SSC TRICKS FOR MATHS

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

Find the remainder when (3^27) is divided by 5? Firstly we will separate 3^27 like this,               (3^26).3  (it means 3^27 only)  Then, We can also write 3^26 as ((3^2)^13)  We just splitted the power of 3 Now,            {((3^2)^13).3}/5.......... (i)  If we divide 3^2 i.e. 9 by 5 remainder comes to be 4 And that means 4^13 and 4 is less than 5 so remainder again comes to be ( -1) and (-1)^13  means  -1 and in eq (i) 3 is also there left to be divided by 5 and hence there is no power on 3 so  remainder will also be 3  Atlast we have {(-1)*3}/5  -3/5 means 5-3=2 (if there is - sign in the numerator then we have to substract the numerator  from denominator only if numerator is smaller than denominator) So,     2 will be the remainder of this question.  FIND THE REMAINDER WHEN 3^1000 IS DIVIDED BY 7 3^1000 = 3^4 * 3^996 3^996 is divided by 7 and gives the remainder 1  by,  996 = 6k  where 6 is the eulers's number for

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

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

10^(3x) = 125 THEN,  10^(-2x) = ? SOLUTION: 10^(3x) CAN BE WRITTEN AS {10^(x)}^3 AND 125 CAN BE WRITTEN AS 5^3 SO, {10^(x)}^3 = 5^3 POWERS ARE SAME ON BOTH THE SIDES WITH DIFFERENT BASE WE ONLY TAKE THE BASE i.e., 10^x = 5 SQUARING BOTH THE SIDES BUT PUT A "minus" SIGN WITH POWER {10^(x)}^(-2) = 5^(-2) THEN, {10^(x)}^(-2) CAN BE WRITTEN AS 10^(-2x) = 5^(-2) SO ANSWER OF 10^(-2x) IS 1/25 AS 5^(-2) IS 1/25

A COMPANY HIRES SOME EMPLOYEES IN A FIX PATTERN. ON THE 1ST DAY IT HIRED 1 MORE PERSON AND ON 3RD 1 MORE AND 1 MORE PERSON INCREASING IN THIS GROUP IN THIS SEQUENCE. THE CAPACITY OF EACH PERSON WAS SAME. IF COMPANY COMPLETE THE PROJECT IN 24TH DAY, THEN OUT OF TOTAL RS 50,000 MAXIMUM EARNING OF A PERSON IS? K+2K+3K+4K+________+24K = 50,000 K(1+2+3+4+5+________24) = 50,000 K * {(24 * 25)/2} = 50,000 K = 500/3 IS MINIMUM AND 500/3 * 24 = 4000 IS MAXIMUM

BY SELLING AN ARTICLE AT 2/3 OF THE MARKED PRICE, THERE IS A LOSS OF 10%. THE PROFIT PERCENT, WHEN THE ARTICLE IS SOLD AT THE MARKED PRICE, IS SUPPOSE MARKED PRICE = x THEREFORE, S.P. = Rs 2x/3 C.P. =(2x/3*90)*100 = 20x/27 PROFIT AT MARKED PRICE   = x - 20x/27 = 7x/27 THEREFORE, PERCENT PROFIT   ={(7x/27)/(20x/27)}*100   = 7x/27 * 27/20x * 100 = 35%

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 PARALLEL TO THE ROAD AND PLACES ANOTHER POLE AT R. THEN THE PERSON MO

IF a = 997, b = 998, c = 999, THEN, FIND, a^3 + b^3 + c^3 - 3abc WITH THE FORMULA OF, a^3 + b^3 + c^3 - 3abc = (a + b + c)/2{(a-b)^2 + (b - c)^2 + (c - a)^2} (997 + 998 + 999)/2{(997 - 998)^2 + (998 - 999)^2 + (999 - 997)^2} 2994/2 * 1 + 1+ 4 2994/2 * 6 2994 * 3 8982

THE SPEED OF A MOTOR-BOAT IS THAT OF THE CURRENT OF WATER AS 36:5. THE BOAT GOES ALONG WITH THE CURRENT IN 5 HOURS 10 MINUTES. IT WILL COME BACK IN ? LET THE SPEED OF MOTOR BOAT BE 36x kmph, AND SPEED OF CURRENT = 5x kmph THE BOAT GOES ALONG WITH THE CURRENT IN 5 HOURS 10 MINUTES i. e. 31/6 HOURS THEREFORE,       DISTANCE = 31/6 * (36x + 5x) = (41x * 31) / 6 km, RATE UPSTREAM = 36x - 5x = 31x kmph THEREFORE,         TIME TAKEN = (41x * 31/6) / 31x =41/6 HOURS OR 6 HOURS 50 MINUTES....

