### Contents of Study Guide SBI PO Preliminary Examination 2019

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

#### 63 Chapters

- 1. Solved Paper (Pre) 08-07-2018
- 2. Solved Paper (Pre) 03-04-2017
- 3. Solved Paper 03-07-2016
- 4. Solved Paper 21-06-2015
- 5. Common Errors
- 6. Phrase Substitution
- 7. Unit Test 1
- 8. Choosing the Appropriate Filler
- 9. Synonyms and Antonyms
- 10. Spelling Test
- 11. Unit Test 2
- 12. Rearrangement of Sentences/Words
- 13. Cloze Test
- 14. Comprehension
- 15. Unit Test 3
- 16. Number System
- 17. Number Series
- 18. Simplification of Numbers
- 19. Percentage Calculation
- 20. Profit and Loss
- 21. Unit Test 1
- 22. LCM and HCF
- 23. Concept of Average
- 24. Ratio and Proportion
- 25. Partnership
- 26. Problems Based on Ages
- 27. Unit Test 2
- 28. Mixed Proportion and Unitary Method
- 29. Problems Based on Distribution
- 30. Simple and Compound Interest
- 31. Time and Work
- 32. Speed, Time and Distance
- 33. Unit Test 3
- 34. Permutations and Combinations
- 35. Probability
- 36. Algebraic Operations
- 37. Mensuration
- 38. Unit Test 4
- 39. Data Sufficiency
- 40. Data Interpretation
- 41. Unit Test 5
- 42. Coding and Decoding
- 43. Ranking Sequence
- 44. Sitting Arrangement
- 45. Blood Relationship
- 46. Direction Sense Test
- 47. Unit Test 1
- 48. Puzzle
- 49. Input-Output
- 50. Inequality
- 51. Syllogism
- 52. Decision Making
- 53. Data Sufficiency
- 54. Unit Test 2
- 55. Statement and Conclusions
- 56. Statement and Assumptions
- 57. Statement and Arguments
- 58. Statement and Courses of Action
- 59. Cause and Effects
- 60. Unit Test 3
- 61. Other Sub-Topics
- 62. Unit Test 4
- 63. Practice Set

#### 3 Topics

English Language

Reasoning Ability

Quantitative Aptitude

#### 3 Topics

English Language

Reasoning Ability

Numerical Ability

#### 3 Topics

English Language

Reasoning Ability

Quantitative Aptitude

#### 3 Topics

English Language

Quantitative Aptitude

Reasoning Ability

#### 13 Topics

Articles

Tense

Noun

Pronoun

Infinitive/Participle/Gerund

Prepositions

Conjunctions

Determiners and Adjectives

Adverbs

Reported Speech

Subject-Verb Agreement

Voice

UnEnglish and Superfluous Expressions

#### 1 Topics

Phrase Substitution

#### 1 Topics

Choosing the Appropriate Filler

#### 2 Topics

Synonyms

Antonyms

#### 1 Topics

Commonly Misspelt Words

#### 1 Topics

Rearrangement of Sentences/Words

#### 3 Topics

Different Types of Numbers

To Find the Unit's Place Digit in the Product of Numbers

Rules for Divisibility

#### 2 Topics

To Find a Missing Term

To Find the Wrong Number

#### 1 Topics

Simplification of Numbers

#### 1 Topics

Percentage Calculation

#### 2 Topics

Based on CP, SP and Profit/Loss Percent

Based on Marked Price (MP), Discount and Discount Per cent

#### 4 Topics

Question Based on LCM and HCF

Question Based on Least or Smallest Number

Question Based on Greatest Number

Question Based on Rings

#### 2 Topics

Ratio

Proportion

#### 1 Topics

Problems Based on Ages

#### 2 Topics

Rules of Mixed Proportion

Unitary Method

#### 1 Topics

Problems Based on Distribution

#### 2 Topics

Simple Interest

Compound Interest

#### 3 Topics

Question Based on Work and Time

Question Based on Work and Wages

Question Based on Pipe and Cisterns

#### 3 Topics

Based on Time, Speed and Distance

Based on Trains

Based on Boats and Streams

#### 2 Topics

Permutation

Combination

#### 5 Topics

Based on Coins

Based on Dice

Based on Playing Cards

Based on Marbles or Balls

Miscellaneous (Choosing a student, hitting a target, etc.)

#### 1 Topics

Algebraic Operations

#### 2 Topics

Area and Perimeter of Plane Figures

Surface Area and Volume of Solid Figures

#### 5 Topics

Based on Tables

Based on Bar Graphs

Based on Pie Charts

Based on Line Graphs

Based on Case Study

#### 4 Topics

Coding Based on Rearrangement of Letters

Number Coding

Symbol Coding Based on Similarity

Coding by Substitution/Word Replacement

#### 2 Topics

Comparison Based Ranking

Position Based Ranking

#### 3 Topics

Linear Arrangement

Circular Arrangement

Quadrilateral Arrangement

#### 1 Topics

Blood Relationship

#### 3 Topics

Based on Direction

Based on Distance

Based on Distance and Direction

#### 3 Topics

Analytical Puzzles

Puzzles Based on Sequential Order of Events

Complex Family Puzzles

#### 2 Topics

Coded Inequality

Direct Inequality

#### 1 Topics

Data Sufficiency

#### 2 Topics

One Statement with Two Conclusions Based

One Statement and More than Two Conclusions Based

#### 2 Topics

Two Assumptions Based

Three Assumptions Based

#### 2 Topics

Two Arguments Based

More than Two Arguments Based

#### 2 Topics

Two Courses of Action Based

More than Two Courses of Action Based

#### 2 Topics

Statement and Direction Based Questions

Direct Cause and Effect Based Questions

#### 5 Topics

Word Classification

Alphabet Classification

Number Classification

Analogy

Alphabet Test

#### 3 Topics

Practice Set - 1

Practice Set - 2

Practice Set - 3

## Experience Tests Tailored to Match the Real Exam

## Practise Questions with Solutions from the Study Guide SBI PO Preliminary Examination 2019

- (a)Quantity I > Quantity II
- (b)Quantity I < Quantity II
- (c)
Quantity I ≥ Quantity II

- (d)Quantity I ≤ Quantity II
- (e)Quantity I = Quantity II or No relation

**Hint**

$x=2,-3$

**Solution:**

Quantity I:

${x}^{2}+x-6=0\phantom{\rule{0ex}{0ex}}{x}^{2}+3x-2x-6=0\phantom{\rule{0ex}{0ex}}x(x+3)-2(x+3)=0\phantom{\rule{0ex}{0ex}}(x-2)(x+3)=0\phantom{\rule{0ex}{0ex}}x=2,-3$

Quantity II:

${y}^{2}+7y+12=0\phantom{\rule{0ex}{0ex}}{y}^{2}+4y+3y+12=0\phantom{\rule{0ex}{0ex}}y(y+4)+3(y+4)=0\phantom{\rule{0ex}{0ex}}(y+3)(y+4)=0\phantom{\rule{0ex}{0ex}}y=-3,-4$

$x=2,y=-3\Rightarrow xy\phantom{\rule{0ex}{0ex}}x=-3,y=-3\Rightarrow x=y\phantom{\rule{0ex}{0ex}}x=2,y=-4\Rightarrow xy\phantom{\rule{0ex}{0ex}}x=-3,y=-4\Rightarrow xy\phantom{\rule{0ex}{0ex}}\therefore x\ge y$

Hence,

Quantity I$\ge $ Quantity II

- (a)Quantity I > Quantity II
- (b)Quantity I < Quantity II
- (c)Quantity I ≥ Quantity II
- (d)Quantity I ≤ Quantity II
- (e)Quantity I = Quantity II or No relation

**Hint**

A's efficiency is 125% of B's efficiency.

**Solution:**

A's efficiency is $25\%$ more than that of B. This means A's efficiency is $125\%$ of B's efficiency.

$\frac{125}{100}=\frac{5}{4}$

Let us assume that A does $5$ units of work daily.

So, B will do $4$ units of work daily.

Let us assume that the total work is $60$ units.

Quantity I:

A does $\frac{5}{6}$th of the work in $x$ days.

So, A will complete $(\frac{5}{6}\times 60)=50$ units of work in $x$ days.

So, $x$$=$$\frac{50}{5}=10$ days.

Quantity II:

B does $\frac{4}{5}$th of the work in $y$ days.

So, B will complete $(\frac{4}{5}\times 60)=48$ units of work in $y$ days.

So, $y=\frac{48}{4}=12$ days.

Thus, $x<y$.

Hence, Quantity I < Quantity II.

- (a)Quantity I > Quantity II
- (b)Quantity I < Quantity II
- (c)Quantity I ≥ Quantity II
- (d)Quantity I ≤ Quantity II
- (e)Quantity I = Quantity II or No relation

**Hint**

Difference between the given quantities is two.

**Solution:**

Let the eight consecutive even numbers be $x,(x+2),(x+4),(x+6),(x+8),(x+10),(x+12)$ and $(x+14)$.

Quantity I:

$(x+2)+(x+14)=2x+16$

Quantity II:

$(x+4)+(x+10)=2x+14$

Clearly, it can be concluded that Quantity I > Quantity II.

- (a)Quantity I > Quantity II
- (b)Quantity I < Quantity II
- (c)Quantity I ≥ Quantity II
- (d)Quantity I ≤ Quantity II
- (e)Quantity I = Quantity II or No relation

**Hint**

$SP=\frac{(100-d\%)}{100}\times MP$

**Solution:**

Let market price, selling price and discount percentage of the article be $MP$, $SP$ and $d\%$, respectively.

$SP=\frac{(100-d\%)}{100}\times MP$

So,

$MP=\frac{SP\times 100}{(100-d\%)}\phantom{\rule{0ex}{0ex}}MP=\frac{1500\times 100}{(100-12.5)}\phantom{\rule{0ex}{0ex}}MP=\frac{150000}{87.5}$

$MP=$ Rs. $1714.29$

So,

Quantity I: Rs. $550$.

Quantity II: Rs. $1714.29$.

Hence, Quantity I < Quantity II.

- (a)Quantity I > Quantity II
- (b)Quantity I < Quantity II
- (c)Quantity I ≥ Quantity II
- (d)Quantity I ≤ Quantity II
- (e)Quantity I = Quantity II or No relation

**Hint**

Upstream speed = Speed of boat - Speed of stream

Downstream speed = Speed of boat + Speed of stream

**Solution:**

Let speed of the current be $ykm/hr$.

So, speed of the boat will be $(y+5y)=6ykm/hr$.

So, downstream speed will be $=(6y+y)=7ykm/hr$.

