observations is x and each observations is divided by a, then new average is a If average of n 1 is Al, & average of is A2, then Average of (111 + 112) is and n 1 Al — n2A2 Average of (n I — 112) is nl —n2 When a person is included or excluded the group, then age/weight of that person = No. of persons in group x (Increase / Decrease) in average ± New average. For example : In a class average age of 15 students is 18 yrs. When the age of teacher is included their average increased by 2 yrs, then find the age of teacher. Sol. Age of teacher = 15 x 2 + (18+2) = 30+ 20 = 50 yrs. When two or more than two persons included or excluded the group, then average age of included or excluded person is No. of person x (Increase / Decrease) in average + New average x (No. of person included or excluded) No. of included or person For example : Average weight of 13 students is 44 kg. After including two new students their average weight becomes 48 kg, then find the average weight of two new students. Sol. Average weight of two new students 52+96 2 2 = 74 kg 2 2xy If a person travels two equal distances at a speed of x km/h and y km/h, then average speed = — km/h 3xyz If a person travels three equal distances at a speed of x km/h, y km/h and z km/h, then average speed = xy + yz + zx RATIO & PROPORTION a (i) If b For example: If — sol. P+Q+R then , then find 7 3+4+7 (ii) Then al If a3 7 a4 K , then a 1 a3 a n an +1
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    46. 47. 48. 49. 50. 51. 52. 101 Shortcuts in Quantitative Aptitude ad bc A number added or subtracted from a, b, c & d, so that they are in proportion = For example : When a number should be subtracted from 2, 3, 1 & 5 so that they are in proportion. Find that number. 10-3 7 sol. Req. No. = If X part of A is equal to Y part of B, then A : B = Y : X. For example: If 20% of A = 30% of B, then find A : B. sol. When Nth part of P, Y th part of Q and Zth part of R are equal, then find A : B : C. Then, A : B : C : zx : xy TIME, DISTANCE AND WORK A can do a/b part of work in tl days and c/d part of work in t2 days, then (i) If A is K times efficient than B, Then T(K+ l) = Kt (ii) If A is K times efficient than B and takes t days less than B t Then T = or K2-1 K 1' B t = ktA (ii) (iii) (i) (ii) If a cistern takes X min to be filled by a pipe but due to a leak, it takes Y extra minutes to be filled, then the time taken by leak X2 + XY min to empty the cistern = If a leak empty a cistern in X hours. A pipe which admits Y litres per hour water into the cistern and now cistern is emptied X+Y+Z in Z hours, then capacity of cistern is = litres. z-x If two pipes A and B fill a cistern in x hours and y hours. A pipe is also an outlet C. If all the three pipes are opened together, xyT the tank full in T hours. Then the time taken by C to empty the full tank is = YT + XT — xy If tl and t2 time taken to travel from A to B and B to A, with speed a km/h and b km/h, then distance from A to B is ab tlt2 ab t2) a—b If 1st part of distance is covered at the speed of a in tl time and the second part is covered at the speed of b in t2 time, then at2 -k the average speed = I
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    101 Shortcuts in Quantitative Aptitude PERCENTAGE 53. 54. 55. 56. Simple Fraction Their Percentage 100% 33.3% 25% 16.67% 14.28% a Simple Fraction 1 8 1 9 1 10 1 11 1 12 (i) (ii) 1 1 2 1 3 1 4 1 5 1 6 1 7 If A is If A is a % less than A. more than B, then B is b a a — less than B, then B is 0 more than A b a—b Their Percentage 12.5% 11.11% 9.09% 8.33% % so that expenditure will be same. if a > b, we take a — b if b > a, we take b— a. b—a If price of a article increase from a to b, then its expenses decrease by b Due to increase/decrease the price x%, A man purchase a kg more in y, then Per kg increase or decrease = IOOxa xy Per kg starting price = (100 ± x) a