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]        = 16000 * 21/100        = RS 3360 THEREFORE, S.I. = 1/2 OF C.I. SO,  P = (S.I. * 100 ) / R * T

CUBE d = side VOLUME : V = d^2 SURFACE AREA : S = 6(d^2) RECTANGULAR SOLID l = length, w = width, h = height VOLUME: V = lwh SURFACE AREA : S = 2lw + 2lh + 2wh SPHERE r = radius VOLUME : V = 4/3*pi*r^3 SURFACE AREA : S = 4*pi*r^2 RIGHT CIRCULAR CYLINDER r = radius, h = height VOLUME : V = pi*r^2*h SURFACE AREA : S = 2*pi*r*h + 2 *pi*r^2 TORUS r = tube radius, R = torus radius VOLUME : V = 2*pi^2*r^2*R SURFACE AREA : S = 4*pi^2*r*R PYRAMID A = area of base , h = height VOLUME : V = 1/3*A*h RIGHT CIRCULAR CONE r = radius, h = height VOLUME : V = 1/3*pi*r^2*h SURFACE AREA : S = pi*r*[sqrt(r^2 + h^2)] + pi*r^2 FRUSTUM OF A CONE r = radius, R = base radius, h = height, s = slant height VOLUME : V = pi/3(r^2 + rR + R^2)h SURFACE AREA : S = pi*s(R+r) + pi*r^2 + pi*R^2 SQUARE PYRAMID s = side, h = height VOLUME : V = 1/3*s^2*h SURFACE AREA :

CIRCLE r = radius; d = diameter DIAMETER : d = 2r AREA : A = (pi) r^2 CIRCUMFERENCE : C = 2(pi)r = (pi)d SECTOR r = radius; angle(AOB) = angle in radius AREA : A = 1/2 (pi) r^2 ARC LENGTH : s = angle(AOB) r ELLIPSE a = semimajor axis b = semiminor axis AREA : A = (pi) ab CIRCUMFERENCE : C = (pi)[3(a+b) - sqrt{(a+3b)(b+3a)} ANNULUS  (COCENTRIC CIRCLES) r = inner radius, R = outer radius AVERAGE RADIUS = 1/2(r+R) WIDTH : w = R-r AREA : A = pi(R^2 - r^2) or A = 2*pi*22/7*average radius*width REGULAR POLYGON s = side length, n = number of sides CIRCUMFERENCE : R = 1/2*s*cos(pi/n) AREA : A = 1/4*n*s^2*cot(pi/n) or A = 1/2*n*R^2*sin(2*pi/n)

SOME RESULTS ON CIRCLES THEOREM 1. IF TWO ARCS OF A CIRCLE ARE CONGRUENT THEN THE CORRESPONDING CHORDS ARE EQUAL. THEOREM 2. THE PERPENDICULAR FROM THE CENTRE OF A CIRCLE TO A CHORD BISECT THE CHORD. THEOREM 3.  THE LINE JOINING THE CENTRE TO THE MID POINT OF A CHORD IS PERPENDICULAR TO THE CHORD. THEOREM 4.  THE PERPENDICULAR BISECTORS OF THE CHORDS OF A CIRCLE INTERSECT AT ITS CENTRE. THEOREM 5.  THERE IS ONE AND ONLY ONE CIRCLE PASSING THROUGH THREE NON COLLINEAR POINTS. (I) AN INFINITE NUMBER OF CIRCLE CAN BE DRAWN TO PASS THROUGH A SINGLE POINT. (II) AN INFINITE NUMBER OF CIRCLE CAN BE DRAWN TO PASS THROUGH TWO GIVEN POINTS. (III) A UNIQUE CIRCLE CAN BE DRAWN TO PASS THROUGH THREE GIVEN NON COLLINEAR POINTS. THEOREM 6.  EQUAL CHORDS OF CONGRUENT CIRCLES ARE EQUIDISTANT FROM THE CORRESPONDING CENTRES ARE EQUAL. THEOREM 7. CHORDS WHICH ARE EQUIDISTANT FROM THE CORRESPONDING CENTRES ARE EQUAL. THEOREM 8.  EQUAL CHORDS OF A CIRCLE ARE EQUIDISTANT F

