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Mensuration- Solids
by Total Gadha - Saturday, 1 March 2008, 11:31 PM

cat 2008 cat 2009 preparation mba 2008 mensuration xat 2008For nearly every CAT 2008 aspirant, the question of ‘what to do and how to do it’ is the most crucial question of all. Besieged by too many topics and pulled in too many directions, a student is dazed and intimidated by the seemingly mammoth task at hand. The trick to winning a long-drawn-out battle like CAT preparation is to get started first and think later. Catch any end of the rope you can lay your hands on and start pulling. Some helpful and cultivated good habits, such as making a timetable for the next day and practicing all the three sections everyday, will also go a long way in preparing you for the D-day. And if you are one of them who forget what they read a month ago, you can certainly move around in TG forums and keep solving problems all the time to keep you in shape! But the most important thing to do is to find that zeal and excitement for your preparations in place of that CAT fear.

Needless to say, CAT preparations will bring you one of the most memorable times of your life. Even with all that stress, those class exercises, those depressing mocks and the fierce competition, CAT preparation is so much fun.

cat 2008 preparation mensuration- 1
cat 2008 preparation mensuration- 2

Given below are some teaser questions for you to try your hands on. Good luck!

cat 2008 preparation, mensuration- questions
Re: Mensuration- Solids
by jai prakash - Sunday, 2 March 2008, 12:02 AM
  Another Nice Stuff to go through.
Expecting the saga to be continued in future.
Re: Mensuration- Solids
by PRIYANKA KESHRI - Sunday, 2 March 2008, 03:50 PM
Re: Mensuration- Solids
by Little Star - Sunday, 2 March 2008, 05:24 PM

4th ans=O option(d)

Little Star

Re: Mensuration- Solids
by ambar patil - Monday, 3 March 2008, 03:06 PM

TG is on total rampage !! It is adding killer material day by day ......

Hats off to TG .... wink

Re: Mensuration- Solids
by Little Star - Monday, 3 March 2008, 03:24 PM
  Nice stuff TG

4th ans-(d)
5th ans-(b)(125pi/sqrt3)
6th ans-(c)= 1 sphere+2cylinder
                  1 sphere+3cone+1 cylinder
                  2 cone+2cylinder
7th ans-(c)(102)
8th ans-(c)
9th ans-240pi

Any one tell me is that right or not?

Little Star
Re: Mensuration- Solids
by ambar patil - Monday, 3 March 2008, 05:19 PM

4 - d

5 -  A 

6 - c .....  but took me too long to come to the final answer .

7 - d

Re: Mensuration- Solids
by ambar patil - Monday, 3 March 2008, 05:36 PM

2 ) can be 9 / sqrt(2) .... not sure though sad

Re: Mensuration- Solids
by jupiter singh - Tuesday, 4 March 2008, 04:00 PM

dude isnt the awnser of 4  c-midpoint

and awnser if 5 ithink is d

Re: Mensuration- Solids
by Rekha Krishnan - Wednesday, 5 March 2008, 03:32 PM
Re: Mensuration- Solids
by ambar patil - Thursday, 6 March 2008, 01:28 AM
  TG , throw some hints for the rest of the problems .....
Re: Mensuration- Solids
by Total Gadha - Thursday, 6 March 2008, 03:05 AM
  Hi Ambar,

Will post solutions in a day or two. smile

Total Gadha
Re: Mensuration- Solids
by JYOTHI PRAKASH GANDHAMANENI - Thursday, 6 March 2008, 07:24 PM

1...17.77 min








9.....240*pi(but not sure)

Re: Mensuration- Solids
by ambar patil - Sunday, 9 March 2008, 12:05 PM

yep , the answer to 3 is 288pi

similarity of triangles is the key smile

Re: Mensuration- Solids
by psp psp - Tuesday, 11 March 2008, 12:15 AM
  2. a/(2*sqrt(2))  not sure of the answer.
Re: Mensuration- Solids
by ambar patil - Tuesday, 11 March 2008, 02:42 AM

psp psp ,

I think it shud be a2/ ( 2 * sqrt(2) )

Re: Mensuration- Solids
by psp psp - Tuesday, 11 March 2008, 04:54 PM
  yes its a2 /(2*sqrt(2)) thanks.
Re: Mensuration- Solids
by psp psp - Tuesday, 11 March 2008, 07:17 PM
  9 )   (4/3)*pi*63 - (1/3)*pi*32*5 =273*pi .. pls show where i am going wrong
Re: Mensuration- Solids
by Manash Saikia - Tuesday, 11 March 2008, 08:52 PM
  Hi Jyothi,

  I have got the ans for question# 1 as (19/27) * 60 mins = 42.2 mins...
( volume of the frustrum is 19/27 times the whole volume of cone)
  Kindly let me know how u have got 17.77 min...

