couldn't understand the explanation much, please elaborate...
Is it C?
Question gives,
x is +ve, is (w-z)> 5( 7^(x-1)-5^x) ?
1. zis (7^x - z) > 5*7^(x-1) - 5^(x+1)
i.e. (7*7^(x-1) - z) > 5*7^(x-1) - 5^(x+1)
7*7^(x-1) is always greater than 5*7^(x-1) for all +ve x
but unless we know the value of z (whether it is +ve or -ve) we cannot find out whether inequality is TRUE.
i.e. If x is small (e.g. 1), then z being +24 and -24 gives TRUE and FALSE result for the inequality.
2. by itself is not sufficient as it does not give values for w & z
combined: take x = 4 in 1st option, inequality becomes:
2401-z >1715-3125
2401-z> -1410
This is always TRUE for all z
Ans seems A to me ...though am not too sure this time ...
Just elaborating your approach ...excellent one btw
lets compare the highlighted text in diff colors :
Red : LHS >RHS
Green : LHS>RHS..
i.eeach of the left terms is greater than its counterpart on the right side ..
Hence, combined LHS > RHS ...
statement in blue seems incorrect to me ...
least value of 5^(x+1) is 25 which is greater than than the max value of z i.e 24
So, always zi.e -z > -5^(x+1) ..
pls correct for mistake(s)
To check : w-z>5(7^x-1 - 5^x)
Statement 1:
let x=1 and z=4 , w=7
put the values in w-z>5(7^x-1 - 5^x)
this gives 3>-25
similiarly, put x=2 and z=5, w=14 we get 9>-90
To confirm, put diff values, x=7 and z=7,w=49 we get
49-7
Hey Neha ...
Statement in bold is incorrect ...i guess in a hurry u have wrongly taken w=7x instead of w=7^x ...if x=2 ; w = 49
atishree Sayscouldn't understand the explanation much, please elaborate...
please write it on paper if you do not understand for the first time... these ^ signs are probably making it hard to understand. see below in this post for explanation of option 1 (my approach).
Ans seems A to me ...though am not too sure this time ...
Just elaborating your approach ...excellent one btw
lets compare the highlighted text in diff colors :
Red : LHS >RHS
Green : LHS>RHS..
i.eeach of the left terms is greater than its counterpart on the right side - not necessarily ..
Hence, combined LHS > RHS ...
statement in blue seems incorrect to me ...
least value of 5^(x+1) is 25 which is greater than than the max value of z i.e 24
So, always zi.e -z > -5^(x+1) ..
pls correct for mistake(s)
will start from LHS>RHS...
in LHS>RHS,
half part... i.e. .... 7*7^(x-1) is always greater than 5*7^(x-1) for all +ve x (like... 7y > 5y for all y>0)
but other part i.e. z (in LHS) and 5^(x+1) (in RHS) cannot be compared... as z can have any value less than 25 (including negative values)
e.g. -->
when z = 25 and x = 1
7*7^(1-1) - 25 > 5*7^(1-1) - 5^(0+1)
i.e.
7-25 > 5-5
-18>0
hence the inequality condition is FALSE
Now take z = -100 and x = 1
7*7^(1-1) - (-100) > 5*7^(1-1) - 5^(0+1)
7+100>5-5
107>0
here the condition becomes TRUE
hence we are getting both possibilities TRUE ans FALSE
hence we cannot answer using just first option.
will start from LHS>RHS...
in LHS>RHS,
half part... i.e. .... 7*7^(x-1) is always greater than 5*7^(x-1) for all +ve x (like... 7y > 5y for all y>0)
but other part i.e. z (in LHS) and 5^(x+1) (in RHS) cannot be compared... as z can have any value less than 25 (including negative values)
e.g. -->
when z = 25 and x = 1
7*7^(1-1) - 25 > 5*7^(1-1) - 5^(0+1)
i.e.
7-25 > 5-5
-18>0
hence the inequality condition is FALSE
Now take z = -100 and x = 1
7*7^(1-1) - (-100) > 5*7^(1-1) - 5^(0+1)
7+100>5-5
107>0
here the condition becomes TRUE
hence we are getting both possibilities TRUE ans FALSE
hence we cannot answer using just first option.
Interesting discussion dude !
to be honest i have not jotted on pen and paper ...found your post interesting and started the discussion ...
Lets take again the highlighted text in red ...
i guess there was 1 small calculaion error in your post ...
we are comparing 7*7^(x-1) - z and 5*7^(x-1) - 5^(x+1)
x=1 and z=25 is not permissible, so lets take z=24
LHS= 7*7^(1-1) - 24 = -17
RHS = 5*7^(1-1) - 5^(1+1)= 5 - 25 = -20
-17>-20 ..
I still maintain that i am not sure of the ans ...but i still haven't got hold of nos such that LHS
Interesting discussion dude !
to be honest i have not jotted on pen and paper ...found your post interesting and started the discussion ...
Lets take again the highlighted text in red ...
i guess there was 1 small calculaion error in your post ...
we are comparing 7*7^(x-1) - z and 5*7^(x-1) - 5^(x+1)
x=1 and z=25 is not permissible, so lets take z=24
LHS= 7*7^(1-1) - 24 = -17
RHS = 5*7^(1-1) - 5^(1+1)= 5 - 25 = -20
-17>-20 ..
I still maintain that i am not sure of the ans ...but i still haven't got hold of nos such that LHS
Damn... I did make a mistake there... :-(
Yes... you are absolutely right... there is no possibility of getting LHS
Hey Neha ...
