Perfect answers montaque !
But coild u pls clarify how u got 2.13 after taking log function ?
In the second question, I didnt get the logic 132000 to be in between the min and max.
KingCat.
Perfect answers montaque !
But coild u pls clarify how u got 2.13 after taking log function ?
In the second question, I didnt get the logic 132000 to be in between the min and max.
KingCat.
here it goes...
1. say X = sq. rt 4 + cub. rt 4 + furth rt. 4
=> Log X = 1/2 LOG 4 + 1/3 log 4 + 1/4 log 4 (here 'log' implies 'log(base 2)')
=> log X = 1/2 log 2^2 + 1/3 log 2^2 + 1/4 log 2^2
=> log X = (2* 1/2)log2 + (2* 1/3)log2 + (2* 1/4)log2 (now, since log 2 = 1)
=> log X = 1 + 2/3 + 1/2
=> Log X= 13/6
=> log X = 2.16 ....( we know that: log 4 (log 2^2) = 2)
therefore X has to be greater than 4 for its LOG to be greater than 2 (in this case it is 2.16)
2. this is quite complicated to explain...but still I will try :smile:
lets analyse option (c) i.e min population is 10900. it means max population that any other district can have is not greater than 11990 (add 10% to the original). And, we are given that total population is 132000. So lets check the upper boundary condition that: all other 10 districts have "max possible pop." i.e 11990 and one district has least i.e 10900. note that total of these population should atleast match the total given population i.e 132000. otherwise, where would the remaining people would go if not in these districts
. And, also if total of the various district adds up to more than the given total population, we can safely say that some districts might have lesser population than the assumed max population (and can remain in 10% range). Now, coming back to the option (C), total of the assumed max population is 130800 - which falls short of total given population. now, what will happen to remaining 1200 people in the city? obviously this can not happen.Now, by applying the same logic for option (D), Max assumed population is 132000 which means some district have lesser number of people to match the given total pop. ; and Min assumed population is 122100 - which means some district can have greater number of people (than the min possible popualtion) to match the total given pop.
Therefore 11000 "could" be a minimum possible population
hope it will clarify ur doubts..
Perfect answers montaque !
But coild u pls clarify how u got 2.13 after taking log function ?
KingCat.
here it goes...
1. say X = sq. rt 4 + cub. rt 4 + furth rt. 4
=> Log X = 1/2 LOG 4 + 1/3 log 4 + 1/4 log 4 (here 'log' implies 'log(base 2)')
=> log X = 1/2 log 2^2 + 1/3 log 2^2 + 1/4 log 2^2
=> log X = (2* 1/2)log2 + (2* 1/3)log2 + (2* 1/4)log2 (now, since log 2 = 1)
=> log X = 1 + 2/3 + 1/2
=> Log X= 13/6
=> log X = 2.16 ....( we know that: log 4 (log 2^2) = 2)
therefore X has to be greater than 4 for its LOG to be greater than 2 (in this case it is 2.16)
Hey montaque...ans is perfect...it is greater than 4...
i'll make a small correction..
statement in bold is incorrect...lucky that u got away in this question using the above formula..
taking log on both sides implies
log x = log (sq. rt 4 + cub. rt 4 + furth rt. 4) and this is not equal to log sq. rt 4 + log cub. rt 4 + log furth rt. 4
i.e log (a+b+c) is not equal to log a + log b + log c...
u can solve this question even without using logs...
just approximate..
sq root of 4 is 2 and fourth root of 4 is sq root of 2 i.e 1.414
hence sq root of 4 + fourth root of 4 is 2 + 1.4 = 3.4
sq root of 4 > cube root of 4 > fourth root of 4 .
2> cube root of 4 > 1.4
add cube root of 4 to 3.4 i.e sum is between 4.8 and 5.4 i.e always above 4...ans E..
here it goes...
