### Contents of ESIC Upper Division Clerks & Multi Tasking Staff Recruitment Exam Phase-I

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

#### 50 Chapters

- 1. Common Errors
- 2. Antonyms
- 3. Synonyms
- 4. Sentence Completion
- 5. One Word Substitution
- 6. Comprehension
- 7. Passage Completion
- 8. Completion of Paragraphs & Sentences
- 9. Percentage
- 10. Profit and Loss
- 11. Ratio and Proportion
- 12. Partnership
- 13. Average
- 14. Simple Interest
- 15. Compound Interest
- 16. Discount, Stock and Shares
- 17. Work and Time
- 18. Speed, Time and Distance
- 19. Unitary Method
- 20. Problems Based on Ages
- 21. Problems Based on Fraction
- 22. Problems Based on Numbers
- 23. Alligation
- 24. Area
- 25. Interpretation of Data
- 26. Paper Folding Type Questions
- 27. Completion of a Figure by its Quadrand
- 28. Embedded Figures
- 29. Figure and Pieces
- 30. To make the Figure with Pieces given
- 31. To Find Out the Fourth Figure
- 32. To Find the Concealed Figure
- 33. Analogy Test
- 34. To Find a Missing Term in a Series
- 35. To Find the Odd Number
- 36. Meaningful Words with the Selected Letters of a Given Word
- 37. Coding-Decoding Test
- 38. Direction Sense Test
- 39. Blood Relation
- 40. Arranging in Proper Sequence
- 41. Assigning Artificial Values and Missing Number
- 42. Matrix Type Test
- 43. Logic
- 44. Problems on Alphabet
- 45. Calendar and Clock
- 46. Venn Diagram Type Test
- 47. To Find Figures From the Picture
- 48. To Find out the Components of the Given Figure
- 49. Grouping of Figures into Classes
- 50. Miscellaneous

#### 5 Topics

Section A (Articles, Nouns, Pronouns) Some Important Rules

Section B (Adjectives, Adverbs, Adverbial order) Some Important Rules

Section C (Verb, Infinitive, Verbal noun, Gerund, Participle) Some Important Rules

Section D (Conjunctions, Prepositions) Some Important Rules

Section E (Miscellaneous Sentences) Some Important Rules

#### 1 Topics

Sentence Completion

#### 1 Topics

One Word Substitution

#### 10 Topics

Passage I

Passage II

Passage III

Passage IV

Passage V

Passage VI

Passage VII

Passage VIII

Passage IX

Passage X

#### 1 Topics

Re-ordering of Sentences, Phrases

#### 1 Topics

Ratio and Proportion

#### 1 Topics

Compound Interest

#### 1 Topics

Discount, Stock and Shares

#### 5 Topics

Relative speed

Average speed

Some Instructions in concern with train

Boats and Stream

Speed, Time and Distance

#### 1 Topics

Problems Based on Ages

#### 1 Topics

Problems Based on Fraction

#### 1 Topics

Problems Based on Numbers

#### 3 Topics

Units of Measuring Length

Units of Measuring Area

Short-Cut Method

#### 1 Topics

Interpretation of Data

#### 1 Topics

Paper Folding

#### 1 Topics

Completion of a Figure by its Quadrand

#### 1 Topics

Embedded Figures

#### 1 Topics

Figure and Pieces

#### 1 Topics

To make the Figure with Pieces given

#### 1 Topics

To Find Out the Fourth Figure

#### 1 Topics

To Find the Concealed Figure

#### 2 Topics

Analogy Test

Number Analogy Test

#### 1 Topics

To Find a Missing Term in a Series

#### 1 Topics

To Find the Odd Number

#### 1 Topics

Meaningful Words with the Selected Letters of a Given Word

#### 1 Topics

Coding-Decoding Test

#### 1 Topics

Direction Sense Test

#### 1 Topics

Arranging in Proper Sequence

#### 1 Topics

Assigning Artificial Values and Missing Number

#### 3 Topics

Logic

First Type

Third Type

#### 1 Topics

Problems on Alphabet

#### 1 Topics

Calendar and Clock

#### 1 Topics

Venn Diagram Type Test

#### 1 Topics

Figures From the Picture

#### 1 Topics

Components of the Given Figure

#### 1 Topics

Grouping of Figures into Classes

#### 2 Topics

Exercise I

Exercise II

## Experience Tests Tailored to Match the Real Exam

## Practise Questions with Solutions from the ESIC Upper Division Clerks & Multi Tasking Staff Recruitment Exam Phase-I

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

**Hint**

In a sentence, the subject and verb must be in agreement.

**Solution:**

'Many' and 'many a' are both adjectives and convey the same meaning. The difference is that 'many' is used with countable, plural nouns followed by plural verbs whereas, 'many a' is used with singular, countable noun and takes a singular verb with it.

In the given sentence, 'many a' is linked with 'man' and 'have'. Therefore, 'have' must be replaced with 'has'.

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

**Hint**

Some collective noun dont take 's' after them, they are already in plural form.

**Solution:**

'Cattle' is a collective noun,which is already in plural form. In the sentence 's' is added to 'cattle' to make it plural,so it is incorrect.

So, change 'cattles' by 'cattle'.

Eg: Cattle are source of income for rural households.

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

**Hint**

'Detail' is used as adverb in the sentence.

**Solution:**

'Details' is used as an adverb here, which is representing the quality of verb "told' in the sentence.

As it is used as 'adverb' it should not be used in plural form.

So, change 'details' to 'detail.'

Eg: (as adverb) He told me the story in detail.

(As noun) Project details were shared to the team.

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

**Hint**

Plural of compound words are formed by adding 's' in the first constituent of the compound words.

**Solution:**

'Sister-in-law' is a compound word and to convert compound words in plural,we need to add 's' to the first constituent of the letter.In this case 's' would be added to 'sister'.

So,change 'sister-in-laws' to 'sisters-in-law'.

Eg: My sisters-in-laws are coming tomorrow.

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

**Hint**

The subject and verb should be in agreement.

**Solution:**

'Many' and 'many a' both are adjectives which are used as quantifiers.

Many is used with plural countable noun where as 'many a' is used in case of singular countable noun.

In the sentence 'many a' is used with 'person' and 'have',which is wrong.

So,change 'have' to 'has'.

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

**Hint**

Opposite of 'not to do something'.

**Solution:**

To abstain means to hold back oneself or not to do something. The clear opposite of 'abstain' is 'to start something' i.e. begin.

The meaning of the other 3 words are:

Hoard means to accumulate something or to preserve something.

Tolerate means to allow the presence of something or to permit.

Forbear means to abstain from or to withhold something.

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

**Hint**

Synonym of profusion is 'abundance'.

**Solution:**

Profusion means to be present in abundant quantity; extravagance.

Hence, scarcity is the clear opposite of it. It means the state of being scarce or in short supply; shortage.

The meaning of the other 3 words are:

Aspersion means to slander i.e. to make derogatory remarks to attack on reputation or integrity.

Aversion means disinclination or dislike towards something or someone.

Confusion means the state of being confused i.e. lack of clearness.

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

**Hint**

Obscure means ambiguous.

**Solution:**

Obscure means something that is hard to perceive; ambiguous or vague.

Explicit means clearly explained or leaving nothing to be implied. Hence, it could be called as the antonym of obscure.

The meaning of the other 3 words are:

Implicit means to imply something rather than explaining completely.

Obnoxious means something that is not pleasant; offensive.

Pedantic means extremely detailing or overly concerned with minute details.

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

**Hint**

Consider this example:

He has a very repulsive nature and cannot be charming at all.

**Solution:**

Repulsive means to keep away or to be at a distance; a tendency to drive away.

Alluring means tempting or something that attracts. Therefore, alluring is opposite of repulsive.

The meaning of the other 3 words are:

Refulgent means to shine brightly; radiant.

Effulgent is a synonym of 'refulgent'. It means shining brightly; radiant.

Meek means to be submissive; docile.

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

**Hint**

Synonym of ambiguous is uncertain.

**Solution:**

Ambiguous means something that is uncertain; having more than one possible interpretations.

Its antonym is 'clear' which means to be transparent; or free from ambiguity.

The meaning of the other 3 words are:

Auxiliary means supplementary; used as a substitute.

Responsible means answerable or accountable.

Salvage means to save something or act of saving something; and if you can save something successfully it is called salvageable.

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- (a)Briskness
- (b)Fear
- (c)Frankness
- (d)Alarm
- (e)None of these

**Hint**

Its other synonyms are smartness, willingness, etc.

**Solution:**

Alacrity means brisk and cheerful readiness.

