### Contents of QUANTITATIVE APTITUDE AND REASONING

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

#### 27 Chapters

- 1. Number System
- 2. Percentages
- 3. Profit and Loss
- 4. Simple Interest and Compound Interest
- 5. Ratio, Proportion, and Variation
- 6. Averages
- 7. Mixture and Alligation
- 8. Work, Pipes, and Cisterns
- 9. Time, Speed, and Distance
- 10. Basic Algebra
- 11. Logarithms and Indices
- 12. Permutation and Combination
- 13. Probability
- 14. Mensuration
- 15. Clock and Calendar
- 16. Series
- 17. Time Sequence, Ranking, and Number Tests
- 18. Direction Sense
- 19. Blood Relations
- 20. Analogies
- 21. Odd One Out
- 22. Logical Consistency
- 23. Deductive Logic
- 24. Assumption
- 25. Conclusion
- 26. Course of Action
- 27. Cause and Effect

#### 7 Topics

Introduction to Numbers

Divisibility Rules

Highest Common Factor

Least Common Multiple

Finding the Unit Place of a Number Using the Cyclicity Rule

Remainders

Dealing with Questions Based on Two-/ Three-Digit Numbers

#### 2 Topics

The Multipliers

Successive Change

#### 4 Topics

Finding the Profit/Loss Percentage

Finding Mark Price, Markup, and Discount

Finding the Profit Margin and the Loss Margin

Solving Questions on False Weights

#### 4 Topics

Simple Interest

Compound Interest

Calculations in Simple Interest and Compound Interest

Effective Rate of Interest (in Case of Compound Interest)

#### 4 Topics

Ratio

Partnership

Proportion

Variations

#### 2 Topics

Average

Weighted Average

#### 1 Topics

Mixture and Alligation

#### 4 Topics

Time and Work

Working on Alternate Days

Work and Wages

Mixed Efficiency Type Work-Based Questions

#### 7 Topics

Relation Between Time, Speed, and Distance

Average Speed

Relation of Proportionality Between Time, Speed, and Distance

Concept of Trains

Boats and Streams

Linear Races

Circular Races

#### 4 Topics

Equations

Quadratic Equation

Arithmetic Progression (A.P.)

Geometrical Progression (G.P.)

#### 2 Topics

Logarithms

Indices and Surds

#### 3 Topics

Fundamental Principle of Counting

Combination

Permutation

#### 2 Topics

Two-Dimensional Structures

Solids or Three-Dimensional Structures

#### 2 Topics

Clocks

Calendars

#### 3 Topics

Time Sequence Test

Ranking Test

Number Test

#### 1 Topics

Logical Consistency

#### 1 Topics

Course of Action

#### 1 Topics

Cause and Effect

## Experience Tests Tailored to Match the Real Exam

## Practise Questions with Solutions from the QUANTITATIVE APTITUDE AND REASONING

- (a)$103$
- (b)$121$
- (c)$171$
- (d)$91$

**Hint**

A prime number is a whole number greater $1$ than whose only factors are $1$ and the number itself.

**Solution:**

A prime number is a whole number greater than $1$, whose only factors are $1$ and the number itself.

In the given question $103$ is a prime number. $121$ is multiple of $11,171$ is divisible by $9$, and $91$ is divisible by $7$.

Hence, option A is correct.

- (a)Positive integer
- (b)Negative integer
- (c)Neither positive nor negative integer
- (d)None of these

**Hint**

Zero is neither positive nor negative. The term non-negative is sometimes used to refer to a number that is either positive or zero. Zero is a neutral number.

**Solution:**

An integer is a whole number but not a fraction. Integer could be positive, negative or zero.

Which means zero is neither positive nor negative.

Hence, option C is correct.

- (a)${n}^{2}+n+2n$
- (b)$n\left(n-1\right)+n-2$
- (c)${n}^{3}+2n$
- (d)${n}^{3}+2n-1$

**Hint**

Odd numbers are whole numbers that cannot be divided exactly into pairs. Hence, when divided by $2$, the number leaves a remainder of $1$.

**Solution:**

According to the question, if '$n$' is even number, then we have to find out the odd number from the given options.

Let us put the value of $n=2$

By putting the value of $n=2$, in option (a)

$={n}^{2}+n+2n$

$={2}^{2}+2+2\left(2\right)$

$=4+2+4$

$=10$

Here, $10$ is an even number

By putting the value of $n=2$, in option (b)

$=n(n-1)+n-2$

$=2(2-1)+2-2$

$=2\left(1\right)+0$

$=2$

Here, $2$ is an even number

By putting the value of $n=2$, in option (c)

$={n}^{3}+2n$

$=(2{)}^{3}+2(2)$

$=8+4$

$=12$

Here, $12$ is even number.

By putting the value of $n=2$, in option (d).

$={n}^{3}+2n-1$

$={2}^{3}+2\times 2-1$

$=11$

Here, $11$ is odd number.

Hence, option (d) is correct answer.

- (a)${n}^{5}+n$
- (b)$n\left(n+1\right)+n{\left(n-1\right)}^{2}$
- (c)$5{n}^{3}-2{n}^{2}$
- (d)${n}^{3}+\left(\frac{n}{2}\right)$

**Hint**

Product of any number which is multiplied with even number will be even only.

**Solution:**

Since '$n$' is the even number then the following will be true-

The product of any number which is multiplied with an even number will be even only.

Sum of the addition of even number to even number will be an even only.

But dividing an even number by an even number may be even or odd.

Here, ${n}^{3}+\left(\frac{n}{2}\right)$ may be even or odd.

Hence, option D is correct.

- (a)$15$
- (b)$12$
- (c)$17$
- (d)None of these

**Hint**

As unit place of the given number ends with $5$ it is divisible by $5$.

**Solution:**

As unit place of the given number ends with $5$ it is divisible by $5$.

And according to divisibility rule, the sum of the digits of the given number is $1+3+4+5+6+5=24$, which is divisible by $3$.

Hence, it must be divisible by $15$ also.

Therefore, option A is correct.

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- (a)$280$
- (b)$300$
- (c)$325$
- (d)$250$

**Hint**

Consider the total quantity of mangoes is $100\%$.

**Solution:**

Let the total mangoes be $100\%$

Bad mangoes$=$ $10\%$Good mangoes $=100\%-10\%=90\%$

She gives $33.33\%$ to poor,$\frac{1}{3}\times 90\%=30\%$

Remaining mangoes $=90\%-30\%=60\%$If she had left with $180$, then $60\%=180$

$\Rightarrow 1\%=3\phantom{\rule{0ex}{0ex}}\Rightarrow 1\times 100\%=3\times 100$$\Rightarrow 100\%=300$

Hence, the correct answer is $300$ mangoes.

- (a)$16.66\%$
- (b)$15.5\%$
- (c)$20\%$
- (d)$14.66\%$

**Hint**

Use the formula: Reduction$\%=\left(\frac{R}{(100+R)}\times 100\right)\%$.

**Solution:**

According to the question: If the price of a commodity increases by $R\%$, the reduction in consumption so as not to increase the expenditure $=\left[\frac{R}{100+R}\times 100\right]\%$

Here, $R=20\%$

$\therefore \left[\frac{20}{100+20}\times 100\right]\%=\frac{20}{120}\times 100=\frac{50}{3}\%$

$\Rightarrow 16.66\%$

Hence, option A is correct.

- (a)$15\%$
- (b)$18\%$
- (c)$12\%$
- (d)$20\%$

**Hint**

Percentage of salt**$=\frac{\mathrm{Amount}\mathrm{of}\mathrm{salt}\mathrm{in}\mathrm{solution}}{\mathrm{Remaining}\mathrm{solution}}\times 100$**

**Solution:**

If $2$ litres of water evaporated from $10$ liters of solution

Remaining solution $=10\u20132=8$ litres

$\therefore 10$litres of $12\%$$=10\times \frac{12}{100}=1.2$ (as only water evaporate from solution)$\therefore $ Solution in $8$ litres is $=\frac{1.2}{8}\times 100$

$=0.6\times 25=15\%$

Salt in the remaining solution is $15\%$

- (a)Increase, $14\%$
- (b)Decrease, $14\%$
- (c)Increase, $12\%$
- (d)Decrease, $12\%$

**Hint**

Percentage Change $=\left(\frac{NewValue-oldValue}{OldValue}\times 100\right)$

**Solution:**

Let the original salary be $\u20b9x$

Salary after increase$=40\%x\phantom{\rule{0ex}{0ex}}=x+\frac{40}{100}x\phantom{\rule{0ex}{0ex}}=\frac{140x}{100}$

$=\frac{7x}{5}$

Salary after decrease

$=\frac{7x}{5}-20\%$ of $\frac{7x}{5}$

$=\frac{7x}{5}-\left(\frac{20}{100}\times \frac{7x}{5}\right)$

$=\frac{7x}{5}-\frac{7x}{25}$

$=\frac{35x-7x}{25}\phantom{\rule{0ex}{0ex}}=\frac{28x}{25}$

Absolute increase

$=\frac{28x}{25}-x$

$=\frac{28x-25x}{25}\phantom{\rule{0ex}{0ex}}=\frac{3x}{25}$

Percentage increase

$=\frac{3x}{25}\times \frac{100}{x}\phantom{\rule{0ex}{0ex}}=12$

So, salary increases by $12\%$.

Hence, the correct answer is 'Increase, $12\%$'.

- (a)A profit of $12$ lakhs
- (b)A profit of $10$ lakhs
- (c)A profit of $9.6$ lakhs
- (d)No profit no loss

**Hint**

Calculate the share of the builder by using formula; $Profit=S.P\u2013C.P.$

**Solution:**

A builder owns a house worth $40$ lakhs.

