SRCC GBO 2024 Question Paper

Option B is True according to the passage.

The passage supports Option B. The other options are either contradicted or not mentioned. For example, the passage states that gardening reduces symptoms of depression and anxiety, not maximizes them (option A). It also mentions that not having access to nature can have negative effects on our health (option C). Lastly, the passage does not compare reactions to real plants versus artificial plants (option D). Therefore, option B is the only true statement according to the passage.

The word ‘resilience’ in the given passage refers to the ability to recover quickly from difficulties or adapt well in the face of adversity. Therefore, the word that stands closest in meaning to ‘resilience’ from the given choices is Option A) Grit

B) Tribulation: Refers to suffering, not recovery.

C) Bummer: A disappointing event, not about resilience.

D) Affliction: Describes hardship, not overcoming it.

The passage's main idea is that houseplants provide multiple benefits, particularly for mental health, and serve as an essential link to nature for those without access to outdoor green spaces. 

Options a, b, and c support this idea by highlighting the negative effects of lacking nature, the positive health impacts of indoor plants, and the emotional benefits of caring for them. 

Option d, however, disapproves the main idea by suggesting that houseplants may not effectively remove indoor pollutants or improve air quality due to impractical requirements, undermining one of the touted benefits and thus challenging the overall positive stance on houseplants.

Hence, Option D is the answer. 

The word ‘cognitive’ in the given passage refers to the mental processes of acquiring knowledge and understanding through thought, experience, and the senses. 

Therefore, rational is the most suitable word among the given options as it relates to the thinking and understanding aspects of cognitive functions.

The passage discusses how houseplants provide a connection to nature for those without access to outdoor green spaces. It mentions the mental health benefits of being around plants, including improved cognitive performance and reduced stress levels. The passage also talks about the potential for houseplants to remove indoor pollutants, although it notes that achieving significant results would require a large number of plants in a brightly lit room. This is captured in Option A

The other options are incorrect because they either negate or ignore key points made in the passage. 

The most appropriate option to fill in the blanks is Option B

This option correctly uses the prepositions to convey the intended meaning. The city is important “for” the Hindus, and the rites are performed “in” the city. 

The other options do not fit grammatically or contextually in the sentence.

The most appropriate antonym for the underlined word “intensify” is Option B. 

In this context, “intensify” means to become more intense or to increase in degree or extent. Therefore, the opposite of “intensify” would be to decrease or lessen, which is what “diminish” means. The other options, “amplify” and “escalate”, are synonyms of “intensify”, not antonyms, and “vorsen” is not a recognized English word. 

Therefore, “diminish” is the correct choice.

The segment that contains a grammatical error is Option C.

The error in this segment is the incorrect use of the indefinite article “a” before a word starting with a vowel sound. The correct phrase should be “as an essential”. The indefinite article “an” is used before words that begin with a vowel sound. 

Therefore, the corrected sentence would be: “In today’s motivational literature, failure is often celebrated as an essential stepping stone to success.” 

The other segments are grammatically correct.

The most appropriate option to fill in the blanks is Option A

So, the completed sentence would be:

Cells also began living together, probably because certain benefits could be obtained. Groups of cells might be able to feed more efficiently or gain protection from simply being bigger. Living collectively, cells began to support the needs of the group by each cell doing a specific job. 

This option makes the most sense in the context of cellular biology and the evolution of multicellular organisms.

Options B,C and D are not suitable because they contain words that do not match the blank in the given context.

The most appropriate option to fill in the blank is Option C

So, the completed sentence would be: 

The crowd was carried away by his passionate speech.

This phrase means that the crowd was deeply moved or swept up in the emotion of the moment. 

The other options (carried off, carried on, carried out) do not fit the context appropriately.

Familiarity breeds contempt, which means extensive knowledge of or close association with someone or something leads to a loss of respect for them or it.

Options B, C and D try to take the idiom's meaning positively, which is the opposite of its actual meaning. 
Option A correctly describes the meaning of the idiom and hence would be our answer. 

Therefore, Option A is the correct answer. 

The sentence essentially conveys that the concept of a heritage and educational district raised/attracted the interest of citizens in the whole district. 

Option C piqued the interest and best fits our sentence. 
When something piques your interest or curiosity, the verb pique means to arouse, stimulate, or excite.

Therefore, option C is the correct answer. 

Pathos is the power of a person, situation, piece of writing, or work of art to cause feelings of sadness, especially because people feel sympathy. So, pity would match best with it.

Paucity: smallness of quantity; scarcity; scantiness:

A pedant is  a person who is excessively concerned with minor details and rules or with displaying academic learning. So, intellectual fits best, Therefore, Option D is the correct answer.

To be a fish in troubled waters is to be involved in a very difficult or delicate situation, which could cause you problems. Therefore, Optionn C best captures the meaning of the idiom.

Option D makes the most sense.

Warm reception: This suggests the hypothesis is being received positively, which aligns with "enjoyed."

Bipartisan support: This indicates support from both sides of the political spectrum.

Disclosure: This refers to making information public, which aligns with the government push.

To surmount something is to to deal successfully with a difficulty or problem

The opposite of this would be to surrender or give up. Therefore, yielded (Option C) is the correct answer.

Option Ddirectly conveys the idea that addressing a small problem early prevents a larger issue later. It emphasizes the importance of being prepared and taking preventative measures to avoid future complications.

Here's why the other options are not as fitting:

Option A: This proverb warns against being overly optimistic about something that hasn't happened yet. While it can be related to caution, it doesn't directly address preparedness.

Option B: This proverb refers to situations where having too many people involved can lead to a negative outcome. It doesn't focus on caution or preparedness.

Option C: This proverb means to reveal a secret. It has no connection to caution or preparedness.

Option C best fits blanks.

Continue: Natural history data and research have always been important, and this phrasing suggests it's an ongoing practice.

Provide: This accurately describes how natural history data offers the foundation for further exploration.

Uncover: This term aligns with the purpose of using natural history data - to discover and reveal the underlying principles governing natural systems.pen_spark

Leaves is the most natural and common way to express a habitual action in the present tense. It indicates a recurring event that happens at a specific time each day.

Has left would be appropriate if your mother had already left for her class today at 9 am, but it wouldn't be the best choice for describing a general habit.

Was leaving implies an action in progress in the past, which doesn't fit the context of a daily routine.

Is leaving could be used if you were talking about right this moment, but the context suggests this is a general routine that happens every day.

Therefore, Option A is the correct answer.

BDCA is the correct sequence of the sentences.

B is the first sentence. The paragraph begins with the description of the streets calming down after an initial period of panic. B is followed by D. Sentence D provides specific details about the reopening of the bazaar and the fishwives’ actions.

Sentence C follows D. After the initial chaos, people resumed their usual activities, emphasizing a return to normalcy. 

A is the conclusive statemen.It highlights a significant change—the absence of foreign faces on the streets.

Therefore, the correct order to form a coherent paragraph is BDCA.

Option C is the correct. This word accurately reflects the meaning of “detrimental.” It implies harm, negative consequences, or adverse effects. Therefore, Option C is the correct choice.

Other options does not match the meaning of the underlined word.

The correct answer is Option D. The passage explains that Christmas trees had a detrimental impact on forests due to monoculture planting, making them vulnerable to pests and harsh weather. Additionally, the development of artificial trees using surplus materials is mentioned.

Option A: "Benign” means harmless or favorable, which doesn’t fit the context. The second blank should be filled with a verb, not a preposition. So this option is not suitable.

Option B : “Prolific” means abundant or productive, but it doesn’t convey the intended meaning of negative impact on forests. The second blank should also be filled with a verb. So this option is incorrect.

Option C: “Differential” doesn’t fit the context, and the verb tense is incorrect. This option is not the right choice.

The segment that contains a grammatical error is Option A.

In English grammar, when we use “one of” followed by a plural noun or pronoun, the verb that follows should be singular, not plural. In this case, “one of these authors” refers to a single author, so the verb “is” should be used instead of “are”. Hence, the segment “One of these authors are” contains a grammatical error. The corrected phrase should be “One of these authors is”. So, the corrected sentence would be:

“One of these authors is going to win the Booker Award for his or her book.”

The segment that contains a grammatical error is Option D.

The correct phrase is “put forth” or “presented” when referring to proposals in a meeting. The phrase “put on” is not typically used in this context. Hence, the segment “put on at the meeting,” contains a grammatical error. The corrected phrase should be “put forth at the meeting,” or “presented at the meeting,”. So, the corrected sentence would be:

“Several proposals were put forth at the meeting, but none was appealing enough.” or “Several proposals were presented at the meeting, but none was appealing enough.”

The most appropriate option to fill in the blanks is Option D.

“Hay provides the bulk of the horse’s ration and may be of varying composition according to locale. Mash is bran mixed with water and with various invigorating additions or medications. It may be given to horses with digestive troubles or deficient eating habits.”

This option makes the most sense in the context of the sentence, as it correctly describes the role of hay and mash in a horse’s diet. The word “bulk” refers to the largest part or most of something, so it accurately describes the role of hay in the horse’s ration. The phrase “according to locale” indicates that the composition of the hay can vary depending on the location. The word “various” in the context of the mash indicates that different additions or medications can be added to the mash.

The most appropriate option to fill in the blanks is Option B.

“The exhibit of the South African men in Boston began with a cacophony of animal sounds made by the ‘savages’ while out of sight. The troupe then leaped onto the stage, barely clad in skins and feathers and clutching spears and clubs, and proceeded to perform several native songs and dances.”

