[RahulDaga89]

Originally Posted by Ankurb

option b)12
L=length, B=width, a= side of tile
x = number of tiles along L = L/a
y = number of tiles along B = B/a
acc. to condition:-
xy = 4(x+y)-8
solving for x:-
x = (8 - 4y)/(4-y)
only y = 12 will give an integer value of x.

hw cum xy = 4(x+y)-8
plz clarify

[ronnie_pg]

Originally Posted by divishth

Minimum Distance = root((4+5)^2 + 6^2) = 3*root13 = 10.81

Please explain the formula that you have used.
Thanks in advance.

[divishth]

Originally Posted by crack_2009

The sum of 20 distinct numbers is 801. What is their minimum LCM possible?

I'm Getting 360

360 = 2^3 * 3^2 * 5

Sum Of Factors = (2^0+2^1+2^2+2^3)*(3^0+3^1+3^2)*(5^0+5^1)
Sum = 15*13*6 = 1170

Total Factors Of 360 = 4*3*2 = 24

Now, Sum = 801, Given of 20 distinct numbers
So, we have to remove 4 numbers from 1170 such that Sum is 801,
So we have to remove 369 from 1170..And 369 must be sum of 4 distinct numbers....

So 369 = 360+1+3+5
Hence LCM = 360..Whats The OA..I think this only would be the OA

PS: Daga..Nice to see you here Man

[divishth]

Originally Posted by ronnie_pg

Please explain the formula that you have used.
Thanks in advance.

I've opened the cuboid..and made it a Rectangle, with sides (4+5) - being minimum and another side which is 6..Hence root(9^2 + 6^2) = 10.8

[RahulDaga89]

ya buddy. mene socha "y shud dev hv all d fun"

[ronnie_pg]

Originally Posted by divishth

I'm Getting 360

360 = 2^3 * 3^2 * 5

Sum Of Factors = (2^0+2^1+2^2+2^3)*(3^0+3^1+3^2)*(5^0+5^1)
Sum = 15*13*6 = 1170

Total Factors Of 360 = 4*3*2 = 24

Now, Sum = 801, Given of 20 distinct numbers
So, we have to remove 4 numbers from 1170 such that Sum is 801,
So we have to remove 369 from 1170..And 369 must be sum of 4 distinct numbers....

So 369 = 360+1+3+5
Hence LCM = 360..Whats The OA..I think this only would be the OA

PS: Daga..Nice to see you here Man

Please explain your approach.How you conclude it to be 360?

[Abhinav90]

Originally Posted by divishth

I'm Getting 360

360 = 2^3 * 3^2 * 5

Sum Of Factors = (2^0+2^1+2^2+2^3)*(3^0+3^1+3^2)*(5^0+5^1)
Sum = 15*13*6 = 1170

Total Factors Of 360 = 4*3*2 = 24

Now, Sum = 801, Given of 20 distinct numbers
So, we have to remove 4 numbers from 1170 such that Sum is 801,
So we have to remove 369 from 1170..And 369 must be sum of 4 distinct numbers....

So 369 = 360+1+3+5
Hence LCM = 360..Whats The OA..I think this only would be the OA

PS: Daga..Nice to see you here Man

In your solution,isnt the 360 you are getting the LCM of the removed factors and not the factors that make the sum 801??
I am also sceptic about why you started with 360? Can you do it by assuming any other number at the start ?

[harshkumar1]

Originally Posted by Lakshya13@@@

Heyyy... anybody help me wid disss....

If x and y are positive real numbers such that

21xy + 15x + 28y = 16

Then find the minimum value of 21(x + y)?
[A]7
[B]17
[C]21
[D]27
[E]11

21xy + 15x + 28y - 16 = 0
3x(7y + 5) + 4(7y + 5) - 20 - 16 = 0
(3x + 4) (7y + 5) = 36
For min x and y even (3x + 4) and (7y + 5) should be min
Hence 3x + 4 = 6
or, x = 2/3
and, 7y + 5 = 6
or, y = 1/7

21(x+y) = 21(2/3 + 1/7)
= 21(17/21) = 17

17(Option B)

[bhupi3149]

Chinese scientists, keeping up with their tradition of making long things like Chinese Wall, worked on a project to make a rope long enough to go around the moon, which is a perfect sphere. When Chinese astronauts finally laid it along moon’s equator, it was realized that it was a meter too long. The two ends of the rope were, however, joined together considering that one meter amongst a length of millions of meters would be insignificant.



If the rope were stiff enough to maintain a uniform distance from the surface all round the equator, how far would it stay away from the surface of the moon?

Choices

A. Approx. 10 mm




B. Approx. 15 cm




C. Approx. 10 m




D. Data insufficient




E. None of these

[shanks4mba]

Originally Posted by bhupi3149

Chinese scientists, keeping up with their tradition of making long things like Chinese Wall, worked on a project to make a rope long enough to go around the moon, which is a perfect sphere. When Chinese astronauts finally laid it along moon’s equator, it was realized that it was a meter too long. The two ends of the rope were, however, joined together considering that one meter amongst a length of millions of meters would be insignificant.
If the rope were stiff enough to maintain a uniform distance from the surface all round the equator, how far would it stay away from the surface of the moon?
Choices
A. Approx. 10 mm
B. Approx. 15 cm
C. Approx. 10 m
D. Data insufficient
E. None of these

is it approx 15 cm???
let the outer radius be R and inner b r
then
2pi(R-r) = 100 cm
hence R-r = 15.9 cm...