Price paid by Rahul for the house $=\frac{6}{5}\times 40=48$ lakhs.Price paid by the builder $=\frac{4}{5}\times 48=38.4$ lakhs

Profit $=48-38.4=9.6$ lakhs

Thus, the profit made by builder is $9.6$ lakhs.

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- (a)$4\%$
- (b)$8\%$
- (c)$10\%$
- (d)$5\%$

**Hint**

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

**Solution:**

According to the question,

Cost price of laptop (CP) $=\u20b91,260$

Selling price of laptop (SP) $=\u20b91,197$

We know that, loss $\mathrm{loss}=\mathrm{CP}\u2013\mathrm{SP}$

$\u20b91,260-\u20b91,197=\u20b963$$\mathrm{loss}\%$ $=\frac{\mathrm{Loss}}{\mathrm{CP}}\times 100$

$\mathrm{loss}\%=\frac{63}{1260}\times 100$

$\mathrm{loss}\%=5\%$

Hence, the correct answer is $5\%$.

- (a)$30\%$
- (b)$33.33\%$
- (c)$35\%$
- (d)None of these

**Hint**

$\mathrm{Profit}\%$ $=\frac{\mathrm{Profit}\times 100}{\mathrm{C}.\mathrm{P}}$

**Solution:**

In the given question, dealer does fraud of $20\%$ while selling and $10\%$ while purchasing. Thus, he gains profit from both sides.

Now, consider $\mathrm{Rs}.100$ be the cost price.

According to question he gains a profit of $10\%$ while purchasing i.e., $10\%$ of $\mathrm{Rs}.100$ is $\mathrm{Rs}.10$. So, the cost price is $\mathrm{Rs}.90$.While selling, he gains a profit of $20\%$ on cost price i.e., $\frac{100\times 20}{100}=\mathrm{Rs}.20$

Dealer gains total profit $=\mathrm{Rs}.\left(10+20\right)=\mathrm{Rs}.30$

Profit percentage $=\frac{30}{90}\times 100=33.33\%$.Hence, dealer gets a profit of $33.33\%$

- (a)$11.11\%$
- (b)$8.25\%$
- (c)$10.33\%$
- (d)$11.5\%$

**Hint**

$\mathrm{Profit}\%$ $=\frac{\mathrm{Profit}}{\mathrm{C}.\mathrm{P}}\times 100$

**Solution:**

According to the given question,

Cost of $40$ mangoes $=\mathrm{Rs}.60$

SP of $12$ mangoes $=\mathrm{Rs}.20$

SP of $40$ mangoes$=$$\frac{20}{12}\times 40=\mathrm{Rs}.66.66$

Using formula,

$\mathrm{Profit}=\mathrm{SP}-\mathrm{CP}$

$=66.66-60=\mathrm{Rs}.6.66$

$\mathrm{Gain}\%=$ $\frac{\mathrm{Profit}}{\mathrm{CP}}\times 100$=$\frac{6.66}{60}\times 100=11.11\%$.

- (a)$66\%$
- (b)$68\%$
- (c)$70\%$
- (d)$64\%$

**Hint**

$\mathrm{Profit}\%=\frac{\mathrm{Profit}}{\mathrm{CP}}\times 100$.

**Solution:**

Let us consider the cost price of a book be $\mathrm{Rs}.1$.

Then, $360$ books each with equal cost will be $\mathrm{Rs}.360$

With $40\%$ of profit, the selling price of books will be$\mathrm{SP}=\mathrm{Profit}+\mathrm{CP}$

$=\frac{360\times 40}{100}+360$

$=144+360=504$

According to the question, the trader sells $160$ books less, Which means he sells only $200$ books.

So, the CP of $200$ books is $\mathrm{Rs}.200$

As he sells books at two-third of earlier SP, then the selling Price of $200$ books is $\frac{2}{3}\times 504=336$Thus,

$\mathrm{Profit}=\mathrm{SP}\u2013\mathrm{CP}$ $=336\u2013200=136$

$\mathrm{Profit}\%$ $=\frac{136}{200}\times 100\%$$=68\%$

Hence, the $\mathrm{Profit}\%$ of the trader is $68\%$.

- (a)$21\%$
- (b)$18\%$
- (c)$19\%$
- (d)$25\%$

**Hint**

$\mathrm{Gain}\%$ $=\frac{\mathrm{Gain}}{\mathrm{C}.\mathrm{P}}\times 100$.

**Solution:**

In the question, it is given that the cost price of $20$ pens is the same as selling price of $16$ pens.

CP $=16$ pens

SP $=20$ pens

$\mathrm{Gain}$ $=20\u201316=4$$\mathrm{Gain}\%$ $=\frac{\mathrm{profit}}{\mathrm{CP}}\times 100$

$\text{Gain\%}$ $=\frac{4}{16}\times 100=25\%$

Hence, the $\mathrm{Gain}\%$ is $25\%$.

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- (a)$20$ years
- (b)$30$ years
- (c)$25$ years
- (d)$40$ years

**Hint**

Profit $=$ SP $-$ CP

**Solution:**

Assuming simple interest, let the amount be $P$ and rate of interest per annum be $R\%$.

Given that $P$ becomes $2P$ in $10$ years.

$\Rightarrow P+P\times 10\times \frac{R}{100}=2P$

$\Rightarrow 1+\frac{R}{10}=2$

$R=10$

Let $P$ triple itself in $N$ years:

$\Rightarrow P+P\times N\times \frac{10}{100}=3P$

$\Rightarrow 1+\frac{N}{10}=3$

$N=20$ years.

- (a)$22$ years
- (b)$23$ years
- (c)$25$ years
- (d)$20$ years

**Hint**

In the question, Sum of money becomes four times itself then we have to assume the principal value be $x$ and the given that amount gets four times will be $4x$ and then find the Simple interest value. Therefore using the simple interest formula we will calculate the time.

**Solution:**

Let the principal be $P$.

Then the amount becomes $4P$.

$\Rightarrow $ S.I. $=4P-P=3P$

$=\frac{P\times 12\times T}{100}$

$\Rightarrow T=\frac{3P\times 100}{P\times 12}$

$T=25$ years

Hence, The correct answer is option (C).

- (a)$24$
- (b)$23$
- (c)$22$
- (d)$21$

**Hint**

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

**Solution:**

According to the question,

Let the principal be $P$.

Then, amount after $12$ years $=8P$

S.I. $=8P-P$

$7P=\frac{P\times R\times 12}{100}$

$R=\frac{175}{3}$

In the second case, the amount will become $15P$.

S.I. $=15P-P$

$14P=\frac{P\times 175\times T}{3\times 100}$

$T=24$ years.

- (a)$Rs.8,200$
- (b)$Rs.6,500$
- (c)$Rs.7,000$
- (d)$Rs.\mathrm{8,500}$

**Hint**

Using Formula:

C.I. $-$ S.I. $=\left[{\left(1+\frac{r}{100}\right)}^{n}-1-\frac{r\times n}{100}\right]$

**Solution:**

According to the question,

Two friends A and B invested equal sums of money for $2$ years at $10\%$ p.a.

A invested amount at C.I. whereas, B invested amount at S.I.

Here A gets $Rs.82$ more than B

C.I. $-$ S.I. for $2$ years,

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

$\Rightarrow P{\left(\frac{1}{10}\right)}^{2}=82$

$\Rightarrow P=Rs.8,200$

Hence the correct answer is $Rs.8,200$.

- (a)$1331\times {10}^{4}$
- (b)$1231\times {10}^{6}$
- (c)$1331\times {10}^{5}$
- (d)$21334\times {10}^{3}$

**Hint**

In the question, annual male Population percentage is given and the present population with these details we have to finds the male population $3$ years ago by using basic formula i.e., $P\times {\left(1+\frac{r}{100}\right)}^{t}$.

**Solution:**

According to the question,

Annual increase of male population $=10\%$

Present population $=17715610$

Let the male population $3$ years ago be $P$

$P=$ Present population $\times {\left(1+\frac{10}{10}\right)}^{3}$

$\Rightarrow P=$ Present population $\times {\left(\frac{11}{10}\right)}^{3}$

$\Rightarrow P=\frac{10\times 10\times 10}{11\times 11\times 11}\times 17715610$

$=13310\times 1000$

$=1331\times {10}^{4}$

Hence option $1331\times {10}^{4}$ is correct.

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- (a)$60$
- (b)$70$
- (c)$140$
- (d)None of these

**Hint**

If a number $\mathrm{C}$ is in ratio $\mathrm{A}:\mathrm{B}$ and then we can write it as $\mathrm{C}=\mathrm{Ax}+\mathrm{bx}$.

**Solution:**

According to the question,

Two numbers are in the ratio $3:4$

So, let the numbers be $3x$and $4x$.

If $10$is added to both the numbers, the ratio becomes $4:5$,

i.e.

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

or,

$15\mathrm{x}+50=16\mathrm{x}+40$

or,

$\mathrm{x}=10$

Therefore, the numbers are

$3\mathrm{x}=3\times 10=30$

and

$4\mathrm{x}=4\times 10=40$

The sum of the numbers

$=30+40=70$.

- (a)$9:10$
- (b)$9:11$
- (c)$9:13$
- (d)None of these

**Hint**

The ratio formula is $\mathrm{a}:\mathrm{b}\Rightarrow \frac{\mathrm{a}}{\mathrm{b}}$.

**Solution:**

According to the question,

$\mathrm{a}:\mathrm{b}=2:3$

So let $a$ and $b$ be $2x$ and $3x$ respectively.