1. say X = sq. rt 4 + cub. rt 4 + furth rt. 4
=> Log X = 1/2 LOG 4 + 1/3 log 4 + 1/4 log 4 (here 'log' implies 'log(base 2)')
=> log X = 1/2 log 2^2 + 1/3 log 2^2 + 1/4 log 2^2
=> log X = (2* 1/2)log2 + (2* 1/3)log2 + (2* 1/4)log2 (now, since log 2 = 1)
=> log X = 1 + 2/3 + 1/2
=> Log X= 13/6
=> log X = 2.16 ....( we know that: log 4 (log 2^2) = 2)
therefore X has to be greater than 4 for its LOG to be greater than 2 (in this case it is 2.16)
Hey montaque...ans is perfect...it is greater than 4...
i'll make a small correction..
statement in bold is incorrect...lucky that u got away in this question using the above formula..
taking log on both sides implies
log x = log (sq. rt 4 + cub. rt 4 + furth rt. 4) and this is not equal to log sq. rt 4 + log cub. rt 4 + log furth rt. 4
i.e log (a+b+c) is not equal to log a + log b + log c...
u can solve this question even without using logs...
just approximate..
sq root of 4 is 2 and fourth root of 4 is sq root of 2 i.e 1.414
hence sq root of 4 + fourth root of 4 is 2 + 1.4 = 3.4
sq root of 4 > cube root of 4 > fourth root of 4 .
2> cube root of 4 > 1.4
add cube root of 4 to 3.4 i.e sum is between 4.8 and 5.4 i.e always above 4...ans E..
Thats absolutely correct Bhavin . Luckily Montagues's answer coincided with the correct answer but derivation of the solution is not correct.
We can rewrite the eqn as below :
sq rt 4 + cube rt 4 + frth rt 4 = 2 + cube rt 4 + 4th rt 4 > 2 + 1 + 1.
simply becoz cube rt 4 > cub rt 1 => cube rt 4 > 1.
similarly frth rt 4 > frth rt 1 => frth rt 4 > 1.
Infact getting to calculating logs was not indeed necessary ,however , this is just a simpler approach to solve such problems
Thats absolutely correct Bhavin . Luckily Montagues's answer coincided with the correct answer but derivation of the solution is not correct.
We can rewrite the eqn as below :
sq rt 4 + cube rt 4 + frth rt 4 = 2 + cube rt 4 + 4th rt 4 > 2 + 1 + 1.
simply becoz cube rt 4 > cub rt 1 => cube rt 4 > 1.
similarly frth rt 4 > frth rt 1 => frth rt 4 > 1.
Infact getting to calculating logs was not indeed necessary ,however , this is just a simpler approach to solve such problems
oh..what a blunder!...thanks a lot guys for correcting me
Perfect answers montaque !
But coild u pls clarify how u got 2.13 after taking log function ?
In the second question, I didnt get the logic 132000 to be in between the min and max.
KingCat.
I solved these using following fundas:
1) X = sq rt 4 + cb rt 4 + 4th rt 4
X = 2 + 1.something + 1.somthing
=> X > 4
2) for the least possible population in one of the state, the only case possible is one of the state has least population and remaining all 10 have extra 10%. Assuming the least population is X
we have
13200 = X + (1+10/100)X*10
This gives x= 11000 directly
HG
Thanks ghitesh, bhavin, rohan and montaques.......
I was having the same problem with the log and did actually believe that there should be a simpler way to tackle the second question.
Great stuff guys !
Apart from approximating it, is there any way to find the answer for the first question ? instead of fourth rt, if they pop a question with 5th or 6th root, what approach should be used ?
Or is that too much for GMAT ?
Thanks once again.
Thanks ghitesh, bhavin, rohan and montaques.......
I was having the same problem with the log and did actually believe that there should be a simpler way to tackle the second question.
Great stuff guys !
Apart from approximating it, is there any way to find the answer for the first question ? instead of fourth rt, if they pop a question with 5th or 6th root, what approach should be used ?
Or is that too much for GMAT ?
Thanks once again.