Therefore, the correct answer is briskness. It means liveliness or readiness.Meanings of other words are:

Fear: to be afraid of

Frankness: the quality of being open, honest, and direct in speech or writing

Alarm: a warning or a warning sound

- (a)Active
- (b)Idle
- (c)Occupied
- (d)Diligent
- (e)None of these

**Hint**

Other synonyms are engaged, employed.

**Solution:**

Busy means having a great deal to do.

Therefore, the correct answer is occupied, which means busy in some work.Meanings of other words are:

Active - agile, energetic

Idle - lazy, slothful

Diligent - hardworking

- (a)Sourness
- (b)Hoarseness
- (c)Acrimony
- (d)Aspersion
- (e)None of these

**Hint**

Other synonyms are resentment, disappointment.

**Solution:**

Bitterness and acrimony are synonyms.

They mean anger and disappointment at being treated unfairly.

Sourness - disappointed

Hoarseness - rough, harsh

Aspersion - an attack on the reputation of someone

- (a)Dry
- (b)Barren
- (c)Childless
- (d)Arid
- (e)None of these

**Hint**

Other synonyms are infertile, fruitless.

**Solution:**

Barren is the right answer.

Sterile or barren means not able to produce children or young ones.

Meanings of other words are:

Dry - free from moisture or liquid

Childless - having no child

Arid - dry, having no or little rain

- (a)Choice
- (b)Charge
- (c)Heated dispute
- (d)Distribution
- (e)None of these

**Hint**

Other synonyms are quarrel, fight.

**Solution:**

Heated dispute is the right answer.

Altercation or heated dispute means a noisy argument or disagreement, especially in public.

Choice - option, alternative

Charge - accuse (someone) of something

Distribution - to share something out among a number of recipients

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- (a)has
- (b)have
- (c)is
- (d)are

**Hint**

Luggage is a collective noun.

**Solution:**

'Luggage' is a collective noun that takes the singular form. Therefore, it is always followed by a singular form of verb and since sentence is in present perfect tense. It takes the form, have/has + the past participle. So, 'has' is the right answer.

- (a)player was
- (b)players were
- (c)players was
- (d)players are

**Hint**

'One of the' is always used with plural form of noun.

**Solution:**

'One of the' is used with plural nouns, so 'players' is right and since the sentence is in past form and the subject is just 'one player' 'was' is used.

- (a)our friend
- (b)our friends
- (c)ours friend
- (d)ours friends

**Hint**

Possessive form of pronoun is not followed by a noun.

**Solution:**

The given expression in C and D is wrong because, the possessive form 'ours' is not followed by any noun. Secondly, the noun should be plural. Thus, option B is correct.

- (a)his seat
- (b)their seat
- (c)their seats
- (d)one's seat

**Hint**

Everyday is an indifinite pronoun.

**Solution:**

Everybody is an indefinite pronoun. As it is indefinite, we cannot determine the gender, and use a plural possessive “they/their”. It is used with singular verb. Thus, the correct answer is part B

- (a)is
- (b)was
- (c)were
- (d)has been

**Hint**

Use the form of auxilary verb which is used in case of imagination.

**Solution:**

'Were' is used as helping verb with any number of subjects, when we imagine, dream or pretend. For example, if I were an old man...

So, C is correct answer.

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

**Hint**

The word is originated from Latin word 'askein' which means ‘to exercise’.

**Solution:**

The correct answer is 'Ascetic'.

Ascetic means characterised by severe self-discipline and abstention from all forms of indulgence, typically for religious reasons.

The meanings of the other words are:

Sceptic means a person inclined to question or doubt accepted opinions.

Devotee means a person who is very interested in and enthusiastic about someone or something.

Antiquarian means relating to or dealing in antiques or rare books.

- (a)Apostate
- (b)Prostate
- (c)Profane
- (d)Agnostic

**Hint**

Prostate is a gland within the male reproductive system that is located just below the bladder.

**Solution:**

The correct answer is 'apostate'. Apostate means a person who renounces a religious or political belief or principle.

The meaning of other words are:

Prostate- a gland surrounding the neck of the bladder in male mammals and releasing a fluid component of semen.

Profane- not relating to that which is sacred or religious

Agnostic- a person who believes that nothing is known or can be known of the existence or nature of God.

- (a)Bibliophile
- (b)Philologist
- (c)Misogynist
- (d)Misologist

**Hint**

The prefix 'biblio-' is related to books.

**Solution:**

The correct answer is misologist which means a person who hates learning and knowledge.

The meaning of other words are:

Bibliophile-a person who collects or has a great love of books.

Philologist-A philologist is someone who studies the history of languages, especially by looking closely at literature.

Misogynist-a person who dislikes, despises, or is strongly prejudiced against women.

- (a)Pun
- (b)Transferred epithet
- (c)Oxymoron
- (d)Alliteration

**Hint**

She sells sea-shells on the sea shore. Consider this sentence while solving the given question.

**Solution:**

The correct answer is alliteration which means the occurrence of the same letter or sound at the beginning of adjacent or closely connected words.

The meaning of other words are

1. Pun-a joke exploiting the different possible meanings of a word or the fact that there are words which sound alike but have different meanings.

2. Transferred epithet-A transferred epithet is a little known—but often used—figure of speech in which a modifier (usually an adjective) qualifies a noun other than the person or thing it is actually describing.

3. Oxymoron-a figure of speech in which apparently contradictory terms appear in conjunction (e.g. faith unfaithful kept him falsely true).

- (a)Theist
- (b)Heretic
- (c)Atheist
- (d)Fanatic

**Hint**

The antonym of this word is theist.

**Solution:**

The correct answer is atheist which means a person who does not believe in the existence of God.

The meaning of other words are:

Thiest: Someone who believes in God.

Heretic: A person believing in or practicing religious heresy.

Fanatic-a person filled with excessive and single-minded zeal.

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- (a)PQSR
- (b)QPSR
- (c)SRQP
- (d)RQPS

**Hint**

The sentence is about the sick Rodin and his desire to see his 'other' wife.

**Solution:**

The passage talks about Rodin, laid old and infirm on a sick bed who wanted to see his wife, hence the correct Statement after 1st part will be P.

Once the subject is clear, we will move further to the answer received by Rodin in response to his previous want. Thus the Statement Q follows.

In further response about Rose Beuret, Rodin’s fretful denial has been projected, hence Statement S follows.

Rodin further asks for the other lady, hence the remaining Statement R will follow.

The correct order after part 1 will be PQSR.

- (a)PSRQ
- (b)SRQP
- (c)RQPS
- (d)QPSR

**Hint**

The sentence is about a nurse, her job and the man for whom she works.

**Solution:**

1. The passage talks about Camilte who was not content to be a nurse and wanted an independent career, hence the correct Statement after 1st part will be P.

2. Further, the Statement S elaborates the characteristics of Camilte, hence Statement P will be followed by Statement S.

3. Statement R further add to the works of Camilte being more contained and intimate and less bombastic, hence Statement R will follow.

4. As Statement Q denotes the fundamental differences of character of the two, hence the remaining Statement Q follows.

he correct order after part 1 will be PSRQ.

- (a)PSRQ
- (b)QPSR
- (c)RQPS
- (d)SRQP

**Hint**

The sentence is about a boy who is spaceman and his interaction with two individuals.

**Solution:**

The passage talks about the boy who says modestly that he is a spaceman.

1. Being unheard once, he repeats himself. Hence Statement R will follow the first part.

2. The boy further denotes being from another planet. Hence Statement Q.

3. The boy further elaborates about the planet which can’t be seen, in Statement P, hence it will follow, which also adds to Statement Q.

4. Statement S denotes the response of George and Cathy after listening to the entire conversation of the boy, hence the remaining Statement S.

The correct order after part 1 will be RQPS.

- (a)RQPS
- (b)QPSR
- (c)PSRQ
- (d)SRQP

**Hint**

The sentence is about a man's efforts to escape and his thinking.

**Solution:**

1. The passage talks about the person who has been in an escape from his enemy and further his bicycle had been seen by the aeroplane, as Statement S plots for the situation. Hence Statement S will follow part 1.

2.Statement S will be followed by Statement R which denotes about the savage thinking by the person, hence R.

3. Statement R will be followed by Statement Q which connects Statement R about the thoughts of the person, hence Q.

4. Statement P further elaborates about the worries of the person of being caught, hence Statement P.

The correct order after part 1 will be SRQP.

- (a)RQPS
- (b)SRQP
- (c)QPSR
- (d)PSRQ

**Hint**

The sentence is about an ascetic who has abjured everything and a girl's concern for him.

**Solution:**

The passage talks about the person in penance abjuring even fruit.

