### Contents of LIC ADO Preliminary & Main Exam

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

#### 63 Chapters

- 1. Previous Paper (Solved), 2019
- 2. Previous Paper (Solved), 2015
- 3. Previous Paper (Solved), 2013
- 4. Previous Paper (Solved), 2010
- 5. Series
- 6. Coding-Decoding
- 7. Symbol Substitution
- 8. Blood Relation
- 9. Direction Sense
- 10. Statement Analysis
- 11. Sitting Arrangement
- 12. Data Sufficiency
- 13. Coded Inequalities
- 14. Input Interpretations
- 15. Drawing Inference
- 16. Syllogism
- 17. Cause & Effect
- 18. Course of Action
- 19. Distinguishing Argument
- 20. Drawing Conclusions
- 21. Statement Assumptions
- 22. Cubes and Dice
- 23. Figure Series
- 24. Analogies or Relationships
- 25. Classification or Odd-One Out
- 26. Simplification
- 27. Powers and Roots Square, Cube, Indices, Surds Squaring
- 28. LCM and HCF
- 29. Ratio and Proportion
- 30. Partnership
- 31. Percentage
- 32. AVERAGE AND AGE RELATED PROBLEMS
- 33. PROFIT & LOSS
- 34. Simple Interest
- 35. Compound Interest
- 36. ALLIGATION OR MIXTURE
- 37. Time and Work
- 38. TIME AND DISTANCE
- 39. STREAMS
- 40. RACES AND GAMES
- 41. MENSURATION - I
- 42. MENSURATION - II
- 43. SERIES
- 44. APPROXIMATE VALUES
- 45. DATA INTERPRETATION
- 46. BAR GRAPHS AND PIE CHARTS
- 47. Spelling Errors
- 48. Spotting Errors
- 49. Articles
- 50. Prepositions
- 51. Synonyms & Antonyms
- 52. One Word Substitutions
- 53. Idioms and Phrases
- 54. Narration
- 55. Voice
- 56. Sentence Completion
- 57. Sentence Improvement
- 58. Reordering Words & Sentences
- 59. Comprehension Passages
- 60. Cloze Test
- 61. INSURANCE KNOWLEDGE
- 62. FINANCIAL MARKETING AWARENESS
- 63. GENERAL KNOWLEDGE

#### 3 Topics

Numerical Ability

English

Reasoning Ability

#### 5 Topics

Reasoning Ability

Numerical Ability

General Knowledge

English

Insurance and Financial Marketing

#### 4 Topics

Reasoning Ability

Numerical Ability

General awareness

English Language

#### 4 Topics

Reasoning Ability

Numerical Ability

GENERAL KNOWLEDGE AND CURRENT AFFAIRS

GENERAL ENGLISH

#### 4 Topics

LETTER SERIES

REPEAT SERIES

NUMBER SERIES

MIXED SERIES

#### 1 Topics

Symbol Substitution

#### 1 Topics

Statement Analysis

#### 1 Topics

Sitting Arrangement

#### 1 Topics

Coded Inequalities

#### 1 Topics

Input Interpretations

#### 1 Topics

Drawing Inference

#### 1 Topics

Course of Action

#### 1 Topics

Distinguishing Argument

#### 1 Topics

Drawing Conclusions

#### 1 Topics

Statement Assumptions

#### 1 Topics

Analogies or Relationships

#### 1 Topics

Classification or Odd-One Out

#### 2 Topics

Short Methods in Squaring

Square Root

#### 2 Topics

Least Common Multiple (LCM)

Highest Common Factor (HCF)

#### 1 Topics

Some other Terms and their Formulae of Ratio and Proportion

#### 1 Topics

To Convert Fraction into Percentage

#### 3 Topics

Type 1(Average Speed)

Type 2(Average Age)

Conditional Trick

#### 1 Topics

Compound Interest

#### 1 Topics

Alligation Rule

#### 1 Topics

Quicker Maths(Direct Formula)

#### 4 Topics

Triangle

Rectangle

Square

Circle

#### 3 Topics

Cuboid

Cube

Cylinder

#### 2 Topics

Arithmetic Series:

Geometric Series:

#### 3 Topics

Multiplication

Division

Percentage

#### 1 Topics

DATA INTERPRETATION

#### 2 Topics

Bar Graphs

Pie Charts

#### 1 Topics

Miscellaneous Errors

#### 6 Topics

Definite and Indefinite Articles

Vowel and Consonant Sounds

Use of Articles 'A' and 'An'

Use of Article 'The'

Omission of Article 'The'

Double Use of Article 'The'

#### 1 Topics

Synonyms & Antonyms

#### 1 Topics

Miscellaneous

#### 1 Topics

Idioms and Idiomatic Phrases

#### 1 Topics

Direct and Indirect Narration/Speech

#### 1 Topics

Sentence Completion

#### 1 Topics

Sentence Improvement

#### 1 Topics

Reordering Words & Sentences

#### 10 Topics

PASSAGE - 1

PASSAGE - 2

PASSAGE - 3

PASSAGE - 4

PASSAGE - 5

PASSAGE - 6

PASSAGE - 7

PASSAGE - 8

PASSAGE - 9

PASSAGE - 10

#### 10 Topics

PASSAGE - 1

PASSAGE - 2

PASSAGE - 3

PASSAGE - 4

PASSAGE - 5

PASSAGE - 6

PASSAGE - 7

PASSAGE - 8

PASSAGE - 9

PASSAGE - 10

## Experience Tests Tailored to Match the Real Exam

## Practise Questions with Solutions from the LIC ADO Preliminary & Main Exam

- (a)$44$
- (b)$12$
- (c)$4$
- (d)$2$
- (e)$5$

**Hint**

Calculate the square root of the inside number first.

**Solution:**

The given expression can be simplified as :

$\sqrt{4+\sqrt{44+\sqrt{10000}}}$

$=\sqrt{4+\sqrt{44+100}}$

$=\sqrt{4+\sqrt{144}}$

$=\sqrt{4+12}$

$=\sqrt{16}$

$=4$

So, $4$ is the correct answer.

- (a)$0$
- (b)$\sqrt{2}$
- (c)$2$
- (d)${2}^{\frac{31}{32}}$
- (e)$3$

**Hint**

Use the formulae : $\sqrt{2\sqrt{2\sqrt{2\sqrt{......n}}}}={2}^{\frac{{2}^{n}-1}{{2}^{n}}}$

**Solution:**

Let's calculate the value of $\sqrt{2\sqrt{2}}$,

$=\sqrt{2\times {2}^{\frac{1}{2}}}={2}^{\frac{3}{4}}$

So, we can say that,

$\sqrt{2\sqrt{2\sqrt{2\sqrt{......n}}}}={2}^{\frac{{2}^{n}-1}{{2}^{n}}}$

Then, $\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}={2}^{\frac{{2}^{5}-1}{{2}^{5}}}={2}^{\frac{31}{32}}$

So, ${2}^{\frac{31}{32}}$ is the correct answer.

- (a)$\sqrt{11}$
- (b)$-\sqrt{11}$
- (c)$11$
- (d)$\sqrt{0.11}$
- (e)$\sqrt{0.011}$

**Hint**

Apply the concept of VBODMAS and Surds.

**Solution:**

The given expression can be simplified as:

$=\sqrt{\frac{0.85\times (0.105+0.024-0.008)}{0.22\times 0.25\times 17}}$

$=\sqrt{\frac{0.85\times (0.129-0.008)}{0.22\times 0.25\times 17}}$

$=\sqrt{\frac{0.85\times 0.121}{0.935}}$

$=\sqrt{\frac{0.10285}{0.935}}$

$=\sqrt{0.11}$

So, the correct answer is $\sqrt{0.11}$.

- (a)$\frac{1}{7}$
- (b)$7$
- (c)$7.1$
- (d)$6.8$
- (e)$6.9$

**Hint**

Assume the value of $x$ and $y$ be $3k$ and $k$.

**Solution:**

Suppose, $x=3k$ and $y=k$

Putting the value in the given expression

$\frac{2x+y}{x-2y}=\frac{6k+1k}{3k-2k}$

$=\frac{7k}{k}=7$

So, the correct answer is $7$.

- (a)$10kmph$
- (b)$20kmph$
- (c)$25kmph$
- (d)$30kmph$
- (e)$35kmph$

**Hint**

Suppose the speed of the boat be $xkm/h$ and the speed of the stream be $ykm/h$.

Then, $x+y=\frac{120}{4}x-y=\frac{40}{4}$

**Solution:**

Let the speed of the boat in still water be $xkmph$

and the speed of the current $=ykmph$

So, downstream speed $=(x+y)kmph$

And upstream speed $=\left(x-y\right)kmph$

According to the question:

$\frac{120}{(x+y)}=4$

$4x+4y=120-----\left(i\right)$

$\frac{40}{(x-y)}=4$

$4x-4y=40-----\left(ii\right)$

Adding $\left(i\right)$ and $\left(ii\right)$

$8x=160$

$x=20$

Put $x=20in\left(ii\right)$

$80-4y=40$

$-4y=-40$

$y=10$

So, the speed of the current is $10kmph.$

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- (a)Charu Chandra Pandey
- (b)Acharya Munishwar Jha
- (c)Mathuradutt Mathpal
- (d)Meenakshi Sundaram K.

**Hint**

He also served as vice-chancellor of Darbhanga’s Kameshwar Singh Sanskrit University.

**Solution:**

Sahitya Academy has recently announced its ‘Bhasha Samman’ award for $2014$. Acharya Munishwar Jha is the recipient of classical and Medieval Literature. The Sahitya Academy Award is an academic award in India. This award annually confers on writers of the most outstanding books of literary merit in the $24$ mentioned languages.

Hence, option B is the correct answer to the question.

- (a)Shillong
- (b)Itanagar
- (c)Kohima
- (d)Imphal
- (e)Aizwal

**Hint**

It is also known as the "Scotland of the East".

**Solution:**

Meghalaya is one of the north-eastern states having its capital at Shillong. Meghalaya means the 'home for clouds'. Moreover, the two wettest places of the Earth-Cherapunji and Mawynsram are located in the Meghalaya state.

Hence, option A is the correct answer to the question.

- (a)Shri Dharmendra Pradhan
- (b)Dr. Jitendra Singh
- (c)Shri Rajiv Pratap Rudy
- (d)Santosh Kumar Gangwar

**Hint**

He has been the Member of Parliament for the city of Bareilly since 1989.