This option makes the most sense in the context of the sentence, as it correctly describes the scene of the exhibit. The word “sounds” accurately describes the noises made by the ‘savages’, “leaped” is a suitable action for the troupe’s entrance onto the stage, “clad” correctly describes their attire, and “proceeded” is the appropriate action for their performance of songs and dances.

The most appropriate option to fill in the blank is Option B.

“A momentous change is evident in transport networking and soon India will join the speed club.”

This option makes the most sense in the context of the sentence, as it correctly indicates a future event or action. In this case, it suggests that India is expected to join the speed club in the near future. The other options do not correctly convey this future intention or assertion.

Option A is the correct answer.

Rejuvenate: This word means to make someone or something look or feel younger, fresher, or more lively. In this context, it is matched with the meaning “Refresh”, which is similar in sense.

Relinquish: This word means to cease to keep or claim voluntarily; to give up. It is matched with the meaning “Renounce”, which also means to declare one’s abandonment of something formally.

Relegate: This word means to consign or dismiss to an inferior rank or position. It is matched with the meaning “Refer”, which in this context could mean to pass a matter to (another body, typically one with more authority or expertise) for a decision.

So, the correct option is a-3, b-1, and c-2, which match each word to its correct meaning.

The segment that contains a grammatical error is Option A.

The correct phrase should be “the quantity of gray matter in your brain”. So, the corrected sentence would be:

“Several studies support the idea that the amount of consistent sleep you receive significantly impacts the quantity of gray matter in your brain.”

This is because “the” is used before a noun when the specific identity of the noun is known to the reader, whereas “a” is used before a noun when the specific identity of the noun is not known to the reader. In this case, we’re referring to a specific quantity of gray matter (the one in your brain), so “the” is the appropriate article to use.

The segment that contains a grammatical error is Option B.

In English grammar, the verb “overcome” is typically followed by the preposition “with” when expressing an emotional state. The phrase “overcome with regret” means that someone is completely filled with regret.

So, in the sentence “At the end, Ashoka was overcome with regret at the bloodshed that had taken place as a result of the Kalinga battle”, “with” is the correct preposition to use after “overcome” to convey the meaning that Ashoka was filled with regret. The preposition “in” is not typically used in this context. Hence, the segment “in regret at the bloodshed” contains a grammatical error.

A porpoise is a type of animal. So Option A is incorrect. 

Option B can be eliminated as it says intentions. 

Option C is incorrect as it implies something entirely different. 

Option D is the correct answer.

To throw in the towel is to abandon a struggle; admit defeat. Therefore, the word that would best fit there is Abandon. So, Option A is the correct answer.

Sentence 1 introduces Nalanda, so it is logical that the next sentence builds on this. Considering this, Sentence A would be best suited to follow Sentence 1. 

Sentence 1 states how  the university "attracted 10,000 students from across Eastern and Central Asia."

Sentence 3 elaborates on this by elaborating why they gathered here. So Sentence C must follow A.

Therefore, the correct answer is Option B.

Ingrained : (of a habit, belief, or attitude) firmly fixed or established; difficult to change.

Renowned: widely acclaimed and highly honoured: celebrated.

Ensued: happen or occur afterwards or as a result.

Renewed: to make like new. 

Considering this, only "renewed" makes sense in the first blank. 

Deeming is to come to think or judge. This fits perfectly in Blank 2 as well. Ineligible is essentially not being eligible (the migrants not being eligible). This fits in Blank 3. The policy is controversial. 

Therefore, Option D is the correct answer.

The phrase hit the hay means to go to bed to sleep. Therefore, Option A is the correct answer. 

CDAB is the correct order.

Sentence C sets the scene by describing the situation with Bala Singh and the luggage. It provides context for the subsequent events.Sentence D adds to the tension. The fact that the other men are watching the narrator intently suggests that something unusual is happening.Sentence A reveals the odd behavior of Bala Singh, who doesn’t greet the narrator and gives a consistent response about not being ill. It builds intrigue and curiosity. Sentence B  provides further details about Bala Singh’s behavior during the march, emphasizing his strange demeanor. It adds to the mystery and sets the tone for the rest of the narrative.

What the passage is trying to say is that because of digital cameras, the sales of the film cameras declined, or they were forced to bring out different models it. Considering this, only Option B makes sense, as simulating, expanding or elevating most film cameras doesn't make sense. 

Also, sales fall, so for the second blank too, fell is the right word. 

Therefore, Option B is the correct answer. 

Mellifluous, perspicacious, and discerning all describe positive mental qualities.

Mellifluous refers to something that is pleasing to the ear (sound).

Perspicacious means having a sharp intellect or good judgment.

Discerning means having or showing good judgment.

Ineffable, on the other hand, means something that cannot be expressed in words.

Therefore, while the other three words describe positive traits, ineffable describes something beyond words.

"Bite the bullet" is an informal phrase that means "to do something unpleasant or painful because it is necessary even though you would like to avoid it." So, Option A is the correct answer.

Sentence 2 is the odd one out. 

Sentences 3 and 4 discuss vegetation in different mountain ranges and woodland types in Northern Spain, respectively.Sentence 1 focuses on esparto grass and its associated alfa grass, which is related to plant species.

However, sentence 2 stands apart because it introduces a different topic: brown forest soils in Galicia and Cantabria.

First, let's create a table with the sales, expenditures, and profits values for each year.

The pie chart shows that 20% of the sales in 2020 were from the US, or 20% of 400 is 80 million.

Similarly, calculating for the other years and calculating, we get the following table:

In the year 2021, the % increase in sales is 16/80 * 100 = 20%; in 2022, it is 4/96 * 100 [less than 5%], whereas in the year 2023, the %change in sales is 24/100 * 100 = 24%

=> 2023 has the highest % increase in sales.

First, let's create a table with the sales, expenditures, and profits values for each year.

The pie chart shows that 20% of the sales in 2020 were from the US, or 20% of 400 is 80 million.

Similarly, calculating for the other years and calculating, we get the following table:

The average profit for the 4 years = (8 + 6 + 10 + 16)/4 = 10.

First, let's create a table with the sales, expenditures, and profits values for each year.

The pie chart shows that 20% of the sales in 2020 were from the US, or 20% of 400 is 80 million.

Similarly, calculating for the other years and calculating, we get the following table:

The profit per rupee of equity calculation is as follows:

In 2020: 8/8 = 1

In 2021: 6/12 = 0.5

In 2022: 10/16 = 0.62

In 2023: 16/28 = 0.57

=> In 2020, it is the highest.

First, let's create a table with the sales, expenditures, and profits values for each year.

The pie chart shows that 20% of the sales in 2020 were from the US, or 20% of 400 is 80 million.

Similarly, calculating for the other years and calculating, we get the following table:

The sales per rupee of expenditure calculation is as follows:

In 2020: 80/72 = 1.111

In 2021: 96/90 = 1.066

In 2022: 100/90 = 1.111

In 2023: 124/108 = 1.15

=> In 2021, it is the lowest of the above values.

From the table, we can see that in the year 2018-2019, Company D shows (25/135)*100% = 18.52% increase in market value, which is higher than any other changes.

The correct option is B

From the table, we can see that in the year 2018-2019, Company D shows (25/135)*100% = 18.52% increase in market value, which is higher than any other changes.

The correct option is C

The absolute change in the market value (year wise) are given in the table below:

From the table, we can see that the greatest absolute change in market share occurred for Company A in the year 2019-20

Hence, the correct option is C

It is given that the ratio of the students in institution A to that of institution B is 7 : 8, which implies if the students in institution A is 7x, then the students in institution B is 8x.

It is also known that the total number of students in both institutions are 1350.

=> (7x+8x) = 1350 => x = 90

Thus, the students in institution A is (7*90) = 630, and the students in institution B is (8*90) = 720

It is given that In institution A, 70% students are boys and the rest are girls, whereas in institution B the ratio of boys to girls is 11 : 7.

Thus, the number of boys in institution A = (70% of 630) = 441, and the number of girls in institution A = (630-441) = 189.

Similarly, the number of boys in institution B = (11/18)*720 = 440, and the number of girls in institution B = (720-440) = 280

It is given that there are only three classes in each Institution, Class X, Class Y and Class Z.

It is known that in institution A, (4/7)th of the total girls joined Class Y, which implies the total number of girls at Class Y in institution A is (4/7)*189 = 108, which implies the remaining number of girls = (189-108) = 81. It is also known that (5/9)th of the remaining girls enrolled in Class Z and rest in Class X.

Thus, the total number of girls at Class Z in institution A is (5/9)*81 = 45 =>  the total number of girls at Class X in institution A is (81-45) = 36

Now, in institution A, the total number of boys at Class X = (3/7)* 441 = 189, which implies the number of boys who are remaining is (441-189) = 252

It is given that (4/9) of the remaining boys are in Class Y = (4/9)*252 = 112, and the boys in class Z in institution A = (252-112) = 140

Now, in Institution B, (4/11)th of the boys are in Class Y = (4/11)*440 = 160, and the number of boys in Class Z is (21/20)*160 = 168, which means the number of boys in Class X = (440-160-168) = 112

Similarly, (1/4)th of the girls are in Class Z = (1/4)*280 = 70, and it is given that the number of girls enrolled in class X is 10% more than the girls enrolled in class Y.

