### Contents of State Bank Probationary Officers Preliminary Examination

This book contains chapter-wise solutions, topic-wise solutions, exercise-wise solutions, and videos.

#### 46 Chapters

- 1. SBI PO Prelims Exam
- 2. Model Paper
- 3. Common Errors
- 4. Antonyms
- 5. Synonyms
- 6. Sentence Completion
- 7. One Word Substitution
- 8. Comprehension
- 9. Passage Completion
- 10. Completion of Paragraphs & Sentences
- 11. Percentage
- 12. Profit and Loss
- 13. Ratio and Proportion
- 14. Partnership
- 15. Average
- 16. Simple Interest
- 17. Compound Interest
- 18. Discount, Stock and Shares
- 19. Work and Time
- 20. Speed, Time and Distance
- 21. Unitary Method
- 22. Problems Based on Ages
- 23. Problems Based on Fraction
- 24. Problems Based on Numbers
- 25. Alligation
- 26. Area
- 27. Interpretation of Data
- 28. Commonsense Reasoning
- 29. Coding and Decoding
- 30. Statement and Argument
- 31. Analytical Reasoning
- 32. Statement and Course of Action
- 33. Statement and Conclusion
- 34. Statement and Assumption
- 35. Direction Sense Test
- 36. Sitting-Arrangement
- 37. Data Sufficiency
- 38. Logical Diagrams
- 39. Ranking Test
- 40. Numerical Puzzles
- 41. Syllogism
- 42. Fictitious Symbols
- 43. Input-Output
- 44. Cause and Effect
- 45. Passage and Inferences
- 46. Word Relationship (Analogy and Classification)

#### 3 Topics

Solved Paper (2018)

Solved Paper (2017)

Solved Paper (2016)

#### 4 Topics

Set-1

Set-2

Set-3

Set-4

#### 5 Topics

Articles, Nouns, Pronouns

Adjective, Adverb, Adverbial Order

Verb, Infinitive, Verbal Noun, Gerund, Participle

Conjunctions, Prepositions

Miscellaneous Sentences

#### 1 Topics

Sentence Completion

#### 1 Topics

One Word Substitution

#### 1 Topics

Passage Completion

#### 1 Topics

Completion of Paragraphs & Sentences

#### 2 Topics

Proportional Division

Compound Proportion

#### 1 Topics

Compound Interest

#### 3 Topics

Discount

Shares

Stock

#### 5 Topics

Main Formulae

Relative Speed

Average Speed

Some Instruction in Concern with Train

Boats and Stream

#### 1 Topics

Problems Based on Ages

#### 1 Topics

Problems Based on Fraction

#### 1 Topics

Problems Based on Numbers

#### 1 Topics

Interpretation of Data

#### 2 Topics

Alphabetical Order of the Words in Dictionary

Series of Letters on the Basis of Some Rhythm

#### 2 Topics

Types of Coding/Decoding

Coding by Shifting Letters

#### 1 Topics

Common Questions Based on Arguments

#### 1 Topics

Analytical Reasoning

#### 1 Topics

Statement and Course of Action

#### 1 Topics

Statement and Conclusion

#### 1 Topics

Statement and Assumption

#### 1 Topics

Direction Sense Test

#### 1 Topics

Sitting-Arrangement

#### 1 Topics

Data Sufficiency

#### 1 Topics

Numerical Puzzles

#### 2 Topics

Fictitious Symbols and Numbers

Fictitious Symbols and Variables

#### 1 Topics

Cause and Effect

#### 1 Topics

Passage and Inferences

#### 2 Topics

Analogy for Meaningful Words

Classification

## Experience Tests Tailored to Match the Real Exam

## Practise Questions with Solutions from the State Bank Probationary Officers Preliminary Examination

- (a)are very different than
- (b)is so much different than
- (c)are very different from
- (d)is very different from
- (e)No correction required

**Hint**

'Than' should be used after a comparative adjective.

**Solution:**

The word ‘than’ is used to compare things. When ‘than’ is used in a sentence, it should be used after a comparative adjective. But, ‘different’ is not a comparative adjective. ‘Than’ can be used after ‘different’ only when ‘different’ is followed by a noun. Here, ‘than’ should be replaced with ‘from’.

The correct form of the sentence is, “Nobody can deny the fact that Indian economy is very different from American economy”.

- (a)Statistic with regards to
- (b)Statistics with regard to
- (c)Statistic with regard to
- (d)Statistics in regards to
- (e)
No correction required

**Hint**

Statistics is plural.

**Solution:**

‘With regard’ is the accurate phrase that has been followed throughout the history. Whenever the preposition ‘with regard’ is used, it indicates the reference to a particular subject/topic. ‘With regards’ is considered as a grammatically incorrect phrase.

The correct form of the sentence is, “Accurate statistics with regard to the area occupied in different forms of cultivation are different to obtain."

- (a)Seldom or never
- (b)Seldom if never
- (c)Seldom or ever
- (d)Seldom has ever
- (e)No correction required

**Hint**

The context of the sentence suggests no possibility for that event to happen.

**Solution:**

The use of ‘seldom if ever’ gives possibility for an event to happen, whereas using ‘seldom or never’ provides no possibility for that event to happen. Therefore, ‘seldom or never’ is the correct usage in the given sentence.

The correct form of the given sentence is, "Seldom or never

was there any training or instructions in such tactics for either the tank crews or the infantry formations."

- (a) then I smelled
- (b)that I smelled
- (c)I smelled
- (d)I smell
- (e)No correction required

**Hint**

Unwanted usage of preposition should be eliminated

**Solution:**

The preposition ‘than’ is used for comparing one thing with another, which is not needed in the given statement. The action of smelling has already been done, therefore the past participle of smell (smelled) should be used.

The correct form of the sentence is, “As soon as I opened the front door of my house,

I smelled the distinctive aroma of fresh coffee."

- (a)he was able
- (b)and he was able
- (c)else he was able
- (d)or he was able
- (e)No correction required

**Hint**

Avoid the unwanted usage of conjunction in the sentence.

**Solution:**

There is no need of usage of any conjunction in the given sentence. 'Although' and 'though' is followed by ‘yet’ and not by ‘but’. 'But', 'else', 'or', etc., are not required.

The correct form of the given sentence is, “Although he had fewer supporters among the governing class, he was able to get the popular vote”.

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- (a)$45$
- (b)$90$
- (c)$180$
- (d)$120$
- (e)None of these

**Hint**

First find the $35\%$ of $740$ and subtract $34$ to find the unknown number and then calculate the two-fifth of the obtained number in order to find the result.

**Solution:**

Given:

Thirty-five percent of $740$ is $34$ more than a number.

Let $x$ be the given number.

Thirty-five percent of$740$is $34$ more than a number can be written as

$35\%\times 740=34+x\phantom{\rule{0ex}{0ex}}\frac{35}{100}\times 740=34+x\phantom{\rule{0ex}{0ex}}3.5\times 74=34+x\phantom{\rule{0ex}{0ex}}259=34+x\phantom{\rule{0ex}{0ex}}x=259-34\phantom{\rule{0ex}{0ex}}x=225$

Now, the number is $225$.

Therefore, the two-fifth of the obtained number is

$\frac{2}{5}\times 225=2\times 45\phantom{\rule{0ex}{0ex}}=90$

Thus, the two-fifth value of the obtained number is $90$.

- (a)$1234km$
- (b)$1144km$
- (c)$1134km$
- (d)$1244km$
- (e)None of these

**Hint**

First, find the speed of the train to calculate the average speed of the given car and then calculate the distance covered by the car using the speed obtained.

**Solution:**

Given:

A train is covering $1235km$ in $19hours$.

By using the formula, $Speed=\frac{\text{Distance travelled}}{\text{Time Taken}}$

Then, the speed of the train $=\frac{1235}{19}=65\mathrm{km}/\mathrm{hr}$

Average speed of a car $=\frac{4}{5}\times 65=52\mathrm{km}/\mathrm{hr}$

By using the formula, $Distance=Speed\times Time$

Then, the car will cover the distance in $22hours$ will be

$=52km/hr\times 22hr=1144km$

Thus, the required distance $=1144km$

- (a)$\u20b910,530$
- (b)$\u20b99,950$
- (c)$\u20b910,350$
- (d)$\u20b911,340$
- (e)None of these

**Hint**

First find the Cost price using the Selling Price and the Loss $\%$ given in the question and then find the Selling Price of the new Profit $\%$ using the obtained Cost Price and the new given Profit $\%$.

**Solution:**

Given that, the selling price is $\u20b96,750$ incurred a loss of $25\%$

$\mathrm{Cost}\mathrm{Price}=\frac{\mathrm{Selling}\mathrm{Price}\times 100}{100-\mathrm{Loss}\%}$On plugging in the given values we get,

$\text{Cost price}=\frac{6750\times 100}{100-25}\phantom{\rule{0ex}{0ex}}\mathrm{CP}=\frac{675000}{75}\phantom{\rule{0ex}{0ex}}\mathrm{CP}=\u20b99000$

$\text{Sellingprice}=\frac{\text{CP}\times (100+\text{Profit\%})}{100}$

We get,

$\text{Selling price}=\frac{9000\times (100+15)}{100}\phantom{\rule{0ex}{0ex}}\mathrm{SP}=\frac{9000\times \left(115\right)}{100}\phantom{\rule{0ex}{0ex}}\mathrm{SP}=90\times 115\phantom{\rule{0ex}{0ex}}\mathrm{SP}=10350$

Therefore, the required selling price of the item is $\u20b910,350$

- (a)$3$
- (b)$6$
- (c)$7$
- (d)$4$
- (e)None of the these

**Hint**

Make the bases same and then equate the power terms to find the question mark in power.

**Solution:**

Given,

$(0.7{)}^{2}\xf7(0.343)=(0.7{)}^{?}\xf7(0.49{)}^{3}$

We can write the above equation like this,

$\frac{(0.7{)}^{2}}{(0.7{)}^{3}}=\frac{(0.7{)}^{?}}{{\left[\right(0.7{)}^{2}]}^{3}}$

$\frac{(0.7{)}^{2}}{(0.7{)}^{3}}=\frac{(0.7{)}^{?}}{(0.7{)}^{6}}$

According to the mathematical protocol, the above step is written like this,

$(0.7{)}^{2-3}=(0.7{)}^{?-6}$

Now to find the question mark,

$?-6=2-3$

We can take RHS number to LHS, so that the negative sign can be changed as positive,

$?=6-1=5$

Thus, the answer is option E.

- (a)$616sqcm$
- (b)$540sqcm$
- (c)$396sqcm$
- (d)Cannot be determined
- (e)None of these

**Hint**

First find the length and the perimeter of the rectangle using the given conditions and the appropriate formulas. Then, find the circumference of the circle using the given conditions and then find the area of the circle to obtain the result.

**Solution:**

We know that, $\mathrm{Area}\mathrm{of}\mathrm{rectangle}=\mathrm{Length}\times \mathrm{Breadth}$

Given that, the area of the rectangle is $112sqcm$ and breadth is $8cm$. On substituting the values in the above formula we get,

$112=\text{length}\times 8\phantom{\rule{0ex}{0ex}}\text{length}=\frac{112}{8}=14cm$Therefore, the length is $14cm$.

Perimeter of the rectangle, $\mathrm{p}=2\times (\mathrm{l}+\mathrm{b})$

On substituting the values we get,

$\mathrm{p}=2\times (14+8)=2\times 22=44$

Therefore, the perimeter is $44cm$.

Given that, the sum of the circumference of the circle and the perimeter of the rectangle is $132cm$.

Let us consider the circumference of the circle as $x$ and we know that the perimeter is $44cm$

$\therefore x+44=132\phantom{\rule{0ex}{0ex}}x=132-44=88$

The circumference is $88cm$ that is,

$2\mathrm{\pi r}=88\mathrm{cm}\phantom{\rule{0ex}{0ex}}2\times \frac{22}{7}\times \mathrm{r}=88\phantom{\rule{0ex}{0ex}}\mathrm{r}=\frac{88\times 7}{2\times 22}=14$

Therefore, the radius is $14cm$.

