CAT Mock Test- 5 (November 7) - CAT MCQ

Through the passage, the author argues against nuclear power generation as an alternative to fossil fuels especially in the context of climate change. The author then gives these two examples to highlight how much damage they did to people and the planet and continue to do so. These two examples are extreme examples of how nuclear power generation can go terribly wrong. Thus, the primary reason to introduce these examples is to highlight the impact of nuclear disasters on the planet and people. Hence, option D.

Though the author does say that with climate change, the nuclear fallout of these disasters is likely to get worse, these examples are not introduced for that purpose. By highlighting the negative impact of these disasters, the author is trying to make the larger point against nuclear power. Hence, we can eliminate option B.

Though option A is the purpose behind the whole passage, it is not the reason why these two examples are introduced. Hence, we can eliminate option A. For the same reason we can eliminate option C.

Through the passage, the author argues against nuclear power. Hence, options B and C which are not negative can be eliminated. Both options A and D are close. But option A misses the primary motivation behind the author's recommendation. The author feels that with climate change, the number of nuclear disasters is likely to increase and thus nuclear power should be avoided. Hence, option D.

The MIT research showed conclusively that the frequent forest fires, surface runoffs and flooding spread radiation. The research concluded that we are still paying for the nuclear by-products that were made in the 20th century. Only option C captures this point and hence is the right answer.
Option D, though true, is not the main conclusion of the study.
Option A contains a distortion. The study does not mention that only areas with radiation will experience acid rain. Hence, we can eliminate option A.
Option B is an exaggeration and hence can be eliminated.
Thus option C is the correct answer. 

Through the passage, the author argues that country music has always had a progressive voice. To support this assertion, the author first gives the example of Blind Alfred Reed who spoke of racial equality and equitable society. Then he gives the example of Loretta Lynn's progressive stand on women's reproductive rights. Thus, through this paragraph, the author is trying to highlight the progressive voices in country music's history to show that country music always had a progressive voice. Thus, option A is closest to the answer.

Option B is incorrect because the author has not claimed that country music facilitated the human rights movement.

Option C is incorrect. Although the author does say that “The Pill” stands out that is not the main point. We must see why the author believes it stands out. Read the lines “In the country-music market, this song stands out as an unabashed and rather radical call for sexual liberation and biological control, challenging the man’s past sexual prerogative and presenting a situation where the woman may also enjoy a variety of sexual liaisons without the social/economic restrictions that come with pregnancy, childbirth and childcare.” Through this example, the author is trying to make a larger point.

Option D is true but only captures a part of what Loretta Lynn’s song indicates. Option A is more accurate.

The second paragraph begins by citing the two approaches used currently to explain why humans dance. Then it states the shortcomings of these theories. Option D summarizes this, so it is correct.
Though the second paragraph hints at options A and C, the main idea behind the paragraph is neither A nor B. Hence option (a) and option (b) are incorrect.
Option C is incorrect because it is not mentioned that the two theories complete each other. Hence option C is incorrect as well.

Only II and IV are mentioned in the passage. The passage refers to Asa as ‘Cambridge botanist’, but that doesn’t mean that he studied at Cambridge University. In the 3rd passage, it is mentioned that Asa wrote for both student and professional naturalist. In the first passage, the author states ‘They lose a wise and sympathetic adviser of great experience and mature judgment to whom they could always have turned with entire freedom and in perfect confidence’, from which we can infer that IV is true. Thus, the answer is C.

In the second para, the author states that 'this power of comprehensive generalization he showed in his paper', from which we can conclude that generalization was one of the aspects of his work. Further in the third para, the author states that 'His long list of educational works have no equals in accuracy and in beauty and compactness of expression'. Thus, B and C are also true. Nowhere in the passage has the author regarded Asa's work as systematically organized. Thus, the answer is D.

Refer to this line from the 4th para, ‘almost alone and single-handed he bore in America the brunt of the disbelief in the Darwinian Theory’. The only takeaway from this sentence is that Gray did believe the Darwinian theory to be true. However, the question asks about Gray’s views on Darwin, which cannot be inferred from the passage. The answer is D.

After reading all the sentences, we know that the paragraph is about the author's dislike for singing and his new found love for spoken poetry. Statement 2 is the opening sentence as it mentions the author's dislike for singing. The author also mentions how discovering spoken poetry changed his understanding of why he disliked singing. Statements 4,3 and 1 together describe the author's experience of the poem recited with the help of a stringed instrument.
Hence, the order is 2-4-3-1.
Hence, 2431 is the correct answer.