**Solution:**

Santosh Kumar Gangwar is the minister for state in charge of Ministry of Labour and Employment (Independent Charge). The Ministry of Labour & Employment is one of the oldest and most important Ministries of the Government of India and is responsible to protect and safeguard the interest of workers in general and the poor, deprived and disadvantaged sections of the society.

- (a)P. T. Usha
- (b)Sania Mirza
- (c)Sushil Kumar
- (d)K. D. Jadhav
- (e)Abhinav Bindra

**Hint**

He was born in Dehradun and picked up shooting at an early age.

**Solution:**

The first Indian to win an individual Olympic Gold Medal is Abhinav Bindra. Abhinav Bindra won gold in the Men's 10-metre air rifle event at the 2008 Beijing Olympics and became the first Indian to win an individual gold medal at the Olympic Games.

- (a)T. H. Vinayakram- Ghatam
- (b)Amjad Ali Khan- Sarod
- (c)Raja Ravi Varma- Painter
- (d)Vasudeo S Gaitonde- Sitar
- (e)Pandit Shivkumar Sharma- Santoor

**Hint**

He received the Padma Shri Award in 1971.

**Solution:**

Vasudeo S Gaitonde is related to painting, not sitar. Vasudeo S Gaitonde was born in Nagpur, Maharashtra in 1924. He received the Padma Shree award by the Government of India in 1971.

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- (a)Pakistan
- (b)England
- (c)India
- (d)Australia
- (e)None of these

**Hint**

It is an Asian country.

**Solution:**

India (Indian cricket blind team) won the inaugural Blind Cricket T20 world cup by defeating team Pakistan by 29 runs. In 2012, the first Blind T20 Cricket World cup was held in Bangalore India.

- (a)4.25 per cent
- (b)7.50 per cent
- (c)10 per cent
- (d)12.5 per cent
- (e)None of these

**Hint**

According to the Tendulkar methodology, the percentage of population below the poverty line was 29.8 per cent at the end of 2009-10.

**Solution:**

(NDC) National Development Council recently approved the 12th Five Year Plan in which they proposes to bring down poverty by 10% points by the end of the 12th Plan.

- (a)Only 1
- (b)Only 2
- (c)Both 1 and 2
- (d)Both 2 and 3
- (e)None of these

**Hint**

Paid-up policy falls into the category of traditional insurance plans.

**Solution:**

Paid up Policy: It is an insurance policy that requires no further premium payments to be made. In such a policy, the sum assured is reduced in accordance with the paid premiums. So, according to the explanation about paid and policy statement 1st and 2nd is correct.

- (a)Only 1
- (b)Only 2
- (c)Both 1 and 2
- (d)Both 2 and 3
- (e)None of these

**Hint**

The agency's headquarters are in Hyderabad, Telangana.

**Solution:**

IRDA aims at the protection of the interests of the policyholders. As laid down in section 14 of the IRDA Act, 1999 it has powers of undertaking inspection of conducting inquiries and investigations into all the companies of the insurance business. Here option (c) is the correct answer.

- (a)Only 1
- (b)Only 2
- (c)Both 1 and 2
- (d)Both 2 and 3
- (e)None of these

**Hint**

Subscription policy is in which multiple insurers share the risk associated with providing the insurance.

**Solution:**

Subscription Policy is a policy shared by two or more insurers. In a subscription policy, each insurer share the risk associated with providing the insurance. Here option c is correct answer.

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- (a)Insurance companies
- (b)Banks
- (c)NABARD
- (d)RBI
- (e)All of these

**Hint**

NIACL is an example of this type of companies.

**Solution:**

Insurance companies issue ‘ULIP’. Unit Linked Insurance Plan is a product offered by insurance companies that, unlike a pure insurance policy, combines the benefits of protection and saving in a single plan.

- (a)NFCG
- (b)IRDA
- (c)CII
- (d)FICCI
- (e)All of these

**Hint**

This organisation was founded in 1999.

**Solution:**

Insurance business in India is regulated by IRDA. It is an agency of GOI for insurance sector supervision and development. IRDA was constituted on 19 April 2000 under the IRDA act 1999. Its headquarter is situated in Kolkata.

- (a)ICICI Prudential
- (b)ING Vysya
- (c)ICICI Lombard
- (d)New India Assurance Company
- (e)National Securities Depository Ltd

**Hint**

It was established in August 1996.

**Solution:**

ICICI Prudential, ING Vysya, ICICI Lombard and New India Assurance Company are insurance companies functioning in India. While National Securities Depository limited is a depository company. Here option E is correct answer.

- (a)Newspaper publications
- (b)Media entertainment
- (c)Car & Automobile production
- (d)Textile
- (e)Heavy engineering & construction

**Hint**

Hydraulic Cylinders is a product of L&T.

**Solution:**

The major business of Larson & Tourbo (L&T) heavy engineering and construction. The CEO of Larson & Tourbo is S. N. Subramanyam. Their headquarters situated in mumbai.

- (a)Death coverage
- (b)Life Insurance
- (c)Savings for future
- (d)Provident Fund
- (e)None of these

**Hint**

Jeevan Pragati Plan is an example of this industry.

**Solution:**

A contract that pledges payment of an agreed-upon amount to the person (or his/her nominee) on the happening of an event covered against is known as Life Insurance. LIC is the only public insurance company in the field of life insurance.

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- (a)$5$
- (b)$24$
- (c)$25$
- (d)$80$
- (e)None of these

**Hint**

Replace the sign with sign given in statement.

**Solution:**

As per Statement:

$\times $ stands for addition,

$\xf7$ stands for subtraction,

$+$ stands for multiplication,

$-$ stands for division,

Now replacing the sign in current scenario to find the solution

$(20\times 6\xf76\times 4)$ that means $20+6\u20136+4=24$.

- (a)$79$
- (b)$-11$
- (c)$91$
- (d)$-48.5$
- (e)None of these

**Hint**

Use BODMAS to solve equation.

**Solution:**

Replace the sign with existing signs as per statement

"$+$" means "$\times $"

"$\xf7$" means "$-$"

"$\times $" means “$\xf7$"

"$-$" means "$+$"

Putting the replaced sign in current question to solve it

$4+11\xf75-50$ that means $4\times 11-5+50=89$.

- (a)$8$
- (b)$36$
- (c)$52$
- (d)$0$
- (e)$16$

**Hint**

Replace the alphabet with numbers.

**Solution:**

As per statement given

$\mathrm{P}=6,\mathrm{J}=4,\mathrm{L}=8,\mathrm{M}=24$.

Now replace the alphabet with numbers in given question

$\mathrm{M}\times \mathrm{J}\xf7\mathrm{L}+\mathrm{J}=24\times 4\xf78+4=16$.

- (a)$AC$
- (b)$A+B2D$
- (c)$A+B2C$
- (d)$A+B2E$
- (e)$D+B2C$

**Hint**

Combine the elements and form a single equation.

**Solution:**

As per given statement $A+BC+D,B+E=2C$ and $C+DB+E$

The combined elements will be $A+BC+DB+E=2C$

So from given statement it is clear that $A+B2C$

It is true because there is a relation of greater than from $\u2018A+B\u2019$ to $\u20182C\u2019$.

- (a)$\mathrm{A}+\mathrm{D}\mathrm{B}+\mathrm{E}$
- (b)$\mathrm{A}+\mathrm{D}\mathrm{B}+\mathrm{C}$
- (c)$\mathrm{A}+\mathrm{B}2\mathrm{D}$
- (d)$\mathrm{B}+\mathrm{D}\mathrm{C}+\mathrm{E}$
- (e)$\mathrm{A}+\mathrm{D}\mathrm{B}+\mathrm{C}$

**Hint**

Take an example by subtituting values according to statement.

**Solution:**

As per given statement, $\mathrm{A}+\mathrm{D}\mathrm{C}+\mathrm{E},\mathrm{C}+\mathrm{D}=2\mathrm{B}$ and $\mathrm{B}+\mathrm{E}\mathrm{C}+\mathrm{D}$.

The combined elements will be

$\mathrm{A}+\mathrm{D}\mathrm{C}+\mathrm{E},\mathrm{B}+\mathrm{E}\mathrm{C}+\mathrm{D}=2\mathrm{B}$

So, $\mathrm{A}+\mathrm{D}\mathrm{B}+\mathrm{C}$.

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- (a)Brother
- (b)Nephew
- (c)Uncle
- (d)Son-in-law

**Hint**

Solve with the help of drawing family tree starting from 'daughter of the father'.

**Solution:**

We consider (+) as male member of the family, (-) as female member of the family, vertical single line as generation gap, horizontal single line as relation of sibling, horizontal double line as relation of husband and wife.

Hence, it is clear that the only possible relation of girl to boy is the relation of siblings.

- (a)Brother
- (b)Uncle
- (c)Cousin
- (d)Father

**Hint**

Solve with the help of drawing family tree starting from 'son'.

**Solution:**

We consider (+) as male member of the family, (-) as female member of the family, vertical single line as generation gap.

With the help of diagram, lets solve this

Clearly, Suresh is the father of that boy.

- (a)$\mathrm{M}-\mathrm{N}\times \mathrm{C}+\mathrm{F}$
- (b)$\mathrm{F}-\mathrm{C}+\mathrm{N}\times \mathrm{M}$
- (c)$\mathrm{N}+\mathrm{M}\u2013\mathrm{F}\times \mathrm{C}$
- (d)$M\times N\u2013C+F$

**Hint**

Replace the sign in the option as per instruction to get the result.

**Solution:**

As per the instruction given,

$\mathrm{A}+\mathrm{B}$ means $\mathrm{A}$ is the brother of $\mathrm{B}$,

$\mathrm{A}-\mathrm{B}$ means $\mathrm{A}$ is the sister of $\mathrm{B}$,

$\mathrm{A}\times \mathrm{B}$ means $\mathrm{A}$ is the father of $B$,

Now from Option A: $\mathrm{N}\times \mathrm{C}$ means $\mathrm{N}$ is the father of $\mathrm{C}$, so it is incorrect.

Option B: $\mathrm{C}+\mathrm{N}\times \mathrm{M}$ means $\mathrm{C}$ is the brother of $\mathrm{N}$ and $\mathrm{N}$ is the father of $\mathrm{M}$, so this is also incorrect.

Option C: $\mathrm{F}\times \mathrm{C}$means $\mathrm{F}$ is the father of $\mathrm{C}$, so it is incorrect.

Clearly, after analysing all the options only in option D. $\mathrm{C}$ is the son of $\mathrm{M}$.

