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

[themystique]

let y1 and y2 be the angles made by the slant height with the base of cones with height 1 and 8 resp.

case 1)
tan y1 = 1/r
tan y2 = 8/r

dividing both, tan y1/ tan y2 = 1/8

case 2)

both heights are increased by 'x'.
vertex angle is unchanged, hence y1 and y2 also no change
==> base radii are changed to r1 and r2 resp.

given volumes are now equal.
==> (r1)^2 * (1+x) = (r2)^2*(8+x)
((r1)^2)/ ((r2)^2) = (8+x)/(1+x) -- 1

and

tan y1= (1+x)/r1
tan y2=(8+x)/r2

==> tan y1/ tan y2 = ((1+x)/(8+x))*(r2/r1) = 1/8

sub from 1,

we get (r2/r1)^3 = 1/8
==> (r2/r1) = 1/2

==> (1+x)/(8+x) = 1/4

solving we get x = 4/3

[mehrozkhn]

Q43
$ol:
let § and ¥ be the respective half vertex angles in the two cones
and let s1 and s2 be the increase in their respective radii after
increasing height.
Then
r/1 = tan § = (r+s1)/(x+1) ....(1)
and r/8 = tan ¥ = (r+s2)/(x+ ....(2)
where r is their initial radii.
Solving
s1 = rx ...(3)
s2 = rx/8 ....(4)

now their volumes equal after increase in height
=> (r + s1)^2 (x + 1) = (r + s2)^2 (x+
solving using (3) and (4)
x = 4/3

[kamalesh123]

Cone 1
R1 = r
H1 = 1
V1 = 1/3*pi*r^2
After increasing the height the vertex angle is still the same=> radius is increased
So new radius = r(1+x)
Cone 2
R2 = r
H2 = 8
V2 = 1/3*pi*r^2*8
After increasing the height the vertex angle is still the same=> radius is increased
So new radius = r(8+x)/8
1/3 *pi*(r(1+x))^2*(1+x) = 1/3 *pi*(r(8+x))^2*(8+x)/64
=> x = 4/3 . Option (2).

[v-factor]

Let the initial radius for both cones be r
V1 = (1/3)*pi*(r1^2)*(1+x)
V2 = (1/3)*pi*(r2^2)*(8+x)
And V1 = V2

But r1 = r(1+x)
r2 = r(8+x)/8

Solving
8+x = 4(1+x)
x= 4/3

[shivam_01]

Originally Posted by smartie

------------------------------------------------------
Quantitative Question # 043
------------------------------------------------------

Consider two cones of heights 1 and 8 units having the same base radii. It is found that their height is increased by x keeping their vertex angle unchanged, their volume becomes equal. Then x equals


(1) 2/3 (2) 4/3 (3) 8/3 (4) 16/3 (5) none of these

hey i got the solution for the problem
let us take the new height of cone 1 and 2 be 1+x,8+x respectively....
now since the vertex angle remains the same hence
tanA remains the same and so does the ratio of radius to height....
For first cone
the r' be the new radius then
r/x=r'/(1+x) and this gives
r'=r(1+x).............1

For second cone
r/8=r''/(8+x) and then
8r''=r(8+x)............2

and now the volume remains constant
1/3pi*r'^2*h'=1/3pi*r''^2*h''

nowsubstituing the values from 1 & 2 we get
r'= r(1+x)
r''= r(8+x)
h'= (1+x)
h''= (8+x)

puting the values we get x=4/3 (b)

[Vishal2704]

QQAD Q##43

New heights are 1 + x and 8 + x respectively....

Since the vertex angle is same...we have...

r1' = r (1+x) and r2' = r(8+x)/8
h1' = 1+x h2'= 8+x

Also New Volumes are equal..therefore..

1/3pi* r1'^2 * h1' = 1/3pi * r2'^2 * h2'

(1+x)^3 = (8+x)^3/64

==> 1+x = (8+x)/4

==> 4 + 4x = 8 + x

=> x = 4/3

Hence the answer option (2) is Correct !!

[shivam_01]

Looks that everyone is getting is the same answer so majority wins
Answer is b congrtas guys for solving the question is such a short time

[Vishal2704]

Originally Posted by Aarav

Let me start this with today's learning

1) Though Y could have been proven with an example, the same was not true for (X). The statement read "for every quadrilateral with sides a, b, c, d" and hence disproving with example wasn't correct. I think few of you assumed "for some quadrilateral with sides a, b, c, d" and made mistake here. Understand the subtle meaning when words like some/any/every are involved.

2) Y could have been sufficient to answer if the question was -> Is 1/x + 1/y + 1/z < 1? Or if the question was not all of x, y, z are equal.

3) Always start a DS question with assumption that examiner had done well to trick you here, and elimination should follow the order -> can it be by X alone? -> can it by by Y alone? -> If answer to previous 2 is yes then you already have the answer else -> It can't be by X and Y also [try to convince yourself on this if possible] -> Last should be by combination of X and Y both.

AARAV Bhai...I have a doubt related to this question...Since it has been mentioned that condition "X" i true for every quad...hence we can have a quad to be a square and in that case a=b=c=d.....

Thus our condition becomes x + y +z > 1

Since x,y,z are real numbers we can take 1st condition say x=y=z=4
In this Case 1/x + 1/y +1/z < 1

IInd Case say x=y=z=1....
In this Case 1/x + 1/y +1/z > 1

Hence from "X" alone we cannot determine..Tell me where i am going wrong

[akshatakrao]

This is the solution to Quant Question QQAD 043.

Since the vertical angle remains the same even after increasing the height by x.
Let this angle be A for cone with height 1 and B for cone with height 8.

tan A = (r1/1) and tan A = (r1'/(1+x)) where r1 was the radius of cone 1 before change and r1' is the radius after change.

so r1' = (1+x)r1/1
Similiarly, r2' = (8+x)r2/8

Since r1 = r2

and now the volumes are the same

(3.142) * r1'^2(1+x) = 3.142 * r2'^2 (6+x)

which boils down to x = 4/3 after substituting values of r1' and r2'

Choice (2)

Tada!!!

Akshata

[sanyo]

Ans 43) option 2) 4/3

Soln: let r be the initial radius for both the cones, and y be the increment in radius for the cone with height 1, and z be the increment in radius for the cone with height 8.

Now (r+y)/(1+x) = r,
and (r+z)/(8+x) = r/8

=> r+y = r(1+x) and r+z = r(8+x)/8

since the cones have equal volume
r^2 (1+x)^3 = r^2 (8+x)^3 /64
=> (8+x)^3 = 4^3(1+x)^3
hence x = 4/3