Let the number of girls in Class Y be 100x, which implies the number of girls in Class X is 110x

Thus, (110x+100x) = 210 => x = 1 => The number of girls in Class Y is 100, and the number of girls in Class X is 110

We know that the number of boys enrolled in class X in institution A is 189, and the number of girls enrolled in class Y in institution B = 100

Thus, the number of boys enrolled in class X in institution A is 89% more than the number of girls enrolled in class Y in institution B.

The correct option is B

It is given that the ratio of the students in institution A to that of institution B is 7 : 8, which implies if the students in institution A is 7x, then the students in institution B is 8x.

It is also known that the total number of students in both institutions are 1350.

=> (7x+8x) = 1350 => x = 90

Thus, the students in institution A is (7*90) = 630, and the students in institution B is (8*90) = 720

It is given that In institution A, 70% students are boys and the rest are girls, whereas in institution B the ratio of boys to girls is 11 : 7.

Thus, the number of boys in institution A = (70% of 630) = 441, and the number of girls in institution A = (630-441) = 189.

Similarly, the number of boys in institution B = (11/18)*720 = 440, and the number of girls in institution B = (720-440) = 280

It is given that there are only three classes in each Institution, Class X, Class Y and Class Z.

It is known that in institution A, (4/7)th of the total girls joined Class Y, which implies the total number of girls at Class Y in institution A is (4/7)*189 = 108, which implies the remaining number of girls = (189-108) = 81. It is also known that (5/9)th of the remaining girls enrolled in Class Z and rest in Class X.

Thus, the total number of girls at Class Z in institution A is (5/9)*81 = 45 => the total number of girls at Class X in institution A is (81-45) = 36

Now, in institution A, the total number of boys at Class X = (3/7)* 441 = 189, which implies the number of boys who are remaining is (441-189) = 252

It is given that (4/9) of the remaining boys are in Class Y = (4/9)*252 = 112, and the boys in class Z in institution A = (252-112) = 140

Now, in Institution B, (4/11)th of the boys are in Class Y = (4/11)*440 = 160, and the number of boys in Class Z is (21/20)*160 = 168, which means the number of boys in Class X = (440-160-168) = 112

Similarly, (1/4)th of the girls are in Class Z = (1/4)*280 = 70, and it is given that the number of girls enrolled in class X is 10% more than the girls enrolled in class Y.

Let the number of girls in Class Y be 100x, which implies the number of girls in Class X is 110x

Thus, (110x+100x) = 210 => x = 1 => The number of girls in Class Y is 100, and the number of girls in Class X is 110

Hence, the total number of students enrolled in class Y of both institutions = (number of boys in class Y in both institutions) + (number of girls in Class Y in both institutions)

=> (112+160)+(108+100) = 480

The correct option is C

It is given that the ratio of the students in institution A to that of institution B is 7 : 8, which implies if the students in institution A is 7x, then the students in institution B is 8x.

It is also known that the total number of students in both institutions are 1350.

=> (7x+8x) = 1350 => x = 90

Thus, the students in institution A is (7*90) = 630, and the students in institution B is (8*90) = 720

It is given that In institution A, 70% students are boys and the rest are girls, whereas in institution B the ratio of boys to girls is 11 : 7.

Thus, the number of boys in institution A = (70% of 630) = 441, and the number of girls in institution A = (630-441) = 189.

Similarly, the number of boys in institution B = (11/18)*720 = 440, and the number of girls in institution B = (720-440) = 280

It is given that there are only three classes in each Institution, Class X, Class Y and Class Z.

It is known that in institution A, (4/7)th of the total girls joined Class Y, which implies the total number of girls at Class Y in institution A is (4/7)*189 = 108, which implies the remaining number of girls = (189-108) = 81. It is also known that (5/9)th of the remaining girls enrolled in Class Z and rest in Class X.

Thus, the total number of girls at Class Z in institution A is (5/9)*81 = 45 => the total number of girls at Class X in institution A is (81-45) = 36

Now, in institution A, the total number of boys at Class X = (3/7)* 441 = 189, which implies the number of boys who are remaining is (441-189) = 252

It is given that (4/9) of the remaining boys are in Class Y = (4/9)*252 = 112, and the boys in class Z in institution A = (252-112) = 140

Now, in Institution B, (4/11)th of the boys are in Class Y = (4/11)*440 = 160, and the number of boys in Class Z is (21/20)*160 = 168, which means the number of boys in Class X = (440-160-168) = 112

Similarly, (1/4)th of the girls are in Class Z = (1/4)*280 = 70, and it is given that the number of girls enrolled in class X is 10% more than the girls enrolled in class Y.

Let the number of girls in Class Y be 100x, which implies the number of girls in Class X is 110x

Thus, (110x+100x) = 210 => x = 1 => The number of girls in Class Y is 100, and the number of girls in Class X is 110

Therefore,  the number of boys enrolled in Class Y in institution B = 160, and the number of girls enrolled in Class X in institution A = 36.

Hence, the difference is (160-36) = 124

The correct option is D

It is given that the ratio of the students in institution A to that of institution B is 7 : 8, which implies if the students in institution A is 7x, then the students in institution B is 8x.

It is also known that the total number of students in both institutions are 1350.

=> (7x+8x) = 1350 => x = 90

Thus, the students in institution A is (7*90) = 630, and the students in institution B is (8*90) = 720

It is given that In institution A, 70% students are boys and the rest are girls, whereas in institution B the ratio of boys to girls is 11 : 7.

Thus, the number of boys in institution A = (70% of 630) = 441, and the number of girls in institution A = (630-441) = 189.

Similarly, the number of boys in institution B = (11/18)*720 = 440, and the number of girls in institution B = (720-440) = 280

It is given that there are only three classes in each Institution, Class X, Class Y and Class Z.

It is known that in institution A, (4/7)th of the total girls joined Class Y, which implies the total number of girls at Class Y in institution A is (4/7)*189 = 108, which implies the remaining number of girls = (189-108) = 81. It is also known that (5/9)th of the remaining girls enrolled in Class Z and rest in Class X.

Thus, the total number of girls at Class Z in institution A is (5/9)*81 = 45 => the total number of girls at Class X in institution A is (81-45) = 36

Now, in institution A, the total number of boys at Class X = (3/7)* 441 = 189, which implies the number of boys who are remaining is (441-189) = 252

It is given that (4/9) of the remaining boys are in Class Y = (4/9)*252 = 112, and the boys in class Z in institution A = (252-112) = 140

Now, in Institution B, (4/11)th of the boys are in Class Y = (4/11)*440 = 160, and the number of boys in Class Z is (21/20)*160 = 168, which means the number of boys in Class X = (440-160-168) = 112

Similarly, (1/4)th of the girls are in Class Z = (1/4)*280 = 70, and it is given that the number of girls enrolled in class X is 10% more than the girls enrolled in class Y.

Let the number of girls in Class Y be 100x, which implies the number of girls in Class X is 110x

Thus, (110x+100x) = 210 => x = 1 => The number of girls in Class Y is 100, and the number of girls in Class X is 110

Hence, the total number of boys enrolled in classes X and Y in institution A = (189+112) = 301, and the number of girls enrolled in classes Y and Z in institution B = (100+70) = 170

Therefore, the total number of boys enrolled in classes X and Y in institution A is (301/170)*100% = 177%

The correct option is A

In the above given bar graphs the population and national income of a country from the fiscal years 2014-15 to 2019-20 are shown.

Let us tabulate the above information from the graph into a table and also calculate per capita income for all the fiscal years.

On the basis of the above table

We need to find the year the increase in per capita income compared to the previous years is the lowest.

Let us check for each option and calculate the increase in per capita income compared to previous years for the following fiscal years.

Option A) 2015-2016

The difference in per capita income for years 2015-2016 and 2014-2015 is approximately equal to 3482-3098 = 384

Option B) 2019-2020

The difference in per capita income for years 2019-2020 and 2018-2019 is approximately equal to 5319-4857 = 462

Option C) 2017-2018

The difference in per capita income for years 2017-2018 and 2016-2017 is approximately equal to 4221-3786 = 435

Option D) 2016-2017

The difference in per capita income for years 2016-2017 and 2015-2016 is approximately equal to 3786-3482 = 304

Hence, the lowest increase in per capita income compared to previous year is in the fiscal year 2016-2017. (Option D)

In the above given bar graphs the population and national income of a country from the fiscal years 2014-15 to 2019-20 are shown.

Let us tabulate the above information from the graph into a table and also calculate per capita income for all the fiscal years.

On the basis of the above table

We need to find the year in which the difference between the percentage increase in per capital income and the percentage increase in population compared to the previous year the highest.

Let us check for each option and calculate the difference of % increase in per capita income and % increase in population compared to previous years for the following fiscal years.

Option A) 2017-2018

The % increase in per capita income in the year 2017-2018 as compared to 2016-2017 is

=> (4221-3786)/3786= 11.49%

The % increase in population in the year 2017-2018 as compared to 2016-2017 is

=> (117.25-115.5)/115.5 = 1.52%

Thus, the difference between them = 11.49-1.52 = 9.97

Option B) 2015-2016

The % increase in per capita income in the year 2015-2016 as compared to 2014-2015 is

=> (3482-3098)/3098 = 12.40%

The % increase in population in the year 2015-2016 as compared to 2014-2015 is

=>(112.5-111)/111 = 1.35%

Thus, the difference between them = 12.40-1.35 = 11.05

Option C) 2016-2017

The % increase in per capita income in the year 2016-2017 as compared to 2015-2016 is

=> (3786-3482)/3482 = 8.73%

The % increase in population in the year 2016-2017 as compared to 2015-2016 is

=> (115.5-112.5)/112.5 = 2.66%

Thus, the difference between them = 8.73-2.66 = 6.07

Option D) 2018-2019

The % increase in per capita income in the year 2018-2019 as compared to 2017-2018 is

=> (4857-4221)/4221 = 15.06%

The % increase in population in the year 2018-2019 as compared to 2017-2018 is

=> (120-117.25)/117.25 = 2.34%

Thus, the difference between them = 15.06-2.34 = 12.72

Thus, the difference between the percentage increase in per capital income and the percentage increase in population compared to the previous year the highest for the year 2018-2019. (Option D)

In the above given bar graphs the population and national income of a country from the fiscal years 2014-15 to 2019-20 are shown.

