Official Quant thread for CAT 2013

MBA CAT 2024
13763
@PrinceSameer said:
Approach for N! is completely divisible by 13^52.what is sum of the digits of the smallest such number N?
Read previous posts properly before submitting the next one.
13764
a+b+c+d+e=8
a^2+b^2+c^2+d^2+e^2=16. [a,b,c,d,e are real numbers]
Find the maximum value of e.
13765
@bs0409 said:
a+b+c+d+e=8a^2+b^2+c^2+d^2+e^2=16. [a,b,c,d,e are real numbers]Find the maximum value of e.
Maximum value of e is 3.
13766
@bs0409 said:
a+b+c+d+e=8a^2+b^2+c^2+d^2+e^2=16. [a,b,c,d,e are real numbers]Find the maximum value of e.
3?
13767
@pavimai said:
N! is completely divisible by 13^52.what is sum of the digits of the smallest such number N
16 is it ?
13768
@doctoriim said:
Maximum value of e is 3.
@mailtoankit said:
3?
Show working......
13769
@bs0409 said:
a+b+c+d+e=8a^2+b^2+c^2+d^2+e^2=16. [a,b,c,d,e are real numbers]Find the maximum value of e.
Is it 16/5??
applied cauchy-swartz inequality theorem...
13770
@bs0409 said:
a+b+c+d+e=8a^2+b^2+c^2+d^2+e^2=16. [a,b,c,d,e are real numbers]Find the maximum value of e.
a+b+c+d = 8-e

(a^2+b^2+c^2+d^2) (1 +1 +1 +1) >=( a+b+c+d)^2 (cauchy scwhartz inequality)
=> (16-e^2).4 >=(8-e)^2
=>64 - 4e^2 >=64 + e^2 - 16e
=> 5e^2 -16e
=> e

P:S @bs0409 .. Hello arijit bro... After months we are back at this thread together...
13771
@Logrhythm said:
Is it 16/5??applied cauchy-swartz inequality theorem...
kaise hoga??
meine to hit n trial lagaya tha..galat hai wo
13772
@mailtoankit said:
kaise hoga??meine to hit n trial lagaya tha..galat hai wo
bhai cauchy lagana chahiye aese mein...

Acc to cauchy-shwartz inequality

(x1^2+x2^2+.....+xn^2)(y1^2+y2^2+.....+yn^2) >= (x1y1+x2y2+.....+xnyn)^2

So for this question

a^2+b^2+c^2+d^2 = 16 - e^2
and a+b+c+d = 8-e
now apply cauchy
(a^2+b^2+c^2+d^2)(1^2+1^2+1^2+1^2) >= (a+b+c+d)^2
(16-e^2)*4 >= (8-e)^2
e =

13773
@allan89 said:
a+b+c+d = 8-e(a^2+b^2+c^2+d^2) (1 +1 +1 +1) >=( a+b+c+d)^2 (cauchy scwhartz inequality)=> (16-e^2).4 >=(8-e)^2=>64 - 4e^2 >=64 + e^2 - 16e=> 5e^2 -16e=> eP:S .. Hello arijit bro... After months we are back at this thread together...
Ki baba....!!! Kemon cholche tomar??
So, u applied to any univ or waiting for CAT'13?
I am assuming your company joining letter will come soon enough.

Find the total number of values of n for which n^2 + 24n + 21 is a perfect square.Where n is natural number.
13774

"If a student weighing 70 kgs joins a group of n students, the average of the group increases by 1 kgs. If the new student weighed 55 kgs, the average of the group would have declined by 2 kgs. Find n"


3, 4, 5, 6, 7
13775
@bs0409 said:
Ki baba....!!! Kemon cholche tomar??So, u applied to any univ or waiting for CAT'13?I am assuming your company joining letter will come soon enough.Find the total number of values of n for which n^2 + 24n + 21 is a perfect square.Where n is natural number.
n^2 + 24n + 21 = p^2
(n+12)^2 - 123 = p^2
(n+12)^2 - p^2 = 123
(n+12+p)(n+12-p) = 123
(n+12+p)(n+12-p) = 3*41 or 1*123

when (n+12+p)(n+12-p) = 3*41
n+12+p = 41 and n+12-p = 3
n+p = 29 and n-p = -9
n = 10 and p = 19

when (n+12+p)(n+12-p) = 123*1
n+12+p = 123 and n+12-p=1
n+p = 111 and n-p = -11
n = 50 and p = 61

so both cases are possible....hence for 2 values of n the expression becomes a prfct sq...

13776
@veertamizhan said:
"If a student weighing 70 kgs joins a group of n students, the average of the group increases by 1 kgs. If the new student weighed 55 kgs, the average of the group would have declined by 2 kgs. Find n"3, 4, 5, 6, 7
(n+1)(x+1) =nx+70
n+x+1=70
(n+1)(x-2)=nx+55
x-2n-2=55
3n =12
n=4
EDITED : Calculation mistake. thanks @Logrhythm

any shortcut via allegation anyone ???
13777
Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number should satisfy the following requirements:
a. The number should be divisible by 9.
b. If the most right digit is removed, the remaining number should be divisible by 8.
c. If then again the most right digit is removed, the remaining number should be divisible by 7.
d. etc. until the last remaining number of one digit which should be divisible by 1.
13778
@veertamizhan said:
"If a student weighing 70 kgs joins a group of n students, the average of the group increases by 1 kgs. If the new student weighed 55 kgs, the average of the group would have declined by 2 kgs. Find n"3, 4, 5, 6, 7
should be 4...

(70+nx)/(n+1) = x+1
=> x+n = 69 ---- 1
(55+nx)/(n+1) = x-2
=> x-2n = 57 ----- 2
Solve 1 and 2
x = 65 and n = 4

13779
@Logrhythm thanks man
13780
@veertamizhan said:
"If a student weighing 70 kgs joins a group of n students, the average of the group increases by 1 kgs. If the new student weighed 55 kgs, the average of the group would have declined by 2 kgs. Find n"3, 4, 5, 6, 7
4?
nx+70=(n+1)(x+2)
n+x=69

nx+55=(n+1)(x-2)
-2n+x=57

n=4
13781
@ankita14 said:
Ah this one didn't consider the vote of the eldest thief, didn't notice. But then I don't think A will ever get the coins. B will never agree to vote for A :/ or does the question mean after A dies, B becomes the oldest thief and hence his vote doesn't count
Ah too many questions..:P
I think I failed to communicate the right information.. The explanation given for 98 seems fine to me.. But the given ans is 97 only.. Will recheck the question..
13782
@bs0409 said:
Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number should satisfy the following requirements:a. The number should be divisible by 9.b. If the most right digit is removed, the remaining number should be divisible by 8.c. If then again the most right digit is removed, the remaining number should be divisible by 7.d. etc. until the last remaining number of one digit which should be divisible by 1.
bahut matha fod liya par nahi hua :banghead:
kisis se hua to please share solution