Answer options in GMAT speak a lot...degree of variations in answer option gives u a hint what degree of approximations can be used...
GMAT is all about saving time and improving accuracy..
for real nos greater than 1, nth root of a no is always greater than 1...
eg 6th root is cube root of square root of no...so u can always approximate to a close value...
A certain clock rings two notes at quarter past the hour, four notes at half past, and six notes at three-quarters past.On the hour it rings eight notes plus an additional number of notes equal to whatever hour it is.How many notes will the clock ring between 1:00 pm and 5:00 pm, including the rings at 1:00 pm and 5:00 pm?
A 87
B 95
C 102
D 103
E 115
Here's a nice one from kaplan 800:
A certain clock rings two notes at quarter past the hour, four notes at half past, and six notes at three-quarters past.On the hour it rings eight notes plus an additional number of notes equal to whatever hour it is.How many notes will the clock ring between 1:00 pm and 5:00 pm, including the rings at 1:00 pm and 5:00 pm?
A 87
B 95
C 102
D 103
E 115
My Answer D) 103
(2+4+6)*4 + 8*5 + (1+2+3+4+5)
HG
thats rite.try this one:
abchekstylo Saysthats rite.try this one:
the answer is none
(as y=30)
let me know if m wrong
guys try this one out
Let f be a factor of 120, then the number of positive integral solutions of xyz = f is
(a) 160 (b) 240 (c) 320 (d) 480 (e) none of the options
guys try this one out
Let f be a factor of 120, then the number of positive integral solutions of xyz = f is
(a) 160 (b) 240 (c) 320 (d) 480 (e) none of the options
will post the answer with solution after 8 hrs
the answer is none
(as y=30)
let me know if m wrong
I believe B) as y=60
450=2*(3^2)*(5^2)
Therefore y=(2^2)*3*5*(N^3) (Where n is any interger)
HG
the answer is none
(as y=30)
let me know if m wrong
answer is B.
the answer is none
(as y=30)
let me know if m wrong
Eric,
The answer is option B.
The smallest value of y for 450y to be a perfect cube is 2^2 x 3 x 5 = 60.
Hence y/(3*2*2*5) is an integer.
Last month 15 homes were sold in town x. The avg sale price of homes was 150,000 Rs and the median sale price was 130,000 rs . Which of following are true?
1.Atleast one of the homes was sold for more than 165,000 rs
2.Atleast one of the homes was sold for more than 130,000 and less than 150,000
3. Atleast one of the homes was sold for less than 130,000
a. 1 only
b. 2 only
c. 3 only
d. 1 and 2
e. 1 and 3
Neo
Last month 15 homes were sold in town x. The avg sale price of homes was 150,000 Rs and the median sale price was 130,000 rs . Which of following are true?
1.Atleast one of the homes was sold for more than 165,000 rs
2.Atleast one of the homes was sold for more than 130,000 and less than 150,000
3. Atleast one of the homes was sold for less than 130,000
a. 1 only
b. 2 only
c. 3 only
d. 1 and 2
e. 1 and 3
Neo
1 only...2 and 3 may not be true...
hey there was a lot of discussion on square root on previous page I would like to clear something square root of a no. is always positive by definition:
let us say we have to find 4^(1/2)
now let us say x = 4^(1/2)------(1)
so how many roots above above eqn. has?
it is polynomial eqn with degree 1 so max. no roots possible(real +nonreal)
can be one and by definition we take it as +2 if no initial condition is metioned
so what about X^2=4 ? -------------------(2)
well solution is +2 and -2 because this eqn is of secomd degree
so u see (1) and (2) r not same but identical so we see whenever u r solving equation by doing various operations the the coming eqn. would be identical to first eqn not same so when u finally arive at roots, actual root would be a subset of the set containing these roots:
but it is confusing is it not? to remove confusion follow following steps:
solve the equation by procedure u know and then check the validity of each root through the first equation( which must include initial condition if any) that would do in most of cases( I believe in all cases)
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