DS Question - 14

How many odd integers are greater than integer X and less than the integer Y?

1). there are 12 even integers greater than X and less than Y
2). there are 24 integers greater than X and less than Y

Answer -- B

From statement 1) -- We cannot determine about both X, Y being even or odd. ... hence insufficient

From statement 2) -- There are 24 integers between X and Y
Let it start with an even integer ...if so then it will end with an odd integer or if it starts with an odd integer then it will end with an even integer ...hence there will always be 12 odd and 12 even integers in the total of 24 consecutive integers
e.g consider X=2, Y=27 thus the series will be is 2 ... 26, 27
X=3, Y=28, thus the series will be 3, ... 27, 28
In each case, the total number of odd integers is the same ... hence sufficient

Problem solving - 14

Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed?

(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040

Answer - A -- generally answer for such questions is the sum of the series of integers from 1 to n+1 where n = number of items to be distributed.
In this ques n = 5, thus the answer is 21.
For n = 6 it will be 28. For n = 7 it will be 36.....

However here is the explanation based on the counting method...

1) 1 person gets all 5 donuts -
Possibility: 3.

2). 2 persons get all 5 donuts -
The one without donuts - possibility - 3.
The other 2 persons have 4 ways.
Hence in total -- 3*4 = 12.

3). All have donuts -
The way to divide can only be - 1, 2, 2; 1, 3, 1
But this arrangement can get interchanged midst 3 persons, hence it becomes --
(3!)/ (2) + (3!)/ (2) = 6.

(3!)/ (2) is needed because donuts are all same without difference.
3! is counting the same arrangement twice.

Hence the answer - 3 + 12 + 6 = 21.

Problem solving - 13

If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?

A) 25%
B) 50%
C) 62.5%
D) 72.5%
E) 75%

Answer - C
n is even - anytime n is even, it is divisible by 8
total nos using sequence theorm 96 = 2 + (#-1) 2,
hence # = 48
n is odd - again 48 no
but in 1-8, only one combination is divisible by 8, when n is 7,15, 22... hence 12 cases
probalility = possible outcomes / total outcomes = 48+12 / 96 = 60/96 = .625 = 62.5%

OR

we need to check for what values of n, n(n+1)(n+2) is divisible 8....
n(n+2) is divisible by 8 for all values of even numbers and there are 48 even nos in 1 to 96....
now the remaining part (n+1) is divisible by 8 for 12 odd numbers....such as 7, 15, 23, 31, 39.... to find this u can divide 96 by 8....
so totally 48 + 12 = 60 numbers are there between 1 to 96 for which n(n+1)(n+2) is divisible 8....
now when calculate the % it would be ( 60 / 96 ) * 100 = 62.5 %....

Problem solving - 12

If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p?

A. 10
B. 12
C. 14
D. 16
E. 18

Answer -- C
P = 1*2*3*.....*30
if u factorize P, what is the power of 3 ?
So we have to find out how many 3's are there in he product.
As is known 3,6,9,12,15,18,21,24,27,30 is total count of 10 numbers
but
9 = 3*3 thus 1 extra 3
18 = 3*3*2 thus 1 extra 3
27 = 3*3*3 thus 2 extra 3
Thus in total 10 + 1 + 1 + 2 = 14 3's are there in product.
Hence k is 14.

Problem solving - 11

A number is selected at random from first 30 natural numbers. What is the probability that the number is a multiple of either 3 or 13?

(A) 17/30
(B) 2/5
(C) 7/15
(D) 4/15
(E) 11/30

Answer -- B
The first 30 natural nos are 1,2,3.....28,29,30.
There are 10 multiples of 3 in the above range and there are 2 multiples of 13 in the same range.
Hence there are total 12 i.e (10+2) nos which can be either a multiple of 3 or 13.
Total numbers = 30.
So the probability is 12/30 = 2/5.

Problem solving - 10

A infinite sequence is 1, 11, 111, 1111, 11111, ..., what is the tens digit of the sum of the first 40 terms?

A) 1
B) 2
C) 3
D) 4
E) 9

Answer -- C
The sum of the units is 40. Then add 4 to the tens , whose sum is 39 and 3 is 43 which gives 3 as tens digit