CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[smartie]

understood now...
but then, I am not able to understand where did I go wrong...was it the isosceles triangle part??

[no signal]

Originally Posted by thebornattitude

QQAD 30!!!

SOLUTION

If we convert the diagram into 2D, it will look like as shown

Attachment 9714

Now in triangle ABC, AD is the portion outside the cylinder.
Now CA=CB due to symmetry of the cylinder inside the cube...hence angle
CAD=CBD= 45 deg..
hence CD= DA tan45=DB tan45= r(radius of the cylinder)

now body diagonol, considering side of cube =a,

ax root(3)= h(height of cylinder) + 2r( two times the portion outside the cylinder )
so axroot(3)= 3root(3) +2root(6)
a= 3+2root(2)

hence answer = none of these

I think the 2-D diagram will be a bit different. When we take a cube and cut it along its diagonal, we get a rectangle not square. And the cylinder will touch only 2 sides of that rectangle. However I wasn't able to get a solution.

@ Aarav: Dude can I change my subscription from text to HTML?

[Aarav]

Originally Posted by no signal

@ Aarav: Dude can I change my subscription from text to HTML?

Kar le yaar, unsubscribe kar and then subscribe with HTML format.

[slam]

Originally Posted by Aarav

The cylinder will touch the face of the cube on its diagonal. Let x be the angle between its diagonal and the diagonal of the cube along the axis of the cylinder. Then sinx = 1/√3, and let l be the length of the side of the cube => √3 l = 3√3 + 2√6 cotx => l = 7 cm.

Damn!
After hours of wondering where I went wrong, I've finally got it! Thanks Aarav Sir, it was your post that cleared the air in the end.
I was taking the triangles to be similar as it is, so took L/R = sqrt(3)/sqrt(2). What I didn't notice was that the right angle in each triangle is at different places, so the equation would actually be:
L/R = sqrt(2)/1 => L = 2sqrt(3)
=> sqrt(3) * a = sqrt(3) * (3 + 2*2)
=> a = 7.

The only reason I've posted this is to convey a sense of victory! And Thank you Aarav Sir.

[abhishekchhajer]

A cylinder of radius √6 cm and height 3√3 cm is inscribed inside a cube such that the axis of cylinder is along a diagonal of the cube. The length of side of the cube is (1) 6 cm (2) 7 cm (3) 8 cm (4) 9 cm (5) none of these Its pretty simple see since height is 3√3=less than 6and since the cylinder is along the diagonal therefore the edge of the cube cannot be greater than the diagonal.So only choice left is none of these

[ravijanjanam]

Consider a rectangular purse of dimension 8 cm X 9 cm. What could be the maximum radius of two identical coins which can be completely put inside the purse without overlapping?

(1) 2 cm (2) 2.25 cm (3) 2.5 cm (4) 2.75 cm (5) none of these


ans: ------------------------
. .
. . 8
. .
------------------------
9
9/2=4.5

4.5 is the diameter of one circle incribed in rectangle

therefore radius =4.5/2=2.75

if radius of cirlcles =2.75
then thay will touch each other

therefore
radius<2.75
from given options radius =2.25 which is maximum

[Rockeeze]

hi
i think the maximum radii is more than 2.25

place the coins on the diagonal of the rect.

2.(2^1/2+1)r=145^1/2

so r=2.5

[smartie]

there are two cases
i)length=8 and breadth=9
-->2r<=9 and 4r<=8
-->r<=2

ii)length=9 and breadth=8
-->2r<=8 and 4r<=9
-->r<=2.25

of these r=2.25 is the maximum
so i would go with option(2)2.25

[srikar2097]

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Quantitative Question # 031
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Consider a rectangular purse of dimension 8 cm X 9 cm. What could be the maximum radius of two identical coins which can be completely put inside the purse without overlapping?

(1) 2 cm (2) 2.25 cm (3) 2.5 cm (4) 2.75 cm (5) none of these

Case 1: When the 2 coins are along the breadth.
Then the sum of the radius (i.e.4r) will equal 8 cm and radius is 2 cm.

Case 2: When the 2 coins are along the length.
Then the sum of the radius (i.e.4r) will equal 9 cm and radius is 2.25cm

Case 3: When the 2 coins are along the diagonal.
The length of the diagonal of the rectangle = (145)^1/2 = 12.1 cm (approx.)
The length of the diagonal in terms of the radius of the coins = [2+2sqrt(2)]r
=> [2+2sqrt(2)]r = 12.1
=> r (4 point eight) = 12.1
=> r = 2.51 cm

Hence the max. radius of the coins could be 2.5 cm or option (3)

[slam]

My method was very similar to srikar's. Taking two identical coins along the diagonal, for them to be of max radius, they will be tangential to the sides of the rectangle.
R^2 + R^2 <= ( sqrt(145)/2 - R )^2
On solving,
R <= ( sqrt(2) - 1 ) * sqrt(145) / 2, which comes out to be around 2.48.

But I'm not convinced with this, as it is not giving an exact value (compared to the answer options given). Are we missing something here?