Hence, the area of the circle is,

$\mathrm{Area}={\mathrm{\pi r}}^{2}=\frac{22}{7}\times 14\times 14$On calculating we get,

$\mathrm{Area}=22\times 2\times 14=44\times 14\phantom{\rule{0ex}{0ex}}\mathrm{Area}=616{\mathrm{cm}}^{2}$

The required area of the circle is $616sqcm$.

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- (a)The establishment of the Third Reich
- (b)influenced events in American history
- (c)by starting a chain of events
- (d)who culminated in war between Germany and the U.S.A.
- (e)No error

**Hint**

The usage of relative pronoun is incorrect in the given sentence.

**Solution:**

The error lies in the fourth part of the sentence. We have to change "who" to "which" to make the sentence meaningful. To refer to inanimate objects or abstract things, we use the relative pronoun 'which'. To refer to human beings, we use the relative pronoun 'who'. Here in the given sentence, we are talking about the chain events, so the proper relative pronoun is 'which'.

- (a)
A weather

- (b)
map is

- (c)
an important tool

- (d)for geographers
- (e)No error

**Hint**

The sentence is a general statement of fact or opinion.

**Solution:**

The statement describes a general statement of facts. In such cases, we need to use the simple present/ present indefinite tense. The present indefinite tense is maintained throughout the sentence. The verb 'is' is also correctly used.

So, the sentence is grammatically correct, and no corrections required.

- (a)
One of the

- (b)youngest independent country
- (c)
in the western Hamisphere, Trinidad and Tobago,

- (d)became a nation on August 31, 1962.
- (e)No error

**Hint**

Check if the indefinite pronoun and noun agree by considering the subject-verb agreement rule.

**Solution:**

The error lies in the second part of the sentence. We have to change 'country' to 'countries' to make the sentence meaningful. As per the rules of subject-verb agreement, the subject following the indefinite pronoun phrase 'one of the' is always a plural noun. The use of verbs as singular or plural will entirely depend upon the subject of the statement.

Hence, this option is the correct answer.

- (a)Make haste
- (b)lest
- (c)you
- (d)should not miss the train.
- (e)No error

**Hint**

Focus on the error of redundancy in the sentence.

**Solution:**

‘Lest' is a conjunction conveying negative meaning. It means 'to avoid the risk of'. Hence, further negation in the form of ‘no’ or ‘not’ is redundant. Change "...should not miss the train" to "... should miss the train".

- (a)The usual form of
- (b)mercury thermometer is used
- (c)for temperatures ranging
- (d)from 40°F to 500°F.
- (e)No error

**Hint**

The sentence describes the property of an object, not an event.

**Solution:**

The error lies in the first part of the sentence. We have to change "usual" to "ordinary" to make the sentence meaningful.

'Usual' is used to refer to something happening frequently or normally. It is usually associated with events or occurrences. 'Ordinary' is an adjective used to describe a natural state of an object. The sentence describes an ordinary thermometer. It does not have any specific equipment attached, and it is the typical instrument used to measure the temperature within a particular range.

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- (a)
Hoard

- (b)Tolerate
- (c)Forbear
- (d)Begin

**Hint**

The correct word means to come into existence.

**Solution:**

Abstain means restrain oneself from doing or enjoying something. And begin means to perform and undergo the first part of (an action or activity). Hence, they are antonyms.

Meanings of some of the other words are:

Forbear means 'to keep away'.

Hoard means 'a hidden supply'.

- (a)Aspersion
- (b)Scarcity
- (c)Aversion
- (d)Confusion

**Hint**

The correct word means shortage of something.

**Solution:**

Profusion means an abundance or large quantity of something.

Scarcity means the state of being scarce or in short supply; shortage.So, they are antonyms of each other.

Meanings of some of the other words are:

Aspersion means an attack on the reputation or integrity of someone or something.

Aversion means a strong dislike or disinclination.

Confusion means uncertainty about what is happening, intended or required.

- (a)Implicit
- (b)Obnoxious
- (c)Explicit
- (d)Pedantic

**Hint**

The correct wors means evident in nature.

**Solution:**

Obscure means not being sure or confirmed.

Explicit means something obvious.So, they are antonyms to each other.

Meanings of some of the other words are:

Pedantic means showy.

Implicit means unspoken.

Obnoxious means extremely offensive.

- (a)Alluring
- (b)Refulgent
- (c)Effulgent
- (d)Meek

**Hint**

The correct word means attracting.

**Solution:**

Repulsive means not so appealing or repel.

Alluring means something appealing or attracting.

So they are opposite of each other.

Meanings of some of the other words are:

Effulgent and Refulgent mean shining.

Meek means humble.

- (a)Auxiliary
- (b)Responsible
- (c)Salvageable
- (d)Clear

**Hint**

The correct word means easy to perceive, understand or interpret.

**Solution:**

Ambiguous means not so clear; confusing.

Its antonym is clear or sure.

Meanings of some of the other words are:

Auxiliary means 'helping'.

Salvageable means 'repair'.

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- (a)
Render

- (b)
Compensate

- (c)Return
- (d)
Restore

- (e)Recover

**Hint**

Identify a word which is the exact opposite of the word 'lose' for the proper synonym of the word in the question.

**Solution:**

In the given question, The word 'Repay' which refers to get our sum of money or any other mode of the transaction back and the word 'Recover' means the same hence the correct synonym is 'Recover'.

For example: My losses have been recovered.

- (a)
Completed

- (b)
Fulfilled

- (c)
Received

- (d)Achieved
- (e)Equipped

**Hint**

Identify a word which means 'feeling of happiness and satisfaction' for the proper synonym of the word in the question.

**Solution:**

In the given question, The word 'Accomplished' means achieving success in a particular task which is the same meaning as the word 'Fulfilled'. Hence, the correct synonym is 'Fulfilled'.

For example: Naruto has accomplished the task with hard training.

- (a)
Worry

- (b)
Scare

- (c)
Dismay

- (d)Horror
- (e)Alarm

**Hint**

Identify a word which refers to an intense feeling of fear for the proper synonym of the word in the question.

**Solution:**

In the given question, The word 'Fright' is referred as the intense feelings of fear and shock hence the correct synonym is 'Horror'.

For example: He is frightened during watching a horror film.

- (a)Competely
- (b)Essentially
- (c)Severely
- (d)
Excessively

- (e)Naturally

**Hint**

Identify a word which is the exact opposite of the word 'mildly' for the proper synonym of the word in the question.

**Solution:**

In the given question, The word 'Radically' which means in a thorough or fundamental way and the meaning of the word 'Severely' is same to some contexts. Hence, the correct synonym is 'Severely'.

For example: The site has never been lost, and the present village Lonawala retains the name radically unchanged.

- (a)Provocation
- (b)Stimulation
- (c)Intuition
- (d)Defamation
- (e)Animation

**Hint**

Identify a word which refers to a method in which figures are manipulated to appear as moving images for the proper synonym of the word in the question.

**Solution:**

In the given question, The word 'Inspiration' means getting enthusiastic to perform a task while animation means liveliness which is the same in one or the other way.

For example: My friend is working on the car design by taking inspiration from Tesla.

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- (a)
nurse

- (b)
doctor

- (c)
delivery

- (d)
husband

- (e)baby

**Hint**

The noun means a very young child.

**Solution:**

Maternity home is made for pregnant women to deliver their babies.

Babyis a very young child, especially one that has not yet begun to walk or talk. Hence, it is the right word.The correct sentence is

She went to the maternity home to deliver her baby.

- (a)Away
- (b)To
- (c)In
- (d)Up
- (e)Down

**Hint**

The adverb means in or into a higher position or level.

**Solution:**

'Take up' is a phrasal verb which means to pick up, lift, or become interested or engaged in a pursuit. Here, in the sentence, the matter was lifted up to the superiors. Hence, it is the correct option for this question.

- (a)
talk

- (b)
address

- (c)
lecture

- (d)
advise

- (e)Welcome

**Hint**

The verb also means to speak or to write someone.

**Solution:**

The noun 'address' means a formal speech to the audience. It also means to say or inform someone. In the sentence, the President will give a speech to the people of his country. Hence, 'address' is the correct word to use.

- (a)Up
- (b)Over
- (c)Away
- (d)On
- (e)Out

**Hint**

The preposition means above something or a higher position.

**Solution:**

The phrasal verb 'take over' means to replace someone or something. In the given sentence, the charge is given to someone else who will be replacing the retiring manager.

Hence, 'over' is the most suitable word for the given sentence.

- (a)
arrive

- (b)
reach

- (c)
provide

- (d)
support

- (e)accord

**Hint**

The verb also means to go to a place or to finalise something.

**Solution:**

'Reach' means here to arrive at a place or to come to a point or to make a decision, agreement, etc. about something. In the sentence, the situation is that the committee failed to make a decision. Hence, 'reach' is the most appropriate word.

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- (a)Iquitos
- (b)Ubiquitous
- (c)Littoral
- (d)Temporal

**Hint**

The word means the same as omnipresent.

**Solution:**

The following are the meanings of the given words:

Iquitos - Grossly fair and morally right.

Littoral - Relating to or situated on the shore of the sea or a lake

Temporal - Relating to worldly as approved to spiritual affairs.

Ubiquitous - Seeming to be everywhere.

- (a)Aggravating
- (b)Declamatory
- (c)Extenuating
- (d)Extirpating

**Hint**

The literal meaning of the word is to make someone or something thin.

**Solution:**

The following are the meanings of the given words:

Aggravating - Making a problem or offers worse or more serious

Declamatory: Vehement or impassioned in expression.

Extirpate - Eradicate or destroy completely.

Extenuating - Serving to lessen the seriousness of an offence partially.

- (a)Insouciance
- (b)Indifference
- (c)Charlatanism
- (d)Euphoria

**Hint**

The word means the same as equanimity.

**Solution:**

The following are the meanings of the given words:

Indifference - Lack of interest.

Charlatanism - A person who pretends to have skills or knowledge that they do not have.

Euphoria - A feeling or state of intense excitement and happiness.

Insouciance - A relaxed and happy way of behaving without feeling worried or guilty.

- (a)Obstreperous
- (b)Obsolescent
- (c)Debilitating
- (d)Preposterous

**Hint**

The word means the same as declining or waning.

**Solution:**

The following are the meanings of the given words:

Obstreperous - Noisy and difficult to control.

Debilitating - Making someone very weak and infirm.

Preposterous - Contrary to reason or common sense, utterly absurd or ridiculous.

Obsolescent - Going out of use.

- (a)Sceptic
- (b)Ascetic
- (c)Devotee
- (d)Antiquarian

**Hint**

The synonym of the word is abstinent or austere.

**Solution:**

The following are the meanings of the given words:

Ascetic - A person who avoids physical pleasures and live a simple life.

Sceptic - A person inclined to question on or doubt accepted opinions.

Devotee - A person very much interested in something/someone.

Antiquarian - Relating to or dealing in antiques or rare books.

That's why the answer is Ascetic.

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- (a)b c d a
- (b)c b a d
- (c)c d a b
- (d)b a d c

**Hint**

These are management strategies followed in a company to rescue it from failures.

**Solution:**

The correct order is:

(b) The process started at the top.

(c) He undoubtedly had the force of personality to impose painful measures.

(d) Within four years, he had rescued the company.

(a) He recognised it and streamlined management.These are management strategies followed in a company or firm to rescue it from failures or losses. When the system is corrected and streamlined from the top to bottom level, then things start moving in the right direction.

- (a)d c a b
- (b)d a b c
- (c)d b a c
- (d)c a b d

**Hint**

These sentences tell us how Gandhiji came to know about his mother’s demise.

**Solution:**

The correct order of sentences:

(d) When Gandhi returned to India, his son Harilal was four.

(a) His mother was dead.

(b) They had not sent him the sad information.

(c) Probably, they knew his deep love for her.

The sentences describe about the Gandhiji’s true and deep love for his mother. This made his family members to hide the news of his mother’s death from him since he was in London pursuing Law.

- (a)b c a d
- (b)c b d a
- (c)b c d a
- (d)c a b d

**Hint**

The sentences talk about one of the Key organizational strategies - teamwork.

**Solution:**

The correct order of sentences:

(c) Teamwork does not just happen.

(a) It results from a carefully revised plan.

(b) Men work together for a cause or purpose.