Statement in bold is incorrect ...i guess in a hurry u have wrongly taken w=7x instead of w=7^x ...if x=2 ; w = 49
gosh!
I don't believe this. :-(
always in DS ques, i do such blunders and it costs me every time.. even in mocks!:banghead:
Yeh toh heights hai!
Btw, thanks Bhavin for pointing that.
gosh!
I don't believe this. :-(
always in DS ques, i do such blunders and it costs me every time.. even in mocks!:banghead:
Yeh toh heights hai!
Btw, thanks Bhavin for pointing that.
To err is human
...Each one of us make silly errors ...infact most of the math errors on GMAT are avoidable ones ...to learn from them n ace our scores is the purpose of the forum ..Cheers
To err is human...
Each one of us make silly errors ...infact most of the math errors on GMAT are avoidable ones ...to learn from them n ace our scores is the purpose of the forum ..Cheers :thumbsup:
Thanks Bhavin. :)
Got your point.
and Yogesh and Bhavin really good discussion. :thumbsup:
my doubts are cleared. 😃
Solve this
If k is a positive constant and y = x - k - x + k, what is the maximum value of y?
(1) x
(2) k = 3
y = x - k - x - k => -2k.
I am trying to explain that value of y won't depend on x, it will just depend on k.
(1) Insuff. because we don't know K
(2) suff. becoz value of K can determine value of y.
So, IMO = B
Solve this
If k is a positive constant and y = x - k - x + k, what is the maximum value of y?
(1) x
(2) k = 3
Solve this
If k is a positive constant and y = x - k - x + k, what is the maximum value of y?
(1) x
(2) k = 3
y = x - k - x - k => -2k.
I am trying to explain that value of y won't depend on x, it will just depend on k.
(1) Insuff. because we don't know K
(2) suff. becoz value of K can determine value of y.
So, IMO = B
IMO also Ans shd be B ...
But y is not necessarily -2k
If, x0 then y=-2k ...in any case, max of 2k and -2k is 2k since k is +ve ...so we need a def value of k to get max value of y ...
St 1 ; we just know y =2k ...no info about k, not suff
St 2 : max y = 2k = 6 ...suff ...Ans B
Ders 3 diff. possible values of y..
y=-2k, 2k & -2x
stmt 1.
xstmt 2.
k=3, nothing said about x.so insuff.
combining both , possible values of y can be -6,6 and -2*(xso insuff..
IMO E
@ sdt83.. can you explain a bit more -- how y = -2x
Ders 3 diff. possible values of y..
y=-2k, 2k & -2x
stmt 1.
xstmt 2.
k=3, nothing said about x.so insuff.
combining both , possible values of y can be -6,6 and -2*(xso insuff..
IMO E
my mistake.. ders r 4 diff values for y..
Lets look @ various possible values of y..
1.(x-k)-(x+k)=-2k, when both x-k & x+k are +ve
2.-(x-k)-(-(x+k))=2K, when both x-k & x+k are -ve.
3.x-k+ x+k=2x, when x-k is +ve & x+k is -ve
4.-x+k-x-k=-2x when x-k is -ve & x+k is +ve.
now both d stmts r not suff to ans.
IMO E
IMO also Ans shd be B ...
But y is not necessarily -2k
If, x0 then y=-2k ...in any case, max of 2k and -2k is 2k since k is +ve ...so we need a def value of k to get max value of y ...
St 1 ; we just know y =2k ...no info about k, not suff
St 2 : max y = 2k = 6 ...suff ...Ans B
IMO Bhavin is right. If the Q asks for a single value of y then it would be E. But the Q asks for max value , so st 2 should be sufficient..so B
Solve this
If k is a positive constant and y = x - k - |x + k, what is the maximum value of y?
(1) x
(2) k = 3
My take:
the maximum value of y will occur when the negative part equates to zero. thus x + k = 0 => x = -k
thus y = |-2k
thus maximum value of y depends on only 2k. therefore B. 😃
Another two,
Is X greater than Y?
1. 6X=5Y
2. X^2
My take:
Stament 1 says 6x=5y. so x/y = 5/6. that's cool we can get it by having both x and y as positive or both negative. So can't tell anything. So Insuff.
St 2 says x2
Another two,
Is X greater than Y?
1. X - Y > X - |Y
2. X|>X AND |Y=Y
My take:
I apply a strategy of looking at St.2 generally. That way, I avoid taking the two stmts together at the start.
We can infer from st.2 that X is NEGATIVE and Y is either ZERO or Positive.
So X is always
St. 1 says X - Y > X| - Y If this is the case, Y has to be necessarily negative. If we try taking X also Negative, then the condition fails. Thus, only Y is negative. X can be zero or any positive number for the situation to hold true. Thus X > Y, a concrete answer. Therfore, st1. suff.
Therefore, D.
Please correct me if I am going wrong somewhere. 😃
If, on a coordinate plane, point A has the coordinates (-3,4), how far is point A from point E?
1) Point E is on the Y axis four units from the origin.
2) If point A were twice as far from point E, it would be the distance from point E as point C is at coordinates (0,-2)
Help needed with this question. It had come on my Kap GMAT. I marked E. OA is C. Please Explain. :)
1) Point E is on the Y axis four units from the origin.
2) If point A were twice as far from point E, it would be the distance from point E as point C is at coordinates (0,-2)
Help needed with this question. It had come on my Kap GMAT. I marked E. OA is C. Please Explain. :)