If m =2^x*5^y*7^z and both 350 and 280 are factors of m, what is the
minimum value of xyz?
(A) 700
(B) 70
(C) 24
(D) 6
(E) 5
The probability that a certain coin will fall heads up is 50%. What is the
probability that it falls heads up on three of six tosses?
(A)41/2^6
(B)35/2^6
(C)1/2^6
(D)21/2^5
(E)5/2^4
Doubts..
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If m =2^x*5^y*7^z and both 350 and 280 are factors of m, what is the
minimum value of xyz?
(A) 700
(B) 70
(C) 24
(D) 6
(E) 5
Solution:
350 = 2 * 5^2 * 7 and 280 = 2^3 * 5 * 7
LCM of 350 and 280 = 2^3 * 5^2 * 7 = 1400 implies x = 3, y = 2, z = 1
So, minimum value of xyz = 3*2*1 = 6
The correct answer is [spoiler](D)[/spoiler].
minimum value of xyz?
(A) 700
(B) 70
(C) 24
(D) 6
(E) 5
Solution:
350 = 2 * 5^2 * 7 and 280 = 2^3 * 5 * 7
LCM of 350 and 280 = 2^3 * 5^2 * 7 = 1400 implies x = 3, y = 2, z = 1
So, minimum value of xyz = 3*2*1 = 6
The correct answer is [spoiler](D)[/spoiler].
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On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
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+91-99201 32411 (India)
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gmatrant wrote: The probability that a certain coin will fall heads up is 50%. What is the
probability that it falls heads up on three of six tosses?
(A)41/2^6
(B)35/2^6
(C)1/2^6
(D)21/2^5
(E)5/2^4
6C3/2^6 ==> 20/2^6 = 5/2^4 E
OA
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1. 350=7.5.2.5=2.5^2.7 and 280=4.7.2.5=2^3.5.7gmatrant wrote:If m =2^x*5^y*7^z and both 350 and 280 are factors of m, what is the
minimum value of xyz?
(A) 700
(B) 70
(C) 24
(D) 6
(E) 5
The probability that a certain coin will fall heads up is 50%. What is the
probability that it falls heads up on three of six tosses?
(A)41/2^6
(B)35/2^6
(C)1/2^6
(D)21/2^5
(E)5/2^4
thus the largest n will be 5^2.7.2^3
so 2.3.1=6
the correct answer is D
2. 500% head so 50% T thus 6 times tosses has its probability is 1/2^6
and we need 3 head time thus 3 tails : HHHTTT=6!/3!3!=20
the probability is 20/2^6=4.5/2^6=5/2^4
the answer is E