(d) It must be clearly known to them.This is about strategies of implementing a good teamwork. Each team should work with a strong motive, common cause or purpose, detailed plan, and execution of plan.

- (a)d c b a
- (b)c b d a
- (c)b d c a
- (d)d b c a

**Hint**

The sentences potray the appreciation of a book by one person.

**Solution:**

The correct order is:

(b) The book was published in New York.

(d) It deals with mankind’s political future.

(c) It is a very interesting book.

(a) I will give you a copy of it.The sentences reveal a conversation where one person is praising a book by giving all its details and willing to give a copy of it to the other person.

- (a)d c a b
- (b)b a d c
- (c)c d a b
- (d)b d c a

**Hint**

The sentences infer about a business deal that happens between the narrator and Mr. Maini.

**Solution:**

The correct order of the sentences:

(d) He agreed to my proposal

(c) It was regarding investing ten thousand rupees in my firm.

(a) He had inherited that money from an uncle.

(b) Mr. Maini was an innocent man.The sentences describe about a business proposal between the narrator and Mr. Maini. The proposal was about Mr. Maini investing the money he inherited from his uncle.

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- (a)$31\%$
- (b)$65\%$
- (c)$50\%$
- (d)$30\%$
- (e)None of these

**Hint**

win$\%=\frac{gameswon}{totalgames}\times 100\%$

**Solution:**

As per the given question,

Games won $=13$

Games lost $=7$

Therefore, the total number of game played will be $=$ Games won $+$ Games lost $=\left(13+7\right)=20$

So,

win$\%=\frac{gameswon}{totalgames}\times 100\%$

$\Rightarrow \frac{13}{20}\times 100$

$\Rightarrow 65\%$

Hence, option B is the correct answer.

- (a)$5\frac{1}{4}\phantom{\rule{0ex}{0ex}}$
- (b)$10\frac{1}{2}$
- (c)$12\frac{1}{2}$
- (d)$6$
- (e)None of these

**Hint**

Use $\mathrm{x}\%=\frac{\mathrm{x}}{100}$

**Solution:**

As per the given question,

$60\%of35=x\%of400\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{60}{100}\times 35=\frac{x}{100}\times 400\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{60\times 35}{400}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{2100}{400}=\frac{21}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow x=5\frac{1}{4}$

Hence, option A is the correct answer.

- (a)$1\%$
- (b)$2\%$
- (c)$99\%$
- (d)$98\%$
- (e)None of these

**Hint**

Calculate the number of people who will not get any prize, then find the required percentage.

**Solution:**

As per the given question,

Number of people who will not get any prize will be $=\left(50-1\right)=49$

So, the required percentage will be,

$\Rightarrow \frac{matcheslost}{totalmatches}\times 100$

$\Rightarrow \frac{49}{50}\times 100$

$\Rightarrow 98\%$

Hence, option D is the correct answer.

- (a)$500\mathrm{cubic}\mathrm{metre}$
- (b)$50\mathrm{cubic}\mathrm{metre}$
- (c)$5\mathrm{cubic}\mathrm{metre}$
- (d)$5000\mathrm{cubic}\mathrm{metre}$
- (e)None of these

**Hint**

$\frac{(\mathrm{Percentage}\mathrm{of}\mathrm{Argon}\mathrm{in}\mathrm{the}\mathrm{air})}{100}\times \mathrm{Volume}\mathrm{of}\mathrm{air}\mathrm{which}\mathrm{contains}1{\mathrm{m}}^{3}\mathrm{of}\mathrm{Argon}=1$

**Solution:**

Let,

The total volume of air is $100$

So, The percentage of argon will be,

$\Rightarrow 100-(79.2+20.7+0.08)=0.02\%$

Let the volume of the air which contains one cubic metre of argon is $V$.

So,

$\Rightarrow 0.02\%\mathrm{of}\mathrm{V}=1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{0.02}{100}\times \mathrm{V}=1\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{V}=5000\mathrm{cubic}\mathrm{metre}.$

Hence, the correct option is D.

- (a)$46913$
- (b)$45913$
- (c)$47913$
- (d)$46000$
- (e)None of these

**Hint**

$\%$ passed students $\times $ Total students $=$ Successful candidates.

**Solution:**

Let the total student appeared be $=x$

As per the question,

$\Rightarrow 49.3$ of $x=23128$

$\Rightarrow \frac{49.3}{100}\times x=23128$

$\Rightarrow 49.3x=23128\times 100$

$\Rightarrow 49.3x=2312800$

$\Rightarrow x=\frac{23128\times 100}{49.3}$

$\Rightarrow x=\frac{2312800}{49.3}$

$\Rightarrow 46912.77\approx 46913$

Hence, the correct option is A.

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- (a)$108\%$
- (b)$90\%$
- (c)$107\%$
- (d)$110\%$
- (e)
None of these

**Hint**

Calculate the selling price of the item, and then find the required percentage.

**Solution:**

As per the question,

Price after raising price will be,

$\Rightarrow 50000+13000=63000$

As he deducted $\frac{1}{7}$ of the new price, So price after deduction will be,

$\Rightarrow (63000-\frac{1}{7}\times 63000)=\u20b954000$

So, the required percentage will be,

$\Rightarrow \frac{54000}{50000}\times 100\phantom{\rule{0ex}{0ex}}\Rightarrow 108\%$

Hence, the correct option is A.

- (a)$45\%$
- (b)$60\%$
- (c)$20\%$
- (d)$75\%$
- (e)None of these

**Hint**

$\mathrm{SP}=\frac{(100+\mathrm{profit}\%)\times \mathrm{CP}}{100}$

**Solution:**

Let the actual value of the chair is $=x$

So, the cost price will be $=(\frac{75}{100}\times x)=0.75x$

So, the selling price will be $=x+\frac{20x}{100}=1.2x$

Hence, the profit percentage will be

$\Rightarrow \frac{1.2x-0.75x}{0.75}\times 100\phantom{\rule{0ex}{0ex}}\Rightarrow 60\%$

Hence, the correct option is B.

- (a)$17.64\%$
- (b)$20\%$
- (c)$30\%$
- (d)$13.04\%$
- (e)None of these

**Hint**

Original formula for $Loss\%=\frac{(CP-SP)}{CP}\times 100$

**Solution:**

Let the selling price of item is $=x$

If there is $15\%$ loss on selling price.

Then the cost price will be

$=(x+\frac{15}{100}\times x)=1.15x$

So, the required loss percentage will be

$\Rightarrow \frac{1.15x-x}{1.15x}\times 100\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{0.15x}{1.15x}\times 100\phantom{\rule{0ex}{0ex}}\Rightarrow 13.04\%$

Hence, the correct option is D.

- (a)$14\%$
- (b)$14.5\%$
- (c)$15\%$
- (d)$14\frac{7}{12}\%$
- (e)None of these

**Hint**

$\mathrm{Required}\mathrm{percentage}=\frac{\mathrm{M}.\mathrm{P}.-\mathrm{C}.\mathrm{P}.}{\mathrm{C}.\mathrm{P}.}\times 100$

**Solution:**

Let the M.P. of the article is $=x$

Discount will be

$\Rightarrow 4\%\mathrm{of}x=0.04x$

So, the selling price will be

$\Rightarrow x-0.04x=0.96x$

Given that, the selling price makes $10\%$profit on C.P.

Let C.P. of the article is $=y$

So,

$\Rightarrow 1.1y=0.96x\phantom{\rule{0ex}{0ex}}\Rightarrow y=0.872x$

Therefore, required percentage added on the price of goods

$\Rightarrow \frac{\mathrm{M}.\mathrm{P}.-\mathrm{C}.\mathrm{P}.}{\mathrm{C}.\mathrm{P}.}\times 100\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x-y}{y}\times 100\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x-0.872x}{0.872x}\times 100\phantom{\rule{0ex}{0ex}}\Rightarrow 14.67\%=14\frac{7}{12}\%$

Hence, the correct answer is option D.

- (a)$20\%$
- (b)$25\%$
- (c)$33\frac{1}{3}\%$
- (d)$30\%$
- (e)None of these

**Hint**

profit$\%=\frac{\mathrm{S}.\mathrm{P}.-\mathrm{C}.\mathrm{P}.}{\mathrm{C}.\mathrm{P}.}\times 100$

**Solution:**

Let S.P. of the item $=x$

Profit percentage on S.P. $=25\%$

So, C.P. of item will be:

$\Rightarrow \left(x-\frac{25}{100}\times x\right)$

$\Rightarrow x-0.25x$

$\Rightarrow 0.75x$

Therefore, the required percentage will be

$\Rightarrow \frac{0.25x}{0.75x}\times 100$

$\Rightarrow 33\frac{1}{3}\%$

Hence, the correct answer is option C.

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- (a)$18:27$
- (b)$2:3$
- (c)$20:30$
- (d)$180:270$
- (e)None of these

**Hint**

As we know that, Numerator of ratio is called antecedent while denominator is called consequent.

**Solution:**

As we know that,

Numerator of ratio is called antecedent while denominator is called consequent.

So, antecedent is $18$

$\frac{18}{x}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}x=\frac{18\times 3}{2}=27$

So, number will be $\frac{18}{27}$

- (a)$\frac{17}{4}$
- (b)$\frac{10}{7}$
- (c)$\frac{5}{3}$
- (d)$\frac{13}{14}$
- (e)None of these

**Hint**

Let $?=\frac{x}{y}$

**Solution:**

Let $?=\frac{x}{y}$

So,

$\frac{2}{3}:\frac{5}{6}::\frac{8}{7}:\frac{x}{y}\phantom{\rule{0ex}{0ex}}\frac{{\displaystyle \frac{2}{3}}}{{\displaystyle \frac{5}{6}}}=\frac{{\displaystyle \frac{8}{7}}}{{\displaystyle \frac{x}{y}}}\phantom{\rule{0ex}{0ex}}\frac{2}{3}\times \frac{x}{y}=\frac{5}{6}\times \frac{8}{7}\phantom{\rule{0ex}{0ex}}\frac{x}{y}=\frac{10}{7}$

- (a)$4:13$
- (b)$16:35$
- (c)$7:24$
- (d)$8:22$
- (e)None of these

**Hint**

Find the product of the ratios.

**Solution:**

So, if A:B, B:C & C:D is given, then A:D can be calculated as,

$\frac{A}{D}=\frac{A}{B}\times \frac{B}{C}\times \frac{C}{D}\phantom{\rule{0ex}{0ex}}\frac{{\displaystyle A}}{{\displaystyle D}}=\frac{{\displaystyle 2}}{{\displaystyle 3}}\times \frac{{\displaystyle 4}}{{\displaystyle 5}}\times \frac{{\displaystyle 6}}{{\displaystyle 7}}\phantom{\rule{0ex}{0ex}}=\frac{16}{35}$

Thus, the ratio of A to D is $16:35$.

- (a)
$1:5$

- (b)$7:17$
- (c)
$3:9$

- (d)$2:5$
- (e)None of these

**Hint**

Let both the glasses are of $\text{'}x\text{'}$capacity.

**Solution:**

Let both the glasses are of $\text{'}x\text{'}$capacity.

So, in first glass,

Amount of milk will be $=0.33x$

Amount of water will be$=0.67x$

In second glass,

Amount of milk will be$=0.25x$

Amount of water will be$=0.75x$

If both are mixed,

Then the ratio of milk and water will be$=\frac{0.33x+0.25x}{0.67x+0.75x}=\frac{0.58x}{1.42x}=\frac{7}{17}$

- (a)
$40$ litre

- (b)
$10$ litre

- (c)
$30$ litre

- (d)
$20$ litre

- (e)None of these

**Hint**

Calculate amount of milk and water in 30 L solution, then find the required value.

**Solution:**

As per question,

In $30$litre mixture,

Amount of milk will be $=\frac{7}{10}\times 30=21$ litre

Amount of water will be $=\frac{{\displaystyle 3}}{{\displaystyle 10}}\times 30=9$ litre

Let water added $=x$litre in order to make this $3:7$

So,

$\frac{21}{9+x}=\frac{3}{7}\phantom{\rule{0ex}{0ex}}=27+3x=147\phantom{\rule{0ex}{0ex}}x=40$

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- (a)$120$
- (b)$20$
- (c)$90$
- (d)$80$
- (e)None of these

**Hint**

Calcute number of days taken by one cattle by grazing the field, and then calculate required answer.