$\frac{3\mathrm{a}+\mathrm{b}}{7\mathrm{a}-\mathrm{b}}=\frac{3\times 2\mathrm{x}+3\mathrm{x}}{7\times 2\mathrm{x}-3\mathrm{x}}$

$=\frac{6\mathrm{x}+3\mathrm{x}}{14\mathrm{x}-3\mathrm{x}}$

$=\frac{9\mathrm{x}}{11\mathrm{x}}$

Therefore,

$3\mathrm{a}+\mathrm{b}:7\mathrm{a}-\mathrm{b}=9:11.$

- (a)$4:1$
- (b)$2:1$
- (c)$16:2$
- (d)None of these

**Hint**

Rate of flow, $R=k\times \frac{1}{{r}^{2}}$, where $k$ is a constant and $r$ is the radius of pipe.

**Solution:**

The radius of two pipes are $4$ $cm$and $8$ $cm$,

so their squares would be $16$ $cm$and $64$ $cm$respectively.

The ratio of the rate of flow of water in pipes varies inversely as the square of the radii of the pipes,

So the ratio of the rates of flow in two pipes is,

$\frac{1}{16}:\frac{1}{64}$ or,

$64:16$ or,

$4:1$

The ratio is $4:1$.

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

**Hint**

Find the total numbers of ball of all types in bag $1$ and bag $2$ individually, then find the ratio.

**Solution:**

Let the balls in Bag$1=x$and Bag$2=y$

First, half the tennis balls from Bag $1$ are transferred to Bag $2$,

So, the Balls present in Bag $1$ $=x-\frac{x}{2}=\frac{x}{2}$ and the Balls in Bag $2=\frac{x}{2}+y=\frac{x+2y}{2}$ Then one-third and two-thirds of the tennis balls from the first and the second bags are transferred to the second and the first bags, respectively,i.e Now Balls in Bag

$1=\frac{2}{3}\times \frac{x}{2}+\frac{2}{3}\times \frac{x+2y}{2}$$=\frac{x}{3}+\frac{x+2y}{3}$

$=\frac{2(x+y)}{3}$

Balls in Bag $2$

$=\frac{1}{3}\times \frac{x}{2}+\frac{1}{3}\times \frac{x+2y}{2}$

$=\frac{x}{6}+\frac{x+2y}{6}$

$=\frac{2(x+y)}{6}$

$=\frac{x+y}{3}$

The ratio of balls present in Bag $1$ and Bag

$2=$$\frac{2(x+y)}{3}:\frac{x+y}{3}$

$=2:1$

- (a)$8$
- (b)$9$
- (c)$7$
- (d)$6$

**Hint**

If $a,bc$ are in continued proportions therefore,$\frac{a}{b}=\frac{b}{c}$

**Solution:**

$3$, $x$ and $27$ are continued proportions as given.

That means,

$\frac{3}{x}=\frac{x}{27}$

or, ${x}^{2}=81$

or, $x=9$

The value of $x=9.$

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- (a)$72$
- (b)$24$
- (c)$36$
- (d)$44$

**Hint**

Let three numbers be $x,yandz$. According to the question $x=2y=3z$.

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

**Solution:**

Let the first, second and third number be $x,y$ and $z$ respectively.

The first is twice the second and thrice the third, i.e.,

$x=2y=3z$

or, $y=\frac{x}{2}$ and $z=\frac{x}{3}$

The average of these three numbers is $44,$ i.e.,

$\frac{x+y+z}{3}=44$

or, $\mathrm{x}+\frac{\mathrm{x}}{2}+\frac{\mathrm{x}}{3}=44\times 3$

or, $6\mathrm{x}+3\mathrm{x}+2\mathrm{x}=44\times 3\times 6$

or, $11\mathrm{x}=44\times 3\times 6$

or, $\mathrm{x}=72$

Thus, the first number is $72.$

- (a)$44.66$
- (b)$40.44$
- (c)$43.67$
- (d)$44.8$

**Hint**

$\mathrm{Average}=\frac{\mathrm{Sum}\mathrm{of}\mathrm{terms}}{\mathrm{Number}\mathrm{of}\mathrm{terms}}.$

**Solution:**

According to the question,

Let the sum of those ten numbers be $\mathrm{x}.$

The average of $10$ numbers is $40.2.$

$\frac{\mathrm{x}}{10}=40.2$

$\mathrm{x}=402$

Two numbers $13$ and $18$ are removed and $31$ is added ($2$ numbers are removed and one is added, so total number of number is $9$),

The new average is

$=\frac{\mathrm{x}-13-18+31}{9}=\frac{402-31+31}{9}=44.66.$

- (a)$11.5$
- (b)$11.4$
- (c)$11.3$
- (d)$11.0$

**Hint**

$\mathrm{Average}=\frac{\mathrm{Sum}\mathrm{of}11\mathrm{numbers}}{11}$

**Solution:**

According to the question,

Let the sum of numbers be $x.$ The average of $11$ numbers is $10.9$ i.e.

$\frac{\mathrm{x}}{11}=10.9$

$\mathrm{x}=119.9$

Let the sum of first six numbers be $y.$ The average of the first six numbers is $10.5.$

$\frac{\mathrm{y}}{6}=10.5$

$\mathrm{y}=63$

And the sum of the last six digits is $11.4.$ Let the sum of the last six numbers be $z,$

$\frac{\mathrm{z}}{6}=11.4$

$\mathrm{z}=68.4$

Then the middle number is

$=$ Sum of first six number $+$ sum of last six numbers $-$ sum of all eleven number

$=63+68.4-119.9$

$=11.5.$

- (a)$73$
- (b)$65$
- (c)$78$
- (d)$64$

**Hint**

$Totalmarks=Averagemarks\times Numberofstudents.$

Average marks of girls $=\frac{Totalmraksoftheclass-Totalmarksofboys}{Numberofgirlsintheclass}$.

**Solution:**

According to the question,

In a class of $100$ students, there are $70$ boys whose average marks in a subject are $75.$ Let the sum of their marks be $x.$

$\frac{\mathrm{x}}{70}=75$

$\mathrm{x}=5250$

The marks of the complete class are $72.$ Let the sum of whole class marks be $y.$

$\frac{\mathrm{y}}{100}=72$

$\mathrm{y}=7200$

Then the average marks of the girls will be,

$=\frac{\mathrm{y}-\mathrm{x}}{100-70}=\frac{7200-5250}{30}=65.$

Hence, the correct answer is $65$.

- (a)$16.66$
- (b)$30$
- (c)$25$
- (d)$17.50$

**Hint**

$\mathrm{Average}=\frac{\mathrm{x}+\mathrm{y}+\mathrm{z}}{3}.$

**Solution:**

The average of $x$ and $y$ is $40.$

$\frac{\mathrm{x}+\mathrm{y}}{2}=40$

$\mathrm{x}+\mathrm{y}=80$

If $z=10,$ then the average of $\mathrm{x},\mathrm{y}$ and $\mathrm{z}$ will be,

$=\frac{\mathrm{x}+\mathrm{y}+\mathrm{z}}{3}$

$=\frac{80+10}{3}$

$=\frac{90}{3}=30.$

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- (a)$3:5$
- (b)$2:7$
- (c)$1:8$
- (d)$1:7$
- (e)None of these

**Hint**

$\mathrm{a}\%$ of a quantity $c$ is written as $\mathrm{c}\times \frac{\mathrm{a}}{100}$.

**Solution:**

According to the question,

$80\%$ of first milk solution and $40\%$ of second milk solution are mixed.

Let $x$ and $y$ be the volumes of each solution be mixed to get the new mixture.

According to question,

$80\%ofx+40\%ofy=45\%of(x+y)\phantom{\rule{0ex}{0ex}}\frac{80}{100}\times x+\frac{40}{100}\times y=\frac{45}{100}\times (x+y)\phantom{\rule{0ex}{0ex}}\frac{1}{100}\left(80x+40y\right)=\frac{1}{100}(45x+45y)\phantom{\rule{0ex}{0ex}}80x+40y=45x+45y\phantom{\rule{0ex}{0ex}}80x-45x=45y-40y\phantom{\rule{0ex}{0ex}}35x=5y\phantom{\rule{0ex}{0ex}}\frac{x}{y}=\frac{5}{35}\phantom{\rule{0ex}{0ex}}\therefore x:y=1:7$

Hence, the correct answer is $1:7$.

- (a)$36$
- (b)$45$
- (c)$72$
- (d)None of these
- (e)$63$

**Hint**

$\mathrm{A}\%$ of $\mathrm{B}=\frac{\mathrm{A}}{100}\times \mathrm{B}.$

**Solution:**

According to the question,

The number of boys $=36$

Percentage of boys passed$=70\%$

Percentage girls passed $=58\%$

Total percentage of boys and girls passed $=64\%$

Let's find the difference between the percent of boys and girls passed with respect to the total percent passed.

Boys: $70-64=6$

Girls: $64-58=6$

Here the difference between both girls and boys passed in the exam are the same which means the ratio will be,

$6:6=1:1$

As the ratio of passed boys and girls is $1:1$, the number of girls will be the same as a number of boys, that is $36$.

Hence the total number of girls$=36$

- (a)$\u20b933$
- (b)$\u20b934$
- (c)$\u20b935$
- (d)None of these
- (e)$\u20b940$

**Hint**

If two ingredients are mixed, then $\frac{Quantityofcheaper}{Quantityofdearer}=\frac{C.P.ofdearer-MeanPrice}{Meanprice-C.P.ofcheaper}.$

**Solution:**

According to the question,

Cheaper price: Juice A $=\u20b920/\mathrm{litre}$

Mean price: Mixture of Juice A and Juice B $=\u20b925/\mathrm{litre}$

Let the Dearer price be $\text{'}x\text{'}$.

