Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 43347

Users' Self Made Questions [#permalink]
Show Tags
05 Dec 2017, 08:04



Math Expert
Joined: 02 Sep 2009
Posts: 43347

Re: Users' Self Made Questions [#permalink]
Show Tags
05 Dec 2017, 08:04



Math Expert
Joined: 02 Sep 2009
Posts: 43347

Re: Users' Self Made Questions [#permalink]
Show Tags
05 Dec 2017, 08:06



Manager
Joined: 27 Oct 2017
Posts: 89
Location: India
Concentration: International Business, General Management
GPA: 3.64
WE: Business Development (Energy and Utilities)

Users' Self Made Questions [#permalink]
Show Tags
05 Dec 2017, 08:28
1
This post received KUDOS
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. Attachment:
Triangle.jpg [ 14.35 KiB  Viewed 1363 times ]
_________________
If you like my post, encourage me by providing Kudos... Happy Learning...



Manager
Joined: 27 Oct 2017
Posts: 89
Location: India
Concentration: International Business, General Management
GPA: 3.64
WE: Business Development (Energy and Utilities)

Re: Users' Self Made Questions [#permalink]
Show Tags
05 Dec 2017, 18:00
Hi math experts, try this question and provide feedback... Janvisahu wrote: 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.
_________________
If you like my post, encourage me by providing Kudos... Happy Learning...



Manager
Joined: 31 Jul 2017
Posts: 142
Location: Malaysia
WE: Consulting (Energy and Utilities)

Re: Users' Self Made Questions [#permalink]
Show Tags
05 Dec 2017, 18:27
Janvisahu wrote: Hi math experts, try this question and provide feedback... Janvisahu wrote: 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 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.



Manager
Status: Searching for something I've been searching..LOL
Joined: 14 Dec 2016
Posts: 64
Location: India
Concentration: Healthcare, Operations
GPA: 3.5
WE: Medicine and Health (Health Care)

Re: Users' Self Made Questions [#permalink]
Show Tags
05 Dec 2017, 19:42
rahul16singh28 wrote: Janvisahu wrote: Hi math experts, try this question and provide feedback... Janvisahu wrote: 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 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



Manager
Joined: 27 Oct 2017
Posts: 89
Location: India
Concentration: International Business, General Management
GPA: 3.64
WE: Business Development (Energy and Utilities)

Re: Users' Self Made Questions [#permalink]
Show Tags
05 Dec 2017, 19:45
Hi Very nice work. I have framed this question myself. If you think, this is a good question, please provide Kudos... \(\) rahul16singh28 wrote: Janvisahu wrote: Hi math experts, try this question and provide feedback... Janvisahu wrote: 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 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.
_________________
If you like my post, encourage me by providing Kudos... Happy Learning...



Manager
Joined: 31 Jul 2017
Posts: 142
Location: Malaysia
WE: Consulting (Energy and Utilities)

Re: Users' Self Made Questions [#permalink]
Show Tags
05 Dec 2017, 19:49
Janvisahu wrote: Hi math experts, try this question and provide feedback... Janvisahu wrote: 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. Sin C = \(8/16/\sqrt{3}\) C = 60.



Manager
Joined: 27 Oct 2017
Posts: 89
Location: India
Concentration: International Business, General Management
GPA: 3.64
WE: Business Development (Energy and Utilities)

Re: Users' Self Made Questions [#permalink]
Show Tags
05 Dec 2017, 19:51
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 ratio1:root 3:2 Janvisahu wrote: 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. [/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]
_________________
If you like my post, encourage me by providing Kudos... Happy Learning...



Director
Joined: 07 Dec 2014
Posts: 884

Users' Self Made Questions [#permalink]
Show Tags
09 Dec 2017, 22:07
2
This post received KUDOS
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



VP
Joined: 22 May 2016
Posts: 1259

Users' Self Made Questions [#permalink]
Show Tags
09 Dec 2017, 22:52
1
This post received KUDOS
gracie wrote: 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 \(A_{n} = A_1 + 2(n1)\) \(A_1 = n\) \(A_2 = n + 2\) . . \(A_6 = n + 10\) \(A_7 = n + 12\) . . \(A_{12} = n + 22\) 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
_________________
At the still point, there the dance is.  T.S. Eliot Formerly genxer123



Manager
Joined: 27 Oct 2017
Posts: 89
Location: India
Concentration: International Business, General Management
GPA: 3.64
WE: Business Development (Energy and Utilities)

Users' Self Made Questions [#permalink]
Show Tags
14 Dec 2017, 06:15
New Question:If from a wellshuffled 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
_________________
If you like my post, encourage me by providing Kudos... Happy Learning...



Manager
Joined: 27 Oct 2017
Posts: 89
Location: India
Concentration: International Business, General Management
GPA: 3.64
WE: Business Development (Energy and Utilities)

Users' Self Made Questions [#permalink]
Show Tags
16 Dec 2017, 07:18
Challenging Question: Earn kudos for correct solution If from a wellshuffled 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
_________________
If you like my post, encourage me by providing Kudos... Happy Learning...



Manager
Joined: 27 Oct 2017
Posts: 89
Location: India
Concentration: International Business, General Management
GPA: 3.64
WE: Business Development (Energy and Utilities)

Re: Users' Self Made Questions [#permalink]
Show Tags
17 Dec 2017, 08:38
genxer123You might like this problem... gmatbusters wrote: Challenging Question: Earn kudos for correct solution If from a wellshuffled 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
_________________
If you like my post, encourage me by providing Kudos... Happy Learning...



Manager
Joined: 27 Oct 2017
Posts: 89
Location: India
Concentration: International Business, General Management
GPA: 3.64
WE: Business Development (Energy and Utilities)

Users' Self Made Questions [#permalink]
Show Tags
19 Dec 2017, 21:52
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.
_________________
If you like my post, encourage me by providing Kudos... Happy Learning...




Users' Self Made Questions
[#permalink]
19 Dec 2017, 21:52






