Bunuel's Algebra DS Diagnostic Test

7. Is \(x = 0\) ?

(1) \(x^2*y^3 > 0\)
(2) \(x^3*y^2 < 0\)

Solution:

(1) \(x^2*y^3 > 0\)

If the expression is not equal to 0, it means that NONE of the terms can be equal to 0 => x NOT EQUAL TO 0

SUFFICIENT

(2) \(x^3*y^2 < 0\)

If the expression is not equal to 0, it means that NONE of the terms can be equal to 0 => x NOT EQUAL TO 0

SUFFICIENT

Answer: D

8. Is \(x > y\)?

(1) \(x + y > 0\)
(2) \(y^2 > x^2\)

Solution:

(1) \(x + y > 0\)

\(x > -y\)

If \(y = 3: x > -3\): If x = -2 NO, If x = 4 YES

Not sufficient

(2) \(y^2 > x^2\)

\(x^2 - y^2 < 0\)
\((x - y)(x + y) < 0\)

Either \(-y < x < y\) OR \(x > y & x < -y\)

In one case, x > y, in other it is not

Not Sufficient

Together: Both statements combined

From statement 2: Either \(-y < x < y\) OR \(x > y & x < -y\)
From statement 1: \(x > -y\)

=> \(-y < x < y\)

SUFFICIENT

Answer: C

9. Is \(|x + 1| < 2\) ?

(1) \((x - 1)^2 < 1\)
(2) \(x^2 - 2 < 0\)

Solution:

Simplifying the question:

\(|x + 1| < 2\)
\(-2 < (x + 1) < 2\)
\(-3 < x < 1\)

Question: Is \(-3 < x < 1\)

(1) \((x - 1)^2 < 1\)

\(-1 < (x - 1) < 1\)

\(0 < x < 2\)

Now, x could be part of the range asked but also it could be out of that range (if x = 1.9 for example)

NOT SUFFICIENT

(2) \(x^2 - 2 < 0\)

\(x^2 < 2\)

\(-\sqrt{2} < x < \sqrt{2}\)

\(-1.4 < x < 1.4\)

Again, x could be part of the range asked but also it could be out of that range (if x = 1.3 for example)

NOT SUFFICIENT

BOTH STATEMENT TOGETHER

Statement 1: \(0 < x < 2\)
Statement 2: \(-1.4 < x < 1.4\)

Combined range: \(0 < x < 1.4\)

Still, for all values of x > 1, it does not satisfy but for rest it does

NOT SUFFICIENT

Answer: E

10. Is \(x > y\) ?

(1) \(x^2 < y^2\)
(2) \(y < 0\)

Solution:

(1) \(x^2 < y^2\)

If \(y = 3, x = 2: x < y\)
If \(y = -3, x = 2: x > y\)

So, if y is NEGATIVE: \(x > y\), and if y is POSITIVE: \(x < y\)

NOT SUFFICIENT

(2) \(y < 0\)

This gives us no information on \(x\)

NOT SUFFICIENT

BOTH STATEMENTS TOGETHER

Statement 1: So, if y is NEGATIVE: \(x > y\), and if y is POSITIVE: \(x < y\)
Statement 2: \(y < 0\)

So, \(x < y\) ALWAYS

Answer: C

11. Is \(|x - y| = ||x| - |y||\) ?

(1) \(xy > 0\)
(2) \(x < y < 0\)

Solution:

Is \(|x - y| = ||x| - |y||\)?

There are 2 rules mentioned in a very interesting and nicely explained modulus topic in the Quantitative Megathread (https://gmatclub.com/forum/properties-o ... 91317.html), and they are:

1. \(|x + y| <= |x| + |y|\) (Equality holds when both have same signs, else inequality holds)
2. \(|x - y| >= |x| - |y|\) (Equality holds when both have same signs and \(|x| >= |y|\) & when y = 0, else inequality holds)

Let us explore rule 2 a little more here:

2. \(|x - y| >= |x| - |y|\)

Since we are checking equal sign, let us test options where equality holds (same signs and \(|x| >= |y|\) & when y = 0)

Case 1: Same sign and \(|x| >= |y|\)

\(x = 4, y = 3: |4-3| = 1 = |4| - |3|\)
So, we can also say that |4-3| = ||4| - |3|| (Since both are same values)

Case 2: y = 0 (We do not need to check this because our statements both specify that y cannot equal to 0)

Let us test cases when both have same signs but \(|x| < |y|\)

Case 3: Same signs (both +) but \(|x| < |y|\)

\(x = 3, y = 4\)
LHS: \(|3-4| = |-1| = 1\)
RHS: \(|3| - |4| = -1 => ||3| - |4|| = |-1| = 1\)

Thus |3-4| = ||3| - |4||

Case 4: Same signs (both -) but \(|x| < |y|\)

\(x = -3, y = -4\)
LHS: \(|-3+4| = |1| = 1\)
RHS: \(|-3| - |-4| = -1 => ||-3| - |-4|| = |-1| = 1\)

Thus |-3-(-4)| = ||-3| - |-4||

As you can see, as long as both x and y have the same sign, the equation \(|x - y| = ||x| - |y||\) will hold

Let us now head to the statements:

(1) \(xy > 0\)

Both POSITIVE: Equation holds true

SUFFICIENT

(2) \(x < y < 0\)