For ratio:

(Dearer price $-$ Mean price) $:$ (Mean price $-$ Cheaper price)

$=\left(x-25\right):\left(25-20\right)=\left(x-25\right):5$

As given in the question, $40$ litres of Juice A and $20$ litres of Juice B mixed with ratio:

$40:20=2:1$

Hence, the ratio will be:

$\left(x-25\right):5=2:1$

$x-25=10$

$x=35$

Therefore, the dearer price or the price of Juice B per litre is $\u20b935$.

- (a)$8:7$
- (b)$8:9$
- (c)$9:8$
- (d)$7:8$
- (e)None of the above

**Hint**

Formula to calculate cost price if selling price and profit percentage are given:

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

**Solution:**

According to the question,

Price of Juice A is $\u20b925/\mathrm{litre}$.

Price of Juice B is $\u20b942/\mathrm{litre}$.

Profit $=10\%$

Selling price $=\u20b936.30/\mathrm{litre}$

Selling price of the mixture is $\u20b936.30$, which includes $10\%$ profit.

Cost price of the mixture $=\frac{36.30}{110}\times 100\phantom{\rule{0ex}{0ex}}=\frac{3630}{110}\phantom{\rule{0ex}{0ex}}=\u20b933$

Now,

$\frac{x}{y}=\frac{42-33}{33-25}=\frac{9}{8}$

Hence, the correct answer is $9:8$.

- (a)$44.44$ litres
- (b)$80$ litres
- (c)$85$ litres
- (d)None of these
- (e)$120$ litres

**Hint**

When $20$ litres of water added the ratios get reversed.

So, equation will be formed as:$\frac{\mathrm{Quantity}\mathrm{of}\mathrm{milk}}{\mathrm{quantity}\mathrm{of}\mathrm{water}+20\mathrm{litres}\mathrm{of}\mathrm{water}\mathrm{added}}=4:5$.

**Solution:**

According to the question,

Ratio of milk and water is $5:4$

Then, let the quanity of milk be $5\mathrm{x}$ and water be $4\mathrm{x}$ .

Hence, the total quantity of solution will be

$5\mathrm{x}+4\mathrm{x}=9\mathrm{x}$.

Here, $20$ litres of water is added, then the ratio becomes:

$\left(5\mathrm{x}\right):(4\mathrm{x}+20)=4:5$ (reversed)

$\frac{5\mathrm{x}}{4\mathrm{x}+20}=\frac{4}{5}$

$25\mathrm{x}=16\mathrm{x}+80$

$25\mathrm{x}-16\mathrm{x}=80$

$9\mathrm{x}=80$

$\mathrm{x}=\frac{80}{9}$.

Now, the initial quantity of solution $=9\times \frac{80}{9}$

$=80$ litres.

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- (a)$40$ days
- (b)$30$ days
- (c)$45$ days
- (d)None of these

**Hint**

If a person A completes a piece of work in a $\text{'}\mathrm{n}\text{'}$ days, then A's one day work $=\frac{1}{\mathrm{n}}$

**Solution:**

According to the question,

Days took by A and B to finish a work $=20$days

B left work after $10$ days.

(A$+$B)'s $10$days' work $\frac{1}{20}\times 10=\frac{1}{2}$

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

Now, $\frac{1}{2}$ work is done by A in $20$ days.

Thus, the entire work will be done in $40$ days.

Alternatively, let A and B together do $1$ unit of work per day.

Thus, work done in $20$ days $=20$ units (total work)

As they work for $10$ days, together, the units of work done $=$ $10$ units.

This means remaining $(20-10)=10$ units of work are done by A in $20$ days.

So, in a day, A can do $\left(\frac{10}{20}\right)=0.5$ units of work.

Therefore, A alone can do it in $\frac{20}{0.5}=40$ days.

Thus, A alone can finish the entire work in $40$ days.

- (a)$60$ days
- (b)$100$ days
- (c)$66.66$ days
- (d)$80$ days

**Hint**

Assume A and B finish the work alone in $x$ days and $y$ days, respectively.

Then together,

$\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}=\frac{1}{\mathrm{n}}$

**Solution:**

According to the question,

Days took by A and B to do a piece of work $=40$ days.

Let A and B finish the work alone in $\mathrm{x}$ days and $\mathrm{y}$ days, respectively.

Then,

$\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}=\frac{1}{40}$$\Rightarrow \frac{40}{\mathrm{x}}+\frac{40}{\mathrm{y}}=1---\left(1\right)$

Now, A works for $10$ days and B works for $60$ days, so

$\Rightarrow \frac{10}{\mathrm{x}}+\frac{60}{\mathrm{y}}=1---\left(2\right)$

Solving (1) and (2), we get,

$\mathrm{x}=100$ and $\mathrm{y}=\frac{200}{3}$

Thus, B can do the work in $\frac{200}{3}=66.66$ days.

- (a)$40$ days
- (b)$45$ days
- (c)$48$ days
- (d)None of these

**Hint**

Calculate $1$ day work $=\frac{1}{\mathrm{Number}\mathrm{of}\mathrm{days}\mathrm{to}\mathrm{complete}\mathrm{the}\mathrm{work}}$.

**Solution:**

Acccording to the question,

Days took by Harshita and Peter to do a piece of work $=16$days

Harshitha can do the same work in $24$ days.

We know that $1$day work $=$ $\frac{1}{\mathrm{Number}\mathrm{of}\mathrm{days}\mathrm{to}\mathrm{complete}\mathrm{the}\mathrm{work}}$

(Harshita$+$Peter)'s $1$ day's work $=\frac{1}{16}$

Harshita's $1$ day's work $=\frac{1}{24}$

Peter's $1$day's work $=\frac{1}{16}-\frac{1}{24}=\frac{1}{48}$

Therefore, Peter can do the entire work in $48$ days.

- (a)$5$ days
- (b)$6$ days
- (c) $11$ days
- (d) $8$ days

**Hint**

●Efficiency $=\frac{\mathrm{total}\mathrm{work}}{\mathrm{no}\mathrm{of}\mathrm{days}\mathrm{does}\mathrm{work}}$.

**Solution:**

According to the question,

Days Chand need to complete a work $=10$days

Days Eshan need to complete a work $=20$days

Total work $=\mathrm{L}.\mathrm{C}.\mathrm{M}.(10,20)=20$ units in $1$ day,

Chand's work $=\frac{20}{10}=2$ units In $1$ day,

Eshan's work $=\frac{20}{20}=1$ unit in $1$ day,

As they work together for $5$ days, work done $=3\times 5=15$ units.

The remaining $5$ units can be done by Eshan in $\frac{5}{1}=5$ days.

Thus, Eshan takes $5$ days to complete the remaining work.

- (a)$4\mathrm{h}$
- (b)$4\frac{7}{13}\mathrm{h}$
- (c)$4\frac{8}{13}\mathrm{h}$
- (d)None of these

**Hint**

Take $1$ hour part to be filled by them alone. Add them to get $1$ hour part to be filled together.

**Solution:**

According to the question,

Time taken by pipes A to fill a tank $=10$hours

Time taken by pipes B to fill a tank $=15$ hours

Time taken by pipes C to fill a tank $=20$hours

Part filled by A in $1$ hour $=\frac{1}{10}$

Part filled by B in $1$ hour $=\frac{1}{15}$

Part filled by C in $1$ hour $=\frac{1}{20}$

Part filled by (A + B + C) in $1$ hour $\frac{1}{10}+\frac{1}{15}+\frac{1}{20}=\frac{13}{60}$

(A + B + C) can fill the tank in $\frac{60}{13}$ hours, i.e.,

$4\frac{8}{13}\mathrm{h}$.

Tank gets filled in $4\frac{8}{13}\mathrm{h}.$

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- (a)$10\mathrm{s}$
- (b)$30\mathrm{s}$
- (c)$40\mathrm{s}$
- (d)$60\mathrm{s}$

**Hint**

Time taken by them to meet each other $=\frac{\mathrm{Distance}}{\mathrm{Relative}\mathrm{Speed}}.$

**Solution:**

Relative speed of Arun and Varun $=10+12$$=22\mathrm{m}/\mathrm{s}$.

Time taken by them to meet each other

$=\frac{\mathrm{Distance}}{\mathrm{Speed}}=\frac{880}{22}=40\mathrm{s}$.

Therefore, they took$40\mathrm{s}$ to meet each other. Hence, option C is correct.

- (a)$10.5\mathrm{s}$
- (b)$8\mathrm{s}$
- (c)$20\mathrm{s}$
- (d)$15\mathrm{s}$

**Hint**

Speed of Kittu is distance covered by Kittu in meters divided by time taken by him in seconds.

**Solution:**

Let the speeds of Kittu, Bittu and Sanu be${\mathrm{S}}_{\mathrm{k}},{\mathrm{S}}_{\mathrm{b}},$ and ${\mathrm{S}}_{\mathrm{s}}$respectively.

Then according the data given,

$\frac{{\mathrm{S}}_{\mathrm{k}}}{{\mathrm{S}}_{\mathrm{b}}}=\frac{3}{5}$

$\frac{{\mathrm{S}}_{\mathrm{b}}}{{\mathrm{S}}_{\mathrm{s}}}=\frac{120}{90}=\frac{4}{3}$

$\frac{{\mathrm{S}}_{\mathrm{k}}}{{\mathrm{S}}_{\mathrm{b}}}\times \frac{{\mathrm{S}}_{\mathrm{b}}}{{\mathrm{S}}_{\mathrm{s}}}=\frac{3}{5}\times \frac{4}{3}$

$\frac{{\mathrm{S}}_{\mathrm{k}}}{{\mathrm{S}}_{\mathrm{s}}}=\frac{4}{5}$

Speed of Kittu $=\frac{1000}{10}=100\mathrm{m}/\mathrm{s}$

${\mathrm{S}}_{\mathrm{s}}=\frac{5}{4}\times {\mathrm{S}}_{\mathrm{k}}=125\mathrm{m}/\mathrm{s}$

Thus, time taken by Sanu to cover

$1\mathrm{km}=\frac{\mathrm{Distance}}{\mathrm{Time}}=\frac{1000}{125}=8\mathrm{s}$

Hence, the correct answer is $8\mathrm{s}$.