Re: Mensuration- Solids
by Naveen Naveen - Wednesday, 12 March 2008, 02:30 PM

Hi Little star,

i got same answer as yours for 4, 5 and 7 but i differ from you on 6...

initially we have 2 S, 1 Cy and 1 Cone...their vol are in the ration of total weight on right is 12 (2 sphere+1 Cone+1Cyl (8+3+1))...

so one way is we have all 4 cyl. on left (3*4=12)...correct me if i am wrong?

Re: Mensuration- Solids
by avijit mohapatra - Thursday, 13 March 2008, 04:19 PM
  Hi all,
        My answers are:-
        1 - 17.77 min
        2 - ((60.5 +1)/ (2*2.5)) a2
3 - 4/3 pi (48/7)3
           4 - A
           5 - A
           6 - C
           7 - C
           8 - C
           9 - 240 pi

Re: Mensuration- Solids
by Tejashree Bhatia - Thursday, 20 March 2008, 10:52 AM
  great article TG...
Re: Mensuration- Solids
by Joel D\'Souza - Thursday, 20 March 2008, 02:31 PM

Great TG Sir. Could you please post some lessons on the very basics of 'Functions'. This chapter really troubles me.



Re: Mensuration- Solids
by ambar patil - Thursday, 20 March 2008, 05:00 PM

echo joel ,

Sir we are waiting for your articles on Functions and Graphs .

Thenks . 

Re: Mensuration- Solids
by mohit kumar - Sunday, 23 March 2008, 03:36 PM
  pls. disclose answer key with detailed solutions.
Re: Mensuration- Solids
by ganesh b - Friday, 11 April 2008, 01:44 PM

hi tg




but pls post the answers soon....




Re: Mensuration- Solids
by Vinay Desai - Wednesday, 16 April 2008, 02:42 PM
  Ya i agree with you manash.......42.222 is the right answer for question #1
Re: Mensuration- Solids
by Swastik Panda - Saturday, 19 July 2008, 11:59 PM
  hv tried 2 do some..
Qn1-> 52min 30sec
Qn2-> a2/4 * (2 * 31/2  +  21/2)
Qn3-> option C
Qn4-> option D
Qn5-> option A(125pi / 31/2)
Qn6-> option C
Qn7-> option C
Qn8-> option C
Qn9-> 240*pi
Re: Mensuration- Solids
by Swastik Panda - Sunday, 20 July 2008, 10:57 AM
  sorry my answer to question no1 was wrong
its 17.77min
Re: Mensuration- Solids
by Ankush Kumar Chakraborty - Monday, 21 July 2008, 05:14 PM
  The answer to Q1 is 42.22 mins.
Re: Mensuration- Solids
by Santosh Sarangi - Monday, 21 July 2008, 11:25 PM

Can someone please help me with the steps for Q3 & Q9 ??


Re: Mensuration- Solids
by Debanshu Sharma - Friday, 25 July 2008, 12:28 PM

Answer no 9)

240pi+9sqrt(2) pi=252.69 pi

The volume of the remaining portion would be vloume of the Sphere(remaining part)+volume of the cone( coz that part is not used too)


Re: Mensuration- Solids
by unnikrishnan p - Monday, 28 July 2008, 03:51 PM
  yep..i also got the answer to question number 1 as (19/27) * 60 which is approx 42.2mn..
Re: Mensuration- Solids
by unnikrishnan p - Monday, 28 July 2008, 04:22 PM

Hi santosh...

take it as a traingle with a circle placed inside it...

since radius and height are given,we can calculate the slant height..

also since radius is 12 and we know tangents to the same circle from the same points are equal..we can calculate the value of all the tangents to the circle..also we know that tangent square = product of the chords to that circle..from this we can calculate the value of the diameter..

dividing it by 2 gives radius which is equal to 6

from which we can calculate the volume

so the answer is 288pi

since i couldn explain this to u with diagram,i dono how much clear my explanation is..anyways..cheers

Re: Mensuration- Solids
by Monika Kadam - Tuesday, 29 July 2008, 07:37 PM

Dear Sir,

It would be immensely helpful if you could post the solutions to all the problems you have given in this article.

