CAT Mock Test- 7 (November 9) - CAT MCQ

The author makes the following observation in the third paragraph- {The robots are now humanity’s saviors, suppressing bad human mass behaviour online with increasingly sophisticated filtering algorithms. We once obsessed about how to restrain machines we could not predict or control — now we worry about how to use machines to restrain humans we cannot predict or control.}
So, humans feared that self-improving machines would go out of control and that it would be difficult for humans to restrain them. However, in a role reversal, the machines today help curb bad human behaviour online. Hence, ironically, the roles have reversed. 
Only option B conveys this inference and hence is the right answer.
Option A is outside the scope of discussion. Machines evolving to differentiate between good and bad has not been mentioned anywhere. Hence they can be eliminated.
Though an irony, Option C has not been mentioned in the passage as such. Hence can be eliminated.
Option D is close but contains the distortion that we cannot do without machines. The passage does not say that machines are indispensable. Hence, D can be eliminated.

Weiner's desire-outcome disparity is about advanced systems which achieve undesired outcomes because of their inherent nature and them acting on misleading substitutes of the objectives fed.
Since the command itself was erroneous, hence the system acted on incorrect information instead of acting on a misleading substitute of the objective. Hence I does not fit the argument.
The AI program was designed to mitigate global warming. But eliminating humans to achieve the same is an undesired outcome, which arises due to the program acting on the misleading substitute of the objective assigned. The substitute being 'achieving the goal without a care for the humans', which the scientists would not have desired. Hence II fits the argument.
Though the author mentions that Weiner did not talk about social media, he also mentions that an analogy can be drawn between the two, hence the argument can be extended to include the misuse of social media, as mentioned in Statement III.
Hence Statements II and III fit Weiner's argument.

"However, behind women participation statistics and progressive policies, gender stereotypes prevail, particularly in the workplace, and women in the region say that there is still a lot of work to be done."
Though referred to as one of the most gender-equal countries, the above lines show that Sweden is still far from being gender-equal. Hence Option A can be inferred.
In the fourth and fifth paragraphs, the author states the following regarding the gender equality situation in Sweden- "Men were also encouraged to take care of the family. “There’s this dual-earner/dual-carer ideal that Swedish gender equality is based on,” says Björklund. Policies in Sweden have since then focused on facilitating that work-family balance."
Option B can be clearly inferred from these lines. 
"The modern concept of gender equality has its foundations in the postwar welfare state. In Sweden, it was motivated by the need for more women in the workforce after the Second World War". Option C has been discussed as well.
Option D is a distortion. In the last paragraph, the author says that the national trait of "Swedishness" has been hijacked by far-right parties. However, we cannot say if Swedishness was introduced by them or that it was introduced to push anti-immigrant policies. As this has not been discussed, we can say that it is the right answer.

{"...However, the expectations on women to be full-time workers, self-sacrificing mothers and still have leisure time have put unrealistic pressure on this ideal. Expectations on men are not as high, and Björklund says that fathers can get away with being less caring than mothers - an idea underpinned by traditional stereotypes and middle-class values..."} In the fifth paragraph, the author talks about how the existing stereotypes lead to a disparity in the expectations from men and women to maintain a work-life balance. The difference in expectation is then highlighted. The author mentions how, in reality, gender stereotypes lead people to expect a lot more from women regarding caregiving responsibilities. Option C conveys the idea correctly.
Option A asserts that men are expected to fare better in the earner role, which has not been implied in the passage. 
Option B states that traditional stereotypes allow men to ignore their caregiving duties. The passage states that men can get away with doing much lesser than women - and not completely ignoring their responsibilities. Moreover, the impact on women due to the imbalanced expectations is missed out in this option. Hence, we can eliminate Option B. 
Option D can be safely eliminated. The author does not discuss the need for a new Swedish dual-earner/dual-giver ideal but talks about the threat to the same.
Hence, Option C is the correct answer.