FATHER OF ALGEBRA IMPORTANT FORMULAE OF ALGEBRA 1. (a+b)^2 = a^2 + b^2 + 2ab 2. (a-b)^2 = a^2 + b^2 - 2ab 3. (a+b)^2 = (a-b)^2 + 4ab 4. (a-b)^2 = (a+b)^2 - 4ab 5. a^2 - b^2 = (a-b)(a+b) 6. a^3 + b^3 = (a+b)(a^2 + b^2 - 2ab) 7. a^3 - b^3 = (a-b)(a^2 + b^2 + 2ab) 8. (a+b)^3 = a^3 + b^3 + 3ab(a+b) 9. (a-b)^3 = a^3 - b^3 - 3ab(a-b) 10. a^3 + b^3 = (a+b)^3 - 3ab(a+b) 11. a^3 - b^3 = (a-b)^3 + 3ab(a-b) 12. a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ac)                                         OR       a^3 + b^3 + c^3 - 3abc = (a+b+c)1/2{2(a^2) + 2(b^2) + 2(c^2) - 2ab -2bc -2ac}                                         OR       a^3 + b^3 + c^3 - 3abc = 1/2(a+b+c)[(a-b)^2 + (b-c)^2 + (c-a)^2] 13. IF a+b+c = 0, THEN a^3 + b^3 + c^3 = 3abc 14. (a+b+c)^3 = a^3 + b^3 + c^3 + 3(b+c)(c+a)(a+b) 15. a^2 + b^2 = (a+b)^2 - 2ab 16. a^2 + b^2 = (a-b)^2 + 2ab 17. (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab +2bc +2ca 18.  a^4 + b^4 +(a^2

SQUARE 1. PERIMETER = 4 * side = 4 * a  WHERE a = SIDE OF SQUARE 2. AREA = (side)^2 3. DIAGONAL = side * sqrt(2) 4. AREA OF THE PATH WHICH IS OUTSIDE OF SQUARE = 4 * (a + x) 5. AREA OF THE PATH WHICH IS INSIDE OF SQUARE = 4 * (a - x) WHERE x = WIDTH OF THE PATH TRAPEZIUM IT IS A QUADRILATERAL WHOSE OPPOSITE SIDES ARE PARALLEL. OTHER TWO OPPOSITE SIDES ARE OBLIQUE. 1. AREA = 1/2 * HEIGHT * (SUM OF PARALLEL SIDES)   HEIGHT IS THE DISTANCE BETWEEN THE TWO PARALLEL SIDES 2. MEDIAN = 1/2 * (SUM OF PARALLEL SIDES) MEDIAN IS THE SEGMENT JOINING THE MIDPOINTS OF OBLIQUE SIDES RHOMBUS IT IS PARALLELOGRAM WHOSE ALL SIDES ARE EQUAL.  ITS DIAGONALS BISECT EACH OTHER AT RIGHT ANGLE. 1. AREA = 1/2 * PRODUCT OF DIAGONALS 2. SIDE = sqrt {(D1/2)^2 + (D2/2)^2} 3. PERIMETER = 4 * SIDE WHERE D1 AND D2 ARE DIAGONALS