**Solution:**

According to the question:

$40$ cattle graze a field in $60$ days

So, $1$ cattle can graze a field in

$\Rightarrow 60\times 40=2400$ days

Thus, $1$ cattle graze double size field in

$\Rightarrow 2\times 2400=4800$ days

Now, $x$ cattle graze double size field in

$\Rightarrow \frac{4800}{x}$ days

So, number of cattle will be:

$\Rightarrow \frac{4800}{x}=40$

$\Rightarrow x=120$ cattle

Hence, the correct answer is option A.

- (a)$\u20b9270,\u20b9369,\u20b9180$
- (b)$\u20b9369,\u20b9270,\u20b9180$
- (c)$\u20b9369,\u20b9180,\u20b9270$
- (d)$\u20b9296,\u20b9370,\u20b9180$
- (e)None of these

**Hint**

Profit ratio will be equal to the ratio of subscription.

**Solution:**

According to the question:

Subscription of A$=\mathrm{subscription}\mathrm{of}\mathrm{B}+700$

Subscription of B $=\mathrm{subscription}\mathrm{of}\mathrm{C}+300$

Let, subscription by C $=x$

So, Subscription by B will be :

$\Rightarrow x+300$

And Subscription by A will be:

$\Rightarrow x+300+700=x+1000$

As per question:

$\Rightarrow \mathrm{A}+\mathrm{B}+\mathrm{C}=4700\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{x}+\mathrm{x}+300+\mathrm{x}+1000=4700\phantom{\rule{0ex}{0ex}}\Rightarrow 3\mathrm{x}+1300=4700\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{x}=\frac{3400}{3}=1133.33$

Thus, Subscription by C will be:

$\Rightarrow \u20b91133.33$

Subscription by B will be:

$\Rightarrow 1133.33+300=\u20b91433.33$

Subscription by A will be:

$\Rightarrow 1133.33+1000=\u20b92133.33$

So, Ratio of subscription of A, B and C will be:

$\Rightarrow 2133.33:1433.33:1133.33\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{6400}{3}:\frac{4300}{3}:\frac{3400}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow 64:43:34$

So, profit ratio among A, B and C will be:

$\Rightarrow 64:43:34$

Therefore, the profit share of A, B and C will be:

$\Rightarrow \frac{34}{141}\times 846,\frac{43}{141}\times 846,\frac{64}{141}\times 846\phantom{\rule{0ex}{0ex}}\Rightarrow \u20b9396,\u20b9270,\u20b9180$

Hence, the correct answer is option B.

- (a)$\u20b91200,\u20b92520$
- (b)$\u20b91680,\u20b91800$
- (c)$\u20b91700,\u20b91780$
- (d)$\u20b91780,\u20b91700$
- (e)None of these

**Hint**

The profit ratio is directly proportional to the product of capital invested and the time for which it is invested.

**Solution:**

According to the question:

As A, B, C invest capital for $12$ month,$7$ months and $5$ months respectively.

As the profit ratio among A, B, C is $4:2:3$

So, their share will be also in same ratio as per their profit ratio.

As per question:

A invest $\u20b91400$for $12$months.

Given that:

$\Rightarrow A:B=4:2\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{A}{B}=\frac{4}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1400\times 12}{B\times 7}=\frac{4}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{1400\times 2\times 12}{4\times 7}=\u20b91200$

Similarly, from B & C ratio:

$\Rightarrow \frac{B}{C}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1200\times 7}{C\times 5}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{1200\times 7\times 3}{5\times 2}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\u20b92520$

Hence, the correct answer is option A.

- (a)
$4$ months

- (b)
$8$ months

- (c)
$12$ months

- (d)
$6$ months

- (e)None of these

**Hint**

Profit is directly proportional to the product of capital invested and months for which it is invested.

**Solution:**

Let B's capital used for $y$ months

Let capital of A & B are $5x$ & $6x$ respectively.

And the profit ratio of A and B is $5:9$

Given that capital used for $8$ months:

So,

$\Rightarrow \frac{5x\times 8}{6x\times y}=\frac{5}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow 5x\times 8\times 9=6x\times y\times 5\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{5x\times 8\times 9}{6x\times 5}=12$

Hence, the correct answer is option C.

- (a)$Rs.7150$
- (b)$Rs.3060$
- (c)$Rs.9180$
- (d)$Rs.1440$
- (e)None of these

**Hint**

$\mathrm{A}\text{'}\mathrm{s}\mathrm{income}=\frac{\mathrm{x}-1440}{2}+2250$

**Solution:**

Let the total profit $=x$

B's salary$=(120\times 12)=1440$

So, net profit to be divided $=x-1440$

A and B equally gets

$\Rightarrow \frac{x-1440}{2}$

Share paid to A as an interest $\Rightarrow 10\%\mathrm{of}22500=2250$

Now,

$\Rightarrow \frac{\mathrm{B}\text{'}\mathrm{s}\mathrm{income}}{\mathrm{A}\text{'}\mathrm{s}\mathrm{income}}=\frac{{\displaystyle \frac{x-1440}{2}+1440-2250}}{\frac{x-1440}{2}+2250}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}=\frac{x-3060}{x+3060}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\u20b99180$

Hence, the correct answer is option C.

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- (a)$14$
- (b)
$8$

- (c)$13$
- (d)
$12$

- (e)None of these

**Hint**

Sum of the numbers is equal to 40.

**Solution:**

Let the missing number is $=x$

As per the question,

$\frac{7+5+x+3+12}{5}=8\phantom{\rule{0ex}{0ex}}27+x=40\phantom{\rule{0ex}{0ex}}x=13$

Hence, the missing number is $13$.

- (a)
$17$ kg

- (b)
$26$ kg

- (c)
$20$ kg

- (d)
$31$ kg

- (e)None of these

**Hint**

The total weight of A, B and C is 135 kg.

**Solution:**

Let the weight of A, B and C are $x,y,zkg$respectively.

As per the question,

$\frac{x+y+z}{3}=45kg\phantom{\rule{0ex}{0ex}}x+y+z=135kg.....\left(1\right)$

Also,

$\frac{x+y}{2}=40kg\phantom{\rule{0ex}{0ex}}\Rightarrow x+y=80kg......\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{y+z}{2}=43kg\phantom{\rule{0ex}{0ex}}\Rightarrow y+z=86kg.....\left(3\right)$

On solving $\left(1\right)-\left(2\right)$,

$z=55kg$

From $\left(3\right)$,

$y=31kg$

Hence, option D is the correct answer.

- (a)$16$
- (b)
$\frac{48}{7}$

- (c)
$20$

- (d)
$18$

- (e)None of these

**Hint**

The sum of the first three numbers is six times the last number.

**Solution:**

Let the numbers be $a,b,cd$.

As per the question,

$\frac{a+b+c}{3}=2d\phantom{\rule{0ex}{0ex}}a+b+c=6d....\left(1\right)$

Also,

$\frac{{\displaystyle a+b+c+d}}{4}=12\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\displaystyle 6d+d}}{4}=12\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{7d}{4}=12\phantom{\rule{0ex}{0ex}}\Rightarrow d=\frac{{\displaystyle 48}}{7}$

Hence, option B is the correct answer.

- (a)$\frac{542}{2}$
- (b)$\frac{539}{7}$
- (c)$\frac{536}{7}$
- (d)$\frac{547}{7}$
- (e)None of these

**Hint**

The total score for the second year is 243.

**Solution:**

As per the question,

The total score for the first year $=76\times 4$

The total score for the second year $=81\times 3$

The overall average will be

$=\frac{(76\times 4)+(81\times 3)}{4+3}\phantom{\rule{0ex}{0ex}}=\frac{547}{7}$

Hence, option D is the correct answer.

- (a)$15.38$ km per hour
- (b)$16$ km per hour
- (c)$15.5$ km per hour
- (d)$16.38$ km per hour
- (e)None of these

**Hint**

$Avgspeed=\frac{totaldis\mathrm{tan}ce}{totaltimetaken}$, use this to find the required value.

**Solution:**

We know that,

$Avg.speed=\frac{totaldis\mathrm{tan}ce}{totaltimetaken}\phantom{\rule{0ex}{0ex}}=\frac{24+36}{{\displaystyle \frac{24}{16}+\frac{36}{15}}}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle 60}}{1.5+2.4}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle 60}}{3.9}=15.38km/hr$

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- (a)
Rs.540

- (b)
Rs. 640

- (c)
Rs. 600

- (d)
Rs.700

- (e)None of these

**Hint**

If the Principal is $Rs.x$, $SI=Rs.720-x$

**Solution:**

Let the Principal is $=Rs.x$

$R=5\%\phantom{\rule{0ex}{0ex}}T=2.5years$

As per the question,

$SI=A-P=Rs.720-x$

Also,

$SI=\frac{P\times R\times t}{100}\phantom{\rule{0ex}{0ex}}(720-x)=\frac{x\times 5\times 2.5}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \left(720-x\right)=\frac{x}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{720\times 8}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow x=Rs.640$

Hence, the principal is $Rs.640$

- (a)
40 years

- (b)
20 years

- (c)
60 years

- (d)
80 years

- (e)None of these

**Hint**

When the amount is doubled, the SI is equal to the principal.

**Solution:**

Let the principal is $=Rs.x$

So, $A=Rs.2x,t=20$

As per the question,

$SI=\frac{P\times R\times t}{100}\phantom{\rule{0ex}{0ex}}(A-P)=\frac{P\times R\times t}{100}\phantom{\rule{0ex}{0ex}}(2x-x)=\frac{x\times R\times 20}{100}\phantom{\rule{0ex}{0ex}}x=\frac{xR}{5}\phantom{\rule{0ex}{0ex}}R=5\%$

Now,

New amount will be $A=Rs.4x$

So,

$SI=\frac{P\times R\times t}{100}\phantom{\rule{0ex}{0ex}}(A-P)=\frac{x\times 5\times t}{100}\phantom{\rule{0ex}{0ex}}(4x-x)=\frac{x\times 5\times t}{100}\phantom{\rule{0ex}{0ex}}3x=\frac{x\times 5\times t}{100}\phantom{\rule{0ex}{0ex}}t=60years.$

- (a)
Rs. 1,00,000

- (b)
Rs. 1,20,000

- (c)
Rs. 1,10,000

- (d)
Rs. 1,30,000

- (e)None of these

**Hint**

Use the formula, $SI=\frac{P\times R\times T}{100}$

**Solution:**

Let the Principal is $=Rs.x$

As, $Rs.600$ is the per month interest.

So, the yearly interest will be $=(600\times 12)=Rs.7200$

As per the question,

$SI=\frac{P\times R\times t}{100}\phantom{\rule{0ex}{0ex}}7200=\frac{x\times 6\times 1}{100}\phantom{\rule{0ex}{0ex}}x=\frac{7200\times 100}{6}=Rs.120000$

Hence, the principal is $Rs.1,20,000$

- (a)$8\frac{1}{2}\%$
- (b)$7\%$
- (c)$6\%$
- (d)$8\%$
- (e)None of these

**Hint**

Let the rate of interest on which $Rs.2000$ and $Rs.1600$ are given is $r\%$ and $\left(r+2\right)\%$ respectively. Form an equation using the data.

**Solution:**

Let the rate of interest on which $Rs.2000$ and $Rs.1600$ are given is $r\%$ and $\left(r+2\right)\%$ respectively.

$S{I}_{1}=\frac{2000\times r\times 3}{100}=60r$

$S{I}_{2}=\frac{1600\times \left(r+2\right)\times 3}{100}=48\left(r+2\right)$

$S{I}_{1}+S{I}_{2}=960$

$60r+48\left(r+2\right)=960$

$60r+48r+96=960$

$108r=864$

$r=\frac{864}{108}=6\%$

Hence, option C is the correct answer.

- (a)
5 : 8

- (b)
8 : 5

- (c)
31 : 6

- (d)
16 : 15

- (e)
None of these

**Hint**

Let money lent at $5\%$ to$8\%$ is $x(1550-x)$ respectively.