We consider $(+)$as male member of the family, $(-)$ as female member of the family, vertical single line as generation gap, horizontal single line as relation of sibling, horizontal double line as relation of husband and wife.

Hence, option D is correct.

- (a)Brother
- (b)Sister
- (c)Nephew
- (d)Cannot be determined

**Hint**

We can draw a tree, consider '+' as male member '_' as a female member.

**Solution:**

We consider (+) as male member of the family, (-) as female member of the family, vertical single line as generation gap, horizontal single line as relation of sibling, horizontal double line as relation of husband and wife.

As per statement given,

A is the brother of B,

B is the sister of C,

C is the father of D,

As the gender of D is not clear, so it cannot be determined.

- (a)$\mathrm{Q}\u2013\mathrm{N}+\mathrm{M}\times \mathrm{P}$
- (b)$\mathrm{P}+\mathrm{S}\times \mathrm{N}-\mathrm{Q}$
- (c)$\mathrm{P}\u2013\mathrm{M}+\mathrm{N}\times \mathrm{Q}$
- (d)$\mathrm{Q}-\mathrm{S}\%\mathrm{P}$

**Hint**

As per instruction replace the sign and check for every option.

**Solution:**

As per instruction given,

$\mathrm{A}+\mathrm{B}$ means $\mathrm{A}$ is the mother of $\mathrm{B}$,

$\mathrm{A}-\mathrm{B}$ means $\mathrm{A}$ is the brother of $\mathrm{B}$,

$\mathrm{A}\%\mathrm{B}$ means $\mathrm{A}$ is the father of $\mathrm{B}$,

$\mathrm{A}\times \mathrm{B}$means $\mathrm{A}$ is the sister of $\mathrm{B}$,

Now from Option A: $\mathrm{Q}-\mathrm{N}+\mathrm{M}\times \mathrm{P}$ means $Q$ is the brother of $\mathrm{N}$, $\mathrm{N}$ is the mother of M and M is sister of $\mathrm{P}$ so $\mathrm{P}$ is not the maternal uncle of $\mathrm{Q}$.

Option B: $\mathrm{P}+\mathrm{S}\times \mathrm{N}-\mathrm{Q}$ means $\mathrm{P}$ is the mother of $\mathrm{S}$, so she cannot be the maternal uncle of $\mathrm{Q}$.

Option C: $\mathrm{P}-\mathrm{M}+\mathrm{N}\times \mathrm{Q}$ means $\mathrm{P}$ is brother of $\mathrm{M}$, $\mathrm{M}$ is mother of $\mathrm{N}$ and $\mathrm{N}$ is sister of $\mathrm{Q}$.

We consider $(+)$ as male member of the family, $(-)$ as female member of the family, vertical single line as generation gap, horizontal single line as relation of sibling

So option C shows $\mathrm{P}$ is the Maternal uncle of $\mathrm{Q}$.

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- (a)East
- (b)West
- (c)North
- (d)South
- (e)South-East

**Hint**

In the morning, the sun will be in the east, and in the evening, the sun will be in the west direction. Use this logic.

**Solution:**

According to the given information in the question, following figure can be drawn,

With the help of above figure, we can find that Vishal is facing south, and Urmila facing north direction.

- (a)North-East
- (b)North-West
- (c)South-East
- (d)South-West

**Hint**

Make the diagram of the given instructions.

**Solution:**

According to the given information in the question, the following figure can be drawn,

It is clear from the diagrams that the new name of the West will become South-East.

- (a)South-East
- (b)South
- (c)North
- (d)West

**Hint**

Read the statement and draw the direction chart and follow the instructions.

**Solution:**

According to the given information in the question, the following figure can be drawn,

Hence, at $9.15\mathrm{P}.\mathrm{M}.$, the minute hand will point towards the west.

- (a)$15\mathrm{m}$ West
- (b)$30\mathrm{m}$ East
- (c)$30\mathrm{m}$ West
- (d)$45\mathrm{m}$ East

**Hint**

The direction is same as in the direction in which the sun rises.

**Solution:**

According to the given information in the question, the following figure can be drawn,

Then, required distance $=\mathrm{AF}=30+15=45.$

From the above figure, F is in the east direction from A.

So the answer is $45\mathrm{m}$ East.

Hence, this option is the correct answer.

- (a)32 m, South
- (b)47 m, East
- (c)42 m, North
- (d)27 m, South

**Hint**

Analyse the statement and follow the steps.

**Solution:**

According to the given information in the question following figure can be drawn,

Hence, Required distance = 20 + 12 = 32 m South.

Therefore, Option A is correct.

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- (a)If statement I is the cause and statement II is it effect
- (b)If statement II is the cause and statement I is its effect
- (c)If both the statements I and II are independent causes
- (d)If both the statements I and II are effect of independent causes
- (e)If both the statements I and II are effect of some common cause

**Hint**

Two different products and two different markets.

**Solution:**

The given statement I have both cause and effect as it states there is an increase in demand for tea that has led to an increase in export to the European market.

Statement II is about the demand for coffee in the domestic market.

As the two statements are about different products and markets, they are effects of independent causes.

- (a)If statement I is the cause and statement II is its effect
- (b)If statement II is the cause and statement I is its effect
- (c)If both the statements I and II are independent causes
- (d)If both the statements I and II are effect of independent causes
- (e)If both the statements I and II are effect of some common cause

**Hint**

The protest has led to intereference.

**Solution:**

According to the statement the parents' protest against high fees being charged by the institutions led the government to interfere and fix the fees at a more affordable level.

Hence, statement II is the cause and statement I is its effect.

- (a)If statement I is the cause and statement II is it effect
- (b)If statement II is the cause and statement I is its effect
- (c)If both the statements I and II are independent causes
- (d)If both the statements I and II are effect of independent causes
- (e)If both the statements I and II are effect of some common cause

**Hint**

The inability of the bank led the Reserve Bank of India make decision.

**Solution:**

According to the statement, the inability of the small banks to compete with the bigger ones shall not ensure security and good service to the customers, which is an essential concomitant that has to be looked into by the Reserve Bank. It seems to be a remedial step for the same.

Hence, statement II is the cause and statement I is its effect.

- (a)If statement I is the cause and statement II is it effect
- (b)If statement II is the cause and statement I is its effect
- (c)If both the statements I and II are independent causes
- (d)If both the statements I and II are effect of independent causes
- (e)If both the statements I and II are effect of some common cause

**Hint**

Try to establish a relation between 'unemployment' and 'large number of applications for the post'.

**Solution:**

According to the statement, an increase in the number of unemployed youth is bound to draw in huge crowds for a single vacancy.

Hence, statement I is the cause and statement II is it effect.

- (a)If statement I is the cause and statement II is it effect
- (b)If statement II is the cause and statement I is its effect
- (c)If both the statements I and II are independent causes
- (d)If both the statements I and II are effect of independent causes
- (e)If both the statements I and II are effect of some common cause

**Hint**

Try to look for the reason for the parents decision.

**Solution:**

According to the statement, it seems quite evident that the parents have instructed their wards to abstain from private tuitions on Sundays and attend special classes organised by the school.

Hence, statement I is the cause and statement II is it effect.

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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**

Financial help is money, while in disaster situation essential items are required.

**Solution:**

From the statement we came to know that there is a natural calamity in the state.

Courses of Action:

1) In the first course of action the government is providing financial assistance but at this situation, affected people only want essential items, not financial help, so this doesn’t follow.

2) Here it is clear that the people and cattle need food, water, and fodder and as per instruction these should be immediately sent to these areas, so this follows.

- (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**

Authorities have to take care for both kinds of people who are infected and who are not.

**Solution:**

From the statement we came to know that in ward J, a large number of people are suffering from a fatal dengue type fever.

Courses of Action:

From the first course of action, the municipal authority should take immediate steps for prevention in ward J and in the second course of action, it is advised to the people to take necessary steps to avoid mosquito bites.

So both the course of action is important and follows.

- (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**

Investing more money is the business does not mean it will give profit.

**Solution:**

From the statement, it is clear that Kingfisher Airlines is in a huge loss.

It is clear from the first action that, it is directed to Airlines to reduce expenditure and increase income by increasing passenger fare, so this is logically following.

In the second course of action, we are talking about spending more amounts and this is not the right way to solve the problem and increase income.

So, the only I follow.

- (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**

One should not take very strict actions.

**Solution:**

From the statement, we came to know that Commercial banks are violating Reserve Bank of India’s guidelines.

To solve the issue the best course of action is to implement the guidelines and act mentioned by Reserve Bank of India and not by suspending officers from commercial banks.

Also, as per the second course of action, it is not logically correct that RBI will stop giving directive to commercial banks. So both the options don’t follow.

- (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**

Awareness of people is very necessary.

**Solution:**

According to the statement, many people are suffering from illness due to drinking polluted water.

Courses of Action:

In the first action, the government is taking adequate steps to provide safe drinking waters to all citizens and it will help in reducing illness.

At the same time, steps should be taken to educate people about the danger of drinking polluted water.

So, both I and II follow.

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- (a)$5$
- (b)$3$
- (c)$4$
- (d)$1$
- (e)None of these

**Hint**

The numbers opposite to $3$ and $5$, will not opposite to $6$.

**Solution:**

From the figures (i), (ii) and (iii) it is clear that the faces adjacent to $3$ have numbers $6,5,1$ and $4$.

So the number on the face opposite to the face $3$ will be $2$.

Similarly, from the given figures it is clear that faces adjacent to $5$ are $6,3,2$ and $4$.

So the number on the face opposite to the face $5$ will be $1$ .

The remaining probable pair of opposite number will be-

$6-4$

Hence, the number on the face opposite the face showing $6$ is $4$.

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

**Hint**

Here, five circles can not be opposite of three circles.

**Solution:**

In both positions, we can see that $2,3,4$ and $5$ can't be the opposite of $1$. So, only $6$ circles remain opposite to $1$.

Also, neither $4$ can be opposite to $2$, nor $3$ can be opposite to $5$.

Further, if the second dice is rotated, such that $1$ circle come at the top, in similarity to the dice in image one.

Now, either $5$ or $3$ is opposite $2$, as $1,6$ and $4$ are adjacent to it.

Hence, it can not be concluded which number of circles is opposite of $2$.

Therefore, the correct answer is none of these.

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

**Hint**

Here, the number $3$ is opposite to $5$ and $2$ is opposite to 6.

**Solution:**

In both positions of Dice, we can see that $1$ is at the top and the numbers $2,3,5$ and $6$, are on adjacent sides. This means $4$ is at the bottom when $1$ is on the top.

