X, Y and Z are three non-negative integers such that X+Y+Z = 36. How many solutions are possible for this equation where \(X >= Y\) (i.e., X is greater than or equal to Y)?

(A) 319

(B) 341

(C) 361

(D) 385

(E) 405

EDIT:

(1) Corrected from "positve integers" to "non-negative integers".
(2) Added: X greater than or equal to Y.

RESERVED................................
_________________

RESERVED................................
_________________

Triangle ABC is a right angled triangle with right angle at Vertex B. Triangle DEF is an equilateral triangle with side 10 cm. The lengths of sides (in cm) are marked in the Drawing. What is the area of the polygon ABFEGA?

(1) The length of CD (overlap of sides) is 3 cm.
(2) The length of altitude (GH) of triangle GDC is 2.598 cm.





_________________

Hi math experts, try this question and provide feedback...


_________________

Hi Janvisahu,

Should be D.

From Stmnt 1: Angle BCA = 60. Hence Triangle, GDC is equilateral Triangle. So, Sufficient.

From Stmnt 2: Triangle ABC is Similar to Triangle GHC. Using this you can find the length of GC. So, Sufficient.

Hope this Helps.

How could you come to the conclusion that angle BCA is 60 degrees from statement 1??

Posted from my mobile device

Hi
Very nice work.
I have framed this question myself. If you think, this is a good question, please provide Kudos...

\(\)
_________________

Sin C = \(8/16/\sqrt{3}\)
C = 60.

Hi
It is easy to prove that angle BCA is 60.
triangle ABC is 30:60:90 triangle as we know by the ration of sides.
Sides are in ratio-1:root 3:2




[/quote]

Hi Janvisahu,

Should be D.

From Stmnt 1: Angle BCA = 60. Hence Triangle, GDC is equilateral Triangle. So, Sufficient.

From Stmnt 2: Triangle ABC is Similar to Triangle GHC. Using this you can find the length of GC. So, Sufficient.

Hope this Helps.[/quote]


How could you come to the conclusion that angle BCA is 60 degrees from statement 1??

Posted from my mobile device[/quote]
_________________

New Question

The square of the median of a sequence of consecutive positive odd integers is equal to the difference between the squares of the first and last terms. If the sequence has twelve terms, what is the sum of the digits of the first and last terms?

A. 12
B. 16
C. 20
D. 24
E. 28


source:self

The median of 12 consecutive odd integers is between the 6th and 7th terms, where if \(n > 1\), then
\(A_{n} = A_1 + 2(n-1)\)

\(A_1 = n\)
\(A_2 = n + 2\)
.
.
\(A_6 = n + 10\)
\(A_7 = n + 12\)
.
.
\(A_{12} = n + 22\)

Median = \(n + 11\). If not sure:

\(\frac{(n+10+n+12)}{2}=\frac{(2n+22)}{2}=(n + 11)\)

Square of median = difference between squares of first and last terms:

\((n + 22)^2 - (n)^2 = (n + 11)^2\)
\(n^2 + 44n + 484 - n^2 = n^2 + 22n + 121\)
\(n^2 - 22n - 363 = 0\)
\((n - 33)(n + 11) = 0\)
\(n = 33\) (prompt says positive)

\(A_1 = 33\)
\(A_{12} = n + 22 = 55\)

Sum of digits of \(33\) and \(55\):
\(3 + 3 + 5 + 5 = 16\)

ANSWER B
_________________

New Question:

If from a well-shuffled deck of 52 cards, five cards are drawn at random one by one with replacement. What is the probability of fifth card being spade?
A) 1/3
B) 1/4
C) 2/5
D) 3/13
E) 3/26

source:self


_________________

Challenging Question:- Earn kudos for correct solution

If from a well-shuffled deck of 52 cards, five cards are drawn at random one by one WITHOUT replacement. What is the probability of fifth card being spade?
A) 1/3
B) 1/4
C) 2/5
D) 3/13
E) 3/26


source:self
_________________

genxer123
You might like this problem...


_________________

This question can be done by two methods, first being conventional method, in which we will find conditional probabilities when the cards drawn in earlier draw is spade or not spade and adding all the cases.
But this method is tedious and lengthy.

My approach:
Since the number of cards of spade, heart, club and diamond is same.
The probability of each group of cards to appear in required draw will be same by symmetry. Let each probability be p.

Now since it is certain that the fifth card will be out of these four groups . TOTAL probability =1
p+p+p+p =1
Hence p =1/4.
Answer is 1/4.

Please provide Kudos if u like my approach.
_________________

Q)How many different numbers can be formed from the products of numbers taken from set of 25 different prime numbers.(consider given is a set of first 25 prime numbers).

A. 2^25
B. 625
C. 25c2
D. 25p2
E. None of these






Please provide Kudos if you liked the questions.
_________________

=25c2+25c3+25c4+...25c25
=25c0+25c1+25c2+25c3+25c4+...25c25-(25c0+25c1)
=2^25 - 26
Answer is E.
_________________

Another approach:
Let the prime numbers are p1,p2,...p25.
To find the product, each of the p1,p2... can be used or not, there is 2choices for each prime number.
2*2*2...2( 25 times) = 2^25
But this includes the cases when no number is selected (1 case) and the case when only one of the prime number is selected (25cases). These are unfavorable cases.
Hence answer is 2^25 -26. Hence answer is E
_________________