Posts

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
Post a Comment
Read more

QUESTION FOR SSC , UPSC , CDS AND ALL GENERAL COMPETITIONS

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
Post a Comment
Read more

STATISTICS FORMULAES FOR FINDING MEAN, MODE AND MEDIAN

Image
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
4 comments
Read more

TRICKY QUESTION

Image
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
Post a Comment
Read more

MIND FREAKING TRICKS

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
Post a Comment
Read more

SSC QUESTION

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
Post a Comment
Read more

COST PRICE AND SELLING PRICE QUESTION

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%
Post a Comment
Read more