As per the question,

$\frac{63}{3}=\left(7y\right)\phantom{\rule{0ex}{0ex}}21=7y\phantom{\rule{0ex}{0ex}}y=3$

Hence, speed of the current and the boat will be $3km/hr$ and $18km/hr$, respectively.

So, upstream speed will be $=x=(18-3)=15km/hr$.

Hence, Quantity I = Quantity II.

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- (a)160 L
- (b)180 L
- (c)200 L
- (d)250 L
- (e)216 L

**Hint**

The initial mixture consists of 162 L of milk.

**Solution:**

Let the initial quantity of milk be $3x$ L, and the initial quantity of water be $x$ L.

Given, $24$ L of mixture is taken out.

Quantity of milk taken out

$=(24\times \frac{3}{4})=18$ L

Quantity of water taken out

$=(24-18)=6$ L

The quantity of milk remaining is $(3x-18)$ L.

After $24$ L of water is added, the quantity of water remaining is $(x-6+24)=(x+18)$ L

Hence,

$\frac{3x-18}{x+18}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}3x-18=4x+36$

$x=54$ L

So, the total quantity of the initial mixture is

$=\left(4x\right)=(4\times 54)=216$ L

- (a)5 km/h
- (b)3 km/h
- (c)4 km/ h
- (d)16 km/h
- (e)8 km/h

**Hint**

$\frac{D}{x+y}=\frac{2(D-56)}{x-y}$

**Solution:**

Let the speed pf Boat and the speed of Current is $xKm/HryKm/Hr$ respectively.

Upstream speed will be $=(x-y)Km/Hr$

Downstream speed will be $(x+y)Km/Hr$

As per Question,

$T=\frac{D}{x+y}$

And,

$\frac{T}{2}=\frac{D-56}{x+y}$

Now, we have,

$\frac{D}{x+y}=\frac{2(D-56)}{x-y}\phantom{\rule{0ex}{0ex}}\frac{D}{x+y}=\frac{2D-112}{x-y}\phantom{\rule{0ex}{0ex}}\frac{(x-y)}{(x+y)}=\frac{2D-112}{D}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}So,\phantom{\rule{0ex}{0ex}}\frac{3}{5}=\frac{2D-112}{D}(As\frac{x-y}{x+y}=\frac{3}{5})\phantom{\rule{0ex}{0ex}}3D=10D-560\phantom{\rule{0ex}{0ex}}7D=560\phantom{\rule{0ex}{0ex}}D=\frac{560}{7}=80Km.\phantom{\rule{0ex}{0ex}}So,\phantom{\rule{0ex}{0ex}}(x-y)=\frac{D-32}{4}=\frac{80-32}{4}=\frac{48}{4}=12Km/Hr\phantom{\rule{0ex}{0ex}}As,\phantom{\rule{0ex}{0ex}}\frac{x-y}{x+y}=\frac{3}{5}\phantom{\rule{0ex}{0ex}}\frac{12}{x+y}=\frac{3}{5}\phantom{\rule{0ex}{0ex}}\frac{12}{12+y+y}=\frac{3}{5}(x-y=12,x=12+y)\phantom{\rule{0ex}{0ex}}\frac{12}{12+2y}=\frac{3}{5}\phantom{\rule{0ex}{0ex}}60=36+6y\phantom{\rule{0ex}{0ex}}6y=24\phantom{\rule{0ex}{0ex}}y=\frac{24}{6}=4Km/Hr.\phantom{\rule{0ex}{0ex}}$

- (a)1500
- (b)800
- (c)600
- (d)1000
- (e)Other than those given as options

**Hint**

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

$CI=P\left[\right(1+\frac{R}{100}{)}^{t}-1]$

**Solution:**

We know that,

$SI=\frac{P\times R\times T}{100}\phantom{\rule{0ex}{0ex}}SI=\frac{4200\times 12\times 2}{100}=1008$

And,

$CI=P\left[\right(1+\frac{R}{100}{)}^{t}-1]\phantom{\rule{0ex}{0ex}}CI=(4200-P\left)\right[(1+\frac{10}{100}{)}^{2}-1]\phantom{\rule{0ex}{0ex}}=(4200-P)\left[\right(\frac{100+10}{100}{)}^{2}-1]\phantom{\rule{0ex}{0ex}}=(4200-P\left)\right[(\frac{110}{100}{)}^{2}-1]\phantom{\rule{0ex}{0ex}}=(4200-P)[\frac{121}{100}-1]\phantom{\rule{0ex}{0ex}}=(4200-P)\left(\frac{121-100}{100}\right)\phantom{\rule{0ex}{0ex}}=(4200-P)\times \frac{21}{100}\phantom{\rule{0ex}{0ex}}=882-\frac{21P}{100}$

Given,

$SI-CI=294$

$1008-(882-\frac{21P}{100})=294\phantom{\rule{0ex}{0ex}}1008-882+\frac{21P}{100}=294\phantom{\rule{0ex}{0ex}}\frac{21P}{100}=294-126\phantom{\rule{0ex}{0ex}}P=\frac{168\times 100}{21}$

$P=$ Rs. $800$

- (a)Rs. 380
- (b)Rs. 280
- (c)Rs. 370
- (d)Rs. 300
- (e)Rs. 340

**Hint**

The profit earned by the sale of article A will be Rs. 92.

**Solution:**

Let the price of article B be Rs. $x$.

Then, the price of article A will be Rs. $(x-140)$.

As per the question,

${P}_{A}+{P}_{B}=18\phantom{\rule{0ex}{0ex}}[40\%\times (x-140\left)\right]+[-20\%\times x]=18\phantom{\rule{0ex}{0ex}}\frac{40}{100}\times (x-140)-\frac{20}{100}\times x=18\phantom{\rule{0ex}{0ex}}\frac{2}{5}(x-140)-\frac{1}{5}x=18\phantom{\rule{0ex}{0ex}}\frac{2x}{5}-56-\frac{x}{5}=18\phantom{\rule{0ex}{0ex}}\frac{(2x-x)}{5}=18+56\phantom{\rule{0ex}{0ex}}\frac{x}{5}=74\phantom{\rule{0ex}{0ex}}x=(74\times 5)$

$x=$ Rs. $370$

- (a)24 yr
- (b)28 yr
- (c)29 yr
- (d)20 yr
- (e)26 yr

**Hint**

A is 48 years old.

**Solution:**

Let the present ages of A, B and C be $A$, $B$ and $C$ respectively.

Given,

$B+18=A\phantom{\rule{0ex}{0ex}}B=A-18$

And,

$A=2C$

Also,

$\frac{B+6}{C}=\frac{3}{2}\phantom{\rule{0ex}{0ex}}\frac{A-18+6}{C}=\frac{3}{2}\phantom{\rule{0ex}{0ex}}\frac{A-12}{C}=\frac{3}{2}\phantom{\rule{0ex}{0ex}}\frac{2C-12}{C}=\frac{3}{2}\phantom{\rule{0ex}{0ex}}4C-24=3C$

$C=24$ years

$A=(2\times 24)=48$ years

$B=48-18=30$ years

Hence, B's age four years ago

$=(30-4)=26$ years

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- (a)31
- (b)27
- (c)29
- (d)25
- (e)33

**Hint**

The average of three consecutive odd numbers is always the middle number.

**Solution:**

Let the numbers be $x,(x+2),(x+4),(x+6),(x+8)$ and $(x+10)$.

As we know, the average of three consecutive odd numbers is always the middle number.

So, the average of the first three consecutive odd numbers will be $(x+2)$.

Square of this average will be

$=(x+2{)}^{2}=({x}^{2}+4x+4)$

Similarly, the average of the last three consecutive odd numbers will be $(x+8)$.

Square of this average will be

$=(x+8{)}^{2}={x}^{2}+16x+64$

As per the question,

$({x}^{2}+16x+64)-({x}^{2}+4x+4)=288\phantom{\rule{0ex}{0ex}}({x}^{2}+16x+64-{x}^{2}-4x-4)=288\phantom{\rule{0ex}{0ex}}(12x+60)=288\phantom{\rule{0ex}{0ex}}12x=288-60\phantom{\rule{0ex}{0ex}}x=\frac{228}{12}=19$

So, the largest number will be

$=(x+10)=(19+10)=29$

- (a)$\frac{29}{35}$
- (b)$\frac{7}{15}$
- (c)$\frac{23}{35}$
- (d)$\frac{17}{35}$
- (e)$\frac{19}{35}$

**Hint**

${}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{r})!\mathrm{r}!}$

**Solution:**

As per the question,

Number of red balls $=6$

Number of green balls $=9$

Total number of balls $=15$

Probability that at least one ball is red

$=1-$ Probability that no ball is red

$=1-$ Probability that both balls are green

$=1-\frac{{}^{9}C_{2}}{{}^{15}C_{2}}\phantom{\rule{0ex}{0ex}}=1-\frac{{\displaystyle \frac{9!}{7!2!}}}{{\displaystyle \frac{15!}{13!2!}}}\phantom{\rule{0ex}{0ex}}=1-\frac{9\times 8}{15\times 14}\phantom{\rule{0ex}{0ex}}=1-\frac{72}{210}\phantom{\rule{0ex}{0ex}}=1-\frac{12}{35}\phantom{\rule{0ex}{0ex}}=\frac{23}{35}$

- (a)27
- (b)45
- (c)24
- (d)36
- (e)42

**Hint**

A can complete the work alone in 12 days.

**Solution:**

A is thrice as efficient as B. So, A takes one-third the time that B takes to complete the work.

Let the number of days taken by A and B to complete the work be $x$ and $3x$ days, respectively.

One day work of A and B will be $\frac{1}{x}$ and $\frac{1}{3x}$ units, respectively.

As per the question,

$(4\times \frac{1}{x})+(15\times \frac{1}{3x})=\frac{75}{100}\phantom{\rule{0ex}{0ex}}(\frac{4}{x}+\frac{15}{3x})=\frac{75}{100}\phantom{\rule{0ex}{0ex}}(\frac{4}{x}+\frac{5}{x})=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\frac{9}{x}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}x=\frac{9\times 4}{3}=12$

So, the time taken by B to complete the work alone will be

$=3x=3\times 12=36$ days

- (a)Rs. 6000
- (b)Rs. 5800
- (c)Rs. 6800
- (d)Rs. 5400
- (e)Rs. 6400

**Hint**

$\mathrm{CI}=\mathrm{P}\left[\right(1+\frac{\mathrm{r}}{100}{)}^{\mathrm{t}}-1]\phantom{\rule{0ex}{0ex}}\mathrm{SI}=\frac{\mathrm{Prt}}{100}$

**Solution:**

As per the question,

$r=20\%$

$t=2$ years

Hence, the compound interest (CI) will be

$=P\left[\right(1-\frac{r}{100}{)}^{t}-1]\phantom{\rule{0ex}{0ex}}=P[(1+\frac{20}{100}{)}^{2}-1]\phantom{\rule{0ex}{0ex}}=P\left[\right(1+\frac{1}{5}{)}^{2}-1]\phantom{\rule{0ex}{0ex}}=P[(\frac{6}{5}{)}^{2}-1]\phantom{\rule{0ex}{0ex}}=P[\frac{36}{25}-1]\phantom{\rule{0ex}{0ex}}=P\left[\frac{36-25}{25}\right]\phantom{\rule{0ex}{0ex}}=P\left(\frac{11}{25}\right)\phantom{\rule{0ex}{0ex}}=\frac{11P}{25}$

Now, this CI is lent at $7.5\%$ on SI for $2$ years.

