Bunuel's Algebra PS Diagnostic Test

Hi Bunuel Thank you for this! Do you have any recommendation on how we should improve our score and increasing our speed? Currently I have 22 correct answer.

Thank you!

We have several topics dedicated to that issue:

5. Strategies and Tactics To Increase Your Score

3 Tips for Handling a Low GMAT Quant Score
Improving from 30-35 to 40-44
Improving from 44 to 50
Improving from 48 to 51 (7 chapters)
The Final Climb: Quest for Q 50+ from Q47/48
3 WAYS TO INCREASE YOUR GMAT SCORE TO A 760
GMAT TIP OF THE WEEK: YOUR 3 STEP PACING PLAN
How to Understand Your GMAT Practice Test Results and Score Higher Next Time

6. Strategies and Tactics To Speedd-Up

Want to speed up? Check this: Timing Strategies on the GMAT
Follow This Strategy to Save Time on the GMAT
How to Improve Your Timing on the GMAT
HOW TO MANAGE YOUR TIME ON THE GMAT
Pacing - No one question matters unless you let it
The Importance of Timing on the GMAT
The Difference Between a 1-Minute Solution and a 4-Minute Solution
Determining How Much Time to Spend on GMAT Quant Questions
Timing is Everything on the GMAT: One Strategy to Help You Succeed
3 Ways to Improve Your Timing on the GMAT

For more check below:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.

Very helpful test. Thank you Bunuel.

Answer to Q22 should be C.
Only statement III must be true.
Can you let me know if this is correct?

Question 26 - Official Solution:

If \(x^x=\sqrt{\frac{\sqrt{2}}{2}}\), which of the following could be a value of \(x\)?

A. \(\frac{1}{16}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{\sqrt{2}}\)
D. \(\frac{1}{2}\)
E. \(\sqrt{2}\)


We need to simplify the expression on the right-hand side of the equation so that it can be written in terms of the same base and exponent as \(x^x\).

\(\sqrt{\frac{\sqrt{2}}{2}}=\)

\(=\sqrt{\frac{1}{\sqrt{2}}}=\)

\(=\sqrt{(\frac{1}{2})^{\frac{1}{2}}}=\)

\(=(\frac{1}{2})^{\frac{1}{4}}=\)

Take to the above expression the fourth power and apply exponentiation to the base, \(\frac{1}{2}\):

\(=((\frac{1}{2})^{\frac{1}{4}})^4=\)

\(=((\frac{1}{2})^4)^{\frac{1}{4}}=\)

\(=(\frac{1}{16})^{\frac{1}{4}}\)

Now, to compensate the previous operation, take the fourth root, but this time apply exponentiation to the exponent, \(\frac{1}{4}\):

\(=((\frac{1}{16})^{\frac{1}{4}})^{\frac{1}{4}}=\)

\(=(\frac{1}{16})^{\frac{1}{16}}\)

Therefor, \(x=\frac{1}{16}\)

Else, after obtaining the expression \(x^x=(\frac{1}{2})^{\frac{1}{4}}\), we can substitute each of the given options for \(x\) to determine which one satisfies the equation.

Option A is a valid solution, as \((\frac{1}{16})^{(\frac{1}{16})}=((\frac{1}{16})^{(\frac{1}{4})})^{(\frac{1}{4})}=(\frac{1}{2})^{\frac{1}{4}}\)


Answer: A

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 14 - Official Solution:


\(x_1+x_2+...+x_{100} = 1\);

\(x_1+x_2+...+x_{99} = 2\);

\(x_1+x_2+...+x_{98} = 3\);

\(...\)

\(x_1= 100\).

What is the value of \(x_1*x_2*...*x_{100}\)?


A. \(-100\)
B. \(-1\)
C. \(0\)
D. \(1\)
E. \(100\)


By subtracting the second equation from the first, we obtain \((x_1+x_2+...+x_{100})-(x_1+x_2+...+x_{99}) = 1-2\), which simplifies to \(x_{100}= -1\).

Similarly, subtracting the third equation from the second yields \((x_1+x_2+...+x_{99}) -(x_1+x_2+...+x_{98}) = 2-3\), which simplifies to \(x_{99}= -1\).

By repeating this process for each successive pair of equations, we can determine that every term up to \(x_1\) is also equal to -1.

Therefore, \(x_1*x_2*...*x_{100}=100(-1)(-1)(-1)...(-1)=100(-1)^{99}=-100\).