Thanking you,


Re: Mensuration- Solids
by Leonidas S - Sunday, 10 August 2008, 08:38 PM

Hi TG,

This is my 1st post in this site.It is really a fascinating place for cat-preparation.I have a problem can u please look into it?

From a cuboid of dimensions 4m*6m*8m , the largest possible cube is cut out.What is the minimum possible number of cubes ,all of equal size,into which the remaining part of the solid can be cut,ensuring that no part of the soild remains??

The answer is 16

Waiting for an 

Re: Mensuration- Solids
by TG Team - Monday, 11 August 2008, 04:47 AM
  Hi Subhajit smile
Largest possible size of cube that has been cut is 4*4*4.
Now just go in to your imagination. See what part of the given cuboid is left.
There are two parts actually cuboids of dimensions:
2*4*4 and 4*4*6.
Now smaller cubes of maximum side 2m can be cut out of them.
So number of smaller cubes cut from the remaining cuboids = (2*4*4/2*2*2) + (4*4*6/2*2*2) = 4 + 12 = 16 smile
Re: Mensuration- Solids
by Gaurav Talwar - Monday, 11 August 2008, 11:03 AM

hi all,

kindly tell where m goin wrong ... in q1 ... the ans m gettin is 20 min ...

as the time wen the ht. wud be half will be wen the upper cone would be 2/3rd of its vol and the lower cone wud be 1/3rd... so the time taken wud be 1/3rd of the total ...

plz.. tell me where m goin wrong i hv solved this using the volume equation ... do i need to do it sm other way ?? ... kindly guide me ...


L - full cone height

l - height wen the cont is met ... i.e. twice

Re: Mensuration- Solids
by Manjunatha G Gopalakrishna - Thursday, 21 August 2008, 09:58 AM
  Can you please post the solutions?
Re: Mensuration- Solids
by Abhishek kumar Singh - Thursday, 21 August 2008, 11:47 AM

HI my answers are

3. 288pi



6. c

7. c.


Re: Mensuration- Solids
by nitin taluja - Tuesday, 9 September 2008, 11:37 PM

Hi TG,

It would be of great help if u post the answer to the teaser problems that you have mentioned.

There is some typo error in the total surface area of the right circular cylinder..?



Re: Mensuration- Solids
by ATOM ANT - Wednesday, 10 September 2008, 04:19 PM
  Ans pls.. 
are these solid answers ?
by nimish vikas - Saturday, 13 September 2008, 03:39 AM
  (x^y is written for x power y)
Q.1. 42.22 secs
let ht of each cone be h
=> depth of sand in upper cone in 2h/3 and lower cone h/3
vol of upper cone emptied = pi.r^2.h/3 - pi.(2r/3)^2.(2h/3)/3
= pi.r^2.h/3 * 19/27 =19V/27 (say)
V takes 1 hr => 19V/27 will take 19/27 hr to empty
=42.22 min
Q.2. 2^(1/2).a^2 /4
cut face is a triangle with base = a cos30 = 3^(1/2).a/2 (=b say)
and height = ht of tetrahedron= 6^(1/2).a/3 {=sqrt(b^2-(b/3)^2)}
=> area = 1/2. [3^(1/2).a/2] . [6^(1/2).a/3]
=2^(1/2).a^2 /4
Q.3. (C) 288pi
H=16 , R=12 , slant edge length L = sqrt(H^2+R^2) =20 .
radius of the sphere will be the radius of the circle inscribed in the triangle which is the face when the cone is vertically sliced into two half.
drop perpendicular from the centre of circle to the slant edge.
this gives two rt angled similar triangle (with half of vertex angle as common angle). let radius of circle=r
=> r/R = (H-r) / L => r=6
volume = 4/3. pi . r^3 = 288 pi
Q.4. apex is point O
Q.5. 125pi /sqrt(3)
let radius of cone be r => 2.pi.r = 2.pi.10 /2 => r=5
slant length l=10 => h=sqrt(l^2-r^2) => h=5.sqrt(3)
vol = pi.r^2.h/3 = 125pi /sqrt(3)
Q.6. (c) 3 ways
let pi.r^3 = m ; (note h=2r given)
vol/m sphere = 4/3=1+1/3 ; cone = 2/3 ; cyl = 2
vol/m on one side= (2*4/3+ 2+ 2/3) = 5+1/3
5 + 1/3 = 2 + 2 + (1+1/3) ................2cyl + 1sph
= 2 + 2 + 2/3 + 2/3 ..............2cyl + 2cone
= 2 + 2/3 + 2/3 + 2/3 + (1+1/3) ..1cyl +3cone + 1sph
time to sleep
let me know if there is any mistake
Re: Mensuration- Solids
by A K - Saturday, 13 September 2008, 10:45 AM

