Official Quant thread for CAT 2013

MBA CAT 2024
14343
@veertamizhan let the numbers be -
n-9, n-7, n-5 , n-3, n-1 , n+1 , n+3, n+5 , n+7, n+9
now
n-9+ n-7+n-5 + n-3 + n-1 + n+1 + n+3+ n+5+ n+7+ n+9 = 180 + (n-9+ n-7+n-5 + n-3 + n-1 + n+1 + n+3+ n+5+ n+7+ n+9)/ 10
=> 10n = 180 + n
=> n = 20
therefore smallest + largest = n-9 + n+9 = 40
14344
@shattereddream said:
i was getting 1336 first
if you include 1 and 3000 , then 1336 !
14345
@bs0409 said:
ANS is 2^(N-1).......
yea so my answer was 8
N was 4
2^(4-1) = 8 = 2^(N-1) = answer...
14346
@albiesriram said:
exactly 60% or more than 60%??
exact 60%
14347
@ravi.theja said:
Which among 2^1/2, 3^1/3, 4^1/4, 6^1/6 and 12^1/12 is the largest?
take log of all

1/2*log 2 ~ 0.150
1/3*log 3 ~ 0.159
4^1/4 = 2^1/2 ~ 0.150
1/6 (log 2 + log 3) = 1/6 (0.307 + 0.477) ~ 0.130
1/12 (2*log 2 + log 3) = 1/12*(2*0.301 + 0.477) ~ 0.089

hence 3^1/3 is the largest...
14348
@Logrhythm said:
take log of all 1/2*log 2 ~ 0.1501/3*log 3 ~ 0.1594^1/4 = 2^1/2 ~ 0.1501/6 (log 2 + log 3) = 1/6 (0.307 + 0.477) ~ 0.1301/12 (2*log 2 + log 3) = 1/12*(2*0.301 + 0.477) ~ 0.089hence 3^1/3 is the largest...
ur solution suits ur name
14349
@IIM-A2013 said:
Every day asha's husband meets her at railway station at 6:00 pm and drives her to their residence.One day she left early from the office and reached railway station by 5:00 pm. She started walking towards her home,met her husband coming from their residence on the way and they reached home 10 mins earlier than usual time.For how long did she walk.(a) 1 hr (b) 50 mins (c) half hour (d) 55 mins
at 5:00 hubby s 1 hr away frm station..
both f dem move 2wards each other...meet..nd go home..time saved=10 min.
=time for to and fro journey of car for the dist covered by asha walkin..

asha walked for=60-[10/2]=60-5=55 min.s
14350
@ravi.theja said:
Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many suchnumbers are perfect squares?
all aabb form numbers ll b multiples f 11...... [a+b=a+b]

trying frm 33 n wards....nly 88..
14351
@bs0409 said:
OA-1250............x,y,z are natural numbers whose sum is 12.If maximum value of X*Y^2*z^3 is M, then how many numbers less than M is a co-prime of M?
x+y+z=12
x+y/2+y/2+z/3+z/3+z/3=12
max valu f x*y^2*z^3/[4*27]=2^6
max valu f x*y^2*z^3=2^6*2^2*3^3=2^8*3^3=M

no f coprimes to M less than M=2^8*3^3[1-1/2][1-1/3]=2^8*3^2=2304
14352
@Logrhythm
@Logrhythm said:
1275?? series -> -50x,-49x,-48x,....0,....48x,49x,50xcase1 - when 0 is picked....we just need to pick 2 complementary tems -> 50 wayscase2 - when 0 is not picked....we need to pick 2 +ve/-ve terms and a terma compensating their weight -> 50c2 ways = 1225so total = 1225+50 = 1275 ways
is dat correct method... may b i am wrong ..but.. u have wriiten 50c2 ..i.e selecting any 2 numbers from the 50 positive numbers or any two from the negative numbers... but wat if someone picks +50 and +49 as two numbers .. then there is no such negative number which can cancel out the sum of these two numbers and vice versa is possible...
OR ELSE i might hv misunderstood ur soln... :)
14353
@bs0409 said:
If the sum of 101 distinct terms in arithmetic progression is zero, in how many ways can three if these terms be selected such that their sum is zero ?
@bs0409
@Logrhythm
@grkkrg
-50,-49,-48,-47......-2,-1,0,1,2,3,4,5,.....46,47,48,49,50
case 1 :-
when one 0f the numbers is 50 , then total ways is 50c1 = 50 ways,
case 2:-
when numners are like two positive and one negative or vice versa.. then
if one of the numbers is 1 (like 1,48 ,-49):---
then with 1 => total ways 48 (out of 50 , 1 already selected and 50 cant be selected, so total 48 ways)
if one of the numbers is 2 :---
then with 2 => total ways 46 (case of one considered earlier, 2 already selected , 49 and 50 cant be selected as the sum then cant be nullified . so total 46 ways )
3 => 44 ways similarly,
4 => 42 ways,
.
.
24 => 2 ways (24, 25, -49) or (24,26,-50)
so , total ways of selcecting two positive numbers and one negative number is 48+46+44+42+...... +6+4+2 = 600 ways but same is possible for two negative and one positive number = 600 ways similarly
therefore total ways 50+600+600 = 1250 ways total.. :)