Answer: A

Question 26 - Official Solution:

If \(x^x=\sqrt{\frac{\sqrt{2}}{2}}\), which of the following could be a value of \(x\)?

A. \(\frac{1}{16}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{\sqrt{2}}\)
D. \(\frac{1}{2}\)
E. \(\sqrt{2}\)


We need to simplify the expression on the right-hand side of the equation so that it can be written in terms of the same base and exponent as \(x^x\).

\(\sqrt{\frac{\sqrt{2}}{2}}=\)

\(=\sqrt{\frac{1}{\sqrt{2}}}=\)

\(=\sqrt{(\frac{1}{2})^{\frac{1}{2}}}=\)

\(=(\frac{1}{2})^{\frac{1}{4}}=\)

Take to the above expression the fourth power and apply exponentiation to the base, \(\frac{1}{2}\):

\(=((\frac{1}{2})^{\frac{1}{4}})^4=\)

\(=((\frac{1}{2})^4)^{\frac{1}{4}}=\)

\(=(\frac{1}{16})^{\frac{1}{4}}\)

Now, to compensate the previous operation, take the fourth root, but this time apply exponentiation to the exponent, \(\frac{1}{4}\):

\(=((\frac{1}{16})^{\frac{1}{4}})^{\frac{1}{4}}=\)

\(=(\frac{1}{16})^{\frac{1}{16}}\)

Therefor, \(x=\frac{1}{16}\)

Else, after obtaining the expression \(x^x=(\frac{1}{2})^{\frac{1}{4}}\), we can substitute each of the given options for \(x\) to determine which one satisfies the equation.

Option A is a valid solution, as \((\frac{1}{16})^{(\frac{1}{16})}=((\frac{1}{16})^{(\frac{1}{4})})^{(\frac{1}{4})}=(\frac{1}{2})^{\frac{1}{4}}\)


Answer: A

Question 28 - Official Solution:


If \(45x = 121y\), which of the following must be true?

I. \(x > y\)

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

III. \(\frac{x}{11}\) is an integer




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


It is important to note that the problem does not specify that \(x\) and \(y\) are necessarily integers, nor does it require \(x\) and \(y\) to be necessarily positive. Therefore, when evaluating each option, it is essential to keep in mind that these variables may take on non-integer or non-positive values!

I. \(x > y\).

This statement is not always true, as \(x\) and \(y\) are not necessarily positive. For instance, consider \(x = -121\) and \(y=-45\), or \(x = 0\) and \(y=0\).

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

This statement is not always true, as \(x\) and \(y\) are not necessarily integers. For example, consider \(x = \frac{1}{45}\) and \(y=\frac{1}{121}\). Additionally, since \(x\) and \(y\) can both be 0, this statement is also not true in that case.

III. \(\frac{x}{11}\) is an integer.

This statement is also not always true, as \(x\) and \(y\) are not necessarily integers. For instance, consider \(x = \frac{1}{45}\) and \(y=\frac{1}{121}\).


Answer: E

Question 40 - Official Solution:

If \(n\) is an integer greater than 1, what is the value of \(10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...}}}}\), where the given expression extends to an infinite number of roots?

A. \(10\)
B. \(10^{\frac{1}{n}}\)
C. \(10^{\frac{n-1}{n}}\)
D. \(10^{\frac{n}{n-1}}\)
E. \(10^{n}\)


Let \(x=10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...}}}}\)

Now, re-write above as \(x=10*\sqrt[n]{(10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...})}}}\).

Since the expression extends to an infinite number of roots, then the expression in brackets would also equal to \(x\). Thus we can replace the expression in brackets with \(x\) and rewrite the expression as: \(x=10*\sqrt[n]{x}\)

Take above to the \(n^{th}\) power:

Take \(n-1^{th}\) root:


Answer: D

question 4 - Official Solution:

If \(7^8 = m\) and \(8^7 = n\), then what is the value of \(56^{56}\) in terms of \(m\) and \(n\) ?

A. \(mn\)
B. \(m^7*n^8\)
C. \(m^8*n^7\)
D. \((mn)^{56}\)
E. \(56^{mn}\)


First, we notice that \(56\) can be written as \(7 *8\), so we can rewrite \(56^{56}\) as \((7 *8)^{56}\)

Next, simplify \((7* 8)^{56}\) to \(7^{56}* 8^{56}\).

Finally, we substitute \(m\) for \(7^8\) and \(n\) for \(8^7\), giving us:

\(56^{56} = (7*8)^{56} = 7^{56} *8^{56} = (7^8)^7* (8^7)^8 = m^7* n^8.\)


Therefore, the value of \(56^{56}\) in terms of \(m\) and \(n\) is \(m^7* n^8\).


