Bunuel's Algebra PS Diagnostic Test

­Question 37 - Official Solution:

If \(x\) and \(y\) are consecutive integers, \(x > y\), and \(x^2 - 1 > y^2 - 4y + x - 1\), which of the following must be true?

A. \(x < 0\)
B. \(x ≠ 2\)
C. \(y > -1\)
D. \(y ≠ 1\)
E. \(x + y > 3\)


Given that \(x\) and \(y\) are consecutive integers, and \(x > y\), we can say that \(x = y + 1\). Substituting this into the given inequality, we get:

Since \(y = x - 1\), we can rewrite this inequality in terms of \(x\): \(x - 1 > 0\), which gives \(x > 1\). \(x\) can be 2, 3, 4, 5, and so on.

Since \(y > 0\), then \(y\) must also be greater than -1, making option C correct.

Even though we have already determined that option C is correct, let's elaborate on the other options for the sake of understanding why they are not necessarily true:

A. \(x < 0\). We found that \(x > 1\), so this option is not true. Discard.

B. \(x ≠ 2\). Since we found that \(x > 1\), this option could be true, but it's not necessarily true. Discard.

D. \(y ≠ 1\). Since we found that \(y > 0\), this option could be true, but it's not necessarily true. Discard.

E. \(x + y > 3\). Since we found that \(y > 0\) and \(x > 1\), the smallest possible value for \(x + y\) is \(1 + 2 = 3\). Therefore, this option is not necessarily true. Discard.


Answer: C­

Question 16 - Official Solution:

A merchant offers a 30% discount on her list price but still manages to make a profit of 19% on her cost price. What was her original profit percentage on the cost price before the discount?

A. 76%
B. 70%
C. 57%
D. 55%
E. 49%


Let the original list price be \(p\) and the cost price be \(c\).

After a 30% discount, the selling price is \(0.7p\). Since she makes a 19% profit on her cost price, this selling price is equal to \(1.19c\). From this, we get \(0.7p = 1.19c\). This simplifies to \(\frac{7}{10}*p = \frac{119}{100}*c\), leading us to \(p = 1.7c\). Therefore, the original profit percentage on the cost price before the discount was 70%.


Answer: B­

Question 17 - Official Solution:

If \((x + y)^2 < x^2\), which of the following must be true?

I. \(y(y + 2x) < 0\)

II. \(y < x\)

III. \(xy < 0\)

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


First, let's simplify the given inequality:
Now, let's evaluate each option:

I. \(y(y + 2x) < 0\). This option is directly true since it matches our simplified inequality.

II. \(y < x\). From \(y(2x + y) < 0\), we can infer that \(y\) and \(2x + y\) must have different signs. Let's try finding a counterexample: if \(y\) is positive, say 1, then \(2x + y\) must be negative, which can be obtained if \(x\) is, for example, -10, making \(y > x\). Hence, this option is not necessarily true.

III. \(xy < 0\). We derived that \(2xy + y^2 < 0\) earlier. Since \(y^2\) is always non-negative, \(2xy\) must be negative for the inequality to hold. This implies that \(xy\) is negative. Thus, this option is always true.

Therefore, only options I and III must be true.


Answer: E­

Question 31 - Official Solution:

If \([x]\) is the greatest integer less than or equal to \(x\), and \([√x] = 5\) and \([√y] = 6\), where \(x\) and \(y\) are positive integers, what is the greatest possible value of \(x + y\) ?

A. \(61\)
B. \(81\)
C. \(82\)
D. \(83\)
E. \(85\)


Some function rounds DOWN a number to the nearest integer. For example:

\([2.7] = 2\) because 2 the largest integer less than or equal to 2.7;

\([3] = 3\) because 3 the largest integer less than or equal to 3;

\([-1.7] = -2\) because -2 the largest integer less than or equal to -1.7.

So, \([√x] = 5\) means that 5 is the largest integer less than or equal to \(√x\), which implies that \(5 \leq √x < 6\). For any number from that range, the largest integer less than or equal that number will be 5. Square: \(25 \leq x < 36\). The greatest integer value of \(x\) is 35.

Similarly, \([√y] = 6\) means that 6 is the largest integer less than or equal to \(√y\), which implies that \(6 \leq √y < 7\). For any number from that range, the largest integer less than or equal that number will be 6. Square: \(36 \leq y < 49\). The greatest integer value of \(y\) is 48.