- (a)$10\mathrm{km}$
- (b)$5.2\mathrm{km}$
- (c)$9.2\mathrm{km}$
- (d)$5.8\mathrm{km}$

**Hint**

Distance of school from his home is = $\mathrm{speed}\times \mathrm{distance}$.

**Solution:**

Let the ratio of speed be $6:4=3:2$.

Ratio of time will be $2:3$.

Difference between time ratio $=1$ part.

Total time$=10+16=26$.

$\Rightarrow 2$ parts $=$ $26\times 2$$=$ $52\mathrm{min}$

When he walks at $6\mathrm{kmph}$, he takes $52\mathrm{min}$ to reach school.

Thus, distance of school from his home is $=\mathrm{Speed}\times \mathrm{Distance}=6\times \frac{52}{60}=5.2\mathrm{km}.$

Distance between Sunil's school and home is $5.2\mathrm{km}$.

Hence, option B is correct.

- (a)$45\mathrm{kmph}$
- (b)$48\mathrm{kmph}$
- (c)$50\mathrm{kmph}$
- (d)None of these

**Hint**

Square has four equal sides, resulting into equal distance,

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

**Solution:**

Let the side of the car speed be $\mathrm{x}\mathrm{km}.$

Let the average speed of the car be $y$ $\mathrm{kmph}$.

Then, total time taken by car $=\frac{\mathrm{Distance}}{\mathrm{speed}}$= $\frac{4x}{y}$.

Time taken individually to cover each side is

$\frac{x}{25},\frac{x}{50},\frac{x}{75},\frac{x}{100}$.

In both case time should be same:

$\frac{x}{25}+\frac{x}{50}+\frac{x}{75}+\frac{x}{100}$=$\frac{4x}{y}$,

$\frac{25x}{300}=\frac{4x}{y}$⇒ $y$ $=48\mathrm{kmph}$.

Therefore, average speed is $48\mathrm{kmph}.$

- (a)$55\mathrm{km}$
- (b)$56\mathrm{km}$
- (c)$65\mathrm{km}$
- (d)$66\mathrm{km}$

**Hint**

$Averagespeed$$=\frac{2\mathrm{xy}}{\mathrm{x}+\mathrm{y}}$.

**Solution:**

Bittu's speed from his home to office,

$\mathrm{x}=40\mathrm{kmph}$.

Bittu's speed from office to home,

$\mathrm{y}=60\mathrm{kmph}$.

Average speed $=\frac{2\mathrm{xy}}{\mathrm{x}+\mathrm{y}}$$=\frac{2\times 40\times 60}{40+60}$$=48\mathrm{kmph}$ .

Distance travelled in $2$ hours $20$ minutes. i.e, $\left(\frac{7}{3}hrs\right)$

$Dis\mathrm{tan}ce=\mathrm{Speed}\times \mathrm{Time}\phantom{\rule{0ex}{0ex}}=48\times \frac{7}{3}\phantom{\rule{0ex}{0ex}}=112\mathrm{km}.$

Therefore, the distance from the office to home is: $\frac{112}{2}=56\mathrm{km}$.

Hence, the correct answer is $56km$.

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- (a)$8,40$
- (b)$9,45$
- (c)$10,50$
- (d)$11,55$

**Hint**

Seven years ago, they both will be seven years younger than their present ages.

**Solution:**

We assume that;

The present age of Ron $=\mathrm{x}$

Then, the present age of Ron’s father $=5\mathrm{x}$

According to the question, we can write as :

$\left(\mathrm{x}-7\right)+(5\mathrm{x}-7)=40$

$\mathrm{x}+5\mathrm{x}-14=40$

$\mathrm{x}+5\mathrm{x}=40+14$

$6\mathrm{x}=54$

$\mathrm{x}=9$

Hence, the present age of Ron’s father

$=5\mathrm{x}=45$

Therefore, option B is correct.

- (a)$50$
- (b)$40$
- (c)$45$
- (d)$55$

**Hint**

Generate two equations as per the condition given and solve both to find two unknowns.

**Solution:**

We assume that;

The no. of tickets sold at $\mathrm{Rs}.60$$=\mathrm{x}$

The no. of tickets sold at $\mathrm{Rs}.80$$=\mathrm{y}$

According to the question,

$\mathrm{x}+\mathrm{y}=100\dots \dots ..1$

$60\mathrm{x}+80\mathrm{y}=7000\dots \dots ...2$

Now, we multiply the eq. $\left(1\right)$ by $6$ and subtract the eq. $\left(2\right)$ from eq. $\left(1\right)$

$-2\mathrm{y}=-100$

$\mathrm{y}=50$

Putting in eq. $\left(1\right)$, we get-

$\mathrm{x}=50$

Thus, the number of tickets sold at

$\mathrm{Rs}.60=50$

Therefore, option A is correct.

- (a)$320$
- (b)$420$
- (c)$520$
- (d)None of these

**Hint**

Generate an equation as per the condition given in the problem and solve to get the assumed unknown.

**Solution:**

We assume that :

Each student receives the number of toffees $=\mathrm{x}$

Then, Saumya has total number of toffees

$=20\mathrm{x}$

According to the question,

$\frac{(20\mathrm{x}+5)}{(20+5)}=\mathrm{x}-4$

Further, we can solve as :

$20\mathrm{x}+5=25\mathrm{x}-100$

$105=5\mathrm{x}$

$\mathrm{x}=21$

Therefore, total number of toffees

$=20\mathrm{x}\phantom{\rule{0ex}{0ex}}=20\times 21\phantom{\rule{0ex}{0ex}}=420$

Hence, the correct answer is $420$.

- (a)$15$
- (b)$16$
- (c)$14$
- (d)None of these

**Hint**

Frame the two algebraic equations based on the condition given and solve these to get the two unknowns.

**Solution:**

We assume that the numbers are $\mathrm{x}\mathrm{y}$.

According to the question;

$\mathrm{x}+\mathrm{y}=20\dots \dots ..1$

${\mathrm{x}}^{2}-{\mathrm{y}}^{2}=240\dots \dots \dots .2$

From equation $2,$

$(\mathrm{x}+\mathrm{y})(\mathrm{x}-\mathrm{y})=240$

Put the value from eq. $1,$

$20(\mathrm{x}-\mathrm{y})=240$

$\mathrm{x}-\mathrm{y}=12\dots \dots 3$

Now, from $1$ and $3,$

$2\mathrm{x}=32$

$\mathrm{x}=16$

$\mathrm{y}=4$

Therefore, the bigger number is $16.$

Hence, the correct answer is $16$.

- (a)$\u20b9500$
- (b)$\u20b9166$
- (c)$\u20b9246$
- (d)$\u20b9512$

**Hint**

Sum of all assumed amount will be equal to the total amount that Pawan has.

**Solution:**

We assume that;

Shyam received the amount $=\mathrm{x}$

Ram received the amount $=3\mathrm{x}$

Krish received the amount $=6\mathrm{x}+20$

According to the question, we can write as :

$\mathrm{x}+3\mathrm{x}+6\mathrm{x}+20=840$

$10\mathrm{x}=820$

$\mathrm{x}=82$

Thus, Krish received the amount

$=(82\times 6)+20\phantom{\rule{0ex}{0ex}}=\u20b9512$

Therefore, $\u20b9512$ is correct answer.

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- (a)$64$
- (b)$81$
- (c)$\frac{1}{64}$
- (d)$-64$

**Hint**

If ${\mathrm{log}}_{b}a=m$ then $a={\mathrm{b}}^{\mathrm{m}}$.

**Solution:**

We know that:

If ${\mathrm{log}}_{b}a=m$ then $a={b}^{m}$.

According to the question,

${\mathrm{log}}_{4}\mathrm{x}=-3$

$\Rightarrow \mathrm{x}={4}^{-3}=\frac{1}{64}$

Hence, the correct answer is $\frac{1}{64}$.

- (a)$2$
- (b)$\frac{1}{2}$
- (c)-$2$
- (d)Cannot be determined

**Hint**

$\mathrm{loga}{}^{\mathrm{a}}=1$.