So, the simple interest (SI) will be

$=\frac{{\displaystyle \frac{11P}{25}\times 7.5\times 2}}{100}\phantom{\rule{0ex}{0ex}}=\frac{11P\times 7.5\times 2}{25\times 100}\phantom{\rule{0ex}{0ex}}=\frac{165P}{2500}$

As per the question,

$(\frac{11P}{25}+\frac{165P}{2500})=3036\phantom{\rule{0ex}{0ex}}\left(\frac{1100P+165P}{2500}\right)=3036\phantom{\rule{0ex}{0ex}}1265P=(3036\times 2500)\phantom{\rule{0ex}{0ex}}P=\frac{3036\times 2500}{1265}$

$P=$ Rs. $6000$

- (a)Rs. 13950
- (b)Rs. 13810
- (c)Rs. 13920
- (d)Rs. 12780
- (e)Rs. 14040

**Hint**

The profits will be shared in the ratio 5 : 2 : 2 by A, B and C respectively.

**Solution:**

As per the question,

Amounts are invested by A, B and C for $12$, $8$ and $6$ months, respectively.

Hence, the ratio of their share of profits will be:

$A:B:C\phantom{\rule{0ex}{0ex}}=(25000\times 12):(15000\times 8):(20000\times 6)\phantom{\rule{0ex}{0ex}}=(25\times 12):(15\times 8):(20\times 6)\phantom{\rule{0ex}{0ex}}=5:2:2$

Hence, A's profit will be

$=\frac{5}{(5+2+2)}\times $ Total Profit

So, Total Profit (in Rs.) $=\frac{7750\times 9}{5}=13950$

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- (a)12 L
- (b)16 L
- (c)18 L
- (d)8 L
- (e)10 L

**Hint**

The final quantity of the mixture is 144 L.

**Solution:**

The quantity of the initial mixture is $140$ L.

After $20$ L of this mixture is given to a customer, the quantity of mixture remaining would be

$=140-20=120$ L

Hence, the quantity of water remaining is

$=30\%\times 120=\frac{30}{100}\times 120=36$ LAnd, the quantity of milk remaining will be

$=120-36=84$ L

Let $x$ L of milk and $x$ L of water be added to the mixture.

$\frac{84+x}{36+x}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}84+x=72+2x\phantom{\rule{0ex}{0ex}}x=12$

So, the quantity of milk added is $12$ L.

- (a)52 min
- (b)44 min
- (c)48 min
- (d)36 min
- (e)54 min

**Hint**

Upstream speed = Speed of boat - Speed of stream

Downstream speed = Speed of boat + speed of stream

**Solution:**

Let the speed of the boat and the stream be $7x$ km/hr and $x$ km/hr, respectively.

Upstream speed will be $=(7x-x)=6x$ km/hr.

Downstream speed will be $=(7x+x)=8x$ km/hr.

So,

$6x=\frac{4.2}{{\displaystyle \frac{14}{60}}}\phantom{\rule{0ex}{0ex}}6x=\frac{42\times 60}{10\times 14}\phantom{\rule{0ex}{0ex}}6x=18$

$x=\frac{18}{6}=3$ km/hr

Hence, the time taken to cover the given distance downstream will be

$=\frac{17.6}{8x}=\frac{17.6}{8\times 3}=\frac{11}{15}$ hr

$=(\frac{11}{15}\times 60)=44$ min

- (a)Rs. 3300
- (b)Rs. 3500
- (c)Rs. 4200
- (d)Rs. 4800
- (e)Rs. 4500

**Hint**

The profits are shared in the ratio of the products of the amounts invested and the durations of investments.

**Solution:**

A invests Rs. $1500$ for $12$ months.

Let the amounts invested (in Rs.) by B and C be $x$ and $y$, respectively. These amounts have been invested for $6$ and $4$ months respectively.

The profits are shared in the ratio of the products of the amounts invested and the durations of investments.

Therefore,

$A:B:C=5:3:3\phantom{\rule{0ex}{0ex}}(1500\times 12):(x\times 6):(y\times 4)=5:3:3\phantom{\rule{0ex}{0ex}}\left(18000\right):\left(6x\right):\left(4y\right)=5:3:3\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{18000}{6x}=\frac{5}{3}\phantom{\rule{0ex}{0ex}}x=\frac{18000\times 3}{5\times 6}=1800\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{18000}{4y}=\frac{5}{3}\phantom{\rule{0ex}{0ex}}y=\frac{18000\times 3}{5\times 4}=2700$

So, the required amount (in Rs.) is $=x+y=1800+2700=4500$.

- (a)549 m
^{2} - (b)545 m
^{2} - (c)557 m
^{2} - (d)560 m
^{2} - (e)500 m
^{2}

**Hint**

Area of square = Side × Side

Area of rectangle = Length × Breadth

**Solution:**

Let the side of the square be $x$ m.

So,$A={x}^{2}$

$x=\sqrt{A}=\sqrt{576}=24$ m

So, length of the rectangle will be $=(x+4)=(24+4)=28$ m

Breadth of the rectangle will be $=(x-4)=(24-4)=20$ m

Hence, area of rectangle will be

$=l\times b=(28\times 20)=560$ sq. m

- (a)Rs. 1240
- (b)Rs. 1250
- (c)Rs. 1440
- (d)Rs. 1450
- (e)Rs. 1400

**Hint**

C paid Rs. 1375 to B for the item.

**Solution:**

Let the CP (in Rs.) of A be $100x$.

So, CP (in Rs.) of B will be $=(\frac{120}{100}\times 100x)=120x$

CP (in Rs.) of C will be $=(\frac{110}{100}\times 120x)=132x$

CP (in Rs.) of D will be

$=132x+(\frac{16}{100}\times 100x)\phantom{\rule{0ex}{0ex}}=132x+16x\phantom{\rule{0ex}{0ex}}=148x$

As per the statement given in the question,

$148x-100x=500\phantom{\rule{0ex}{0ex}}48x=500\phantom{\rule{0ex}{0ex}}x=\frac{500}{48}$

So, the amount (in Rs.) paid by B to A

$=120x\phantom{\rule{0ex}{0ex}}=(120\times \frac{500}{48})\phantom{\rule{0ex}{0ex}}=1250$

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

**Hint**

In the given sentence, a person is being referred to as 'chief minister'.

**Solution:**

When a person is referred by his designation, instead of his name, then definite article 'The' is used before his designation. Therefore, the correct answer will be- "The Chief Minister will decide the matter soon".

- (a)Kshitiz
- (b)is
- (c)a
- (d)honest boy
- (e)No error

**Hint**

Refer to the use of indefinite articles.

**Solution:**

Indefinite article 'an' is used before a singular noun starting with a vowel sound. In the word 'Honest', h is silent. Therefore, its pronunciation starts with sound 'o'. So, 'a' will be replaced by 'an'.

- (a)The Sun
- (b)rises in
- (c)the East
- (d)is an universal truth
- (e)No error

**Hint**

Refer to the use of indefinite articles.

**Solution:**

The indefinite article 'an' is used before singular nouns starting with a vowel sound. In this sentence, the word 'universal' is starting with a consonant sound 'yoo' and not with a vowel sound 'uh'. Therefore, 'an' will be replaced by 'a'.

- (a)Times of India
- (b)is a popular
- (c)newspaper
- (d)of these days
- (e)
No error

**Hint**

Refer to the use of the definite article.

**Solution:**

The error lies in first part of the sentence. The definite article 'The' is used before the names of newspapers. For example, The Times, The Hindu, etc. Therefore, in this sentence, The will be used before 'Times of India'.

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

**Hint**

The bus service was unknown.

**Solution:**

There is no error in the given sentence. The speaker here is talking about some bus accident of which the listener is not aware. There is no information about the bus service. Therefore, instead of using the definite article 'the', the indefinite article 'a' has been used before the noun bus in the sentence.

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- (a)even though it was
- (b)though it was not
- (c)as if it was
- (d)despite it was not
- (e)No correction required

**Hint**

This phrase includes a conditional clause.

**Solution:**

The sentence "He behaved though it was his fault, we knew he was not responsible for it" is not correct as of the phrase 'though it' used here is incorrect. The phrase 'though it' does not only make it grammatically incorrect but doesn't add any meaning to the sentence either.

Option (c) here whereas makes the sentence meaningful and grammatically correct as well. Hence, the correct sentence will be, "He behaved as if it was his fault, we knew he was not responsible for it".

- (a)were not of her business
- (b)was none of her business
- (c)was of not her business
- (d)was not of her businesses
- (e)No correction required

**Hint**

The answer does not include the word 'not' at all.

**Solution:**

The sentence "She never felt that it was not her business to get involved in somebody else's family matter" is not correct due to the phrase ‘was not her business’. The use of 'not her' makes the sentence grammatically incorrect. Instead of 'not her', 'none of her' should be used to make the sentence grammatically meaningful.

Therefore, the second option will be the correct option and the correct sentence will be, "She never felt that it was none of her business to get involved in somebody else's family matter".

- (a)Be born
- (b)Taking born
- (c)By birth
- (d)Being borned
- (e)No correction required

**Hint**

Being is present participle form including progressive tenses.

**Solution:**

The sentence, "Being born in a certain family is not in our control" is grammatically correct. ‘Being’ is a participle that is in perfect coherence with the word ‘born’. It relates to the birth of a person in a family over which one has no choice. Therefore, option (e) is the correct answer. There is no need to make any sort of amendment to the sentence.

- (a)would be taken back by
- (b)was taken backwards by
- (c)was taken back for
- (d)was taken aback by
- (e)No correction required

**Hint**

The correct answer here will be a phrase with the meaning 'startled'.