Both NEGATIVE: Equation holds true

SUFFICIENT

Answer: D

12. What is the value of positive integer n ?

(1) \(n^{5!} = 1\)
(2) \(n^5 = n!\)

Solution:

(1) \(n^{5!} = 1\)

\(n^{120} = 1\)

Only possible if \(n = 1\) or \(n = -1\)

But since, its specified that \(n\) is a positive integer, there is only 1 definite answer

SUFFICIENT

(2) \(n^5 = n!\)

This is only possible when \(n = 1\). A definite answer

SUFFICIENT

Answer: D

13. If m and n are positive integers, and \(x = 2^m3^n\), is m < n ?

(1) x is divisible by 144
(2) x is not divisible by 648

Solution:

(1) x is divisible by 144

x is divisible by 144 => x is divisible by \(2^43^2\) (x can be any power of 2 and 3 either equal to or greater than current values)

If \(x = 2^43^2: m > n\)
If \(x = 2^43^5: m < n\)

NOT SUFFICIENT

(2) x is not divisible by 648

x is NOT DIVISIBLE by 648 => x is NOT DIVISIBLE by \(2^33^4\) (x can be any power of 2 and 3 LESS than the current values)

If \(x = 2^23^1: m > n\)
If \(x = 2^23^3: m < n\)

NOT SUFFICIENT

Together (Both Statements Combined)

\(\frac{x}{2^43^2} = \) Integer &
\(\frac{x}{2^33^4} \neq\) Integer

=> \(m\) IS DEFINITELY >= 4 (to satisfy 1st condition) and \(n\) HAS TO BE < 4 in that case because if \(n\) is even = 4, then the second condition falls apart

Therefore, \(m < n\) NEVER

SUFFICIENT

Answer: C

where are the explanations for questions 14 to 40?

Hey buddy! Actually I'm not from the GMAT club team, I'm just a member trying to do my part for the community
Did post all solutions for the PS Diagnostic test
Doing the same for this DS one as well

Actually we need to get 100 replies here as well as in PS and then Bunuel has promised to post such diagnostic tests for a number of other topics as well

Posted from my mobile device

Hi are there explanations on how to solve these?

Hi Bunuel

This is an amazing resource, thank you for this!

Just one question, are there official answer explanations for the questions in this series? If yes, where can I find them?

Thanks.

These questions do not have OE's yet. You are welcome to post your solutions though!
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Update:

• Bunuel's Word Problems PS Diagnostic Test
• Bunuel's Word Problems DS Diagnostic Test

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Bunuel question 13 has an error

Please post your detailed analysis of the question and we'll see. Thank you!
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35. If \(a\) and \(b\) are single-digit positive numbers and \(\frac{a}{b}\) is NOT a recurring decimal, what is the value of \(a\)?

Given that \(a\) and \(b\) are single-digit positive numbers, the possible values for each are 1, 2, 3, 4, 5, 6, 7, 8, or 9.

(1) \(-\frac{1}{3} > -\frac{a}{b} > -\frac{4}{5}\)

Multiply by -1 to reverse the inequalities and simplify into: \(\frac{1}{3} < \frac{a}{b} < \frac{4}{5}\)

Convert to decimals: \(0.\overline{3} < \frac{a}{b} < 0.8\).

Since \(\frac{a}{b}\) is NOT a recurring decimal, the possible values can be 0.4 (\(a = 2\), \(b = 5\)), 0.5 (\(a = 1\), \(b = 2\)), ... Not sufficient.

(2) \(b\) is equal to the sum of its positive factors, excluding \(b\) itself.

Of the single-digit numbers, only 6 meets this condition: \(6 = 1 + 2 + 3\). Given that \(\frac{a}{b}\) is NOT a recurring decimal, then \(\frac{a}{6}\) can be either \(\frac{3}{6} = 0.5\), \(\frac{6}{6} = 1\), or \(\frac{9}{6} = 1.5\). Not sufficient.

(1)+(2) Using statement (2), we know that \(\frac{a}{b}\) can be \(\frac{3}{6} = 0.5\), \(\frac{6}{6} = 1\), or \(\frac{9}{6} = 1.5\). Among these, only 0.5 lies between \(0.\overline{3}\) and 0.8. Therefore, \(a = 1\). Sufficient.


Answer: C
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Bunuel , i think there is some kind of mistake in Q13

There was a typo in that question. Edited. Thank you for noticing. Here is a correct version of that question with a solution:

13. If m and n are positive integers, and \(x = 2^m3^n\), is m < n ?

(1) x is divisible by 144

Factoring 144, we get 144 = 2^4 * 3^2.
Hence, this statement implies that m ≥ 4 and n ≥ 2. However, without knowing the upper limits of m and n, we cannot determine whether m < n. Not sufficient.

(2) x is not divisible by 648

Factoring 648, we get 648 = 2^3 * 3^4.
Hence, this statement implies either m < 3 or n < 4, or both. For instance, if m = 2 and n = 10, then m < n. But if m = 10 and n = 3, m > n. Not sufficient.

(1)+(2) From (1) we know m ≥ 4, so the condition m < 3 from (2) cannot be true. Therefore, based on (2), n < 4 must be true. This means m ≥ 4 and n < 4, leading to the conclusion that m > n. Sufficient.

Answer: C.

M02-18
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Bunuel : Where to find the solutions or discussion for all the 40 Questions, I need to know whether my thought process and approach to the solution is okay or not. Probably, I can also learn something new.

Please respond..