The author concludes the passage on a foreboding note, highlighting how certain discussions now assume gender equality has gone too far. In the same paragraph, the author discusses how the disparity among different women groups, especially white, middle-class and immigrant women has been used as a tool against immigration. Hence, any discussion about how gender equality has not gone too far and how it is needed for all women would be the right way to continue the line of thought. Option C gives a rebuttal of sorts to the xenophobic idea presented in the last paragraph. Hence, C is the aptest answer.
The author does not explicitly discuss diversity in workplaces. It does not connect with any previous discussion. Hence, option A can be eliminated.
Option B is narrow. The author does not discuss anything related to intersectional groups in the passage and is focused on the larger group of women. 
Option D is extreme and has not been implied in the passage. Moreover, the focus would shift from gender equality to minorities which is inconsistent with the passage.

The last three paragraphs talk about how genetic research aided by SPARK has helped in the treatment and mitigation of various kinds of autism. The author further says that the addition of rich data to this will help provide better analysis and support. So the author tries to show the benefits of SPARK and how the addition of rich data will help provide better support. Option C captures the points correctly.
Option A contains a distortion in the second half. The passage does not imply that Spark can reach its true potential only through greater participation. Hence, A can be eliminated. 
The author does not talk about various developments but the achievements due to a particular method. Additionally, the need for more academic participation has not been emphasised. Hence Option B is a distortion.
The author does not focus on the plight of children suffering from autism. Instead, he highlights how SPARK helps these children and how promoting participation in genetic research could render valuable data. Hence Option D is incorrect.

"..we need not only the largest but also the most diverse group of participants."
From this line, we can infer that along with a large dataset, we also need diversity to make big data effective. But is has not been mentioned that the size of the data is not important. Hence Statement I is a distortion and cannot be inferred.
"We now know that genes play a central role in the causes of these autisms,...”
From this line, we can infer that the role of genes in causing autism is prominent. Hence Statement II can be inferred.
In paragraph 7, it is implied that a diet containing low levels of phenylalanine was given to stabilize the level of amino acid in the blood. It has not been implied that the level of amino acid is decreased because of it. Hence Statement III cannot be inferred.

Sentence 2 is the opening sentence since it highlights the damaging effect of the pandemic, especially with regard to the loss of jobs. The author draws a parallel between this group and the industries in general by continuing the idea in 1, which mentions the measures taken by certain industries. 
34 is a bloc. 3 describes the industrial scenario at large - how large-scale industries are doing well, despite laying off employees and not doing enough to arrest the pandemic's effects. Sentence 4 talks about the tax incentives these large-scale industries get, further critiquing the current situation. Hence, 2134 form a coherent paragraph.

After reading all the sentences, it can be inferred that the passage discusses data brokerage and the threats it poses to the proper functioning of a democracy.
Sentence 3 is independent and is the opening sentence. 12 is a bloc as it discusses the situation in the absence of strong data safeguards. 5 has to follow 2 as it describes the larger consequences- the word 'this' is the link. Hence, 3125 form a coherent paragraph.
Sentence 4 does not fit the context as it talks about data brokers lobbying more aggressively. Though it is not an independent sentence, none of the other sentences logically lead to the idea discussed in sentence 4. Hence 4 is the odd one out here.

The proper sequencing of the sentences is:
2. The integration of advanced algorithms and everyday technology has streamlined countless processes, improving efficiency exponentially.
4. Such an environment, teeming with both subtle surveillance and convenience, highlights the double-edged nature of progress.
1. However, the very ubiquity of such technology poses critical questions about data privacy and individual autonomy.
3. As we entrust more of our lives to digital gatekeepers, the parameters of the debates shift from technical concerns to ethical dilemmas.