All the five sentences talk about the relation between psychology and teaching. But the structure of sentence 4 indicates that the topic of "ingenuity" and "tact" has already been discussed in the preceding lines. Since, there is no mention of "ingenuity" and "tact" in any other sentences, sentence 4 cannot be a continuation of any of the sentences given. The other 4 sentences can be arranged in the sequence 3-1-5-2.

The paragraph logically progresses from a general statement about revolutions, particularly the digital one (2), relating it historically and conceptually with how revolutions bring change (3), the underestimation of this current revolution (1), and finally the drawbacks it entails (4). The sequence interlinks historical context, the nature of the digital revolution, and its contradictory consequences.

The paragraph begins by introducing the impact of automation and AI on the workforce (2), prompting a reconsideration of economic norms (1), , and concluding with a speculative, positive outlook on the future of work (4), followed by the challenges faced by nations due to these advancements (3)

Let the number of people who followed all but CSK be x.
Now, 
I + II + III + IV = 1307.
I + 2 II + 3 III + 4 IV = 2155
Hence,
II + 2 III + 3 IV = 2155 - 1307 = 848
Now it is given that II = 0
Also, III = 49 + 53 + 37 + x = 139 + x
Hence, 2(139 + x) + 3 IV = 848
278 + 2x + 3 IV = 848
2x + 3 IV = 570
Now, IV = (570 - 2x)/3
IV = 190 - 2x/3
Since, IV must be an integer, x must be a multiple of 3 for sure. 
Hence, 63 is a possibility and none other option is possible. 

First, we would try to determine the relative positions of A to H for the first session. We know in hint 5 that all the people are facing the table.
Using hint 6,

Using hint 4 and hint 3 together, we get the following 2 arrangements.

Using hint 2, we can add F to the arrangements as follows:

Using hint 1, we can add G to the arrangements as follows:

We are yet to figure out the positions of D and H. So let us consider all possible alternatives for their positions:

Now let us use the hints given to determine the positions in the second session. We know that they are facing the table. 
Using hints 5 and 4 in all the four configurations, we get their arrangements in the second sessions as follows:

In the second session, A moves 3 places to his left, but in arrangements 1 and 2, that place is already occupied.
So, those are invalidated. We can process the remaining information on arrangements 3 and 4. Also, we can add this information (new position of A) in 3 and 4.

Using hint (2) and the fact that all people but H are in different seats, we get the following:

G is not adjacent to H.


Therefore the 2 different possible seating arrangements in the first and second sessions are as follows:

H is definitely a neighbour of A, because H sits beside A in both possible cases.

First, we would try to determine the relative positions of A to H for the first session. We know in hint 5 that all the people are facing the table.
Using hint 6,

​Using hint 4 and hint 3 together, we get the following 2 arrangements.

Using hint 2, we can add F to the arrangements as follows:

Using hint 1, we can add G to the arrangements as follows:

We are yet to figure out the positions of D and H. So let us consider all possible alternatives for their positions:

Now let us use the hints given to determine the positions in the second session. We know that they are facing the table. 
Using hints 5 and 4 in all the four configurations, we get their arrangements in the second sessions as follows:

In the second session, A moves 3 places to his left, but in arrangements 1 and 2, that place is already occupied.
So, those are invalidated. We can process the remaining information on arrangements 3 and 4. Also, we can add this information (new position of A) in 3 and 4.

Using hint (2) and the fact that all people but H are in different seats, we get the following:


G is not adjacent to H.


Therefore the 2 different possible seating arrangements in the first and second sessions are as follows:

E is sitting opposite C in session 2 in either arrangement.

In such questions, we can draw a table as shown below. But the digit as shown in the third row is not necessarily the sum of the digits in the first two columns, as we know that in addition, we have carry-forward. This means that if the sum of two digits yields a two-digit sum, the tens digit is carried forward to the forward column and only the ones digit is written.

The alphabets in the units place are S S and O. So we have units digit of S + S = O or S + O = S 

Case 1: S + O = S
⇒ O = zero and S,T,P are non zero. Hence SCORPIO will be a 7 digit number and TAURUS and PISCES will be 6 digit numbers. We can't get a 6 digit number by adding a 7 digit number and a 6 digit number. Hence this case case is not possible.

