"ANY SQUARE IS A RHOMBUS BUT ANY RHOMBUS IS NOT A SQUARE".
Is quadrilateral ABCD a rhombus?
(1) Line segments AC and BD are perpendicular bisectors of each other.
(2) AB = BC = CD = AD
It should be D.. Both statements alone are sufficient.
"ANY SQUARE IS A RHOMBUS BUT ANY RHOMBUS IS NOT A SQUARE".
"ANY SQUARE IS A RHOMBUS BUT ANY RHOMBUS IS NOT A SQUARE".
Neither statement alone is suff.Even combining both the statement we can't say if the quadrilateral is either a square or a rhombus.
I will go with option E
It should be D.. Both statements alone are sufficient.
"ANY SQUARE IS A RHOMBUS BUT ANY RHOMBUS IS NOT A SQUARE".
The ans is D.
But one doubt Nipun..I wonder how statement A alone is sufficient.
Perpendicular bisector-> the quadrilateral can be a kite as well.
So how does this be a sufficient condition to conclude its a rhombus!?.
Please clarify this.
Mathematical proof:-
Lets take a quadrilateral ABCD, AC and BD are 2 diagonals which intersect at 0.
Now, AB, BC, CD, DA are four sides.
AO = OC and BO = OD and all angles at the center of quad. are 90 (Perp. bisector)
Take any two adjacent trianges say tri. AOB and tri. BOC:-
1) OB = OB (Common side)
2) AO = OC (PErpendicuar bisector)
3) Angle AOB = Angle AOC (90 degree)
So, By SAS Theorm, Triangle AOB is concurrent to Triangle BOC.
U remember -- SSS, RHS, SAS, ASA concurrency theorms in class 10th
So, Rest all angles and sides will be same to their adjacent angles n sides
hence, AB=BC
This proof will never be used but just for ur satisfaction..
Now, think of that kite. Kite can only be formed in square. And defination of square is:- any rhombus with its side angles as 90 is square. so, all squares are rhombus.
The ans is D.
But one doubt Nipun..I wonder how statement A alone is sufficient.
Perpendicular bisector-> the quadrilateral can be a kite as well.
So how does this be a sufficient condition to conclude its a rhombus!?.
Please clarify this.
The ans is D.
But one doubt Nipun..I wonder how statement A alone is sufficient.
Perpendicular bisector-> the quadrilateral can be a kite as well.
So how does this be a sufficient condition to conclude its a rhombus!?.
Please clarify this.
Dude...for a kite, the diagonals are perpendicular to each other, but do not bisect. Visualize again.
Cheers.
Ankit.
guys,
cud u help me out with this one(with explanation)? i think the answer given is wrong...just wanted to confirm:
What fractional part of the total surface area of cube C is red?
(1) Each of 3 faces of C is exactly red.
(2) Each of 3 faces of C is entirely white.
guys,
cud u help me out with this one(with explanation)? i think the answer given is wrong...just wanted to confirm:
What fractional part of the total surface area of cube C is red?
(1) Each of 3 faces of C is exactly red.
(2) Each of 3 faces of C is entirely white.
From 1 only or 2 only we cant say anything about other 3 faces. If we combine 1 nd 2 then 3 faces are entirely white and other 3 are half red (Since white cant come with red as it is entirley covering the face).
Hence if we combine 1 nd 2 then 25% of the surface area is red.
What is the answer?
I would go with C.
1st one says 3 faces are half red. So, we are left with 3 more faces.
2nd one says 3 faces are total white.
If we combine these two, One which are half red cannot be complete white. So, we have a cube with 3 faces half red and half white and rest 3 faces total white. So, C
1st one says 3 faces are half red. So, we are left with 3 more faces.
2nd one says 3 faces are total white.
If we combine these two, One which are half red cannot be complete white. So, we have a cube with 3 faces half red and half white and rest 3 faces total white. So, C
guys,
cud u help me out with this one(with explanation)? i think the answer given is wrong...just wanted to confirm:
What fractional part of the total surface area of cube C is red?
(1) Each of 3 faces of C is exactly red.
(2) Each of 3 faces of C is entirely white.
Dude...for a kite, the diagonals are perpendicular to each other, but do not bisect. Visualize again.
Cheers.
Ankit.
Ankit, sorry if im ignorant on this..but cant it be a rectangle?even in a rectangle the diagonals bisect each other
In a rectangle, diagonals can bisect each other but they are never be perpendicular.
for a kite, 1st condition I feel is diagonals perpendicular to each other.
yserious SaysAnkit, sorry if im ignorant on this..but cant it be a rectangle?even in a rectangle the diagonals bisect each other
yserious SaysAnkit, sorry if im ignorant on this..but cant it be a rectangle?even in a rectangle the diagonals bisect each other
No, a kite can not be a rectangle. If that happens, the following property of kite will be defied:
"One diagonal divides a kite into two isosceles triangles; the other (the axis of symmetry) divides the kite into two congruent triangles."
I was earlier wrong in assuming that a kite can not have its diagonals perpendicularly bisecting each other. I just now realized that even rhombus and square can be categorized as a kite.
