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[Nareshkumar j]

If an integer n is to be chosen at random from the integers 1 to 96,
inclusive, what is the
probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

OA is "D". Is there any shorcut to slove these kind of plms?

[Nareshkumar j]

If n and k are positive integers, is n divisible by 6?
(1) n = k(k + 1)(k - 1)
(2) k – 1 is a multiple of 3.

Is there any shorcut for these kind of plms?

[Nareshkumar j]

At a garage sale, all of the prices of the items sold were different.
If the price of a radio
sold at the garage sale was both the 15th highest price and the 20th
lowest price among
the prices of the items sold, how many items were sold at the garage
sale?
A. 33
B. 34
C. 35
D. 36
E. 37

Could anyone help me in sloving the above pblm?

[Nareshkumar j]

The numbers x and y are three-digit positive integers, and x + y is a
four-digit integer.
The tens digit of x equals 7 and the tens digit of y equals 5. If x <
y, which of the
following must be true?
I. The units digit of x + y is greater than the units digit of either
x or y.
II. The tens digit of x + y equals 2.
III. The hundreds digit of y is at least 5.
A. II only
B. III only
C. I and II
D. I and III
E. II and III

[Nareshkumar j]

[NEETHU]

Case 1 :If n is even, n(n+1)(n+2) will be divisible by 8.

Number of even numbers between 1 and 96 = 48

Case 2: If n is odd, and n+1 is a multiple of 8, then n(n+1)(n+2) will be divisible by 8 (Eg: numbers like 7*8*9, 15*16*17)

Number of such groups between 1and 96 = 12

Therefore probability = (48 + 12)/96 = 5/8

Thanks

Neethu

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[Kripa Tewari]

good question....
n(n+1)(n+2) will be divisible by 8 for all even numbers, i.e. total 48 numbers
also cases in which (n+1) is a multiple of 8 will be divisible by 8
for ex: n=7,15,23.. i.e. total 12 numbers

so total number of such cases is 48+12=60
probability = 60/96 = 5/8

Regards

Kripa

On Sat, Dec 17, 2011 at 11:11 PM, Nareshkumar j <j.nares...@gmail.com> wrote:

[NEETHU]

Answer to garage problem:

Case1: X is the 15th highest price. This means if we arrange everything in decreasing order of prices, there are 14 values greater  than X

Case2 : X is the 20th lowest price. This means if we arrange everything in increasing order of prices, there are 19 values less than X.

Therefore number of items = 14+ 19 + 1 =  34. (Adding 1 because, you have to count X as well.)

Thanks

Neethu

[Nareshkumar j]

The infinite sequence a1, a2,…, an,… is such that a1 = 2, a2 = -3, a3
= 5, a4 = -1, and an =
an-4 for n > 4. What is the sum of the first 97 terms of the sequence?
A. 72
B. 74
C. 75
D. 78
E. 80

I don't know why iam taking more time for these kind of problems!!!!
could any one help me on these?

[NEETHU]

Answer to "Divisibility by 6" question.

General Rule: Any 3 consecutive numbers will be divisible by 6. So answer is A!!

[amith kumar]

an =an-4 .This is the crux of the problem so a1=a5 and a2=a6 a3=a7 a4=a8 a5=a6

So the sequence is basically a repetition of a1 a2 a3 a4 The sum is 2+(-3)+5+(-1)=3.So there are 97 terms this sequence of 4 numbers will be repeated for 24 time + the first term in the sequence ie the 97th term
So the sum of the 4 terms* 24 times + the 97th term(which is a1=2)

3(which is the sum of 4 terms)*24(times the sequence is repeated)+2(the 97th term)=74

Hope this is clear

On Sat, Dec 17, 2011 at 11:32 PM, Nareshkumar j <j.nares...@gmail.com> wrote:

The infinite sequence a1, a2,…, an,… is such that a1 = 2, a2 = -3, a3

= 5, a4 = -1, and  for n > 4. What is the sum of the first 97 terms of the sequence?

A. 72
B. 74
C. 75
D. 78
E. 80

I don't know why iam taking more time for these kind of problems!!!!
could any one help me on these?

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Rgds
Amith Ravindra

[amith kumar]

What would the ans we if (2) is N-1 instead of k-1

--
Rgds
Amith Ravindra

[NEETHU]

If that is the case then n-1 = 3x (x can be any integer)

ie. n= 3x +1

3x+1  is not divisible by 6. Hence the question can be answered from stmnt 2 alone.(But then there is a conflict between the answers we got from stmt 1 and stmt 2. So this cannot be true. Stmt 2 should be k-1 and not n-1)

Thanks

Neethu