Isn't the curved area for a frustum (pi)(R-r)l ? [ As cone's curved area is (pi)rl ]

Re: Mensuration- Solids
by naren suggula - Saturday, 13 September 2008, 01:17 PM

Hi TG and all,

Please explain the solution for the Q9. Is it solved with some concept of steradians?

Thanx in advance

Re: Mensuration- Solids
by Total Gadha - Saturday, 13 September 2008, 01:26 PM
  Hi AK,

the 'l' (slant height) is not the same in the two cones if I am guessing right what you are thinking. smile

Total Gadha
Re: Mensuration- Solids
by A K - Saturday, 13 September 2008, 09:01 PM seems I have goofed up with my concepts.
Yes, I misinterpreted the 'l.'
Re: Mensuration- Solids
by nimish vikas - Saturday, 13 September 2008, 11:46 PM
  TG sir
pls tell me whether this and my earlier answers are correct.
Q.9. 9pi( 16 + 9sqrt(3) )
instead of applying much brain i have done it thru integration which is quiet straight forward. let r= radius of any slice and h=distance to the slice from centre.
vol of elementary (slice+dh) = pi.r^2.(dh) = pi.(6^2-h^2).(dh)
..........................dv = 36pi.(dh) - h^2.(dh)
required vol = integration of dv
.............= 36pi[h] - pi.[h^3]/3 limits h= -6 to +6cos30
.............= 9pi( 16 + 9sqrt(3) )

kindly tell me a method without integration

Re: Mensuration- Solids
by nimish vikas - Tuesday, 16 September 2008, 03:25 PM
  Hi TG
Kindly tell whether these are sound alternatives to integration for Q.9.
Since you have given the angle as 60 degree i cannot help but imagine that there must be a very simple alternative way such as inscribing a hexagon (in 3D, regular dodecahedron) or some other solid figure in the sphere. But these figures will not have circular base of cut portion. hence this approach will only give approx. answer.
Another alt is : the cut portion is divided into several congruent pyramids( each of which in one plane cut across centre subtend an angle pi/n at the centre where n is the no. of pyramids in that plan.)
here centre is the centre of the base circle of the cut portion.
Now taking the limit n tends to infinity should give the answer.

Integration was much simpler and shorter than this.
Kindly tell us the simplest way.

Re: Mensuration- Solids
by abhishek chawla - Tuesday, 23 September 2008, 12:27 AM


Can you please explain the first question?????????

Re: Mensuration- Solids
by abhishek chawla - Tuesday, 23 September 2008, 12:27 AM


Can you please explain the first question?????????

Re: Mensuration- Solids
by abhishek chawla - Thursday, 2 October 2008, 03:03 PM

TG we are not getting replies too often from ur side???????