**Solution:**

According to the question,

${\mathrm{log}}_{4}{\mathrm{log}}_{25}625={\mathrm{log}}_{4}{\mathrm{log}}_{25}{\left(25\right)}^{2}$

Now,

${\mathrm{log}}_{25}{\left(25\right)}^{2}=2{\mathrm{log}}_{25}25=2\times 1=2$

$\Rightarrow {\mathrm{log}}_{4}{\mathrm{log}}_{25}{\left(25\right)}^{2}={\mathrm{log}}_{4}2$

${\mathrm{log}}_{4}2=2{\mathrm{log}}_{{2}^{2}}2=\frac{1}{2}{\mathrm{log}}_{2}2=\frac{1}{2}$

- (a)$2$
- (b)$-\frac{8}{3}$
- (c)$\frac{8}{3}$
- (d)$4$

**Hint**

$\sqrt{\mathrm{a}}\times \sqrt{\mathrm{a}}=\mathrm{a}$ and $\sqrt{\mathrm{a}}\times \sqrt{\mathrm{a}}\times \sqrt{\mathrm{a}}=\mathrm{a}\sqrt{\mathrm{a}}.$

**Solution:**

According to the question,

${\mathrm{log}}_{2\sqrt{2}}16={\mathrm{log}}_{{\left(\sqrt{2}\right)}^{3}}{\left(\sqrt{2}\right)}^{8}$

{since $16={2}^{4}={\left(\sqrt{2}\right)}^{8}$ and $2\sqrt{2}={\left(\sqrt{2}\right)}^{3}$

Therefore,

$=\frac{8}{3}{\mathrm{log}}_{\sqrt{2}}\sqrt{2}$

$=\frac{8}{3}$

Hence, the correct answer is $\frac{8}{3}.$

- (a)${\mathrm{log}}_{120}60$
- (b)$\mathrm{log}1$
- (c)$1$
- (d)$0$

**Hint**

It is related to, $\mathrm{loga}\mathrm{xy}=\mathrm{loga}\mathrm{x}+\mathrm{loga}\mathrm{y}.$

**Solution:**

According to the question,

${\mathrm{log}}_{120}2+{\mathrm{log}}_{120}3+{\mathrm{log}}_{120}4+{\mathrm{log}}_{120}5$

We know that, (

$\mathrm{log}\mathrm{ab}=\mathrm{log}\mathrm{a}+\mathrm{log}\mathrm{b}$)

$\Rightarrow {\mathrm{log}}_{120}\left(2\times 3\times 4\times 5\right).$

$\Rightarrow {\mathrm{log}}_{120}120=1$

$\left(\mathrm{as}{\mathrm{log}}_{\mathrm{a}}\mathrm{a}=1\right)$

- (a)$1.699$
- (b)$0.699$
- (c)$1.3010$
- (d)Cannot be determined

**Hint**

${\mathrm{log}}_{\mathrm{a}}\frac{\mathrm{x}}{\mathrm{y}}={\mathrm{log}}_{\mathrm{a}}\mathrm{x}-{\mathrm{log}}_{\mathrm{a}}\mathrm{y}.$

**Solution:**

According to the question,

${\mathrm{log}}_{10}50={\mathrm{log}}_{10}\left(\frac{100}{2}\right)$

$={\mathrm{log}}_{10}100-{\mathrm{log}}_{10}2$

$={\mathrm{log}}_{10}{10}^{2}-{\mathrm{log}}_{10}2$

$=2{\mathrm{log}}_{10}10-{\mathrm{log}}_{10}2$

$=2\times 1-0.3010$

$=1.699.$

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- (a)$180$
- (b)$210$
- (c)$190$
- (d)$200$

**Hint**

Since six points are collinear, therefore, ${}^{6}C_{3}$ combinations are not possible as these would lie on collinear points.

**Solution:**

Three points are required to form a triangle. So, ${}^{12}C_{3}$ triangles can be formed from $12$ points.

Since six points are collinear, therefore, ${}^{6}C_{3}$ combinations are not possible as these would lie on collinear points.

So, total possible triangles

$={}^{12}C_{3}-{}^{6}C_{3}=220-20=200.$

- (a)$144$
- (b)$288$
- (c)$121$
- (d)None of these

**Hint**

Take vowels as one set. So total number of sets become $4$.

**Solution:**

Considering all vowels, i.e., A, E, U as one, the count of the remaining letters $=3$.

So, these four letters, i.e., $(3+1)$ letters can be arranged in $4!$ ways.

And, the three vowels can be arranged in $3!$ ways among themselves.

Hence, total ways $=4!\times 3!=24\times 6=144$.

Hence, the correct answer is $144$.

- (a)$2160$
- (b)$7120$
- (c)${}^{8}C_{2}\times {}^{6}C_{4}$
- (d)None of these
- (e)$4380$

**Hint**

$2$ women can select $2$ chairs from $4$ chairs and sit on them in ${}^{4}C_{2}\times 2!={}^{4}P_{2}$ ways.

**Solution:**

Two women can be arranged in four seats (numbered $1,2,3,4$) in ${}^{4}P_{2}$ ways, and four men can be arranged in the remaining six seats in ${}^{6}P_{4}$ ways.

Total number of possible arrangements $={}^{4}P_{2}\times {}^{6}P_{4}$

$=\frac{4!}{(4-2)!}\times \frac{{\displaystyle 6!}}{{\displaystyle (6-4)!}}$

$=\frac{4!}{2!}\times \frac{{\displaystyle 6!}}{{\displaystyle 2!}}$

$=\frac{4\times 3}{1}\times \frac{{\displaystyle 6\times 5\times 4\times 3}}{{\displaystyle 1}}$

$=4\times 3\times 6\times 5\times 4\times 3=4320$.

Thus, the total number of arrangements is $4320$.

- (a)${2}^{7}$
- (b)${}^{6}C_{2}$
- (c)${2}^{6}$
- (d)None

**Hint**

Each coin can go in either of the two pockets.

**Solution:**

Every coin has two possibilities, i.e., it can go to any of the two pockets.

So, total number of ways $=$ $2\times 2\times 2\times 2\times 2\times 2={2}^{6}$.

Hence, option C is correct.

- (a)$250$
- (b)$600$
- (c)$900$
- (d)None of these

**Hint**

We can select seven question from two groups in the following way:

${}^{6}C_{4}={}^{6}C_{2}$

**Solution:**

He can select seven questions from two groups in the following ways: $(4,3),(3,4)$.

Number of ways$=$ ${}^{6}C_{4}\times {}^{6}C_{3}$ + ${}^{6}C_{4}\times {}^{6}C_{3}$ $=15\times 20+20\times 15$$=300+300=600.$

Hence, option B is correct.

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- (a)$\frac{1}{216}$
- (b)$\frac{91}{216}$
- (c)$\frac{91}{36}$
- (d)None of these

**Hint**

To find the probability of at least one of something, calculate the probability of none and then subtract that result from $1$.

**Solution:**

In the simultaneous throw of three dice, the probability that the resultant will have at least one result as $6$

$=1-$Probability that no die shows $6$

$=1-\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}$

$=1-\frac{125}{216}$

$=\frac{91}{216}$

Therefore, the correct answer is option 'B'.

- (a)$0.35$
- (b)$0.30$
- (c)$0.41$
- (d)None of these

**Hint**

First, we calculate the value of $\mathrm{P}\left(\mathrm{X}\mathrm{or}\mathrm{Y}\right)=\mathrm{P}\left(\mathrm{X}\right)+\mathrm{P}\left(\mathrm{Y}\right)-\mathrm{P}\left(\mathrm{X}\mathrm{and}\mathrm{Y}\right)$, then calculate the value of $\mathrm{P}\left(\mathrm{none}\right)$.

**Solution:**

Let the two events be $\mathrm{X}$ and $\mathrm{Y}$.

Now, given: $\mathrm{P}\left(\mathrm{X}\right)=0.35$

$\mathrm{P}\left(\mathrm{Y}\right)=0.60$

$\mathrm{P}\left(\mathrm{X}\mathrm{and}\mathrm{Y}\right)=0.25$

We know that,

$\mathrm{P}\left(\mathrm{X}\mathrm{or}\mathrm{Y}\right)=\mathrm{P}\left(\mathrm{X}\right)+\mathrm{P}\left(\mathrm{Y}\right)-\mathrm{P}\left(\mathrm{X}\mathrm{and}\mathrm{Y}\right)$

Substituting the given values we get:

$\mathrm{P}\left(\mathrm{X}\mathrm{or}\mathrm{Y}\right)=0.35+0.60-0.25$

$\Rightarrow $ $0.70$

$\mathrm{P}\left(\mathrm{none}\right)=1-\mathrm{P}\left(\mathrm{X}\mathrm{or}\mathrm{Y}\right)$

$\Rightarrow 1-0.70=0.30$

Therefore, the correct answer is option 'B'.

- (a)$\frac{1}{16}$
- (b)$\frac{6}{16}$
- (c)$\frac{4}{16}$
- (d)None of these

**Hint**

No. of total outcomes on tossing $n$ coins simultaneously $={2}^{n}$

**Solution:**

Total outcomes on tossing four coins $={2}^{4}$

As four coins are tossed, and we need exactly two tails,

$\Rightarrow $Number of ways to get exactly two tails out of four $=$${}^{4}C_{2}$

Therefore, required probability $=\frac{{}^{4}{C}_{2}}{{2}^{4}}=\frac{6}{16}$

Therefore, the correct answer is option 'B'.

- (a)$\frac{20}{{2}^{20}}$
- (b)$\frac{19}{{2}^{20}}$
- (c)$\frac{20}{{2}^{17}}$
- (d)$\frac{5}{{2}^{17}}$

**Hint**

Tossing a coin can give $2$ outcomes.

So, tossing a coin $20$ times can give $\left({2}^{20}\right)$ outcomes.

**Solution:**

In a toss of $20$ coins one after the other, the number of ways of getting exactly two heads on consecutive throws $=$ (head on the first coin, second coin, and tail on all the remaining $18$ coins) or (head on the second coin, third coin, and tail on all the remaining $18$ coins) or (head on nineteenth coin, twentieth coin, and tail on all the remaining $18$ coins) $=19$ ways .

Total outcomes $={2}^{20}$

Thus, required probability $=\frac{19}{{2}^{20}}$

Therefore, the correct answer is option 'B'.

- (a)$\frac{1}{3}$
- (b)$\frac{19}{20}$
- (c)$\frac{1}{20}$
- (d)$\frac{9}{20}$
- (e)None of these

**Hint**

Calculate compliment of $P,Q$ and $R$.