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    s-10 57. 58. 59. 6(). 61. 62. 63. 101 Shortcuts in Quantitative Aptitude For two articles, if price: 1st Increase (x%) Increase (x%) Decrease (x%) Increase (x%) Overall Increase Increase (y%) Decrease (y%) Decrease (y%) Decrease (x%) 100 100 If +ve (Increase) If —ve (Decrease) Decrease x + Y 100 2 x Decrease 100 If the side of a square or radius of a circle is x% increase/decrease, then its area increase/decrease = If the side of a square, x% increase/decrease then x% its perimeter and diagonal increase/decrease. 2 x 2x ± 100 (i) (ii) t IOO±R If population P increase/decrease at r% rate, then after t years population = P 100 If population P increase/decrease rl % first year, increase/decrease second year and % C/o increase/decrease third year, then after 3 years population = P 1+— 100 100 100 If increase we use (+), if decrease we use (—) If a man spend x% of this income on food, y% of remaining on rent and z% of remaining on cloths. If he has P remaining, then total income of man is = (100 -x) (100 - y) (100 -z) [Note: We can use this table for area increase/decrease in mensuration for rectangle, triangle and parallelogram]. PROFIT AND LOSS If CP of x things = SP of y things, then Profit/Loss — If +ve, Profit; If —ve, Loss If after selling x things P/L is equal to SP of y things, x 100 then P/L = Profit — Loss
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    101 Shortcuts in Quantitative Aptitude 64. 65. 66. 67. 68. 69. 70. If CP of two articles are same, and they sold at 1st (x%) Profit (x%) Profit (x%) Loss (x%) Profit (y%) Profit (y%) Loss (y%) Loss (y%) Loss Overall % Profit 2 Profit, if x > y 2 Loss, if x < y % Loss 2 No profit, no loss If SP of two articles are same and they sold at 1st Profit (x%) Profit (x%) Overall 2 x Loss Loss(x%) Loss (y%) 100 IOO(x-y) — 2xy 200+x-y or 200+x-y s-11 If + ve, then Pr ofit% -100 % If — ve, then Loss% After D% discount, requires P% profit, then total increase in C.P.= (100 + P) (100 -D) (M.p. - c.p.) x 100 Profit % = 100 -D 100 -k r2 + r3 (i) (ii) (i) (ii) C.P. 100 + rl For discount r 1% and r2%, successive discount 100 For discount r 1 and % 0/0, successive discount 100 100 100 1 100 x 100 100 P = Principal, T = Time in years, A = Amount PRT 100 A=P+SI SIMPLE AND COMPOUND INTEREST R = Rate per annum, SI = Simple interest, 100
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    s-12 71. 72. 73. 74. 75. 76. 77. 78. 79. 101 Shortcuts in Quantitative Aptitude If P = Principal, A = Amount in n years, R = rate of interest per annum. A-P 1+— 100 (i) (ii) (ii) , interest payable annually , interest payable half-yearly 100 4n , interest payable quarterly; 400 is the yearly growth factor; 400 is the yearly decay factor or depreciation factor. 400 When time is fraction of a year, say 4— , years, then, Amount 4 100 CI = Amount — Principal 3 4 1 100 100 -1 When Rates are different for different years, say RI, R2, for 1 st, 2nd & 3rd years respectively, then, Amount = P 1+— 100 100 100 In general, interest is considered to be SIMPLE unless otherwise stated. (i) (ii) (iii) (iv) (v) (vi) GEOMETRY Sum of all the exterior angle of a polygon = 3600 3600 Each exterior angle of a regular polygon = n Sum of all the interior angles of a polygon = (n -2) x 1800 x 1800 Each interior angle of a regular polygon = n No. of diagonals of a polygon = , n no. of sides. 2 The ratio of sides a polygon to the diagonals of a polygon is 2 : (n — 3) (vii) Ratio of interior angle to exterior angle of a regular polygon is (n — 2) : 2 Properties of triangle: (i) When one side is extended in any direction, an angle is formed with another side. This is called the exterior angle. There are six exterior angles of a triangle. (ii) Interior angle + corresponding exterior angle = 1800
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    101 Shortcuts in Quantitative Aptitude AE AB EC' AD s-13 (iii) (iv) (v) (vi) (vii) An exterior angle = Sum of the other two interior opposite angles. Sum of the lengths of any two sides is greater than the length of third side. Difference of any two sides is less than the third side. Side opposite to the greatest angle is greatest and vice versa. A triangle must have at least two acute angles. Triangles on equal bases and between the same parallels have equal areas. (viii) If a, b, c denote the sides of a triangle then (i) if c2 < a2 + b2, Triangle is acute