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rakpachu
please someone help me to solve Bunuel's Algebra Diagnostic Test level 6 problem number 26

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
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Question 20 - Official Solution:

If \(x > 0\) and \(x^{(3*x^{12})}=4\), what is the value of \(x\) ?

A. \(\sqrt[12]{2}\)
B. \(\sqrt[6]{2}\)
C. \(\sqrt[3]{2}\)
D. \(\sqrt{3}\)
E. \(\sqrt{2}\)


\(x^{(3*x^{12})}=4\);

Take to the fourth power:

\((x^{(3*x^{12})})^4=4^4\);

\(x^{(12*x^{12})}=4^4\);

\((x^{12})^{(x^{12})} =4^4\);

\(x^{12}=4=2^2\);

\(x=(2^2)^{(\frac{1}{12})}\);

\(x=2^{(\frac{1}{6})}\);

\(x=\sqrt[6]{2}\)


Answer: B
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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:

\(x^n=10^n*x\)

\(x^{n-1}=10^n\)

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

\(x=10^{\frac{n}{n-1}}\)


Answer: D
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Question 23 - Official Solution:

What is the value of \(4^{3^{2^{1^{2^{3^4}}}}}\)?

A. \(2,144\)
B. \(262,142\)
C. \(262,144\)
D. \(262,146\)
E. \(262,148\)


To evaluate \(4^{3^{2^{1^{2^{3^4}}}}}\), we need to remember the rule for working with stacked exponents, which is to begin with the highest exponent and work our way down. For example, \(a^{m^n}\) means we first compute \(m^n\) and then use that result as the exponent for \(a\). Therefore, \(a^{m^n} = a^{(m^n)}\). On the other hand, if we have \((a^m)^n\), we compute \(a^m\) first and then raise the result to the power of \(n\), so \((a^m)^n=a^{mn}\).

Let's evaluate the exponents of 4 in our expression. Since \(1^{2^{3^4}}\) equals 1, we can ignore it completely, which means we will be left with \(4^{3^2}\).

Using the exponent rule, \(4^{3^2} = 4^9\).

Since the GMAT is a timed test, we may not have the luxury of actually computing the value of \(4^9\). Instead, we need to use some shortcuts to arrive at the answer. Fortunately, we can notice that the units digit of each answer choice is different, and if we can determine the units digit of \(4^9\), we can identify the correct answer.

The units digit of \(4^k\), where \(k\) is a positive integer, alternates between 4 and 6 for odd and even values of \(k\), respectively. Since 9 is an odd number, the units digit of \(4^9\) is 4.

We can now eliminate any answer choices that do not end in 4. The only answer choice that ends in 4 is C, so our answer is \(4^{3^{2^{1^{2^{3^4}}}}} = 262,144\).


Answer: C
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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:

\(1!+2!+3!+4!+5!+6!+7!+8!+9!+10!+11!\).

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
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Question 27 - Official Solution:

A cloth merchant professes to sell at a loss of 5%. However, the merchant falsifies the meter scale and gains 25%. What is the measure of the scale, in centimeters, that he uses when measuring 1 meter of cloth? (1 meter = 100 centimeters)

A. 70
B. 72
C. 75
D. 76
E. 77


Let's assume that the cost of 1 centimeter of cloth is $1. Therefore, for a meter (100 centimeters), the base cost would be $100. However, the merchant claims to sell 1 meter of cloth at a 5% loss, which means he is selling it for $95 (95% of the original cost).

In reality, this $95 still incorporates a 25% gain for the merchant, which means \(1.25 * ($1 * \text{actual length in meters}) = $95\). This yields \(\text{actual length in meters} = 0.76\). So, the actual length of cloth the merchant claims to be 1 meter is, in fact, only 0.76 meters.


Answer: D­
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Question 8 - Official Solution:

If \(\sqrt{2+\sqrt{2+x}} +\sqrt{2-\sqrt{2+x}}=\sqrt{2x}\), what is the value of \(x\) ?

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

APPROACH 1 - USING ANSWER OPTIONS AND ESTIMATION:

To succeed on the GMAT, it's essential to master working backward and estimation techniques. Let's see how we can apply these tools to this problem.

Firstly, note that since all numbers on the GMAT are real numbers, the even roots of negative numbers are undefined. Therefore, options A and B can be ruled out immediately because if \(x\) is negative, \(\sqrt{2x}\) will not be defined.

