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mysterymanrog
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mysterymanrog
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mysterymanrog
I don’t think the GMAT will ever require you to memorize such a formula, so I would try solving it without using the formula
Could you show another way of solving please
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mysterymanrog
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Kenykruzy
Could you show another way of solving please
A few ways come to mind -> you can calculate the total number of ways that the children do not each get one candy and subtract from the total number of ways and so on

but I don’t think I am able to solve such a problem
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Anki111
The cost of a mixture that contains candy-coated chocolate pieces and salted peanuts in the ratio of 2 pounds of chocolate pieces to 3 pounds of peanuts is $3.80 per pound. If the chocolate pieces cost $2.00 more per pound than the peanuts, how much do the peanuts cost per pound?
Is this 3?
2(2+p)+3p=3.8×5
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AG2907
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the-weights-of-a-set-of-moving-boxes-on-a-truck-have-a-mean-of-22-poun-373314.html

can someone please help with this question?

i took random values which same situations, i see the SD decreases
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AG2907
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if-x-is-positive-integer-is-x-evenly-divisible-by-413103.html --- in this how is x=7 evenly divisible by 7?
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AG2907
https://gmatclub.com/forum/if-x-is-posi ... 13103.html --- in this how is x=7 evenly divisible by 7?
1)
(147+2x)/7=k (k is some integer)
147+2x=7k
2x=7(k-21)
2x/7=(k-21)
Integer-Integer=integer, so 2x/7 is integer, and since 2 is not a factor of 2, x must be some multiple of 7 and therefore divisble by 7.

2)
(154+7x)/7=k
7(22+x)/7=k
22+x=k
x can be anything, no info

A
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mysterymanrog
1) (147+2x)/7=k (k is some integer) 147+2x=7k 2x=7(k-21) 2x/7=(k-21) Integer-Integer=integer, so 2x/7 is integer, and since 2 is not a factor of 2, x must be some multiple of 7 and therefore divisble by 7. 2) (154+7x)/7=k 7(22+x)/7=k 22+x=k x can be anything, no info A
but doesnt evenly divisible mean that it should be an even multiple only?
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Evenly divisible means that the remainder should be zero. The quotient can be either odd or even
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so evenly divisible is same as divisible?
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AG2907
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The answer and explanations say B and on taking random values, I am getting C, please help
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mysterymanrog
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AG2907
The answer and explanations say B and on taking random values, I am getting C, please help
You’re over thinking

SD is dispersion from the mean

If you add two points to a set which are different to the mean, the SD must necessairly inceease as the data set is now more dispersed
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So shoulnt that hold true for any random values I take?

shouldn’t*
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AG2907
So shoulnt that hold true for any random values I take?
In theory, yes

Where are you running into the issue?
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I took values say 3,5,6,14 - calculated mean and SD - then added to more values 9,5 and observed SD decreased
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AG2907
I took values say 3,5,6,14 - calculated mean and SD - then added to more values 9,5 and observed SD decreased
But you added one value that is already in the set (5) which decreases the SD, and one value that is outside the set (9)
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Omg, thank u so much, sorry for the silly mistake!

wait, but if i add say, 4 and 10, SD still decreases, sorry if i am still making a silly error again
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AG2907
wait, but if i add say, 4 and 10, SD still decreases, sorry if i am still making a silly error again
because 4 and 10 are still within the bounds of the set

So they dont increase the dispersion
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