Bunuel's Algebra PS Diagnostic Test

28. If 45x = 121y, which of the following must be true?

I. x > y
II. x^2 > y^2
III. x/11 is an integer

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

Solution:

MUST BE TRUE: Should be always true irrespective of any condition. Let us now examine the statements (Find the condition that makes the statement false and MUST BE TRUE cannot happen)

I. x > y

If x = y = 0, then 45x = 121y, but x is NOT > y

II. x^2 > y^2

If x = y = 0, then 45x = 121y, but x^2 NOT > y^2

III. x/11 is an integer

If x = 121/45 and y = 45/121, then 45x = 121y, but x/11 is NOT an integer

So, None of the above is MUST BE TRUE

Answer: E

29. If x and y are integers 35x = 69y, which of the following must be true?

I. x > y
II. y/7 is an integer
III. x/23 is an integer

(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III

Solution:

MUST BE TRUE: Find the condition that can break the statement and that statement cannot be MUST BE TRUE

I. x > y

If x = y = 0: Both are integers, 35x = 69y, BUT x NOT > y

II. y/7 is an integer

Only conditions where 35x = 69y where x and y are integers are:

a) x = y = 0
b) x = 69, y = 35
c) x = -69, y = -35

In any of these cases, y/7 is always an integer. MUST BE TRUE

III. x/23 is an integer

Similar to II, the same three cases: In all 3 cases, x/23 will always be an integer. MUST BE TRUE

II & III: MUST BE TRUE

Answer: D

30. 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

Solution:

Root(15) = 3.88
Root(6) = 2.45
Root(10) = 3.16

=> Expression = \((\sqrt{0.12})(7.88)(2.45 - 3.16)\)

\(\sqrt{0.12} = \sqrt{\frac{12}{100}} = \frac{2\sqrt{3}}{10}\)

=> Expression = \((\frac{2\sqrt{3}}{10})(7.88)(2.45 - 3.16)\)
=> On solving, we get an approximate value of -2

Answer: B

p.s. Since I knew the values of roots, that is why I was able to solve it, but still not the GMAT way, because GMAT does not require such calculations. Different approaches are welcome!

31. 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

Solution:

[x] = Greater integer less than or equal to x => [5.1] = 5, and even [5.9] = 5 but [6] = 6

So, [√x] = 5 => 25 <= x < 36 (Because is x is anywhere in this range, sqrt(x) will be between 5 and 6 and hence value will be 5)
Similarly, [√y] = 6 => 36 <= y < 49

x + y = 35 + 48 = 83

Answer: D

32. If \(\frac{|x| - 2}{|x - 2|}= 1\), 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

Solution:

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

The equation will hold for all values of x > 2: Let us now examine the statements

I. \(|x| > 2\)

If x > 2, then \(|x| > 2\) ALWAYS (MUST BE TRUE)

II. \(x^2 < 1\)

If x > 2, then \(x^2\) will NEVER be < 1

III. \(x^3 > 0\)

If x > 2, then \(x^3\) ALWAYS > 0 (MUST BE TRUE)

I and III: MUST BE TRUE

Answer: D

34. Mbappe withdraws \(\frac{100}{x}\%\) of his money each time he visits the bank, where \(x > 1\). If after 5 visits, he has less than \(\frac{1}{x}^{th}\) of the initial amount in the bank, what is the range of all possible value of \(x\) ? (Assume x is an integer)

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

Solution:

Value substitution in \(\frac{100}{x}\) is the best way to go ahead in this question: Since x > 1, we will start with x=2

x=2 (50%): After 5 visits, Mbappe will have 1/32 of initial amount, which is less than 1/2 of amount. YES
x=3 (33.33%): After 5 visits, Mbappe will have approx 6.6^5/10^5 of initial amount, which is less than 1/3 of amount. YES
x=4 (25%): After 5 visits, Mbappe will have 243/1024 of initial amount, which is slightly less than 1/4 of amount. YES
x=5 (20%): After 5 visits, Mbappe will have 1024/3125 of initial amount, which is MORE than 1/5 of amount. NO

Possible values x = {2,3,4}
Therefore Range = 4-2 = 2

Answer: B

35. If the sum of the first 11 terms of an arithmetic progression consisting 16 terms is zero, then which of the following COULD 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. 6th largest term is zero
III. All terms are non negative

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

Solution:

This is a COULD BE TRUE question meaning that if a statement is true for even one condition, it qualifies for COULD BE TRUE

I. 11th smallest term is zero

Consider an AP where every term = 0 and common difference = 0.

Therefore, 11th smallest term = 0

II. 6th largest term is zero

Consider an AP where every term = 0 and common difference = 0

Therefore, 6th largest term = 0

III. All terms are non negative

Consider an AP where every term = 0 and common difference = 0

All terms non negative

I, II, and III COULD BE TRUE

Answer: E

36. If \(\frac{x+500}{510}+\frac{x+510}{515} + \frac{x+520}{520} + \frac{x+530}{525}+ \frac{x+540}{530}=10\), what is the value of x ?