14354
@bs0409 said:
If the sum of 101 distinct terms in arithmetic progression is zero, in how many ways can three if these terms be selected such that their sum is zero ?
@bs0409
@Logrhythm
@grkkrg
-50,-49,-48,-47......-2,-1,0,1,2,3,4,5,.....46,47,48,49,50
case 1 :-
when one 0f the numbers is 50 , then total ways is 50c1 = 50 ways,
case 2:-
when numners are like two positive and one negative or vice versa.. then
if one of the numbers is 1 (like 1,48 ,-49):---
then with 1 => total ways 48 (out of 50 , 1 already selected and 50 cant be selected, so total 48 ways)
if one of the numbers is 2 :---
then with 2 => total ways 46 (case of one considered earlier, 2 already selected , 49 and 50 cant be selected as the sum then cant be nullified . so total 46 ways )
3 => 44 ways similarly,
4 => 42 ways,
.
.
24 => 2 ways (24, 25, -49) or (24,26,-50)
so , total ways of selcecting two positive numbers and one negative number is 48+46+44+42+...... +6+4+2 = 600 ways but same is possible for two negative and one positive number = 600 ways similarly
therefore total ways 50+600+600 = 1250 ways total.. :)

14355
@swapnil4ever2u http://www.pagalguy.com/posts/4435426

chk this....answer shld be 1250...
14356
@Logrhythm said:
@Logrhythm thats wat i am saying 1250 ..
14357
@Logrhythm how can you only get 25 cases which you have subtracted where there are no such complacent terms... in that case .. the case that u shoud have subtacted should have been 25c2 ie from 26 to 50... not only 25 cases ..
14358
@sunnychopra89 : we basically have to find the remainder of
1^1 + 2^2 + 3^6 + 4^24

from rest onwards it will be divisible by 2^3*5 i.e by 40

1 +4+16+16 = 37
14359
@ravi.theja said:
Which among 2^1/2, 3^1/3, 4^1/4, 6^1/6 and 12^1/12 is the largest?
multiply the powers by the lcm of their dinominators i.e lcm(2,3,4,6,12)numbers become 2^6 , 3^4 , 4^3, 6^2 , 12^1 .. clearly 3^4 is largest ..
therefore 3^1/3 is largest
14360
Find the sum of the remainders obtained when a number n is divided by 9 and 7 given that n is the smallest number that leaves remainders of 4,6,9 when divided by 13,11 and 15

1. 4
2. 5
3. 9
4. 6
14361
@shattereddream said:
Find the sum of the remainders obtained when a number n is divided by 9 and 7 given that n is the smallest number that leaves remainders of 4,6,9 when divided by 13,11 and 151. 4 2. 5 3. 9 4. 6
Solve using chinese remainder theorem....

If a number N, upon division with 13,11,15 leaves remainder 4, 6, 9 then We can find that number using chinese remainder theorem:

N can be written as 13x + 4 = 11y + 6 = 15z + 9

i) 13x+4= 11y+6 = N1

13x -11 y = 2

x = (11y +2 ) / 13=>

x is integer when y = 1

and so here x is 1

N1 = 17

ii) LCM (13, 11 ) *m + N1 = 15z + 9

143m + 17 = 15z + 9

z= (143m +8 ) /15
z = (8m + 8) 15 => 8(m +1 ) / 15=> m = 14

so we get z= 143

=> N = 15*134 + 9 = 2019

so, 2019 is the least such number.

dividing by 9 remainder will be 3 and dividing by 7 remainder will be 3...Sum of remainder will be 6.
14362
Alphonso, on his deathbed, keeps half his property for


his wife and divides the rest equally among his three
sons Ben, Carl and Dave. Some years later Ben dies
leaving half his property to his widow and half to his
brothers Carl and Dave together, shared equally. When
Carl makes his will he keeps half his property for his
widow and the rest he bequeaths to his younger brother
Dave. When Dave dies some years later, he keeps half
his property for his widow and the remaining for his
mother. The mother now has Rs 1,575,000.
74. What was the worth of the total property?
a. Rs. 30 lakh
b. Rs. 8 lakh
c. Rs. 18 lakh
d. Rs. 24 lakh



75. What was Carls original share?
a. Rs. 4 lakh
b. Rs. 12 lakh
c. Rs. 6 lakh
d. Rs. 5 lakh
76. What was the ratio of the property owned by
the widows of the three sons, in the end?
a. 7 : 9 : 13
b. 8 : 10 : 15
c. 5 : 7 : 9
d. 9 : 12 : 13