Official Quant thread for CAT 2013

MBA CAT 2024
19766
@pavimai said:
four horses are tethered at four corners of a square plot of side 14 metres so that the adjacent horses can just reach one another .there is a small circular pond of area 20m^2 at the centre..find the ungrazed area??
196 - 49pi -20..
19767
@amresh_maverick

Yeah it will be 56..

Given, Rem[2^2^2003/100]= ?

Now, Rem[1^20/50] = 1...Hence, Rem[2^2^(20k + r)/100] = Rem[2^2^(20k + 3)/100]

=>Rem[2^8/100]= 56..
19768
@amresh_maverick said:
first correct2nd OA : 56
2^2^2003=((2^2^10)^200)^2^3
=((2^1024)^200)^8
=2^76*8
=2^4608
=( 1024)^460 *2^ 8
=76*56=56
19769
@pavimai said:
how many natural numbers are there with 24 factors where 2,3,5 are the only prime factors??
2power a (3 power b ) (5power c) Let this be n..

(a+1)(b+1)(c+1) = 24...
Now we will have to see how many values satisfy this
In office so unable to the full question....
19770
@pyashraj said:
@amresh_maverickYeah it will be 56.. Given, Rem[2^2^2003/100]= ?Now, Rem[1^20/50] = 1...Hence, Rem[2^2^(20k + r)/100] = Rem[2^2^(20k + 3)/100]=>Rem[2^8/100]= 56..
Rem[2^2^(20k + r)/100] = Rem[2^2^(20k + 3)/100] - plz explain this
19771
A husband and wife has a combined age of 91.the husband is now twice as old as his wife was when he was as old as she is now. How old are they?
19772
@amresh_maverick said:
1> RIghtmost non zero digit of 15!2>last two digits of 2^2^2003
2) 8.??
19773
The age of a father was three times that of his son when the father was as old as the son is now.find ratio of present ages of father and son????

plzz xplain them not good at solving ages problems
19774
@jaswik said:
A husband and wife has a combined age of 91.the husband is now twice as old as his wife was when he was as old as she is now. How old are they?
52 and 39...ratio 4:3 hoga.

regards
scrabbler
19775
@scrabbler yah thats the oa but how??
19776
@jaswik said:
The age of a father was three times that of his son when the father was as old as the son is now.find ratio of present ages of father and son????plzz xplain them not good at solving ages problems
3 : 2 should be

So basically if the son then was x then father today is 3x. Now the middle age i.e.2x would be the age of father then or son now.

Re-reading that looks incoherent 😞 Can't frame it better though.

regards
scrabbler
19777
@scrabbler yah even i got it to be as 3:2 but the oa was 5:3 so confused between these two questions
19778
@scrabbler checked the oa again it was a qsn from aimcat
19779
@jaswik said:
The age of a father was three times that of his son when the father was as old as the son is now.find ratio of present ages of father and son????plzz xplain them not good at solving ages problems
2:5..?
19780
@jaswik said:
@scrabbler yah thats the oa but how??
Same logic as I wrote for the other problem.

The description boils down to: if wife's age then is a, wife's now (and husband's then) is b and husband now is c, then a, b, c are in AP. So b is average of a and c.

For this specific problem, what I did is: let the two values (H now and W earlier) be 2x and x. Henc the number in between, 1.5x, is the age of H then of W now. Hence H:W now is 2 : 1.5 = 4 : 3. Total 7 parts is 91, so 4 and 3 parts are 52 and 39.

regards
scrabbler
19781
@scrabbler said:
52 and 39...ratio 4:3 hoga.regardsscrabbler
bhai yeh thoda explain kar do.. whats the logic here..
19782
@jaswik said:
@scrabbler yah even i got it to be as 3:2 but the oa was 5:3 so confused between these two questions
OA could also be wrong. It has been known to happen?

regards
scrabbler
19783
@iLoveTorres said:
bhai yeh thoda explain kar do.. whats the logic here..
Explained just above your post. Not very coherent, kind of intuitive/no-pen-paper approach. Hard to put those into words :(

regards
scrabbler
19784
@sbharadwaj oa:5:3
19785
@scrabbler source aimcat 1305..