This sequence creates a coherent paragraph by first introducing the integration of advanced algorithms and everyday technology and how it has improved efficiency. Then, it raises the issue of the double-edged nature of progress, emphasizing the presence of both surveillance and convenience. This leads to a discussion about the critical questions regarding data privacy and individual autonomy that arise due to the ubiquity of technology. Finally, it concludes by stating that as we entrust more of our lives to digital gatekeepers, the focus of the debates shifts from technical concerns to ethical dilemmas.

Since Mr. Wilson plotted only ten points and connected consecutive points using straight lines, wherever there is a change in the slope of the line, Mr. Wilson must have made a count.
Hence, Mr. Wilson must have counted the number of animals born on months 1, 4, 7, 10, 14, 18 and 20 and the number of animals that were born were 300, 330, 270, 240, 280, 360 and 340 respectively.
The differences in number of animals born between these counting are 30, 60, 30, 40, 80 and 20 respectively.
From (1), he must have made at least one count between month 4 and month 7 and at least one count between month 14 and month 18. These account for nine counts in total.
From (3), Mr. Wilson's second count is the same as the eighth count. The second count can be on month 2 or month 3 or month 4.
In any of the three cases, for the eighth count to be the same as the second count, the eighth count must have been made from month 14 to month 18.

Case 1: If second count was on month 2:

If second count was on month 2, (i.e., of 310 animals), the eighth count cannot be on any month because no month from month 14 to month 18 have the number of animals born as 310.

Case 2: If second count was on month 4:

Similarly, If second count was on month 4, (i.e., of 330 animals), the eighth count cannot be on any month because no month from month 14 to month 18 have the number of animals born as 310.

Case 3: If second count was on month 3:

The second count on month 3 (of 320 animals) is possible because month 16. has the number of animals born as 320.
So, eighth count must be month 16.

So, the third count is on month 4 (of 330 animals).

From (2), the number of animals born on fourth count must be either 290 or 370 (inferable from the graph).
So, the fourth count must be on month 6.

The following table presents the counting that Mr. Wilson made:

Average number of animals born between second count and third count = 

Since Mr. Wilson plotted only ten points and connected consecutive points using straight lines, wherever there is a change in the slope of the line, Mr. Wilson must have made a count.
Hence, Mr. Wilson must have counted the number of animals born on months 1, 4, 7, 10, 14, 18 and 20 and the number of animals that were born were 300, 330, 270, 240, 280, 360 and 340 respectively.
The differences in number of animals born between these counting are 30, 60, 30, 40, 80 and 20 respectively.
From (1), he must have made at least one count between month 4 and month 7 and at least one count between month 14 and month 18. These account for nine counts in total.
From (3), Mr. Wilson's second count is the same as the eighth count. The second count can be on month 2 or month 3 or month 4.
In any of the three cases, for the eighth count to be the same as the second count, the eighth count must have been made from month 14 to month 18.

Case 1: If second count was on month 2:

If second count was on month 2, (i.e., of 310 animals), the eighth count cannot be on any month because no month from month 14 to month 18 have the number of animals born as 310.

Case 2: If second count was on month 4:

Similarly, If second count was on month 4, (i.e., of 330 animals), the eighth count cannot be on any month because no month from month 14 to month 18 have the number of animals born as 310.

Case 3: If second count was on month 3:

The second count on month 3 (of 320 animals) is possible because month 16. has the number of animals born as 320.
So, eighth count must be month 16.

So, the third count is on month 4 (of 330 animals).

From (2), the number of animals born on fourth count must be either 290 or 370 (inferable from the graph).
So, the fourth count must be on month 6.

The following table presents the counting that Mr. Wilson made:

The difference between the fifth count and the ninth count that Mr. Wilson took = 360 - 270 = 90

Since Mr. Wilson plotted only ten points and connected consecutive points using straight lines, wherever there is a change in the slope of the line, Mr. Wilson must have made a count.
Hence, Mr. Wilson must have counted the number of animals born on months 1, 4, 7, 10, 14, 18 and 20 and the number of animals that were born were 300, 330, 270, 240, 280, 360 and 340 respectively.
The differences in number of animals born between these counting are 30, 60, 30, 40, 80 and 20 respectively.
From (1), he must have made at least one count between month 4 and month 7 and at least one count between month 14 and month 18. These account for nine counts in total.
From (3), Mr. Wilson's second count is the same as the eighth count. The second count can be on month 2 or month 3 or month 4.
In any of the three cases, for the eighth count to be the same as the second count, the eighth count must have been made from month 14 to month 18.