**Solution:**

The sentence, "I was taken back by his sudden comment on this issue" is grammatically incorrect. The term 'taken back' means literally taken back somewhere (to a different period of time) which is a total misfit in this sentence. Instead of using 'taken back', the term 'taken aback' can be used, which actually means being startled.

Therefore, the correct answer is option d and the correct sentence will be, "I was taken aback by his sudden comment on this issue".

- (a)came to know of
- (b)come to be known of
- (c)come to know off
- (d)are coming to know of
- (e)no correction required

**Hint**

A gerund will make the sentence correct.

**Solution:**

The sentence "In a matter of seconds, we come to know of what is happening anywhere in the world" is not correct as the phrase 'come to know of' used here is incorrect. The phrase 'come to know of' does not only make it grammatically incorrect due to the wrong use of tense but also doesn't add any meaning to the sentence either. Option d here whereas makes the sentence meaningful and grammatically correct.

Hence, the correct sentence will be, "In a matter of seconds, we are coming to know of what is happening anywhere in the world".

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- (a)I have been
- (b)living in Hyderabad
- (c)at my uncle
- (d)since my birth
- (e)No error

**Hint**

The person is being accompanied by another person.

**Solution:**

'At' is a preposition used to indicate a point or place occupied by someone or something. However, 'with' means accompanied by another person or thing.

Here, the speaker is accompanied by his uncle. Therefore, in the given sentence preposition 'with' should be used.

Hence, the correct option is 3.

- (a)He is the man
- (b)who I know
- (c)has helped my son
- (d)in the final examination.
- (e)No error

**Hint**

Revisit the difference between who and whom.

**Solution:**

'Who' is the doer of the action and generally used as the subject of a sentence. However, 'whom' is used instead of ‘who’ as the object of a verb or preposition.

So, in the given sentence 'whom' should be used.

Hence, the error is in this part.

- (a)The number of employees
- (b)reporting sick
- (c)has reduced significantly
- (d)because of the incentive
- (e)No error

**Hint**

The given sentence is in the present perfect tense.

**Solution:**

The use of has + the third form of the verb i.e., 'reduced'. It tells that the sentence is in the present perfect tense.

The sentence has a main and subordinate clause, connected by 'because' which is a subordinate conjunction.

Therefore, there are no errors in this sentence.

- (a)Everyone of us know
- (b)that he is not capable of
- (c)remaining under water
- (d)for such a long time.
- (e)No error

**Hint**

Try to recall the basic rule of Subject-verb agreement.

**Solution:**

In this sentence, the indefinite pronoun 'everyone' is singular. Therefore, it requires a singular verb 'knows'. 'Know' is used for the plural subject.

Hence, the correct sentence will be: Everyone of us knows that he is not capable of remaining underwater for such a long time.

- (a)
The leader was so shrewd

- (b)that he could not deceive
- (c) by the words of
- (d)the sycophant courtiers.
- (e)No Error

**Hint**

The sentence is not in active voice.

**Solution:**

The given sentence is in the passive voice. A passive sentence always takes the 'be + the third form of the verb'. So, the third form of the verb 'deceive' is 'deceived'.

Therefore, 'that he could not deceive' will be replaced with 'that he could not be deceived'.

Hence, the correct option is 2.

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- (a)look
- (b)peep
- (c)sight
- (d)gaze
- (e)None of these

**Hint**

The word that means 'to try to find'.

**Solution:**

In the given sentence, the speaker is saying that seeing the people in a specified manner will help the person to know who is the culprit so the correct filler is the word 'look'.

'Look' means to see in a specified direction.

Hence, option (a) is the correct answer.

- (a)draw
- (b)buy
- (c)remove
- (d)take
- (e)None of these

**Hint**

Place the suitable word which refers to the action of buying something.

**Solution:**

In the given sentence completion question, the speaker is saying that they should buy the ticket in advance. So, the correct filler is 'buy'.

'Buy' means to purchase.

Therefore, the correct option is 2.

- (a)spread
- (b)increase
- (c)spurt
- (d)augment
- (e)None of these

**Hint**

Place the suitable word which refers to an act of becoming larger or greater.

**Solution:**

In the given sentence completion question, the speaker is saying that family planning is important to prevent the rapid growth in population so the correct filler is the word 'increase'.

'Increase' means to become or make something greater in amount or degree.

Hence, the correct answer is option 2.

- (a)deposit
- (b)deficit
- (c)debit
- (d)demote
- (e)None of these

**Hint**

Place a word that refers to put something in a specific place for safekeeping.

**Solution:**

In the given sentence completion question, the speaker is saying that his friend put the first pay in the bank so the correct filler is the word 'deposit'.

'Deposit' means to put something in a specific place for safekeeping.

Hence, the correct answer is option 1.

- (a)placid
- (b)contractual
- (c)tacit
- (d)verbal
- (e)None of these

**Hint**

Place the suitable word which relates as understand but not directly expressed.

**Solution:**

In the given sentence completion question, the speaker is saying that they were having similar characteristics to understand each other without expressing it and it is not verbal so the correct filler is the word 'tacit'.

'Tacit' means understood without being expressed directly.

Hence, the correct answer is option 3.

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- (a)Gradient

- (b)Mania
- (c)Psychosis

- (d)Idiocy
- (e)None of these

**Hint**

A thought that preoccupies a person’s mind.

**Solution:**

Fetish is a noun, that means an obsession. Mania means an excessive desire that is similar to fetish in meaning thus it is a synonym. Whereas, gradient means a slope; psychosis means a severe mental disorder and idiocy means madness, these words do not relate to the given word.

- (a)Sordid

- (b)Goodwill

- (c)Aversion
- (d)Malevolence

- (e)None of these

**Hint**

A feeling of intense hatred.

**Solution:**

Antipathy is a noun, that means an opposition. Aversion means a strong dislike that is similar to antipathy in meaning thus it is a synonym. Whereas sordid means immoral actions; goodwill means friendly and malevolence means hatred, these words do not relate to the given word.

- (a)Construction
- (b)Catastrophe
- (c)Saviour

- (d)Demolition
- (e)None of these

**Hint**

It means a disaster to the environment.

**Solution:**

Apocalypse is a noun, that means judgment day or the final destruction of the world or an environmental disaster. Catastrophe means sudden damage or disaster to the environment, which is similar to Apocalypse in meaning thus it is a synonym. Whereas, construction means creation; saviour means a person who saves someone and Demolition means damaging, these words do not relate to the given word.

- (a)Lucid
- (b)Magnetism
- (c)Cogent
- (d)Fascination
- (e)None of these

**Hint**

It means the ability to attract and charm people.

**Solution:**

Charisma is a noun, that means a charm or compelling attractiveness. Magnetism means attraction, which is similar to charisma in meaning thus it’s synonym. Whereas, lucid means easy to understand; cogent means a convincing argument and fascination means interest or passion, these words do not relate to the given word.

- (a)Push
- (b)Hold on
- (c)Convincing
- (d)Putting forth
- (e)None of these

**Hint**

The capability of making someone believe that something is real.

**Solution:**

Persuasive is an adjective that means influential. Convincing means capable of making someone believe that something is real, that is similar to persuasive in meaning thus it is a synonym. Whereas, push means to apply force; hold on means wait and putting forth

means to make or create, these words do not relate to the given word.

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- (a)Comission
- (b)Commisson
- (c)Comession
- (d)Commission
- (e)None of these

**Hint**

One of its synonyms is ‘council’.

**Solution:**

Commission is the correctly spelt word. It is used as a noun or a verb. It refers to ‘an instruction’ or ‘ a group of people given an authority by the government to do something’. And its synonyms are-project, task, committee, board and council.

For example:

A commission is set up in Delhi.

- (a)Comittee
- (b)Committee
- (c)Coomittee
- (d)Commettee
- (e)None of these

**Hint**

One of its synonyms is a task force.

**Solution:**

'Committee' is the correctly spelt word. It is used as a noun. It refers to ‘a group of people appointed for a specific function’. And its synonyms are board or panel or bureau.

For example:

The committee is ready with its guidelines.

- (a)Colleage
- (b)Colleagu
- (c)Colleague
- (d)Colleugue
- (e)None of these

**Hint**

One of its synonyms is a co-worker.

**Solution:**

'Colleague' is the correctly spelt word. It is used as a noun. It refers to ‘a person with whom one works in a profession’. Its synonyms are teammate or co-worker.

For example: Mohan is my colleague, we have been working for this company for more than 5 years.

- (a)Corruptive
- (b)Corupptive
- (c)Corrupptive
- (d)Corraptive
- (e)None of these

**Hint**

One of its synonyms is dishonesty.

**Solution:**

'Corruptive' is the correctly spelt word. It is used as an adjective that refers to ‘willingness to act dishonestly’. Its synonyms are-dishonest or untrustworthy.

For example: The corruptive officials levied heavy taxes on people.

- (a)Exorbitant
- (b)Exhorbitant
- (c)Exorbitent
- (d)Exorbeetent
- (e)None of these

**Hint**

One of its synonyms is sky-high.

**Solution:**

'Exorbitant' is the correctly spelt word. It is used as an adjective. It refers to 'the unreasonably high price of something’. Its synonyms are excessive and sky-high.

For example: The bill for lunch was exorbitant.

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- (a)have to
- (b)would
- (c)will
- (d)destined
- (e)use to

**Hint**

It is the past tense form of 'will'.

**Solution:**

'Would' is a past-tense form of will.

If past events are being discussed then 'would' can be used to indicate something that was in the future at that point in time, but is not necessarily in the future right now. In other words, 'would' is used to preserve the future aspect when talking about the past.

The completed sentence is- When I was training for the marathon, I would run over 100 km a week.

- (a)that I knows
- (b)I know
- (c)who I know
- (d)which I know
- (e)I had known

**Hint**

The sentence in the present tense.

**Solution:**

"I know" is the best suitable option. The sentence is in the present tense, hence, the blank should also be in the present tense. With a plural pronoun, a plural verb should be used in option (a). "Who" and "Which" cannot be used before "I know" in option (c) and (d).

The completed sentence is- My colleague is one of the kindest people I know.

- (a)behavior
- (b)fees
- (c)advice
- (d)impact
- (e)approval

**Hint**

The synonym of the word is "consultation".

**Solution:**

Option (C) is the most suitable answer. Advice means guidance or recommendations offered with regard to prudent future action.

In the given question, the lawyer's role was to solve a particular problem. 'Advice' fits the blank perfectly.

The completed sentence- The lawyer's advice led to the resolution of the problem.

- (a)budget
- (b)finance
- (c)installments
- (d)decrease
- (e)competition

**Hint**

It is the activity or condition of striving to gain or win something by defeating or establishing superiority over others.