**Solution:**

Given, probability of hitting the target:

Probability of $P$ hitting the target,

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

Probability of $Q$ hitting the target, $P\left(Q\right)$ $=\frac{1}{3}$

Probability of $R$ hitting the target, $P\left(R\right)$$=\frac{2}{5}$

Therefore, the probability that $P$ does not hit the target, $P(P\text{'})=1-$$P\left(P\right)$ $=\frac{3}{4}$

Probability that $Q$ does not hit the target, $P(Q\text{'})$ $=1-$$P\left(Q\right)=\frac{2}{3}$

Probability that $R$ does not hit the target, $P(R\text{'})=1-$$P\left(R\right)=\frac{3}{5}$

Therefore, the probability of exactly one of them hitting the target

$=P\left(P\right)\times P(Q\text{'})\times P(R\text{'})+P\left(Q\right)\times P(P\text{'})\times P(R\text{'})+P\left(R\right)\times P(P\text{'})\times P(Q\text{'})$

$=\frac{1}{4}\times \frac{2}{3}\times \frac{3}{5}+\frac{1}{3}\times \frac{3}{4}\times \frac{3}{5}+\frac{2}{5}\times \frac{3}{4}\times \frac{2}{3}$

$=\frac{9}{20}$

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- (a)$432$
- (b)$308$
- (c)$478$
- (d)None of these

**Hint**

To find the area of a rectangle, multiply the length by the breadth.

**Solution:**

We know that,

Area of the rectangular room

$=$Lenght $\times $ Spred

Area of the rectangular room

$=24\times 18=432\mathrm{sq}.\mathrm{m}.$

Therefore, $432\mathrm{sq}.\mathrm{m}.$ is the correct answer.

- (a)$24\text{\hspace{0.17em}}\mathrm{m}$
- (b)$30\mathrm{m}$
- (c)$32\mathrm{m}$
- (d)None of these
- (e)$35\mathrm{m}$

**Hint**

Length of the diagonal is find out by the help of pythagoras theorem that is $\mathrm{Diagonal}=\sqrt{{\mathrm{Length}}^{2}+{\mathrm{Breadth}}^{2}}$

**Solution:**

We know that the maximum possible length of a rod that can be put on the floor of the room is equal to the length of diagonal.

Length of diagonal $=\sqrt{{\mathrm{length}}^{2}+{\mathrm{breadth}}^{2}}$

$=\sqrt{{24}^{2}+{18}^{2}}=\sqrt{576+324}=\sqrt{900}=30\mathrm{m}$

Therefore, option B is correct.

- (a)$86\mathrm{cm}$
- (b)$94.8\mathrm{cm}$
- (c)$31.6\mathrm{cm}$
- (d)None of these
- (e)$80\mathrm{cm}$

**Hint**

Find the diagonal using pythagoras theorem, in terms of assumed quantity.

**Solution:**

Let the breadth of rectangle be $\mathrm{x}\mathrm{cm}$

According to given question,

Length of rectangle $=3\mathrm{x}\mathrm{cm}$

Length of diagonal$=\sqrt{{\mathrm{breadth}}^{2}+{\mathrm{length}}^{2}}$

$100=\sqrt{{\mathrm{x}}^{2}+{\left(3\mathrm{x}\right)}^{2}}=\sqrt{10{\mathrm{x}}^{2}}=\sqrt{10}\mathrm{x}$

$\Rightarrow \mathrm{x}=\frac{100}{\sqrt{10}}\times \frac{\sqrt{10}}{\sqrt{10}}$

$\mathrm{x}=10\sqrt{10}$

$\mathrm{x}=31.6$

Hence, length of rectangle

$=3\times 31.6=94.8\mathrm{cm}$

Therefore, option B is correct.

- (a)$15\mathrm{m}$
- (b)$20\mathrm{m}$
- (c)$9\mathrm{m}$
- (d)$6\mathrm{m}$
- (e)None of these

**Hint**

Get the height using formula of calculating area and with the relation given get length also.

**Solution:**

Let the height of wall be $\mathrm{x}\mathrm{m}$

According to given question

Length of wall$=\frac{2}{3}\mathrm{x}\mathrm{m}$

Area$=\mathrm{length}\times \mathrm{height}$

$54=\frac{2}{3}\mathrm{x}\times \mathrm{x}$

$\Rightarrow 2{\mathrm{x}}^{2}=162$

${\mathrm{x}}^{2}=81$

$\mathrm{x}=9$

Length of wall$=\frac{2}{3}\times 9=6\mathrm{m}$

Therefore, sum of length and height$=6+9=15\mathrm{m}$

Therefore, option A is correct.

- (a)$4000$
- (b)$4440$
- (c)$5400$
- (d)$5440$

**Hint**

Required no of tiles $=\frac{\mathrm{Area}\mathrm{of}\mathrm{hall}}{\mathrm{Area}\mathrm{of}\mathrm{one}\mathrm{tile}}$

**Solution:**

We know that, length of hall $=12\mathrm{m}=1200\mathrm{cm}$

Breadth $=10\mathrm{m}=1000\mathrm{cm}$

So, required number of tiles$=\frac{\mathrm{Area}\mathrm{of}\mathrm{hall}}{\mathrm{Area}\mathrm{of}\mathrm{one}\mathrm{tile}}$$=\frac{1200\times 1000}{15\times 20}=4000$

Therefore, option A is correct.

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- (a)$10$
- (b)$12$
- (c)$30$
- (d)$15$

**Hint**

The angle at any time is given as, $\theta =\left|\frac{11}{2}m-30h\right|$

**Solution:**

Given:

Time $=4:20$

According to question,

We know that,

$\theta =\left|\frac{11}{2}m-30h\right|$

Here, $m=20$ and $h=4$

$\theta =\left|\frac{11}{2}\left(20\right)-30\left(4\right)\right|$

$\theta ={10}^{\mathrm{o}}$

The angle between the two hands of the clock at $4:20$ is ${10}^{\mathrm{o}}$

- (a)$120$
- (b)$130$
- (c)$125$
- (d)$132$

**Hint**

Every minute, the hour hand moves half a degree (there are $720$ minutes in half a day).

The difference between is $\left(30h+\frac{m}{2}\right)-6m=30h-\left(\frac{11}{2}\right)m$

**Solution:**

Given:

Time $=3:40$

According to question,

$\theta =\left|\frac{11}{2}m-30h\right|$

Here, $m=40$ and $h=3$

$\theta =\left|\frac{11}{2}\left(40\right)-30\left(3\right)\right|$

$\theta ={130}^{\mathrm{o}}$.

The angle between the two hands of the clock at $3:40$ is ${130}^{\mathrm{o}}$.

- (a)$327$
- (b)$323$
- (c)$317$
- (d)$332$

**Hint**

Every minute, the hour hand moves half a degree (there are $720$ minutes in half a day).

The difference between them is $\left(30h+\frac{m}{2}\right)-6m=30h-\left(\frac{11}{2}\right)m$.

**Solution:**

Given:

Time $=7:46$

According to question,

We know that angle between hands of a clock can be calculated as,

$\theta =\left|\frac{11}{2}m-30h\right|$

Here, $m=46$ and $h=7$

$\theta =\left|\frac{11}{2}\left(46\right)-30\left(7\right)\right|$

$\theta ={317}^{\mathrm{o}}$

The angle between the two hands of a clock at $7:46$. is ${317}^{\mathrm{o}}$.

- (a)$6:49\frac{1}{11}$
- (b)$6:16\frac{4}{11}$
- (c)Both (a) and (b)
- (d)None of these

**Hint**

$T=\frac{2}{11}\times (30H\pm \theta )$

**Solution:**

We know that,

$T=\frac{2}{11}\times (30H\pm \theta )$

Here, $H=6$ and $\theta ={90}^{\mathrm{o}}$

Plugging the values, we get

$T=\frac{2}{11}\times \left(30\right(6)\pm 90)$

$T=\frac{540}{11}\mathrm{min}$ or $T=\frac{180}{11}\mathrm{min}$

$T=49\frac{1}{11}\mathrm{min}$ or $T=16\frac{4}{11}\mathrm{min}$

Therefore, two hands are perpendicular at

$6:49\frac{1}{11}$ and $6:16\frac{4}{11}$ .

- (a)$8:11\frac{10}{11}$
- (b)$8:10\frac{10}{11}$
- (c)$8:10\frac{1}{11}$
- (d)None of these

**Hint**

$\theta =$ angle (degree of two hands of a clock) $=180$

**Solution:**

We know that,

$\mathrm{T}=\frac{2}{11}\times \left(30\mathrm{H}\pm \theta \right)$

Here, $\mathrm{H}=8$ and $\theta ={180}^{\mathrm{o}}$

Plugging the values, we get:

$\mathrm{T}=\frac{2}{11}\times \left(30\left(8\right)\pm 180\right)$

$\mathrm{T}=\frac{120}{11}\mathrm{min}$ or $\mathrm{T}=\frac{840}{11}\mathrm{min}$

$\mathrm{T}=10\frac{10}{11}\mathrm{min}$

Therefore, the time when the two hands of a clock are opposite to each other will be $8:10\frac{10}{11}$.

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- (a)E
- (b)C
- (c)S
- (d)G
- (e)None of these

**Hint**

The series contain the vowels written in reverse.

**Solution:**

The series contains the vowels A, E, I, O, U in alphabetical order but written in reverse.

Therefore the missing term is E.

- (a)N, J
- (b)M, L
- (c)J, R
- (d)O, M
- (e)None of these

**Hint**

On subtracting two from the current letter, you get the next term.

**Solution:**

In the given series, we can observe that the following pattern is observed:

Y -2 = W

W -2 = U

U- 2 = S

S- 2 = Q

Q -2 = O

O -2 = M

Hence, the missing terms are O and M.

- (a)$M$
- (b)$L$
- (c)$K$
- (d)$H$
- (e)None of these

**Hint**

The series follows a definite pattern, by adding $+1,+2,+3,+4$ and so on to the current letter we get the next letter.