A. 515
B. 520
C. 525
D. 530
E. 535

Solution:

\(\frac{x+500}{510}+\frac{x+510}{515} + \frac{x+520}{520} + \frac{x+530}{525}+ \frac{x+540}{530}=10\)

If you observe each term of the LHS, you will notice that with each subsequent term, the denominator is increasing by 5 and the numerator is increasing by 10

We can observe a pattern forming where the denominator can be the average of the two terms in the numerator. In such a case, the fraction will always equal 2 and since we have five terms, sum will equal 10

So what can be that term:

500 and x: Average = 510 => x = 520
510 and x: Average = 515 => x = 520
520 and x: Average = 520 => x = 520
530 and x: Average = 525 => x = 520
540 and x: Average = 530 => x = 520

Putting x = 520 in the LHS will give us 2 as each term and therefore sum = 10 = RHS

Answer: B

38. If \(\frac{m^4}{|m|}<\sqrt{m^2}\), then which of the following must be true?

I. \(m < \pi\)
II. \(m^2<1\)
III. \(m^3>-8\)

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

Solution:

The Range of values that satisfy the inequality is:

\(-1 < m < 0\) & \(0 < m < 1\)

Now, let us examine the following statements:

I. \(m < \pi\)

Range of all values of m: \(-1 < m < 0\) & \(0 < m < 1\)

So, m will always be less than Pie

MUST BE TRUE

II. \(m^2<1\)

Range of all values of m: \(-1 < m < 0\) & \(0 < m < 1\)

So, \(m^2\) will always be less than 1

MUST BE TRUE

III. \(m^3>-8\)

Range of all values of m: \(-1 < m < 0\) & \(0 < m < 1\)

So, \(m^3\) will always be > -8

MUST BE TRUE

I, II, III MUST BE TRUE

Answer: E

39. The infinite sequence of integers \(a_1\), \(a_2\), …, \(a_n\), … is such that \(a_1 = -17\) and \(a_{n-1} - 7 < a_n < a_{n-1} - 2\) for \(n > 1\). If \(a_x=-53\), then how many different values can x take?

A. 1
B. 4
C. 5
D. 6
E. 7

Solution:

a1=-17 then as per the inequality of the question stem

-24 < a2 < -19 => a2 = {-23,-22,-21,-20} Because sequence consists of integers only

Now, if you consider a3 will have 4 different inequality ranges based on the value of a2 which will be:

-30 < a3 < -25, -29 < a3 < -24, -28 < a3 < -23, -27 < a3 < -22

Combining the entire inequality will yield: -30 < a3 < -22 => a3 = {-29,-28,-27,-26,-25,-24,-23} All inclusive

Using this logic, we can see that the inequality range of any term of the sequence can be determined by taking adding -6 to the highest term and -3 to the lowest term
And the inclusive integers of the term can be determined by adding 1 to the lowest term of the inequality and subtracting 1 from the highest term of the inequality

So, based on this logic, we can determine:

a4 = {-35 to -26} All inclusive
a5 = {-41 to -29} All inclusive
a6 = {-47 to -32} All inclusive
a7 = {-53 to -35} All inclusive : x=7 can = -53 (1)
a8 = {-59 to -38} All inclusive : x=8 can = -53 (2)
a9 = {-65 to -41} All inclusive : x=9 can = -53 (3)
a10 = {-71 to -44} All inclusive : x=10 can = -53 (4)
a11 = {-77 to -47} All inclusive : x=11 can = -53 (5)
a12 = {-83 to -50} All inclusive : x=12 can = -53 (6)
a13 = {-89 to -53} All inclusive : x=13 can = -53 (7)

Now since -53 is the last term of a13, we know that following terms will be less than -53 in their entire range so no use going on from here
We have 7 possible terms where the term can = -53

Answer - E

40. 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. \(10^{n}\)

Solution:

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

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

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

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

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

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

Answer: D

Bunuel

Thanks a ton for posting such an amazing set of questions..

It provides a well organized set of Practice Questions, stacked as per increasing difficulty levels.



Will be waiting for questions pertaining to balance 4 areas, highlighted in one of your posts :-

1. Number Properties;
2. Geometry;
3. Combinatorics, Probability and Statistics;
4. Word Problems.

May I also request to include the following topics as well :-

5. Absolute Values and Modulus
6. inequalities
7. Functions and Graphs



Also, request if the 'Timer' functionality be introduced for this set of questions...

Best Regards

We need a 100 replies on this and the DS diagnostic test for Bunuel to post the same for all the other topics

If people solving can post their solutions we can reach there and have awesome diagnostics for the remaining topics

Posted from my mobile device

singhaz


For Question No 8

Though we get two answers --> 1 and 2 ie, both satisfying the Equation,

On a closer look,

1. For x =1,

root (2+root(3)) > root (2) ...............(1)
0< root (2-root(3)) < root (2) .............(2)

(1) + (2)

we get (root (2+root(3)) + (root (2-root(3))) is always greater than (root(2))

So, the Equation is not satisfied.

2. For x=2, however

(root (2+root(4)) + (root (2-root(4))) = root (2)

Here, the Equation is satisfied


Note :- This happens due to assignment of only a positive value to root(x) where x>=0...by convention

(E) is the CORRECT answer

For Q32 why is I also correct?

Two typos in the questions.
Q 36. Answer is 520
Q 37. Option A is x<0 and not x>0.
Same are edited now.

Enjoy the awesome questions

32. If \(\frac{|x| - 2}{|x - 2|}= 1\), 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


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

Now, \(\frac{|x| - 2}{|x - 2|}= 1>0\)

\(\frac{|x| - 2}{|x - 2|}>0\)

|x-2|>0, so the numerator also has to be >0.
Hence, \(|x|-2>0…….|x|>2\)
Option I must be true.