**Solution:**

Competition in the telecommunication industry will mean a lower price for customers because industries will try to earn more and more customers and for that, they will lower the prices which will benefit the customers.

Option E is the correct choice.

The completed sentence is- The government claims that competition in the telecommunication industry will mean lower prices for customers.

- (a)when
- (b)as
- (c)while
- (d)then
- (e)however

**Hint**

It means "at or during the time that".

**Solution:**

'When' is used as a conjunction meaning ‘at the time that’.

The clause with 'when' is a subordinate clause and needs a main clause to complete its meaning. If the 'when-clause' comes before the main clause, we use a comma.

The completed sentence is: Sarah was walking along the street when she tripped over.

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

**Hint**

Identify the opening sentence then find the clue from it and arrange the remaining sentences.

**Solution:**

The correct sequence after rearrangement is CEAFDB. From the above sentences we can understand that this passage is speaking about the person named Raman who had problems with his rented house and his conversation about the same with landlord. So the first sentence should be about the person Raman.

C. Raman once lived in a rented house

From the first sentence we can understand that he is living in a rented house so the 2nd sentence should be about his rented house.

E. The building was old and whenever there was a strong wind, its rafters would creak and squeak.

From the second sentence we can understand that he has been facing problems with his rented house. So, to complain about his problems he needs to speak to the landlord. So the third sentence should speak about landlord.

A. One day the landlord came to collect the rent from Raman.

From the third sentence we can understand that the landlord came to collect rent from Raman. Here Raman needs to complain, so the fourth sentence should speak about it.

F. Raman took the opportunity to tell him about the alarming noises that the building was making.

When Raman complains, the landlord needs to answer him. So, the fourth sentence should speak about that.

D. “Don’t let that worry you,“ said the landlord, airily. ” Those noises are nothing but the praises the old building is singing to the Almighty”.

Here after arranging all five sentences, the last(sixth) is obviously the one remaining.

B. “Oh, I’m not worried about the hymns,” said Raman.” But what if it decided to kneel down and worship him?

From above, we can now easily identify the fifth sentence after rearranging the sentences.

So the correct answer is D i.e. Option 2.

- (a)B
- (b)c
- (c)E
- (d)A
- (e)F

**Hint**

Identify the topic from given sentences so that related sentences can be arranged.

**Solution:**

The second sentence after rearrangement is E. From the above sentences the passage explains about a person named Raman and his complaint about problems of a rented house to the landlord. From this we can understand that the first sentence introduces Raman and his rented House. Logically the next sentence talks about his rented house, so we find the sentence related to the rented house problems is E. So the correct answer is Option 3.

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

**Hint**

Identify the first sentence, from that sentence you can drag clues to remaining sentences.

**Solution:**

The fourth sentence after rearrangement is F. By observing the given sentences we can say that Raman had problems with his rented house and he wanted to complain to the landlord. Relating to that, logically the sequence could be Raman-Rented House problems-Landlord introduction-complaining. So the sentence that is related to complaining has to be searched is F. Therefore, the correct answer is F i.e. Option 4.

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

**Hint**

Identify the topic the passage is speaking about and from that identify first sentence.

**Solution:**

The first sentence after rearrangement is C. Here the first sentence should be like an introduction to the topic so that we can get a clue for remaining sentences from it by relating to each sentence logically. By clearly observing the given six sentences, we can say that sentence C tells us about the person Raman and where he lives, so the first sentence ideally should be C. So the correct answer is C i.e option 1.

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

**Hint**

Identify the first sentence and from that sentence identify the clue for sentence similarly for other sentences.

**Solution:**

The second sentence after rearrangement is A. From the above sentences we can understand that passage is speaking about a Wrestler named O-nami and his problems with the game. By coherently reading the first sentence we are to be introduced about a person named O-nami and his wrestling skills so the second sentence should be related to the game wrestling. Logically from the given sentences we can say sentence A is speaking about O-nami and the art of wrestling and his problem. So the second sentence is A.

The correct sequence is: D,A,F,B,E,C.

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

**Hint**

Just before this statement, the author explains how one of the servants tried to escape by cutting the stick.

**Solution:**

The sixth sentence of the paragraph is ‘D’, ‘In this way, the wise priest caught the thief’. It is a concluding statement where the author says that the wise priest caught the thief. Just before this statement, the author explains how one of the servants tried to escape by cutting the stick.

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

**Hint**

Just after this statement the author marks the end of the story.

**Solution:**

The fifth sentence of the paragraph is ‘A’, ‘The next day, the priest discovered that one of the servants cut his stick shorter by two inches, fearing that it would grow’. In this statement, the author explains how one of the servants tried to escape by cutting the stick. Just after this statement, the author says that the wise priest caught the thief.

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

**Hint**

Just before this statement, the author says that a rich merchant's house was robbed and he doubted one of his servants.

**Solution:**

The second sentence of the paragraph is ‘E’, ‘He approached the wise priest in the village and asked for help on the matter’. In this statement, the author mentions that a person reached out to a wise priest to help him on some matter. Just before this statement, the author says that a rich merchant's house was robbed and he doubted one of his servants.

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

**Hint**

Just after this statement, the author mentions that the merchant reached out to a wise priest to help him with this matter.

**Solution:**

The second sentence of the paragraph is ‘C’, ‘One day a rich merchant's house was robbed, and he suspected one of his servants’. In this statement, the author says that a rich merchant's house was robbed, and he doubted one of his servants. Just after this statement, the author mentions that the merchant reached out to a wise priest to help him with this matter.

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

**Hint**

Just after this statement, the author explains how one of the servants tried to escape by cutting the stick.

**Solution:**

The fourth sentence of the paragraph is ‘F’, ‘The priest then gave them each a stick of equal length and said that the stick of the real thief would grow by two inches the next morning’. In this statement, the author says that the priest gave one stick of equal length to the servants and added that whoever is the real thief will get his stick increased by two inches the next morning. Just after this statement, the author explains how one of the servants tried to escape by cutting the stick.

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- (a)$1518$
- (b)$1471$
- (c)$1618$
- (d)$1406$
- (e)None of these

**Hint**

The product of $n$ and $n+1$ gives the sum of $n$ even numbers.

**Solution:**

According to the question:

Upto $74$, there $37$ even integers.

Here $n=37$

We know that;

Sum of first $n$ even natural numbers $=n(n+1)$

$=37(37+1)$

$=1406$

Hence, $1406$ is the correct answer.

- (a)$351$
- (b)$378$
- (c)$392$
- (d)$364$
- (e)None of these

**Hint**

Let the number be $x$. Then: $\frac{x\times 51}{100}=714$

**Solution:**

Let the number be $x$.

Now, according to the question:

$\frac{x\times 51}{100}=714$

$\Rightarrow x=\frac{714\times 100}{51}=1400$

So,

$28\%\text{of}1400=\frac{28}{100}\times 1400\phantom{\rule{0ex}{0ex}}=392$

Hence, $392$ is the correct answer.

- (a)$15$
- (b)$11$
- (c)$14$
- (d)$10$
- (e)None of these

**Hint**

It is a Arithmetic Progression for which nth term is the number less than 80 and divisible by 6.

**Solution:**

The numbers divisible by $6$ between $19$and $80$ are $24,30,..........,78$

This is Arithmetic Progression because difference is same

Here $a=$ first term $=24$

$d=$difference$=30-24=6$

$l=$ last term $=78$

$l=a+(n-1)\times d$

$78=24+(n-1)\times 6$

$\frac{78-24}{6}=n-1$

$9=n-1$

$n=10$

- (a)$17850$
- (b)$16820$
- (c)$18020$
- (d)$19450$
- (e)None of these

**Hint**

41% multiplied by the number is equal to 8610.

**Solution:**

Let the number be $x$

As per given condition ,

$(55-14)\%$ of $x=8610$

$41\%$ of $x=8610$

$x=21000$

Now ,

$85\%$ of $x=85\%$ of $21000=17850$

- (a)$360$
- (b)$1080$
- (c)$120$
- (d)$40$
- (e)None of these

**Hint**

35% multiplied by the number is equal to 126.

**Solution:**

Let the number be $x$,

Then, $\left(x\times \frac{15}{100}\right)+\left(x\times \frac{20}{100}\right)=126$

$x\times \frac{35}{100}=126$

$x=\frac{126\times 100}{35}$

$x=360$

Now, one third of the number is $=$ $\frac{1}{3}\times 360=120$

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- (a)$3430$
- (b)$3340$
- (c)$40320$
- (d)$43240$
- (e)None of these

**Hint**

Pattern is $\times 8,\times 7,\times 6,\times 5,....$

**Solution:**

Given that,$2,16,112,672,3360,13440,?$

The pattern will be :

$2\times 8=16\phantom{\rule{0ex}{0ex}}16\times 7=112\phantom{\rule{0ex}{0ex}}112\times 6=672\phantom{\rule{0ex}{0ex}}672\times 5=3360\phantom{\rule{0ex}{0ex}}3360\times 4=13440$After analyzing the pattern, we get the next number as:

$13440\times 3=40320$

- (a)$59$
- (b)$39$
- (c)$49$
- (d)
_{$29$} - (e)None of these

**Hint**

In this series, the pattern is first a number is multiplied with a specific number and then a different number is added to it. For example: $(4\times A)+B=9$

**Solution:**

Given that,

$4,9,19,?,79,159,319$The pattern will be:

$(4\times 2)+1=9\phantom{\rule{0ex}{0ex}}(9\times 2)+1=19\phantom{\rule{0ex}{0ex}}(19\times 2)+1=?\phantom{\rule{0ex}{0ex}}(?\times 2)+1=79\phantom{\rule{0ex}{0ex}}(79\times 2)+1=159\phantom{\rule{0ex}{0ex}}(159\times 2)+1=319$After analyzing the pattern, we get:

(?) =$(19\times 2)+1=39$

- (a)$80$
- (b)$65$
- (c)$62.5$
- (d)$83.5$
- (e)None of these

**Hint**

In this series, the number is divided by a specific number to get the next number.

**Solution:**

Given that,

$4000,2000,1000,500,250,125,?$The pattern will be:

$4000\xf72=2000\phantom{\rule{0ex}{0ex}}2000\xf72=1000\phantom{\rule{0ex}{0ex}}1000\xf72=500\phantom{\rule{0ex}{0ex}}500\xf72=250\phantom{\rule{0ex}{0ex}}250\xf72=125$So, the next number will be:$125\xf72=62.5$

- (a)$500$
- (b)$496$
- (c)$494$
- (d)$490$
- (e)None of these

**Hint**

In this series, the pattern is:

$588-A=563,563-(A-2)=540,540-(A-4)=519..$

**Solution:**

Given that,

$588,563,540,519,?,483,468$The pattern will be:

$588-\left(25\right)=563\phantom{\rule{0ex}{0ex}}563-\left(23\right)=540\phantom{\rule{0ex}{0ex}}540-\left(21\right)=519$So, next will be:

$519-\left(19\right)=500\phantom{\rule{0ex}{0ex}}500-\left(17\right)=483\phantom{\rule{0ex}{0ex}}483-\left(15\right)=468$

- (a)92
- (b)114
- (c)98
- (d)100
- (e)None of these

**Hint**

Series of squares of consecutive numbers.

