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Example Questions
Example Question #791 : Psat Mathematics
Given the inequality above, which of the following MUST be true?
Subtract
from both sides:
Subtract 7 from both sides:
Divide both sides by :
Remember to switch the inequality when dividing by a negative number:
Since is not an answer, we must find an answer that, at the very least, does not contradict the fact that
is less than (approximately) 4.67. Since any number that is less than 4.67 is also less than any number that is bigger than 4.67, we can be sure that
is less than 5.
Example Question #792 : Psat Mathematics
A factory packs cereal boxes. Before sealing each box, a machine weighs it to ensure that it is no lighter than 356 grams and no heavier than 364 grams. If the box holds grams of cereal, which inequality represents all allowable values of
?
The median weight of a box of cereal is 360 grams. This should be an allowable value of w. Substituting 360 for w into each answer choice, the only true results are:
and:
Notice that any positive value for w satisfies the second inequality above. Since w must be between 356 and 364, the first inequality above is the only reasonable choice.
Example Question #793 : Psat Mathematics
What values of x make the following statement true?
|x – 3| < 9
–12 < x < 6
–3 < x < 9
x < 12
6 < x < 12
–6 < x < 12
–6 < x < 12
Solve the inequality by adding 3 to both sides to get x < 12. Since it is absolute value, x – 3 > –9 must also be solved by adding 3 to both sides so: x > –6 so combined.
Example Question #794 : Psat Mathematics
If –1 < w < 1, all of the following must also be greater than –1 and less than 1 EXCEPT for which choice?
3w/2
w/2
|w|
|w|0.5
w2
3w/2
3w/2 will become greater than 1 as soon as w is greater than two thirds. It will likewise become less than –1 as soon as w is less than negative two thirds. All the other options always return values between –1 and 1.
Example Question #795 : Psat Mathematics
Solve for .
Absolute value problems always have two sides: one positive and one negative.
First, take the problem as is and drop the absolute value signs for the positive side: z – 3 ≥ 5. When the original inequality is multiplied by –1 we get z – 3 ≤ –5.
Solve each inequality separately to get z ≤ –2 or z ≥ 8 (the inequality sign flips when multiplying or dividing by a negative number).
We can verify the solution by substituting in 0 for z to see if we get a true or false statement. Since –3 ≥ 5 is always false we know we want the two outside inequalities, rather than their intersection.
Example Question #796 : Psat Mathematics
If and
, then which of the following could be the value of
?
To solve this problem, add the two equations together:
The only answer choice that satisfies this equation is 0, because 0 is less than 4.
Example Question #141 : Algebra
If , which of the following could be a value of
?
-
In order to solve this inequality, you must isolate on one side of the equation.
Therefore, the only option that solves the inequality is .
Example Question #21 : Inequalities
What values of make the statement
true?
First, solve the inequality :
Since we are dealing with absolute value, must also be true; therefore:
Example Question #797 : Psat Mathematics
If –1 < n < 1, all of the following could be true EXCEPT:
n2 < n
16n2 - 1 = 0
n2 < 2n
(n-1)2 > n
|n2 - 1| > 1
|n2 - 1| > 1
Example Question #798 : Psat Mathematics
(√(8) / -x ) < 2. Which of the following values could be x?
-1
-2
-4
-3
All of the answers choices are valid.
-1
The equation simplifies to x > -1.41. -1 is the answer.
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