Let us tabulate the above information from the graph into a table and also calculate per capita income for all the fiscal years.

On the basis of the above table

In the above given bar graphs the population and national income of a country from the fiscal years 2014-15 to 2019-20 are shown.

Let us tabulate the above information from the graph into a table and also calculate per capita income for all the fiscal years.

On the basis of the above table

We need to find the year which witnessed the highest increase in per capita income compared to the previous year.

Let us check for each option and calculate the increase in per capita income compared to previous years for the following fiscal years.

Option A) 2016-2017

The difference in per capita income for years 2016-2017 and 2015-2016 is approximately equal to 3786-3482 = 304

Option B) 2017-2018

The difference in per capita income for years 2017-2018 and 2016-2017 is approximately equal to 4221-3786 = 435

Option C) 2015-2016

The difference in per capita income for years 2015-2016 and 2014-2015 is approximately equal to 3482-3098 = 384

Option D) 2018-2019

The difference in per capita income for years 2018-2019 and 2017-2018 is approximately equal to 4857-4221 = 636

Hence, the highest increase in per capita income compared to previous year is in the fiscal year 2018-2019. (Option D)

First, let us note down the table and find out the sum of the costs across the rows and columns:

If the exigencies double, the expense increases by 37.8 => The required % increase in expenses = 37.8/1088.5 * 100 = 3.4726 [Option-B]

First, let us note down the table and find out the sum of the costs across the rows and columns:

For the budget to be below 1050 lakhs, we must reduce the management budget by 1088.5 - 1050 = 38.5 => 9.625 per year.

In the year 2021, the % decrease = 9.625/22.5 * 100 = 42.777% = 42.78% [Option-A]

First, let us note down the table and find out the sum of the costs across the rows and columns:

The cost of materials increases by 5% each year from 2022 onwards:

=> We need to consider the costs of Bolster, Brace, Slabs and the other materials.

=> The increase in over cost is 5% of [120 + 67.5 + 18 + 27 + 112.5 + 90 + 24 + 31.5] = 24.525 [close to option-D]

First, let us note down the table and find out the sum of the costs across the rows and columns:

The total cost of materials and labour for the four years = 375 + 262.5 + 64.5 + 96 + 97.7 = 897.5
By the third year, the cost = 897.5 - 112.5 - 90 - 24 - 31.5 - 27 = 612.5

=> The % of project completed = 612.5/897.5 = 0.682 [Option-B]

In the given pie charts , the percentage of revenue generated in four quarters of the year 2020 for a company is given.

The total revenue generated by the company was 225 million rupees in the year 2020. So, we have tabulated the revenue generated by the company in each quarter below.

Also, the second pie chart shows theprofit as a percentage of revenue generated for that respective quarter in 2020.

We have tabulated the profit generated by the company in each quarter below-

Now, on the basis of the information filled in the above table-

The profit was maximum in the first quarter. (Option B)

In the given pie charts , the percentage of revenue generated in four quarters of the year 2020 for a company is given.

The total revenue generated by the company was 225 million rupees in the year 2020. So, we have tabulated the revenue generated by the company in each quarter below.

Also, the second pie chart shows theprofit as a percentage of revenue generated for that respective quarter in 2020.

We have tabulated the profit generated by the company in each quarter below-

Now, on the basis of the information filled in the above table-

The average of profit (in million rupees generated in the $$1^{st}$$ and $$4^{th}$$ quarters is equal to

(17.43+12.39)/2 = 14.91 million rupees (Option A)

In the given pie charts , the percentage of revenue generated in four quarters of the year 2020 for a company is given.

The total revenue generated by the company was 225 million rupees in the year 2020. So, we have tabulated the revenue generated by the company in each quarter below.

Also, the second pie chart shows theprofit as a percentage of revenue generated for that respective quarter in 2020.

We have tabulated the profit generated by the company in each quarter below-

Now, on the basis of the information filled in the above table-

The profit (in rupees) of the company in the $$2^{nd}$$ quarter of the year 2020 is equal to 8.66 million rupees (Option B)

In the given pie charts , the percentage of revenue generated in four quarters of the year 2020 for a company is given.

The total revenue generated by the company was 225 million rupees in the year 2020. So, we have tabulated the revenue generated by the company in each quarter below.

Also, the second pie chart shows theprofit as a percentage of revenue generated for that respective quarter in 2020.

We have tabulated the profit generated by the company in each quarter below-

Now, on the basis of the information filled in the above table-

If x is the profit generated in the $$1^{st}$$ and $$2^{nd}$$ quarters and y is the profit generated in the $$3^{rd}$$ and $$4^{th}$$ quarters,

The value of x = 17.43+8.6 = 26.09

The value of y = 11.81+12.39 = 24.20

Hence, x > y (Option B is the correct answer.)

Marks obtained by Sumit in Hindi, English and Maths = 75+85+75 = 235
These marks were obtained out of 120+130+150 = 400

Giving the percentage marks obtained by Sumit to be$$\frac{235}{400}\times\ 100$$

Marks obtained by Sandhya in Hindi, English and Maths = 85+110+120 = 315
Giving the percentage marks obtained by Snadhya to be $$\frac{315}{400}\times\ 100$$

This would give their ratio to be, $$\frac{235}{400}\times\ 100\ :\ \frac{315}{400}\times\ 100$$
$$235:315$$
47:63

Therefore, Option C is the correct answer.

Marks obtained by Lokesh and Sumit in Hindi, English and Maths = (55+115+60) + (75+85+75) = 465

Marks obtained by Aman and Kartik in Economy, Accountancy and Geography would be (100+120+90) + (95+100+80) = 585

We are to find out what percentage of the second number is approximately the first number. 
We calculate this using$$\frac{465}{585}\times\ 100=\frac{31}{39}\times\ 100$$

The denominator is very close to 40, approximating the denominator to be 40; we get the percentage value to be 77.5%

Since we have approximated up the denominator, the value which we have gotten is actually smaller than the real value. Therefore, the real value would be a little higher than 77.5%

Of the given options, only 78.8% satisfy our requirements. 

Therefore, Option C is the correct answer. 

Total marks obtained by Aman in all subjects: 95+65+115+100+125+90 = 590
Total marks obtained by Kartik in all subjects: 60+95+130+95+100+80 = 560
Total marks obtained by Mukesh in all subjects: 105+90+95+110+140+85 = 625
Total marks obtained by Lokesh in all subjects: 55+115+60+65+75+70 = 440

The average of these total marks would then be $$\frac{590+560+625+440}{4}=\frac{2215}{4}=553.75$$

Therefore, Option D is the correct answer. 

Total marks obtained by Sumit in all subjects: 75+85+75+80+95+75 = 485

Total marks obtained by Mukesh in all subjects: 105+90+95+110+140+85 = 625

The difference between these two would be $$\left|625-485\right|=140$$

Therefore, Option D is the correct answer. 

Total runs scored by X in the given six years: 470+510+780+525+880+450 = 3615

Total runs scored by Z in the given six years: 310+980+450+795+220+690 = 3445

Giving the ratio of the two to be 3615:3445, which is 723:689

Therefore, Option A is the correct answer. 

The total runs scored by the three players in the following years would be 
2018:   310+470+670 = 1450
2020:   230+450+780 = 1460
2023:   450+690+840 = 1980
2019:   290+510+980 = 1780

The lowest total runs were scored in the year 2018. 

Therefore, Option A is the correct answer, 

Total runs scored by X from the year 2020 to the year 2023 would be 780+525+880+450 = 2635

Total runs scored by Z from the year 2018 to the year 2021 would be 310+980+450+795 = 2535

Giving the difference between the two scores to be $$\left|2635-2535\right|=100$$

Therefore, Option D is the correct answer. 

The total runs scored by the three batsmen in the year 2021 would be 525+795+1170 = 2490

Giving the average of the runs scored to be $$\frac{2490}{3}=830$$

Therefore, Option C is the correct answer. 

The total revenue generated by companies A, B and C in the year 2015 would be 455+210+325 = 990 million 

Giving the average to be $$\frac{990}{3}=330$$ million

Therefore, Option A is the correct answer. 

Total Revenue generated by company A in the five years would be: 455+225+305+340+440 = 1,765

Total Revenue generated by company B in the five years would be 210+245+350+280+320 = 1,405

The ratio of these two value would be 1765:1405, which is 353:281

Therefore, Option B is the correct answer. 

Revenue generated by companies A and B in the year 2016: 225+245= 470

Revenue generated by companies B and C in the year 2018: 280+310 = 590

We are asked that first number is what percentage of the second number; this can be calculated by $$\frac{47}{59}\times\ 100$$

Here, the denominator is very close to 60. Approximating the value of the denominator to 60, we get the percentage value to be 78.33%

Since we have approximated up the value of the denominator, the  percentage value would be a little higher than 78.33%

Of the given options, 79.66% is the only number that would satisfy our requirements. 