angled. (ii) if c2 = a2 + b2, Triangle is right angled. (iii) if c2 > a2 + b2, Triangle is obtuse angled. Ox) (i) (ii) (iii) (x) (xi) (xii) If 2 triangles are equiangular, their corresponding sides are proportional. In triangles ABC and XYZ, if ZX, LB = zy, ZC = ZZ, then AB AC BC XY XZ YZ In AABC, LB = 900 BD L AC 1 1 1 2 AB BC BD2 = AD x DC c The perpendiculars drawn from vertices to opposite sides (called altitudes) meet at a point called Orthocentre of the triangle. The line drawn from a vertex of a triangle to the opposite side such that it bisects the side is called the Median of the triangle. A median bisects the area of the triangle. When a vertex of a triangle is joined to the midpoint of the opposite side, we get a median. The point of intersection of the medians is called the Centroid of the triangle. The centroid divides any median in the ratio 2 : 1. (xiii) Angle Bisector Theorem— In the figure if AD is the angle bisector (interior) of Z BAC. Then, c 1. 2. AB/AC = BD/DC. AB -BD x DC = AD2 (xiv) Midpoint Theorem — In a triangle, the line joining the mid points of two sides is parallel to the third side and half of it. (w) Basic Proportionality Theorem A line parallel to any one side of a triangle divides the other two sides proportionally. If DE is parallel to BC, then D E c AD AC AE AD DE AB = — and so on. BC
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    s-14 80. Properties of circle — 101 Shortcuts in Quantitative Aptitude (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) Ox) (x) (xi) (xii) (xiii) (xxv) (xv) (xvi) (xvii) Only one circle can pass through three given points. There is one and only one tangent to the circle passing through any point on the circle. From any exterior point of the circle, two tangents can be drawn on to the circle. The lengths of two tangents segment from the exterior point to the circle, are equal. The tangent at any point of a circle and the radius through the point are perpendicular to each other. When two circles touch each other, their centres & the point of contact are collinear. If two circles touch externally, distance between centres = sum of radii. If two circles touch internally, distance between centres = difference of radii Circles with same centre and different radii are concentric circles. Points lying on the same circle are called concyclic points. Measure of an arc means measure of central angle. m(minor arc) + m(major arc) = 3600. Angle in a semicircle is a right angle. Only one circle can pass through three given If ON is L from the centre O of a circle to a chord AB, then AN = NB. o (L from centre bisects chord) If N is the midpoint of a chord AB of a circle with centre O, then ZONA= 900 (Converse, L from centre bisects chord) Two congruent figures have equal areas but the converse need not be true. A diagonal of a parallelogram divides it into two triangles of equal area. (xviii) Parallelograms on the same base and between the same parallels are equal in area. (xix) Triangles on the same bases and between the same parallels are equal in area. 6«) If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to the half of the parallelogram. If PT is a tangent to the circle, then 01)2 = PT2 = OT2 O If PT is tangent and PAB is secant of a circle, then PT2 = PA.PB O'
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    101 Shortcuts in Quantitative Aptitude If PB & PD are two secant of a circle, then PA.PB = PC.PD If two circles touch externally, then distance between their centres If two circles touch internally, then distance between their centres — r 1 — r2 where r 1 > r . MENSURATION Area of triangle = — x base x altitude s-15 81. 82. 83. (i) (ii) Area of triangle using heron's formula — S/S-a(S c) , where S = In an equilateral triangle with side a, then where A Area of triangle p Perimeter h Height In an isosceles triangle PQR arAPQR=— 4a Height = b
