@rkshtsurana said:bhai approach bata...
32.1^1/5 is a little more than 2... 1.022^-1/4 is a little less than 1..so closest option 1.005 hai.. maine to aise he kia.. what's d ans?
@gs4890 said:izz it 1.455 ? approx
1.005 he yaar...IIFT approx puchte hi nhi..kese krnge isse. CAT hi achaa tha
@ScareCrow28 - ans sahi he tera..
@sonamaries7 said:A no N expressed to the base 5 is 232323...a total of 100 digits. What is the rem when N^4231 is divided by 4?0124
3+5*2+5^2*3+5^3*2+5^4*3+-------------------+5^99*2
Remainder is 3+2+3+2+-------------------------------100 terms
5*50=250/4
remainder is 2
@gnehagarg said:3+5*2+5^2*3+5^3*2+5^4*3+-------------------+5^99*2Remainder is 3+2+3+2+-------------------------------100 terms5*50=250/4remainder is 2
0 OA
@sujamait said:An ant starts from a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Find the probability that ant moves to the starting vertex on its tenth move.
total 10 moves, After 10 moves it has to be at starting vertex again
required ways =>
5 clockwise moves, 5 anticlockwise => 10!/5!*5! = 252
8 clock wise, 2 anticlockwise => 10!/8!*2! = 45
2 clockwise, 8 anticlockwise => 10!/2!*8! = 45
=> 342 ways
total = 2^10 = 1024 ways (as for each move ant has 2 alternatives)
prob = 342/1024 = 171/512
@rkshtsurana said:find the value of (32.1)^1/5 - ( 1.022)^-1/4options0.9851.3251.4551.005p.s - Time iift mocks @gs4890
(32+.1)^1/5-(1+.022)^-1/4
2(1+.1/32)^1/5-(1+.022)^-1/4
2(1+.1/32*5)-(1-.022/4)
2+1/800-1+11/2000
1+3/1000
1.003
which is close to 1.005....hence option4..
edit:typo
@pankaj1988 said:(32+.1)^1/5-(1+.022)^-1/42(1+.1/32)^1/5-(1-.022/4)^-1/42(1+.1/32*5)-(1+.022/4*4)2+1/800-1+11/20001+3/10001.003which is close to 1.005....hence option4
best approach 
@sonamaries7 said:A no N expressed to the base 5 is 232323...a total of 100 digits. What is the rem when N^4231 is divided by 4?0124
sum of digits is 500 so its gives remainder of 0..be it raised to any power
a boy has 80 choc with him,he plays a gmae so that he eats every second choc start from 1,what is the original value of the choc that he eats at the last
77,31,74,33,71
@Ashmukh said:a boy has 80 choc with him,he plays a gmae so that he eats every second choc start from 1,what is the original value of the choc that he eats at the last77,31,74,33,71
ques is not framed properly. can u elaborate ?
@shanu18 said:Ramesh sells rasgulla at Rs.15 per kg.A rasgulla is made up of flour and sugar in the ratio 5:3.The ratio of price of sugar and flour is 7:3 per kg.Thus he earns 66 2/3 profit.what is the cost price of sugar ?1)10/kg2)9/kg3)18/kg4)14/kg
this problem is a direct application of alligation:..since sp=15,profit=66.66%, so cp=9
sugar:...................flour
7x.........................3x
..........9
9-3x...................7x-9
9-3x/7x-9=3/5=>x=2
hence cost of sugar=7x=7*2=14
@rkshtsurana a boy has 80 choc,he decides to play a game with his choc .he keeps all his choc on table and takes out every second choc and eats it.he starts from the first choc and continues till only last choc is left,what was the original position of last choc left
@Ashmukh said:@rkshtsurana a boy has 80 choc,he decides to play a game with his choc .he keeps all his choc on table and takes out every second choc and eats it.he starts from the first choc and continues till only last choc is left,what was the original position of last choc left
I am not able to comprehend the question..last choc left will be 1 na... ??
@Ashmukh said:@rkshtsurana a boy has 80 choc,he decides to play a game with his choc .he keeps all his choc on table and takes out every second choc and eats it.he starts from the first choc and continues till only last choc is left,what was the original position of last choc left
2^6 + 16 = 80
2(16) +1 = 33rd ?
@ScareCrow28 said:I am not able to comprehend the question..last choc left will be 1 na... ??
iske kehna ka matlab yeh he...har doosri choclate uthayi usne..jese 1 then 3 thn 5 thn 7...79 , 2 and so on