Or, when $4$ is at the bottom then the opposite side of $4$ is $1$.

Hence, $1$ is at the top.

- (a)$4$
- (b)$8$
- (c)$12$
- (d)$16$
- (e)None of these

**Hint**

Cube with two sides paint = $12(n-2)$

**Solution:**

Consider, we have a cube:

Let, the sides, EGAC, ABCD and BDFH are painted red.

The sides, ABEF and EFGH are painted yellow.

The side CDGH is painted green.

Since, it is cut into 64 cubes, it means,

$64=4\times 4\times 4$

Number of cuts required is 4.

We know,

Cubes with two sides painting = $12\times (n-2)$

Therefore,

Cubes painted on two sides,

$=12(4-2)\phantom{\rule{0ex}{0ex}}=12\left(2\right)\phantom{\rule{0ex}{0ex}}=24$

So, $24$ cubes will have $2$ side painted, in which $8(24\xf73)$ cubes are of each type (red- yellow, yellow-green and green-red).

Here, in the above diagram, the sides GC, CD and DH are adjacent to red colour sides, while, GH is adjacent to the yellow colour side. Hence, the cubes marked from $1$ to $8$ represent the cubes with only one red and one green face.

Therefore, only $8$ cubes are there which have $2$ side painted with only one red and one green face.

- (a)S
- (b)D
- (c)Y
- (d)W
- (e)None of these

**Hint**

Capture the letters on adjacent sides and find opposites.

**Solution:**

The information drawn from each image:

Image $1$: M and S are on adjacent sides.

Image $2$: M, and W are on adjacent sides.

Image $3$: U and D are on adjacent sides.

From this, it can be concluded that the dice is being rotated clockwise each time.

Such that, in the last image:

The letter on the sides are U, Y, M and S.

Hence, in the fourth image, the letter adjacent of U will be Y.

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

**Hint**

Odd one out is the mirror image of the others.

**Solution:**

All other figures are identical, the two line segments are to the left of the line with a circle in this option.

So, this is an odd one out as this is a mirror image while others are rotated images.

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

**Hint**

Observe the movement followed by the elements of the image.

**Solution:**

Observe there are two circles and a square on each other.

Figures A and E and figures B and C form opposite pairs because the part which is shaded in A is unshaded in E and the part which is unshaded in A is shaded in E, same as in B and C. So only figure D is left single.

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

**Hint**

It includes roman numerals and rotation of elements.

**Solution:**

All other figures are Roman Numerals. They are being rotated $90$ degrees in the anti-clockwise direction with a line.

Thus,

The above figure is different from the rest of the figures.

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

**Hint**

Observe the images for their mirror images.

**Solution:**

All other figures have identical squares shaded with a mirror image pair A -B and D-E. In figure C, one shaded square is on the diagonally opposite corner which makes this an odd one out.

Hence, the answer is

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

**Hint**

The shading in the identical figures is from right to left.

**Solution:**

According to the figure, we analyse that the shaded triangle in each of the figures rises from the right corner of the base. For E, it rises from the left corner of the base.

So, E is the odd one out.

Hence, 'figure E' is the correct answer.

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- (a)$5.78$
- (b)$4.718$
- (c)$4.818$
- (d)$5.818$
- (e)None of these

**Hint**

Solve step by step LHS and RHS and equate to find the value of $?$.

**Solution:**

Given,

$13.243+5.409+?=24.71$

Let the question mark$(?)$ be $x$.

$13.243+5.409+x=24.71$

Now, solve step by step using VBODMAS rule.

$18.652+x=24.71$

$x=24.71-18.652$

$x=6.058$

Hence, the correct answer is $6.058$.

- (a)$6.40$
- (b)$0.640$
- (c)$640$
- (d)$450$
- (e)None of these

**Hint**

$\mathrm{a}\%\mathrm{of}\mathrm{x}=\frac{\mathrm{a}}{100}\times \mathrm{x}$

**Solution:**

It is given;

$16\%$ $\mathrm{of}$ $40=?\%$ of $1$

$\frac{16}{100}\times 40=\frac{?}{100}\times 1$

$\frac{640}{100}=\frac{?}{100}$

$?=640$

- (a)$5$
- (b)$7$
- (c)$4$
- (d)$9$
- (e)None of these

**Hint**

$x\%$ is written as $\frac{x}{100}$.

**Solution:**

$\frac{?}{100}\times 346=10.38$

$?=\frac{1038}{346}$

$?=3$

- (a)$2\frac{7}{9}$
- (b)$\frac{7}{9}$
- (c)$1\frac{1}{7}$
- (d)$1\frac{17}{18}$
- (e)None of these

**Hint**

Apply the concept of BODMAS.

**Solution:**

$3\frac{1}{3}+?-2\frac{2}{4}=2\frac{29}{26}$

$\frac{10}{3}+?-\frac{10}{4}=\frac{81}{26}$

$?=\frac{81}{26}+\frac{10}{4}-\frac{10}{3}$

$?=\frac{486+390-520}{156}$

$?=\frac{876-520}{156}$

$?=\frac{356}{156}$

$?=2.28$

- (a)$2\frac{1}{5}$
- (b)$1\frac{1}{10}$
- (c)$\frac{2}{3}$
- (d)$1\frac{1}{3}$
- (e)None of these

**Hint**

$\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{c}}{\mathrm{d}}+\mathrm{e}=\frac{\mathrm{ad}+\mathrm{bc}+\mathrm{bde}}{\mathrm{bd}}$

**Solution:**

It is given,

$\frac{1}{2}+\frac{1}{3}+?=\frac{3}{2}$

$\frac{5}{6}+?=\frac{3}{2}$

$?=\frac{3}{2}-\frac{5}{6}$

$?=\frac{9-5}{6}$

$?=\frac{4}{6}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}$

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- (a)$99999$
- (b)$99764$
- (c)$99976$
- (d)$99856$
- (e)None of these

**Hint**

Apply the concept of largest digit of a number and square root.

**Solution:**

Largest number of $5$ digits is $99999$, but it is not a perfect square.

If we subtract $143$ from $99999$ we will get a perfect square.

$99999-143=99856$

Square root of $99856=316$

- (a)$1.410$
- (b)$1.412$
- (c)$1.413$
- (d)$1.414$
- (e)None of these

**Hint**

Apply the concept of Non- Perfect Square Root.

**Solution:**

$\sqrt{2}$ is a non perfect square.

The value of $\sqrt{2}$ up to three places of decimals is $1.414$.

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

**Hint**

Use the concept of Perfect Square Root.

**Solution:**

We will solve this question by options

Option $1:$ $294\times 2=548$ is not a perfect square

Option $2:$ $294\times 3=882$ is not a perfect square

Option $3:$ $294\times 6=1764$ is a perfect square

Option $4:$ $294\times 5=1470$ is not a perfect square

- (a)$31$
- (b)$16$
- (c)$7$
- (d)$20$
- (e)None of these

**Hint**

Apply the concept of square root.

**Solution:**

$269$ is nearest to $289$.

$289$ is the square root of $17$.

If we add $20$ to $269$ then it becomes $289$, which is a perfect square.

- (a)$9$
- (b)$50$
- (c)$300$
- (d)$450$
- (e)None of these

**Hint**

Find the Prime Factor of the given number and divide the given number by multiplication of the number which is not making a cube.

**Solution:**

Prime factors of $3600$

$3600=2\times 2\times 2\times 2\times 3\times 3\times 5\times 5$

$=\left(2\times 2\times 2\right)\times \left(2\times 3\times 3\times 5\times 5\right)$

To make $3600$ perfect cube we will divide $3600$ by

$2\times 3\times 3\times 5\times 5=450$

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- (a)$0.0011$
- (b)$0.121$
- (c)$0.1331$
- (d)$12.21$
- (e)None of these

**Hint**

The HCF of fraction numbers is = $\frac{\mathrm{HCF}\mathrm{of}\mathrm{numerator}}{\mathrm{LCM}\mathrm{of}\mathrm{denominator}}$.

**Solution:**

In this question, we have to change all the numbers into fraction format to apply the formula:

HCF of fraction $=\frac{\mathrm{HCF}\hspace{0.17em}\mathrm{of}\mathrm{numerators}}{\mathrm{LCM}\hspace{0.17em}\mathrm{of}\mathrm{denominators}}$.

So, the numbers in the fraction format are

$\frac{11}{1},\frac{121}{1000},\frac{1331}{10000}$.

Now, we have to find the factors of the numerators.

So, $11=11\times 1$

$121=11\times 11\times 1$

$1331=11\times 11\times 11\times 1$

The HCF of these numbers will be the highest common factor that divides each number completely. So, here the HCF is $11$.

Now, the LCM of $11000,10000$ is $10000$.

So, applying the formula, the HCF is $\frac{11}{10000}=0.0011$

- (a)$330$
- (b)$1980$
- (c)$5940$
- (d)$11880$
- (e)None of these

**Hint**

Finding the prime factors and then choosing the maximum power of distinct primes is the key to solve this question.

**Solution:**

To find the LCM of the given numbers, we need to find their factors.

$22=2\times 11$

$54=2\times 3\times 3\times 3=2\times {3}^{3}$

$108=2\times 2\times 3\times 3\times 3={2}^{2}\times {3}^{3}$

$135=5\times 3\times 3\times 3=5\times {3}^{3}$

$198=2\times 3\times 3\times 11=2\times {3}^{2}\times 11$

Now, LCM $=$ (max power of $2$) $\times $ (max power of $3$) $\times $ (max power of $5$) $\times $ (max power of $11$)

$={2}^{2}\times {3}^{3}\times 5\times 11=5940$

- (a)${8}^{-2}$
- (b)${8}^{-3}$
- (c)${8}^{-4}$
- (d)${8}^{-5}$
- (e)None of these

**Hint**

HCF of a fraction $=\frac{\mathrm{HCF}\hspace{0.17em}\mathrm{of}\mathrm{numerators}}{\mathrm{LCM}\hspace{0.17em}\mathrm{of}\mathrm{denominators}}$.

**Solution:**

HCF is the highest common factor. So, here, we can rewrite these as

$\frac{1}{{8}^{2}},\frac{1}{{8}^{3}},\frac{1}{{8}^{4}},\frac{1}{{8}^{5}}$

Now, HCF of fractions $=\frac{\mathrm{HCF}\hspace{0.17em}\mathrm{of}\mathrm{numerators}}{\mathrm{LCM}\hspace{0.17em}\mathrm{of}\mathrm{denominators}}$

HCF of $1,1,1$ is $1$.