Let's check the other answer options.

If \(x = 1\), the right-hand side becomes \(\sqrt{2x}=\sqrt{2}\), while the left-hand side's first term will be \(\sqrt{2+\sqrt{2+1}} =\sqrt{2+\sqrt{3}}\). Since this expression is already greater than \(\sqrt{2}\), adding a positive value (\(\sqrt{2-\sqrt{2+x}}\)) to it will definitely result in a greater value than \(\sqrt{2}\). We can eliminate this option.

If \(x = \sqrt{2}\), the right-hand side becomes \(\sqrt{2x}=\sqrt{2*\sqrt{2}}\approx \sqrt{2*1.4}=\sqrt{2.8}\). Meanwhile, the left-hand side's first term will be \(\sqrt{2+\sqrt{2+x}} =\sqrt{2+\sqrt{2+\sqrt{2}}}\approx \sqrt{2+\sqrt{2+1.4}}= \sqrt{2+\sqrt{3.4}}\). Since \(\sqrt{3.4}\) is definitely greater than 1, we can conclude that \(\sqrt{2+\sqrt{3.4}} > \sqrt{2+1} > \sqrt{2.8}\). Therefore, adding some positive value (\(\sqrt{2-\sqrt{2+x}}\)) to this expression will definitely result in a greater value than \(\sqrt{2.8}\). We can eliminate this option as well.

Therefore, the answer must be E. Nonetheless, for practice sake, we can still test \(x = 2\):

The right-hand side becomes \(\sqrt{2x}=\sqrt{2*2}=2\), and the left-hand side becomes:

\(\sqrt{2+\sqrt{2+x}} +\sqrt{2-\sqrt{2+x}}=\)

\(=\sqrt{2+\sqrt{2+2}} +\sqrt{2-\sqrt{2+2}}=\)

\(=\sqrt{2+2} +\sqrt{2-2}=\)

\(=2+0=2\). The answers match!

APPROACH 2 - ALGEBRA:

We can start by squaring both sides of the given equation:

\((\sqrt{2+\sqrt{2+x}} +\sqrt{2-\sqrt{2+x}})^2=(\sqrt{2x})^2\)
Expanding the left-hand side and simplifying, we get:

\((2+\sqrt{2+x}) + 2*(\sqrt{2+\sqrt{2+x}})(\sqrt{2-\sqrt{2+x}})+(2-\sqrt{2+x})=2x\)

\(4 + 2*\sqrt{ (2+\sqrt{2+x}) (2-\sqrt{2+x}) }=2x\)

\(4 + 2*\sqrt{2^2-(\sqrt{2+x})^2}=2x\)

\(4 + 2*\sqrt{4-(2+x)}=2x\)

\(4 + 2*\sqrt{2-x}=2x\)

\(2*\sqrt{2-x}=2x-4\)

\(\sqrt{2-x}=x-2\).
Since the square root of a number cannot be negative, \(x-2\) must be positive or zero. This implies that \(x \geq 2\).

Squaring both sides of the equation \(\sqrt{2-x}=x-2\), we get:

\((\sqrt{2-x})^2=(x-2)^2\)

\(2-x=x^2-4x+4\)

\(x^2-3x+2=0\)
\(x = 1\) or \(x = 2\). Since we found that \(x \geq 2\), the only valid solution is \(x = 2\).

Answer: E­
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Question 25 - Official Solution:

If \(9^x + 9^{-x} = 62\), then what is the value of \(3^x + 3^{-x}\)?

A. \(\frac{1}{64}\)
B. \(\frac{1}{8}\)
C. \(8\)
D. \(62\)
E. \(64\)


We are given that \(9^x + 9^{-x} = 62\). To find the value of \(3^x + 3^{-x}\), we can square it and simplify:

\((3^x + 3^{-x})^2 = \)

\(=(3^x)^2 + 2*3^x*3^{-x} +(3^{-x})^2=\)

\(=9^x + 2*3^x*\frac{1}{3^{x}} +9^{-x}=\)

\(=9^x + 2+ 9^{-x}\)

\(=62 + 2 = 64\)

Now, since \(3^x + 3^{-x}\) squared is 64, then \(3^x + 3^{-x}\) is 8 (notice that it cannot be -8 since \(3^x + 3^{-x}\) is positivefor all values of \(x\)).