1. Statement P will follow part 1 as it enhances the level of abjurement.

2. Further Statement S will follow which denotes about the girl being sorrowful about the state of the man who has abjured everything. She thinks what she can do.

3. Further Statement R will follow which marks girl gathering the wild blossoms.

4. Further Q also talks about blossoms as they remain rotting at his feet, untouched, being pecked by the birds.

Hence Statement Q will follow.

The correct order after part 1 will be PSRQ.

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- (a)$35$
- (b)$65$
- (c)$50$
- (d)$30$
- (e)None of these

**Hint**

Use the given formula:

Winning$\%=\frac{Gamewon}{Totalgame}\times 100$

**Solution:**

They have won $13$ games and lost $7$ games.

So, $W=13$

$L=7$

Total games $=13+7=20$

They have won $13$ of their games.

Winning percentage $=\frac{13}{20}\times 100=65\%$

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

**Hint**

Find $35\%$ of $60$, then convert it into $\%$ with respect to $400$.

**Solution:**

According to the question, we have:

$60\%$ of $35=\left(\frac{60}{100}\right)\times 35=21$

Now, converting $21$ in $\%$ of $400$

$=\frac{21}{400}\times 100=5\frac{1}{4}$

Hence, $5\frac{1}{4}$ is the correct answer.

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

**Hint**

Use the given formula:

$x\%$ of $y=\frac{x}{y}\times 100$

**Solution:**

Given that,

$1$ win, then $49$ lose.

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

$=98\%$

- (a)$500cu.$ metre
- (b)$50cu.$ metre
- (c)$5cu.$ metre
- (d)$5000cu.$ metre
- (e)None of these

**Hint**

Get the value of argon using :

$\%$ of Nitrogen $+$ $\%$ of Oxygen $+$ $\%$ of light gases $+$ $\%$ of Argon $=100$

**Solution:**

Argon $\%=100-\left(79.2+20.7+0.08\right)$

Argon $=0.02\%$ cubic meter of air

Now, we have $0.02$ cubic metre of Argon in $100$ cubic metre of air.

For $1$ cubic meter of Argon we should have

$=\frac{100}{0.02}=5000$ cubic metre.

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

**Hint**

Let the total students be $x$, and $49.3\%$ of $x=23128$

**Solution:**

Let the total students be $x$.

Students passed $=49.3\%$

So,

$x\times 49.3\%=23128$

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

$=46912.77$

$=46912.77$

$=46913$ (approx)

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- (a)$14$
- (b)$11$
- (c)$10$
- (d)$12$
- (e)None of these

**Hint**

Let $SP$ of oranges in first case be $x$, then $CP=0.88x$. Use this $CP$ to calculate $SP$, when profit $=12\%$.

**Solution:**

Let $SP$ of oranges be $x$.

Loss $=12\%$

Loss$\%=\frac{CP-SP}{CP}\times 100$

$12=\frac{CP-x}{CP}\times 100$

$CP=\frac{x}{0.88}$

To gain $12\%$

$SP=CP$ $+$ Profit

$=\frac{x}{0.88}+0.12\times \frac{x}{0.88}$

$=1.2727x$

So, in $Rs.1.2727=14$ oranges

$x=\frac{14}{1.2727}=11$ oranges

- (a)$21\frac{1}{2}\%$
- (b)$22\%$
- (c)$23\frac{3}{4}\%$
- (d)$26\frac{1}{7}\%$
- (e)None of these

**Hint**

According to question: $MP=1.3CP$ and $SP=(1-.0625)MP$

**Solution:**

Since price is marked $30\%$ above

So, $MP=1.30CP$

Discount $=6\frac{1}{4}\%=6.25\%$

So,

$SP=(1-.0625)MP=(1-.0625)1.3CP$

$SP=1.21875\hspace{0.17em}CP$

Gain$\%=21.875\%$

- (a)$10\%$
- (b)$11\%$
- (c)$20\%$
- (d)$21\%$
- (e)None of these

**Hint**

He earns profit in two stages, one at the time of buying and the other one while selling.

**Solution:**

Since, he defaults $10\%$ in buying, his earning $=1.10CP$

He also defaults $10\%$ in selling, new earning $=1.1CP\times 1.1$

$=1.21CP$

So, he gains $=21\%$

- (a)Gain of $Rs.90$
- (b)Loss of $Rs.90$
- (c)Gain of $Rs.150$
- (d)None of these
- (e)No loss no profit

**Hint**

Since, profit on first horse is $15\%$, $SP=1.15CP$.

Similarly, price of second horse $=0.85CP$.

**Solution:**

First horse $SP=CP\times 1.15$

Second horse $SP=CP\times .85$

Total cost price $=\frac{1955}{CP\times 1.15}+\frac{1955}{CP\times 0.85}$

$CP=4000$

Loss $=$ Cost price $-$ Selling price

Loss $=4000-3910=Rs.90$

- (a)$8\%$
- (b)$5\%$
- (c)$10\%$
- (d)No profit No loss
- (e)None of these

**Hint**

Calculate the cost of two varieties separately and add them to get the total cost price.

**Solution:**

Price of $26kg$ tea $=26\times 2=Rs.52$

Price of $30kg\times 3.6=Rs.108$

Price of mixture $=52+108=Rs.160$

Selling price of mixture $(26+30)\times 3=168$

Profit $=168-160=Rs.8$

Profit$\%=\frac{8}{160}\times 100=5\%$

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

**Hint**

Use the formula:

Ratio $=\frac{Antecedent}{Consequent}$

**Solution:**

Let the consequent be $x$.

According to question:

$\frac{2}{3}=\frac{18}{x}$

$x=27$

Ratio $=18:27$

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

**Hint**

To be in proportion$(::)$, ratio of both sets should be equal.

**Solution:**

Let the number be $x$.

$\frac{{\displaystyle \frac{3}{2}}}{{\displaystyle \frac{5}{6}}}=\frac{{\displaystyle \frac{8}{7}}}{x}$

Multiply both side by $\frac{5}{6}$

$\frac{{\displaystyle \frac{2}{3}}}{{\displaystyle \frac{5}{6}}}\times \frac{5}{6}=\frac{{\displaystyle \frac{8}{7}}}{{\displaystyle x}}\times \frac{5}{6}$

$\Rightarrow \frac{2}{3}x=\frac{40}{42}$

$\Rightarrow x=\frac{10}{7}$

Hence, $\frac{10}{7}$ is the correct answer.

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

**Hint**

To find $A:D$, we have to make one value common in both the ratios.

**Solution:**

To find $A:D$, we have to make one value common in the two ratios.

$A:B=2\times 8:3\times 8=16:24$

$B:C=4\times 6:5\times 6=24:30$

$C:D=6\times 5:7\times 5=30:35$

$A:B:C:D=16:24:30:35$

$A:D=16:35$

- (a)$1:5$
- (b)$7:17$
- (c)$3:9$
- (d)$2:5$
- (e)None of these

**Hint**

Assume the quantity of glass such that it is divisible by both $3$ and $4$. Find the quantity of water and milk, and consequently ratio.

**Solution:**

Assume both glass to be of $12$ litre.

Milk in first glass $=\frac{1}{3}\times 12=4$ litre

Water in first glass $=8$ litre

Milk in second glass $=\frac{1}{4}\times 12=3$ litre

Water in second glass $=9$ litre

Total volume $=24$ litres

Total volume of water $=9+8=17$litre

Total volume of milk $=4+3=7$ litre

Ratio $=7:17$

- (a)$40$ litres
- (b)$10$ litres
- (c)$30$ litres
- (d)$20$ litres
- (e)None of these

**Hint**

Find the initial quantity of water and milk from the ratio given, then add the quantity of water according to question to make it in ratio $7:3$.

**Solution:**

Total volume of mixture $=30$ litres

Ratio of milk and water $=7:3$

Volume of water $=\frac{3}{10}\times 30=9$ litres

Volume of milk $=\frac{7}{10}\times 30=21$ litres

Let the total volume be $x$, when ratio changes to $3:7$

Volume of milk will remain same.

So, $\frac{3}{10}\times x=21$

$x=70$ litres

So, Volume of water in new mixture $=70-21=49$ litres

Volume of water added $=49-9=40$ litres

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- (a)$Rs.270,396,180$
- (b)$Rs.396,180,270$
- (c)$Rs.396,270,180$
- (d)$Rs.296,370,180$
- (e)None of these

**Hint**

Profit will be shared in the ratio of investment.