Hence, find the interest using $SI=\frac{PRT}{100}$

**Solution:**

Let the money lent at $5\%$ and $8\%$ is $Rs.xRs.(1550-x)$ respectively.

So,

$(SI{)}_{1}=\frac{x\times 5\times 3}{100}=\frac{15x}{100}\phantom{\rule{0ex}{0ex}}(SI{)}_{2}=\frac{(1550-x)\times 8\times 3}{100}=\frac{(1550-x)\times 24}{100}$

Also,

$(SI{)}_{1}+(SI{)}_{2}=300\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{15x}{100}+\frac{(1550-x)\times 24}{100}=300\phantom{\rule{0ex}{0ex}}\Rightarrow 15x+37200-24x=30000\phantom{\rule{0ex}{0ex}}\Rightarrow 7200=9x\phantom{\rule{0ex}{0ex}}\Rightarrow x=800$

The money lent at $5\%$ $=Rs.800$

So, money lent at $8\%$ will be $=1550-800=Rs.750$

Hence, the required ratio will be $=\frac{800}{750}=16:15$

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- (a)$Rs.560.00$
- (b)$Rs.561.80$
- (c)$Rs.572.70$
- (d)$Rs.512.40$
- (e)None of these

**Hint**

$A=P{\left(1+\frac{r}{100}\right)}^{t}$

**Solution:**

As per the question,

$P=Rs.500,r=6\%,t=2$

So,

$A=P(1+\frac{r}{100}{)}^{t}\phantom{\rule{0ex}{0ex}}=500(1+\frac{6}{100}{)}^{2}\phantom{\rule{0ex}{0ex}}=500(\frac{100+6}{100}{)}^{2}\phantom{\rule{0ex}{0ex}}=Rs.561.80$

- (a)6 years
- (b)8 years
- (c)12 years
- (d)10 years
- (e)None of theses

**Hint**

Let, the Principal is $P=x$

**Solution:**

Let, the Principal is $P=x$

As per question,

$A=2x,t=4$

So,

$A=P(1+\frac{r}{100}{)}^{t}\phantom{\rule{0ex}{0ex}}2x=x(1+\frac{r}{100}{)}^{4}\phantom{\rule{0ex}{0ex}}1+\frac{r}{100}={2}^{\frac{1}{4}}\phantom{\rule{0ex}{0ex}}\left(\frac{100+r}{100}\right)={2}^{\frac{1}{4}}\phantom{\rule{0ex}{0ex}}r=12years$

- (a)$Rs.50000$
- (b)$Rs.60000$
- (c)$Rs.65000$
- (d)$Rs.55000$
- (e)None of these

**Hint**

Calculate the compound interest for $2$ years using the formula. The compound interest for the second year will be the total compound interest minus the compound interest for the first year.

**Solution:**

Let,

$P=Rs.x,R=4\%,T=1$ year

$A=P{\left(1+\frac{R}{100}\right)}^{T}=x{\left(1+\frac{4}{100}\right)}^{1}=Rs.1.04x$

So, ${\mathrm{CI}}_{1}=1.04x-x=Rs.0.04x$

Now,

$P=Rs.x,t=2$ yr

$A=P{\left(1+\frac{r}{100}\right)}^{t}=x{\left(1+\frac{4}{100}\right)}^{2}=x(1.04{)}^{2}$

$A=Rs.1.0816x$

$\mathrm{CI}=(1.0816x-x)=Rs.0.0816x$

This is the compound interest for $2$ years.

Hence, the compound interest for the second year

$={\mathrm{CI}}_{2}=0.0816x-0.04x=0.0416x$

According to the question,

${\mathrm{CI}}_{2}-{\mathrm{CI}}_{1}=88$

$0.0416x-0.04x=88$

$x=Rs.55000$

Hence, the principal is $Rs.55000$

- (a)$46000$
- (b)$46200$
- (c)$48020$
- (d)$48320$
- (e)None of these

**Hint**

New population = Old population$(1\pm \frac{R}{100}{)}^{n}$, where $n$ denotes the no. of years.

**Solution:**

As the population decreases by$20$ per thousand each year,

$R=\frac{20\times 100}{1000}=2\%$

So, the population after$2$ years will be

$=50,000(1-\frac{2}{100}{)}^{2}\phantom{\rule{0ex}{0ex}}=50000\times {\left(\frac{49}{50}\right)}^{2}\phantom{\rule{0ex}{0ex}}=48020$

Hence, option C is the correct answer.

- (a)$10000$
- (b)
$9000$

- (c)$11000$
- (d)$8000$
- (e)None of these

**Hint**

Let the population before$3$years $=x$. Hence, use the formula:

New population = Old population$(1\pm \frac{R}{100}{)}^{n}$.

**Solution:**

Let the population before$3$years $=x$

As, the population increases by$10\%$ per year

So,

$10648=x(1+\frac{10}{100}{)}^{3}\phantom{\rule{0ex}{0ex}}10648=x(1.1{)}^{3}\phantom{\rule{0ex}{0ex}}x=\frac{10648}{1.1\times 1.1\times 1.1}=8000$

Hence, $3$ years ago the population was $8000$.

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- (a)$Rs.50$
- (b)$Rs.55$
- (c)$Rs.40$
- (d)$Rs.45$
- (e)None of these

**Hint**

$Income=\frac{investment\times 100\times interest}{numberofshares}$

**Solution:**

As per the question, the stock is at$84$ in$3\%$for $Rs.1260$

So, income will be = $\frac{investment\times 100\times interest}{numberofshares}$

$\Rightarrow \frac{1260\times 100\times 3}{84\times 100}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{378000}{8400}\phantom{\rule{0ex}{0ex}}\Rightarrow Rs.45$

Hence, the correct option is D.

- (a)$\u20b91300,\u20b91400$
- (b)$\u20b91250,\u20b91450$
- (c)$\u20b912600,\u20b914400$
- (d)$\u20b91360,\u20b91340$
- (e)None of these

**Hint**

Let he invest $\u20b9x$ in $6\%$ stock at $\u20b9126$ and $\u20b9(27000-x)$ in $5\%$ stock at $\u20b9120$.

**Solution:**

Let he invest $\u20b9x$in $6\%$ stock at $\u20b9126$and $\u20b9(27000-x)$ in $5\%$ stock at $\u20b9120$.

As per the question,

$\Rightarrow \frac{x\times 100\times 6}{126\times 100}=\frac{(2700-x)\times 100\times 5}{120\times 100}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{6x}{126}=(27000-x)\frac{5}{120}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{144x}{126}=(27000-x)\phantom{\rule{0ex}{0ex}}\Rightarrow 144x=3402000-126x\phantom{\rule{0ex}{0ex}}\Rightarrow 270x=3402000\phantom{\rule{0ex}{0ex}}\Rightarrow x=12600$

So, the value of investment will be $12600,(27000-12600)=\u20b912600,\u20b914400$

Hence, the correct option is $\u20b912600,\u20b914400$.

- (a)$\mathrm{Rs}47$ increase
- (b)$\mathrm{Rs}50$ increase
- (c)$\mathrm{Rs}48$ increase
- (d)$\mathrm{Rs}45$ increase
- (e)None of these

**Hint**

Solve the question by using the concept of conversion of percentage.

**Solution:**

As per question,$(Rs.98\frac{1}{4}+25paise)=Rs.98.5cash$

So, we can buy stock of $Rs.100$

So, for $Rs.12805$, we can buy stock of $=\frac{12805\times 100}{98.5}=Rs.13000$Again,

$Rs.102\frac{5}{8}-25paise=Rs.\frac{819}{8}$So, for $Rs.13000$ we will get stock of

$=\frac{819}{8}\times \frac{13000}{100}=Rs.\frac{106470}{8}$

Now,$(Rs.105\frac{3}{8}+25paise)=Rs.\frac{845}{8}$As, in $\frac{1064700}{8}cash$ we can buy stock of $=\frac{106470}{8}\times \frac{100}{{\displaystyle \frac{845}{8}}}=Rs.12600$

So, income from first stock will be $\frac{12600}{100}\times 4.5=Rs.567$Hence, the change in income will be $=(567-520)=Rs.47$

The correct option is A.

- (a)$Rs.35200$
- (b)$Rs.105600$
- (c)$Rs.70400$
- (d)$Rs.17600$
- (e)None of these

**Hint**

Let the person invest $Rs.x$, then solve the equation, according to the question by converting percentages.

**Solution:**

Let the person invest $Rs.x$.

So, income at $4\frac{1}{2}\%$ stock at $Rs.96$ will be

$\Rightarrow (x\times \frac{100}{96}\times \frac{4.5}{100})=\frac{4.5x}{96}$

Income at $4\%$ stock at $Rs.88$ will be

$\Rightarrow (x\times \frac{100}{88}\times \frac{4}{100})=\frac{4x}{88}$

So, as per the question,

$\Rightarrow \frac{4.5x}{96}=\frac{4x}{88}+100$

$\Rightarrow \frac{4.5x}{96}-\frac{{\displaystyle 4x}}{{\displaystyle 88}}=100$

$\Rightarrow \frac{12x}{96\times 88}=100$

$\Rightarrow x=Rs.70400$

Hence, the correct option is C.

- (a)$Rs.60$
- (b)$Rs.50$
- (c)$Rs.40$
- (d)$Rs.30$
- (e)None of these

**Hint**

Calculate the value of total share and use it to find the income.

**Solution:**

As per the question,

The face value of $1$ share is $Rs.50$.

So, the value of total share will be $=50\times 20=Rs.1000$

So, income of $Rs.1000$ at $5\%$ will be

$\Rightarrow \frac{5\times 1000}{100}=Rs.50$

Hence, the correct option is B.

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- (a)$12$
- (b)$7$
- (c)$9$
- (d)$5$
- (e)None of these

**Hint**

Let $1$ person does $1$ unit work daily. So work completed by $8$ people is $96$ units.

**Solution:**

Let $1$ person does $1$ unit work daily.

So, total work will be $=(14\times 16\times 1)=224$ units

As per the question,

$8$ people started the work for $12$ days,

So, the amount of work done will be $=(8\times 12\times 1)=96units$

So, the remaining units of work will be $=(224-96)=128$ units

After that, $8$ more people joined them.

So, the total number of people will be $=(8+8)=16$

Hence, the required number of days will be $=\frac{128}{16}=8$ days

Hence, option E is the correct answer.

- (a)$10$ days
- (b)$8$ days
- (c)$12$ days
- (d)$9$ days
- (e)None of these

**Hint**

Total work $=\mathrm{number}\mathrm{of}\mathrm{men}\times \mathrm{number}\mathrm{of}\mathrm{days}\times \mathrm{one}\mathrm{day}\mathrm{work}\mathrm{of}1\mathrm{man}$

**Solution:**

As per the question,

$15$ men can complete a work in $10$ days while $20$ boys can complete the same work in $15$ days.

So,

$\Rightarrow (15M\times 10)=(20B\times 15)\phantom{\rule{0ex}{0ex}}\Rightarrow 150M=300B\phantom{\rule{0ex}{0ex}}\Rightarrow M=2B$

Let one man and one boy does $1unit2unit$ respectively.

So, total work will be $=(15\times 1\times 10)=150units$

Hence, time taken by $(10M+10B)=(10M+5M)=15M$ will be $=\frac{150}{15\times 1}=10$ days.

Therefore, the correct option is A.

- (a)$2$ days
- (b)$3$ days
- (c)$4$ days
- (d)$3.5$ days
- (e)None of these

**Hint**

Let one man does one unit of work in a day. Solve by using unitary method.

**Solution:**

In $5$ days $14$ workers make $1400$ toys.

So, in one day$14$ workers make

$\Rightarrow \frac{1400}{5}=280$ toys

Number of remaining toys $=1400-280=1120$

Number of total workers

$=(14+14)=28$ $14$ workers make$1400$toys in $5$ days.

So, $28$ workers make$1120$ toys

$\Rightarrow \frac{5\times 14\times 1120}{28\times 1400}=2$ days.

Hence, the correct option is A.

- (a)$9$
- (b)$5$
- (c)$6$
- (d)$7$
- (e)None of these

**Hint**

Let one man does one unit of work in a day. So, the amount of work done by 14 men in 12 days is 168 units.