Case 2: S + S = O
The sum will look like as below:

In the last column, we can see that S+S=O. Hence S can't be zero as O will also be zero then. 
We know that the maximum sum of any two distinct digits can be 17, and the minimum can be 3. Since S cannot be zero S can only attain a value of 1, which is carried forward. Since there is a carry forward, neither of T and P can be zero.
In the last column, we can see that S+S=O. Hence the value of O is 2.
Also, UR and US are consecutive numbers. As S=1, R can be either 0 or 2. Since O has the value 2 already, R will be 0. The sum looks like this:

Here we can see that in the third last column, 0 is added to C to get P. Now since each letter represents a unique number, 1 must have been carried forward from the sum of U and E. So we have the following equations:
U+E=10+I
C+1=P or C+1=10+P
Now C+1 cannot be 10+P as only 1 is being added to C here. So to give a two digit number, C will be 9 and P will be 0, which is not possible as 0 is already assigned to R. Hence C+1=P.
Hence no carry-forward is there from the third last column. This means that U+1=0, and U=9. The distribution of digits to letters looks as follows:

Now, we need to do hit and trial to ascertain the values of other letters. Since it has been given that there is a unique combination, we need to do trials only till we find a solution.
As P=C+1, we will place them in consecutive positions on the above distribution. We will try with 3 and 4:

Now, since the sum in the 4th column is 10, so 1 will get carried forward to column 3. Hence the equation for column 3:
A+I+1=10+2, as sum of A and I cannot be 2, as 0 and 1 are already taken. So the sum is a two-digit number.
We have A+I=11.
In the distribution, we have only 5 and 6 left which can give a sum of 11.

Since the sum in column 3 is 11, 1 will get carried forward to column 2, and we get:
T+P+1=13, since C=3.
Hence T+P=12. Now P=4, so T=8 and E=7.
Since U+E=10+I, I= 9+7-10= 6 and A=5.
The final distribution:

We can see that E has the highest value among the four equal to 7.

In such questions, we can draw a table as shown below. But the digit as shown in the third row is not necessarily the sum of the digits in the first two columns, as we know that in addition, we have carry-forward. This means that if the sum of two digits yields a two-digit sum, the tens digit is carried forward to the forward column and only the ones digit is written.

The alphabets in the units place are S S and O. So we have units digit of S + S = O or S + O = S 

Case 1: S + O = S
⇒ O = zero and S,T,P are non zero. Hence SCORPIO will be a 7 digit number and TAURUS and PISCES will be 6 digit numbers. We can't get a 6 digit number by adding a 7 digit number and a 6 digit number. Hence this case case is not possible.

Case 2: S + S = O
The sum will look like as below:

In the last column, we can see that S+S=O. Hence S can't be zero as O will also be zero then. 
We know that the maximum sum of any two distinct digits can be 17, and the minimum can be 3. Since S cannot be zero S can only attain a value of 1, which is carried forward. Since there is a carry forward, neither of T and P can be zero.
In the last column, we can see that S+S=O. Hence the value of O is 2.
Also, UR and US are consecutive numbers. As S=1, R can be either 0 or 2. Since O has the value 2 already, R will be 0. The sum looks like this:

Here we can see that in the third last column, 0 is added to C to get P. Now since each letter represents a unique number, 1 must have been carried forward from the sum of U and E. So we have the following equations:
U+E=10+I
C+1=P or C+1=10+P
Now C+1 cannot be 10+P as only 1 is being added to C here. So to give a two digit number, C will be 9 and P will be 0, which is not possible as 0 is already assigned to R. Hence C+1=P.
Hence no carry-forward is there from the third last column. This means that U+1=0, and U=9. The distribution of digits to letters looks as follows:

Now, we need to do hit and trial to ascertain the values of other letters. Since it has been given that there is a unique combination, we need to do trials only till we find a solution.
As P=C+1, we will place them in consecutive positions on the above distribution. We will try with 3 and 4:

Now, since the sum in the 4th column is 10, so 1 will get carried forward to column 3. Hence the equation for column 3:
A+I+1=10+2, as sum of A and I cannot be 2, as 0 and 1 are already taken. So the sum is a two-digit number.
We have A+I=11.
In the distribution, we have only 5 and 6 left which can give a sum of 11.

Since the sum in column 3 is 11, 1 will get carried forward to column 2, and we get:
T+P+1=13, since C=3.
Hence T+P=12. Now P=4, so T=8 and E=7.
Since U+E=10+I, I= 9+7-10= 6 and A=5.
The final distribution:

Since S is the leftmost digit, only the change in its value matter. This means that we have to either replace S with the highest digit possible or the lowest digit possible. In any case, if S is replace by the value of U, that is 9, the difference would be the largest: |1320462-9320462|=8000000. Hence the answer is A.