**Solution:**

Given that,

$121,?,81,64,49,36,25$

The pattern will be,

$121=(11{)}^{2}\phantom{\rule{0ex}{0ex}}?\phantom{\rule{0ex}{0ex}}81=(9{)}^{2}\phantom{\rule{0ex}{0ex}}64=(8{)}^{2}\phantom{\rule{0ex}{0ex}}49=(7{)}^{2}\phantom{\rule{0ex}{0ex}}36=(6{)}^{2}\phantom{\rule{0ex}{0ex}}25=(5{)}^{2}\phantom{\rule{0ex}{0ex}}(?)=(10{)}^{2}=100\phantom{\rule{0ex}{0ex}}$

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- (a)$396.1$
- (b)$6433.6$
- (c)$10052.6$
- (d)$402.1$
- (e)None of these

**Hint**

Bring all the known values to one side and solve carefully using BODMAS rule.

**Solution:**

Given that,

$5\times ?=\frac{8042}{4}$

Bring the known values to one side and the unknown value on another side.

$?=\frac{8042}{5\times 4}$

$?=\frac{2010.5}{5}$

$?=402.1$

Hence, the correct answer is $402.1$

- (a)$2546$
- (b)$2654$
- (c)$2564$
- (d)$2645$
- (e)None of these

**Hint**

Apply BODMAS Rule.

**Solution:**

Given that,

$206\times 71\u201312080=?$

First multiply the number and then subtract $12080$ from it.

$=14626\u201312080$

$=2546$

Hence, the correct answer is $2546$.

- (a)$272$
- (b)$136$
- (c)$102$
- (d)$68$
- (e)
None of these

**Hint**

Apply BODMAS rule to get the missing numbers.

**Solution:**

First divide the numbers and then apply appropriate operation to get the missing number

$7632\xf732+9248\xf7?=306.5$

$238.5+\frac{9248}{?}=306.5$

$\frac{9248}{?}=306.5-238.5$

$\frac{9248}{?}=68$

$?=\frac{9248}{68}$

$?=136$

- (a)$5.5$
- (b)$6.5$
- (c)$4.5$
- (d)$3.5$
- (e)
None of these

**Hint**

Use the BODMAS rule and solve the bracket first.

**Solution:**

Given that,

$2073.5\xf7(22\times 14.5)=?$

First multiply the two numbers of denominator and divide the numerator by the resultant.

$22\times 14.5=319$

Now, $\frac{2073.5}{319}=6.5$

Hence, the correct answer is $6.5$.

- (a)$17$
- (b)$26$
- (c)$13$
- (d)$34$
- (e)None of these

**Hint**

Apply BODMAS in both numerator and denominator and then divide.

**Solution:**

Apply the rule of BODMAS and solve the numerator and denominator

$\frac{18+51-1}{8-15\xf73-1}$

$=\frac{18+51-1}{8-5-1}$

$=\frac{18+50}{2}$

$=\frac{68}{2}$

$=34$

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- (a)$47$
- (b)$48$
- (c)$48.10$
- (d)$47.10$
- (e)None of these

**Hint**

$46\%$ of $156$ is $71.76$

**Solution:**

On simplifying:

$\frac{46}{100}\times 156-23.76$

$=0.46\times 156-23.76$

$=71.76-23.76$

$=48$

- (a)$25$
- (b)$36$
- (c)$20$
- (d)$40$

**Hint**

$25\%$ of $420$ is $105.$

**Solution:**

On simplifying:

$\frac{25}{100}\times 420-\frac{?}{100}\times 140=77$

$105-?\times \frac{7}{5}=77$

$?\times \frac{7}{5}=105-77$

$?=28\times \frac{5}{7}$

$?=20$

- (a)$820$
- (b)$810$
- (c)$790$
- (d)$800$

**Hint**

Mulitply $90$ by $9$ and subtract $10$ from it.

**Solution:**

On simplifying:

$\frac{36}{100}\times 250\times \frac{18}{100}\times 50=?+10$

$90\times 9=?+10$

$?=810-10=800$

- (a)$21$
- (b)$14$
- (c)$20$
- (d)$30$
- (e)None

**Hint**

Convert percentage into fraction and then simplify.

**Solution:**

On simplifying:

$\frac{7}{36}\times \frac{20}{100}\times 540$

$=\frac{7\times 15}{5}$

$=21$

- (a)$31$
- (b)$33$
- (c)$35$
- (d)$37$
- (e)None of these

**Hint**

$a\%ofb=\frac{ab}{100}$

**Solution:**

On simplifying:

$\frac{?}{100}\times 924+39=371.64$

$\frac{?}{100}\times 924=371.64-39=332.64$

$?=\frac{33264}{924}=36$

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- (a)$13800$
- (b)$12500$
- (c)$11200$
- (d)Cannot be determined
- (e)None of these

**Hint**

Calculating cost price by incurring $25\%$ loss and then find the selling price with $25\%$ profit.

**Solution:**

Let cost price = $CP$

$SP=CP\left(\frac{100-Loss}{100}\right)$

$7500=CP\left(\frac{100-25}{100}\right)$

$CP=7500\times \frac{100}{75}$

$CP=100\times 100=10000$

Selling price to gain $25\%$ profit

$=10000+\left(10000\times \frac{25}{100}\right)$

$=10000\left(1+\frac{1}{4}\right)$

$=\frac{10000\times 5}{4}=12500$

- (a)$5725$
- (b)$4080$
- (c)$5250$
- (d)$4400$
- (e)
None of these

**Hint**

Equate $130\%$ of CP to SP and calculate CP.

**Solution:**

SP $=6110,$ Profit $=30\%$

Let the cost price = CP

$CP=\frac{6110\times 100}{100+30}$

$=4700$

- (a)$27700$
- (b)$25600$
- (c)$21250$
- (d)$29000$
- (e)
None of these

**Hint**

$25\%$ of CP is equal to SP.

**Solution:**

The cost price of an article

$=\frac{6800\times 100}{100-75}$

$=\frac{680000}{25}$

$=27200$

- (a)$\u20b915680$
- (b)$\u20b916220$
- (c)$\u20b914540$
- (d)Cannot be determined
- (e)
None of these

**Hint**

Multiply $140\%$ and divide $75\%$ onto the previous selling price.

**Solution:**

According to the question:

SP$=\u20b98400$

Loss $\%=25\%$

Required price of the item

$=\u20b98400\times \frac{100+40}{100-25}$

$=8400\times \frac{140}{75}$

$=\u20b915680$

Hence, the correct answer is $\u20b915680$.

- (a)$14656$
- (b)$11856$
- (c)$13544$
- (d)Cannot be determined
- (e)
None of these

**Hint**

Find cost price by dividing selling price by $70\%$ and then find new sellling price by multiplying by $130\%$

**Solution:**

Cost prices of the item

$=\frac{6384\times 100}{100-30}$

$=\frac{638400}{70}$

$=9120$

Hence, at a $30\%$ profit selling price of the item

$9120\times \frac{100+30}{100}$

$=\frac{9120\times 130}{100}$

$=11856$

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- (a)$18$
- (b)$282$
- (c)$40$
- (d)$106$
- (e)$183$

**Hint**

It is a series of difference of odd multiples of 11.

**Solution:**

$7+11\times 1=18\phantom{\rule{0ex}{0ex}}18+11\times 3=51\phantom{\rule{0ex}{0ex}}51+11\times 5=106\phantom{\rule{0ex}{0ex}}106+11\times 7=183\phantom{\rule{0ex}{0ex}}183+11\times 9=282\phantom{\rule{0ex}{0ex}}282+11\times 11=403$

So, the wrong number is $40$.

- (a)$3360$
- (b)$3460$
- (c)$3440$
- (d)$3406$
- (e)None of these

**Hint**

Train fare is equal to three-fourth of twice of bus fare.

**Solution:**

Let the bus fare be $B$ for one person and let the train fare be $T$ for more than one person.

$B=420$,

$=T=\frac{3}{4}\times 2B=\frac{3}{4}\times 2\times 420=630$

$2B+4T$$=2\times 420+4\times 630=840+2520=3360$.

- (a)$144$
- (b)$136$
- (c)$134$
- (d)$146$
- (e)None of these

**Hint**

It is series with difference of prime numbers.

**Solution:**

Pattern of the series is prime numbers subtracted in decreasing order.

$248-31=217$

$217-29=188$

$188-23=165$

$165-19=146$

$146-17=129$

$129-13=116$

So missing number of the series is $146.$

- (a)$255$
- (b)$218$
- (c)$243$
- (d)$225$
- (e)None of these

**Hint**

It is the series of difference of multiples of 12.

**Solution:**

The pattern followed in this series is given below:

$3+12\times 1=15$

$15+12\times 2=39$

$39+12\times 3=75$

$75+12\times 4=123$

$23+12\times 5=183$

$183+12\times 6=183+72=255$

So correct answer is $255.$

- (a)$425$
- (b)$520$
- (c)$500$
- (d)$625$
- (e)None of these

**Hint**

Find passing marks of boys and then equate that to the percentage and then find passing marks for girls.

**Solution:**

Passing marks for boys $=483+117=600$

Passing $\%$ for boys $=40\%$

Total marks $=600\times \frac{100}{40}=1500$

Passing $\%$for girls $=35\%$

Total marks for girls$=\frac{35}{100}\times 1500=525$

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- (a)$3$
- (b)$9$
- (c)$13$
- (d)$1$
- (e)None of these

**Hint**

Calculate HCF by using concept of finding HCF of prime numbers.

**Solution:**

As they are prime numbers, their HCF will be $1$ because they are divisible by $1$ and itself. The number $143$ is a multiple of $11$ and $13$ and as we can see both are prime numbers, So, therefore, their HCF will be $1$.

- (a)$189$
- (b)$190$
- (c)$289$
- (d)$198$
- (e)$200$

**Hint**

Product of LCM and HCF of two numbers = product of two numbers.

**Solution:**

Let the second number is $x$.

