Well Bhavin is absolutely correct! These are basics. sqrt(25) is only and only +5 and nver... -5. In the question we are taking x to ensure that it stays +ve, irrespective of what x we choose. When u r asked what is the value of x, given that x^2 = 25, then x=+5, and -5 are correct. There is no convention and all. And not to forget that sqrt of a negative number is not defined at all.
i disagree to your statement in bold i know these are basics but bro what do you mean when you say that square root of 25 is only +5 i compeltely disagree. what will happen to bechaaara -5 is bechaare ka kya kasoor hai
when you take square of -5 you write it as 25 but you say when you will take square root you will forget about -5 . you are doing wrong with poor bechaara -5 Please consider again. I guess -5 also has the right to be called the square root of 25
i disagree to your atatement in bold i know these are basics but bro what do you mean when you say that square root of 25 is only +5 i compeltely disagree. what will happen to bechaaara -5 is bechaare ka kya kasoor hai
when you take square of -5 you write it as 25 but you say when you will take square root you will forget about -5 . you are doing wrong with poor bechaara -5 Please consider again. I guess -5 also has the right to be called the square root of 25
Well u r mistaking in formulating the question. finding sqrt(25) is not same as finding all those x for which x^2 = 25. Well I cannot reply any more to this discussion. If u don agree, u have every right to disagree. Probably u surf a bit and make ur own decision.
Well u r mistaking in formulating the question. finding sqrt(25) is not same as finding all those x for which x^2 = 25. Well I cannot reply any more to this discussion. If u don agree, u have every right to disagree. Probably u surf a bit and make ur own decision.
well i again disagree with your statement in bold
can you tell me the diff in the statement i mean you have written 2 ways of doing the same finding square root of 25 what is the diff? can you please explain? if you dont wanna post you can PM me
Well Bhavin is absolutely correct! These are basics. sqrt(25) is only and only +5 and nver... -5. In the question we are taking x to ensure that it stays +ve, irrespective of what x we choose. When u r asked what is the value of x, given that x^2 = 25, then x=+5, and -5 are correct. There is no convention and all. And not to forget that sqrt of a negative number is not defined at all.
i disagree to your atatement in bold i know these are basics but bro what do you mean when you say that square root of 25 is only +5 i compeltely disagree. what will happen to bechaaara -5 is bechaare ka kya kasoor hai
when you take square of -5 you write it as 25 but you say when you will take square root you will forget about -5 . you are doing wrong with poor bechaara -5 Please consider again. I guess -5 also has the right to be called the square root of 25
hey bro...jha is correct....sqrt of non negative integer is never negative..
u can also look at this way... A implies B does not mean B implies A. i.e square of -5 implies 25....does not mean root of 25 implies -5...
hi sillyfool, like i already posted long back in this thread, we do have to consider the positive as well as the negative root of a positive integer. In fact that's a common GMAT trap.There were some questions that checked just this fact and as u can guess, even those books(good ones like Kaplan 800) highlighted the fact that there are 2 roots and we must consider them.
So 25 has 2 roots:+5 and -5(like u already know) :) You can take my word for it.
ok sillyfool let me try to clear it if the Qn says x^2=25 then x= +/- srt25 which implies x=+/-5 but if the Qn is x=srt25 then x=5 you must understand here that the +/- comes when you shift a square across the equality and is in no way associated with Square root as such. this is a mathematical convention just as 0!=1 u cannot deny this convention. Its a mathematical law dude. hope it helps..........
and should i jump in too let x=square root of 25 so that x2-25=0 ....(1) taking f(x)=x2-25 , newton raphson method gives xn+1 = xn - [ f(xn) /f '(xn) ] =xn - [ x2n - 25 /2xn ] = [ xn + (25 / xn ) ] .....(2) now since f(4) = -9 , f(6) = 11 , and also since f(-4) = -9 and f(-6) = 11 a root of (1) lies between 4 and 6 and -4 and -6 Case 1 therefore , taking x0 = 4.5 , equ (2) gives x1= [ x0 + (25 /x0 ) ] = [ 4.5 +(25/4.5)] = 5.0277778 x2= [ x1 + (25 /x1 ) ] = [5.0277778 +(25/5.0277778 )] = 5.0000767 x3= [ x2 + (25 /x2 ) ] = [5.0000767 +(25/5.0000767 )] = 5 x4= [ x3 + (25 /x3 ) ] = [5 +(25/5 )] = 5 since x3=x4 we take square root of 25 = 5 Case 2 therefore , taking x0 = -4.5 , equ (2) gives x1= [ x0 + (25 /x0 ) ] = [ -4.5 +(25/-4.5)] = -5.0277778 x2= [ x1 + (25 /x1 ) ] = [-5.0277778 +(25/-5.0277778 )] = -5.0000767 x3= [ x2 + (25 /x2 ) ] = [5.0000767 +(25/5.0000767 )] = -5 x4= [ x3 + (25 /x3 ) ] = [-5 +(25/-5 )] = -5 since x3=x4 we take square root of 25 = -5
Hence there are 2 cases -5 and 5 . Thus, every positive number has two square roots, one positive and the other negative.Hope Helps !! Silly,ab,eric lets stop the love story here, now , in GMAT its important to refer to both the roots. So this is very clear .