Therefore, Option C is the correct answer. 

Revenue generated by company A in the year 2019 is given to be 440 million
Revenue generated by company B in the year 2018 is given to be 280 million
And finally, Revenue generated by company C in the year 2016 is given to be 450 million

Adding them up and dividing by 3 in order to get the average, we get $$\frac{440+280+450}{3}=\frac{1170}{3}=390$$

 Therefore, Option C is the correct answer. 

Join the points O and P, the slope of this line is 1/2/ Join the points O and Q, the slope of this line is -2. The product of slopes of two perpendicular lines will be -1. These two lines are perpendicular. The point R lies on OP and S on OQ. Since these two lines are perpendicular, the line joining R and S will be the hypotenuse. So it is $$\sqrt{\ 30^2+40^2}=50$$.

Option B and D are both correct. The sum of interior and exterior angles is 180. Since the ratio is 9:2, let the exterior angle be 9x, and the interior angle be 2x.

$$11x=180$$

$$x=\frac{180}{11}$$

The exterior angle of a regular polygon is 360/n. The derived exterior angle from the x value is 360/11.

So, the n value is 11.

So option B, D are correct.

In this question, we need to check with each option.

A) If 1/6 part of the mixture is replaced with water.

The initial part of the water is 3/8, and that of milk is 5/8. When 1/6 of the mixture is replaced with water, the milk taken out will be added as water. $$\frac{1}{6}\cdot\frac{5}{8}=\frac{5}{48}$$ is the amount of milk that is taken out, and that will also be the amount of water added.

So, the new ratio will be $$\frac{x}{y}=\frac{\left(\frac{3}{8}+\frac{5}{48}\right)}{\frac{5}{8}-\frac{5}{48}}=\frac{23}{25}$$. But in the option it is given that x>y. So, it is the incorrect option.

B) If $$\frac{1}{5}$$, part of the mixture is replaced with water.

The initial part of the water is 3/8, and that of milk is 5/8. When 1/5 of the mixture is replaced with water, the milk taken out will be added as water. $$\frac{1}{5}\cdot\frac{5}{8}=\frac{1}{8}$$ is the amount of milk that is taken out, and that will also be the amount of water added.

So, the new ratio will be $$\frac{x}{y}=\frac{\left(\frac{3}{8}+\frac{1}{8}\right)}{\frac{5}{8}-\frac{1}{8}}=\frac{\left(\frac{4}{8}\right)}{\left(\frac{4}{8}\right)}=1:1$$. In the option it is given that x=y. So, it is a correct option.

C) If $$\frac{1}{7}$$, part of the mixture is replaced with water.

The initial part of the water is 3/8, and that of milk is 5/8. When 1/7 of the mixture is replaced with water, the milk taken out will be added as water. $$\frac{1}{7}\cdot\frac{5}{8}=\frac{5}{56}$$ is the amount of milk that is taken out, and that will also be the amount of water added.

So, the new ratio will be $$\frac{x}{y}=\frac{\left(\frac{3}{8}+\frac{5}{56}\right)}{\frac{5}{8}-\frac{5}{56}}=\frac{26}{30}$$. In the option it is given that xy. So, it is a correct option.

D) If $$\frac{1}{4}$$, part of the mixture is replaced with water.

The initial part of the water is 3/8, and that of milk is 5/8. When 1/4 of the mixture is replaced with water, the milk taken out will be added as water. $$\frac{1}{4}\cdot\frac{5}{8}=\frac{5}{32}$$ is the amount of milk that is taken out, and that will also be the amount of water added.
So, the new ratio will be $$\frac{x}{y}=\frac{\left(\frac{3}{8}+\frac{5}{32}\right)}{\frac{5}{8}-\frac{5}{32}}=\frac{17}{15}$$. In the option it is given that x>y. So, it is a correct option.

$$\log_{10}50 + \left(\frac{\log_{0.5}5}{1+\log_{2}5}\right)$$

$$\log_{0.5}5\ $$ can be written as $$\frac{\log_25}{\log_20.5}$$

It becomes -$$\log_25$$ as $$\log_20.5\ $$ is -1.

$$1+\log_25\ $$ will be $$\log_210\ $$ since 1 can be written as $$\log_22$$.

After these calculations $$ \left(\frac{\log_{0.5}5}{1+\log_{2}5}\right)$$ will trun into -$$\frac{\log_25}{\log_210}$$ which is -$$\log_{10}5$$

Now we have $$\log_{10}50 - \log_{10}5$$ which is nothing but $$\log_{10}10$$. So the answer is 1.

I) When a number has to be divisible by 52, basically it has to be divisible by 4 and 13.

Taking $$2^{10\ }$$common from the expression $$2^{10\ }*\left(2^{10}+3^{10}\right)$$. $$2^{10\ }$$ is divisible by 4 but not by 13.

So lets check if $$2^{10\ }+3^{10}$$ is divisible by 13.

Remainder of$$2^{10\ }$$when divided by 13.

$$2^{10\ }=2^5\cdot2^5$$

$$2^5\%13=32\%13=6$$

$$2^{10\ }=2^5\cdot2^5 = \left(13k+6\right)\cdot\left(13k+6\right)=13K+36=13K+10$$

So the remainder is 10.

Remainder of $$3^{10\ }$$when divided by 13.

$$3^{10\ }=3^5\cdot3^5$$

$$3^5\%13=243\%13=9$$

$$3^{10\ }=3^5\cdot3^5 = \left(13k+9\right)\cdot\left(13k+9\right) = 13K+81 = 13K+3$$

The remainder of $$2^{10 }+3^{10}$$when divided by 13 = 10+3 = 13, so it is divisible by 13.

II. $$7^{15} + 64^{5}$$ is divisible by 11.

Check for $$7^3$$ %11

343%11=2

$$7^{15}=\left(7^3\right)^5=\left(11k+2\right)^5=11K+32=11K+10$$

This gives the remainder 10.

Remainder of $$64^5$$ %11

This is same as$$\left(2^6\right)^5\%11=2^{30}\%11$$

$$2^{10}\%11=1024\%11=1$$

$$2^{30}\%11=\left(2^{10}\right)^3\%11=1^3\%11$$

This gives the remainder 1.

$$\left(7^{15}+64^5\right)\%11=\left(10+1\right)\%11=0$$

So, it is divisible by 11.

III.$$2^{20} - 49^{10}$$ is divisible by 9.

Re-write it as $$\left(2^{10}\right)^2-\left(7^{10}\right)^2$$

It is in the form of $$a^2-b^2$$=$$(a+b)\cdot(a-b)$$.

$$\left(2^{10}+7^{10}\right)\cdot\left(2^{10}-7^{10}\right)$$

Apply the same for $$\left(2^{10}-7^{10}\right)$$

$$\left(2^5+7^5\right)\cdot\left(2^5-7^5\right)\cdot\left(2^{10}+7^{10}\right)$$

Now, check each of the terms to see if they are divisible by 9. If atleast one is divisible, then the whole term is divisible by 9.

Check $$\left(2^5+7^5\right)$$

$$\left(32+343\cdot49\right)\%9$$

$$\left(9k_1+5+\left(9k_2+1\right)\cdot\left(9k_3+4\right)\right)\%9$$

where $$k_1,\ k_2,\ k_3\ are\ constant$$

$$\left(9k_1+5+\left(9k_4+4\right)\right)\%9$$

$$\left(9K+5+4\right)\%9$$

$$\left(9K+9\right)\%9$$

Since this term is divisible by 9, statement 3 is true.

IV. $$3^{15} - 8^{5}$$ is divisible by 5.

$$3^{15} - 8^{5}$$ = $$3^{15}-2^{15}$$

Rewriting it as $$\left(3^5\right)^3-\left(2^5\right)^3$$

This is in the form $$a^3-b^3=\left(a-b\right)\cdot\left(a^2+b^2+a\cdot b\right)$$

$$\left(3^5-2^5\right)\cdot\left(3^{10}+2^{10}+6^5\right)$$

$$\left(243-32\right)\cdot\left(243\cdot243+32\cdot32+6^5\right)$$

$$\left(211\right)\cdot\left(243\cdot243+32\cdot32+6^5\right)$$

When divided by 5, the remainder will depend on the unit digit. If it is below 5, that will be the remainder. If it is above 5, subtract the number from 5, the result will be the remainder.

$$\left(1\right)\cdot\left(3\cdot3+2\cdot2+1\right)=\left(1\right)\cdot\left(9+4+1\right)$$

Any power of six will have the unit digit as 6; hence, the remainder will be 1.

14%5=4.

So, the number will not be divisible by 5.

Hence option B is correct.

Convert the ratio into integers by multiplying it by the LCM of denominators; now, the ratio becomes 40:105:36.

Let the shares held by A, B, and C be 40x, 105x, and 36x, respectively.

A, B and C hold shares for one year. Other than that, A holds 108.75% more than he holds, which is 43.5x more from the 5th month.

Now multiply shares, and the time each one holds equates to 17208.

$$40x\cdot12+105x\cdot12+36x\cdot12+43.5x\cdot8=2520\cdot x$$

$$2520x=17208$$

$$x=\frac{17208}{2520}$$

B's share is $$105x*12$$, which gives $$105\times 12\times \frac{17208}{2520}= 8604$$

Thus, the correct answer is option A.