**Solution:**

Let one man does one unit of work in a day.

So,

Amount of work done by $14$ men in $12$ days will be $=(14\times 12)=168units$

As per the question,

Amount of work done by $14$ men in $4$ days will be $=(14\times 4)=56units$

The remaining unit of work is $=(168-56)=112units$

After that $2$ more men join, so the total number of workers will be $=(14+2)=16$

Hence, the number of days will be $=\frac{112}{16}=7$

Hence, option D is the correct answer.

- (a)$\frac{8}{3}$
- (b)$\frac{4}{3}$
- (c)$3$
- (d)$2$
- (e)None of these

**Hint**

Time taken to complete the work together $=\frac{xy}{x+y}$

**Solution:**

Time taken to complete the work together

$=\frac{xy}{x+y}$

Where, $x=8$ and $y=4$

So,

Time taken $=\frac{8\times 4}{8+4}=\frac{32}{12}=\frac{8}{3}$ hr

Hence, option A is the correct answer.

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- (a)
(a) and (b) together are not sufficient

- (b)
(a) and (b) together are needed

- (c)
(a) alone is sufficient

- (d)
(b) alone is sufficient

- (e)None of these

**Hint**

Let the speed of the boat in still water is $xkm/hr$. So, upstream speed $=\left(x-1\right)km/hr$.

**Solution:**

Let the speed of the boat in still water be $xkm/hr$.

From statement (a) and (b),

Speed of river $=1km/hr$

Downstream speed $=\left(x+1\right)km/hr$

Upstream speed $=\left(x-1\right)km/hr$

Time taken to go and return is $3hours$

So,

$\Rightarrow 3=\frac{4}{x+1}+\frac{4}{x-1}\phantom{\rule{0ex}{0ex}}\Rightarrow 3=\frac{4(x-1+x+1)}{{x}^{2}-1}\phantom{\rule{0ex}{0ex}}\Rightarrow 3{x}^{2}-3=4x-4+4x+4\phantom{\rule{0ex}{0ex}}\Rightarrow 3{x}^{2}-8x-3=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3{x}^{2}-9x+x-3=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3x(x-9)+1(x-3)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (3x+1)(x-3)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=3,\frac{-1}{3}$

Hence, both the statements are required to answer the question.

Therefore, the correct option is B.

- (a)$8km$
- (b)$6km$
- (c)$12km$
- (d)
Data inadequate

- (e)
None of these

**Hint**

Apply the concept of boat and stream.

**Solution:**

As per the question, the boat travels from P to Q along the current and from Q to P against the current in$3hours$. If the speed of the boat in still water is $4km/hr$.

Since, the speed of the stream is not given, the question is incomplete.

Hence, the question cannot be solved.

Therefore, D is the correct option.

- (a)$250\mathrm{m}$
- (b)$150\mathrm{m}$
- (c)$900\mathrm{m}$
- (d)Data inadequate
- (e)None of these

**Hint**

speed$=\frac{\mathrm{distance}}{\mathrm{time}}$

**Solution:**

As per question,

Speed of the train will be $=90\mathrm{km}/\mathrm{hr}=\frac{90\times 5}{18}$$=25\mathrm{m}/\mathrm{sec}$

As, speed$=\frac{\mathrm{distance}}{\mathrm{time}}$

So,

$\Rightarrow 25=\frac{\mathrm{Length}\mathrm{of}\mathrm{train}}{\mathrm{Time}}$$\Rightarrow 25=\frac{\mathrm{Length}}{10}$

$\Rightarrow \mathrm{Length}=\left(25\times 10\right)=250\mathrm{m}$

Hence, A is the correct option.

- (a)$1\frac{3}{4}km$
- (b)$1\frac{1}{2}km$
- (c)$2\frac{1}{2}km$
- (d)$2\frac{3}{4}km$
- (e)None of these

**Hint**

The distance of the office from his house $=\frac{x\times y}{y-x}\times \frac{{t}_{1}+{t}_{2}}{60}km$ where $x,y$ are the speeds respectively and ${t}_{1},{t}_{2}$ are time taken respectively.

**Solution:**

In these type of question,

Distance $=\frac{\mathrm{product}\mathrm{of}\mathrm{speed}}{\mathrm{difference}\mathrm{of}\mathrm{speed}}\times \mathrm{time}\mathrm{in}\mathrm{hours}$So, distance $=\frac{{\displaystyle \frac{5}{2}}\times {\displaystyle \frac{7}{2}}}{{\displaystyle \frac{7}{2}}-{\displaystyle \frac{5}{2}}}\times \frac{6+6}{60}km$

$\Rightarrow \frac{35}{4}\times \frac{1}{5}=\frac{7}{4}km=1\frac{3}{4}km$

Hence, the correct option is A.

- (a)$60km/h$
- (b)$65km/h$
- (c)$75km/h$
- (d)$70km/h$
- (e)None of these

**Hint**

$\mathrm{Speed}=\frac{\mathrm{distance}}{\mathrm{time}}$

**Solution:**

Distance $=300km$

Time $=\frac{4}{5}\times 5=4h$

Speed $=\frac{\mathrm{distance}}{\mathrm{time}}$

So, the speed $=\frac{300}{4}=75km/h$

Hence, the correct option is C.

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- (a)$7.2$ kg
- (b)$10.8$ kg
- (c)$12.4$ kg
- (d)$18.0$ kg
- (e)None of these

**Hint**

Weight of one metre rod is $1.8$ kg

**Solution:**

As per the question,

The weight of a $13$ m long rod is $23.4$ kg.

So, the weight of a $1$ m rod will be $=\frac{23.4}{13}=1.8$ kg

Hence, the weight of a $6$ m rod will be $=(6\times 1.8)=10.8$ kg

Thus, option B is the correct answer.

- (a)Rs.$2,200$
- (b)Rs.$2,000$
- (c)Rs.$2,400$
- (d)Rs.$2,600$
- (e)None of these

**Hint**

The cost of one mango is Rs.$6.85$

**Solution:**

As per question,

As cost of $36$ mangoes is Rs.$245$

So, the weight of $1$ mango will be $=\frac{245}{36}=$Rs.$6.85$

Hence, the cost of $353$ mangoes will be $=(6.85\times 353)=2402.36\approx $Rs.$2400$

Thus, option C is the correct answer.

- (a)$42$ paise
- (b)$48$ paise
- (c)$40$ paise
- (d)$50$ paise
- (e)None of these

**Hint**

The cost of one $\mathrm{kg}$ is $\mathrm{Rs}.2.4$

**Solution:**

As per the question,

If the cost of $\frac{1}{4}\mathrm{kg}$ item is $\mathrm{Rs}.0.60$.

So, the cost of $1\mathrm{kg}$ item will be $=(0.60\times 4)=\mathrm{Rs}.2.4$

Hence, the cost of $200\mathrm{gm}$ item will be $=\frac{2.4}{1000}\times 200=\mathrm{Rs}.0.48=48$ paise

Hence, option (B) is the correct answer.

- (a)$\mathrm{Rs}.500$
- (b)$\mathrm{Rs}.400$
- (c)$\mathrm{Rs}.800$
- (d)$\mathrm{Rs}.600$
- (e)None of these

**Hint**

The cost of a table is $2.5$ times the cost of a chair.

**Solution:**

As per question,

The cost of $2$ tables is equal to the cost of $5$ chairs.

Let the cost of $1$ table and the cost of $1$ chair be $\mathrm{Rs}.x$ and $y$ respectively.

So,

$2x=5y$

$x=\frac{5}{2}y$

And,

$(x-y)=1200$

So,

$\left(x-y\right)=1200$

$\left(\frac{5}{2}y-y\right)=1200$

$\frac{5y-2y}{2}=1200$

$3y=2400$

$y=Rs.800$

Hence, option (C) is the correct answer.

- (a)$52$
- (b)$68$
- (c)$62$
- (d)$58$
- (e)None of these

**Hint**

Calculate the length of one piece by dividing it into equal pieces.

**Solution:**

As per the question,

When a $192$ metre long rod is cut down into small pieces of length $3.2$ metre each.

Then total number of pieces will be $=\frac{192}{3.2}=60$

Hence, option E is the correct answer.

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- (a)
$18$years

- (b)
$20$years

- (c)
$22$years

- (d)
$21$years

- (e)None of these

**Hint**

Let the present ages of Sudhir and Madan be $4x\text{and}5x$, respectively.

As per the question,

$\frac{4x+5}{5x+5}=\frac{5}{6}.$

**Solution:**

Let the present ages of Sudhir and Madan be $4x\text{and}5x$, respectively.

Then, after five years, the ages of Sudhir and Madan will be $\left(4x+5\right)\text{and}\left(5x+5\right)$, respectively.As per the question,

$\frac{4x+5}{5x+5}=\frac{5}{6}\phantom{\rule{0ex}{0ex}}24x+30=25x+25\phantom{\rule{0ex}{0ex}}\Rightarrow x=5$So, the present age of Sudhir $=4x=4\times 5=20$ years

Hence, the correct answer is $20$ years.

- (a)$20$ years
- (b)$22$ years
- (c)$24$ years
- (d)$18$ years
- (e)None of these

**Hint**

If the present age of a person is $n$ years, $m$ years ago, his age would have been $\left(n-m\right)$ years.

**Solution:**

Let the present ages of Lata and Arun be $x\text{and}y$, respectively.

Five year ago, the age of Lata and Arun would have been $\left(x-5\right)\text{and}\left(y-5\right)$, respectively.As per question,

$\left(x-5\right)=2\left(y-5\right)$

$\Rightarrow x-5=2y-10$

$\Rightarrow 2y-x=5...\left(i\right)$

After ten years, the ages of Lata and Arun will be $\left(x+10\right)\text{and}\left(y+10\right)$, respectively.

So,

$(x+10)=\frac{4}{3}(y+10)$

$3x+30=4y+40$

$4y-3x=-10...\left(ii\right)$

Now, from $\left(i\right)$:

$2y-x=5$

$\Rightarrow 4y-2x=10$

From $\left(ii\right)$:

$4y-3x=-10$

$\Rightarrow x=20$

Hence, the present age of Lata will be $=x=20$ years

- (a)$17$years
- (b)$19$years
- (c)$18$years
- (d)$16$years
- (e)None of these

**Hint**

Let the present ages of Kunal and Ganesh be $3x\text{and}5x$, respectively.

The age difference between two persons is always constant.

**Solution:**

Let the present ages of Kunal and Ganesh be $3x\text{and}5x$, respectively.

After four years, the ages of Kunal and Ganesh will be $\left(3x+4\right)\text{and}\left(5x+4\right)$, respectively.

As per question,

$3x+4=5x+4-12\phantom{\rule{0ex}{0ex}}2x=12\phantom{\rule{0ex}{0ex}}x=6$So, the present age of Kunal $=3x=\left(3\times 6\right)=18$ years

Hence, the correct answer is $18$ years.

- (a)$82$years
- (b)$88$years
- (c)$83$years
- (d)$78$years
- (e)None of these

**Hint**

If the age of a person is $P$ years at present, then his age after $x$ years will be $P+x$ years.

**Solution:**

The father was $45$ years old $4$ years ago.

So, the present age of the father will be $=\left(45+4\right)=49$ years

So, the present age of the son will be $=\left(49-25\right)=24$ years

So, the sum of ages of father and the son after five years will be $=(49+5+24+5)=83$ years

Hence, the correct answer is $83$ years.

- (a)$32$years
- (b)$24$years
- (c)$34$years
- (d)$42$years
- (e)None of these

**Hint**

Let the ages of Yogesh, Prakash and Sameer before $10$ years be $2x,3x\text{and}4xyears$, respectively. The sum of present ages $2x+10,3x+104x+10$ is equal to $93$ years.

**Solution:**

Let the ages of Yogesh, Prakash and Sameer before $10$ years was$2x,3x\text{and}4x$ years, respectively.

So, the present ages of Yogesh, Prakash and Sameer will be $\left(2x+10\right),\left(3x+10\right)\text{and}\left(4x+10\right)$years, respectively.