Case 1: If second count was on month 2:

If second count was on month 2, (i.e., of 310 animals), the eighth count cannot be on any month because no month from month 14 to month 18 have the number of animals born as 310.

Case 2: If second count was on month 4:

Similarly, If second count was on month 4, (i.e., of 330 animals), the eighth count cannot be on any month because no month from month 14 to month 18 have the number of animals born as 310.

Case 3: If second count was on month 3:

The second count on month 3 (of 320 animals) is possible because month 16. has the number of animals born as 320.
So, eighth count must be month 16.

So, the third count is on month 4 (of 330 animals).

From (2), the number of animals born on fourth count must be either 290 or 370 (inferable from the graph).
So, the fourth count must be on month 6.

The following table presents the counting that Mr. Wilson made:

The average number of animals born counted by Mr. Wilson =

Total number of defective units sold in 2015 = (3,24,000 × 0.20 × 0.45) + (3,24,000 × 0.30 × 0.4) + (3,24,000 × 0.30 × 0.35) + (3,24,000 × 0.20 × 0.25)
= 29,160 + 38,880 + 34,020 + 16,200
= 1,18,260

It is given that Earl bought 1020 candies. He could have bought the packets of type 20/30/170. Amy bought 3570 candies. He could have bought the packets of type 30/70/170. Spencer bought 3990 candies. He could have bought the packets of type 30/70/190. Tyler bought 2380 candies. He could have bought the packets of type 20/70/170. Nicole bought 1470 candies. He could have bought the packets of type 30/70. Since only Spencer could have bought the packets of type 190, he must have bought the packets of type 190 candies. The number of packets that he would have bought will be 21. If Nicole bought packs of 70 candies, he must have bought 21 packets. This is not possible. Hence, Nicole bought the packets containing 30 candies each. Amy could not have bought the packets of type 170, as she would have bought 21 packets. Hence, Amy must have bought packets containing 70 candies each. He must have bought 51 packets. Earl could not have bought packets containing 20 candies each as he would have then bought 51 packets. Hence, Earl must bought packets containing 170 candies. Tyler must have bought packets containing 20 candies each.
The following table provides the type of packet and the number of packets bought by each of the five persons:

Total number of packets that the five friends bought from the shopkeeper = 246

It is given that Earl bought 1020 candies. He could have bought the packets of type 20/30/170. Amy bought 3570 candies. He could have bought the packets of type 30/70/170. Spencer bought 3990 candies. He could have bought the packets of type 30/70/190. Tyler bought 2380 candies. He could have bought the packets of type 20/70/170. Nicole bought 1470 candies. He could have bought the packets of type 30/70. Since only Spencer could have bought the packets of type 190, he must have bought the packets of type 190 candies. The number of packets that he would have bought will be 21. If Nicole bought packs of 70 candies, he must have bought 21 packets. This is not possible. Hence, Nicole bought the packets containing 30 candies each. Amy could not have bought the packets of type 170, as she would have bought 21 packets. Hence, Amy must have bought packets containing 70 candies each. He must have bought 51 packets. Earl could not have bought packets containing 20 candies each as he would have then bought 51 packets. Hence, Earl must bought packets containing 170 candies. Tyler must have bought packets containing 20 candies each.
The following table provides the type of packet and the number of packets bought by each of the five persons:

Statement I and III are correct.