TRIANGLE 1. AREA = 1/2*BASE*HEIGHT   OR, √[s(s-a)(s-b)(s-c)] WHERE a, b, c  ARE THE LENGTHS OF THE SIDES OF TRIANGLE AND  s=(s+b+c)/2 2. AREA OF AN EQUILATERAL TRIANGLE =  √3/4 * (side)^2 3. AREA OF AN ISOSCELES TRIANGLE = b/4  √4(a^2)-(b^2) 4. PERIMETER OF AN EQUILATERAL TRIANGLE = 3 * SIDE RECTANGLE 1. AREA = LENGTH * BREADTH 2. PERIMETER = 2 (LENGTH + BREADTH) 3. DIAGONAL =  √(LENGTH)^2 + (BREADTH)^2 4. AREA OF PATH (OUTSIDE THE RECTANGLE) = 2 * (LENGTH + BREADTH + 2x) WHERE x= WIDTH OF THE PATH 5. AREA OF THE PATH INSIDE OF THE RECTANGLE = 2 * (LENGTH + BREADTH -2x) WHERE x= WIDTH OF THE PATH

IF SAME NUMBER IS THERE IN A SERIES LIKE 88888 OR 9999 UP TO AS MUCH AS YOU CAN.... IF YOU WANT TO FIND THEIR SUM LETS TAKE AN EXAMPLE 8888+888+88+8=? WE START FROM THE LEFT MULTIPLY 8 BY 1, 2, 3 & 4. U WILL GET YOUR ANSWER THOSE NUMBER WHOSE ALL DIGITS ARE 9 (99)^2=9801 (999)^2=998001 (9999)^2=99980001 (99999)^2=9999800001 THOSE NUMBER WHOSE ALL DIGITS ARE 3 (33)^2=1089                   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-1). WHERE n= NUMBER OF TIMES 3 IS REPEATED (333)^2=110889 (3333)^2=11108889 TO CHECK 401 IS A PRIME NUMBER APPROXIMATE SQUARE ROOT = 20 PRIME NO < 20 ARE 2, 3, 5, 7, 11, 13, 17, 19 401 IS NOT DIVISIBLE BY ANY OF THESE PRIME NUMBERS SO 401 IS A PRIME NUMBER.

175 X 157 = ? THE RESULT OF MULTIPLICATION OF THREE DIGIT NUMBER IS 175 X 157 = 27475 STEP 1:  MULTIPLY  (5 X 7) = 35 (NOTE DOWN 5 CARRY 3) STEP 2: THEN DO CROSS MULTIPLICATION (7 X 7 + 5 X 5 + 3 (ADD CARRY)) = 77 (NOTE DOWN 7 CARRY 7) STEP 3: AGAIN (1 X 7 + 1 X 5 + 7 X 5 + 7 (ADD CARRY)) = 54 (NOTE DOWN 4 CARRY 5) STEP 4: DO CROSS MULTIPLICATION AND ADD CARRY  (1 X 5 + 1 X 7 + 5 (ADD CARRY)) = 17 (NOTE DOWN 7 CARRY 1) STEP 5: AGAIN (1 X 1 + 1) = 2, NOTE IT DOWN. AND FINALLY THE RESULT WE GET 27475

LET ME XPLAIN THIS RULE BY TAKING SOME EXAMPLES.... CONSIDER NUMBER 44679387, WE HAVE TO CALCULATE THE REMINDER ON DIVIDING THIS NUMBER BY 7 11 AND 13 RESPECTIVELY... MAKE TRIPLETS AS WRITTEN BELOW STARTING FROM UNITS PLACE 44.........679.........387 NOW ALTERNATE SUM = 44+387 = 431 AND 679 AND DIFFERENCE OF THESE SUMS = 679-431= 248 DIVIDE IT BY 7 WE GET REMINDER AS 3 DIVIDE IT BY 11 WE GET REMINDER AS 6 DIVIDE IT BY 13 WE GET REMINDER AS 1

LET'S LEARN HOW TO FIND SQUARE ROOT BY TAKING DIFFERENT EXAMPLES.... EXAMPLE: FIND THE SQUARE ROOT OF 4489 WE GROUP THE LAST PAIR OF DIGITS, AND THE REST OF THE DIGITS TOGETHER. NOW, SINCE THE UNIT DIGIT OF 4489 IS 9. SO WE CAN SAY THAT UNIT DIGIT OF ITS SQUARE ROOT WILL BE EITHER 3 OR 7. NOW CONSIDER FIRST TWO DIGITS i.e. 44. SINCE 44 COMES BETWEEN THE SQUARES OF 6 ND 7 (i.e. 6^2 < 44< 7^2),SO WE CAN DEFINITELY SAY THAT THE TEN'S DIGIT OF THE SQUARE ROOT OF 4489 WILL BE 6. SO, FAR WE CAN SAY THAT THE SQUARE ROOT WILL BE EITHER 63 OR 67. NOW WE WILL FIND THE EXACT UNIT DIGIT. TO FIND THE EXACT UNIT DIGIT, WE CONSIDER THE TEN'S DIGIT i.e, 6 ND THE NEXT TERM i.e, 7. MULTIPLY THESE TWO TERMS SINCE, 44 IS GREATER THAN 42. SO SQUARE ROOT OF 4489 WILL BE THE BIGGER OF THE TWO OPTIONS i.e, 67.