As per the question,

$(2x+10)+(3x+10)+(4x+10)=93years\phantom{\rule{0ex}{0ex}}9x+30=93\phantom{\rule{0ex}{0ex}}9x=63\phantom{\rule{0ex}{0ex}}x=\frac{63}{9}=7years$

So, the present age of Sameer $=\left(4x+10\right)=4\times 7+10=38$ years

Hence, the correct answer is None of these.

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- (a)100 litre
- (b)120 litre
- (c)50 litre
- (d)110 litre
- (e)none of these

**Hint**

$\frac{3}{4}\left(\text{Capacity}\right)+5=\frac{4}{5}\left(\text{Capacity}\right)$

**Solution:**

Let the capacity of tank be $x$.

As per the question,

$\frac{3x}{4}+5=\frac{4x}{5}\phantom{\rule{0ex}{0ex}}\frac{4x}{5}-\frac{3x}{4}=5\phantom{\rule{0ex}{0ex}}\frac{x}{20}=5\phantom{\rule{0ex}{0ex}}x=100$Therefore, the capacity of the tank is $100$ litres.

- (a)50
- (b)62
- (c)45
- (d)40
- (e)none of these

**Hint**

$\frac{1}{3}\left(160-\text{Marksingeography}\right)=\text{Marksingeography}$

**Solution:**

Let the marks of Gauri in geography be $x$.

Total marks in geography and history $=160$

Marks obtained in history $=(160-x)$

As per the question,

$\left(160-x\right)\frac{1}{3}=x\phantom{\rule{0ex}{0ex}}160-x=3x\phantom{\rule{0ex}{0ex}}160=3x+x\phantom{\rule{0ex}{0ex}}160=4x\phantom{\rule{0ex}{0ex}}x=40$Therefore, her marks in geography is $40$.

- (a)₹450
- (b)₹500
- (c)600
- (d)₹380
- (e)none of these

**Hint**

Cost of the kurta will be $=\frac{2}{3}\times 180=\u20b9120$

**Solution:**

Cost of the shirt$=\u20b9180$

So, the cost of the kurta $=\frac{2}{3}\times 180=\u20b9120$

The cost of the saree $=120\times \frac{5}{2}=\u20b9300$

Total expenditure $=180+120+300=\u20b9600$

- (a)₹1,50,000
- (b)₹1,50,250
- (c)₹1,40,350
- (d)₹1,40,200
- (e)none of these

**Hint**

Numbers of male donors will be $=\frac{1}{3}\times 150=50$

**Solution:**

The number of male donors $=\frac{1}{3}\times 150=50$

So, the number of female donors $=150-50=100$

Donation of a female donor $=\frac{1}{5}\times 2000=\u20b9400$

Total donation $=(50\times 2000)+(100\times 400)\phantom{\rule{0ex}{0ex}}=100000+40000\phantom{\rule{0ex}{0ex}}=\u20b9140000$

- (a)16 cm
- (b)45 cm
- (c)22 cm
- (d)18 cm
- (e)none of these

**Hint**

$\text{Lengthofthesecondpiece}+\frac{2}{5}\left(\text{Lengthofthesecondpiece}\right)=63$ cm

**Solution:**

Let the length of the second piece be $x$.

So, the length of the first piece $=\frac{2}{5}\times x=\frac{2x}{5}$

As per the question,

$x+\frac{2}{5}x=63\phantom{\rule{0ex}{0ex}}\frac{7x}{5}=63\phantom{\rule{0ex}{0ex}}7x=315\phantom{\rule{0ex}{0ex}}x=45$

Therefore, the length of the small piece $=63-45=18$ cm

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- (a)7
- (b)8
- (c)12
- (d)9
- (e)none of these

**Hint**

Let the numbers be$x,x+2\text{and}x+4$. Equate their sum to $x+20$.

**Solution:**

Let the odd numbers be$x,x+2\text{and}x+4$.

As per the question,

$x+x+2+x+4=20+x\phantom{\rule{0ex}{0ex}}2x+6=20\phantom{\rule{0ex}{0ex}}2x=20-6=14\phantom{\rule{0ex}{0ex}}x=7$So, the middle number $=x+2=7+2=9$

- (a)52
- (b)45
- (c)54
- (d)data is inadequate
- (e)none of these

**Hint**

$\text{Firstnumber}:\text{Secondnumber}:\text{Thirdnumber}=4:9:16$

**Solution:**

Let the numbers be $x,y\text{and}z$.

As per the question,

$x:z=1:4\phantom{\rule{0ex}{0ex}}y:z=9:16$Multiplying the first ratio by $4$,

$x:z=4:16$So,

$x:y:z=4:9:16$Let the three numbers be $4k,9k\text{and}16k$.

Then,

$4k+9k+16k=174\phantom{\rule{0ex}{0ex}}29k=174\phantom{\rule{0ex}{0ex}}k=\frac{174}{29}=6$Therefore, the second number $=9x=9\times 6=54$

- (a)5:6
- (b)3:2
- (c)2:3
- (d)1:2
- (e)none of these

**Hint**

$30\%\text{offirstnumber}+\text{secondnumber}=\text{secondnumber}+\frac{1}{5}\left(\text{secondnumber}\right)$

**Solution:**

Let the first and second numbers be $x\text{and}y$, respectively.

As per the question,

$30\%\text{of}x+y=y+\frac{1}{5}\left(y\right)\phantom{\rule{0ex}{0ex}}\frac{30x}{100}=\frac{y}{5}\phantom{\rule{0ex}{0ex}}\frac{x}{y}=\frac{100}{30\times 5}=\frac{2}{3}$Therefore, the required ratio is $2:3$.

- (a)4
- (b)5
- (c)6
- (d)7
- (e)none of these

**Hint**

If the unit's and ten's digits of a two-digit number are $y\text{and}x$, respectively. Then, the two-digit number can be written as $10x+y$.

**Solution:**

Let the unit's and ten's digits of the number be $y\text{and}x$, respectively.

The two-digit number will be $\left(10\times x\right)+y=10x+y$.

So, the number formed by interchanging the position of the digits will be $\left(10\times y\right)+x=10y+x$.

As per the question,

$\left(10x+y\right)-\left(10y+x\right)=45\phantom{\rule{0ex}{0ex}}\left(9x-9y\right)=45\phantom{\rule{0ex}{0ex}}x-y=\frac{45}{9}=5$Therefore, the difference between the digits is $5$.

- (a)18
- (b)24
- (c)22
- (d)36
- (e)none of these

**Hint**

Let the number is $x$

**Solution:**

Let the number be $x$.

As per the question,

$x-28=\frac{x}{3}\phantom{\rule{0ex}{0ex}}x-\frac{x}{3}=28\phantom{\rule{0ex}{0ex}}\frac{2x}{3}=28\phantom{\rule{0ex}{0ex}}x=14\times 3\phantom{\rule{0ex}{0ex}}x=42$Hence, th e required number is $50\%\text{of}42=\frac{50}{100}\times 42=21$.

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- (a)$Rs.510$
- (b)$Rs.700$
- (c)$Rs.190$
- (d)$Rs.290$
- (e)None of these

**Hint**

The total quantity of wheat$=50+20=70kg$

**Solution:**

Cost of the first quality of wheat will be$=(50\times 7)=Rs.350$

Cost of the second quality of wheat will be$=(20\times 8)=Rs.160$

Hence, the total costing of wheat will be $=(350+160)=Rs.510$

As per the given question,

The selling price of the total mixture will be$=(70\times 10)=Rs.700$

Therefore, profit will be$=(700-510)=Rs.190$

Hence, option C is the correct answer.

- (a)$Rs.9.00$
- (b)$Rs.12.00$
- (c)$Rs.8.50$
- (d)$Rs.8.00$
- (e)None of these

**Hint**

$Costperkgofrice=\frac{Total\mathrm{cos}tofrice}{Totalquantityofrice}$

**Solution:**

Cost of the first quality of rice will be$=(30\times 8.5)=Rs.255$

Cost of the second quality of rice will be$=(20\times 8)=Rs.160$

So, the total cost of rice will be$=255+160=Rs.415$

The total quantity of rice will be $=30+20=50kg.$

So, the cost of per kg rice will be$=\frac{415}{50}=Rs.8.3perkg.$

As per the given question,

Profit$=20\%$

Therefore, the selling price of rice will be

$\Rightarrow (100+20)\%\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{120}{100}\times 8.3\phantom{\rule{0ex}{0ex}}\Rightarrow Rs.9.96perkg\approx Rs.10perkg$

Hence, option E is the correct answer.

- (a)$Rs.6.80$
- (b)$Rs.7.00$
- (c)$Rs.8.00$
- (d)$Rs.6.00$
- (e)
None of these

**Hint**

$Totalsellingprice=Total\mathrm{cos}tprice+Profit$

**Solution:**

Cost of the first quality of rice will be$=(20\times 6.5)=Rs.130$

Cost of the second quality of rice will be$=(30\times 7)=Rs.210$

So, the total cost will be $=130+210=Rs.340$

As per the given question,

Profit$=Rs.60$

Therefore, the total selling price of rice will be$=(340+60)=Rs.400$

So, the selling price of one kg of rice will be$=\frac{400}{50}=Rs.8$

Hence, option C is the correct answer.

- (a)$27\mathrm{litre}$
- (b)$35\mathrm{litre}$
- (c)$25\mathrm{litre}$
- (d)$24\mathrm{litre}$
- (e)None of these

**Hint**

$\mathrm{Cost}\mathrm{per}\mathrm{litre}\mathrm{of}\mathrm{mixture}=\frac{\mathrm{Cost}\mathrm{of}\mathrm{total}\mathrm{quantiy}}{\mathrm{Total}\mathrm{quantity}}$

**Solution:**

Cost of $175L$ will be $=(175\times 8)=Rs.1400$

Let the volume of water added is$=x\mathrm{litre}$

As per the given question,

New quantity of water will be$=(175+x)L$

So,

$\Rightarrow \frac{1400}{175+x}=7\phantom{\rule{0ex}{0ex}}\Rightarrow 1400=1225+7x\phantom{\rule{0ex}{0ex}}\Rightarrow 7x=175\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{175}{7}\phantom{\rule{0ex}{0ex}}\Rightarrow x=25L$

Therefore, the quantity of water added$=25L$

Hence, option C is the correct answer.

- (a)$63kg$
- (b)$53kg$
- (c)$52kg$
- (d)$60kg$
- (e)None of these

**Hint**

$\mathrm{Cost}\mathrm{per}\mathrm{kg}\mathrm{of}\mathrm{mixture}=\frac{\mathrm{Cost}\mathrm{of}\mathrm{total}\mathrm{quantiy}}{\mathrm{Total}\mathrm{quantity}}$

**Solution:**

Let the first type of wheat is$=xkg.$

So, the cost of the first type of wheat will be$=(4\times x)=Rs.4x$

So, the cost of the second type of wheat will be$=(42\times 6)=Rs.252$

As per the given question,

$\Rightarrow 4x+252=(x+42)\times 4.80\phantom{\rule{0ex}{0ex}}\Rightarrow 4x+252=4.8x+201.6\phantom{\rule{0ex}{0ex}}\Rightarrow 0.8x=50.4\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{50.4}{0.8}\phantom{\rule{0ex}{0ex}}\Rightarrow 63kg.$

Hence, option A is the correct answer.

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- (a)$1\frac{4}{5}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
- (b)$2$
- (c)$3\frac{2}{5}$
- (d)$4\frac{1}{5}$
- (e)None of these

**Hint**

Increased area = $1.5\times \text{length}+1.2\times \text{breadth}$

**Solution:**

Let the original length and breadth of plot be $x$ and $y$, respectively.

So, the original area $=xy$

As per the question, the length and breadth are increased by $50\%$ and $20\%$, respectively.

So, the new length and breadth of the plot will be $1.5x$ and $1.2y$, respectively.

The new area $=1.5x\times 1.2y=1.8xy\phantom{\rule{0ex}{0ex}}=\frac{9}{5}=1\frac{4}{5}xy$

Hence, the new area will be $1\frac{4}{5}$ times of the old area.

- (a)10 cm
- (b)8 cm
- (c)6 cm
- (d)9 cm
- (e)None of these

**Hint**

Area of the rectangle $=\text{length}\times \text{breadth}=36{\mathrm{cm}}^{2}$

**Solution:**

Area of the square = Area of the rectangle.