It is given that Earl bought 1020 candies. He could have bought the packets of type 20/30/170. Amy bought 3570 candies. He could have bought the packets of type 30/70/170. Spencer bought 3990 candies. He could have bought the packets of type 30/70/190. Tyler bought 2380 candies. He could have bought the packets of type 20/70/170. Nicole bought 1470 candies. He could have bought the packets of type 30/70. Since only Spencer could have bought the packets of type 190, he must have bought the packets of type 190 candies. The number of packets that he would have bought will be 21. If Nicole bought packs of 70 candies, he must have bought 21 packets. This is not possible. Hence, Nicole bought the packets containing 30 candies each. Amy could not have bought the packets of type 170, as she would have bought 21 packets. Hence, Amy must have bought packets containing 70 candies each. He must have bought 51 packets. Earl could not have bought packets containing 20 candies each as he would have then bought 51 packets. Hence, Earl must bought packets containing 170 candies. Tyler must have bought packets containing 20 candies each.
The following table provides the type of packet and the number of packets bought by each of the five persons:

Total number of packets bought by Amy and Spencer = 51 + 21 = 72
Number of packets bought by Tyler = 119
So, 'the total number of packets bought by Amy and Spencer is less than that bought by Tyler' is true.

The two players who played Classic got highest increase in their FIDE Ratings, the three players who played Rapid got increase in their FIDE Ratings less than Classic players and the three players who played Biltz got increase in their FIDE Ratings less than Rapid players.
Hence, we can order the eight players from 1 to 8 such that 1st and 2nd are Classic, and 3rd, 4th and 5th are Rapid, and 6th, 7th and 8th are Biltz.
Carlson got the highest increase in rating in Biltz. So, his order is 6.

From conditions 1, 2 and 5, it is clear that Anand, Harikrishna, Caruana, Vidit, Boris not played Biltz. So, Carlson, Nakamura and Taplov played Biltz.
From condition 4, Nakamura did not get the lowest increase in his rating. So, Nakamura's order is 7 and Taplov's order is 8.

From conditions 1 and 5, at least two players increase their ratings more than Anand did, and Anand did not increase his ratings less than Harikrishna did, and Caruana increased his rating less than Harikrishna did. So, the order of Anand, Harikrishna and Caruana is 3, 4 and 5, respectively.
From condition 2, Vidit, who did not played Biltz, has his increase of rating at just less than the increase of Boris. So, the order of Boris and Vidit is 1 and 2, respectively.

Let us draw a rectangle on the coordinate axis. With (0,0) be one of the edge A. B can be (6,0), C(6,12) and D(0,12).  E and F divide the AC in 3 equal part. E has to be (2,4) and F (4,8)
G is  centroid of DEF, Coordinates of G will be 

Upon finding G the required diagram looks like this

Area DGBC = Area GBC + Area DCG
Area DGBC = 
Area DGBC = 24+12 = 36 sq units

The divisibility rule of 3 states that the sum of digits has to be divisible by 3 in order to be divisible by 3.
Such numbers are 1002,1005,1008.....9999 : 3000 numbers.
The divisibility rules of 8 imply that the last 3 digits have to be divisible by 8: Thus "dcba" has to be divisible by 8. Such number are : 1000,1008,..9992 : 1125.
We have included the number which are divisible by 24 twice. Thus we have to remove those extra numbers.
Such number are : 1008,1032,...9984: 375 numbers.
The numbers of 4-digit number  which are divisible by 3 or 8: 3000+1125-375 = 3750.

Since ABC is an equilateral triangle and DEF is the midpoint. Thus they will divide the ABC in 4 equilateral triangle of side half of that of ABC
Let side of ABC = 2a units. Then the side of the smaller triangle is a unit.
Shaded region = Area of 3 small circles + Area DEF - Area of 1 small circle
Shaded region = Area of 2 small circles + Area DEF 
Circles are an incircle of equilateral triangle of side "a"
Radius of this incircle 
Area of 1 small circle = 
Area of 2 small circle 
Area of triangle DEF = 
Shaded area = 
Unshaded area = Total area - Shaded area
Unshaded area =  - Shaded area
Unshaded area = 
Unshaded area = 
Ratio = 

Let the profit be P and cost price be C.P
Then 10 P = C.P ....(I)
Selling Price = Cost Price + Profit = 10 P +P = 11P
Discount given  = 3P.
Marked price = Selling Price + Discount = 11P+3P = 14P.
Let the discount given to ensure profit of 20% be d%.

