[sandeep jha]

anyone can help me on this....

Consider a cylinder of height h cms and radius r = 2/π cms as shown in the figure (not drawn to scale). A string of a certain length, when wound on its cylindrical surface, starting at point A and ending at point B, gives a maximum of n turns (in other words, the string?s length is the minimum length required to wind n turns.)



What is the vertical spacing in cms between two consecutive turns?


a. h / n b. h / √n c. h / n2 d. Cannot be determined with given information(cat-04) 4.

[hameed]

Originally Posted by sandeep jha

anyone can help me on this....

Consider a cylinder of height h cms and radius r = 2/π cms as shown in the figure (not drawn to scale). A string of a certain length, when wound on its cylindrical surface, starting at point A and ending at point B, gives a maximum of n turns (in other words, the string?s length is the minimum length required to wind n turns.)



What is the vertical spacing in cms between two consecutive turns?


a. h / n b. h / √n c. h / n2 d. Cannot be determined with given information(cat-04) 4.

I guess it has to be h/n only. because any other value of dist bet windings will result in a total hieght (dist between windings*n) not equal to h.

For the length of the string we can do the below..

For each turn the length shd be atleast 2*pie*r = 2*pie*(2/n) = 4*pie/n
For n number of turns the length shd be = 4*pie*n/n = 4*pie

Please do tell me if I am wrong.

Rock on!
hameed

[krishor]

Originally Posted by prasad1981

Hi Kishore,

Even I am shocked when I refered to CL Key. The answer for this question is as follows.

Sum of all 3 digit numbers = 494550 (We calcld the same)

Number of three digit numbers such that there is no odd digit = 4*5*5 = 100 (This makes sense!)

Sum = (2+4+6+*25*100 + (2+4+6+*20*10 + (2+4+6+*20*1 = 54400

Answer = 440150

How and why? Please explain. Also anyone who happens to see this post, please respond.

Wowww..... Thats such a nice way .....

(Sum of all possible digits in 100s place)*100*no of repetition cycles +
(Sum of all possible digits in 10s place)*10*no of repetition cycles +
(Sum of all possible digits in unit place)*1*no of repetition cycles

No of repetition cycle = 100/4 for 100s place
= 100/5 for remaining places

Hope you understood.


Where did we go wrong?

Step1. number of 3 digit numbers = 5p3 = 60 ( including first digit as 0)

here we assumed that a digit cannot repeat in 100s, 10s and unit digits.
actually it should be 5*5*5 = 125. We did this mistake througout our discussion.

CL method looks easiest.

PS: I have just started taking CL tests, so i do not have these questions.

~krishor

[jimmygoogle]

1. RAJU HAS 64 SMALL CUBES OF 1CM^3.HE WANTS TO ARRANGE ALL OF THEM IN A CUBOIDAL SHAPE,SUCH THAT THE SURFACE AREA WILL BE MINIMUM.WHAT IS THE DIOGONAL OF THIS LARGER CUBOID?
a)8 root 2 b)root 273 c)4root3 d)root129

2.THE RADIUS OF A CONE IS 2^0.5 TIMES THE HEIGHT OF THE CONE.A CUBE OF MAXIMUM POSSIBLE VOLUME IS CUT FROM THE SAME CONE.WHAT IS THE RATIO OF THE VOLUME OF THE CONE TO THE VOLUME OF THE CUBE?
A/3.18PI B/2.25PI C/2.35 D/CAN'T BE DETERMINED

[krishor]

Originally Posted by jimmygoogle

1. RAJU HAS 64 SMALL CUBES OF 1CM^3.HE WANTS TO ARRANGE ALL OF THEM IN A CUBOIDAL SHAPE,SUCH THAT THE SURFACE AREA WILL BE MINIMUM.WHAT IS THE DIOGONAL OF THIS LARGER CUBOID?
a)8 root 2 b)root 273 c)4root3 d)root129

minimum area = 16*6 = 96. i.e. it is a 4*4*4 cube.
diagonal = √[ (4√2)^2 + 4^2 ] = 4√3.

Was just wondering, what will be the maximum surface area possible?
is it when all cubes arranged side by side? 4*64 + 2 = 258 ?

[jimmygoogle]

Originally Posted by krishor

Wowww..... Thats such a nice way .....

(Sum of all possible digits in 100s place)*100*no of repetition cycles +
(Sum of all possible digits in 10s place)*10*no of repetition cycles +
(Sum of all possible digits in unit place)*1*no of repetition cycles

No of repetition cycle = 100/4 for 100s place
= 100/5 for remaining places

Hope you understood.


