Official Quant thread for CAT 2013

MBA CAT 2024
12201
@hari_bang

Let t be the time taken for Bilu to read any one of the caselet..

Then, Rate of answering the questions=t/12 ques/min..

Thus, Total Time taken by Bilu to finish off the paper=27*t/12 + 4t = 25/4*t min..

Now. New Time=25/4*t*9/10 = 45/8*t min

Let t' be the decreased time taken for Bilu to read any one of the caselet..

Thus, Total Time= 4t' + 25/4*t = 45/8*t

=>t'= 27/32*t

Required % change in reading speed= [(1/t' - 1/t)/(1/t)]*100= 500/27= 18.5%..

12202
@pyashraj said:
@hari_bangLet t be the time taken for Bilu to read any one of the caselet..Then, Rate of answering the questions=t/12 ques/min..Thus, Total Time taken by Bilu to finish off the paper=27*t/12 + 4t = 25/4*t min..Now. New Time=25/4*t*9/10 = 45/8*t minLet t' be the decreased time taken for Bilu to read any one of the caselet..Thus, Total Time= 4t' + 25/4*t = 45/8*t=>t'= 27/32*tRequired % change= [(t-t')/t']*100= 500/27= 18.5%..PS: Par mujhe lagta hai (t-t')/t*100 = 500/32 = 15.625% should be the answer..
why?
12203
@hari_bang

Coz the % change in time has to be calculated on the original time taken by Bilu..n not the new time.. Ya phir main galat hun..
12204
@pyashraj said:
@hari_bangCoz the % change in time has to be calculated on the original time taken by Bilu..n not the new time..
yr chodo
main bahut jaldi confuse ho jata huin...:):):)

12205
@hari_bang

Srry Yaar..Its said change% of reading speed..

Hence=(1/t' - 1/t)/(1/t)*100 = (32/27 - 1)*100 = 18.51%..
12206
@pyashraj said:
@hari_bangSrry Yaar..Its said change% of reading speed..Hence=(1/t' - 1/t)/(1/t)*100 = (32/27 - 1)*100 = 18.51%..
both sol is same...so i hve no prblm
12207
1. Find the sum and the no. of divisors of 540 excluding those which are perfect squares????
12208
@hari_bang said:
1. Find the sum and the no. of divisors of 540 excluding those which are perfect squares????
24 , 4..?
12209

540= 2^2*5*3^3

excluding perfect squares..so 5*3
Sum 24 number 2*2=4..
12210
@saurav5517 said:
540= 2^2*5*3^3excluding perfect squares..so 5*3Sum 24 number 2*2=4..
its nt ans?
12211
@hari_bang said:
its nt ans?
Wats the OA then?
12212
@saurav5517 said:
Wats the OA then?
this q is copied frm dis site
bt sol smjh nhi aaya?
12213
@hari_bang said:
1. Find the sum and the no. of divisors of 540 excluding those which are perfect squares????
Is it 1630 and 20?
12214
@deedeedudu said:
Is it 1525 and 20?
1630 and 20:)

12215
@deedeedudu said:
Is it 1525 and 20?
bro
hw did solve it?
12216
@hari_bang said:
1630 and 20
Yes yaar calculation mistake ho gayi thi its coming 1630 alright edited the post :)
12217

solution is dis


540 = 2^2*3^3*5

Total number of factors = 3*4*2 = 24.
number of factors which are perfect squares = 2*2 = 4.
Number of factors excluding squares = 24-4 = 20.

Sum of factors which are squares = (2^0+2^2)(3^0+3^2)(5^0) = 5*10*1 = 50

Sum of all factors = 7*40*6 = 1680.
Sum of factors excluding squares = 1630.
12218
@deedeedudu said:
Yes yaar calculation mistake ho gayi thi its coming 1630 alright edited the post
provide sol...
12219
@hari_bang said:
brohw did solve it?
540=2^2*3^3*5
Total factors=3*4*2=24
But we have to subtract those factors that r squares, those will be
1,2^2,3^2,(2*3)^2
Hence total factors that r not squares r 24-4=20
Now similarly sum=(1+2+4)*(1+3+9+27)*(1+5)=1680
But subtract sum of squares
1680-(1+4+9+36)=1630
12220
@deedeedudu said:
540=2^2*3^3*5Total factors=3*4*2=24But we have to subtract those factors that r squares, those will be1,2^2,3^2,(2*3)^2Hence total factors that r not squares r 24-4=20Now similarly sum=(1+2+4)*(1+3+9+27)*(1+5)=1680But subtract sum of squares1680-(1+4+9+36)=1630
got it:)thnx