Let the cost price of an article be C. Given the profit percentage is 25%. The selling price of the article will be 1.25C. Given that the cost price of 25 articles is equal to the selling price of m articles, let's frame an equation from this.

$$25\cdot C\ =\ m\cdot1.25\ C=>m=20$$

This is a concept of weighted average; assuming that there are equal men and women, the average would be 1150. But since there are more men, the average is more than 1150. Also, the average can't go beyond 1300 and below 1000. So, option C is the

Sum to n terms in GP is $$a\cdot\frac{\left(r^n-1\right)}{r-1}$$

The sum of the first 18 terms is equal to 22 terms.

So $$a\cdot\frac{\left(r^{18}-1\right)}{r-1} = a\cdot\frac{\left(r^{24}-1\right)}{r-1}$$

This will be reduced to $$\left(r^{18}-1\right)=\ r^{22}-1$$

Reducing this we get $$r^{4\ }=1$$

Now r has 4 posibilities 1, -1, i,-i.

1 is impossible as that would lead to a zero denominator in the sum of terms formulae.

In all the other cases of r, the sum of the first four values would be zero.

$$\frac{x^\frac{1}{2} + x^{-\frac{1}{2}}}{1-x}+\frac{1-x^{-\frac{1}{2}}}{1-\sqrt{x}}$$

$$\frac{\left(\sqrt{\ x}+\frac{1}{\sqrt{\ x}}\right)}{1-x}+\frac{\left(1-\frac{1}{\sqrt{\ x}}\right)}{1-\sqrt{\ x}}$$

$$\frac{1}{\sqrt{\ x}}\cdot\frac{\left(x+1\right)}{1-x}+\left(\frac{1}{\sqrt{\ x}}\cdot\frac{\left(\sqrt{\ x}-1\right)}{1-\sqrt{\ x}}\right)$$

$$\left(\frac{1}{\sqrt{\ x}}\cdot\frac{x+1}{1-x}\right)-\frac{1}{\sqrt[\ ]{x}}$$

$$\frac{1}{\sqrt{\ x}}\cdot\left(\frac{\left(x+1\right)}{1-x}-1\right)$$

$$\frac{1}{\sqrt{\ x}}\cdot\frac{\left(x+1-1+x\right)}{1-x}$$

$$\frac{1}{\sqrt{\ x}}\cdot\frac{2x}{1-x}$$

$$2\ \frac{\sqrt{\ x}}{1-x}$$

Substitute 7 in place of x.

$$2\ \frac{\sqrt{\ 7}}{1-7}$$

That will be $$-\frac{\sqrt{\ 7}}{3}$$

Let us find out how many subsets of B are subsets of A. For this, we must find A and B's intersection, {6,12}. The number of subsets of this set is $$2^2=4$$. Now, find the total subsets of B.

$$2^n\ where\ n\ is\ no.\ of\ elements$$

$$2^4=16$$

The answer is to subtract the four subsets that are in common with A and B. 16 - 4 =12.

I. $$(4523)_{6} = (1059)_{10}$$

$$\left(4523\right)_6=4\cdot6^3+5\cdot6^2+2\cdot6^1+3\cdot6^0=1059$$

So, this is correct.
II. $$(4523)_{6} = (1069)_{10}$$

$$\left(4523\right)_6=4\cdot6^3+5\cdot6^2+2\cdot6^1+3\cdot6^0=1059$$

So, this is incorrect.
III. $$(0.203)_{5} = (0.424)_{10}$$

$$\left(0.203\right)_5$$ = $$2\cdot\frac{1}{5}+0\cdot\frac{1}{25}+3\cdot\frac{1}{125}=0.424$$

That will be $$\left(0.424\right)_{10}$$

So, this is correct.

IV. $$(0.203)_{5} = (0.406)_{10}$$

$$\left(0.203\right)_5$$ = $$2\cdot\frac{1}{5}+0\cdot\frac{1}{25}+3\cdot\frac{1}{125}=0.424$$

So, this is incorrect.

Here, we have to find the area of the  quadrilateral EQFS.

Lets draw a line from S to Q.

From the image we can say that the area of  SFQE = area of $$\triangle\ $$SFQ + area of $$\triangle\ $$SEQ

area of $$\triangle\ $$SFQ= $$\frac{1}{2}\cdot b\cdot h$$

We know that SR and RQ are perpendicular(given in question). So, height of $$\triangle\ $$SFQ is 11.

$$=\frac{1}{2}n\cdot11$$=$$\frac{11n}{2}$$.

area of $$\triangle\ $$SEQ = $$\frac{1}{2}\cdot b\cdot h$$

We know that SP and PQ are perpendicular(given in question). So, height of $$\triangle\ $$SEQ is 7.

$$=\frac{1}{2}2n\cdot7$$=$$7n$$.

$$\frac{11n}{2}+7n=225\ \ \Rightarrow\ 25n=225*2\ \Rightarrow n=18$$

Formulae for the compound interest is $$p\cdot\left(1+\frac{r}{100}\right)^n$$

Here n=3, p=5500. The amount he receives after three is equal to $$5500\cdot\left(1+\frac{10}{100}\right)^3$$

It is given that he receives an interest of 33.1% after three years. So, the total sum would be 133.1% of the initial investment.

$$5500\cdot\left(1+\frac{10}{100}\right)^3=5500\cdot\frac{133.1}{100}$$

$$\left(1+\frac{r}{100}\right)^3=1.331$$

$$\left(1+\frac{r}{100}\right)^3=\left(1.1\right)^3$$

From this, r=10.

As per the question, x and y sum up to 20. Assume that there are 72 units of work. Kamla does three units of work a day, and Nirmala does four. We got another equation as well: 3x+4y=72. Solving both gives us x=8, y=12. 

Let's verify each option:

$$\frac{4x}{y}+1\ is\ even\ $$

$$\frac{4x}{y}+1\ $$ = $$\frac{4*8}{12}+1\ $$ $$=\frac{8}{3}+1$$

2X + Y = 22 $$\Rightarrow 2*8+12=28$$

X - Y is an odd integer. 8-12=-4 it is even.

$$x^{2} + 1$$ is a multiple of 13

$$8^{2} + 1=65$$ is a multiple of 13.

So, only option D is correct.

$$\log_42$$ is 1/2. So the denominator will be $$x-12$$. In numrator if x goes below 10 or above 15 the value under root goes negative. X can't be 12 as the denominator goes to 0. So the intervel is $$\left[10,12\right)\ \ U\ \left(12,15\right]$$

The formulae for the sum of first n terms in AP is $$\frac{n}{2}\cdot\left(2\cdot a+\left(n-1\right)\cdot d\right)$$.

$$\frac{12504}{25}=\frac{24}{2}\cdot\left(2\cdot a+23\cdot d\right)$$

$$\frac{1042}{25}=\left(2\cdot a+23\cdot d\right)$$

Lets generate other equation, given the sum of next 24 terms is $$\frac{17112}{25}$$.

So the first 48 terms is $$\frac{29616}{25}=\frac{48}{2}\cdot\left(2a+\left(47\cdot d\right)\right)$$

Solving both we get a=429/25 d=8/25

Find out the LCM of the ropes' length. The LCM of 450, 990,1620 is 89100 cm. So, the side length can be any multiple of 891m, given that the area is less than 10,00,000. So, it has to be 891.

Given the ratio and two points on the line, we can find out the point that divides the line in that ratio using internal section formulae $$\left(\frac{\left(mx_1+nx_2\right)}{m+n},\frac{\left(my_1+ny_2\right)}{m+n}\right)$$.

The divider of line MN(point A) is as follows.

$$\left(\frac{\left(3\cdot1+2\cdot-1\right)}{3+2},\frac{\left(3\cdot1+2\cdot3\right)}{3+2}\right)$$

A$$\left(\frac{1}{5},\frac{9}{5}\right)$$

Similarly, for line ST (point B)

$$\left(\frac{\left(3\cdot2+2\cdot0\right)}{3+2},\frac{\left(3\cdot7+2\cdot-4\right)}{3+2}\right)$$

B$$\left(\frac{6}{5},\frac{13}{5}\right)$$

Also, do the same for the other line; we get points A and B. Now, find the line equation using the formula y=mx+c.

m=$$\frac{y_2-y_1}{x_2-x_1}\ =\ \frac{\left(\frac{13}{5}-\frac{9}{5}\right)}{\frac{6}{5}-\frac{1}{5}}\ =\ \frac{4}{5}$$

The euation becoms $$y=\frac{4}{5}\cdot x+c$$

Substotue one of those two points to get c value.$$\frac{13}{5}=\frac{4}{5}\cdot\frac{6}{5}+c=>c=\frac{41}{25}$$

The equation will be $$y=\frac{4}{5}\cdot x+\frac{41}{25}=>20x-25y+41=0$$

Rama and Jaya differ by a percentage of 10 and marks of 8. Jaya's share is 45%, and 55% is Rama's. Each percent corresponds to 4/5 marks. From this, rams's mark is 44, and that of Jaya is 36.

The coefficient of the x term in a quadratic equation represents the sum of roots, while the constant term indicates the products of roots. Directly for the P(x) $$\alpha,\ \beta\ $$ are the roots. In q(x), the x coefficient can be written as $$\alpha\ +1+\beta\ +1$$. By this, we say $$\alpha\ +1,\beta\ \ +1$$ are the roots of q(x).

Now, check with the options where the root of p(x) is one less than q(x). It is option C.

The distance between the two opposite sides is 18 which is just the length of shorter diagonal. That is $$ \sqrt{\ 3}a$$. Hence a is $$6\sqrt{\ 3}$$.

