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Users' Self Made Questions [#permalink]
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 05 Dec 2017, 09:04
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Re: Users' Self Made Questions [#permalink]
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05 Dec 2017, 09:04
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05 Dec 2017, 09:06
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Re: Users' Self Made Questions [#permalink]
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05 Dec 2017, 09:28
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 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 3038 times ]
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Re: Users' Self Made Questions [#permalink]
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05 Dec 2017, 19:00
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. _________________
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Re: Users' Self Made Questions [#permalink]
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05 Dec 2017, 19:27
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.
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Re: Users' Self Made Questions [#permalink]
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05 Dec 2017, 20:42
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
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Re: Users' Self Made Questions [#permalink]
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05 Dec 2017, 20:45
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. _________________
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Re: Users' Self Made Questions [#permalink]
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05 Dec 2017, 20:49
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.
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Re: Users' Self Made Questions [#permalink]
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05 Dec 2017, 20:51
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 ratio-1: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]
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Re: Users' Self Made Questions [#permalink]
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09 Dec 2017, 23:07
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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
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Re: Users' Self Made Questions [#permalink]
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09 Dec 2017, 23:52
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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 \(A_{n} = A_1 + 2(n-1)\) \(A_1 = n\) \(A_2 = n + 2\) . . \(A_6 = n + 10\) \(A_7 = n + 12\) . . \(A_{12} = n + 22\) 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
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Re: Users' Self Made Questions [#permalink]
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14 Dec 2017, 07:15
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
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Re: Users' Self Made Questions [#permalink]
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16 Dec 2017, 08:18
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
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Re: Users' Self Made Questions [#permalink]
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17 Dec 2017, 09:38
genxer123You might like this problem... gmatbusters wrote: 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 _________________
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Re: Users' Self Made Questions [#permalink]
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19 Dec 2017, 22:52
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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.
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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 Attachment: 2c0 and 2c1 are not products Please provide Kudos if you liked the questions.
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Re: Users' Self Made Questions [#permalink]
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05 Feb 2018, 06:03
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05 Feb 2018, 06:14
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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
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Only one of ten pens in a box is defective. Three pens are randomly dr [#permalink]
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Only one of ten pens in a box is defective. Three pens are randomly drawn from the box, one at a time. What is the probability that the next pen randomly drawn is defective? (A) 1/10 (B) 1/7 (C) 7/10 (D) 6/7 (E) 9/10 Source: Inspired from a Bunuel sir's question.
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Only one of ten pens in a box is defective. Three pens are randomly dr
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