SSC CGL TRICKS FOR SQUARE ROOT QUESTIONS


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STATISTICS FORMULAES FOR FINDING MEAN, MODE AND MEDIAN

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MEAN FORMULA:- MEAN = (ADDITION OF f i x i) / f i MEAN FORMULA:- MEAN = a + { (ADDITION OF  f i x i) /  f i} * h a = assumed mean  h = class size MODE FORMULA:- MODE = l + [( f1 - f0) / (2f1 - f0 - f2) ] * h l = lower limit of class of which frequency comes f1 = frequency       f1 = n/2 f0 = preciding f0 f2 = after f0 MEDIAN FORMULA:- MEDIAN = l + {(n/2 - cf) / f} * h l = lower limit of class of which frequency comes n = addition of frequency or last number cumulative frequency h = class size SOME OTHER EXAMPLES ARE: SO MEAN OF THIS QUESTION IS: BY DIRECT METHOD:                             = fixi / fi                             = 3620/140                             = 25.86 BY ASSUMED MEAN METHOD: MEAN = a + fidi/fi              = 25 + 12/140              = 25.86
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MISCELLANEOUS QUESTIONS

THERE IS 40% INCREASE IN AN AMOUNT IN 8 YEARS AT SIMPLE INTEREST. WHAT WILL BE THE COMPOUND INTEREST IN RS OF RS 30000 AFTER 2 YEARS AT THE SAME RATE? ACCORDING TO THE QUESTION, IF THE PRINCIPLE = RS 100 THEN INTEREST  = RS 40 THEREFORE RATE = ( I * 100 ) / ( P * T ) = 40 * 100 * 1/100 * 1/8 = 5 % p. a. CASE II, A = P[1+R/100]^T A = 30000[1+5/100]^2 A = 30000[1+1/20]^2 A = 30000 * 21/20 * 21/20 A = RS 33075 THEREFORE, = RS [33075 - 30000] = RS 3075 WHAT IS THE MEASURE OF THE CENTRAL ANGLE OF THE ARC WHOSE LENGTH IS 11 CM AND RADIUS OF THE CIRCLE IS 14 CM? ARC LENGTH IS 11 CM AND IF WE CALCULATE THE CIRCUMFERENCE OF THE CIRCLE, = 2 * pi * r = 2 * 22/7 * 14 = 88 cm AND NOW WE HAVE TO FIND THE ANGLE OF THE ARC, WE KNOW THAT IN A CIRCLE THE ANGLE IN THE MIDDLE IS 360 deg SO, 88cm = 360 deg 1 cm = 360 / 88 AND IN ARC WHICH IS 11 cm, 11 cm = 360/88 * 11 = 45 deg IF THE SHOPKEEPER SELLS AN ITEM AT RS 960 WHICH IS MARKED AS RS 1200, WH
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TRICKY QUESTION

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A STRAIGHT HIGHWAY LEADS TO THE FOOT OF A TOWER. A MAN STANDING AT THE TOP OF THE TOWER OBSERVES A CAR AT AN ANGLE OF DEPRESSION OF 30 deg, WHICH IS APPROACHING THE FOOT OF THE TOWER WITH A UNIFORM SPEED. SIX SECONDS LATER, THE ANGLE OF THE DEPRESSION OF THE CAR IS FOUND TO BE 60 deg. FIND THE TIME TAKEN BY THE CAR TO REACH THE FOOT OF THE TOWER FROM THIS POINT. TAKE DB = x, SO, AB = 6 + x, TAN 30 = BC / AB 1 / sqrt(3) = BC / 6 +x BC = (6 + x) / sqrt(3).....................(i) THEN, TAN 60 = BC / DB sqrt(3) = BC / x BC = x * sqrt(3)............................(ii) BY (i) AND (ii) WE GET, x * sqrt(3) = (6 + x) / sqrt(3) x * sqrt(3) * sqrt(3) = 6 + x x * 3 = 6 + x 3x = 6 + x 3x - x = 6 2x = 6 x = 3sec A CUBE OF EDGE 6 cm IS PAINTED ON ALL SIDES AND THEN CUT INTO UNIT CUBES. THE NUMBER OF UNIT CUBES  WITH NO SIDES PAINTED IS, USUALLY THIS QUESTION TAKES LOT OF TIME IN THE EXAM UP TO 2 MINUTES BUT THIS THE TRICK BY WHICH YOU CAN DO THIS QUESTIO
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