Answer: C
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Question 22 - Official Solution:

If the sum of the first 31 terms of an arithmetic progression consisting of 46 terms is zero, then which of the following MUST be true? (An arithmetic progression is a sequence of numbers such that the difference between consecutive terms is constant.)

I. The 31st smallest term is zero.

II. The 16th largest term is zero.

III. The sum of the largest and smallest terms of the sequence is positive.


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

Firstly, note that we don't know whether the sequence is increasing or decreasing.

If the sequence is increasing, then the fact that the sum of the first 31 terms of the arithmetic progression is zero would imply that:

\(a_1 \leq ... \leq (a_{16} = 0) \leq ... \leq a_{31} \leq ... \leq a_{46}\)
If the sequence is decreasing, then the fact that the sum of the first 31 terms of the arithmetic progression is zero would imply that:

\(a_1 \geq ... \geq (a_{16} = 0) \geq ... \geq a_{31} \geq ... \geq a_{46}\)
As we can see, none of the options MUST be true.

Answer: E­
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Question 34 - Official Solution:

Mbappe withdraws \(\frac{100}{x}%\) of the amount in his account each time he visits the bank, where \(x\) is an integer greater than 1. If, after 5 visits, he has less than \(\frac{1}{x}^{th}\) of the initial amount left in the bank, what is the range of possible values for \(x\)?

A. 1
B. 2
C. 3
D. 4
E. 5


Since \(\frac{100}{x}\%\) of the money is withdrawn each time, after each visit, \((1 - \frac{100}{x}*\frac{1}{100})= (1 - \frac{1}{x})\) of the previous amount will remain in the bank. For example, if 10% is withdrawn each time, after each visit, \((1 - 10*\frac{1}{100})= (1 - \frac{1}{10})=\frac{9}{10}\) of the previous amount will remain in the bank.

Assuming the initial amount in the bank is A, then after 5 visits, the amount would be \(A(1 - \frac{1}{x})^5\). We are told that this is less than \(\frac{1}{x}\) of the initial amount, so we have:

\(A(1 - \frac{1}{x})^5 < A*\frac{1}{x}\)

\((1 - \frac{1}{x})^5 < \frac{1}{x}\)

\((\frac{x-1}{x})^5 < \frac{1}{x}\)

\(\frac{(x-1)^5}{x^5} < \frac{1}{x}\)

\((x-1)^5 < x^4\)
As \(x\) increases, the left-hand side of the inequality grows faster than the right-hand side, so the inequality will only hold for smaller values of \(x\). Let's test some values. For \(x = 5\), the inequality does not hold: \(4^5 > 5^4\). However, if \(x\) is 4 or less, it does hold. Since \(x\) is an integer greater than 1, it can be 2, 3, or 4, making the range of possible values for \(x\) equal to \(4 - 2 = 2\).


Answer: B­
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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 also not always true. For example, consider \(x = y = 0\).

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
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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
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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
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Question 24 - Official Solution:

If \(x\) is an integer and \(\frac{(x^2 - 2)!}{x^2-2} = 33!\), what is the range of all possible values of \(x\)?

A. 0
B. 2
C. 6
D. 11
E. 12


Since \(\frac{n!}{n} = \frac{(n-1)!*n}{n}=(n-1)!\), then:

\(\frac{(x^2 - 2)!}{x^2-2} = \)

\(=\frac{(x^2 - 2-1)!*(x^2 - 2)}{x^2-2} =\)

\(=(x^2 - 2-1)!=\)

\(=(x^2 - 3)!\)
Thus, we'd have:

\((x^2 - 3)!=33!\)

\(x^2 - 3=33\)

\(x^2=36\)

\(x=-6\) or \(x=6\)
Thus, the range of all possible values of \(x\) is (the largest value) - (the smallest value) \(= 6 - (-6) = 12\).

Answer: E­
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Question 29 - Official Solution:


If \(x\) and \(y\) are integers \(35x = 69y\), which of the following must be true?

I. \(x > y\)

II. \(\frac{y}{7}\) is an integer

III. \(\frac{x}{23}\) is an integer


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


It is important to note that the problem does not specify that \(x\) and \(y\) are necessarily positive. Therefore, when evaluating each option, it is essential to keep in mind that these variables may take non-positive values!