Now, LCM of ${8}^{2},{8}^{3},{8}^{4}$ and ${8}^{5}$ is ${8}^{5}$

since the highest power of $8$ among these is $5$.

Now, according to the formula stated above,

HCF $=\frac{\mathrm{HCF}\mathrm{of}\mathrm{numerators}}{\mathrm{LCM}\mathrm{of}\mathrm{denominators}}=\frac{1}{{8}^{5}}={8}^{-5}$

- (a)${4}^{5}$
- (b)${4}^{11}$
- (c)${4}^{15}$
- (d)$4$
- (e)None of these

**Hint**

The highest common factor which divides all three completely is the required HCF.

**Solution:**

HCF of similar bases being raised to different powers is the one which has been raised to least power.

Here, among

${4}^{5},{4}^{11},{4}^{15}$,

The least power is $5$.

Hence, the answer is ${4}^{5}$.

- (a)${2}^{3}$
- (b)${3}^{2}$
- (c)$1$
- (d)$360$
- (e)None of these

**Hint**

We have to see the common factor between the numbers given.The highest common factor of all numbers is HCF.

**Solution:**

According to the question:

HCF of ${2}^{3},{3}^{2},4,15$ is:

By factorisation method, we see that

${2}^{3}=2\times 2\times 2$

${3}^{2}=3\times 3$

$4=2\times 2$

$15=3\times 5$

No common factor among these numbers except $1$.

Hence, the answer is $1$.

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- (a)$100$
- (b)$75$
- (c)$150$
- (d)$50$
- (e)All of these

**Hint**

After the increase, the total number of students in the three classes is $160$.

**Solution:**

According to the question, the initial ratio is $2:3:5$.

Let us take $x$ as the proportionality constant.

Then, the number of students in the three classes are $2x,3x,$ and $5x$.

Now, $20$ students are increased in each class and we are given the new ratio i.e. $4:5:7$.

Let us take $y$ as the new proportionality constant.

Therefore,

$2x+20=4y$ .... (A)

$3x+20=5y$ .... (B)

$5x+20=7y$ .... (C)

Subtracting equation (A) from equation (B), we get:

$(3x+20)-(2x+20)=5y-4y$

$x=y$ .... (D)

Substituting equation (D) in equation (C), we get:

$5y+20=7y$

$2y=20$

$y=10$ .... (E)

Substituting equation (E) in equation (D), we get:

$x=10$ .... (F)

So, the total number of students in all the classes before the increase

$=2x+3x+5x$

$=10x$

$=10\times 10$

$=100$

So, the answer is $100$.

- (a)$12,16$
- (b)$16,12$
- (c)$12,15$
- (d)$13,14$
- (e)None of these

**Hint**

Let, the two numbers be $3x,4x$ then, $\frac{3x+2}{4x+2}=\frac{7}{9}$, solve this equation.

**Solution:**

According to the question, the ratio between two numbers is $3:4$. When each number is increased by $2$, the ratio becomes $7:9$.

Let numbers be $3x$ and $4x$.

$\frac{3x+2}{4x+2}=\frac{7}{9}$

$\Rightarrow 27x+18=28x+14$

$\Rightarrow x=4$

Hence, the numbers are $3\times 4=12$ and $4\times 4=16$.

- (a)$285,330,830$
- (b)$280,330,930$
- (c)$280,330,980$
- (d)$330,380,980$
- (e)None of these

**Hint**

The sum is divided among $A,B,$ and $C$ in the ratio $28:33:93$.

**Solution:**

According to the question,

$A$'s share $(B+C)$'s share $=2:9$ ... $\left(1\right)$

$B$'s share $(A+C)$'s share $=3:11$ ... $\left(2\right)$

Now dividing $Rs.1540$ in the ratio of $2:9$ and $3:11$,

$A$'s share $=\frac{2}{11}$ of $Rs.1540=Rs.280$

$B$'s share $=\frac{3}{14}$ of $Rs.1540=Rs.330$

$C$'s share

$=Rs.1540-(Rs.280+Rs.330)=Rs.930$

- (a)$45$ days
- (b)$35$ days
- (c)$44$ days
- (d)$53$ days
- (e)None of these

**Hint**

For the same provision, the number of men is inversely proportional to the number of days the said provision will last.

**Solution:**

According to the question,

Let the required number of days be $x$.

When there are more men, the provision will last for less days.

When there are less men, the provision will last for more days.

So,

$275:x::\left(275-125\right):\left(40-16\right)$

So, $x=\frac{275\times 24}{150}=44$ days.

- (a)$1225$
- (b)$1572$
- (c)$1125$
- (d)$1229$
- (e)None of these

**Hint**

For the same amount of provisions available, the number of men is inversely proportional to the number of days the said provisions will last.

**Solution:**

Let, the number of persons initially in the fort be $x.$

According to the question,

$\left(35-5\right)x=25\left(x+225\right)$

$\Rightarrow 30x-25x=25\times 225$

$\Rightarrow x=\frac{25\times 225}{5}$

$\Rightarrow x=1125$ persons

Hence, the correct answer is $1125$.

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- (a)$3$ months
- (b)$4$ months
- (c)$5$ months
- (d)$6$ months
- (e)$8$ months

**Hint**

Suppose $A$ invests $Rs.x$ for $p$ months and $B$ invests $Rs.y$ for $q$ months then,

($A$'s share of profit) $:$($B$' s share of profit) $=xp:yq$.

**Solution:**

Bhavana began a business with $Rs.2100$ and is joined afterwards by $Y$ with $Rs.3600$.

According to the question,

$\frac{2100\times 12}{3600\times \mathrm{x}}=\frac{1}{1}$

Hence, $Y$ joins after $12-7=5$ months

- (a)$7$ months
- (b)$5$ months
- (c)$8$ months
- (d)$4$ months
- (e)None of these

**Hint**

Suppose $A$ invests $Rs.x$ for $p$ months and $B$ invests $Rs.y$ for $q$ months then,

($A$'s share of profit) $:$ ($B$'s share of profit) $=xp:yq$

**Solution:**

According to the question,

$A$ contributes $Rs.3500$ for $8$ months $B$ contributes $Rs.4,000$

Now,

$\frac{3500\times 8}{4000\times x}=\frac{1}{1}\Rightarrow x=7$ months

Hence, the correct answer is $7$ months

- (a)$Rs.20000$
- (b)$Rs.35000$
- (c)$Rs.30000$
- (d)$Rs.19000$
- (e)None of these

**Hint**

Suppose $A$ invests $Rs.x$ for $p$ months and $B$ invests $Rs.y$ for $q$ months then,

($A$'s share of profit) $:$ ($B$'s share of profit) $=xp:yq$.

**Solution:**

According to the question,

$A$'s capital $=30000$

$B$ joins four months later and profits are in the ratio $9:4$.

To get the capital of $B$

$\frac{30000\times 12}{x\times 8}=\frac{9}{4}$

So, $x=20000$

Hence, the correct answer is $20000$.

- (a)$\mathrm{Rs}.2000$
- (b)$\mathrm{Rs}.1800$
- (c)$\mathrm{Rs}.3600$
- (d)$\mathrm{Rs}.4500$
- (e)None of these

**Hint**

Suppose $A$ invests $\mathrm{Rs}.x$ for $p$ months and $B$ invests $\mathrm{Rs}.y$ for $q$ months then,

($A$'s share of profit) $:$ ($B$'s share of profit) $=xp:yq$

**Solution:**

According to the question,

$A$ invests $3000$ for one year in the business.

The profit after one year is divided in the ratio $2:3$.

We know that, suppose $A$ invests $\mathrm{Rs}.x$ for $p$ months and $B$ invests $\mathrm{Rs}.y$ for $q$ months then,

($A$'s share of profit) $:$ ($B$'s share of profit) $=xp:yq$.

Let, the investment of $B$ is $\mathrm{Rs}.x$.

To get investment of $B$

$\frac{3000\times 12}{x\times 12}=\frac{2}{3}$

$\Rightarrow x=4500$

Hence, the correct answer is $4500$.

- (a)$\mathrm{Rs}.10$
- (b)$\mathrm{Rs}.15$
- (c)$\mathrm{Rs}.20$
- (d)$\mathrm{Rs}.25$
- (e)None of these

**Hint**

If $A$'s share is $x$ then, $B$'s share is $x-20$ and $C$'s share is $x+20$.

**Solution:**

Given that $\mathrm{Rs}.120$ was divided between $A,B$ and $C$.

Let $A$'s share be $x$, then $B$'s share will be $x-20$ and $C$'s share will be $x+20$

We know that $A$'s share $+B$'s share $+C$'s share $=120$

$\Rightarrow x+(x-20)+(x+20)=120$

$\Rightarrow 3x=120$

$\Rightarrow x=40$

Hence, $B$'s share $=\left(x-20\right)=40-20=\mathrm{Rs}.20$

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- (a)$11$
- (b)$81.5$
- (c)$5.5$
- (d)$815$
- (e)None of these

**Hint**

Take the value which is to be found as y and then start solving by converting the percentage into fractional form.

**Solution:**

Let the value of '$?$' be $y$.

Now, as per the question.

$y\times 15=37.5\%$ of $220$

$15y=\frac{37.5}{100}\times 220$

$15y=82.5$

$y=\frac{82.5}{15}=5.5$

- (a)$5163$
- (b)$5963$
- (c)$0$
- (d)$1$
- (e)None of these

**Hint**

First find the individual percentages of the numbers and divide both of them

**Solution:**

First we calculate, both the percentages individually

$67\%$ of $89$

$\frac{67}{100}\times 89=59.63$

$89\%$ of $67$

$\frac{89}{100}\times 67=59.63$

Now, on dividing the two results we get

$\frac{67\%\mathrm{of}89}{89\%\mathrm{of}67}=\frac{59.63}{59.63}=1$

- (a)$10.1$
- (b)$0.101$
- (c)$101$
- (d)$1.01$
- (e)None of these

**Hint**

Use the concept of converting percentages into fractions.

**Solution:**

Given expression:

$10\%\mathrm{of}?=0.101$

Let the number be $y$.

Now, we can write the equation as $10\%$ of $y=0.101$

$\frac{10}{100}\times y=.101$

$y=.101\times \frac{100}{10}=1.01$

Hence, The correct answer is $1.01$.

