Official Quant thread for CAT 2013

MBA CAT 2024
25608
@Narci said:
cbd ?
@nole said:
A shipping clerk has five boxes of different but unkown weights each weighing less than 100 kg,the clerk weighs the boxes in pairs.The weights obtained are 110,112,113,114,115,116,117,118,120 and 121 kgs.What is the weight,in kgs of the heaviest box ?a) 60b) 62c) 64d) can't be determined.
62
25609
@nole said:
A shipping clerk has five boxes of different but unkown weights each weighing less than 100 kg,the clerk weighs the boxes in pairs.The weights obtained are 110,112,113,114,115,116,117,118,120 and 121 kgs.What is the weight,in kgs of the heaviest box ?a) 60b) 62c) 64d) can't be determined.
25610
@viewpt can u share your solution ?
25611
@nole said:
@jai.tuteja answer is 62.@Dexian : can u explain what u did after getting value of c ?
A+B will always give u min weight... so A+B=110
similarly D+E will give u max weight... so D+E=121..
(A+B+C+D+E)=289
289-(110+121)=58.
25612
@Dexian said:
bhai iska solution post karo. na.......
15X^2 + 8Y^2 = 22XY

15X^2 - 22XY + 8Y^2 = 0

solving... we get X/Y = 4/5 or 2/3

now 0

for X/Y = 4/5.. so Y = 5,10,15,20....100 and X = 4,8,12....80... 20 ordered pairs..

for X/Y = 2/3.. so Y = 3,6,9... 99 and X = 2,4,6.... 66 .... 33 ordered pairs

so total 53 pairs..
25613
@nole said:
@viewpt can u share your solution ?
by equations.
=====================================

Q: 1*1!+ 2*2!+............................+ 20*20!=Y ; find the value of Y?
25614
@Dexian for D and E u tried two combinations.why didn't u try further combinations ? wanna know, coz that will save lot of time.
25615
@albiesriram I am only partially able to deduce the solution of this p-articular question...
Please share the options if you know...

AB > CD , and
25616
@albiesriram I am only partially able to deduce the solution of this p-articular question...
Please share the options if you know...

AB > CD , and
25617
@viewpt said:
by equations.=====================================Q: 1*1!+ 2*2!+............................+ 20*20!=Y ; find the value of Y?
21! - 1 ?
25618
@nole said:
@Dexian for D and E u tried two combinations.why didn't u try further combinations ? wanna know, coz that will save lot of time.
actually max is 121....
and C is 58... so they are the only 2 combinations possible for D and E...
any way it is only asked to find the max weight... so didnot bother to find A and B...
exact sequence shud be 54 56 58 59 62
25619
@viewpt said:
by equations.=====================================Q: 1*1!+ 2*2!+............................+ 20*20!=Y ; find the value of Y?
today only we discussed a problem of sililar kind... :)
21! - 1
25620
@nole said:
A shipping clerk has five boxes of different but unkown weights each weighing less than 100 kg,the clerk weighs the boxes in pairs.The weights obtained are 110,112,113,114,115,116,117,118,120 and 121 kgs.What is the weight,in kgs of the heaviest box ?a) 60b) 62c) 64d) can't be determined.
a1+a2=110
a1+a3=112
a4+a5=121
a3+a5=120
a3=a2+2
a4=a3+1
a4=a2+3
a2=110-a1
a3=112-a1
a4=113-a1
a5=8+a1
a1 has to be at max 54
a5 has to be atleast 61

2 possibilities
54,56,58,59,62
53,57,59,60,61

60+59=119(not present in the weights)

hence 62


25621
@viewpt sorry,but i don't know the answer to ur question.
25622
@viewpt said:
by equations.=====================================Q: 1*1!+ 2*2!+............................+ 20*20!=Y ; find the value of Y?
21! - 1
25623
8 litres are drawn from a cask full of wine and is then filled with water.This operation is performed three more times.The ratio of the quantity of wine now left in cask to that of water is 16:81.How much wine did the cask hold originally ?
25624

One from my side: Three cubes of edge lengths 8cm, 6m and 2cm are placed face to face to form a solid object. Find the difference between the maximum and minimum possible surface area of the solid object formed.

Team BV - Kamal Lohia

25625
@pakkapagal said:
A string of a certain length is cut at random at two points.Find the probabilty that the three pieces that result can be the three sides of a triangle.
Another way:

Let the string be of length 1.

The first cut has to be within 0 to 1/2.

If the first cut is very close to 0, say at 0.01, the probability that a triangle can be formed is near 0 (as the other cut has to be very precisely located, close to the centre, for the triangle to be possible)

If the first cut is close to 1/2, say at 0.49, then the probability that a triangle is formed is close to 1 (as the other cut can be almost anywhere and the three sides will still form a triangle!)

So we are going from 0 to 1 over the range. The average probability will be 1/2, and since the length of the range is also 1/2, we get 1/2 * 1/2 = 1/4.

regards
scrabbler
25626
@bodhi_vriksha said:
One from my side: Three cubes of edge lengths 8cm, 6m and 2cm are placed face to face to form a solid object. Find the difference between the maximum and minimum possible surface area of the solid object formed.Team BV - Kamal Lohia
72 cm^2?

regards
scrabbler
25627
@jain4444 said:
8 litres are drawn from a cask full of wine and is then filled with water.This operation is performed three more times.The ratio of the quantity of wine now left in cask to that of water is 16:81.How much wine did the cask hold originally ?
24 ?