**Solution:**

Assume, $C$ subscribed at amount $X$

$B$ subscribed at $X+500$

$A$ subscribed at $X+500+700$

So,

$X+X+500+(X+500)+700=4700$

$3X+1700=4700$

$3X=4700-1700$

$3X=3000$

$X=\frac{3000}{3}$

$X=1000$

So,

$C$ subscribed $1000$

$B$ subscribed $1500$

$A$ subscribed $2200$

Profit ratio will be $A:B:C=22:15:10$

So, the profit of $A,B$ and $C$ will be

$A=\frac{2200}{4700}\times 846=396$

$B=\frac{1500}{4700}\times 846=270$

$C=\frac{1000}{4700}\times 846=180$

- (a)$Rs.1800,1680$
- (b)$Rs.1680,1800$
- (c)$Rs.1700,1780$
- (d)$Rs.1780,1700$
- (e)None of these

**Hint**

Investment $x$ Time $=$ Profit sharing ratio

**Solution:**

$A$ $B$ $C$ Investment $1400$ Months invested $12$ $7$ $5$ Total investment for $A=1400\times 12=16800$

Assume the profit of $A,B$ and $C$ as $X$.

So, according to profit ratio it is $4X:3X:2X$.

For $A$, $4X=16800$

So, $X=4200$

Profit Share of $B=3\times 4200=12600$

Profit Share of $C=2\times 4200=8400$

Capital invested by $B=\frac{12600}{7}$ $=Rs.1800$

Capital Invested by $C=\frac{8400}{5}$ $=Rs.1680$

- (a)$4$ months
- (b)$8$ months
- (c)$12$ months
- (d)$6$ months
- (e)None of these

**Hint**

Profit Sharing ratio $=$ Investment $\times $ Time

**Solution:**

Assume the capital of $B$ was used for $X$ months.

According to the question:

$\frac{5}{6}\times \left(\frac{8}{X}\right)=\frac{5}{9}$

$X=12$ months

- (a)$Rs.7,150$
- (b)$Rs.3,060$
- (c)$Rs.9,180$
- (d)$Rs.1,440$
- (e)None of these

**Hint**

Assume profit as $X$ and Use formula :

$SI=\frac{P\times R\times T}{100}$

**Solution:**

Assume profit as$Rs.X$.

Amount paid to $B$ as salary $=\left(120\times 12\right)=Rs.1440$

Share of each $=X-\frac{1440}{2}$

Interest paid by $B=\frac{22500\times 10}{100}=Rs.2250$

Total money received by $A=X-\frac{1440}{2}+2250=Rs.X+\frac{3060}{2}$

Total money received by

$B=\left[\left(X-\frac{1440}{2}\right)+1440-2250\right]=\left(X-\frac{3060}{2}\right)\frac{1}{2}-\left(X+\frac{3060}{2}\right)-X-\left(\frac{3060}{2}\right)$

$X=Rs.9,180$

- (a)$Rs.4,000$
- (b)$Rs.5,000$
- (c)$Rs.3,837.50$
- (d)$Rs.3,937.50$
- (e)None of these

**Hint**

Use the concept:

Ratio of investment $=$ Ratio of profit.

**Solution:**

Ratio of the capital of two partners $=12,500:8,500=25:17$

Now, the interest has to be divided in the ratio

$=25:17$

Difference $=25-17=8$

If $Rs.8$ is the difference

Then total interest $=25+17=Rs.42$

But, The actual difference is $Rs.300$

Total interest $=\frac{42\times 300}{8}=Rs.1575$

This interest is $40\%$ of the total income

Hence, complete profit $=\frac{1575\times 100}{40}$

$=Rs.3937.50$

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- (a)$14$
- (b)$8$
- (c)$13$
- (d)$12$
- (e)None of these

**Hint**

Average $=\frac{\mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{the}\mathrm{numbers}}{\mathrm{Number}\mathrm{of}\mathrm{entries}}$

**Solution:**

Let the missing number be $x$.

Average $=\frac{\mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{numbers}}{\mathrm{Number}\mathrm{of}\mathrm{entries}}$

Sum of $5$ numbers $=$ Average $\times 5=8\times 5=40$

Sum of given $4$ numbers $=7+5+3+12=27$

Missing number $=40-27=13$

- (a)$17kg$
- (b)$26kg$
- (c)$20kg$
- (d)$31kg$
- (e)None of these

**Hint**

Use the formula:

Average $=\frac{sumofalltheentries}{numberofentries}$

**Solution:**

Let $A,BC$ represent their respective weights.

Then, we have

$A+B+C=(45\times 3)=135$ .....(i)

$A+B=(40\times 2)=80$ ..... (ii)

$B+C=(43\times 2)=86$ ....(iii)

Adding (ii) and (iii),

$A+2B+C=166$ .... (iv)

Subtracting (i) from (iv),

$B=31$ (i.e., $166-135$)

Therefore, $B$'s weight $=31kg$.

- (a)$16$
- (b)$\frac{48}{7}$
- (c)$20$
- (d)$18$
- (e)None of these

**Hint**

Use the formula:

Average $=\frac{sumofalltheobservations}{no.ofobservations}$

**Solution:**

Let the $4$ numbers be$A,B,C,D$.

According to the question,

$\frac{(A+B+C+D)}{4}=12.....\left(i\right)$ also,

$\frac{(A+B+C)}{3}=2D$

$A+B+C=6D$

Putting the value of $A+B+C$ in $\left(i\right)$,

$\frac{7D}{4}=12$

$D=\frac{48}{7}$

Hence, option $\left(B\right)$ is correct.

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

**Hint**

Use the formula :

Average $=\frac{sumofalltheobservations}{numberofobservations}$

**Solution:**

The average marks that Anil got in the first year is $76$.

Therefore, the total marks that Anil got in the first year $=\left(76\times 4\right)=304$

Average marks that Anil got in the ${2}^{{}_{nd}}$ year is$81$.

So, the total marks of ${2}^{{}_{nd}}$ year is $=(81\times 3)=243$

Total marks in both the years $=304+243=547$

The average marks of Anil for both the years $=\frac{547}{7}$

Hence, option (D) is correct.

- (a)Man $Rs.2.75$, Woman $Rs.2.25$
- (b)Man $Rs.3.25$, Woman $Rs.2.75$
- (c)Man $Rs.3$, Woman $Rs.2.50$
- (d)Man $Rs.2.50$, Woman $Rs.2$
- (e)None of these

**Hint**

Total wage of man and woman $=1000x-200$

**Solution:**

Let the wage of man $=Rs.x$

Wage of woman $=Rs.x-.50$

Total wage of man and woman $=600x+400x-200$

Average $=\frac{Totalwage}{Totalnumberofmenandwomen}$

$2.55=\frac{\left(1000x-200\right)}{1000}$

$2550=1000x-200$

$1000x=2750$

$x=Rs.2.75$ per day

Wage of woman $=Rs.2.75-0.50$ $=Rs2.25$ per day

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- (a)$Rs.540$
- (b)$Rs.640$
- (c)$Rs.600$
- (d)$Rs.700$
- (e)None of these

**Hint**

Use the formula:

SI $=\frac{P\times R\times T}{100}$

**Solution:**

Simple interest $=\frac{P\times R\times T}{100}$

Amount $=$ Principal $+$ SI

$720=P+\frac{P\times 5\times 2.5}{100}$

$720=P+\frac{25P}{200}$

$720=1.125\times P$

$P=\frac{720}{1.125}$

$P=Rs.640$

- (a)$40$ years
- (b)$20$ years
- (c)$60$ years
- (d)$80$ years
- (e)None of these

**Hint**

When the amount gets doubled, interest is equal to the principal.

**Solution:**

$SI=\frac{P\times R\times T}{100}$

In $20$ years it is doubled.

So, Amount $=2P$

Amount $=SI+P$

$2P=SI+P$

$P=SI$

Time $=20$ years

$P=\frac{P\times R\times T}{100}$

$R=5\%$

Amount after becoming $4$ times $=4P$

Amount $=SI+P$

$4P=SI+P$

$SI=3P$

$3P=\frac{P\times 5\times T}{100}$

$T=60$ years

- (a)$Rs.1,\mathrm{00,000}$
- (b)$Rs.1,\mathrm{20,000}$
- (c)$Rs.1,\mathrm{10,000}$
- (d)$Rs.1,\mathrm{30,000}$
- (e)None of these

**Hint**

Divide the yearly interest by $12$ to find the monthly interest.

**Solution:**

We know that,

$SI=\frac{P\times R\times T}{100}$

$600\times 12=\frac{P\times 6}{100}$

$P=Rs.1,20,000$

- (a)$8\frac{1}{2}\%$
- (b)$8\frac{1}{3}\%$
- (c)$8\%$
- (d)$8\frac{2}{3}\%$
- (e)None of these

**Hint**

Find the total interest in terms of rate and equate it to the total interest given.

