Official Quant thread for CAT 2013

MBA CAT 2024
16063
@swapnil4ever2u said:
actually i dont have the OA .. but can u explain ur approaches ..
231 * 319 =3*7*11^2 * 29

no f factos ( f ) = 2 * 2 * 3 * 2 = 24

for its square (231 * 319)^2

no f factor ( F ) = 3^3 * 5 = 135

Some points.

1 ------------ N ----------------- N^2

DOTTED LINE REPRESENTS NO F FACTORS bw 1 to N to N^2
Also N is the middle factor of N^2 when all factors are arranged in increasing or decresing order.

so just find the
no f factors of N^2 less than N = (numbers of factors of N^2 - 1) / 2---------- EQ 1
factors of N less than N = N-1 ----- EQ 2

just subtract eq 1 - eq 2

here it is, (135 -1)/2 - (24-1)

= 67 - 23 = 44
16064
@techgeek2050 said:
answer is 22. see the attached fileED = d DB = cDC = aDF = bfrom the given areassince the heights are same, area will be in ratio of basesa/b = 2c/d = 5/4area of AED = y and ADF = x using the same concept as above(5 + x) / y = 5/4(8+y)/ x = 2solving them.. x = 10 and y = 12area remaining part = x + y = 10 + 12 = 22
how that ratio comes 5/4 it has to be 5/8 i thnk so ??
16065
@arnabdutta77 said:
1000!/10^n .What can be the largest possible integral value of n??
249?
1000!/10^n = 1000!/(2*5)^n
highest power of 5 = 249
16066
@mailtoankit said:
249?1000!/10^n = 1000!/(2*5)^n highest power of 5 = 249
Yes.
16067
@arnabdutta77 said:
1000!/10^n .What can be the largest possible integral value of n??
10 = 2*5

just find the max power of 2 or 5 that is present in 1000!
clearly powers of 2 greater than powers of 5.

So just find the power of 5 present in 1000!
this can be done by
1000/5 + 1000/25 + 1000/125 + 1000/625

= 200 + 40 + 8 + 1
= 249. ??
16068
@The_Loser said:
10 = 2*5just find the max power of 2 or 5 that is present in 1000!clearly powers of 2 greater than powers of 5.So just find the power of 5 present in 1000!this can be done by1000/5 + 1000/25 + 1000/125 + 1000/625= 200 + 40 + 8 + 1= 249. ??
Yes O.A. is 249 !!
16069
@ishu1991 said:
how that ratio comes 5/4 it has to be 5/8 i thnk so ??
10/8 = 5/4
16070
Let 20 x 21 x 22 x.... x 30 = A. If A is divisible by 10x, then find the maximum value of x.
A.3
B.4
C.5
D.6
16071
@techgeek2050 said:
answer is 22. see the attached fileED = d DB = cDC = aDF = bfrom the given areassince the heights are same, area will be in ratio of basesa/b = 2c/d = 5/4area of AED = y and ADF = x using the same concept as above(5 + x) / y = 5/4(8+y)/ x = 2solving them.. x = 10 and y = 12area remaining part = x + y = 10 + 12 = 22
shandaarrrr yaar...
16072
@The_Loser said:
231 * 319 =3*7*11^2 * 29no f factos ( f ) = 2 * 2 * 3 * 2 = 24for its square (231 * 319)^2no f factor ( F ) = 3^3 * 5 = 135Some points.1 ------------ N ----------------- N^2DOTTED LINE REPRESENTS NO F FACTORS bw 1 to N to N^2Also N is the middle factor of N^2 when all factors are arranged in increasing or decresing order.so just find the no f factors of N^2 less than N = (numbers of factors of N^2 - 1) / 2---------- EQ 1factors of N less than N = N-1 ----- EQ 2just subtract eq 1 - eq 2here it is, (135 -1)/2 - (24-1) = 67 - 23 = 44
bhai ye question mujhe kyun bta rha h.. maine nhi pucha.. maine wo area wala pucha tha..
16073
@ishu1991 said:
how that ratio comes 5/4 it has to be 5/8 i thnk so ??
bhai .. 10/8 => 5/4
16074
@techgeek2050 bhai ye mass point geometry se nhi ho sakta kya .. waise ur soln was better.. but mass point geo. kisi aur jagah kaam aa sakta h ..
16075
@ishu1991 said:
Let 20 x 21 x 22 x.... x 30 = A. If A is divisible by 10x, then find the maximum value of x.A.3B.4C.5D.6
4?
10^x hai ya 10*x?
16076
@negiSannu bhai samjha sakta h kya ?? ladder theorm
16077
@swapnil4ever2u said:
@techgeek2050 bhai ye mass point geometry se nhi ho sakta kya .. waise ur soln was better.. but mass point geo. kisi aur jagah kaam aa sakta h ..
sorry bhai.. no idea about mass point geometry here..
16078
@swapnil4ever2u said:
@negiSannu bhai samjha sakta h kya ?? ladder theorm
http://mathworld.wolfram.com/CrossedLaddersTheorem.html have a look at this.
16079
@mailtoankit said:
4?
10^x hai ya 10*x?
10^x hi hona chaie...
20*21*22*23*24*25*26*27*28*29*30
20 has one 5
25 has two 5's
30 has one 5
total four 5's
hence 10^4
16080
@mailtoankit said:
4?10^x hai ya 10*x?
same doubt i have thts y i put it here if it is 10^x thn OA is 4 but wth 10x i thnk this qstn is wrng
16081
@ishu1991 said:
same doubt i have thts y i put it here if it is 10^x thn OA is 4 but wth 10x i thnk this qstn is wrng
well agar 10^x hai then highest power of 5 is coming 4
30!/5^x ---> x = 7
19!/5^x ---> x = 3
7-3 = 4
16082
@TootaHuaDil said:
If K = 231 — 319, then how many positive divisors of K^2 are less then K but do not divide K?
some one please elaborate the solution.
K=3*7*11^2*29
K^2=3^2*7^2*11^4*29^2
Factors of K^2=3*3*5*3=135
factors of K=2*2*3*2=24
Factors of K^2 which are less than K but do not divide K=(135-1)/2 - (24-1)=67-23=44