dear puys... wanted some suggestion(s)... i 'm thru with the OG-11 prob solving section..with >86% accuracy( my weak areas would be - TSD& probs on mixtures-ratios and prob.) to tell you prob solving section is my weak area... so my question is 1. is the level of gmat questions same as given in OG 2. what extra material (from my weak area pt of view) to refer that would be pertinent to the level of gmat prob solving 3. i'm referring to OG and Kaplan only (for now) and would consider Manhattan question bank too...so would it suffice my prep... regards Jakhar P.S.- i understand that it is not a relevant thread to ask such a question but i'm of the opinion tht a diverse view on the questions above would help me ...but my sincere apology to those who felt otherwise
puys wat ll be the curve of f(x) = x/lxl
how ll it look in the graph
hi there fellow puy,
the key to draw the graph of y/f(x)=x/lxl
, is to understand how the function lxl behaves....
as we know mo of a positive no is the number itself, for eg- l2l=2, l5l=5
but with -ve integers the function behaves like this...l-2l= -(-2)=2...i.e. mod of a negative integer is the = -( the no.).
so now talking about the functionj y=x/lxl
take different values of x you'll get diff values of y...
x=1 ; y=1
x=2;y=2
x=3;y=3
x=-1; y= 1/-(-1)=1
x=-2; y=2
x=-3;y=3
plot the values mate...you'll get the desired graph....hope it clarifies your doubt..
regards
Jakhar
puys wat ll be the curve of f(x) = x/lxl
how ll it look in the graph
@ muru_chn,
he graph would look like
regards
Jakhar
puys wat ll be the curve of f(x) = x/lxl
how ll it look in the graph
x=1 f(x)=x/x|=1=y
x=2 f(x)=x/x=1=y
x=3 f(x)=x/x=1=y
x=-1 f(x)=x/x=-1=y
x=-2 f(x)=x/x=-1=y
x=-3 f(x)=x/x=-1=y
The curve should be a line y=1 in I quadrant and y=-1 in III quadrant.
I am not sure but it should not be touching y axis...coz touching y axis will make f(x)=0/|0
a set of 15 different integers has a median of 25 and a range of 25.what is the greatest possible integer that can be in the set?
Is the answer 38?
rd880 Saysa set of 15 different integers has a median of 25 and a range of 25.what is the greatest possible integer that can be in the set?
=== >> 43
hw come 43?
vibhu.guptta Sayshw come 43?
Median is 25...this means the middle number is 25...
There are 15 number...7 lie before 25...7 lie after 25...
Here's the series : 18 19 20 21 22 23 24 ( 7 numbers )
Next Number : 25
Greatest possible : First Number + Range : 18+25=43
Hope this helps...
dont know where my mind was when i posted 38. the thot process was same, a minor carelessness cost me dearly..but nevr mind...m back on the circuit again.. thanks again..
How many different positive integers exist between 10^6 and 10^7, the sum of whose digits is equal to 2?
- 6
- 7
- 5
- 8
- 18
There Are 7 they are 1---------1 to 110000 and 200000 since there are 7 digits after the one can take any of these position and the 2*10^6
Median is 25...this means the middle number is 25...
There are 15 number...7 lie before 25...7 lie after 25...
Here's the series : 18 19 20 21 22 23 24 ( 7 numbers )
Next Number : 25
Greatest possible : First Number + Range : 18+25=43
Hope this helps...
yes that is correct.
But can some one explain abt the case where the series is some thing like 16, 16, 18, 20, 21,22,25 ...
if n and y are positive integers such that 450y = n^3
which of the following must be an integer
1 y/(3 *2*2*5)
2 y/(3*3*2*5)
3 y/(3*2*5*5)
a none
b 1 only
c 2 only
d 3 only
e 1 2 and 3
if n and y are positive integers such that 450y = n^3
which of the following must be an integer
1 y/(3 *2*2*5)
2 y/(3*3*2*5)
3 y/(3*2*5*5)
a none
b 1 only
c 2 only
d 3 only
e 1 2 and 3
450y=n^3
450y=(y/x)^3
450=y^2/x^3
3^2*5^2*2*x^3=y^2
that means 2*x^3 must be perfect square
None satisfy these condition so answer is none
How many different positive integers exist between 10^6 and 10^7, the sum of whose digits is equal to 2?
- 6
- 7
- 5
- 8
- 18
the answe is 7.
we sustitute 1 for 0 in each of the six cases in the numger 1000000 ie 1100000, 1010000, 1001000, 1000100, 1000010, 1000001 and one more case wd be 2000000. so totally 7 cases.
if n and y are positive integers such that 450y = n^3
which of the following must be an integer
1 y/(3 *2*2*5)
2 y/(3*3*2*5)
3 y/(3*2*5*5)
a none
b 1 only
c 2 only
d 3 only
e 1 2 and 3
the answer is B. coz for 450y to be a perfect cube it has to be a multiple of 2*2*3*5 else n will not be a natural number. it wll become a fraction. note here that only those multiples of 2*2*3*5 are acceptable whic are perfect cubes again like 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197 and so on. The anwer is a case whre the multiple is 1.
Please look at what i posted
eric.segal1 Saysthe answer is B. coz for 450y to be a perfect cube it has to be a multiple of 2*2*3*5 else n will not be a natural number. it wll become a fraction. note here that only those multiples of 2*2*3*5 are acceptable whic are perfect cubes again like 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197 and so on. The anwer is a case whre the multiple is 1.
yes that is correct.
But can some one explain abt the case where the series is some thing like 16, 16, 18, 20, 21,22,25 ...
That will be a different question...here you can't have such a series "16, 16, 18, 20, 21,22,25" coz the question states "15 distinct numbers"
if n and y are positive integers such that 450y = n^3
which of the following must be an integer
1 y/(3 *2*2*5)
2 y/(3*3*2*5)
3 y/(3*2*5*5)
a none
b 1 only
c 2 only
d 3 only
e 1 2 and 3
450y = 3*3*5*5*2*y = n^3
For the above to be true y has to be : 2*2*5*3, which makes y/2*2*5*3 an integer.
Answer for me : B (1 only)