The greatest integer value of \(x+y=35+48=83\)


Answer: D­

Question 1 - Official Solution:


If the average of \(a\), \(b\), \(c\), 14 and 15 is 12. What is the average value of \(a\), \(b\), \(c\) and 29?


A. \(12\)
B. \(13\)
C. \(14\)
D. \(15\)
E. \(16\)


The average of \(a\), \(b\), \(c\), 14 and 15 is 12 means:

We want to find the average of \(a\), \(b\), \(c\) and 29, so the value of \(\frac{a+b+c+29}{4}\).

Since from above we have that \(a+b+c+29=60\), then \(\frac{a+b+c+29}{4}=\frac{60}{4}=15\).


Answer: D

Question 3 - Official Solution:


Jeeves prepares a hangover cure using four identical cocktail shakers. The first shaker is \(\frac{1}{2}\) full, the second shaker is \(\frac{1}{3}\) full, the third shaker is \(\frac{1}{4}\) full and the last one is empty. If Jeeves redistributes all the content of the shakers equally into the four shakers, what fraction of each shaker will be filled?


A. \(\frac{13}{48}\)
B. \(\frac{4}{13}\)
C. \(\frac{13}{36}\)
D. \(\frac{9}{13}\)
E. \(\frac{35}{48}\)


Let's denote the capacity of each shaker as \(x\) and calculate the total amount of liquid in them:

\(\frac{x}{2} + \frac{x}{3} + \frac{x}{4} = \frac{13x}{12}\).

Now, if we redistribute the liquid equally among the four shakers, each shaker will receive 1/4 of the total amount of liquid, which is:

\(\frac{1}{4}*\frac{13x}{12} = \frac{13x}{48}\).

Therefore, each shaker will be filled with \(\frac{13}{48}\) of its capacity.


Answer: A

Question 11 - Official Solution:


Jeeves prepares a hangover cure using four identical cocktail shakers. The first shaker is \(\frac{1}{2}\) full, the second shaker is \(\frac{4}{5}\) full, the third shaker is \(\frac{1}{k}\) full and the last one is empty. After Jeeves redistributed all the content of the shakers equally into the four shakers, each shaker became \(\frac{31}{80}\) full. What is the value of \(k\)?


A. \(2\)
B. \(3\)
C. \(4\)
D. \(5\)
E. \(6\)


Let's denote the capacity of each shaker by \(x\).

Then, the amount of liquid in the first shaker is \(\frac{1}{2}*x\), the amount of liquid in the second shaker is \(\frac{4}{5}*x\), the amount of liquid in the third shaker is \(\frac{1}{k}*x\), and the amount of liquid in the fourth shaker is 0.

After redistributing the liquid equally among the four shakers, each shaker will contain \(\frac{1}{4}^{th}\) of the total amount of liquid, which is given to be \(\frac{31}{80}*x\), so we have:


Answer: C

Question 5 - Official Solution:

5. If \(ab^2 > 0\) and \(ac < 0\), which of the following must be true?

I. \(ab >0\)

II. \(b^2c < 0\)

III. \(a*c^2 > 0\)

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

The first condition, \(ab^2 > 0\), to hold, \(a\) and \(b^2\) must have the same sign and since \(b^2\) cannot be negative, then both must be positive. Hence, \(ab^2 > 0\) implies that \(a > 0\) and \(b \neq 0\).

The second condition, \(ac < 0\) to hold, \(a\) and \(c\) must have opposite signs. Since from above we established that \(a > 0\), then \(c < 0\).

Thus, we have established that \(a > 0\), \(b \neq 0\), and \(c < 0\). Let's evaluate each option:

I. \(ab >0\). Since we don't know the sign of \(b\) and only know that \(b \neq 0\), this option may or may not be true.

II. \(b^2c < 0\). Since \(b \neq 0\) and \(c < 0\), this option must be true.

III. \(a*c^2 > 0\). Since \(a > 0\) and \(c \neq 0\), this option must be true.

Answer: D­

Question 2 - Official Solution:

If \(22 - |y + 14| = 20\), what is the sum of all possible values of y ?