Where did we go wrong?

here we assumed that a digit cannot repeat in 100s, 10s and unit digits.
actually it should be 5*5*5 = 125. We did this mistake througout our discussion.

CL method looks easiest.

PS: I have just started taking CL tests, so i do not have these questions.

~krishor

now i am totally confused.
your formula works good to find out the sum of 2/3 digit numbers formed using the given set.

while calculating sum of all 3 digit numbers using {2,4,6,8,0}

we do this way:
number of 3 digit numbers formed with the repetition of digits allowed: 4*5*5 = 100
now average of given numbers * 100 = 400
0 carry 40
0 carry 44
4 carry 4
for hundred's digit we have to calculate avg of {2,4,6,8} => 5 * 10
so we get:
54 (50+4)
Ans: 54400 (am i right here?)

now the final ans: 494550 - 54400 = 440150

[sandeep jha]

Originally Posted by hameed

I guess it has to be h/n only. because any other value of dist bet windings will result in a total hieght (dist between windings*n) not equal to h.

For the length of the string we can do the below..

For each turn the length shd be atleast 2*pie*r = 2*pie*(2/n) = 4*pie/n
For n number of turns the length shd be = 4*pie*n/n = 4*pie

Please do tell me if I am wrong.

Rock on!
hameed

u r right hameed bhai.... bas do me one more favour....tell me how to solve below mentioned problem????

The same string, when wound on the exterior four walls of a cube of side n cms, starting at point C and ending at point D, can give exactly one turn (see figure, not drawn to scale).
The length of the string, in cms, is

a. √2n b. √17 n c. n d. √13 n

[krishor]

Originally Posted by jimmygoogle

now i am totally confused.
your formula works good to find out the sum of 2/3 digit numbers formed using the given set.

while calculating sum of all 3 digit numbers using {2,4,6,8,0}

we do this way:
number of 3 digit numbers formed with the repetition of digits allowed: 4*5*5 = 100
now average of given numbers * 100 = 400
0 carry 40
0 carry 44
4 carry 4
for hundred's digit we have to calculate avg of {2,4,6,8} => 5 * 10
so we get:
54 (50+4)
Ans: 54400 (am i right here?)

now the final ans: 494550 - 54400 = 440150

this method works best, when there is no repetition of digits.

But let me try for this scenario.
Sum of 3 digit number numbers formed using "2/4/6/8/0"
Step1. number of 3 digit numbers = 5^3 = 125 ( including first digit as 0)
Step2. average of the given digits * 125 = 500.
Step3.
Unit digit = 0, 50 carry forward
Tens digit = 0, 55 carry forward
100s digit = 5, 55 carry forward
Ans1 = 55500.

But we have included '3 digit' numbers starting with 0 also.
Now we need to subtract all 2 digit numbers possible from {2,4,6,8}
Step1. number of 2 digit numbers = 4^2 = 16
Step2. average of the given digits * 16 = 80.
Step3.
Unit digit = 0, 8 carry forward
Tens digit = 8, 8 carry forward
100s digit = 8, 8 carry forward
Ans2 = 880.

But sadly we have not yet excluded all two digit numbers whoes first digit is zero, as well as all two digit numbers whoes second digit is zero.

Sum of all 'Two digit' numbers whoes first digit is zero = 2+4+6+8 = 20.
Sum of all two digit numbers whoes second digit is zero = 20+40+60+80 = 200.
Ans3 = 220.

Ans = Ans1-Ans2-Ans3 = 54400

Did you get it Jimmy?

But CLs method rocks.

[krishor]

Originally Posted by sandeep jha

u r right hameed bhai.... bas do me one more favour....tell me how to solve below mentioned problem????

The same string, when wound on the exterior four walls of a cube of side n cms, starting at point C and ending at point D, can give exactly one turn (see figure, not drawn to scale).
The length of the string, in cms, is

a. √2n b. √17 n c. n d. √13 n

If you observe carefully, you need to find out 4 similar hypotenuses.
Lets take each dimension as n = 4.
in each right angled triangle height = 4/4 = 1.
Base = 4.
Hypotenuse = √17.
Ans 4√17 = √17n.

Hope you got it.

~krishor

[krishor]

Originally Posted by jimmygoogle

for hundred's digit we have to calculate avg of {2,4,6,8} => 5 * 10
so we get:
54 (50+4)
Ans: 54400 (am i right here?)

now the final ans: 494550 - 54400 = 440150

Did not observe before.
Yes it looks good for me. Looks like u just mastered the method.