The growth percentage in 11 years is $$(224500-150000)/150000$$. That is 49.6666. Divide it by 11 to get an average percentage. That is 4.52

This is an AP with a common difference of 0.5.Use the formulae of $$n^{th}\ term\ =\ a+\left(n-1\right)\cdot d$$.

$$-0.25+\left(n-1\right)\cdot0.5\ =\ 17.25$$

n will 36.

Slit the given numbers as follows.

321 - 107*3

48 - $$2^4$$*3

66 - 11*2*3

To have a perfect square, we need to have even powers to the prime numbers. That is $$2^4\cdot107^2\cdot11^2\cdot3^2$$. That will be $$16 \times 1089 \times 11449 $$.

There are four different equations possible from this given equation. They are 

x+y=2

x-y=2

-x+y=2

-x-y=2

These lines forms a rhombus having diogonal length 4. The area of rhombus is $$\frac{1}{2}\cdot d_1\cdot d_2$$.

$$\frac{1}{2}\cdot4\cdot4\ =\ 8$$

Lets expand the given equation.

$$91\left(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}\right)$$

$$91\left(1+\frac{x+y}{xy}+\frac{1}{xy}\right)$$

x+y is constant, 91 is constant.

To get the minimum value we have to minimixe the xy term.

That happens when both are equal.

So x=26, y=26.

$$91\left(\left(1+\frac{1}{26}\right)\left(1+\frac{1}{26}\right)\right)$$

On simiplifing we get $$\frac{5103}{52}$$

Here we need to determine to which quadrant each of the g1, g2, g3, g4 goes to. For that we need to find the range and domain of these functions. The domain of g1 is same as that of f as the value is directly inserted in to the function. Since the value of g1(x) is 2 less than the f(x) the range would be(-1,0). So g1 goes to 3rd quadrant. Similarly for g2 the domain should be negative of what f has as the input for f is 

Check with each of the options,

$$7^{141} + 1$$ can be written as $$\left(7^{47}\right)^3+1^3$$

$$a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)$$

So, $$7^{141} + 1$$ is $$\left(7^{47}+1\right)\left(7^{94}+1-7^{47}\right)$$

Hence, option B is the correct answer.

Assume that there are x horses and y sheep. They both add up to 100%. If 50% of horses were sheep, there would have been 50% more sheep than the number of horses. So, considering horses count as h/2 and sheep count as h/2+s, we can frame the equation (h/2+h/4) = s+h/2. Solving this, we get s=(h/4). Since both add to 100%, we get horses as 80%.

Factorize 1080 - $$2^3\cdot3^3\cdot5$$. Here, HCF has to be between 3 and 12; the only possible number would be 6. The remaining thing that exists is only 5. So there are two ways: one, 5, belong to x, and two, five, belong to y.

Given a chord of length r, and the lines joining the centre and the ends of the chord are also radii, these lines form an equilateral triangle.

The area of the minor part will be the area of sector OAB minus the $$\triangle\ $$le OAB will be the required answer.

The area of sector = $$\frac{\theta}{360^o}\cdot\pi\ r^2$$

Where $$\theta\ $$=$$60^o$$

Area of sector = $$\frac{60^o}{360^o}\cdot\pi\ \cdot42^2$$ = $$\frac{\pi}{6}\ \cdot42^2$$

Area of triangle OAB = $$\frac{1}{2}\cdot b\cdot h\ =\ \frac{1}{2}\cdot42\cdot\frac{42\sqrt{\ 3}}{2}\ =\ 42^2\cdot\frac{\sqrt{\ 3}}{4}$$

Area of shaded region = Area of sector OAB - area of $$\triangle\ $$le OAB = $$\frac{\pi}{6}\ \cdot42^2 - \ 42^2\cdot\frac{\sqrt{\ 3}}{4}$$ = $$42^{2} (\frac{\pi}{6} - \frac{\sqrt{3}}{4})$$

The domain of the log is (0, $$\infty\ $$). So the x could go from (-1,$$\infty\ $$). In general, the value of a log goes negative when the argument of a log is between 0 and 1 when the base is 10. In this question, the argument stays in that range when x is between -1 and 0. Hence, the graph will pass through the third quadrant when x is in that range. And when the x is positive, the log value will be positive. Hence, the graph passes through the first quadrant.

In this, there are two kinds of separations. Men and women, Indian and Non-Indian. It is given that 2/3 of all are Indians. From this, we deduce that 1/3 are Non-Indians. Of these 1/3 Non-Indians, 3/8 are women. Now we can derive the Non-Indian man, it is (1/3)*(1-(3/8))=5/24. 5/24 of all are Non-Indian men. It is given that 1/3 of all are women, so 2/3 are men. Of 2/3 men, 5/24 are Non-Indians, so Indian men are 11/24. The probability of a picked man being Non-Indian is (5/24)/(2/3). That is 5/16.

The longest chord means diameter. The given lines make a right-angled triangle. The circumcircle will touch all three vertices of the triangle. Since there is a right angle at a vertex, the side opposite to that will be the diameter. The line is 4x + 3y = 24. This line touches the x-axis at(0, 8) and the y-axis at (6,0). The distance between the points is the diameter.

The distance between the two points is $$\sqrt{\ 8^2+6^2}=10$$

I) $$ab(a + b) + bc(b + c) + ca(c +a) \leq 6abc $$

$$\frac{ab(a+b)}{abc}+\frac{bc(b+c)}{abc}+\frac{ca(c+a)}{abc}\le6$$

$$\frac{(a+b)}{c}+\frac{(b+c)}{a}+\frac{(c+a)}{b}\leq6$$

We shall rearrange the terms

$$\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{a}\le6$$

Here, we have six terms: three fractions and their reciprocals.

$$t+\frac{1}{t}\ge2$$ this is always true.

So three such equations add to $$\ge6$$

So, it is only sometimes valid.

II. $$\frac{a^{2} + b^{2} + c^{2}}{abc} \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$$

$$\frac{a^2+b^2+c^2}{abc}\le\frac{a\cdot b+b\cdot c+c\cdot a}{abc}$$

$$a^2+b^2+c^2\le a\cdot b+b\cdot c+c\cdot a$$

Consider (a,b,c) as (1,2,3)

Substituting them in the above equation,

$$1^2+2^2+3^2\le1\cdot2+2\cdot3+3\cdot1$$

$$14\le11$$ is not true, so this is also not true.

Hence, option D is correct.

Let the speed of car A be A and car B be B. Given that the speed of A is 60kmph. Let us assume that they both meet at Y after a time of t. A travels a distance of 400 in 6hr 40m. This distance is covered by B in t time.

400= $$B\cdot t$$. --->1

Again, the distance travelled by A in t time will be covered by B in 3hr 45m, which can be written as$$\ A\cdot t=B\cdot\ \frac{15}{4}$$ -->2

Substitute B =$$400/t$$ from eq 1. We know that A is 60.

$$60\cdot t=\left(\frac{400}{t}\right)\cdot\frac{15}{4}$$

$$t^2=25$$

t=5

The question is asked for atleast two, but the answer is provided for three.

First, arrange two out of four girls in the last two positions in 4*3 ways. Now consider three boys as a unit and two unarranged girls as two units. Arranging these three units can be done in 3! ways. Again, those three boys can be arranged in 3! ways.

So, the answer is 432.

From statement 1, we know that there is at least some overlap between sheds and caves. From statement 2, we know that all sheds are nests, so the overlap between sheds and caves must also be within the category of nests. Statement 4 confirms that there is an overlap between sheds and dens, but since all sheds are nests, this also implies an overlap between nests and dens.

Conclusion I: Some caves are nests. This conclusion logically follows from the given statements. Since some sheds are caves and all sheds are nests, it's logical to conclude that some caves are nests. So, Conclusion I is true.

Conclusion II: No shed is stable. This conclusion does not logically follow from the given statements. Statement 3 states that only a few nests are stables, but it doesn't exclude the possibility that some sheds might also be stables. Therefore, we cannot conclude that no shed is stable based on the given statements. So, Conclusion II is false.

Therefore, Conclusion I logically follows from the given statements, but Conclusion II does not.

The correct option is A

Here, "SUN RISES IN THE EAST" can be decoded using the rule mentioned below:

First Reverse the order of the word.  "SUN RISES IN THE EAST" => "EAST THE IN RISES SUN"

Now count the number of letters exist in each of the word, and add plus 2 in that. 

Hence, The number corresponds to EAST is (4+2) = 6, and the number of  Corresponds to 'THE' is (3+2) = 5, and so on.

Now, in every word take the last letter of it. Hence 'EAST' can be represented as 6T, and 'THE' can be represented as 5E, and so on.

By this logic, "SUN RISES IN THE EAST" can be decoded as 6T 5E 4N 7S 5N

By the same logic, 'RAM GOES TO SCHOOL' can be written as 'SCHOOL TO GOES RAM' => (6+2)L (2+2)O (4+2)S (3+2)M => 8L 4O 6S 5M

The correct option is A

The number of unmarried literate people are highilted in the below image.

The count is 2003+367=2370

So, statement 1 is not true.

The number of not literate working persons are highlighted in the below picture.

The count is 154+1003+356=1513.

So, statement 2 is correct.

Each term's first letter changes at a difference of 2, 4, and 6.

The second letter is changing at a difference of 1, 2, and 3.

The third letter is changing at a difference of 2, 4, and 5. But it has to be 2, 4, 6.

