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|6 - |x + 2|| = 10

| - 10| = 10 and |10| = 10. So we need 6 - |x + 2| = 10 or -10.

=> 6 - |x + 2| = 10 for this 6 - 10 = -4 = |x + 2| (Not possible)

=> 6 - |x + 2| = -10 for this 6 + 10 = 16 = |x + 2|

=> x + 2 = 16 and x = 14 and x + 2 = -16 and x = -18

'x' is a non-negative and hence x = 14 is the value considered.

Answer B
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Kritisood
If x is a non-negative integer and |6 - |x + 2|| = 10, then find the number of values of x that satisfy the given absolute value inequality?

a. 0
b. 1
c. 2
d. 3
e. 4


The question simply asks us to find whether |x+2| = 16

=>which is possible in two scenarios that being at x=14 since it's positive it can take this posiibilities

next we have x=-18 which it cannot be included since we require only positive values

Therefore IMO only value satisfies
Hence B
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When should we be solving it the way described in the explanations vs substituting | x+ 2 | with a and solving based on that? Would that even work? Why or why not?

I tried it and got the right answer.. but I think I stumbled onto it through luck
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egmat KarishmaB gmatophobia Can you please check whether my approach is correct ?­
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Kritisood
If x is a non-negative integer and |6 - |x + 2|| = 10, then find the number of values of x that satisfy the given absolute value inequality?

a. 0
b. 1
c. 2
d. 3
e. 4

i did as follows: |6-x-2|=10 (x+2 will be +ve since x is a non neg integer) => |4-x|=10 (again 4-x should be +ve since x is a non neg integer) 4-x=10 => -6=x hence there should be zero values that satisfy the inequality as x has to be a non neg integer. could someone help pls?
­|6 - |x + 2|| = 10

Since x is non negative, (x+2) will always be positive. So |x + 2| = (x + 2)
|6 - x - 2| = 10
which is equivalent to:
|x - 4| = 10
x is a point at a distance 10 away from 4. There are only 2 such points: -6 and 14

Answer (C)

Absolute values as distances is discussed here:
https://youtu.be/oqVfKQBcnrs
 
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Can,Will
egmat KarishmaB gmatophobia Can you please check whether my approach is correct ?­
­It is correct though too many unnecessary calculations. ­
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KarishmaB
If p is a non-positive number, then for what value of p does the expression |77 – 6p| holds the minimum value?

A. 77/12
B. 77/6
C. 12
D. 0
E. -77/6

in this problem can we say that since p is a non positive I77-6pI is negative i.e -77+6p
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Can,Will
KarishmaB
If p is a non-positive number, then for what value of p does the expression |77 – 6p| holds the minimum value?

A. 77/12
B. 77/6
C. 12
D. 0
E. -77/6

in this problem can we say that since p is a non positive I77-6pI is negative i.e -77+6p
|77 - 6p| = |6p - 77| gives two cases:
6p - 77 >= 0 which means p >= 77/6 (Not possible because p is not positive) 
OR
6p - 77 < 0 which means p < 77/6 (which includes 0 and all negative values)
So |6p - 77| = - (6p - 77)

You should check out this post:
https://anaprep.com/algebra-the-why-beh ... questions/

This is how you will arrive at the answer here though:

­p is non positive (though this term isn't really used) means p can be 0 or negative. 
If p = 0,  |77 – 6p| = 77
If p is negative, – 6p becomes positive and when added to 77 is gives us something more than 77.
So the minimum value of |77 – 6p| happens when p = 0.­
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KarishmaB
If p is a non-positive number, then for what value of p does the expression |77 – 6p| holds the minimum value?

A. 77/12
B. 77/6
C. 12
D. 0
E. -77/6

in this problem can we say that since p is a non positive I77-6pI is negative i.e -77+6p
­This question is discussed here:

https://gmatclub.com/forum/if-x-is-a-no ... l#p3389087
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