hi,
pls. solve this
log0.3125 to the base 2 is...
hi,
pls. solve this
log0.3125 to the base 2 is...
my method.... but i get a feeling that there will be a better method to do this but anyway....(it may be an option oriented question)
log0.3125 to the base 2 = log0.3125/ log 2
=(log 3.125 - log 10) / log 2
now we know the value of log 3 =0.477 so we can assume log 3.125= 0.49
= (.49 - 1)/.301
=-0.51/.301
=-1.7 (approx)
pls someone suggest a better method.....
hi,
pls. solve this
log0.3125 to the base 2 is...
(log10^-4 + log5^4) to the base 2
4log5/log2 - 4
4( (log5 - log2)/log2)
4*(0.69 - 0.30)/0.3010)
4*(0.39/0.3)
4*1.3 = 5.2
i dont understand the part in bold letter please help :|
the if you form a subset of integers chosen from between 1 to 3000,such that no 2 integers add up to a multiple of 9, what can be the max no of elements in the subset?
2,3,4,8,9,11,12,13,17,...2990,2991,2992,2996,2999
see,it follows a pattern
(x-7),(x-6),(x-5),(x-1) here,x is 9 or multiple of 9.
from 9*0 to 9*332=>333*4=1332
9 comes at the start so,1 more
last number is 2999
1334 terms total.dunno where 1 term goes :|
no of even nos=2*(multiple of 4). last post edited. sorry for the mistakes..
ne ways the logic is as follows: whenever u have 4 or a multiple of 4 when u reduce by 2 u will get an even.but in this case either n or (n+1) can be a multiple of 4. for eg: n, n+1 can be (3,4) or (4,5). so we will need to multipy 44/4 with 2 that gives 22 even nos.
So total no of odd nos. will be 44-22=22.
odd nos r always the difference between consecutive primes. so 22 is the answer... hope am clear.
someone plz explain how traingular no's are 44................i m confused very much
in a hurry im solved 2 qns:
answer for the first:
only when the two digits have a 1 or 0 as one of the digits then sum>product of factorial of nos. also if both digits r same then sum=product. both cases can be subtracted from 90 (no of 2 digit nos.)
1st case: 17+9=26
2nd case: only 1. so total=27. so total nos: 90-27=63. ans: c
answer for third:
sum of n consecutive nos= n(n+1)
total nos of triangular nos= 44
now the diff between two consecutive squares is always an odd.
here no of even nos is 2*(a multiple of 4) since the multiple can be n or n+1.
so total nos= 44*2/4=22
so the ans= option 'c'.
hope the answrs r correct...
plz someone let me know how triangular no's r 44.........
Hi puys,
Please help me out with this prob --> find inverse of f(t) = (10^t-10^-t)/(10^t + 10^-t) is
(A) 1/2log{(1-t)/(1+t)}
(B)0.5log{(t-1)/(t+1)}
(C)1/2log(base10){(2^t-1)}
(D)None of the above.
I am getting answer as 0.5log{-(1+t)/t-1)} which is not in the option.please tell where i am doing the mistake.it is a ques no16 of functions LOD 3
What is the answer.I guess even i am making some mistake
ashwini2tyagi Saysplz someone let me know how triangular no's r 44.........
@ashwini2tyagi:
since n(n+1)/2 or n(n+1)n^2+n-2000
n is an integer..which means in n=/2,the largest number in square root must be 89^2 or 7921...44
somebody please clarify
kano_kyosuke SaysWhat is the answer.I guess even i am making some mistake
apply componendo and dividendo.
10^t/10^-t = 1+k/1-k
t = 1/2 log(1+k/1-k)
f inverse = 0.5log(1+t/1-t)
but i guess she says that is not the answer..even i got the same
kano_kyosuke Saysbut i guess she says that is not the answer..even i got the same
Some answers of arun sharma are wrong.so don't worry about it.:cheerio:
kano_kyosuke SaysWhat is the answer.I guess even i am making some mistake
Is the ans 2
amitprakash SaysSome answers of arun sharma are wrong.so don't worry about it.:cheerio:
cool
KAC8L591 SaysIs the ans 2
the answer that is got is 1/2log((1+t)/(1-t))
find the remainder when 51^203 is divided by 7.
a.4 b.2 c.1 d.6
find the remainder when 51^203 is divided by 7.
a.4 b.2 c.1 d.6
51^203 = 2^203 = 4*2^201 = 4*8^67 = 4*1 = 4
find the remainder when 51^203 is divided by 7.
a.4 b.2 c.1 d.6
Solution:
Re (51^203) = Re ((51^4)^50) X Re (51^3)
= 2 X 1
= 2
2 shud be the ans..correct me if i m wrong..
find the remainder when 51^203 is divided by 7.
a.4 b.2 c.1 d.6
Solution:
Re (51^203) = Re ((51^4)^50) X Re (51^3)
= 2 X 1
= 2
2 shud be the ans..correct me if i m wrong..
51^203 mod 7
2^203 mod 7
2^5 mod 7
32 mod 7
answer is 4.
in your solution,the first part,it is
(51^4)^50
(2^4)^50
2^50 mod 7
2^2 mod 7
answer is 4.
hope that helps.
Hi, i a newbie on this site and have just started preparing for MAT, wanted to know if the arun sharma book for DO and LR will be as per MAT question stndards or is it much more advanced?
Please do help
HEY plz. sm1 help....
q no.93 lod-2
how many integer values of x and yare there such that 4x+7y=3while x,ya.144 b.141 c.143 d.142
Ans is 143
eq can b written
y=(3-4x)7
=(7-4(1+x))/7
=1-4/7*(1+x)
so x takes values frm -498 to 496