Side of the square $=6\mathrm{cm}$

So, the area of the square $={\mathrm{side}}^{2}={6}^{2}=36{\mathrm{cm}}^{2}$

Let the length and breadth of the rectangle be $l$ and $b$, respectively.

Given that $36=l\times b$

Breadth of the rectangle, $b=6-2=4\mathrm{cm}$

Therefore, the length of the rectangle, $l=\frac{36}{4}=9\mathrm{cm}$

- (a)$5.2\mathrm{m}$
- (b)$6.8\mathrm{m}$
- (c)$5.8\mathrm{m}$
- (d)$6.2\mathrm{m}$
- (e)None of these

**Hint**

$\text{Breadth}\times 1.3\left(\text{Breadth}\right)=20.8{\mathrm{m}}^{2}$

**Solution:**

Let the breadth of the rectangular plot be $x$.

The length of the rectangular plot $=x+30\%\text{of}x=1.3x$

Area of the rectangular plot $=(l\times b)=(x\times 1.3x)=1.3{x}^{2}$

As per the question,

$1.3{x}^{2}=20.8{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}{x}^{2}=\frac{20.8}{1.3}=16{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\sqrt{16}=4\mathrm{m}$

Hence, the length of the rectangle $=1.3x=1.3\times 4=5.2\mathrm{m}$

- (a)1:3
- (b)13:10
- (c)3:1
- (d)4:7
- (e)None of these

**Hint**

The new length of the plot is $1.3$ times the original plot and the breadth remains unchanged.

**Solution:**

Let the original length and breadth of the plot be $l$ and $b$, respectively.

The original area of the plot $=lb$

As per the question, the length is increased by $30\%$ and the breadth remains unchanged.

Increased length of the plot $=l+30\%\text{of}l=1.3l$

Area of the plot thus formed $=1.3l\times b=1.3lb$

Hence, the required ratio $=1.3lb:lb=13:10$

- (a)$20\mathrm{cm}$
- (b)$25\mathrm{cm}$
- (c)$15\mathrm{cm}$
- (d)$30\mathrm{cm}$
- (e)None of these

**Hint**

Area of the circle is equal to area of square.

$\pi \times {\text{radius}}^{2}={\text{side}}^{2}$

**Solution:**

Area of the circle = Area of the square

The radius of the circle $=14\mathrm{cm}$

Area of the circle $={\mathrm{\pi r}}^{2}=\frac{22}{7}\times {14}^{2}\phantom{\rule{0ex}{0ex}}=616{\mathrm{cm}}^{2}$

So, the area of the square $={\mathrm{side}}^{2}$

Hence, the side of the square $=\sqrt{616}=24.81\mathrm{cm}\approx 25\mathrm{cm}$

Hence, The correct answer is $25cm.$

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- (a)$110\%$
- (b)$260\%$
- (c)$320\%$
- (d)$95\%$
- (e)None of these

**Hint**

Percentage $=\frac{\mathrm{Increased}\text{'}\mathrm{No}\mathrm{Supervision}\text{'}}{\mathrm{Increased}\mathrm{Efficiency}\mathrm{Increment}}\times 100\%$

**Solution:**

Increase from January to February in 'No Supervision'.

$=30-15=15$ thousandIncrease front January to February in 'Efficiency Increment' $=25-15=10$ thousand

Percentage $=\frac{15\times 100\%}{10}=150\%$

Hence, the correct answer is none of these.

- (a)Febraury and June
- (b)February and April
- (c)March and April
- (d)February and March
- (e)None of these

**Hint**

Compare total production of all the three groups in every month.

**Solution:**

For this question we need to check with options:

According to first option:

Total production of all the three groups in February $=15+25+30=70\text{thousand}$

Total production of all the three groups in April $=12.5+15+25=62.5\text{thousand}$

The total production in February and June are not same.

From the second option:

Total production of all the three groups in February$=15+25+30=70\text{thousand}$

The total production of all the three groups in April $=15+20+35=70\text{thousand}$

In both months the total production is same.

Hence, the correct answer is February and April.

- (a)16000
- (b)10000
- (c)20000
- (d)31000
- (e)None of these

**Hint**

$Average=\frac{Sumoftheterms}{Numberofterms}$

**Solution:**

Production in Feb. in 'No supervision' = 30 thousand

and average production in 'No supervision' from Feb. to June

$=\frac{30+35+35+30+25}{5}=31thousand$

Increase = 31 - 30 = 1 thousand

- (a)No supervision
- (b)All the three
- (c)Lunch Allowance and No supervision
- (d)Lunch Allowance
- (e)Efficiency Increment and No supervision

**Hint**

Compare the data of June month with May month.

**Solution:**

From the data given in the question,

The impact on launch allowance in the month of May is $17500$.

The impact on launch allowance in the month of June is $12500$.

Therefore, the decrease in launch allowance is $17500-12500=5000$.

The impact on no supervision in the month of May is $30000$.

The impact on no supervision in the month of June is $25000$.

Therefore, the decrease in no supervision is $30000-25000=5000$.

The impact on efficiency increment in the month of May is $20000$.

The impact on efficiency increment in the month of June is $15000$.

Therefore, the decrease in efficiency increment is $20000-15000=5000$

Therefore, in the month of June, the decrease of each is $5000$.Hence, the correct answer is All the three.

- (a)$120000$
- (b)$100000$
- (c)$105000$
- (d)$150000$
- (e)None of these

**Hint**

Add all production from Feb to June in 'Efficiency Increment'.

**Solution:**

According to the question:

Total production from February to June in 'Efficiency Increment'

$=(25+25+20+20+15)$ thousand

$=105000$

Hence, the correct answer is $105000$.

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- (a)OFILGAE
- (b)EOAILGF
- (c)FGLAIOE
- (d)FILOGAE

**Hint**

Take the given word and compare it with the coded word and rearrange the places of odd numbered letters for the word and leave the even numbered letters the same.

**Solution:**

First, take the word and compare it with NEISNOP.

From here, we can depict that the first letter is interchanged with the last letter and then the third letter is being interchanged with the third from the end. The position of odd numbers letters is being interchanged and even number letters remain the same. Hence, this rule is followed by the word FOLIAGE.

This is the word formed after applying the rule, the answer is EOAILGF.

- (a)ROTAGILA
- (b)ROTAGIAL
- (c)ROTAGILE
- (d)None of these

**Hint**

Take the given word and compare it with the code given and for the other word inverse the letters in the way it is done for the given code word.

**Solution:**

First take the word TELEPHONE which is written as ENOHPELET and find the code which is used.

This gives us the coded word ENOHPELET. Now, use this principle for the word ALIGATOR. The word needs to be just taken and inverse it, which means the code is written in the reverse order.

ALIGATOR becomes ROTAGILA.

- (a)6198
- (b)6395
- (c)6217
- (d)6285

**Hint**

First, decode the words and find out which letter has been coded to which number and then find the code for each letter of MALE and join it.

**Solution:**

PLAY has the code 8123

RHYME has the code 49367

Since the number of letters of the word and numbers of the code are the same which means each letter is assigned to the number.

P-8, L-1, A-2, Y-3 and similarly R-4, H-9, Y-3, M-6, E-7.

Now, we know the number codes for M, A, L, E.

Hence, the code for MALE is 6217.

- (a)863584
- (b)863063
- (c)863903
- (d)863203

**Hint**

Corresponding letters are assigned with unique number.

**Solution:**

In this case take the words TWENTY, ELEVEN and check the number of that with the code if it is equal. Then assign each letter to a number as per the code. Now, compare the letters and code to get each value.

TWENTY - 863985 which means T-8, W-6, E-3, N-9, Y-5.

Similarly, ELEVEN has the code 323039 which means E-3, L-2, V-0, N-9.

Now, T has code 8, W has code 6, E has code 3, L has code 2, V has code 0. Hence, the code for TWELVE is 863203.

- (a)needle
- (b)nib
- (c)lead
- (d)thread

**Hint**

Nib is used in a pen to write.

**Solution:**

According to the question, the names of things are being coded to other words. In this case, the code of lead is stick, code of stick is nib and code name of nib is needle, code of needle is rope and rope is thread.

Nib is used in a pen to write. But the nib is coded as needle.

Hence, the answer is needle.

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- (a)A
- (b)B
- (c)C
- (d)D
- (e)E

**Hint**

Investing on advancement of technology related machine is essential for growth.

**Solution:**

Argument (I) is weak, because investing money on computers are not waste of money, it's upgrading the work with the help of technology which make task easier and faster.

Argument (II) does not produce any fact. It is just conveying that it is found to be used in advanced countries, India should also follow. This does not justify fact or right reason.

Hence, none of the statements are strong.

- (a)A
- (b)B
- (c)C
- (d)D
- (e)E

**Hint**

It is the rights of every citizen to contribute his opinion in selection of members of the constitution and there is no partiality between educated and uneducated people.

**Solution:**

Argument (I) is completely negative and there are no facts pointed out to argue with the given statement.

Whereas Argument (II) talks about the fact. That is, constitutional rights, which provides rights for every citizen to vote. Both illiterate and literate has the power to make decision, analyze the right and wrong. So illiterate are also eligible for voting. So only Argument (II) is strong.

- (a)1
- (b)2
- (c)3
- (d)4
- (e)5

**Hint**

Connecting remote places is advantageous to the people.

**Solution:**

An argument should always quote the fact. Argument I is completely invalid because connecting remote parts will ease the communication and movements. Argument II is positive but there is no fact or reason pointed out, but the action is expected to happen immediately. Hence, it is weak.

None of the given arguments are strong.

- (a)1
- (b)2
- (c)3
- (d)4
- (e)5

**Hint**

Sex education is necessary to protect children or youth from harassment and deeper knowledge is not required at the earlier stage.

**Solution:**

Argument I is right and it is also associated with the fact considering the impact on the young minds and blooming buds. Whereas when we consider Argument II, it is entirely false. The education of adult featured film is not necessary for youths and children at this age. So, Argument I is strong and Argument II is weak.

Hence, option A is correct.

- (a)1
- (b)2
- (c)3
- (d)4
- (e)5

**Hint**

Effect of pollution can be anywhere.

**Solution:**

According to the question:

Argument I and Argument II are expressed with proper facts which concentrate on pollution.

Installation of chimneys anyway creates pollution. Argument I is strong and Argument II is also strong. Both are right in their own way.

Hence, option C is correct.

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- (a)If only I follows
- (b)If only II follows
- (c)If either I or II follows
- (d)If neither I nor II follows
- (e)If both I and II follow

**Hint**

People working in a certain process must tackle their problems themselves, as they know the best about their situation.

**Solution:**

The second statement does not provide any solution to the main statement as it talks about increase in air export which is a good symbol but it does not affect the problem.

On the other hand Statement (I) is a follow up of the problem given in the main statement and tries to give a solution to the problem by advising that the airlines and the agents should work together and find a way to solve the problem such that the work becomes easier for the agents.

Hence, option A is correct.

- (a)If only I follows
- (b)If only II follows
- (c)If either I or II follows
- (d)If neither I nor II follows
- (e)If both I and II follow

**Hint**

Exports help in increasing the economy of a country.

**Solution:**

In this case, Statement 2 is not a solution or follow up to the main statement as reducing export would further destroy the economy as more the exports, more will be the economy.

While the Statement 1 is a follow up to the main statement and tries to give a proper solution in order to increase share in global agricultural market by increasing the overall agricultural production in the country as it would also increase the export.

Hence, option A is correct.

- (a)If only I follows
- (b)If only II follows
- (c)If either I or II follows
- (d)If neither I nor II follows
- (e)If both I and II follow

**Hint**

Tourism must not be restricted to certain areas and must be fully explored.

**Solution:**

In this case, the second statement is a good follow up as it is mentioned that India is endowed with tourist spots so definitely it should be exploited to its full potential.

While the first statement is not a proper follow up as building resorts only in the coastal areas would not utilise the full potential of tourism of India and the other important destinations will lack proper accommodations.

Hence, option B is correct.

- (a)only 1 follows
- (b)Only 2 follows
- (c)Only 3 follows
- (d)Only 1 and 3 follow
- (e)Only 2 and 3 follow