LETS DISCUSS SOME BASIS IDEAS IN MATHEMATICS WHETHER YOU LOVE MATHS, OR DESPITE IT, THE FOLLOWING TRICKS CAN TURN YOU INTO A MATH WHIZ-OR AT LEAST WILL HELP TO SPEED UP SOME OF THE CALCULATIONS YOU NEED TO DO IN YOUR HEAD. I'VE INCLUDED FIVE OF MY FAVORITES HERE........ QUICK SQUARE IF YOU NEED TO SQUARE A 2 DIGIT NUMBER ENDING IN 5, MULTIPLY THE FIRST DIGIT BY ITSELF PLUS 1, AND PUT 25 ON THE END. THAT'S ALL! (25)^2=(2*(2+1))&25 2*3=6 & 25 625 MULTIPLYING BY 9 TO MULTIPLY ANY NUMBER BETWEEN 1 AND 9 BY 9, HOLD BOTH HANDS IN FRONT OF UR FACE, WITH FINGERS EXTENDED. NOW DROP THE FINGER THAT CORRESPONDS TO THE NUMBER YOU R MULTIPLYING (FOR EXAMPLE, FOR 9*3, DROP UR THIRD FINGER). NOW COUNT THE FINGERS BEFORE THE DROPPED FINGER(IN THE CASE OF 9*3 IT IS 2)-THAT'S UR FIRST DIGIT. THEN, COUNT THE FINGERS AFTER (AGAIN IN THE CASE, IT'S 7). THE ANSWER IS 27. DIVIDING BY 5 TO DIVIDE A LARGE NUMBER BY 5, ALL YOU NEED TO DO IS MULTIPLE

LE TS TAKE AN EXAMPLE OF ANY 2 DIGIT NUMBER Let us take 57 so, (57)^2 first of all square the unit place number i.e, 7 and its square is 49 Then square tens place number i.e, 5 and its square is 25 and then take 5 and 7 and multiply them 7*5=3 and  also multiply 35 by 2 i.e,35*2=70 now write in this format  25  70  49 now take 4 of 49 and add it to the left of 49 i.e, 70 but you have add in 0 i.e, 0+4=4 and now take 7 of 70 and add it to 25 i.e, 25+7=32 so,  the square is 3249.

1. TIME TAKEN TO CROSS A POLE OR A MAN = LENGTH OF A TRAIN  /  SPEED OF A TRAIN 2. TIME TAKEN TO CROSS A PLATFORM = LENGTH OF ( TRAIN + PLATFORM ) / SPEED OF TRAIN 3.   TIME TAKEN TO CROSS A BRIDGE = LENGTH OF (TRAIN + BRIDGE) / SPEED OF TRAIN 4. WHEN 2 TRAINS MOVE TOWARDS EACH OTHER THEN, TIME TAKEN TO CROSS EACH OTHER = SUM OF LENGHTHS OF BOTH TRAINS / SUM OF THEIR SPEEDS 5. WHEN 2 TRAINS ARE MOVING IN THE SAME DIRECTION THEN, TIME TAKEN TO CROSS EACH OTHER = SUM OF LENGTHS OF BOTH TRAINS / DIFFERENCE OF THEIR SPEEDS 6. WHEN 2 TRAINS ARE MOVING AT SPEED OF x km/hr AND y km/hr  (i) RELATIVE SPEED WHEN MOVING TOWARDS EACH OTHER=(x+y) km/hr (ii) RELATIVE SPEED WHEN MOVING IN THE SAME DIRECTION=(x-y) km/hr WHERE x>y