- (a)$7$
- (b)$\frac{2}{7}$
- (c)$\frac{1}{7}$
- (d)$\frac{3}{7}$
- (e)None of these

**Hint**

First find $20\%$ of $740$ and then apply appropriate mathematical adjustments which are used to solve simple equations.

**Solution:**

Let the number be $y$

So, we can write the equation as

$\frac{20\%\mathrm{of}740}{y}=1036$

$20\%$ of $740$

$=\frac{20}{100}\times 740=148$

Now, multiply both sides by $y$

$\frac{148}{y}\times y=1036\times y$

$\Rightarrow 1036\times y=148$

$\Rightarrow y=\frac{148}{1036}=\frac{1}{7}$

- (a)$24$
- (b)$25$
- (c)$27$
- (d)$28$
- (e)$23$

**Hint**

Convert the percentage into fractional form and then solve the equation

**Solution:**

Given expression:

$12\%$ of $200$

It can be written mathematically as

$\frac{12}{100}\times 200\phantom{\rule{0ex}{0ex}}=12\times 2=24$

Hence, the required answer is $24$.

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- (a)$66kg$
- (b)$75kg$
- (c)$67.6kg$
- (d)$76kg$
- (e)None of these

**Hint**

To determine the weight of the new person, just start by taking the product of total number of persons and new average weight and then take the sum of the weight obtained by that product and the weight of the old person who is replaced.

**Solution:**

Total number of persons $=8$

And their average weight is increased by $2.5kg$.

Now, total weight increased $=$ (total number of persons) $\times $(increase in average weight)

Total weight increased $=8\times 2.5=20kg$

Hence, the weight of new man $=56+20=76kg$

- (a)$28$
- (b)$50$
- (c)$48$
- (d)$49$
- (e)None of these

**Hint**

Use the formula, Average of first $n$ consecutive natural numbers is $\frac{n+1}{2}$

**Solution:**

Average of first $n$ natural numbers is given by,

Average $=\frac{n+1}{2}$

From the question, we know that $n=49$

So, average of consecutive natural numbers from $1$ to $49$ is given by

Average $=\frac{49+1}{2}=\frac{50}{2}=25$

- (a)$16$ approximately
- (b)$18$ approximately
- (c)$19.50$ approximately
- (d)$17$ approximately
- (e)None of these

**Hint**

Use the formula, Average daily expenditure $=\frac{Totalexpenditure}{Totaldays}$

**Solution:**

Since the average of daily expenditure is given for September, October and November, which are $15,20$ and $13$ respectively, we can find the average daily expenditure by the formula

Average daily expenditure

$=\frac{\mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{average}\mathrm{expenditure}\mathrm{for}\mathrm{individual}\mathrm{months}}{\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{months}}$

$=\frac{15+20+13}{3}=\frac{48}{3}=16$

- (a)$15$
- (b)$9$
- (c)$12$
- (d)$3$
- (e)None of these

**Hint**

The average of the first five multiples of $3$ $=\frac{\mathrm{Sum}\mathrm{of}\mathrm{first}\mathrm{five}\mathrm{multiples}of3}{5}$.

**Solution:**

The average of the first five multiples of $3$ will be given by

$\frac{\mathrm{Sum}\mathrm{of}\mathrm{first}\mathrm{five}\mathrm{multiples}}{5}$

First five multiples of $3$ are $3,6,9,12,15$.

$\Rightarrow $ Average

$=\frac{3+6+9+12+15}{5}\phantom{\rule{0ex}{0ex}}=\frac{45}{5}\phantom{\rule{0ex}{0ex}}=9$

Hence, the correct answer is $9$.

- (a)$25$ years
- (b)$20$ years
- (c)$28$ years
- (d)$30$ years
- (e)None of these

**Hint**

Ritu is $45$ years old.

**Solution:**

Let, the age of Ram be $y$ years and the age of Ritu be $z$ years.

As per the question, difference between their ages is $15$ years.

$\Rightarrow \left(z-y\right)=15$

$\Rightarrow z=\left(15+y\right)$

It is also given that the ratio of their ages is $3:2$.

$\frac{3}{2}=\frac{z}{y}$

$3y=2z$

$3y-2z=0$

Now put the value of $z$,

$3y-(2\times \left(15+y\right))=0$

$3y-30-2y=0$

$y-30=0$

$y=30$

So, $30$ years is the required answer.

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- (a)$33\frac{1}{2}\%$
- (b)$33\frac{1}{3}\%$
- (c)$33\%$
- (d)$34\frac{1}{3}\%$
- (e)None of these

**Hint**

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

**Solution:**

Let the cost price of $1$ metre cloth be $Rs.1$.

CP of $66m$ cloth be $Rs.66$.

According to the question,

after selling $66$ metres of cloth he gained the cost of $22$ metres

Gain $\%=\frac{22}{66}\times 100=33\frac{1}{3}\%$

Hence, the correct answer is $33\frac{1}{3}\%$.

- (a)$55.5\%$
- (b)$50.5\%$
- (c)$40\%$
- (d)$45\%$
- (e)None of these

**Hint**

Gain $\%=\frac{\mathrm{Selling}\mathrm{price}-\mathrm{Cost}\mathrm{price}}{\mathrm{cost}\mathrm{price}}\times 100$

**Solution:**

Since we know from the question that the cost price of buying $87$ goods is same as the selling price of $60$ goods i.e. $890$ rupees.

CP of $1$ unit of goods is $\frac{890}{87}$ rupees.

SP of $1$ unit of goods is $\frac{890}{60}$ rupees.

So, gain percentage can be calculated as

$\frac{{\displaystyle \frac{890}{60}}-{\displaystyle \frac{890}{87}}}{{\displaystyle \frac{890}{87}}}\times 100=45\%$

- (a)$330.50$
- (b)$337.50$
- (c)$300.70$
- (d)$331.50$
- (e)None of these

**Hint**

Selling price $=\mathrm{CP}-\frac{\mathrm{Loss}\%\times \mathrm{CP}}{100}$

**Solution:**

Let the cost price of the article be $y$.

Now, when the article is sold at a loss of $17.8\%$, selling price is calculated by

Selling price

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

${\mathrm{SP}}_{1}=y-\frac{17.8\times y}{100}$

${\mathrm{SP}}_{1}=y-0.178y=0.822y,$ this is the selling price of the loss of $17.8\%$

Now, for the loss of $34.8\%$, selling price is given by,

${\mathrm{SP}}_{2}=\mathrm{CP}-\frac{\mathrm{Loss}\%\times \mathrm{CP}}{100}$

${\mathrm{SP}}_{2}=y-\frac{34.8\times y}{100}$

${\mathrm{SP}}_{2}=y-0.348y$

${\mathrm{SP}}_{2}=0.652y$

It is also mentioned in the question that as compared to the loss of $17.8\%$, seller got $Rs.19.50$ less for the loss of $34.8\%$.

So, this can be written as

${\mathrm{SP}}_{1}-{\mathrm{SP}}_{2}=19.5$

$0.822y-0.652y=19.5$

$0.17y=19.5$

$y=\frac{19.5}{0.17}=114.7$

So, cost price of the article is $Rs.114.7$

- (a)$750,600$
- (b)$650,500$
- (c)$700,650$
- (d)$600,750$
- (e)None of these

**Hint**

Equate the profit percent of one horse and the loss percent of second horse.

**Solution:**

Let the cost of first horse be $Rs.y$ and the cost of second horse be $Rs.z$.

It is given in the question that the man buys two horses at $Rs.1350$. This can be written as,

$y+z=1350$

$z=1350-y$

Also, it is given that the man neither loses nor gains when he sells one horse at a loss of $6\%$ and other at a profit of $7.5\%$. This can be written as

$6\%$ of $y=7.5\%$ of $z$

$0.06y=0.075z$

$0.06y-0.075z=0$

Now, putting the value of $z$ from step $1$

$0.06y-0.075\times (1350-y)=0$

$0.06y-101.25+0.075y=0$

$0.135y=101.25$

$y=\frac{101.25}{0.135}=750$

$z=1350-y$

$z=1350-750=600$

So, cost of one horse is $Rs.750$ and cost of other horse is $Rs.600$.

- (a)$\frac{160}{206}\%$ gain
- (b)$\frac{160}{203}\%$ gain
- (c)$\frac{160}{205}\%$ loss
- (d)$\frac{160}{203}\%$ loss
- (e)None of these

**Hint**

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

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

**Solution:**

It is given in the question that the selling price of both the cars is same.

Let us assume that the selling price is $Rs.100$.

Now it is given in the question that first car is sold at a profit of $10\%$.

So, cost price of first car will be given by

$C{P}_{1}=\frac{100}{1.1}=Rs.90.9$

Similarly, is given that the second car is sold at a loss of $7\%$.

So, cost price of second car will be

$C{P}_{2}=\frac{100}{0.93}=Rs.107.52$

Also, $S{P}_{1}=S{P}_{2}=Rs.100$

Here subscript $1$ and $2$ represent the first and second car respectively.

Now, to determine combined gain or loss,

$C{P}_{\mathrm{total}}=C{P}_{1}+C{P}_{2}=90.9+107.52=Rs.198.42$

Similarly, $S{P}_{\mathrm{total}}=S{P}_{1}+S{P}_{2}=100+100=Rs.200$

Since, $SP>CP$, there will be gain

Gain $\%=\frac{S{P}_{\mathrm{total}}-C{P}_{\mathrm{total}}}{C{P}_{\mathrm{total}}}\times 100$

Gain $\%=\frac{200-198.42}{198.42}\times 100$

Gain $\%=\frac{1.58}{198.42}\times 100=0.79\%=\frac{160}{203}\%$

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- (a)$25\%$
- (b)$6\%$
- (c)$6.66\%$
- (d)$8\%$
- (e)None of these

**Hint**

Use the formula, Simple interest $=\frac{Principal\times Rate\times time}{100}$

**Solution:**

When the amount doubles itself in a specified time, we can calculate its rate by

Rate $=\frac{100}{time}$

In the question, it is given that time is $15$ years.

Rate $=\frac{100}{15}=6.66\%$

Hence, the required answer is $6.66\%$.