**Solution:**

Let the rate of interest $=x\%$

Total interest in $3$ years $=Rs.960$

$SI=\frac{P\times R\times T}{100}$

Total interest $=$ Interest on $2000+$ Interest on $1600$

$\left(\frac{2000\times x\times 3}{100}\right)+\left(\frac{1600\times \left(x+2\right)\times 3}{100}\right)=960$

$108x+96=960$

$x=\frac{864}{108}$

$x=8\%$

- (a)$5:8$
- (b)$8:5$
- (c)$31:6$
- (d)$16:15$
- (e)None of these

**Hint**

Use formula:

$SI=\frac{P\times R\times T}{100}$

**Solution:**

Let the sum invested at $5\%=Rs.x$

Sum invested at $8\%=Rs.1550-x$

Total interest $=Rs.300$

$\frac{x\times 5\times 3}{100}+\frac{(1550-x)\times 8\times 3}{100}=300$

$x=Rs.800$

$1550-800=Rs.750$

Required ratio $=800:750=16:15$

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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**

Use formula:

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

**Solution:**

Using the compound interest formula for amount,

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

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

$A=500{\left(1+\frac{6}{100}\right)}^{2}=500{\left(\frac{106}{100}\right)}^{2}=Rs.561.80$

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

**Hint**

In the next $4$ years, it will become $4$ times.

**Solution:**

Suppose $Rs.100$ be the principal.

In $4$ years, it doubles and becomes $Rs.200$.

In next $4$ years, $Rs.200$ double to $Rs.400$.

In next $4$ years, $Rs.400$ doubles to $Rs.800$.

Hence, the period for $8$ times is $12$ years.

- (a)$Rs.50,000$
- (b)$Rs.60,000$
- (c)$Rs.65,000$
- (d)$Rs.55,000$
- (e)None of these

**Hint**

Difference in interest on $Rs.100$ is $Rs.0.16$.

**Solution:**

Let the money be $Rs.100$.

Interest on $Rs.100$ in first year $=4$

Interest on $Rs.100$ in second year $=100{\left[1+\frac{4}{100}\right]}^{2}=4.16$

Difference in interest $=4.16-4=.16$

$Rs.0.16$ is the difference when principal is $Rs.100$.

Difference when principal is $Rs.88$

$=100\times \frac{88}{0.16}=Rs.55,000$

- (a)$Rs.8,000$
- (b)$Rs.8,400$
- (c)$Rs.8,200$
- (d)$Rs.10,000$
- (e)None of these

**Hint**

Amount in both the cases is same.

**Solution:**

Let the share of younger and older son be $X$ and $16400-X$.

Interest time for younger son $=3$ years

Interest time for older son $=2$ years.

Amount in both the cases is same.

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

$X(1+\frac{5}{100}{)}^{3}=(16400-X\left)\right(1+\frac{5}{100}{)}^{2}$

$\Rightarrow X(1.05)=16400-X$

$\Rightarrow 2.05X=16400$

$X=Rs.8,000$

- (a)$4\%$
- (b)$3\%$
- (c)$5\%$
- (d)$6\%$
- (e)None of these

**Hint**

For the first year, $SI=CI$

**Solution:**

For the first year, CI and SI is same $=Rs.200$

$CI=A-P$

For the first year,

$200=P\left(1+r\%\right)-P=P\left(\left(1+r\%\right)-1\right).....\left(i\right)$

For the second year,

$410=P(1+r\%{)}^{2}-P=P((1+r\%{)}^{2}-1)$

$=P\left(\right(1+r\%)+1)\left(\right(1+r\%)-1).....\left(ii\right)$

Dividing $\left(ii\right)$ by $\left(i\right)$

$\Rightarrow 2.05=2+r$

$r=.05=5\%$

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

**Hint**

Use the formula:

$\frac{Totalinvestment}{Invetsmentof1share}=\frac{{\displaystyle Totalincome}}{{\displaystyle Incomeof1share}}=\frac{{\displaystyle Totalfacevalue}}{{\displaystyle Facevalueof1share}}$

**Solution:**

Assume face value $=Rs.100$

Dividend per share $=Rs.3$

By investing $Rs.1260$, he earns $Rs.84$

Investment needed to earn $Rs.3$ $=1260\times \frac{3}{84}=Rs.45$

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

**Hint**

One worker can make $20$ toys in one day.

**Solution:**

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

In $1$ day $14$ workers make $\frac{1400}{5}=280$ toys.

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

and number of total workers $=14+14=28$

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

$28$ workers make $1120$ toys $=\frac{(5\times 14\times 1120)}{(28\times 1400)}=2$ days

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

**Hint**

Use formula: $\frac{man\left(M\right)\times day\left(d\right)\times hour\left(H\right)}{work\left(W\right)}=$ constant

**Solution:**

$14$ men complete the work in $12$ days.

Work done by $14$ men in $4$ days $=\frac{4}{12}=\frac{1}{3}$

Remaining work $=\frac{2}{3}$

No. of men $=14+2=16$

So, $\frac{2}{3}$ of the work will be completed by $16$ men in $=\frac{12\times 14}{16}\times \frac{2}{3}$

$=7$ days.

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

**Hint**

Use the formula:

Together they can finish the work in: $\frac{xy}{(x+y)}$, where $x$ and $y$ is days taken by them to complete the work individually.

**Solution:**

Rohan can complete the work in $8$ hour

One hour work of Rohan $=\frac{1}{8}$

Similarly, one hour work of Sunil $=\frac{1}{4}$

When they work together, part of the work done in one hour $=\frac{1}{8}+\frac{1}{4}$

$\therefore $ Required time when they work together $=\frac{8\times 4}{8+4}=\frac{32}{12}=2\frac{2}{3}$ hours

- (a)$42$ days
- (b)$24$ days
- (c)$36$ days
- (d)$48$ days
- (e)None of these

**Hint**

If $A$ can complete the work in $x$ days, then its one day's work is the reciprocal of the time taken.

**Solution:**

Work of (Ganesh, Ram and Sohan) for $1$ day $=\frac{1}{16}$

Work of Ganesh and Ram for $1$ day $=\frac{1}{24}$

Thus,

Work of Sohan for $1$ day $=\frac{1}{16}-\frac{1}{24}=\frac{1}{48}$

Hence, Sohan alone will complete the work in $48$ days.

- (a)$12$
- (b)$20$
- (c)$24$
- (d)$28$
- (e)None of these

**Hint**

If they can finish the work individually in $x$ and $y$ days.

Together they can finish in $=\frac{1}{\left(\frac{1}{x}+\frac{1}{y}\right)}$

**Solution:**

If they can finish the work individually in $x$ and $y$ days.

Together they can finish in days

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

$\Rightarrow 8=\frac{1}{\left(\frac{1}{12}+\frac{1}{y}\right)}$

$\Rightarrow \frac{1}{y}=\frac{1}{8}-\frac{1}{12}=\frac{1}{24}$

$y=24$

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- (a)Either (a) or (b) alone is sufficient
- (b)Only (b) is sufficient
- (c)(a) and (b) both together are needed
- (d)(a) and (b) together is also not sufficient
- (e)Only (a) alone is sufficient

**Hint**

When two trains cross each other in opposite direction, then their speeds are added, to find the relative speed.

**Solution:**

Let the speed of the first train be $\mathrm{x}km/\mathrm{hr}$.

$\because $ Time $=\frac{\mathrm{Total}distance}{\mathrm{Speed}}$

$\therefore \frac{12}{60}=\frac{\left(\mathrm{Length}\mathrm{of}\mathrm{first}train+\mathrm{length}\mathrm{of}\mathrm{second}train\right)}{\left(80+\mathrm{x}\right)}$

Hence, in order to find the value of $x$, both the statements together are necessary.

- (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)Either (a) or (b) is sufficient

**Hint**

Speed of boat in upstream $=$ Speed of boat in still water $-$ Speed of current.

**Solution:**

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

Speed of current $=1km/hr$

Time taken $=3$ hour

$\because 3=\frac{4}{x+1}+\frac{4}{x-1}$

$\Rightarrow 3=\frac{4(x-1+x+1)}{{x}^{2}-1}$

$\Rightarrow \frac{3}{4}=\frac{2x}{{x}^{2}-1}$

$\Rightarrow 3{x}^{2}-8x-3=0$

$\Rightarrow \left(3x+1\right)\left(x-3\right)=0$

$\therefore x=3km/hr$

$\therefore $ Both statements together are needed.

