Official Quant thread for CAT 2013

MBA CAT 2024
8201
@bullseyes said:
If the sum of ten non-negative integers is 48, what is the minimum value of the sum of the cubes of the ten integers?
All the ten integers should be as close as possible, so there will be eight 5 and two 4

So, lest value will be:- 8*5^3 + 2*4^3 = 1128
8202
@Torque024 said:
Maximize the repeating digit such that remaining digits to add be min. 9(5)+3 fits min sum=1152 In any other case it is comming more than that
you sure..? I also did it in same way.

5^3*8+4^3+4^3=1000+128=1128
8203
@Torque024 said:
Maximize the repeating digit such that remaining digits to add be min. 9(5)+3 fits min sum=1152 In any other case it is comming more than that

1128 is the answer
8204
@vijay_chandola said:
you sure..? I also did it in same way.5^3*8+4^3+4^3=1000+128=1128
@chillfactor sir nai clear kr dia, though we both were close. :/
8205
@gautam22 said:
sir i did same sir in dis d ques says 1978^n-1.....i cud not do dis
Still same approach, as 1977 is multiple of 3 but not 9
8206
N is the smallest number such that N/2 is a perfect square and N/3 is a perfect cube. Find the number of divisors of N.
8207
@vijay_chandola said:
N is the smallest number such that N/2 is a perfect square and N/3 is a perfect cube. Find the number of divisors of N.
20?
8208
@vijay_chandola said:
N is the smallest number such that N/2 is a perfect square and N/3 is a perfect cube. Find the number of divisors of N.
20
N= 2^3*3^4
8209
@vijay_chandola said:
N is the smallest number such that N/2 is a perfect square and N/3 is a perfect cube. Find the number of divisors of N.
20
num=6*6*3*3*2
8210
@vijay_chandola said:
N is the smallest number such that N/2 is a perfect square and N/3 is a perfect cube. Find the number of divisors of N.
smallest value of N =(2^3)*(3^4)

=>no. of divisors = (1+3)*(1+4) = 20.

ATDH.
8211
@bullseyes said:
If the sum of ten non-negative integers is 48, what is the minimum value of the sum of the cubes of the ten integers?
in such cases we want to make all the 10 integers as close as possible.

best way to accomplish this is 5*8+2*4 i.e 8 times 5 and 2 times 4

hence sum of cubes = 8*5^3 + 2*4^3 = 1000+128 = 1128.

ATDH.
8212
@chillfactor said:
As, both points are on the same side of the line y = xFrom any point on the y = x, point B and B' (mirror image) are equidistant.Now, in the question we have to minimize AP + PB, as PB = PB', we can say thatWe have to minimize AP + PB' (or AB'), so follow the straight line path and find the least possible distance.Required point will be intersection point of AB' and y = xAnother question on similar logic (check the attachment)Euler's number of 11 is 10and 51^52 when divided by 10 remainder is 1So, remainder is 50^1 or 6
angle of incidence =angle of reflection
6/x=y/(16-x)=(18-y)/z=4/(12-z)=k
x=6/k
y=16k-6
z=(24-16k)/k
put z value in
4=k(12-z)

k=1
x=6
y=10
z=8
P1W1=6rt2
W1W2=10rt2
W2W3=8rt2
W3P2=4rt2
Total=28rt2
8213
Weights of the 5 boxes are positive integers and their mean as well as median is 6. If the only mod is 8, then find the sum of weights of heaviest box and lightest box.
8214
@vijay_chandola said:
Weights of the 5 boxes are positive integers and their mean as well as median is 6. If the only mod is 8, then find the sum of weights of heaviest box and lightest box.
only possible distribution in this case

3,5,6,8,8


Hence sum (lightest+heaviest) = (3+8) = 11.

ATDH.
8215
@gautam22 said:
Dragons have to meet for a brainstorm in a convention center. The delegates have to be selected to provide the maximum efficiency of the brainstorm session. A dragon has any amount of heads, and for any N, any amount of N-headed dragons is available if needed. The problem is that the size of the convention center is limited so no more than 1000 heads can fit into the assembly hall. The intellectual power of a dragon pack is the product of head numbers of dragons in the pack. How should an optimum pack look like (the total number of dragons, the head number distribution)?
if I'm not misinterpreting then total no of dragons=40
30 dragons having 30 heads each
and 10 dragons having 10 heads each
8216
@vijay_chandola said:
Weights of the 5 boxes are positive integers and their mean as well as median is 6. If the only mod is 8, then find the sum of weights of heaviest box and lightest box.
3,5,6,8,8
req sum =3+8=11
8217
@vijay_chandola 11 ?
8218
@gautam22 said:
Dragons have to meet for a brainstorm in a convention center. The delegates have to be selected to provide the maximum efficiency of the brainstorm session. A dragon has any amount of heads, and for any N, any amount of N-headed dragons is available if needed. The problem is that the size of the convention center is limited so no more than 1000 heads can fit into the assembly hall. The intellectual power of a dragon pack is the product of head numbers of dragons in the pack. How should an optimum pack look like (the total number of dragons, the head number distribution)?
if i m getting question correct. there is no point in choosing 1 headed dragon as it will not increase the brainstrom. Brain strom will be maximum when we choose 3 headed dragon and 2 headed dragon.
coz no greater than that will be converted into smaller values
suppose x>3, can have 2 headed and (x-2) headed dragon
therefore, 2(x-2)> x

hence, optimum solution possible only with 3 and 2 headed dragon,
In this case too, we need 3 headed dragon more as two dragon with three head can be written as 3 dragon with 2 head, but brainstrom 3*3> 2*2*2

looking at it we have, 332 3-headed dragon and 2 2-headed dragon.

8219
@vijay_chandola said:
Weights of the 5 boxes are positive integers and their mean as well as median is 6. If the only mod is 8, then find the sum of weights of heaviest box and lightest box.
x,y,6,8,8
Now x+y+6+8+8=30
x+y=8
Only possible values such that x,yis 3,5
Hence sum=3+8=11
8220
@gautam22 said:
i don't think that number of n headed dragons shud be n..i guess dis is wat u r asking for.....but i m not sure.....plz correct me if i m wrong


YA thats why I was in the confused zone while giving the abuv solution.........btw in which direxion u r heading?