Collection of 8 DS questions

OA's are given here: collection-of-8-ds-questions-85290.html#p639290 OA fot this question is E, not B.

Is the measure of one of the interior angles of quadrilateral ABCD equal to 60?

Sum of inner angels of quadrilateral is 360 degrees. (Sum of inner angles of polygon=180*(n-2), where n is # of sides)

(1) Two of the interior angles of ABCD are right angles --> angles can be 90+90 + any combination of two angels totaling 180. Not sufficient.

(2) The degree measure of angle ABC is twice the degree measure of angle BCD --> <ABC=2<BCD. Not sufficient

(1)+(2) Angles can be 90+90+45+135 Or 90+90+60+120. Not sufficient.

Answer: E.

By combining both we can get answer.

Two of the interior angles of ABCD are right angles. which sum of two interior angles = 180

Now fom B

The degree measure of angle ABC is twice the degree measure of angle BCD. i.e. 1:2,

X+2X + 90 + 90 = 360
3X=180
X=60

Again OA is E, not C. Check the solution above: Angles can be 90+90+45+135 Or 90+90+60+120. Not sufficient.

Bunuel, thanks for the wonderful set, i had a question

. Is x^4 + y^4 > z^4 ?
(1) x^2 + y^2 > z^2
(2) x+y > z

x^4 + y^4 = ((x^2) + (y^2))^2 - 2*x^2*y^2

So basically we want to find out if ((x^2) + (y^2))^2 - 2*x^2*y^2 > z^4

OR if

((x^2) + (y^2))^2 > z^4 + 2*x^2*y^2

Now by (1), we know that x^2 + y^2 > z^2

Or

((x^2) + (y^2))^2 > z^4 ( squaring both sides, both sides are absolutely positive)...... (a)

We also know that 2*x^2*y^2 will be less than ((x^2) + (y^2))^2......(b)

Combining (a) and (b), can we not conclude that ((x^2) + (y^2))^2 > z^4 + 2*x^2*y^2 and hence making (1) sufficient

Is \(x^4+y^4>z^4\)?

The best way to deal with this problem is plugging numbers. Remember on DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

(1) \(x^2+y^2>z^2\)
It's clear that we get YES answer very easily with big x and y (say 10 and 10), and small z (say 0).

For NO answer let's try numbers from Pythagorean triples:
\(x^2=3\), \(y^2=4\) and \(z^2=5\) (\(x^2+y^2=7>5=z^2\)) --> \(x^4+y^4=9+16=25=z^4\), so we have answer NO (\(x^4+y^4\) is NOT more than \(z^4\), it's equal to it).

Not sufficient.

(2) \(x+y>z\). This one is even easier: again we can get YES answer with big x and y, and small z.

As for NO try to make z some big enough negative number: so if \(x=y=1\) and \(z=-5\), then \(x^4+y^4=1+1=2<25=z^4\).

Not sufficient.

(1)+(2) As we concluded YES answer is easily achievable. For NO try the case of \(x^2=3\), \(y^2=4\) and \(z^2=5\) again: \(x+y=\sqrt{3}+\sqrt{4}>\sqrt{5}\) (\(\sqrt{3}+2\) is more than 3 and \(\sqrt{5}\) is less than 3), so statement (2) is satisfied, we know that statement (1) is also satisfied (\(x^2+y^2=7>5=z^2\)) and \(x^4+y^4=9+16=25=z^4\). Not sufficient.

Answer: E.

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-4-y-4-z-101358.html

Ok Bunuel, i shall stick to the plugging number approach, please can you confirm if both terms in an equality are positive, we can square both sides or even raise to the power of 3, without changing the inequality sign?

1. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality)

2. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).

Adding/subtracting/multiplying/dividing inequalities: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html

Answer is C. X can also be 1. 1 is factor of all the numbers.