Official Quant thread for CAT 2013

MBA CAT 2024
22007
@Dexian said:
How many 4 digit numbers of the form AABB r there where AABB is a square of a natural number?
7744 cat 2008 question if m not wrong.
22008
@viewpt said:
7744 cat 2008 question if m not wrong.
haan.... u r ri8
22009
@albiesriram said:
1??options..a) 1478b)1296c) 1728d)1925e) 3850
Bro what is the OA for this? is it c)1728? i am getting a weird pattern
22010
@albiesriram said:
1??options..a) 1478b)1296c) 1728d)1925e) 3850
Bro what is the OA for this? is it c)1728? i am getting a weird pattern
22011
@albiesriram said:
In one of the scams, 12 ministers agree to share the money. They agree that the kth minister will walk away with k/12 of the money remaining. Each minister comes and takes away his/her share in a diplomatic way. What is the share for 12th minister ,if all the ministers walk away with a Whole number of money, and the money gained through the scam is the smallest satisfying the above condition. ?
11*5*5*7 = 1925
22012
@jain4444 said:
11*5*5*7 = 1925
bro can you explain your approach please?
22013
@nole said:
@scrabbler I understood about joining diagonals and proving congruent ( since it is standard figure like kite,we will prove congruency by aplying SSS rite?) but after that u said we get the formulae area = 1/2 * product of diagonals .how ?
Arre try and prove it for yourself 😃 That way you'll remember it.

(Hint: Start with any quad with perp diags. Draw 4 lines, 2 parallel to each diagonal, through the 4 vertices. It will form a rectangle. Aage soch samajhke kar le!)

regards
scrabbler
22014
@Dexian said:
isme the last guy will take away the 12th of the remaining or will take away whole of the sum left after 11th.....
n/12 = 12/12 = whole of the sum. This is the last multiplier that gave me 12!/12^12...

regards
scrabbler
22015
@iLoveTorres said:
bro can you explain your approach pleas

it seems @scrabbler bhai got the nerve of this question

scrabbler bhai explain karo yaar
22016
@scrabbler said:
n/12 = 12/12 = whole of the sum. This is the last multiplier that gave me 12!/12^12...regardsscrabbler
bro mujhe ek alag pattern aaya hai.. shall i type it out so that v can figure out the logic behind it?
22017
@jain4444 said:
it seems @scrabbler bhai got the nerve of this question scrabbler bhai explain karo yaar
From your answer it seems same logic as mine...1/12 * 2/12 *3/12 *4/12....*12/12 i.e. 12!/12^12 is the fraction...on cancelling numerator is 5 * 7 * 5 * 11 and denominator is 12^7 (I think) so the minimum possible integer values are (hopefully) 12^7 total and 5 * 7 * 5 * 11 = 1925 for the last guy. Can't prove, exactly 😞 but given answer choices woh le liya...

regards
scrabbler
22018
@iLoveTorres said:
bro mujhe ek alag pattern aaya hai.. shall i type it out so that v can figure out the logic behind it?
let say our answer = A

money till 12th = A

(A + x)*11/12 = x
=> x = 11A
so , money till 11th = 11A + A = 12A

like this we need to multiply resultant number by 5 at 7th minister when (12 - 7) = 5 , then by 7 at 5th minister(12 - 7 = 5) , then by 5 at 2nd minister(12 - 2 = 10) and then at last by 11 at 1st minister(12 - 1 = 11)

so , 5*7*5*11 = 1925
22019
@iLoveTorres said:
bro mujhe ek alag pattern aaya hai.. shall i type it out so that v can figure out the logic behind it?
Please do...I will try....

regards
scrabbler
22020
If, 4^y=5.
3^(2x)=4.
5^z=8.
Find value 9^(yxz) *3^2=...?
22021
@scrabbler said:
n/12 = 12/12 = whole of the sum. This is the last multiplier that gave me 12!/12^12...regardsscrabbler
the first person will take 1/12th of the whole.. the second person will take away 11/144th part .. third will take away 133/1728th part..
meine isse thoda analyse kiya toh it is falling in a pattern like
from second person onwards the part the guy takes is 12-1/12^2
3rd person (12^2-(12-1)/12^3
so for the 12th person it will be (12^11-(12^10-(12^9-(12^8-(12^7-(12^6-(12^5-(12^4-(12^3-(12^2-(12-1))/12^12

(12^11-12^10+12^9-12^8+12^7-12^6+12^5-12^4+12^3-12^2+12-1)/12^12
(12^10(11)+12^8(11)+12^6(11)+12^4(11)+12^2(11)-1)/12^12
(12^8(11)(144+1)+12^4(11)(144+1)+1583)/12^12

iske baad i was wondering what have i done.. so would love to kno any mistake or if i have applied wrong logic
22022
@jain4444 said:
If, 4^y=5.3^(2x)=4.5^z=8.Find value 9^(yxz) *3^2=...?
72
22023
@jain4444 said:
If, 4^y=5.3^(2x)=4.5^z=8.Find value 9^(yxz) *3^2=...?
72????
22024
@jain4444 said:
If, 4^y=5.3^(2x)=4.5^z=8.Find value 9^(yxz) *3^2=...?
72?
22025
@jain4444 said:
11*5*5*7 = 1925
OA is 1925
22026
@iLoveTorres said:
the first person will take 1/12th of the whole.. the second person will take away 11/144th part .. third will take away 133/1728th part..meine isse thoda analyse kiya toh it is falling in a pattern like from second person onwards the part the guy takes is 12-1/12^23rd person (12^2-(12-1)/12^3so for the 12th person it will be (12^11-(12^10-(12^9-(12^8-(12^7-(12^6-(12^5-(12^4-(12^3-(12^2-(12-1))/12^12(12^11-12^10+12^9-12^8+12^7-12^6+12^5-12^4+12^3-12^2+12-1)/12^12(12^10(11)+12^8(11)+12^6(11)+12^4(11)+12^2(11)-1)/12^12(12^8(11)(144+1)+12^4(11)(144+1)+1583)/12^12iske baad i was wondering what have i done.. so would love to kno any mistake or if i have applied wrong logic
if we tend to size down the problem for 3 ppl thsn acc to ur approach... 3rd guy will take away 3^2-(3-1)/3^3....

if the total sum is 3^3 = 27 then its 7....

but 27/3=9 I Guy
18/3=6 II guy
12 III guy....

kuchh to missing hai na......
is assumption of 27 as total right?????