So, as per question,

$LCM\times HCF=productoftwonumbers$

So,

$(2079\times 27)=(297\times x)\phantom{\rule{0ex}{0ex}}x=\frac{2079\times 27}{297}\phantom{\rule{0ex}{0ex}}x=189$

- (a)$75$
- (b)$30$
- (c)$210$
- (d)$420$
- (e)None of these

**Hint**

Let the first number be $3x$and other number be $4x$

**Solution:**

Let the first number be $3x$and other number be $4x$.

Now,

$(LCM\times HCF)=(3x\times 4x)\phantom{\rule{0ex}{0ex}}12{x}^{2}=10800\phantom{\rule{0ex}{0ex}}{x}^{2}=\frac{10800}{12}=900\phantom{\rule{0ex}{0ex}}x=\sqrt{900}=30$

So, the sum of number will be $=(3x+4x)=7x=(7\times 30)=210$

- (a)$10$
- (b)$46$
- (c)$70$
- (d)$90$
- (e)None of these

**Hint**

Let the first number be$x$ and other number is $y$.

**Solution:**

Let the first number be$x$.

Other number be $y$.

As per given question,

$x+y=100\phantom{\rule{0ex}{0ex}}y=100-x....\left(1\right)$

As,

$LCM\times HCF=productoftwonumbers\phantom{\rule{0ex}{0ex}}LCM\times HCF=x\times y\phantom{\rule{0ex}{0ex}}495\times 5=xy\phantom{\rule{0ex}{0ex}}495\times 5=x\times (100\u2013x)\phantom{\rule{0ex}{0ex}}2475=100x\u2013{x}^{2}\phantom{\rule{0ex}{0ex}}{x}^{2}\u2013100x+2475=0\phantom{\rule{0ex}{0ex}}{X}^{2}\u201345x\u201355x+2475=0\phantom{\rule{0ex}{0ex}}x(x-45)\u201355(x\u201345)=0\phantom{\rule{0ex}{0ex}}(x\u201355)(x\u201345)=0\phantom{\rule{0ex}{0ex}}x=55orx=45$So, required difference will be $(55-45)=10$

- (a)$115$
- (b)$122$
- (c)$124$
- (d)$138$
- (e)
None of these

**Hint**

Let the LCM and HCF be $x$ and $y$ respectively.

**Solution:**

Let the LCM and HCF be $x$ and $y$ respectively.

According to the question,

$x=12y\phantom{\rule{0ex}{0ex}}x+y=403\phantom{\rule{0ex}{0ex}}12y+y=403\phantom{\rule{0ex}{0ex}}13y=403\phantom{\rule{0ex}{0ex}}y=31\phantom{\rule{0ex}{0ex}}x=12y=(12\times 31)=372$

LCM $\times $ HCF $=$ Product of two numbers

$372\times 31=93\times {n}_{2}\phantom{\rule{0ex}{0ex}}{n}_{2}=\frac{372\times 31}{93}=124$

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- (a)$403$
- (b)$396$
- (c)$428$
- (d)$383$
- (e)None of these

**Hint**

$Average=\frac{Sumofalltheterms}{No.ofterms}$

**Solution:**

Given that,

Set of scores $=152,635,121,423,632,744,365,253,302$

Required average will be $=\frac{\mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{the}\mathrm{terms}}{\mathrm{Number}\mathrm{of}\mathrm{terms}}$

$=\frac{152+635+121+423+632+744+365+253+302}{9}\phantom{\rule{0ex}{0ex}}=\frac{3627}{9}\phantom{\rule{0ex}{0ex}}=403$

Hence, the correct answer is $403$.

- (a)$2499$
- (b)$2352$
- (c)$2450$
- (d)$2550$
- (e)None of these

**Hint**

The numbers following each other from the smallest number to the largest number order are the Consecutive numbers.

**Solution:**

Let the four consecutive numbers A, B, C and D be $x,(x+1),(x+2),(x+3)$

As per question,

The Average of four number is $49.5$

So,

$\frac{x+(x+1)+(x+2)+(x+3)}{4}=49.5$

$x+(x+1)+(x+2)+(x+3)=(49.5\times 4)$

$4x+6=198$

$4x=192$

$x=\frac{192}{4}=48$

So, Product of B and D will be,$=(x+1)\times (x+3)=(48+1)\times (48+3)=(49\times 51)=2499$

- (a)$35$
- (b)$52$
- (c)$53$
- (d)$63$
- (e)None of these

**Hint**

The numbers following each other from the smallest number to the largest number order are the Consecutive numbers.

**Solution:**

Let the consecutive numbers be$x,(x+2),(x+4),(x+6),(x+8),(x+10)(x+12)$

So,

$\frac{x+(x+2)+(x+4)+(x+6)+(x+8)+(x+10)+(x+12)}{7}=47$

$7x+42=(47\times 7)$

$7x=329-42=287$

$x=\frac{287}{7}=41$

So, Required number will be $=(x+12)=(41+12)=53$

- (a)$2916$
- (b)$2988$
- (c)$3000$
- (d)$2800$
- (e)None of these

**Hint**

In consecutive even integers, starting with an even number, each number in the sequence is 2 more than the previous number.

**Solution:**

Let the consecutive even numbers be$x,(x+2),(x+4),(x+6),(x+8)$

As per question,

$x+(x+2)+(x+4)+(x+6)+(x+8)=(52\times 5)$

$5x+20=260$

$5x=240$

$x=\frac{240}{5}=48$

So, Product of B and E will be $=(x+2)\times (x+8)=(48+2)\times (48+8)=(50\times 56)=2800$

- (a)$483$
- (b)$675$
- (c)$621$
- (d)$525$
- (e)None of these

**Hint**

To calculate the average of n odd numbers. Use formula $\frac{(n+1)}{2}$

**Solution:**

Let the numbers be $x,(x+2),(x+4),(x+6)$

So as per the question,

$x+(x+2)+(x+4)+(x+6)=(24\times 4)\phantom{\rule{0ex}{0ex}}4x+12=96\phantom{\rule{0ex}{0ex}}4x=84\phantom{\rule{0ex}{0ex}}x=\frac{84}{4}=21$

So, Required product of B and D will be $=(x+2)\times (x+6)=(21+2)\times (21+6)=(23\times 27)=621$

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- (a)$6:4:8:10$
- (b)$6:8:9:10$
- (c)$8:6:10:9$
- (d)$4:6:8:10$

**Hint**

Equalise the ratio, by analysing the values of A, B, C and D.

**Solution:**

As per the question-

$A:B=\frac{1}{2}:\frac{3}{8}...\left(1\right)\phantom{\rule{0ex}{0ex}}B:C=\frac{1}{3}:\frac{5}{9}...\left(2\right)\phantom{\rule{0ex}{0ex}}C:D=\frac{5}{6}:\frac{3}{4}..\left(3\right)$

After equalising the ratio, we will get-

$A:B=\frac{1}{2}:\frac{3}{8}=(\frac{1}{2}\times 8):(\frac{3}{8}\times 8)=4:3....\left(4\right)\phantom{\rule{0ex}{0ex}}B:C=\frac{1}{3}:\frac{5}{9}=(\frac{1}{3}\times 9):(\frac{5}{9}\times 9)=3:5...\left(5\right)\phantom{\rule{0ex}{0ex}}C:D=\frac{5}{6}:\frac{3}{4}=(\frac{5}{6}\times 12):(\frac{3}{4}\times 12)=10:9...\left(6\right)$

From equations 4 and 5-

$A:B:C=[4:3:5]...\left(7\right)$

Equating the value of C in equation 6 and 7. We multiply equation 7 by 2.

$A:B:C:D=[2\times (4:3:5\left)\right]:9\phantom{\rule{0ex}{0ex}}=8:6:10:9$

- (a)$29$
- (b)$25$
- (c)$27$
- (d)$49$
- (e)$20$

**Hint**

Let, the value of fourth proportional is $x$. So,

$(6:10)::(12:x)\phantom{\rule{0ex}{0ex}}$

**Solution:**

Let, the value of fourth proportion is $x$.

So,

$(6:10)::(12:x)\phantom{\rule{0ex}{0ex}}\frac{6}{10}=\frac{12}{x}\phantom{\rule{0ex}{0ex}}x=\frac{12\times 10}{6}=20$

So, the value of fourth proportion will be $=x=20$.

- (a)37
- (b)40
- (c)45
- (d)31
- (e)None of These

**Hint**

Let, the value of third proportional $=x$

Now, we have-

$(5:15)::(15:x)$

**Solution:**

Let, the value of third proportional will be $x$

So,

$(5:15)::(15:x)\phantom{\rule{0ex}{0ex}}\frac{5}{15}=\frac{15}{x}\phantom{\rule{0ex}{0ex}}x=\frac{15\times 15}{5}=\frac{225}{5}=45$

So, the value of third proportional will be -

$x=45$

- (a)$12$
- (b)$13$
- (c)$18$
- (d)$21$
- (e)$15$

**Hint**

Let, the value of mean proportion is $=x.$

So, $(8:x)::(x:18)$

**Solution:**

Let, the value of mean proportion is $=x$

As per the question,

$(8:x)::(x:18)\phantom{\rule{0ex}{0ex}}\frac{8}{x}:\frac{x}{18}\phantom{\rule{0ex}{0ex}}{x}^{2}=18\times 8=144\phantom{\rule{0ex}{0ex}}x=\sqrt{144}=12$

- (a)$(x-y)$
- (b)$({x}^{2}-{y}^{2})$
- (c)${x}^{2}+{y}^{2}$
- (d)$(x+y)$
- (e)None of these

**Hint**

Let, the fourth proportional is $a$.

**Solution:**

Let, the fourth proportional is $a$

So,

$\frac{(x+y)}{(x+y{)}^{2}}=\frac{(x-y)}{a}$

$a=\frac{(x+y{)}^{2}(x-y)}{(x+y)}\phantom{\rule{0ex}{0ex}}a=(x+y)(x-y)\phantom{\rule{0ex}{0ex}}a={x}^{2}-{y}^{2}$

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- (a)$Rs.14133$
- (b)$Rs.15000$
- (c)$Rs.13460$
- (d)Cannot be determined
- (e)None of these

**Hint**

Profit share of $B$ will be $=\frac{B\text{'}sratio}{Totalratio}\times $Total profit

**Solution:**

Let, the ratio of the amount invested by $A,B$ and $C$ in business will be-

$A:B:C$

$=(35,000\times 12):(20,000\times 5):(15,000\times 7)$

$=420000:100000:105000$

$=420:100:105$

$=84:20:21$

Hence, the share of $B$ will be:

$=\frac{B\text{'}sshare}{Total\hspace{0.17em}share}\times $total profit

$=\frac{20}{84+20+21}\times 84125$

$=\frac{20}{125}\times 84125$

$=Rs.13460$

- (a)$Rs60000$
- (b)$Rs180000$
- (c)$Rs30000$
- (d)$Rs90000$
- (e)None of these

**Hint**

Profit is $15\%$ of total investment

$=4500000=\left(\frac{15}{100}\right)x$

**Solution:**

As per the question,

Profit is $15\%$ of total investment

Therefore, the total investment will be-

$=45000=\left(\frac{15}{100}\right)x$

$=x=45000\times \frac{100}{15}$

$=Rs300000$

Let, the investment of Avinash, Manoj and Arun will be $3x,2x$ and $5x$.

