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help required

by swapna » Sat Jan 23, 2010 8:30 pm
In a certain game, a large container is filled with red,yellow,green and blue beads worth,respectively,7,5,3 and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147000, how many red beads were removed?
5
4
3
2
0

Answer is D. This is from the official guide 11th edition. Any value can be gvn to r and still by manipulating the value of y,g and b answer can be obtained as 147000.Can some explain??
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by chaitanya.sonavale » Sun Jan 24, 2010 2:04 am
Hi,

147000 = 147 * 10 * 10 * 10 = (7 * 7 * 3) * (5*2) * (5*2) * (5*2)

but r = 7, y = 5, g = 3 , b = 2

Hence 2 red, 3 yellow, 1 green and 3 blue were removed

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by bedazzled » Sun Jan 24, 2010 7:35 am
What if we try to do in this way


147000 = 35 * 21 * 20 * 10
= (7*5) * (3 * 7) * (5*4 ) * (2*5)

here we get red =5 , yellow=7, green=4, blue=5.

We get red=5 which is indeed an answer.

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by chaitanya.sonavale » Sun Jan 24, 2010 7:59 am
7,5,3,2 are values assigned to red,yellow,green and blue beads respectively. So we are trying to find how many beads of each color are used. In other words, trying find the prime factors of 147000 and comparing them to given values to get the answer.

e.g. we get 7 twice and 7 is value of red beads. Hence there have to be 2 red beads.

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by bedazzled » Sun Jan 24, 2010 8:45 am
I think I missed the essence of question. Thanks Chaitanya.

we have to consider here the no. the time that value is occurring like 7 occurred 2 times hence red=2 not the product value of it.

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by shashank.ism » Fri Feb 19, 2010 6:05 am
This question is really tricky ,....factorizing is the main concept in this problem...
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by abhi.genx7 » Fri Feb 19, 2010 12:37 pm
This is the method i tried , could be useful.
I looked for 147 . that is 3 less than 150. so 49*3 so two 7's are used.
The rest are 5*2=10.

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by shashank.ism » Sat Feb 20, 2010 12:38 am
abhi.genx7 wrote:This is the method i tried , could be useful.
I looked for 147 . that is 3 less than 150. so 49*3 so two 7's are used.
The rest are 5*2=10.


thats a nice approach..abhi.genx
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