help!

[AC2009]

hey! got my gre in a few days, so would be grateful if u could just help clear some doubts i have. ( as the D- Day is approaching ive started gettin confused in petty quant Q ( ) these questions are from the PP practice questions and in most of them i am not looking for any answer but some "confusions" that ive got -

1) n = 7(19 ^ 3)

Col A: distinct positive factors of n
Col B: 10

---- all i want to know here is that is the col A = 8?

2) n is a multiple of both 5 and 9

which of the following statements hold true:

I n is odd
II n is 45
III n is a multiple of 15

---- here, i understand that n can have many values, but why can we not say that II is true? afterall it is one of the values :S

3) There is a figure given witha parallelogram ABCD, such that angle ABC = 125, AD = 4, DC = 6. Have to find the area and compare it with col B which is 24.

---- the area of the parallelogram is smaller because its not a rectangle? am i right? as in a parallelogram having lenght and width equal to that of a rectangle will have a smaller area. please confirm.

---- also if we are just given the length and width of a parallelogram. is it possible to find its area?

4) ok there is a circle given in a fig, with centre (0, a). (d,e) and (b, c) are points lying onthe circle.
Col A: b^2 + c^2
Col B: d^2 + e^2

I guess you must have seen this question but just in caseyou have not ...the points (d,e) and (b,c) are positioned like X and Y respectively ... both in the quadrant I.

....................._
............._................X
........_.........................Y
......_.............O..............._
........_......................._
............._................_
......................_

please tell me how exactly to make the comparison,

i started off with OX = OY
used the distance formula and came to the equation
d^2 + e^2 - 2ae = b^2 + c^2 - 2 ac
and since e > c
hence d^2 +e^2 is greater.
i did not get it right in the first turn and maybe there is an even simpler explanation for it. so kindly help!


thanks in advance!

[fromcttoupenn]

#3
Area of a parallelogram equals product of any two sides* cos of the angle included between them. If the sides are given as 4 and 6 then the area would be 4*6*cos(angle). Since the maximum value of sin theta = 1 for theta =90, area would never exceed 24

[tommytommy]

Hey AC, calm down and think with a cool mind

1) yes, it is 8.

2) the question is 'must be true' ie has to be true in every case. so only III fits every case. 45 is true for only one case.

3) area of parallelogram = base * height.
let base = 6
now no matter what (unless angle ABC= 90, which it is not) the height has to be lesser than 4 (other side)
so area of paralellogram has to be lesser than 24.

4) You got it right. I dont know of a shortcut. Kudos for deriving the eqn. of a circle.

[fromcttoupenn]

The equation of a circle is not always x^2+y^2=k. It depends where the centre is located.

[Abaabeel]

3) area of parallelogram = base * height.
let base = 6
now no matter what (unless angle ABC= 90, which it is not) the height has to be lesser than 4 (other side)
so area of paralellogram has to be lesser than 24.

I got confused a bit with this explanation. Why the height cant be 4 or greater than 4? Didnt get the point. please explain

[tommytommy]

Originally Posted by Abaabeel

3) area of parallelogram = base * height.
let base = 6
now no matter what (unless angle ABC= 90, which it is not) the height has to be lesser than 4 (other side)
so area of paralellogram has to be lesser than 24.

I got confused a bit with this explanation. Why the height cant be 4 or greater than 4? Didnt get the point. please explain

In a right angle triangle the hypotenuse is always greater than the other 2 sides. In this case suppose we drop a perpendicular from point A onto side CD on a point X. Then AD is the hypotenuse and and AX(Altitude) & DX are the sides. So AX<AD. ie Hieght < Side.

[fromcttoupenn]

#4 No, there is no simple explanation for this. You just have to remember that the equation of a circle is (x-h)^2+(y-k)^2=k^2

In this case it is;
(x-0)^2+(y-a)^2=a^2

x^2+y^2-2ay=0 or x^2+y^2=2ay

Hence its A.

[Elnaz_b]

#4 No, there is no simple explanation for this. You just have to remember that the equation of a circle is (x-h)^2+(y-k)^2=k^2

In this case it is;
(x-0)^2+(y-a)^2=a^2

x^2+y^2-2ay=0 or x^2+y^2=2ay

Hence its A.

How did you assume that the radius is a?
The problem above doesn't give the radius.

[fromcttoupenn]

Oh I was just giving a general equation in which we denote the radius as 'a'. Anyway, it doesn't make any difference because even if its k^2(instead of a^2), we are left with 2ay on the right hand side apart from ( k^2-a^2) which is constant. So ,A is greater.

[tommytommy]

hmm ..i just thought of something for #4. I wish i had an image to explain it of of.

A(0,a) is a point on y-axis.
X(d,e) & Y(b,c) are points on the circle with center A(0,a)
O(0,0) is the origin.

Now join OX & AX and OY & AY, we get 2 triangles OAX & OAY.

sqrt( d² + e²) is the distance of X from the origin O(0,0), and
sqrt( b² + c²) is the distance of Y from the origin O(0,0)

So we need to compare OX to OY

We know, OA = OA, AX = AY
Now since angle OAX > angle OAY
OX > OY.

done.