• Let us start by writing the N in empirical form which is 2⁵ x 3⁸ x 2⁴ x 5³ x 2 x 3 x 7²
• Which equals 2¹⁰ x 3⁹ x 5³ x 7²
• Let A = number of factors not divisible by 9
• B = Number of Factors which are odd
• We have to find 
• (A∩B) = Number of factors not divisible by 9 and are odd. It implies it does not have 2 as a factor and the highest power of 3 can be  1. Such type of factors are = (1+1)(3+1)(2+1) = 24
• A = Number of factors that are not divisible by 9. Thus the maximum power of 3 they can have is 1. Such number are = (10+1)(1+1)(3+1)(2+1) = 264
• Probability = 24/264 = 1/11
• m+n = 12.

Let the scheme be Scheme I and Scheme II. Money spent on scheme I = 
Money on scheme II = 50,000
We don't know which of the scheme has an interest rate of 20%. Thus there will be 2 cases
Case I: Scheme I has a 20% return rate, In this case, let r be return rate for scheme II

Here we can see that the RHS is less than 1, the value of rr has to be negative. Which is not possible. Thus this case is rejected
Case II: Scheme II has a 20% return rate, In this case, let r be return rate for scheme I

Therefore r = 10. Which is an integer

t/3 = 14 Let the original speed of the car be  xx km/hr
Since he was originally supposed to cover the distance in 17 hours, The total distance of the journey is 17x km.
After the malfunction, his speed will be x/3. Let him take time "t" hours to complete after the vehicle malfunction. Equating the distances we get
3x + (x/3)t = 17x
Cancelling x from both sides
We get t/3 = 14
t = 42 hours.
Initial travelling time = 17 hours. 
Final travelling time = 3+42 = 45 hours.
% increase in time = 

A and B start on bike and A drops B at R. The time taken is T₁. Hence , PR 40T₁.
At the same time, C walks from P to Q. Distance covered = 5T₁.
Distance between A and C = 40T₁ - 5T₁ = 35T₁.
Relative speed between A and C = 40 + 5 = 45.
Time taken to meet = 35/45T₁ = 7/9T₁
Now, in the meantime, B also walks from R to T. His speed is same as C, so distance travelled is same, i.e, 5 x 7/9T₁ = 35/9T₁.
Hence, the distnce between the bike and B when the bike starts from S is ST, i.e, 35T₁.
Now, let they meet at Z after time T₂.
Relative speed = 40 - 5 = 35.
Hence, 35 T₂ = 35T₁
T₂ = T₁
Hence, time taken by the overall journey = T₁ + 7/9T₁ + T₁ = 25/9T₁
Time spent by B walking = Toatl time - T₁ = 25/9T₁-T₁ = 16/9T₁.
Hence, percentage time = 16/25 x 100 = 64%.
 