- (a)$3\%$
- (b)$5\%$
- (c)$12\%$
- (d)$2.40\%$
- (e)None of these

**Hint**

Simple interest $=\frac{principal\times rate\times time}{100}$

**Solution:**

According to the question,

A sum becomes $\frac{28}{25}$ of itself in $5$ years

Suppose, principal amount $=P$

Amount (after $5$ years)

$=\frac{28}{25}P$

Simple interest,

$\mathrm{SI}=\frac{28}{25}P-P=\frac{3P}{25}$

We know,$\mathrm{SI}=\frac{P\times r\times t}{100}$

$\frac{3P}{25}=\frac{P\times 5\times r}{100}$

$\Rightarrow r=\frac{3\times 100}{25\times 5}=\frac{12}{5}=2.4\%$

- (a)$Rs.1,020$
- (b)$Rs.1,050$
- (c)$Rs.1,080$
- (d)$Rs.1,200$
- (e)None of these

**Hint**

Simple interest $=\frac{principal\times rate\times time}{100}$

**Solution:**

According to the question,

If $P=900$,

Amount $=Rs.1,080$ ; $T=4$ years

$\mathrm{SI}=1080-900=Rs.180$

Rate, $r=\frac{\mathrm{SI}\times 100}{\mathrm{P}\times \mathrm{T}}$

$r=\frac{180\times 100}{4\times 900}=5\%$

Let $P=Rs.100$

$\mathrm{SI}=100+25=125$

When amount is $125$ then, $P=100$

When amount is $1275$ then,

$P=\frac{100}{125}\times 1275$

$=Rs.1020$

- (a)$Rs.650$
- (b)$Rs.625$
- (c)$Rs.500$
- (d)$Rs.475$
- (e)None of these

**Hint**

The formula to calculate sum is given by

Sum $=\frac{100\times \mathrm{Amount}}{100+rT}$

**Solution:**

From the question, we can conclude that time for the sum of money to be amounted to $Rs.575$ is same as the time in which $Rs.750$ is amounted to $Rs.840$.

So,

$\mathrm{SI}=840-750=Rs.90$

Now, $T=\frac{100\times \mathrm{SI}}{P\times r}$

$T=\frac{100\times 90}{750\times 4}=3$ years

Now, the formula to calculate sum is given by

Sum $=\frac{100\times \mathrm{Amount}}{100+rT}$

Sum $=\frac{100\times 575}{100+(5\times 3)}$

Sum $=\frac{57500}{115}=Rs.500$

- (a)$Rs.600$
- (b)$Rs.650$
- (c)$Rs.700$
- (d)$Rs.675$
- (e)None of these

**Hint**

Simple Interest $=\frac{\mathrm{Principal}\times \mathrm{rate}\times \mathrm{time}}{100}$

**Solution:**

According to the question,

$\mathrm{S}.\mathrm{I}.=806-767=39$

We know,

Principal $\left(P\right)=\frac{100\times \mathrm{SI}}{\mathrm{Rate}\times \mathrm{time}}$

$P=\frac{39\times 100}{1\times 6}=Rs.650$

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- (a)$4\%$
- (b)$5\%$
- (c)$6\%$
- (d)$8\%$
- (e)$2.5\%$

**Hint**

Compound interest $\left(\mathrm{CI}\right)=$ Amount $-$ Principal,

Where Amount $=P{\left(1+\frac{R}{100}\right)}^{n}$, $n=$ time in years.

**Solution:**

We know, in compound interest,

Amount $=P\times {\left(1+\frac{R}{100}\right)}^{N}$

Where $P=$ Principal amount, $R=$ rate and $N=$ time in year

Compound interest $\left(\mathrm{CI}\right)=$ Amount $-P$

$41=P{\left(1+\frac{R}{100}\right)}^{2}-P$

$41=P\left(1+\frac{{R}^{2}}{{10}^{4}}+\frac{2R}{100}-1\right)$

$41=P\left(\frac{{R}^{2}}{{10}^{4}}+\frac{2R}{100}\right)$

Now, we know simple interest $=\frac{P\times R\times T}{100}$

$40=\frac{P\times R\times 2}{100}$

$4000=2PR$

$P=\frac{2000}{R}$

Now putting this value of $P$ in the expression obtained in ${1}^{st}$ step $41$

$=P\left(\frac{{R}^{2}}{{10}^{4}}+\frac{2R}{100}\right)$

$41=\frac{2000}{R}\left(\frac{{R}^{2}}{{10}^{4}}+\frac{2R}{100}\right)$

$41=\frac{2000R}{10000}+\frac{4000}{100}$

$41=0.2R+40$

$0.2R=1$

$R=\frac{1}{0.2}=5\%$

- (a)$18$ years
- (b)$12$ years
- (c)$9$ years
- (d)$6$ years
- (e)None of these

**Hint**

Compound interest $\left(\mathrm{CI}\right)=P{\left(1+\frac{R}{100}\right)}^{n}-P$

**Solution:**

It is given in the question that the amount triples in $3$ years.

So, we can conclude that it becomes $9$ times itself in $(3+3)=6$.

So, $6$ years is the correct answer.

- (a)$\frac{{\left(\frac{r}{100}\right)}^{2}}{P}$
- (b)$\frac{{\left(\frac{100}{r}\right)}^{2}}{P}$
- (c)$\frac{P}{{\left(\frac{100}{r}\right)}^{2}}$
- (d)$P{\left(\frac{100}{r}\right)}^{2}$
- (e)None of these

**Hint**

Compound interest $=P{\left(1+\frac{r}{100}\right)}^{n}-P,$ $n=$ time in years.

**Solution:**

As per the question,

Required difference

$=P\left[{\left(1+\frac{r}{100}\right)}^{2}-1\right]-\frac{P\times r\times 2}{100}$

$=P\left[1+{\left(\frac{r}{100}\right)}^{2}+\left(\frac{2r}{100}\right)-1\right]-\frac{2Pr}{100}$

$=P\left[{\left(\frac{r}{100}\right)}^{2}+\frac{2r}{100}-\frac{2r}{100}\right]$

$=P{\left[\frac{r}{100}\right]}^{2}=\frac{P}{{\left(\frac{100}{r}\right)}^{2}}$

- (a)$16$ years
- (b)$8$ years
- (c)$12$ years
- (d)$20$ years
- (e)None of these

**Hint**

Compound Interest, $\mathrm{CI}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}-\mathrm{P}$.

**Solution:**

It is given in the question that the amount doubles itself in $4$ years when placed at a compound interest.

Now, we can conclude that the amount will be $4$ times of itself in $8$ years.

Similarly, it will be $8$ times of itself in $12$ years.

Hence, $12$ years is the required answer.

- (a)$\mathrm{Rs}.600$
- (b)$\mathrm{Rs}.500$
- (c)$\mathrm{Rs}.400$
- (d)$\mathrm{Rs}.300$
- (e)None of these

**Hint**

$\mathrm{CI}-\mathrm{SI}=\mathrm{P}{\left(1+\frac{\mathrm{r}}{100}\right)}^{\mathrm{n}}-\frac{\mathrm{P}\times \mathrm{r}\times \mathrm{t}}{100}.$

**Solution:**

According to the question,

$\Rightarrow $Compound interest $-$ Simple interest

$\Rightarrow \mathrm{P}{\left(1+\frac{\mathrm{r}}{100}\right)}^{\mathrm{n}}-\mathrm{P}-\frac{\mathrm{P}\times \mathrm{r}\times \mathrm{t}}{100}=1.50\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{P}{\left(1+\frac{5}{100}\right)}^{2}-\mathrm{P}-\frac{\mathrm{P}\times 5\times 2}{100}=1.50\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{441\mathrm{P}}{400}-\mathrm{P}-\frac{\mathrm{P}}{10}=1.50\phantom{\rule{0ex}{0ex}}\Rightarrow 441\mathrm{P}-40\mathrm{P}-400\mathrm{P}=600\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{P}=600\phantom{\rule{0ex}{0ex}}$

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- (a)$35\%$
- (b)$30\%$
- (c)$40\%$
- (d)$45\%$
- (e)None of the these

**Hint**

Use the formula, $S.P.=C.P.+\left(\mathrm{Profit}\%\times C.P.\right)$

**Solution:**

Given that:

Cost Price $=Rs.450$

One-third of the goods are sold at a loss of $10\%$.

$CP$ of one-third goods $=\frac{1}{3}\times 450=Rs.150$

$SP$ of one-third goods $=CP$ of one-third goods $-$ Loss $\%$

$=150-10\%\mathrm{of}150$ $=150-(0.1\times 150)=150-15=Rs.135$.

$SP$ of total goods for $20\%$ profit $=CP+$ Profit $\%$

$=450+20\%\mathrm{of}450$ $=450+(0.2\times 450)=Rs.540$.

Hence, $SP$ of two-third goods $=540-135=Rs.405$.

$CP$ of two-third goods $=\frac{2}{3}\times 450=Rs.300$

Profit of two-third goods $=SP$ of two-third goods $-CP$ of two-third goods $=405-300=Rs.105$.

Profit $\%$ of remaining two-third goods

$=\left(\frac{\mathrm{Profit}\hspace{0.17em}\mathrm{of}\hspace{0.17em}\mathrm{two}\hspace{0.17em}\mathrm{third}\hspace{0.17em}\mathrm{goods}}{\mathrm{CP}\hspace{0.17em}\mathrm{of}\hspace{0.17em}\mathrm{two}\hspace{0.17em}\mathrm{third}\hspace{0.17em}\mathrm{goods}}\right)\times 100$

$=\left(\frac{105}{300}\right)\times 100=35\%$

Hence, the required answer is $35\%$.

- (a)$Rs.7.50$
- (b)$Rs.9$
- (c)$Rs.8.20$
- (d)$Rs.8.85$
- (e)None of these

**Hint**

$SP=CP\times \frac{gainover100}{100}$

**Solution:**

According to the question, $50kg$ sugar worth $6.75$ per $kg$ is mixed with $120kg$sugar worth of $8$ per $kg$.

$20\%$ profit means gain over $Rs.100$ is $Rs.120$.

$CP=50\times 6.75+120\times 8=Rs.1500$

$SP$ per $kg$ $=\frac{120\times 1500}{100\times 200}=Rs.9$

- (a)$20kg$
- (b)$30kg$
- (c)$15kg$
- (d)$35kg$
- (e)None of these

**Hint**

$S.P.=C.P.+(Profit\%ofC.P.)$

**Solution:**

Let $C.P.$ per $kg$ of sugar be $Rs.x$

Therefore, $C.P.$ of $50kg$ of sugar $=50x$

Profit on $50kg$ of sugar $=14\%$

Hence, $S.P.$ on $50kg$ of sugar

$=50x+(14\%\mathrm{of}50x)=50x+7x=57x$

Let the quantity sold at $18\%$ profit be $ykg$.

Hence, the quantity sold at $8\%$ profit is $(50-y)kg$.