So, the last term is not in the correct order.

278926534292897242592976479273

The bold 2's are the only 2's in the series that are preceded by 9 and followed by an even number.

There are 12 divisions in a clock constituting 360 degree angle, which means that the movement by the hand of the clock in going from one division to the next is $$\frac{360^{\circ\ }}{12}=30^{\circ\ }$$

Our clock hand is going from 9 to 7; the hand would travel from 9 to 12, 3 divisions and then from 12 to 7, 7 more divisions; i.e., in total, the then would travel 10 divisions on the clock, which would make the distance travelled in degrees to be $$30^{\circ\ }\times\ 10=300^{\circ\ }$$

Therefore, Option B is the correct answer. 

Option A:

N is father of T, T is sister of W, W is brother of X, X is daughter of M.

In this case N is husband of M.

Option B:

N is father of T, T is sister of W, W is daughter of X, X is daughter of M.

In this case M is mother in law of N.

Option C:

N is father of T, T is sister of W, W is brother of X, X is sister of M.

In this case N is father of M.

 Option D:

N is son of T, T is sister of W, W is brother of X, X is daughter of M. 

In this case N is grand son of M.

So, only option A is correct.  

We are given that the ranking system is from top to bottom, meaning the lower values are better. 

We are given that Kunal (rank 17) is ahead of Pinki by 6 ranks, which would mean that Pinki's rank is 6 more than Kunal, $$17+6=23$$

we are then given that Pinki (rank 23) is ahead of Balwan by 7 ranks, which would mean that Blawan's rank is 7 more than Pinki, 23+7 = 30

Therefore, Option D is the correct answer. 

The general pattern being followed here is that the nth term is defined as $$n^2+1$$
$$T_1=1+1$$
$$T_2=2^2+1$$
$$T_3=3^2+1$$
$$T_4=4^2+1$$
$$T_5=5^2+1$$

Therefore, the question mark would be best replaced by $$4^2+1=17$$

Hence, Option C is the correct answer. 

256435295653 16495624215549672 14755496425

The 5's that are preceded by odd number and not succeeded by odd number are highlighted in bold. There are 5 such 5's.

Looking at the relationship between the first two terms, we see that F is the sixth alphabet from the beginning, whereas U is the sixth alphabet from the end. 

Similarly, K is the 11th alphabet from the start, and P is the 11 alphabet from the end. 

Looking at the two terms in between, H is the 8th alphabet from the start, and S is the 8th alphabet from the end. 
Finally, J is the 10th alphabet from the start, and Q is the 10th alphabet from the end. 
And the terms in the centre are switched with each other. 

In order to find the same pattern for the third term, we would begin with the first alphabet from the end: Z

The term would end with the 12th alphabet from the end (since L is the twelfth alphabet from the start), which would be O)
The second character in the term would be opposite of the third character in the third term which would be D, therefore, we want the fourth term from the end, which would be W. 
Finally, for the third character, we would want the opposite of the second character of term three, which would be C. Therefore, we would want the third character from the end, which would be X

Therefore, the correct term would be ZWXO

Hence, Option D is the correct answer. 

99 can be factorized as 9*11

120 can be factorized as 10*12 = (9+1)*(11+1)

143 can be written as 13*11

So, 14*12=168

 H O L Ω Y & 4 Z 2 * 3 M & 7 S W # 8 2 p H * L is the given sequence.

After removing all the numbers from sequence, the resulting sequence will be  H O L Ω Y & Z * M & S W # p H * L

Seventh from the right will be S.

I. People engaged in tertiary sector from urban area is more than non-urban area.

103+67>76+84

170>160 True.

So, statement 1 is true.

II. People engaged in primary sector from urban area is more than non-urban area.

103+106+93>83+96+84

302>263 true

This statement is true.

The given sequence is PRG, VLJ, BFM.

PRG, VLJ, BFM

First letter pattern P, V, B

P+6 = V

V+6 = B

B+6 = H

Second letter pattern R, L, F

R-6 = L

L-6 = F

F-6 = Z

Third lette G, J, M

G+3 = J

J+3 = M

M+3 = P

So, the next term in the sequence is HZP.

JXG : ZNW

J+16 = Z

X+16 = N

G+16 = W

Similarly,

TRC

T+16 = J

R+16 = H

C+16 = S

JHS is the answer.

Number of odd days till 2000 is 0.

Number of odd days from 2001 to 2009 is 10( 2 leap years +6 normal years).

Number of odd days in 2010 till sept 10 is (31+28+31+30+31+30+31+31+30 = 274) 274

Total odd days is 274+10=284%7=4.

So, the day is thursday.

Option A: GIHJ : FLGK

F = G-1

L = I+3

G = H-1

K = J+1

The pattern here is (-1,+3,-1,+1)

Option B: ZYAX : YZZY

Y = Z-1

Z = Y+1

Z = A-1

Y =X+1

The pattern here is (-1,+1,-1,+1)

Option C: EHFG : DIEH

D = E-1

I = H+1

E = F-1

H =G+1

The pattern here is (-1,+1,-1,+1)

Option D: PVQU : OWPV

O = P-1

W = V+1

P = Q-1

V =U+1

The pattern here is (-1,+1,-1,+1)

Option 1 is different form other options.

The pattern in this sequence is as follows:

if w,x,y,z is the sequence.

x=w*2+3

y=x*2+4

z=y*2+5

Let's look at the original sequence.

11 = 4*2+3

26 = 11*2+4

? = 26*2+5

120 = ?*2+6

247 = 120*2+7

$$\Rightarrow$$ ? = 26*2+5=57

India has its own culture, but due to introduction of multiple cultures India lost its own culture. So, two is the cause, one is the effect.

Somu's daughter in law's father's only daughter is Somu's daughter in law.
The man is the son of Somu's daughter in law which means that he is Somu's grand son.
So, Somu is the man's Paternal Grandfather i.e. Option A.

Option A)SHAR : UJCT

S+2 = U

H+2 = J

A+2 = C

R+2 = T

Option B)BEYO : DGBQ

B+2 = D

E+2 = G

Y+2 = A (But it is B)

O+2 = Q

Option C)MIGH : OKIJ

M+2 = O

I+2 = K

G+2 = I

H+2 = J

Option D)PHOB : RJQD

P+2 = R

H+2 = J

O+2 = Q

B+2 = D

Option B is different from the others.

4 = 4*1

12=4*3

? = 4*x

60=4*15

124=4*31

252=4*63

Here, every number is a multiple of 4. The multiple in the second term is one more than the twice of the previous multiple.

3 = 1*2+1

31 = 15*2+1

63 = 31*2+1

x=3*2+1

x=7

The formed right-angled triangle is having base 18(10+8) and height 7(12-5).

The distance is $$\sqrt{\ 18^2+7^2}=\sqrt{\ 373}$$

80 can be written 8*10

99 can be written as 9*11

Option A)25 : 122

25=1*25, 5*5

The possible value is 2*26=52, 6*6=36

This option is not satisfying the relation.

Option B)79 : 90

79 = 1*79

The possible value is 2*80=160.

This option is not satisfying the relation.

Option C)110 : 169

110 = 1*110, 2*55, 5*22, 10*11

The possible values are 2*111=222, 3*56=168, 6*23=138, 11*12=132.

This option is not satisfying the relation.

Option D)120 : 143

120 = 1*120, 2*60, 3*40, 4*30, 5*24, 6*20, 8*15, 10*12

 The possible values are 2*121=242, 3*61=183, 4*31=124, 6*25=150, 9*16=144, 11*13=143.

This option is satisfying the relation.

The pattern in this sequence is as follows:

Of the three letters the first is the third next letter in the alphabetical sequence of the first letter in the previous term.

The second letter is the second next letter in the alphabetical sequence of the second letter in the previous term.

Similarly the third letter is the 5th next letter in the alphabetical sequence of the third letter in the previous term.

Let's find the second term from the first term.

JWM: _ _ _

Third letter after J is M

Second letter after W is Y.

Fifth letter after M is R.

So, the second term is MYR.

In the same way, the next term of SCB is 

Third letter after S is V

Second letter after C is E.

Fifth letter after B is G.

VEG is the correct answer.

We know that Urvi got less than Hina and Tina. We also know that Kiran got greater than or equal to Hina and Tina. So, Kiran marks are more than Urvi even if Kiran marks are equal to the Hina's marks.

So, Statement 1 is definitely true.

We know that Ira's marks are greater than or equal to Urvi's marks. So, we don't know by how much if Ira's marks are greater than to the Urvi's marks. We know that Tina marks are more than Urvi's. So, we cannot compare the marks of Tina with Urvi's marks.

So, statement 2 cannot be determined.

* 5 U £ $$\pm$$ 2 W @ 3 K 9 $$\div$$ # 6 M & 4 A $$\alpha$$ 5 G $$\delta$$ 7 B % R $$\varnothing$$

In this sequence 3 K 9 $$\div$$ # 6 M & 4 A $$\alpha$$ is the middle 9 terms.

If we reverse this we get $$\alpha$$ A 4 & M 6 #  $$\div$$ 9 K 3

* 5 U £ $$\pm$$ 2 W @ $$\alpha$$ A 4 & M 6 # $$\div$$ 9 K 3 5 G $$\delta$$ 7 B % R $$\varnothing$$

The twenty-first element from the right end will be W. Then seventh to the right of that W will be 6.

08 : 62

Here the pattern is two subtracted from number square.

$$8^2-2=62$$

Similarly,

$$15^2-2=223$$.