Can anyone tell me if thr's a formula for this kind of ques
Q1.The sum of 2 positive numbers is 1215. HCF is 81. How many pairs of such integers are possible? a. 4 b. 6 c. 7 d. 8 e. 14
If HCF and sums appear in such problems , its good to rewrite expression A + B = 1215 as 81m + 81n = 1215 ( where m and n are positive intergers but dont have any common factors as we are taking into HCF here by rewriting A = 81m and B = 81n)
so 81(m + n) = 1215 m + n = 15 now it becomes fairly simple to calculate how many such pair of m and n exist.
14, 1 = 15 13,2 = 15 12,3 = reject --- as 3 is a common factor here 11, 4 = 15 10,5 = reject --- same reason as 5 is common factor here 9,6 = reject as 3 and 2 are common factors. 8,7 = 15
If HCF and sums appear in such problems , its good to rewrite expression A + B = 1215 as 81m + 81n = 1215 ( where m and n are positive intergers but dont have any common factors as we are taking into HCF here by rewriting A = 81m and B = 81n)
so 81(m + n) = 1215 m + n = 15 now it becomes fairly simple to calculate how many such pair of m and n exist.
14, 1 = 15 13,2 = 15 12,3 = reject --- as 3 is a common factor here 11, 4 = 15 10,5 = reject --- same reason as 5 is common factor here 9,6 = reject as 3 and 2 are common factors. 8,7 = 15
So we come down to 4 such pairs..
I did follow the same technique but after getting X+Y=15, was not able to get down to '4' as the ans coz my confusion here is why are we rejecting the 3 pairs ? 😞
I did follow the same technique but after getting X+Y=15, was not able to get down to '4' as the ans coz my confusion here is why are we rejecting the 3 pairs ? :-(
The reason to reject these 3 pairs is they have a common factor. If we consider these pairs HCF will not be 81 but something else such as if 5 is common factor HCF will be 81*5.
So these pairs will make our factual information(HCF is 81) fall apart. I hope this explaination helps.
The reason to reject these 3 pairs is they have a common factor. If we consider these pairs HCF will not be 81 but something else such as if 5 is common factor HCF will be 81*5. So these pairs will make our factual information(HCF is 81) fall apart. I hope this explaination helps.
1. sq.rt4 + cu.rt4 + fourthroot of 4 = ? a) less than 3 b) equal to 3 c) between 3 & 4 d) equal to 4 e) greater than 4
2. A city with a population of 132000 is divided into 11 voting districts in such a way that no district is to have a population more than 10% greater than the population of any other district. What is the minimum possible population that the least populated district could have ? a) 10700 b)10800 c) 10900 d) 11000 e) 11100
1. sq.rt4 + cu.rt4 + fourthroot of 4 = ? a) less than 3 b) equal to 3 c) between 3 & 4 d) equal to 4 e) greater than 4
2. A city with a population of 132000 is divided into 11 voting districts in such a way that no district is to have a population more than 10% greater than the population of any other district. What is the minimum possible population that the least populated district could have ? a) 10700 b)10800 c) 10900 d) 11000 e) 11100
Happy solving !!!
Hi kingCat,
1. answer is e) greater than 4 .. taking a log of function=> x = sqr. rt 4 + cub. rt 4 + fourth rt. 4 ---> log (base 2) X = 2.13 hence X will always be greater than 2^2 i.e 4
2. answer is d) 11000
to clarify, lets assume 10900 is the least, than the max cannot be greater than 11990. Now, if even all the other 10 district have population of 11990 each. total population of all the 11 district will be: 11990*10 + 10900 = 130800. but given population is 132000. hence wrong
Now take d) 11000. max population can be 12100. if we proceed like above, tota population will exeed the given 1320000, hence can be a probable answer. now, if we consider that all the ten district have least pop. i.e 11000, and 1 dist. has max pop i.e 12100, the total population of all the 11 districts will be 122100. hence actual populatiuon is coming in between the range of min and max population as per option d)