Let the multiplier of Healthcare, Population, and Infrastructure be a, b and c respectively.
Let us start by finding out which parameter has a multiplying factor of 1 and which parameter has a multiplying factor of 2.
Case I: If all of the parameters have a multiplying factor of 2, then the cumulative score would always be even. But we see that the cumulative score of H is odd, hence it is not possible.
Case II: If every country has a multiplying factor of 1 then the maximum possible cumulative score can be 8+8+8 =24. But H has a cumulative score of 27, thus this is also not the case.
Thus there is at least 1 parameter that has a multiplying factor of 1 and at least 1 parameter which has a multiplying factor of 2.
Let us look at the country E. 
Since rank 8 has a cumulative score of 15, E should have a cumulative score ≥ 17 to have a rank of 6.
Thus 3a + 5b + 3c ≥ 17. It is only possible when atleast 2 of a,ba,b and cc has value of 2.
Thus 1 parameter has a multiplier of 1 and 2 parameters have a multiplier of 2
Country B has value for a score for Population parameter = 6. If we assume the score for other parameters to be xx and yy respectively and b=1, then a = c = 2. Thus 2x + 6 + 2y = 15 or 2x + 2y = 9 which is not possible. Hence b = 2.
For country F we see that the cumulative score = 25 which is odd. And score for healthcare is 5. It is only possible when a = 1 and b = c = 2. Else the sum will be even.
Thus a = 1,b = 2 and c = 2. The table can be updated as follows:

The score of B for Healthcare and Infrastructure has to be 1 so that cumulative becomes 15. Thus score of A for Population is also 1.
Cumulative of E = 19
The score of F for the population has to be 2 for cumulative to be 25
Score of 19 has a rank of 6 and score of 25 has a rank of 4. Thus C has a rank of 5

The score of A for Population is 1. The maximum score for Healthcare can be 8 and for infrastructure can be 7. The maximum cumulative score of A can be 24. But we know that at 25 the rank is 4 and at rank 5 the score is 22. Thus rank of A has to be worse than 5. Therefore only possibility is rank 7.
Then the rank of D can be either 2 or 3. Thus the score of D has to be greater than 25
The score for D for Health can be 2 or 4 or 6 or 8. The score of D can be thus 20 or 22 or 24 or 26 respectively. Thus score of D for Healthcare should be 8 with a cumulative score of 26. Thus D has rank 3 and H will have rank 2

Exactly 3 countries have the same Healthcare and Infrastructure score. B and E two of these countries. The third country can be A or C or G or H.
For A, the common score can be  4 only. In the case of which cumulative becomes. 14 But the score of A has to be 16 or more as it has rank 7. Thus rejected.
For C, since the score of Infrastructure is 6, the score of Healthcare needs to be 6. For cumulative to be 22 the score of Population has to be 2 but, F has that score so it is also not possible.
For G, the common number can be 4. In the case of which if G gets 8 score in population, still the cumulative will be 26. Which is a contradiction as G should have score 28 or above
If the score of H for Infrastructure = 7 then the Population score has to be 3. This is the only possible case

The score of Population for C can be 4 or 8. If the score is  8 then the cumulative will exceed 22. Thus THE score of C for the population is 4. Its score for Healthcare has to be 2. G has a population score = 8

Since the score of Healthcare for A can be 4 or 6. The cumulative score will be even. Thus a cumulative score of A can be 16 or 18.
Case I: 
If A has a Healthcare score of 4 then A needs either an Infrastructure score to be 5 or 6. The only possibility is 5. Thus is cumulative will be 16. It will imply that score for G has to be 6,8,4. The cumulative score, in this case, will be 30.

