Few Numericals

**sporty.arjun** · 07-23-2008, 06:13 PM

1.) For a party, 2 types of dishes are to be selected. For the first type of dish 2 are to be selected from 4 and for second type of dish 4 are to be selected from 5. In how many ways can these both(1st & 2nd) type of dishes selected?

2.)If (x + 1/x) / (1 + 1/x) = 99 then what is the value of (x + 1/x) / (1- 1/x) = ?

3) Given a series k, k-1, k+3, k+1, k+2. Find the ratio of mean and standard deviation?
A. k+1
B. k
C. 1
D. 1/k
E. 1/(k+1)

4)Find the total no. of ways in which a black square and a white square can be selected from a chess board such that neither of them lie in the same row nor in the same column.

5.)Find the total of only rectangles on a 5X5 square board
How many ways are there to split 100 one dollar bills to 3 people?
When 10^(22) + 1 is divided by 11, what is the remainder

6) 53 students are divided into P and Q groups. P has 7 batches of n students each. Q has x students of 5 batches or 6 batches with x students in 5 batches and x+1 students in 6th batch
Col A: x
Col B: n
Ans: C

7) If 0 > 7/m > m/7, then
Col A: m
Col B: 7

**taminkeu** · 07-24-2008, 01:12 PM

1) 2× (4×3) × (5×4×3×2)
2) Here is the farthest I could get. 99(x+1)/(x-1) I strongly believe that the answer should be depending on x. It cannot, to the best of my knowledge be a standalone value.
3) Mean = (k+ k-1+ k+3+k+1+k+2)/5 = k+1
(K – (k+1))2= 1, (k-1 – (k+1))2= 4, (k+3 – (k+1))2 = 4, (k+1 – (k+1))2 = 0, (k+2 – (k+1))2=1
Summing the above values we obtain 1+4+4+0+1 = 10
Sd= sqrt( 10/5)=sqrt(2)[/font]
Ans: mean/sd = sqrt(2)/k+1[/font]
Even though it doesn’t match any of the suggested answers, I think nothing was left out in computing the sd.
4) 64×20
Considering the first white square on the right corner of the chess board, there are 20 black squares that can be coupled with it and since there are 64 squares on the board, therefore there 20×64 ways.
5) Apparently this section has three questions:
a. The total only rectangles can be found only if the dimensions of the desired rectangles are given
b. There is only 3 ways to share that $100 bill.
c. (10^1 +1)=11+0, (10^2+1)=11×9+2, (10^3+2)=11×91+0, (10^4+1)=11×909+2 for any even power of 10, the remainder (10^22+1)=q×11+2
6) 7n+5x+(x+1)=53 x and n are integers consequently, x=4 and n=4
7)0>7/m means m is a negative number. Col B rules

**chestnut.cc** · 07-24-2008, 03:09 PM

1.) For a party, 2 types of dishes are to be selected. For the first type of dish 2 are to be selected from 4 and for second type of dish 4 are to be selected from 5. In how many ways can these both(1st & 2nd) type of dishes selected?

6*5 = 30 ways...

4)Find the total no. of ways in which a black square and a white square can be selected from a chess board such that neither of them lie in the same row nor in the same column.

We could select successively fewer squares from each row and column... 2*8! ways...

5.)Find the total of only rectangles on a 5X5 square board

The total no of rectangles would be 10C2 - 5... 45 -5... 40 rectangles...

**taminkeu** · 07-24-2008, 04:51 PM

sorry chestnut I misinterpreted the question. I thought of 1 dish that needs to be selected.
Still no clear enough about the number of rectangles in question 5

**chestnut.cc** · 07-24-2008, 06:45 PM

count all possible rectangles that can be formed from the square of finite dimensions... ie take all combinations of its dimensions... subtract those dimensions that correspond to squares... there will only be 5... (1,1) (2,2) (3,3) (4,4) (5,5)... hence the remaining no are all rectangles... of course these are rectangles of integral dimensions... otherwise there are an infinite no...

**sporty.arjun** · 07-24-2008, 07:10 PM

hey taminkeu....even i was havin d same problem....but very clearly the option doesnt match...so ther definitely is sumthin else... and 4 dat $100 question...its not a $100 bill but 100 bills of $1...

**sporty.arjun** · 07-24-2008, 07:23 PM

hey chestnut....isnt this d way 2 solve d rectangle sum....a 5X5 board is made of 6 horizontal and 6 vertical lines....2 form a rectangle(or any 4 sided figure(includeing square)) we select 2 horizontal and 2 vertical lines....i.e. 6C2 x 6C2=225...this is the total number of rectangles+squares....now eliminate number of squares...1x1 are 25 and 5x5 are 1....my only doubt is how do u find no. of 2x2 3x3 and 4x4 squares....is there any shortcut.....

**chestnut.cc** · 07-24-2008, 07:42 PM

what you have described is the same thing... just take combinations of all possible dimensions... subtract the combinations that correspond to squares...

**scaredOfQuant** · 07-25-2008, 09:18 AM

@ sporty.arjun

to find the number of square in n*n dimensional figure simply do the following

1^2+2^2+3^2.....n^2 = [n*(n+1)*(2n+1)] /6

I am not sure how to find the number of rectangles. Chestnut can u please elaborate on that.

**prithibi** · 07-27-2008, 09:26 AM

3) Mean = (k+ k-1+ k+3+k+1+k+2)/5 = k+1
(K – (k+1))2= 1, (k-1 – (k+1))2= 4, (k+3 – (k+1))2 = 4, (k+1 – (k+1))2 = 0, (k+2 – (k+1))2=1
Summing the above values we obtain 1+4+4+0+1 = 10
Sd= sqrt( 10/5)=sqrt(2)[/font]
Ans: mean/sd = sqrt(2)/k+1[/font]
Even though it doesn’t match any of the suggested answers, I think nothing was left out in computing the sd.

i also agree with u