- (a)Only (a) is sufficient
- (b)Only (b) is sufficient
- (c)Either (a) or (b) is sufficient
- (d)(a) and (b) together are also not sufficient
- (e)(a) and (b) both together are necessary

**Hint**

Use concept of equivalent distance and use formula:

Distance $=$ Time $\times $ Speed

**Solution:**

Speed of train

$=\frac{Length\mathit{}\mathit{of}\mathit{}\mathit{the}\mathit{}\mathit{train}+length\mathit{}\mathit{of}\mathit{}\mathit{platform}}{Time\mathit{}\mathit{taken}\mathit{}\mathit{to}\mathit{}\mathit{cross}\mathit{}\mathit{the}\mathit{}\mathit{platform}}$

Hence, both the information is necessary.

- (a)$75$
- (b)$180$
- (c)$200$
- (d)$150$
- (e)None of these

**Hint**

When man is moving in opposite direction,

Equivalent speed $=$ Speed of train $+$ Speed of man

**Solution:**

The train and the man are moving in opposite directions.

So, speed $=$ speed of train $+$ speed of man

$=84+6=90km/hr$

$=90\times \frac{5}{18}m/s$

$=25m/s$

$\therefore $ Time taken to cross the man $=4$ second

$\therefore $ speed $\times $ time

$=25\times 4=100m$

- (a)$132km/hr$
- (b)$83km/hr$
- (c)$60km/hr$
- (d)$50km/hr$
- (e)None of these

**Hint**

Use the formula :

Distance $=$ Speed $\times $ Time

**Solution:**

Let the speed of the second train be $xkm/hr$.

And sum of lengths of both the trains

$=100+120=220$ meter

$=\frac{220}{1000}=\frac{11}{50}km$

Since, both the trains are moving in opposite directions.

$\therefore $ Relative Speed

$=(50+x)km/hr$

Time $=6$ seconds

$\frac{6}{60\times 60}=\frac{6}{3600}\mathrm{hr}$

$\frac{1}{600}=\frac{{\displaystyle \frac{11}{60}}}{50+x}$

$\left(50+x\right)=\frac{11}{50}\times 600=132$

$\therefore x=132-50$

$=82km/hr$

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

**Hint**

Weight of 10 m is 10 kg, then the weight of 1 m is 1 kg.

**Solution:**

Weight of a $13m$ long rod $=23.4kg$

Weight of $1m$ long rod $=\frac{23.4}{13}=1.8kg$

So, weight of a $6m$ long rod

$=6\times 1.8=10.8kg$

- (a)$\u20b92200$
- (b)$\u20b92000$
- (c)$\u20b92400$
- (d)$\u20b92600$
- (e)None of these

**Hint**

$1\mathrm{dozen}=12\mathrm{units}$

**Solution:**

We know that,

$1\mathrm{dozen}=12\mathrm{units}$

$\therefore 3\mathrm{dozen}=36\mathrm{units}$

Thus,

$36$ mangoes cost $=\u20b9245$

Cost of $1$ mango $=\frac{245}{36}=\u20b96.8$

$\therefore 353$ mangoes cost $=353\times 6.8\approx \u20b92400$

Thus, the approximate cost of $353$ mangoes is $\u20b92400$.

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

**Hint**

$1kg=1000$ grams

**Solution:**

$\frac{1}{4}kg=250gm$

$250gm$ costs $=Rs.0.60$

$1gm$ costs $=\frac{0.60}{250}$

$200gm$ costs

$=\left(\frac{0.60}{250}\right)\times 200=Rs.0.48=48$ paise

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

**Hint**

Determine the cost of table in terms of chair.

**Solution:**

Let, cost of one chair be $Rs.x$

Cost of $2$ tables $=$ Cost of $5$ chairs

Cost of $1$ table $=\frac{(Costof5chairs)}{2}$

$=Rs.\frac{5x}{2}$

According to the question,

$=Rs.1200$

$\Rightarrow \frac{5x}{2}-x=1200$

$\Rightarrow \frac{3x}{2}=1200$

$\Rightarrow x=Rs.800$

- (a)$52$
- (b)$68$
- (c)$62$
- (d)$587$
- (e)None of these

**Hint**

Total length is the product of no. of units and length of each unit.

**Solution:**

Let $192m$ rod be cut into $x$ pieces.

Each of $x$ pieces measure $=3.2m$

$3.2x=192$

$x=\frac{192}{3.2}=60$

$=60$ pieces

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- (a)$18$ years
- (b)$20$ years
- (c)$22$ years
- (d)$21$ years
- (e)None of these

**Hint**

The difference of ratio scale is one and actual difference is five.

**Solution:**

Let Sudhir age be $x$$y$.

Present age ratio $=x:y=4:5$

$x=\frac{4y}{5}$

After $5$ years:

$\frac{(x+5)}{(y+5)}=\frac{5}{6}$

$\Rightarrow 6x+30=5y+25$

Substituting value of $x$ in second equation

$=\frac{24y}{5}-5y=-5$

$5=5y\u2013\frac{24y}{5}$

$25=y$

$x=20$

Present age of sudhir is $20$ years.

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

**Hint**

The present of Arun is $12.5$ years.

**Solution:**

Let the present age of Arun be $x$.

$5$ years ago his age is $=x-5$

So, $5$ years ago, the age of Lata

$=2(x-5)=2x-10$

Present age of Lata is

$=2x-10+5=2x-5$

$10$ years from now, the age of Arun $=x+10$

$10$ years from now, the age of Lata

$=2x-5+10=2x+5$

According to the question,

$2x+5=\left(\frac{4}{3}\right)(x+10)$

$\Rightarrow 6x+15=4x+40$

$\Rightarrow x=\frac{25}{2}=12.5$ years

Present age of Lata

$=2x-5=2\times 12.5-5=20$ years.

Hence, the correct answer is $20$ years.

- (a)$18$ years
- (b)$19$ years
- (c)$22$ years
- (d)$16$ years
- (e)None of these

**Hint**

Let their current age be $3x$ and $5x$ respectively and add years after which age needs to be calculated.

**Solution:**

Let age of Ganesh be $x$ and age of Kunal be $y$.

Ratio between present age of Kunal and Ganesh $=3:5$

$\Rightarrow \frac{y}{x}=\frac{3}{5}$

$4$ years later age of Kunal $=y+4$

$4$ years later age of Ganesh $=x+4$

According to question:

$y+4=x+4-12$

$x\u2013y=12$

Substituting value of $y=\frac{3x}{5}$ in above equation.

$x\u2013\frac{3x}{5}=12$

$\frac{2x}{5}=12$

$x=30$ years

$y=\frac{330}{5}=18$ years

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

**Hint**

Age of father is $25$ years more than the age of the son.

**Solution:**

Let father present age be $x$.

Son’s present age $=x-25$

After $4$ years ago, age of father $=x-4$

According to question

$x-4=45$

$x=49$ yrs

Father's age $=49$ yrs

Son’s age $=49-25=24$ yrs

Total age of both son and father $=54+29=83$ yrs.

- (a)$33$ years
- (b)$24$ years
- (c)$34$ years
- (d)$42$ years
- (e)None of these

**Hint**

Sum of ages of Yogesh, Prakash and Sameer before $10$ years is $63$ years.

**Solution:**

Present age of Yogesh $=X$

Present age of Prakash $=Y$

Present age of Sameer $=Z$

$X+Y+Z=93$ ... (i)

Also,

$(X-10):(Y-10):(Z-10)=2:3:4$

$3X-30=2Y-20$

$3X-2Y=10$ ... (ii)

$4Y-40=3Z-30$

$4Y-3Z=10$ ... (iii)

$2X-20=Z-10$

$2X-Z=10$ ... (iv)

Put the value in equation (i)

$\left(\frac{Z+10}{2}\right)+\left(\frac{3Z+10}{4}\right)+Z=93$

$Z=38$ years

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- (a)$100$ litres
- (b)$120$ litres
- (c)$50$ litres
- (d)$110$ litres
- (e)None of these

**Hint**

Let capacity of tank be $X$ then water in the tank be $\frac{3X}{4}$, form equations accordingly.

**Solution:**

Let capacity of tank be $X$.

Water in the tank be $\frac{3X}{4}$

According to question,

$\Rightarrow \frac{3X}{4}+5=\frac{4X}{5}$

$\Rightarrow \frac{4X}{5}\u2013\frac{3X}{4}=5$

$\Rightarrow \frac{X}{20}=5$

$X=100$

Capacity of tank $=100$ litres

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

**Hint**

Given: $\frac{X}{3}=Y$

Solve accordingly by making equation.

**Solution:**

Let her marks in History be $X$.

And her marks in Geography be $Y$.

Given:

$\frac{X}{3}=Y$

$X=3Y$

$X+Y=160$

$4Y=160$

$Y=40$

Her marks in Geography is $40$.