$C.P.$ of $ykg$ of sugar $=yx$

$S.P.$ on $ykg$ of sugar

$=yx+(18\%\mathrm{of}yx)=yx+0.18yx=1.18yx$

$C.P.$ of $(50-y)kg$ of sugar $=(50-y)\times x$

$S.P.$ on $(50-y)kg$ of sugar

$=(50-y)\times x+(8\%\mathrm{of}(50-y)\times x)$

$=(50-y)\times x+0.08(50-y)\times x=1.08(50-y)\times xy$

Hence, total $S.P.$

$=1.18yx+1.08(50-y)\times x$$=57\times 1.18yx+54x-1.08yx=57\times 0.1yx=3x$

$y=\frac{3x}{0.1x}=30kg$

Hence, the quantity of sugar sold at $18\%$ profit is $30kg$.

- (a)$52km$
- (b)$48km$
- (c)$36km$
- (d)$44km$
- (e)None of these

**Hint**

Total time $=$ Sum of times taken to travel by bicycle and by foot.

Time $=$ Distance / Speed

**Solution:**

Let the distance travelled by bicycle be $xkm$.

$\therefore $ The distance travelled on foot will be $\left(60-x\right)km$.

According to question,

$\frac{60-x}{8}+\frac{x}{16}=7$

$\Rightarrow \frac{120-2x+x}{16}=7$

$\Rightarrow x=8$

Distance travelled on foot $60-8=52km$

- (a)$1:5$
- (b)$1:6$
- (c)$1:7$
- (d)$1:9$
- (e)None of these

**Hint**

$\frac{water}{wine}=gain\%$

**Solution:**

According to the question

Selling the mixture at the cost price a profit of $20\%$

To get the ratio of water and wine

Water : Wine $=20:100=1:5$.

Hence, The correct OPTION is A i.e $1:5$.

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- (a)$10$ days
- (b)$11$ days
- (c)$20$ days
- (d)$33\frac{1}{11}$ days
- (e)None of these

**Hint**

If a men can do a work in $\mathrm{A}$ days then one day work$=\frac{1}{\mathrm{A}}$.

**Solution:**

According to the question,

$\mathrm{A}$ can do piece of work in $25$ days

$\mathrm{B}$ can finish it in $25$ days.

They work together for $5$ days and then $\mathrm{A}$ goes away

To get in how many days will $\mathrm{B}$ finish the work

$\left(\mathrm{A}+\mathrm{B}\right)$'s $5$ day’s work $=\left(\frac{1}{25}+\frac{1}{20}\right)\times 5$

$=\frac{9}{20}$

Remaining work $=1-\frac{9}{20}=\frac{11}{20}$$\mathrm{B}$ will finish it in $=\frac{11}{20}\times 20=11$ days.

- (a)$15$ days
- (b)$12\frac{1}{2}$ days
- (c)$16\frac{1}{2}$ days
- (d)$14\frac{2}{9}$ days
- (e)None of these

**Hint**

$\mathrm{Remaining}\mathrm{work}\mathrm{done}\mathrm{by}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{together}=1-(\mathrm{work}\mathrm{done}\mathrm{by}\mathrm{A}\mathrm{for}10\mathrm{days}).$

**Solution:**

According to the question,

$\mathrm{A}$ can do a piece of work in $25$ days which $\mathrm{B}$ alone can do in $20$ days

$\mathrm{A}$'s $10$ day’s work $=\frac{10}{25}=\frac{2}{5}$

Remaining work $=1-\frac{2}{5}=\frac{3}{5}$

$\left(\mathrm{A}+\mathrm{B}\right)$'s $1$ day’s work $\frac{1}{25}+\frac{1}{20}=\frac{9}{100}$

So, $\left(\mathrm{A}+\mathrm{B}\right)$ complete $\frac{3}{5}$ of the work in

$=\frac{100}{9}\times \frac{3}{5}=6\frac{2}{3}$ daysSo, work lasted for $=10+6\frac{2}{3}=16\frac{2}{3}$ days.

Hence, the correct answer is $16\frac{2}{3}$ days.

- (a)$13$ days
- (b)$12\frac{1}{2}$ days
- (c)$15$ days
- (d)$18\frac{1}{7}$ days
- (e)None of these

**Hint**

$\mathrm{B}\text{'}\mathrm{s}\mathrm{part}\mathrm{of}\mathrm{work}=1-(\mathrm{work}\mathrm{done}\mathrm{by}\mathrm{in}5\mathrm{days}).$

**Solution:**

According to the question,

$\mathrm{A}$ can do a piece of work in $40$ days.

He worked at it for $5$ days and $\mathrm{B}$ finished the remaining work in $21$ days.

$\mathrm{A}$’s $5$ day’s work $=\frac{5}{40}=\frac{1}{8}$

Remaining work $=1-\frac{1}{8}=\frac{7}{8}$

$\mathrm{B}$ can do a piece of work in $=\frac{8}{7}\times 21=24$ days

i.e. $\mathrm{A}=\frac{1}{40},\mathrm{B}=\frac{1}{24}$

So, $\left(\mathrm{A}+\mathrm{B}\right)$ can complete the work in $=\frac{40\times 24}{40+24}=15$ days

Hence, the correct answer is $15$ days.

- (a)$60$ days
- (b)$45$ days
- (c)$20$ days
- (d)$40$ days
- (e)None of these

**Hint**

Consider $\mathrm{A}=3\mathrm{B}$ and calculate work done by $\mathrm{B}$.

**Solution:**

According to the question,

$\mathrm{A}$ is thrice as good a workman as $\mathrm{B}$ i.e $\mathrm{A}=3\mathrm{B}$.

Together they can do a job in $15$ days then one day work,

$\frac{1}{\mathrm{A}}+\frac{1}{\mathrm{B}}=\frac{1}{15}$

But $\mathrm{A}$ is thrice as good as $\mathrm{B}$ then,

$\frac{1}{\mathrm{B}}+\frac{1}{3\mathrm{B}}=\frac{1}{15}\phantom{\rule{0ex}{0ex}}\frac{4}{3\mathrm{B}}=\frac{1}{15}\phantom{\rule{0ex}{0ex}}\mathrm{B}=20$

So,

$3\mathrm{B}=3\times 20=60$.

Hence, the correct answer is $60$ days.

- (a)$5$ days
- (b)$7$ days
- (c)$6\frac{3}{4}$ days
- (d)$7\frac{1}{2}$ days
- (e)None of these

**Hint**

Number of days $=\frac{\mathrm{Total}\mathrm{work}}{\mathrm{Efficiency}}$

**Solution:**

According to the question,

$A$ can do a work in $8$ days.

$B$ can do in $6$ days.

$A$ and $B$ can do the work on alternate days.

$A$ and $B$ together can do $=\frac{1}{8}+\frac{1}{6}=\frac{14}{48}$ of total work in a day.

If $48$ be the total work and $A$ and $B$ takes $2$ days to do $14$ of it. They together complete $42$ of the total work in $6$ days and after that $A$ will complete the remaining work in $7$th day.

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- (a)$9$ a.m.
- (b)$10$ a.m.
- (c)$11$ a.m.
- (d)$10:30$ a.m.
- (e)None of these

**Hint**

Total distance $=$ sum of distance covered by train $A$ and train $B$ by the time they meet

**Solution:**

According to the question

Let they meet in $\mathrm{x}$ hours after $7$ a.m.

Distance covered by $A$ in $\mathrm{x}$ hours $=20\mathrm{x}\mathrm{km}$

Distance covered by $B$ in $\left(\mathrm{x}-1\right)$ hours $=25\left(\mathrm{x}-1\right)\mathrm{km}$

To get the time they meet

$\because 20\mathrm{x}+25\left(\mathrm{x}-1\right)=110$

$\Rightarrow 45\mathrm{x}=135$

$\Rightarrow \mathrm{x}=3$

So, they meet at $10$ a.m.

Hence, the correct answer is $10$ a.m.

- (a)$70\mathrm{m}$
- (b)$80\mathrm{m}$
- (c)$85\mathrm{m}$
- (d)$90\mathrm{m}$
- (e)None of these

**Hint**

$\mathrm{Distance}=\mathrm{Speed}\times \mathrm{Time}$.

**Solution:**

According to the question,

The train crosses a pole in $8$ seconds. It means the distance covered by the train in $8$ seconds is equal to length of the train.

Speed of the train

$=36\mathrm{km}/\mathrm{hr}\phantom{\rule{0ex}{0ex}}=36\times \frac{5}{18}\phantom{\rule{0ex}{0ex}}=10\mathrm{m}/\mathrm{s}$

Distance $=\left(8\times 10\right)=80\mathrm{m}$

Hence, the correct answer is $80\mathrm{m}$.

- (a)Only the length of the train
- (b)Only the length of the engine
- (c)Either the length of the train or the length of the engine
- (d)Both the length of the train and the length of the engine
- (e)None of these

**Hint**

Use the formula, $\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}$.

**Solution:**

According to the question:

Train running at certain speed crosses a stationary engine in $20$ seconds

Since the sum of the lengths of the train and the engine are needed. So, both the lengths must be known.

- (a)$72m$
- (b)$54m$
- (c)$50m$
- (d)$45m$
- (e)None of these

**Hint**

Total time taken by train is sum of times taken by train to cross both person's.

time $=\frac{lengthoftrain}{relativespeedoftrain}$

**Solution:**

According to the question

Train overtakes two persons who are walking in the same direction in which the train is going, at the rate of $2kmph$ and $4kmph$ and passes them completely in $9$ and $10$ seconds

To get the length of the train

$2kmph=\left(2\times \frac{5}{18}\right)m/\mathrm{sec}=\frac{5}{9}m/\mathrm{sec}$

and

$4kmph=\left(4\times \frac{5}{18}\right)m/\mathrm{sec}=\frac{10}{9}m/\mathrm{sec}$

Let the length of the train be $x$ metres and its speed be $ym/\mathrm{sec}$

Then $\frac{x}{\left(y-{\displaystyle \frac{5}{9}}\right)}=9$ and $\frac{x}{\left(y-{\displaystyle \frac{10}{9}}\right)}=10$

So,

$9y-5=x$ and $10\left(9y-10\right)=9x$

So, $9y-x=5$ and $90y-9x=100$

On solving we get: $x=50$

So, length of the train is $50m$.

Hence, the correct answer is $50m$.

- (a)$100m$
- (b)$200m$
- (c)$300m$
- (d)$400m$
- (e)None of these