SSC TRICKS FOR MATHS

FROM A POINT A ON THE GROUND, THE ANGLE OF ELEVATION OF THE TOP OF A 20 m TALL BUILDING IS 45. A FLAG IS HOSTED AT THE TOP OF THE BUILDING AND THE ANGLE OF ELEVATION OF THE TOP OF THE FLAGSTAFF FROM A IS 60. FIND THE LENGTH OF THE FLAGSTAFF AND THE DISTANCE OF THE BUILDING FROM THE POINT A. LET "BC" BE THE BUILDING AND "BD" BE A FLAG GIVEN,       BC=20 m   ANGLE(BAC)=45     ANGLE(DAC)= 60 LET      BD= x m IN TRIANGLE(BAC),       tan45 = BC/AC = 20/AC AC = 20 m NOW IN TRIANGLE(DCA),     DC = DB + BC = x + 20 tan 60 = DC/AC sqrt(3) = (x + 20)/20 x + 20 = 20*sqrt(3) = 20 * 1.732 = 34.64 x = 34.64 - 20 = 14.64 m HENCE, THE LENGTH OF THE FLAGSTAFF IS 14.64 M AND DISTANCE OF POINT A FROM THE BUILDING IS 20 m. TWO POLES ARE PLACED AT P AND Q ON EITHER SIDE OF A ROAD SUCH THAT THE LINE JOINING P AND Q IS PERPENDICULAR TO THE LENGTH OF THE ROAD. A PERSON MOVES x METRE AWAY FROM P ...

FIND THE RATE PERCENT PER ANNUM, IF RS. 2000 AMOUNTS TO RS. 2,315.25 IN A YEAR AND A HALF, INTEREST BEING COMPOUNDED HALF YEARLY. TIME = 3/2 years = 3 half years RATE = 2R% per annum = R% for half year THEREFORE AMOUNT,  = principal (1 + (Rate/100))^time  2315.25 = 2000 (1 + (R/100))^3 231525/200000 = (1 + (R/100))^3 9261/8000 = (1 + (R/100))^3 (21/20)^3 = (1 + (R/100))^3 CANCELLING THE CUBE ON BOTH SIDES 21/20 = 1 + R/100 21/20 - 1 = R/100 1/20 = R/100\ R = 100/20 R = 5% per half year THEREFORE REQUIRED RATE IS, 10% p. a......... SIMPLE INTEREST ON A CERTAIN SUM OF MONEY FOR 3 YEARS AT 8% PER ANNUM IS HALF THE COMPOUND INTEREST ON RS 16000 FOR 2 YEARS AT 10% PER ANNUM. THE SUM PLACED ON THE SIMPLE INTEREST IS C.I  = P[(1 + R/100)^T - 1]        = 16000[(1 + 10/100)^2 - 1]        = 16000[(11/10)^2 - 1]        = 16000[ 121/100 - 1]   ...