A. \(-28\)
B. \(-16\)
C. \( -12\)
D. \(-4\)
E. \(4\)
Therefore, the sum of all possible values of \(y\) is \(-12 + (-16) = -28.\)

Answer: A­

Question 6 - Official Solution:

What is the value of \(\sqrt{(49\%)} -\frac{7}{100}\) ?

A. 0
B. 0.63%
C. 0.063
D. 63%
E. 6.3


Worth knowing that "per cent" from Latin literally means "per one hundred" or "out of one hundred", so for example, \(x\%\) is \(\frac{x}{100}\) (\(x\) per one hundred) and say \(5\%\) is \(\frac{5}{100} = 0.05\) (5 out of one hundred). On the other hand, we can write 0.05 as \(0.05*100\% = 5\%\) and say 20 as \(20*100\% = 2000\%\).

Basically, "%" symbol just means "per 100", or algebraically, "/100". Thus:
Back to the question..
Answer: D­

Question 7 - Official Solution:

What is the value of \((12\%)^2\)?

A. \(0.144\%\)
B. \(1.44\%\)
C. \(0.144\)
D. \(144\%\)
E. \(14.4\)


Worth knowing that "per cent" from Latin literally means "per one hundred" or "out of one hundred", so for example, \(x\%\) is \(\frac{x}{100}\) (\(x\) per one hundred) and say \(5\%\) is \(\frac{5}{100} = 0.05\) (5 out of one hundred). On the other hand, we can write 0.05 as \(0.05*100\% = 5\%\) and say 20 as \(20*100\% = 2000\%\).

Basically, "%" symbol just means "per 100", or algebraically, "/100". Thus:
Back to the question..
Answer: B­

Question 12 - Official Solution:

If the sum of all 21 terms of an arithmetic progression is zero, then which of the following MUST be true ? (An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant)

I. 11th smallest term is zero.

II. 11th largest term is zero.

III. The range of the numbers is 20.

A. I only
B. II only
C. III only
D. I and II only
E. None of the above


The sum of ALL 21 terms of an arithmetic progression being zero, implies that 10 terms are less than 0, 10 corresponding terms, with the same magnitude are greater than 0 and the middle term is 0.

The above would make the 11th smallest term equal to zero and the 11th largest term equal to zero. However, since we can stretch or shrink the sequence any way we'd like, the range—the positive difference between the largest and smallest terms—could be anything, not necessarily equal to 20.

Therefore, only options I and II must be true.

Answer: D­

Question 15 - Official Solution:

What is the value of \((\sqrt[3]{800\%})^2\)?

A. \(0.04\%\)
B. \(0.4\)
C. \(100\%\)
D. \(2\)
E. \(400\%\)


Worth knowing that "per cent" from Latin literally means "per one hundred" or "out of one hundred", so for example, \(x\%\) is \(\frac{x}{100}\) (\(x\) per one hundred) and say \(5\%\) is \(\frac{5}{100} = 0.05\) (5 out of one hundred). On the other hand, we can write 0.05 as \(0.05*100\% = 5\%\) and say 20 as \(20*100\% = 2000\%\).

Basically, "%" symbol just means "per 100", or algebraically, "/100". Thus:


Back to the question..


Answer: E­

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

Question 33 - Official Solution:

\(\{x, \ y, \ z\}\) is an increasing sequence of numbers such that the ratio between the consecutive terms is constant.

\(\{x + 2y, \ 2x + y + z, \ x + 3y + z\}\) is an increasing sequence of numbers such that the difference between the consecutive terms is constant.

What is the value of \(\frac{x}{z}\) ?

A. \(\frac{1}{4}\)
B. \(\frac{1}{2}\)
C. \(1\)
D. \(2\)
E. \(4\)


The first statement implies that \(\frac{x}{y}=\frac{y}{z}\), which yields \(y^2 = xz\).

The second statement implies that \((2x + y + z) - (x + 2y) = (x + 3y + z) - (2x + y + z)\), which yields \(3y = z + 2x\).

Squaring \(3y = z + 2x\) yields \(9y^2= (z + 2x)^2\). Substituting \(y^2 = xz\) gives us \(9xz = (z + 2x)^2\).
\(4x - z = 0\) or \(x - z = 0\), which is not possible since it would give \(x = z\), and this is not possible given that \(\{x, y, z\}\) is an increasing sequence.

Therefore, \(4x = z\), and thus \(\frac{x}{z}=\frac{1}{4}\).

Answer: A­

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