Case II: A has a healthcare score of 6. Thus Infrastructure score has to be 4 or 5. If it is 4 then the cumulative score of A will be 16. The score of G, in that case, will be 4,8,5 which has a total cumulative of 30.
If the infrastructure score of A is 5, then its cumulative will be 18. The score of G, in that case, will be 4,8,4 with a cumulative of 26. Which is not possible as it has to be above 28. This is rejected

The above mentioned are the only possible 2 cases. Both the cases the score of G is 30

Let the multiplier of Healthcare, Population, and Infrastructure be a, b and c respectively.
Let us start by finding out which parameter has a multiplying factor of 1 and which parameter has a multiplying factor of 2.
Case I: If all of the parameters have a multiplying factor of 2, then the cumulative score would always be even. But we see that the cumulative score of H is odd, hence it is not possible.
Case II: If every country has a multiplying factor of 1 then the maximum possible cumulative score can be 8+8+8 =24. But H has a cumulative score of 27, thus this is also not the case.
Thus there is at least 1 parameter that has a multiplying factor of 1 and at least 1 parameter which has a multiplying factor of 2.
Let us look at the country E. 
Since rank 8 has a cumulative score of 15, E should have a cumulative score ≥ 17 to have a rank of 6.
Thus 3a + 5b + 3c ≥ 17. It is only possible when atleast 2 of a,ba,b and cc has value of 2.
Thus 1 parameter has a multiplier of 1 and 2 parameters have a multiplier of 2
Country B has value for a score for Population parameter = 6. If we assume the score for other parameters to be xx and yy respectively and b=1, then a = c = 2. Thus 2x + 6 + 2y = 15 or 2x + 2y = 9 which is not possible. Hence b = 2.
For country F we see that the cumulative score = 25 which is odd. And score for healthcare is 5. It is only possible when a = 1 and b = c = 2. Else the sum will be even.
Thus a = 1,b = 2 and c = 2. The table can be updated as follows:

The score of B for Healthcare and Infrastructure has to be 1 so that cumulative becomes 15. Thus score of A for Population is also 1.
Cumulative of E = 19
The score of F for the population has to be 2 for cumulative to be 25
Score of 19 has a rank of 6 and score of 25 has a rank of 4. Thus C has a rank of 5

The score of A for Population is 1. The maximum score for Healthcare can be 8 and for infrastructure can be 7. The maximum cumulative score of A can be 24. But we know that at 25 the rank is 4 and at rank 5 the score is 22. Thus rank of A has to be worse than 5. Therefore only possibility is rank 7.
Then the rank of D can be either 2 or 3. Thus the score of D has to be greater than 25
The score for D for Health can be 2 or 4 or 6 or 8. The score of D can be thus 20 or 22 or 24 or 26 respectively. Thus score of D for Healthcare should be 8 with a cumulative score of 26. Thus D has rank 3 and H will have rank 2

Exactly 3 countries have the same Healthcare and Infrastructure score. B and E two of these countries. The third country can be A or C or G or H.
For A, the common score can be  4 only. In the case of which cumulative becomes. 14 But the score of A has to be 16 or more as it has rank 7. Thus rejected.
For C, since the score of Infrastructure is 6, the score of Healthcare needs to be 6. For cumulative to be 22 the score of Population has to be 2 but, F has that score so it is also not possible.
For G, the common number can be 4. In the case of which if G gets 8 score in population, still the cumulative will be 26. Which is a contradiction as G should have score 28 or above
If the score of H for Infrastructure = 7 then the Population score has to be 3. This is the only possible case

The score of Population for C can be 4 or 8. If the score is  8 then the cumulative will exceed 22. Thus THE score of C for the population is 4. Its score for Healthcare has to be 2. G has a population score = 8

Since the score of Healthcare for A can be 4 or 6. The cumulative score will be even. Thus a cumulative score of A can be 16 or 18.
Case I: 
If A has a Healthcare score of 4 then A needs either an Infrastructure score to be 5 or 6. The only possibility is 5. Thus is cumulative will be 16. It will imply that score for G has to be 6,8,4. The cumulative score, in this case, will be 30.

Case II: A has a healthcare score of 6. Thus Infrastructure score has to be 4 or 5. If it is 4 then the cumulative score of A will be 16. The score of G, in that case, will be 4,8,5 which has a total cumulative of 30.
If the infrastructure score of A is 5, then its cumulative will be 18. The score of G, in that case, will be 4,8,4 with a cumulative of 26. Which is not possible as it has to be above 28. This is rejected

The above mentioned are the only possible 2 cases. In both the cases the score of E is 19

Q1 cost is $15 Million.
So, Q1 profit is ($20 - $15) Million = $5 Million.
Thus, the profits are $5 Million, $10 Million, $15 Million and $10 Million respectively in Q1, Q2, Q3 & Q4 as per the ratio (1 : 2 : 3 : 2).
The cost in Q2 = ($18 - $ 10) Million = $8 Million
The cost in Q3 = ($22 - $ 15) Million = $7 Million
The cost in Q4 = ($24 - $ 10) Million = $14 Million
Thus, total cost in Q2 & Q4 is $22 Million