I. \(x > y\).

This statement is not always true, as \(x\) and \(y\) are not necessarily positive. For instance, consider \(x = -69\) and \(y=-35\), or \(x = 0\) and \(y=0\).

II. \(\frac{y}{7}\) is an integer

From \(35x = 69y\), we can infer that since the left-hand side, \(35x = 7(5x)\), is a multiple of 7, the right-hand side, \(69y\), must also be a multiple of 7. Since 69 is not a multiple of 7 (prime number), \(y\) must be, making \(\frac{y}{7}\) an integer. Hence, this option is always true.

III. \(\frac{x}{23}\) is an integer

From \(35x = 69y\), we can infer that since the right-hand side, \(69y = 23(3y)\), is a multiple of 23, the left-hand side, \(35x\), must also be a multiple of 23. Since 35 is not a multiple of 23 (prime number), \(x\) must be, making \(\frac{x}{23}\) an integer. Hence, this option is always true.

Answer: D­
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Question 32 - Official Solution:

If \(x \neq2\) and \(\frac{|x| - 2}{|x - 2|} = 1\), then which of the following must be true?

I. \(|x| > 2\)

II. \(x^2 < 1\)

III. \(x^3 > 0\)


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


Note that \(\frac{|x| - 2}{|x - 2|} = 1\) implies that both \(x\) and \(x - 2\) must be positive. In this case, the left-hand side becomes \(\frac{x - 2}{x - 2}\), which equals 1. For both \(x\) and \(x - 2\) to be positive, \(x\) must be greater than 2. Therefore, \(\frac{|x| - 2}{|x - 2|} = 1\) implies that \(x > 2\).

Alternatively, we can derive the same conclusion using the following algebraic manipulations:

    \(\frac{|x| - 2}{|x - 2|} = 1\)

Cross-multiply:

    \(|x| - 2= |x - 2|\);

    \(|x|= |x - 2|+2\)

Square both sides (we can do this safely since both sides of the equation are positive):

    \(x^2 = x^2 - 4x + 4 + 4|x-2| + 4\)

    \(x - 2 = |x-2|\)

The above implies that \(x - 2 > 0\) (recall that \(|a| = a\) when \(a \geq 0\)).

Now, \(x - 2 > 0\) means that \(x > 2\).

So, the question essentially asks if \(x > 2\), then which of the following must be true?

I. \(|x| > 2\). Since the possible values of \(x\) are such that \(x > 2\), \(|x| > 2\) will always be true. So, this option is always true.

II. \(x^2 < 1\). This option is never true because if \(x > 2\), then \(x^2\) will be greater than 1, not less than 1.

III. \(x^3 > 0\). Since \(x\) is positive (\(x > 2\)), the cube of \(x\) will always be greater than 0. Thus, this option is always true.


Answer: D

Check other similar questions from Trickiest Inequality Questions Type: Confusing Ranges (part of our Special Questions Directory).

Hope it helps.­
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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:

A. \(|\frac{m - n}{n}|=|\frac{2 - (-1)}{-1}|=|-3|=3\).

B. \(|\frac{m - n}{m}|=|\frac{2 - (-1)}{2}|=|1.5|=1.5\).

C. \(|\frac{m + n}{m - n}|=|\frac{2 + (-1}{2 - (-1)}|=|\frac{1}{3}|=\frac{1}{3}\)

D. \(|\frac{m + n}{n}|=|\frac{2 + (-1)}{-1}|=|-1|=1\).

E. \(|\frac{m + n}{m}|=|\frac{2 + (-1)}{2}|=|\frac{1}{2}|=\frac{1}{2}\)

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:

(i) \(|m - n| > |m + n|\)

(this is because it's given that \(m\) is a positive number and \(n\) is a negative number. For example, \((|5 - (-1)| = 6) > (|5 + (-1)| = 4)\)).



(ii) \(|\frac{a}{b}|=\frac{|a|}{|b|}|\). So:



A. \(|\frac{m - n}{n}|=\frac{|m - n|}{|n|}\)

B. \(|\frac{m - n}{m}|=\frac{|m - n|}{|m|}\)

C. \(|\frac{m + n}{m - n}|=\frac{|m + n|}{|m - n|}\)

D. \(|\frac{m + n}{n}|=\frac{|m + n|}{|n|}\)

E. \(|\frac{m + n}{m}|=\frac{|m + n|}{|m|}\)



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
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