Kite (geometry) - Wikipedia, the free encyclopedia
Cheers.
Ankit.
Hey guys please solve these challanging ds problems:
If a 1) z-a| 2) y-a|
1) x-y > -2
2) x-2y
Hey guys please solve these challanging ds problems:
If xy >0 ?
1) x-y > -2
2) x-2y
Statement 1:
=========
x - y = -1 ==> x=3 ,y=4 xy>0 or x=-2 , y=1 , xy
Statement 2:
==========
x - 2y = -7 ==> x = -13 ,y=-3 , xy>0 or x=-1 y = 3, xy
Combining both the statments
x - y = -1 & x-2y = -7 ===> x & y as positive.Even changing the eqns also gives both the values of x & y as +
I will go with option C
Hey guys please solve these challanging ds problems:
If a 1) z-a| 2) y-a|
I will go with option D.Each statement alone is suff to ans the q
Hey guys please solve these challanging ds problems:
If a 1) z-a| 2) y-a|
If xy >0 ?
1) x-y > -2
2) x-2y
IMO: D and C
Q1: Place a, y, z and b on a number line starting left to right.
y-a|, |y-b|, |z-a| and z-b are nothing but distance between any two points on the number line.
Looking at the number line, we see that if the distance b/w z and a is less than the distance between z and b, and that y lies between a and z, then clearly distance between a and y has to be less than the distance between z and b.
Similarly, second point can also be validated.
Q2:
Option 1 is the area below the line x-y=2, which will cover infinite number of points. Hence either x or y could be negative/positive/both. Not sufficient.
Option 2 is the area above the line x-2y=-6, which will cover infinite number of points. Hence either x or y could be negative/positive/both. Not sufficient.
Now, if we combine both the options, and see the common area covered by these inequalities, we see that the common area lies in the first quadrant only, indicating that x and y both are positive.
A General Rule that I follow: For all inequality Qs, draw the lines on paper. This might take a few seconds more, but surely hits the right answer.
On the same lines, you may want to try this one...
Is m + z > 0 ?
1) m - 3z > 0
2) 4z - m > 0
Cheers.
Ankit.
Somebody please help me with these ones. I have been scratching my head since long now and not able to justify the OA.
Q1: If there are more than two numbers in a certain list, is each of the numbers in the list equal to 0 ?
1) The product of any two numbers in the list is equal to 0.
2) The sum of any two numbers in the list is equal to 0.
Q2: On the number line shown, is 0 halfway between r and s ?
1) s is to the right of 0.
2) The distance between t and r is the same as the distance between t and -s.
(This is a number line with r, s and t placed on it 
)
.
Q2: On the number line shown, is 0 halfway between r and s ?
1) s is to the right of 0.
2) The distance between t and r is the same as the distance between t and -s.
(This is a number line with r, s and t placed on it)
Statement 1:
==========
Only talks about s.Noting is said about r.So this is nt suff
statement2:
=========
r lies to the left of t. distance b/w r and t = distance b/w -s and t.So r=-s.
so 0 must lie halfway b/w -s and s.
I will go with option B for the above DS..Watz the OA?
Somebody please help me with these ones. I have been scratching my head since long now and not able to justify the OA.
Q1: If there are more than two numbers in a certain list, is each of the numbers in the list equal to 0 ?
1) The product of any two numbers in the list is equal to 0.
2) The sum of any two numbers in the list is equal to 0.
statement 1:
==========
only product is zero.The set may consists of entire zero's or can also be filled with only one integer.
s= { 1,0,0,0...}
s = {0,0,0,0,0..}
so nt suff.
statement 2:
==========
Sum of any two nos is zero.This can be achieved only when all the nos are zero.
s = {0,0,0,0..}
So I will go with option B for the above DS.watz the OA?
Statement 1:
==========
Only talks about s.Noting is said about r.So this is nt suff
statement2:
=========
r lies to the left of t. distance b/w r and t = distance b/w -s and t.So r=-s.
so 0 must lie halfway b/w -s and s.
I will go with option B for the above DS..Watz the OA?
OA is C.
--> Taken from GMAT Prep Practice Test
statement 1:
==========
only product is zero.The set may consists of entire zero's or can also be filled with only one integer.
s= { 1,0,0,0...}
s = {0,0,0,0,0..}
so nt suff.
statement 2:
==========
Sum of any two nos is zero.This can be achieved only when all the nos are zero.
s = {0,0,0,0..}
So I will go with option B for the above DS.watz the OA?
Ok, I got that...
OA is infact B.
I dont know, for some reason, I was considering this as a valid set for statement 2: {0, -5, 5, 0, 0}
Q2: On the number line shown, is 0 halfway between r and s ?
1) s is to the right of 0.
2) The distance between t and r is the same as the distance between t and -s.
(This is a number line with r, s and t placed on it
Yes...Ans for this one has to be C ..
St 1 : tells us nothing...not suff...
St 2 : t-r = t+s
Hence, either r = -s OR 2t = r-s ...not suff ...
Combined : t+s is +ve and t-r is +ve ...
Hene, r=-s only ...
Ans C