Hence,

Manoj's investment will be-

$=\frac{2x}{3x+2x+5x}\times 300000$

$=\frac{2x}{10x}\times 300000$

$=Rs60000$

- (a)$Rs280$
- (b)$Rs270$
- (c)$Rs120$
- (d)$Rs100$
- (e)None of these

**Hint**

Let, the total capital invested and total time period is $xt$ respectively.

**Solution:**

Let, the total capital invested and the total time period is $xt$ respectively.

Capital and time period for which A invests will be $\frac{x}{2},\frac{3t}{4}$ respectively.

Capital and time period for which B invests will be $\frac{x}{3},\frac{t}{2}$ respectively.

Capital and time period for which C invests will be $[1-\frac{x}{3}-\frac{x}{2}]t=\frac{x}{6}t$ respectively.

As per the question,

The ratio of investment will be

$=\left[\left(\frac{x}{2}\times \frac{3t}{4}\right):\left(\frac{x}{3}\times \frac{t}{2}\right):\left(\frac{x}{6}\times t\right)\right]$

$=\frac{3xt}{8}:\frac{xt}{6}:\frac{xt}{6}$

$=9:4:4$

Hence, the profit of A will be $=\frac{9}{9+4+4}\times Totalprofit$

$=\frac{9}{17}\times 510$

$=Rs270$

- (a)$Rs60000$
- (b)$Rs120000$
- (c)$Rs90000$
- (d)$Rs750000$
- (e)None of these

**Hint**

Let, the amount invested by Pankaj, Ashwin, and Tarun is $6x,5x8x$respectively.

$Profit=\frac{15}{100}\times Totalinvestment$

**Solution:**

Let, the amount invested by Pankaj, Ashwin and Tarun is $6x,5x$ and $8x$ respectively.

As per the question-

Profit is $15\%$ of total investment

So,

$Profit=\frac{15}{100}\times Totalinvestment\phantom{\rule{0ex}{0ex}}Totalinvestment=\frac{100\times 85500}{15}=Rs570000$

Hence,

Investment of Ashwin will be

$=\frac{5x}{6x+5x+8x}\times 570000\phantom{\rule{0ex}{0ex}}=\frac{5x}{19x}\times 570000\phantom{\rule{0ex}{0ex}}=Rs150000$

- (a)$15\frac{2}{3}\%$
- (b)$16\frac{2}{3}\%$
- (c)$20\%$
- (d)$18\%$
- (e)None of these

**Hint**

Increased price$=\mathrm{Old}\mathrm{price}\left(1+\frac{\mathrm{increased}\mathit{\%}}{100}\right)$

**Solution:**

Let, the old price of rice is $\u20b9100x$ per kg.

The value of the new price will be $20\%$ more than of old price.

Increased price of rice$=100x\left(1+\frac{20}{100}\right)$

$=100x\times \frac{120}{100}\phantom{\rule{0ex}{0ex}}=\u20b9120x$

So, the percentage decrease in consumption, if he wants to keep the expenditure fixed$=\frac{\mathrm{Increased}\mathrm{price}-\mathrm{Old}\mathrm{price}}{\mathrm{Increased}\mathrm{price}}\times 100$

$=\frac{120x-100x}{120x}\times 100$

$=\frac{20x}{120x}\times 100\%$

$=16\frac{2}{3}\%$

Hence, consumption should be reduced by $16\frac{2}{3}\%$.

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- (a)$10$ yr
- (b)$12$ yr
- (c)$14$ yr
- (d)$16$ yr
- (e)None of these

**Hint**

Let, the present age of Raghav and Amit is $7x$ and $8x.$ Four years ago, as per the question-

$\frac{7x-4}{8x-4}=\frac{5}{6}$

**Solution:**

Let, the present age of Raghav is $7x$

And the present age of Amit is $8x$

Four years ago,

Age of Raghav will be $(7x-4)yr.$

Age of Amit will be $(8x-4)yr.$

As per the question,

$\frac{7x-4}{8x-4}=\frac{5}{6}$

$42x-24=40x-20$

$2x=4$

$x=\frac{4}{2}=2$

Hence, the present age of Amit will be $=8x=(8\times 2)=16$ yr.

- (a)$46$ yr
- (b)$40$ yr
- (c)$32$ yr
- (d)$16$ yr
- (e)None of these

**Hint**

Let, Rupa’s age before six years is $4x$ and Seema's age before six years is $2x.$

**Solution:**

Let, the Rupa’s age before six years is $4x$

And Seema’s age before six years will be $2x$

After four years,

Rupa's age will be $(4x+6+4)yr$

Seema's present age will be $(2x+6+4)yr$

As per the question,

$\frac{2x+10}{4x+10}=\frac{3}{5}$

$10x+50=12x+30$

$2x=20$

$x=10$ yr

Hence, the present age of Rupa will be $=4x+6=(4\times 10+6)=46$ yr

- (a)$12$ yr
- (b)$24$ yr
- (c)$26$ yr
- (d)$42$ yr
- (e)$36$ yr

**Hint**

Let, the present age of A is $13x$ and the present age of B is $7x.$ Before six years-

$\Rightarrow \frac{3x-6}{7x-6}=\frac{2}{1}$

**Solution:**

Let, the present age of A is $13x$ and the present age of B is $7x.$

Before six years,

Age of A will be $(13x-6)$ yr

Age of B will be $(7x-6)$ yr

$\frac{13x-6}{7x-6}=\frac{2}{1}$

$13x-6=14x-12$

$x=6$

Hence, the difference between the ages of A and B will be $(13x-7x)=6x=36$ yr

- (a)$50$ yr
- (b)$28$ yr
- (c)$42$ yr
- (d)$36$ yr
- (e)None of these

**Hint**

Let, the present age of Purvi is $x$ yr and the present age of Anil is $1.5x$ yr.

**Solution:**

Let, the present age of Purvi is $x$ yr and the present age of Anil is $1.5x$ yr.

Eight years hence,The age of Purvi will be $(x+8)$ yr.

The age of Anil will be $(1.5x+8)$ yr.As per the question,

$\frac{1.5x+8}{x+8}=\frac{25}{18}$

$27x+144=25x+200$

$2x=56$

$x=\frac{56}{2}=28$ yr

- (a)$35$ yr
- (b)$45$ yr
- (c)$40$ yr
- (d)$50$ yr
- (e)None of these

**Hint**

Let, the Saroj’s age before $10$ years $=x.$

And the Rakhi's age before $10$ years $=\frac{x}{3}.$

**Solution:**

Let the Saroj’s age before $10$ years $=x$

Therefore, Rakhi’s age before $10$ years $=\frac{1}{3}x$

Hence, the present age of Saroj is $\left(x+15\right)$ yr

So, the present age of Rakhi is $\left(\frac{x}{3}+10\right)$ yrAs per the question,

$\frac{x+10}{{\displaystyle \frac{x}{3}+10}}=\frac{5}{3}$

$\frac{x+10}{{\displaystyle \frac{x+30}{3}}}=\frac{5}{3}$

$\frac{(x+10)\times 3}{x+30}=\frac{5}{3}$

$\frac{3x+30}{x+30}=\frac{5}{3}$

$9x+90=5x+150$

$4x=60$

$x=\frac{60}{4}=15$ yr

Hence, the required total will be-

$=(15+10)+\left(\frac{15}{3}+10\right)$

$=25+15=40$ yr

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- (a)$159.86cm$
- (b)$158.54cm$
- (c)$159.56cm$
- (d)$158.74cm$
- (e)None of these

**Hint**

Calculate the total height of all the girls and then find the actual averages.

**Solution:**

As per question,

Total height of $35$ girls will be $=(35\times 160)=5600cm$

Hence,

Corrected average height of $35$ girls $=\frac{5600+(104-114)}{35}\phantom{\rule{0ex}{0ex}}=\frac{5600-10}{35}\phantom{\rule{0ex}{0ex}}=\frac{5590}{35}\phantom{\rule{0ex}{0ex}}=159.71cm$

- (a)$8\mathrm{min}$
- (b)$10\mathrm{min}$
- (c)$3\mathrm{min}$
- (d)$6\mathrm{min}$
- (e)$12\mathrm{min}$

**Hint**

Calculate the LCM of individual times, then convert time into minutes.

**Solution:**

Time to meet again will be equal to LCM of their individual times.

So,

LCM of $54s,42s,$ and $63\mathrm{s}$ will be,

$54=2\times 3\times 3\times 3$$42=2\times 3\times 7$ $63=3\times 3\times 7$

LCM $=2\times 3\times 3\times 3\times 7=378\mathrm{sec}$

Now, On Converting time into minutes $\frac{378}{60}\approx 6\mathrm{min}$.

- (a)$25$ yr
- (b)$20$ yr
- (c)$30$ yr
- (d)$24$ yr

**Hint**

Let the present age of Tina and Rakesh as $9x10x$.

**Solution:**

Let the present age of Tina and Rakesh is $=9x10x$ respectively.

Ten years ago, ages of Tina and Rakesh will be $(9x-10)(10x-10)$ respectively.

As per question,

$\frac{9x-10}{10x-10}=\frac{4}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow 45x-50=40x-40\phantom{\rule{0ex}{0ex}}\Rightarrow 5x=10\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{10}{5}=2$

So, the present age of Rakesh will be

$=10x\phantom{\rule{0ex}{0ex}}=(10\times 2)\phantom{\rule{0ex}{0ex}}=20yr$

Hence, the correct answer is $20$ yr.

- (a)$196$
- (b)$288$
- (c)$324$
- (d)Cannot be determined
- (e)None of these

**Hint**

Calculate the LCM of$18,24$ and $32$.

**Solution:**

Time for meeting again at the starting point will be equal to LCM of their individual times.

So,

LCM of$18,24$ and $32$ will be,

$18=2\times 3\times 3$

$24=2\times 3\times 2\times 2$

$32=2\times 2\times 2\times 2\times 2$

LCM$=2\times 3\times 3\times 2\times 2\times 2\times 2=288$

Hence, they will meet after$288\mathrm{sec}$