Q1 cost is $15 Million.
So, Q1 profit is ($20 - $15) Million = $5 Million.
Thus, the profits are $5 Million, $10 Million, $15 Million and $10 Million respectively in Q1, Q2, Q3 & Q4.
The cost in Q2 = ($18 - $ 10) Million = $8 Million
The cost in Q3 = ($22 - $ 15) Million = $7 Million
The cost in Q4 = ($24 - $ 10) Million = $14 Million
Thus, the answer is= 75%

Total revenue in 2022 is $(20 + 18 + 22 + 24) Million = $ 84 Million.
The projected revenue for 2023 is $84 MillionMillion
The ratio of the projected revenues in Q1, Q2, Q3 & Q4 is 2 : 3 : 3 : 4
So, the revenue in Q1Million.
The revenue in Q2 = revenue in Q3 = $ 18 Million.
The revenue in Q4 = $72 - $12 - $18 - $18 = $24 Million.
Thus, the answer is $24 Million.

Total revenue in 2022 is $(20 + 18 + 22 + 24) Million = $ 84 Million.
The projected revenue for 2023 is $84 MillionMillion.
The ratio of the projected revenues in Q1, Q2, Q3 & Q4 is 2 : 3 : 3 : 4
So, the revenue in Q1 
= $ 12 Million.
The revenue in Q2 = revenue in Q3 = $ 18 Million.
The revenue in Q4 = $72 - $12 - $18 - $18 = $24 Million.
Let’s assume that the cost for quarter 1 is $x Million, then we can say that
⇒ (18 – x) = (24 – 2x)
⇒ x = 6
So, the savings in Q1 = $12 Million - $ 6 Million
= $ 6 Million.
The savings in Q2 = $18 Million - $ 12 Million
= $ 6 Million
The savings in Q3 = $18 Million - $ 6 Million
= $ 12 Million
The savings in Q4 = $24 Million - $ 12 Million
= $ 12 Million
Total savings from Q1 & Q4 = $(6 + 12) Million
= $18 Million.

The problem figure in the question can be drawn as:

If we construct an equilateral triangle with base CD, we will get:

 

AM and DK are parallel to each other (makes 60 degrees with the base) and AD and MK are parallel to each other. 
∴AMKD is a parallelogram and hence AM= DK and AD= MK.
But since AM=AB=AD.
So, AMKD is a rhombus. 
Also, since MK= KD and M and D are the points on the circle, K must be centre of the circle and so, the radius = KD = CD = 12 cm.

We can depict all the directions in the given information as follows:



AJ = 45 + 45 + 50 = 140
FJ = 50 + 45 - 20 - 50 = 25

Let's convert everything to cm.
Volume of the hemisphere

Volume of each hollow cylinder = π R²h − π r²h = π (R² − r²)h = π (49 − 9)2.5 = 100π 
Hence, number of hollow cylinders that can be formed

Let |x| = z.
|x|³ + 8x² + 5|x| - 50 > 0
z³ + 8z² + 5z - 50 > 0
Using hit and trial, we can see that z = 2 satisfies the equation
Hence, z-2 is a factor.
Using algebraic division, we can derive, 

z³ + 8z² + 5z - 50 = (z − 2)(z² + 10z + 25) = (z − 2)(z + 5)²  

For z > 2, the condition is satisfied. Hence for -5 < x < 2, the expression is less than zero. SInc the power of z + 5 is even, the expression < 0 for z < - 5 a well.
So, the condition is satisfied for z > 2 |x| > 2.
Hence, the condition is not satisfied for x = -2, -1, 0, 1, 2.

log 25 = a
∴ 2 log 5 = a
∴ log 5 = a/2
log 180 = b
∴ log 5 x 6 x 6 = b
∴  log 5 + 2 log 6 = b
∴ a/2 + 2 log 6 =b
∴ log 6 = b/2 - a/4
log 750 = log 5 x 5 x 5 x 6 = 3log 5 + log 6 = 3a/2 + b/2 - a/4 = b/2 + 5a/4
m = 5/4, n = 1/2
m + n = 5/4 + 1/2 =7/4

2⁵3⁶4⁷ can be written as 2¹⁹3⁶
Hence it has 20 x 7 =140 factors. 
Hence, it can be written as a product of 2 numbers in 140/2 = 70 ways
Out of these 70 ways, in 2 of these ways, one of the numbers is prime, as follows

Hence, total number of ways = 70 - 2 = 68.