Alternate solution:
Let us divide all the activities into certain time periods for simplification. Let all of them start at time t = 0.
(1) From time t = 0 to t₁. A and B will move towards Z on bike and C will walk towards Z. At exactly t₁, A will drop B and start moving back to pick up C. Distance covered by A = 40t₁, Distance covered by B = 40t₁ and Distance covered by C = 5(t₂ - t₁) and distance by A in this time period = - 40(t₂ - t₁). We have taken"-" as A is moving in opposite direction 3) From time t = t₂ to t₃. A has picked up C and now both are on bike moving towards city Z. 0. B is walking towards city Z. They all reach at city Z at same time t₃ Distance travelled by A = 40(t₃ - t₂). Distance covered by B = 5(t₃ - t₂). Distance travelled by C = 40 (t₃ - t₂)
Let the overall distance be d
Distance travelled by A = 40t₁ - 40 (t₂ - t₁) + 40(t₃ - t₂) = d
Thus 40(2t₁ + t₃ - 2t₂) = d ...(I)
Distance travelled by B = 40t₁ + 5(t₂ - t₁)+5(t₃-t₂)
35t₁ + 5t₃ = d ...(II)
Distance travelled by C = 5t₁ + 5(t₂ - t₁) + 40(t₃ - t₂) = d
40t₃ - 35t₂ = d ...(III)
Equating (II) and (III) 
35t₁ + 5t₃ = 40t₃ - 35t₂
t₁ = t₃ - t₂ ...(IV)
Equating (I) and (III)
80t₁ + 40t₃ - 80t₂ = 40t₃ - 35t₂
⇒ 80t₁  = 45t₂
Or t₂ = (16/9)t₁ ...(V)
Putting (V) in(IV)

From time t₁ to t₃ B was walking and total journey time was t₃
% time on walking = 
% time on walking = 
% time walking = 

Let us look at the series 8, 24, 62, 122
Their first level difference = 24-8 = 16, 62-24= 38 and 122-62= 60
Their second level difference = 38-16 = 22, 60-38 = 22 which are same. So this series varies with power of 2
Let T_n  represent the term of series. Then T_n can be assumed as an² + bn + c

From (III)-(II) we get 5a+b = 38
From (II)-(I) we get 3a+b =16
Eliminating b we get 2a = 22 or a=11. Thus b = -17 and c = 14

Sum = 31570-3570+280 = 28280

Let the amount of work done by a man in 1 day be "m" units and the amount of work done by a woman in 1 day be "w" units. Let's assume they take N days to complete
As per the first line we can say that 
(2m + w) x N = (N/8)(13m + 12w)
On cancelling the common factors and rearranging we get
(2m + w) x 8  = (13 m + 12w)
Which implies 16m + 8w = (13m + 12w)
or 3m = 4w ...(I)
It is also given that it is equal to 4 men and 27 children. Let a child to "c" units of work in a day
Then
(N/8) (13m + 12w) = (N/8)(4m +27c)
Or, 13m + 12w = 4m + 27c
Putting value of w from (I)
13m+9m=4m+27c
or, 18m=27c or 2m = 3c...(II)
Given that another task takes 12 days to be completed by 2 children and a man. Let us assume that it takes x days to be completed by 3 women and 1 child. Equating the amount of work done

Cancelling m and cross multiplying

9.6 days is needed. 10 is the next biggest number .

Let 100kg of fresh grapes be there. Then 25 kgs is the pulp and 75 kg is the weight of water,
Initial fraction of pulp = 25% = 1/4
FInal fraction of pulp = 
SInce the weight of pulp will not change 
Putting pulp = 25kgs

This gives updated weight of water as 50 kgs.
Evaporated water =  75-50 = 25
Ratio = 25:25 = 1:1 = m:n
a) |n-5m| =|1-5| = 4
b) |n+m| = 1+1 =2
c) |6m-3n| = 6-3 = 3
d)  |m-3n| = |1-3| = 2
A has the greatest value

From curve C₁ we can see that x ≥ 0 and y can be either positive or negative. Thus either intersection happens in the first quadrant or fourth quadrant.
x^1/2 = |y| or x = y² ...(I)) from C₁
Putting this in the C₂ equation
y²(y) = -4y² + y
On rearranging we get y³ + 4y² - y  = 0
y(y² + 4y - 1) = 0
Discriminate of (y² + 4y - 1) is positive so it must have 2 distinct roots ≠ 0
y(y² + 4y - 1) = 0 has 3 distinct roots. 
For each value of y there will be unique x such that x^1/2 = |Y|.