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Posted a solution to question 7 here:

https://gmatclub.com/forum/bunuel-s-alg ... l#p3447632

Hope it helps.
­
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Posted a solution to question 12 here:

https://gmatclub.com/forum/bunuel-s-alg ... l#p3447662

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Posted a solution to question 15 here:

https://gmatclub.com/forum/bunuel-s-alg ... l#p3447669

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Posted a solution to question 16 here:

https://gmatclub.com/forum/bunuel-s-alg ... l#p3447671

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Posted a solution to question 17 here:

https://gmatclub.com/forum/bunuel-s-alg ... l#p3447674

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Posted a solution to question 21 here:

https://gmatclub.com/forum/bunuel-s-alg ... l#p3447678

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Posted a solution to question 31 here:

https://gmatclub.com/forum/bunuel-s-alg ... l#p3447680

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Posted a solution to question 32 here:

https://gmatclub.com/forum/bunuel-s-alg ... l#p3447682

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Posted a solution to question 33 here:

https://gmatclub.com/forum/bunuel-s-alg ... l#p3447721

Hope it helps.­

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This can't be be right. First term in the expression is negative, second is positive and third is negative. Product should be positive. Am I missing something here?
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Question 30 - Official Solution:

What is the value of \((\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})\) ?

A. \(-4\)
B. \(-2\)
C. \(-1\)
D. \(1\)
E. \(2\)


We want to find the value of the expression \((\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})\). To simplify this expression, we can start by squaring it.

\(=((\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10}))^2=\)

\(=(4-\sqrt{15})(4 + \sqrt{15})^2(\sqrt{6} - \sqrt{10})^2=\)

\(=(4-\sqrt{15})(4 + \sqrt{15})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})^2=\)
Applying the difference of squares identity \((a - b)(a+b)=a^2-b^2\) to \((4-\sqrt{15})(4 + \sqrt{15})\) we can simplify to:

\(=(16 -15)(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})^2=\)
Now using the identity \((a - b)^2=a^2-2ab+b^2\) to simplify the squared term::

\(=(4 + \sqrt{15})(6 - 2\sqrt{60}+10)=\)

\(=(4 + \sqrt{15})(16 - 4\sqrt{15})=\)

\(=(4 + \sqrt{15})4(4 - \sqrt{15})=\)

\(=4(16-15)=4\).
Since the square of the expression is 4, the expression itself must be either 2 or -2. However, since the product of two positive terms and one negative term is negative (\((\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})=positive*positive*negative=negative\)), we know that \((\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})\) is negative. Therefore, the final answer is -2.


Answer: B­
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This can't be be right. First term in the expression is negative, second is positive and third is negative. Product should be positive. Am I missing something here?
Bunuel
Question 30 - Official Solution:

What is the value of \((\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})\) ?

A. \(-4\)
B. \(-2\)
C. \(-1\)
D. \(1\)
E. \(2\)


We want to find the value of the expression \((\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})\). To simplify this expression, we can start by squaring it.

\(=((\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10}))^2=\)

\(=(4-\sqrt{15})(4 + \sqrt{15})^2(\sqrt{6} - \sqrt{10})^2=\)

\(=(4-\sqrt{15})(4 + \sqrt{15})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})^2=\)
Applying the difference of squares identity \((a - b)(a+b)=a^2-b^2\) to \((4-\sqrt{15})(4 + \sqrt{15})\) we can simplify to:

\(=(16 -15)(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})^2=\)
Now using the identity \((a - b)^2=a^2-2ab+b^2\) to simplify the squared term::

\(=(4 + \sqrt{15})(6 - 2\sqrt{60}+10)=\)

\(=(4 + \sqrt{15})(16 - 4\sqrt{15})=\)

\(=(4 + \sqrt{15})4(4 - \sqrt{15})=\)

\(=4(16-15)=4\).
Since the square of the expression is 4, the expression itself must be either 2 or -2. However, since the product of two positive terms and one negative term is negative (\((\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})=positive*positive*negative=negative\)), we know that \((\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})\) is negative. Therefore, the final answer is -2.


Answer: B­

The first multiple, \( \sqrt{4-\sqrt{15}} \), is the square root of a number, which by definition cannot be negative:


\( \sqrt{4-\sqrt{15}} = \sqrt{4-3.87} \approx 0.36 \)
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What if all five numbers were -ve?
if we consider the above scenario, isn't option E the correct answer?
Bunuel
Question 21 - Official Solution:


If \(a > b > c > d > e\) and \(abcde > 0\), then which of the following must be true ?

I. \(ab > 0\)

II. \(bc > 0\)

III. \(de > 0\)



A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III


Given that the product of five numbers, \(a\), \(b\), \(c\), \(d\), and \(e\), is positive, there must be an even number of negative numbers among them (0, 2, or 4 negative numbers). Since it is also given that \(a > b > c > d > e\), we can have the following three cases:

\(a\; |\; b\; |\; c\; |\; d\; |\; e\)

\(+ | + | + | + | +\) (no negative numbers)

\(+ | + | + | - | -\) (two negative numbers)

\(+ | - | - | - | -\) (four negative numbers)

Let's analyze each option, taking into consideration that the question asks which of them MUST be true, not COULD be true.

I. \(ab > 0\)

If we have the third case, then this option is not true. Eliminate.

II. \(bc > 0\)

This option is true for each of the three cases. Therefore, this option is always true.

III. \(de > 0\)

This option is true for each of the three cases. Therefore, this option is always true.

Consequently, only options II and III are always true, and thus the answer is D.


Answer: D­
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raziqsid
What if all five numbers were -ve?
if we consider the above scenario, isn't option E the correct answer?
Bunuel
Question 21 - Official Solution:


If \(a > b > c > d > e\) and \(abcde > 0\), then which of the following must be true ?

I. \(ab > 0\)

II. \(bc > 0\)

III. \(de > 0\)



A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III


Given that the product of five numbers, \(a\), \(b\), \(c\), \(d\), and \(e\), is positive, there must be an even number of negative numbers among them (0, 2, or 4 negative numbers). Since it is also given that \(a > b > c > d > e\), we can have the following three cases:

\(a\; |\; b\; |\; c\; |\; d\; |\; e\)

\(+ | + | + | + | +\) (no negative numbers)

\(+ | + | + | - | -\) (two negative numbers)

\(+ | - | - | - | -\) (four negative numbers)

Let's analyze each option, taking into consideration that the question asks which of them MUST be true, not COULD be true.

I. \(ab > 0\)

If we have the third case, then this option is not true. Eliminate.

II. \(bc > 0\)

This option is true for each of the three cases. Therefore, this option is always true.

III. \(de > 0\)

This option is true for each of the three cases. Therefore, this option is always true.

Consequently, only options II and III are always true, and thus the answer is D.


Answer: D­
All 5 numbers cannot be negative because the product of 5 (odd) negative numbers would be negative, not positive as given in the stem.
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Hey Bunuel, I might need help in order to understand question 2? Can you explain it to me? Thanks in Advance
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Hey Bunuel, I might need help in order to understand question 2? Can you explain it to me? Thanks in Advance

Please try checking the topic for solutions: https://gmatclub.com/forum/bunuel-s-alg ... l#p3447611
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