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    s-16 sin0= r sm 84. 85. 86. 87. 88. 89. 90. (i) (ii) (iii) Area of A Area of A Area of A = —bc SinP where ZP — - ZQPR 2 b 101 Shortcuts in Quantitative Aptitude b = —ac SinQ 1 = —ab SinR 2 Q a 2 2 a 2 c a CosP CosR - Sine Rule : cosQ 2bc 2ab a SinP 2ac Area of square 2 b SinQ SinR Perimeter of square 4 Square Diagonal of square 4 In a circle with radius r. A c D — where A - Area of circle 4 C - Circumference of circle D - Diameter of circle IfO -600, arAAOB= 32 4 12 r O r 12 IfO,arAAOB 0 . cos 2 2 side of square 2 x area of AABC (i) A circle with largest area inscribed in a right angle triangle, then r Perimeter of AABC r
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    101 Shortcuts in Quantitative Aptitude Ita (ii) If ABC is an equilateral triangle with side a, then Area of circle = 12 Ita (iii) If ABC is an equilateral triangle with side a, then area of circle = (iv) If AABC is an equilateral triangle, and two circles with radius r and R, then s-17 Ttr — and TtR 2 4 91. ( 2Nfi It). r (v) Three equal circle with radius r and an equilateral triangle ABC, then area of shaded region = area of square 7 ABCD is a square placed inside a circle with side a and radius of circle r, then area of circle 11
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    s-18 92. 93. 94. 95. 96. 97. 98. 101 Shortcuts in Quantitative Aptitude Diagonal of a cube = x side I +192 +112 ; where C •Length, b —5 breadth, h *height Diagonal of a cuboid = For two cubes al dl where Al, A2 Area of cubes VI, v2 volume al, a2 Sides dl, d2 —5 Diagonals Units of Measurement of Area and Volume The inter-relationships between various units of measurement of length, area and volume are listed below for ready reference: Length 1 Centimetre (cm) 1 Decimetre (dm) 1 Metre (m) 1 Decametre (dam) 1 Hectometre (hm) 1 Kilometre (km) 1 Myriametre Area I cm2 = I cm x 1 cm I dm2 = I dm x I dm Im2=lmxlm I dam2 or 1 are 1 hm2 = I hectare 1 km2=1kmx1km Volume I cm3= I ml 1 litre= 1000 ml 1 m3 = 1 mxl mxlm 1 dm3 = 1000 cm3, 1m3 10 milimetre (mm) 10 centimetre 10 dm = 100 cm= 1000 mm 10m = 1000 cm 100m 1000m = 100 dam = 10 hm 10 kilometre 10mmx 100mm2 10 cm x 10 cm = 100 cm2 10 dm x 100 dm2 1 dam x Idam= 10m x 10m= 100m2 1 hm x 1 hm = 100m x 10000m2 = 100 dm2 10 hm x 10 hm = 100 hm2 or 100 hectare 1 cm x 1 cm x 1 cm= 10mmx I()mmx 10mm= 1000mm3 1000 cm3 100 cm x 100 cm x 100 cm = 106 cm3 1000 litre = 1 kilometre 1000 dm3, 1 km3 = 109 m3 If a, b, c are the edges of a cuboid, then 2 The longest diagonal (i) If the height of a cuboid is zero it becomes a rectangle. (ii) If "a" be the edge of a cube, then (iii) The longest diagonal = ax13 1 Volume of pyramid Base Area x height (H) 3 (i) If A denote the areas of two similar figures and Il & 12 denote their corresponding linear measures, (ii) If V & V 2 denote the volumes of two similar solids and Il, 12 denote their corresponding linear measures, Total volume of objecÉ submerged or taken out (iii) The rise or fall of liquid level in a container Cross sectional area of container then then 2 1 2 3 1 2
  • 19
    101 Shortcuts in Quantitative Aptitude 99. If a largest possible cube is inscribed in a sphere of radius 'a' cm, then s-19 (i) (ii) (iii) (iv) (v) 2a the edge of the cube If a largest possible sphere is inscribed in a cylinder of radius 'a' cm and height 'h' cm, then for h > a, the radius of the sphere = a and the radius h —— (for a > h) 2 If a largest possible sphere is inscribed in a cone of radius 'a' cm and slant height equal to the diameter of the base, then a the radius of the sphere If a largest possible cone is inscribed in a cylinder of radius 'a' cm and height 'h' cm, then the radius of the cone = a and height = h. If a largest possible cube is inscribed in a hemisphere of radius 'a' cm, then the edge of the cube 2 3
  • 20
    s-20 100. 101. In any quadrilateral 101 Shortcuts in Quantitative Aptitude 1 2 (i) (ii) 1 = — x one diagonal x (sum of perpendiculars to it from opposite vertices) = — x d (dl + d2) Area 2 (s - a)(s - b)(s - c)(s - d) Area of a cyclic quadrilateral where a, b, c, d are sides of quadrilateral and s = semi perimeter — 2 If length, breadth & height of a three dimensional figure increase/decrease by x%, y% and z%, then 100 ± x 100 ± y Change in area = Change in Volume = 100 100 ± X 100 100 ± y 100 ± z 100 100

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