- (a)$Rs.450$
- (b)$Rs.500$
- (c)$Rs.600$
- (d)$Rs.380$
- (e)None of these

**Hint**

Let the cost of kurta be $Y$ and solve the fractions accordingly and find the sum of kurta, shirt and saree.

**Solution:**

Let the cost of kurta be $Y$

$Y=\frac{2}{3}$ of $180=Rs.120$

Cost of saree $\frac{7}{3}$ times of shirt

$=\frac{7}{3}$ of $180=420$

Total cost of all items $=180+120+420=Rs.720$

- (a)$Rs.1,50,000$
- (b)$Rs.1,50,250$
- (c)$Rs.1,40,350$
- (d)$Rs.1,40,200$
- (e)None of these

**Hint**

First find the number of Male and female donor and calculate the amount.

**Solution:**

Each male donor donates $Rs.2000$

Each female donor donation

$=\frac{1}{5}$ of $2000=Rs.400$

Total men $=\frac{1}{3}$ of $150=50$

Total women $=150-50=100$

Total yearly collection

$=50\times 2000+100\times 400=Rs.1,40,000$

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

**Hint**

Let the longer piece be $X$ and other piece be $Y$. Solve the fraction given between them.

**Solution:**

Let the longer piece be $X$.

The other piece be $Y$.

$Y=\frac{2X}{5}$

$\Rightarrow Y+X=63$

$\Rightarrow \frac{7X}{5}=63$

$X=45cm$

$Y=63\u201345=18cm$

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

**Hint**

Let first number be $n+1$ second number $n+3$ third number be $n+5$ and solve them by making equations.

**Solution:**

Let first number be $n+1$

Second number be$n+3$

Third number be $n+5$

According to question,

$n+1+n+3+n+5\u2013(n+1)=20$

$2n+8=20$

$2n=12$

$n=6$

So, numbers are $7,9$ and $11$.

Middle number is $9$.

- (a)$52$
- (b)$45$
- (c)$54$
- (d)Data is inadequate
- (e)None of these

**Hint**

Let numbers be $a,b$ and $c$ and form equations according to given ratio and solve.

**Solution:**

Let numbers be $a,b,c$.

$a+b+c=174$ ... (1)

$\frac{b}{c}=\frac{9}{16}$

$b=\frac{9c}{16}$

$\frac{a}{c}=\frac{1}{4}$

$a=\frac{c}{4}$

Substituting these values in equation (1),

$\frac{c}{4}+\frac{9c}{16}+c=174$

$4c+9c+16c=174\times 16$

$29c=2784$

$c=96$

$a=\frac{96}{4}=24$

$b=\frac{996}{16}=54$

Hence, second number is $54$.

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

**Hint**

Let first number be $a$ and according to question $30\%$ of this number is $0.3$.

**Solution:**

Let first number be $a$.

So,

$0.3a+b=\frac{6b}{5}$

$\frac{3a}{10}=\frac{b}{5}$

$b=\frac{3a}{2}$

$\frac{a}{b}=\frac{2}{3}$

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

**Hint**

Let two-digit number xy, change them according to given condition and solve the formed equation.

**Solution:**

Let two-digit number $xy$.

Number $=10x+y$

Number obtained by interchanging digits $=yx$

$10y+x$

$10x+y\u201310y-x=45$

$9x\u20139y=45$

$x-y=5$

So, difference of the digit is $5$.

- (a)$23$
- (b)$24$
- (c)$22$
- (d)$36$
- (e)None of these

**Hint**

Let the number be x, then solve according to given condition.

**Solution:**

Let the number be $x$.

$x-28=\frac{x}{3}$

$3x-84=x$

$2x=84$

$x=42$

$50\%$$=\frac{42}{2}=21$

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

**Hint**

Calculate the price of two quality of wheat seperately and add them to get total cost of wheat.

**Solution:**

Trader spent on wheat having cost $Rs.7/kg$

$=50\times 7=Rs.350$

Trader spent on wheat having cost $Rs.8/kg$

$=20\times 8=Rs.160$

He sold the mixture at $Rs.10/kg$

He sold $70kg$ of wheat Profit at

$70\times 10=Rs.700$

So profit $=700-(350+160)=Rs.190$

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

**Hint**

First calculate overall cost price of the rice then calculate the selling price on which rice should be sold if those are sold at $20\%$ profit then find the price per KG by dividing S.P. by total weight of rice.

**Solution:**

Alok spent on rice $=308\times 50+20\times 8=415$

To make $20\%$ profit

Profit $=0.2\times 415=83$

Profit $=SP-CP$

$83=SP-415$

$SP=Rs.498$.

This $SP$ is of $50kg$ rice i.e $Rs.498$

Cost of per $kg$ rice $=\frac{498}{50}=Rs.9.96$

- (a)$Rs.6.80$
- (b)$Rs.7.00$
- (c)$Rs.8.00$
- (d)$Rs.60.00$
- (e)None of these

**Hint**

First calculate the C.P. of all rice then calculate the value of selling price at profit of 60 Rs. then divide S.P. by amount of rice.

**Solution:**

Trader spent on wheat

$=20\times 6.5+30\times 7=Rs.340$

In order to make profit

$=Rs.60$

Profit $=SP-CP$

$60=SP-340$

$SP=Rs.400$

This selling price is of total wheat

$=20+30=50kg$

Cost of per wheat $=\frac{400}{50}=Rs.8$

- (a)$200ml$
- (b)$250ml$
- (c)$150ml$
- (d)$20ml$
- (e)None of these

**Hint**

Let the volume added be $x$ and calculate the difference in price of pure milk and mixture,

**Solution:**

Suppose the man buys $1$ litre milk at $Rs.1$ per litre.

Then, total cost price of $1$ litre milk $=Rs.1$

Suppose he mixes $x$ litre water in $1$litre milk and sells the mixture at $Rs.1$ per litre

Total quantity of the mixture $=(1+x)$ litre

Total selling price $=Rs.(1+x)$

Profit $=(1+x)-1=Rs.x$

Given, Profit $\%$

$=\frac{x}{1}\times 100=20$

$x=2$

So he mixes $0.2$ litre water in $1$ litre milk

Thus $200ml$ in $1$ litre.

- (a)$1:2$
- (b)$3:2$
- (c)$3:1$
- (d)$2:3$
- (e)None of these

**Hint**

CP would be sum of price of ghee and oil, equate the difference of SP and CP with profit.

**Solution:**

Let he adds $x$ oil in $y$ ghee,

So total mixture $=x+y$

Pure ghee of $ykg$ cost $=100y$

Mixture price $=100y+50x=CP$

$SP=96(x+y)=96x+96y$

Profit $=20\%$

Using formula:

Profit $=SP-CP$

$0.2\times (100y+50x)=96x+96y\u2013100y-50x$

$20y+10x=46x-4y$

$24y=36x$

$x:y=36:24=3:2$

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

**Hint**

The new area is 1.8 times the initial area.

**Solution:**

Let the length and breadth be $l$ and $b$.

Original area $=l\times b$

New length, $l=l+0.5l=1.5l$

New breadth, $b=b+0.2b=1.2b$

New Area $=1.5\times 1.2lb=1.8lb$

$=\frac{9lb}{5}=1\frac{4}{5}lb$

- (a)$5.2m$
- (b)$6.8m$
- (c)$5.8m$
- (d)$5m$
- (e)None of these

**Hint**

Area of the rectangle is the product of its length and breadth.

**Solution:**

Let the breadth of rectangle is $bm$.

Then its length will be:

$b+30\%\times b=1.3bm$

Area of rectangular plot $=$ length $\times $ breadth

$20.8=1.3b\times b$

${b}^{2}=16$

$b=4m$

Since length is $30\%$ more than breadth.

$l=1.3\times 4=5.2m$

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

**Hint**

Area is directly proportional to the length.

**Solution:**

Let area of original rectangle $=A$

New length is $1.3L$.

As the length is increased, area of formed figure $=1.3A$

Ratio of formed figure to original $=1.3A=13:10$

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

**Hint**

Equate the area of square and circle.

**Solution:**

Area of square $=$ Area of circle

$Sid{e}^{2}=3.14{r}^{2}$

${S}^{2}=3.14\times 14\times 14$

$S=14\times (3.14{)}^{0.5}$

$S\approx 25cm$

- (a)$4000{m}^{2}$
- (b)$4200{m}^{2}$
- (c)$4550{m}^{2}$
- (d)$4800{m}^{2}$
- (e)None of these

**Hint**

Use this formula: Area $=$ Length $\times $ Breadth.

Diagonal $=\sqrt{{B}^{2}+{L}^{2}}$.

**Solution:**

Let the length and breadth be $L$ and $B$ respectively.

