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marbles

by cans » Wed Jun 08, 2011 8:11 am
Barry plays a game in which he has a jar of marbles, some blue, some pink, some orange, and some yellow, to which he assigns point values of 2, 4, 5, and 7, respectively. After removing some marbles, Barry finds that the product of the point values of the marbles he has removed is 56,000. What could be the total number of blue and orange marbles he removed?
a)3
b)4
c)6
d)10
e)11
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by Frankenstein » Wed Jun 08, 2011 8:19 am
Hi,
b-2, p-4, o-5, y-7
56000 = (2^6).(5^3)(7)
2^6 can be written as 2^2.4^2 or 2^4.4 or 4^3
Number of orange balls = 3
Number of blue balls is 6 or 4 or 2 or 0.
So, number of blue and orange balls can be 9,7,5,3

Hence, A
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by manpsingh87 » Wed Jun 08, 2011 8:37 am
cans wrote:Barry plays a game in which he has a jar of marbles, some blue, some pink, some orange, and some yellow, to which he assigns point values of 2, 4, 5, and 7, respectively. After removing some marbles, Barry finds that the product of the point values of the marbles he has removed is 56,000. What could be the total number of blue and orange marbles he removed?
a)3
b)4
c)6
d)10
e)11
blue=2,pink=4,orange=5,yellow=7;
56000=2^6*5^3*7;
maximum value of blue and orange marbles can be 6+3=9; hence option d,e are straight away out;
also 2^6 can be written as 2^6=2^0*4^3; orange+blue=3+0=3
2^6=2^2.4^2;orange+blue=3+2=5;
2^6=2^4.4; orange+blue=3+4=7;

Hence A

Also if we observe,we will notice that no. of blue balls will always be even because no. of pink=4=2^2; so we have a algebraic expression a+2b=6; here a represents the value of blue balls and b represents the value of pink balls; also since sum of a+2b is even, therefore 'a' must be even(because 2b will always be even and even+even=even), hence sum orange+blue= even+odd=odd, and out of a,b,c, only option A is odd..!! hence answer should be A
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