Answer: B

Question 9 - Official Solution:

If \(x > 0\) and \(\sqrt{17^{\sqrt{x}}} = 17^{\frac{1}{\sqrt{x}}}\), what is the value of \(x\) ?

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


To tackle this problem, we'll start by simplifying the equation so that the bases on both sides are the same. Once we've achieved this, we can equate the powers and solve for \(x\). To equate the bases, we can square the equation.

\((\sqrt{17^{\sqrt{x}}})^2 = (17^{\frac{1}{\sqrt{x}}})^2\)

\(17^{\sqrt{x}} = 17^{\frac{2}{\sqrt{x}}}\)

\(\sqrt{x} = \frac{2}{\sqrt{x}}\)

Cross-multiply:

\(x = 2\)


Answer: D

Question 10 - Official Solution:


If \(m\) is a positive number and \(n\) is a negative number, and \(|m| > |n|\), then which of the following has the greatest value ?




A. \(|\frac{m - n}{n}|\)
B. \(|\frac{m - n}{m}|\)
C. \(|\frac{m + n}{m - n}|\)
D. \(|\frac{m + n}{n}|\)
E. \(|\frac{m + n}{m}|\)


Probably the easiest wat to solve this problem would be plug the values. Let \(m=2\) and \(n=-1\) (this satisfies conditions given in the stem), then:

As we can see the answer is option A.



Still, to test our absolute value skills, let's also solve the question using absolute value properties. Let's establish two points:







A and B have the same numerator so let's compare these two options first (both the denominator and numerator are positive, so the one with smaller denominator will have the larger value). Since given that \(|m| > |n|\), the A > B.

B, C and E have the same numerator, so let's compare these three options next (again, both the denominator and numerator are positive, so the one with smaller denominator will have the larger value). The denominator of E is less then that of D (it's given that \(|m| > |n|\)) and since also given that \(m\) is a positive number and \(n\) is a negative number, the denominator of E is also less then that of C.

So, we are left to compare A and E. The numerator of A is greater than that of E (check (i) above) plus the denominator of A is less than that of E (it's given that \(|m| > |n|\)), so A > E. Therefore, A has the greatest value.


Answer: A

Question 18 - Official Solution:


The infinite sequence \(a_1, \ a_2, \ …, \ a_n, \ …\) is such that \(a_n = n!\) for \(n > 0\). What is the sum of the first 11 terms of the sequence?




A. \(43954414\)
B. \(43954588\)
C. \(43954675\)
D. \(43954713\)
E. \(43954780\)


According to the sequence formula given, the sequence is: \(1!, \ 2!, \ 3!, \ ... \). We are asked to find the value of:

Direct calculation will take long time and will be quite tiresome. So, what to?

First notice that the units digits of answer choices are different. Next, notice that from 5! onwards the units digit of all terms will be 0 (because each of them will have a 2 and a 5 in it). So, to get the units digit of the sum we only need to calculate the units digit of \(1!+2!+3!+4!\), which is 3 (\(1!+2!+3!+4!=1+2+6+24=33\)).

Thus, the units digit of \(1!+2!+3!+4!+5!+6!+7!+8!+9!+10!+11!\) will also be 3. The only option which has a number with the units digit of 3 is D.


Answer: D

Question 19 - Official Solution:


What is the value of \(|\sqrt{3} - |\sqrt{5}-\sqrt{7}|| - ||\sqrt{3} - \sqrt{5}|-\sqrt{7}|\) ?


A. \(2\sqrt{5}-2\sqrt{7}-2\sqrt{3}\)
B. \(2\sqrt{5}-2\sqrt{7}\)
C. \(2\sqrt{7}-2\sqrt{5}\)
D. \(2\sqrt{5}-2\sqrt{7}+2\sqrt{3}\)
E. \(2\sqrt{7}+2\sqrt{5}\)


To answer this question we should recall the property of the absolute value:

So, we should evaluate the expressions in the modulus to see whether they are positive or negative.

STEP 1:

Thus, \(|\sqrt{3} - |\sqrt{5}-\sqrt{7}|| - ||\sqrt{3} - \sqrt{5}|-\sqrt{7}|\) will become:

STEP 2:

Thus, \(|\sqrt{3} +\sqrt{5}-\sqrt{7}| - |\sqrt{5} -\sqrt{3}-\sqrt{7}|\) will become:


Answer: B