Hop u r not busy with classes

Re: Mensuration- Solids
by ravi mr. - Wednesday, 8 October 2008, 01:00 PM

1:  17 7/9 minute
2:  [[2√3+√(3+√3)]a^2]/4
4:  (d).o
5:  A
6:  (b)2  (1sphere,2cylinder);(1sphere,1cylinder,3cone)
further soln. later.......for explanation(if needed) write  here: :-)
Re: Mensuration- Solids
by ravi mr. - Wednesday, 8 October 2008, 02:29 PM
  7: (c)102
8: (c)
9: (128-9√3)π

hey TG!!!
by ritika bajaj - Wednesday, 8 October 2008, 10:14 PM
  hii tg

i need ur m gearing up myslf for CAT 09 but the problem  iam facing is how to start and from whr to start.. wen i read ur forums an thn awsm replies for each, i rly get disappointd tht y cn they solve it an nt me? sad sad sad

cn u plz telme wat to do an hw to do??

i rly want to increase my lvl

need u



Re: hey TG!!!
by amogh ranade - Friday, 17 October 2008, 09:47 AM
  sir, please post the solutions
Re: Mensuration- Solids
by shanti gattineni - Thursday, 21 May 2009, 03:34 PM
  the correct answers??
Re: Mensuration- Solids
by tulasi ram - Friday, 22 May 2009, 04:56 PM
  Hi my answers are :
1) 42.22 mins   [volume in lower cone is 19/27 times of total volume cool
9)272.7 (i got different answer here from all of u,
volume of portion cut out is:
i got these values in this way:
now from total if subtract this i m gettin 272.7,
plzz help some one  )
Re: Mensuration- Solids
by Rajesh Kumar - Sunday, 24 May 2009, 06:01 PM

Hi TG,

Answers please....

Re: hey TG!!!
by Abhishek Goyal - Tuesday, 26 May 2009, 10:53 AM
  just try continuously in building ur basics
u will be easily able to do it
Re: Mensuration- Solids
by nimish vikas - Monday, 29 June 2009, 11:34 AM
  Ram, You are subtracting cone from sphere which is not asked.
In a circle if you subtract area of triangle (or hexagon) you will get the required area. But in sphere you will not get it. Either used solid angles or integration. i think the answer given by me above is correct.
Re: Mensuration- Solids
by Pravin Vaidya - Thursday, 20 May 2010, 09:39 PM

Answer to the first question is (19/27)*60  min.....

Please find below the approach:

Question says that , how long does it take for the depth of the sand in lower cone to be half of the depth of sand i upper cone, it indirecty means how much time will it take for the depth in lower cone to be (1/3)rd of total depth (height of cone).

flow rate = V (m3/hr)    ---------------(1)


where, V= Total volume of cone [ (1/3) Pi *R2 * h]

where R- radius of the base of cone

      h - height of cone

at height = h/3 , the radius of cone is 'r'

thus , at height = h/3, volume of sand in lower cone will be

v= [ (1/3) Pi *(R2 +r2+rR)* (h/3)] ----- (2)

therefore, total time required will be,

T = v/ Flow rate

  = (R2 +r2+rR)/(R2 *3) ------ (3)


From the similiraty of triangle , we can deduce that,

R/r = 3/2


if we put this condition in equation 3  , we get

T = 19/27  hr

= (19/27)*60  minutes..

Re: Mensuration- Solids
by vasu r - Thursday, 8 September 2011, 11:38 AM

only if the dimensions of the cuboid are same a sphere can circumscribe it..does it mean only in a cube a sphere can be circumscribed? the formula for the radius of the circumscribed sphere is given as sqrt(l2+b2+c2) l=b=h?

Re: Mensuration- Solids
by neha aggarwal - Thursday, 19 July 2012, 08:30 PM
  Hi Sir smile,

Please help me with this:
A vertical cylinder vessel contains water in it up to height of √3 unit. The cylinder is then tilted till its axis make 30 degrees with the vertical and the level of water just covers the base of the vessel. The radius of the base of the vessel is

(a) 1 (b) 3 (c) √3/2 (d) 2/√3 (e) none of the foregoing

Pls explain wit diagram if possible..

Re: Mensuration- Solids
by dhwani bhatt - Wednesday, 8 August 2012, 05:55 PM
  TG Sir ,
Where I can find the solutions to above questions ?
Re: Mensuration- Solids
by Shubham Sharma - Wednesday, 3 July 2013, 06:19 AM
  Sir can u provide pdf on each quant lessons...
Pls sir,
thanks in advance...