Now, according to the information given in the question,

$B=75\%L$

$B=0.75L=\frac{3}{4}L$

We know that,

$Diagona{l}^{2}={B}^{2}+{L}^{2}$.

$100=\sqrt{{B}^{2}+{L}^{2}}=\sqrt{{\left(\frac{3}{4}\right)}^{2}+1}L$

$100=\frac{5L}{4}$

$L=80$

$B=\frac{3}{4}\times 80=60$

Area $=4800{m}^{2}$

Hence, the correct answer is $4800{m}^{2}$.

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

**Hint**

Hexagonal shapes will be formed.

**Solution:**

Here, the paper-sheet is folded half from right to left and cut according to the diagram, after opening we will get four hexagonal shapes on the paper-sheet. Hence, option (C) is correct.

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

**Hint**

Four circular shape will be formed.

**Solution:**

Here, paper-sheet has been folded half from right to left and again folded half from top to bottom and cut according to the diagram. On opening, we get four circular shapes in a line at the middle of the paper-sheet. Hence, option (A) is correct.

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

**Hint**

After opening the paper sheet, more than one rectangular shape will be formed on the paper sheet.

**Solution:**

Here the paper-sheet has been folded diagonally from the top left and again folded from the top right and is cut according to the diagram given. After opening, two rectangular shapes will form diagonally from top left to the bottom right. Hence, option (C) is correct.

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

**Hint**

After opening the paper-sheet, six square shapes will be formed on the diagonal line.

**Solution:**

Here the paper-sheet has been folded diagonally from the top left and again from the top right and cut according to the given diagram. On opening, we get six square shapes, diagonally from top left to bottom right. Hence, option (E) is correct.

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

**Hint**

Since, two triangular cuts are made on the radius of the circular paper-sheet so, after opening more than two square shapes will be formed.

**Solution:**

Here, the paper-sheet has been folded and cut according to the given diagram. On opening, four square shapes will be formed in a straight line in the middle of the circular sheet. Hence, option (A) is correct.

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

**Hint**

In the problem figure, each corner has a quarter circle.

**Solution:**

We need a quarter circle at the bottom-right corner of the missing piece. Here, only the answer figure (B) completes the pattern of the problem figure.Hence, (B) is the correct option.

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

**Hint**

Find the water image of the figure in the 1^{st} quadrant.

**Solution:**

The missing part image needs to be the water image of the figure in the 1

^{st}quadrant. After observing the options we get that, the answer figure (A) completes the pattern of the problem figure. Hence, (A) is the required answer.

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

**Hint**

An obtuse-angled triangle on the base of the hypotenuse is formed at each corner.

**Solution:**

An obtuse-angled triangle on the base of the hypotenuse is formed at each corner. In each figure, there are two lines emerging from the top of the triangle and meeting the opposite vertices of the quadrant. After observing the options we get that, the answer figure (C) completes the pattern of the problem figure.

Hence, option (C) is the required answer.

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

**Hint**

An arrow, meeting the bottom right corner is to be formed.

**Solution:**

An arrow, meeting the bottom right corner is to be formed. After observing the option we get that, the answer figure (C) completes the pattern of the problem figure.

Hence, option (C) is the required answer.

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

**Hint**

A triangle is to be formed, with one of the sides being the diagonal of the square.

**Solution:**

A triangle is to be formed, with one of the sides being the diagonal of the square. After observing the options we get that, the answer figure (A) completes the pattern of the problem figure.

Hence, option (A) is the required answer.

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

**Hint**

Observe the right half of the problem figure from the top.

**Solution:**

This figure is first figure from the right in the first row. Hence, option (B) is correct.

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

**Hint**

Observe the right part of the problem figure.

**Solution:**

This figure is the first figure from the right in the middle. Hence, option (A) is correct.

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

**Hint**

Observe the triangle in lower right part.

**Solution:**

This figure is first figure from the right in the lower part.

Hence, option (D) is correct.

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

**Hint**

Observe the left, lower half of the given problem figure.

**Solution:**

This figure is second figure from the left in the lower part. Hence, option (B) is correct.

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

**Hint**

Observe the right half of the problem figure from the top.

**Solution:**

This figure is first figure from the right in the second row.

Hence, option (B) is correct.

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

**Hint**

Observe the movement of the plus or cross sign in the problem figure.

**Solution:**

In each figure a square is divided into three parts and small lines within a part are shifting form left to right and from right to left alternately The small lines are reducing also in number by one. Besides, the cross or plus sign moves one arm ahead in the clockwise direction.

Hence option (A) is correct.

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

**Hint**

Observe the given figure carefully and mark the answer correctly.

**Solution:**

A small circle is rotating outside of the large circle in anticlockwise order and is changing from black to white and vice versa in each subseuqent figure. Fig. (I) has two lines while fig. (II) has one line and so on. Besides, the lines are changing their positions from horizontal to vertical and so on.

Hence option (B) is correct.

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

**Hint**

Observe the change in the number of dashes in the problem figure.

**Solution:**

On each side of the triangle there are two dashes in fig. (I) and three dashes in fig. (II), but again one dash in fig (III) so there should be two dashes in fig. (IV). Since the longest side of the triangle is dotted in fig. (I) and (III) but continuous in fig. (II).

Hence in fig. (IV) the longest side should be continuous. In fig. (I) inside is a cross sign in fig. (II) a plus sign and in fig. (Ill) there is a circle.

Hence in fig. (IV) but continuous in fig. (II). Hence in fig. (IV) the longest side should be continuous.

In fig. (I) inside is a cross sign in fig. (II) a plus sign and in fig. (Ill) there is a circle.

Hence in fig. (IV) there should be a cross sign.

So option (D) is correct.

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

**Hint**

Observe the arc and cross, plus signs in the problem figure.

**Solution:**

The arc is concave in fig. (I) but convex in fig. (II) and so on.

Fig. (I) has cross signs but fig. (II) has plus sign and so on.

The signs are two in fig. (I), one in fig. (II) and three in fig. (Ill), there is a dash in the lower part of each figure, but it is on both sides in fig. (I) and (III) and one side in fig. (II).

Hence option (A) is correct.

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

**Hint**

Observe the change in the number of dashes in problem figures.

**Solution:**

Vertical dashes are reducing by one in each subsequent figure. In other respects fig. (I) and (III) are same. So, fig. (IV) would be like as fig. (II).

So option (B) is correct.

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

**Hint**

Observe the upper part of the problem figure and find same design in the answer figures.

**Solution:**

If from the fourth answer figure horizontal line from the bottom, vertical line from the left side and upper inclined line are removed, what is left is exactly the top figure.

Hence option (D) is correct.

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

**Hint**

Observe a triangle inside the rectangle in each answer figure.

**Solution:**

If from the third answer figure, two inclined lines are removed from the central part, what is left, is exactly the top figure.

So option (C) is correct.

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

**Hint**

Find the lower part of the problem figure in each answer figure.

**Solution:**

If the sides of the pentagon are removed from the third answer figure what is left, is exactly the top figure

So option (C) is correct.

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

**Hint**

Find the upper part of the problem figure that is T-shape design in each answer figure.

**Solution:**

Observe the central portion of the second answer figure with the top and the base lines in the figure.

Hence, option (B) is correct.

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

**Hint**

Find the triangular part of the problem figure in each answer figure.

**Solution:**

If from the first answer figure, parallelogram portions are removed, the top figure comes out.

So option (A) is correct.

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- (a)Animal
- (b)Tree
- (c)Fruit
- (d)Garden

**Hint**

The first one is formed by the combinations of the second one.

**Solution:**

Just as a train is made up of a combination of several bogies, a forest is the result of a combination of many trees.

So, the correct option is (B).

- (a)Cloth
- (b)Sewing
- (c)Machine
- (d)Needle

**Hint**

Words are related to the worker and the raw material used by them.

**Solution:**

A carpenter makes different types of furniture using wood.

In the same way,

A tailor sews many types of pants, shirts and suits using clothes.

So, the correct option is (A).

- (a)Earth
- (b)Daughter
- (c)Sister
- (d)Sun

**Hint**

The second is the parent of the first.

**Solution:**

Just as a child's parent is a mother, in the same way, Mars is a part of the solar system where the Sun acts as a parent.

So, the correct option is (D).

- (a)Cheerful
- (b)Joy
- (c)Gloomy
- (d)Anger

**Hint**

Words are related to antonyms.

**Solution:**

As, heavy is opposite to light, in the same way, gloomy is opposite to glad.

Cheerful and joy are the synonyms of glad and anger is not the opposite of glad.

So, the correct option is (C).

- (a)Spectator
- (b)Captain
- (c)Player
- (d)Playground