**Solution:**

After observing we find that the following pattern is followed.

$A\stackrel{+1}{\to}B\stackrel{+2}{\to}D\stackrel{+3}{\to}G\stackrel{+4}{\to}K$

Therefore the missing term is $K$.

- (a)$I$
- (b)$K$
- (c)$M$
- (d)$N$
- (e)None of these

**Hint**

The series follows a definite pattern, on subtracting$-5,-4,-3,-2$ from the current letter we get the next letter.

**Solution:**

After observing we find that on subtracting $-5,-4,-3,-2$ from the current letter we get the next letter.

$Z\stackrel{-5}{\to}U\stackrel{-4}{\to}Q\stackrel{-3}{\to}N\stackrel{-2}{\to}L$

Hence the missing term is $N$.

- (a)$L$
- (b)$K$
- (c)$J$
- (d)$I$
- (e)None of these

**Hint**

It's an incremental series, we add $+2,+3$ alternatively to get the successive terms.

**Solution:**

On observing the series, we find that the below pattern is followed:

$A\stackrel{+2}{\to}C\stackrel{+3}{\to}F\stackrel{+2}{\to}H\stackrel{+3}{\to}K\stackrel{+2}{\to}M$

Therefore the missing term is $K$.

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- (a)14
^{th}June - (b)15
^{th}June - (c)16
^{th}June - (d)None of these

**Hint**

According to Sukhinder, his father's birthday is on : 14^{th} or 15^{th}

According to his sister, their father's birthday is on : 15^{th} or 16^{th} or 17^{th}

**Solution:**

According to Sukhinder, his father's birthday is on either 14

^{th}or 15^{th}of June. While according to his sister, her father's birthday is on one of the days among 15, 16 and 17 June.Since 15

^{th}June is common to both of their prediction, we can say that their father’s birthday is on 15^{th}of June.

- (a)Monday
- (b)Tuesday
- (c)Wednesday
- (d)None of these

**Hint**

$5$^{th} January $1992$ will be on a Sunday because it has only $1$ odd day.

**Solution:**

The year $1991$is an ordinary year. Therefore it has only $1$ odd day. Thus $5$

^{th}January $1992$ will be a day after Saturday, i.e. Sunday.Now, in January $1992$, there are $26$days left, which results in $5$odd days. In February $1992$, there are $29$days which leads to $1$ odd day.

In March $1992$, there are $3$ odd days.

Total number of odd days after $5$

^{th}January $1992=5+\hspace{0.17em}1+3=9$days i.e. $2$ odd days.Hence, $3$

^{rd}March $1992$will be $2$ days after Sunday, i.e. Tuesday.

- (a)$31$
- (b)$30$
- (c)$29$
- (d)None of these

**Hint**

Class strength can be calculated by adding the number of students whose ranks are higher than Kamini, lower than Kamini and Kamini herself.

**Solution:**

$15$ students who have a rank higher than that of Kamini, $14$ students who have a rank lower than Kamini and Kamini herself.

Therefore, the total students in her class are

$=15+14+1=30$

- (a)Four
- (b)Five
- (c)Six
- (d)Three

**Hint**

Identify the $7$s that are preceded by $5$and then check if it is followed by$3$.

**Solution:**

The conditions mentioned in the question can be shown as below:

$37\mathbf{\left(}57\right)4573978\left(57\right)78971\left(57\right)6\left(57\right)4375738$

Thus, there are four such $7$.

- (a)$8:00$ hrs
- (b)$8:05$ hrs
- (c)$8:10$ hrs
- (d)$8:45$ hrs
- (e)None of these

**Hint**

Anuj reached $15$ minutes before $8:30$ hrs which means that he reached at $8:15$ hrs.

**Solution:**

According to the question, Anuj reached the place of meeting at $8:15$ hrs.

The man was $40$ minutes late to the meeting and Anuj was $30$ minutes earlier than him,

which means that he was $10$ minutes late to the meeting. Therefore the scheduled time of the meeting was $8:05$ hrs.

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- (a)$40\mathrm{m}$
- (b)$10\mathrm{m}$
- (c)$50\mathrm{m}$
- (d)$30\mathrm{m}$
- (e)None of these

**Hint**

The minimum distance, which is required to reach Aman's original point. Draw the diagram by using the direction diagram.

**Solution:**

He walks from $\mathrm{A}$ to $\mathrm{B}30\mathrm{m}$ towards the north.

$\mathrm{B}$ to $\mathrm{C}=40\mathrm{m}$

$\mathrm{C}$ to $\mathrm{D}=30\mathrm{m}$

Now, he is at point $\mathrm{D}$ in the west direction from there, he walks $50\mathrm{m}$ towards the east and reached at point $\mathrm{E}$.

Hence, $10\mathrm{m}$ is the correct answer.

- (a)$10km$
- (b)$25km$
- (c)$20km$
- (d)$15km$
- (e)None of these

**Hint**

The minimum distance, which is required to reach Ram's house. Draw the conditions with the help of a direction diagram.

**Solution:**

Starts from point A

Distance covers from A to B $=15km$

B to C $=20km$

C to D $=25km$

Finally, D to E (End point) $=20km$

At point E, he is facing the north direction.

Hence, he is $10km$ far from his house.

- (a)$12\mathrm{km}$
- (b)$10\mathrm{km}$
- (c)$8\mathrm{km}$
- (d)$16\mathrm{km}$
- (e)None of these

**Hint**

We'll use the concept of Pythagoras theorem to solve the questions on directions.

${\mathrm{c}}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}$

**Solution:**

Draw the diagram, according to the given conditions.

In the above diagram $\mathrm{BCDE}$ is a square

$\mathrm{BE}=2\mathrm{km}:$ $\mathrm{DE}=2\mathrm{km}$

$\mathrm{AE}=8\mathrm{km}:$ $\mathrm{FE}=6\mathrm{km}$

Hence $\mathrm{AF}$:

${8}^{2}+{6}^{2}={\mathrm{x}}^{2}$

$64+36={\mathrm{x}}^{2}$

$100={\mathrm{x}}^{2}$

$x=10$

Thus, Shyam's distance from the starting point

$=10\mathrm{km}$.

- (a)West
- (b)North
- (c)South
- (d)East
- (e)None of these

**Hint**

After taking the first right turn, she is facing the north direction.

**Solution:**

Kamu walks $5\mathrm{km}$ straight from her house towards the west, then turns right and walks $3\mathrm{km}$. Now she is facing the north direction. Thereafter, she takes a left turn and walks $2\mathrm{km}$. Further, she turns left and walks $3\mathrm{km}$. Now she is facing the south direction. Finally, she turns right and walks $3\mathrm{km}$.

It is clear from the diagram that Kamu is to the west of her house.

Hence, the required answer is the west direction.

- (a)$17\mathrm{km}$
- (b)$2\mathrm{km}$
- (c)$15\mathrm{km}$
- (d)$32\mathrm{km}$
- (e)None of these

**Hint**

After taking the second left turn, he is facing the west direction.

**Solution:**

Hari travelled $17\mathrm{km}$ to the east, turned left, now he is facing the north direction and went $15\mathrm{km}$ : he again turned left and went$17\mathrm{km}$ and facing the west direction.

The distance from his starting point to the final point is $15\mathrm{km}$.

Hence, $15\mathrm{km}$ is the answer.

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- (a)$\mathrm{L}*\mathrm{R}\u2022\mathrm{M}\beta \mathrm{K}\beta \mathrm{O}$
- (b)$\mathrm{R}*\mathrm{L}\Delta \mathrm{P}\beta \mathrm{K}\u2022\mathrm{O}$
- (c)$\mathrm{L}*\mathrm{M}\Delta \mathrm{R}*\beta \mathrm{O}$
- (d)$\mathrm{L}*\mathrm{R}\beta \mathrm{M}*\mathrm{K}\beta \mathrm{O}$

**Hint**

Check where M is the father of K.

**Solution:**

According to the question, we can clearly see that:

$\Delta \mathrm{and}*\to $mother - father relation (So, there is the generation gap of one.)

Similarly,

$\u2022\mathrm{and}\beta \to $brother - sister (so, there is no generation gap)

Now checking by the option, here we will replace ' $\Delta $ and $*$' by '$+1$' and '$\u2022$and $\beta $' by '$0$'.

In option (a), we have:

$\mathrm{L}(+1)\mathrm{R}\left(0\right)\mathrm{M}\left(0\right)\mathrm{K}\left(0\right)\mathrm{O}\Rightarrow +1+0+0+0=+1$

Similarly, in option (c), we have:

$\mathrm{L}(+1)\mathrm{M}(+1)\mathrm{R}(+l)\mathrm{K}\left(0\right)\mathrm{O}\Rightarrow +1+1+1+0=+3$

But, the generation gap between a grandfather and his grandchildren should be '$+2$'.

Therefore, both options are wrong.

Again, option (b) is also wrong because, here, L is the mother (a female) of P.So, by using the 'check the sex' rule, option (b) is also eliminated.

This leaves us with options (d) and (e).On trial, we find that option (d) is indeed the right answer as it says:

L is the father of R. R is the brother of M. M is the father of K. K is the brother of O.

This implies, L is the paternal grandfather of O.

- (a)Sister
- (b)Grandfather
- (c)Grandmother
- (d)Father
- (e)None of these

**Hint**

Son of A is father of C and D.

**Solution:**

The given equation can be decoded as:

A+B= A mother of B

B$\xf7$C = B is the father of C

C-D = C is the brother of D

It can be represented in the family tree as:

Here, A is the mother of B who is the father of C and D.

Therefore, A is D's grandmother.

- (a)Granddaughter
- (b)Great-grandson
- (c)Grandson
- (d)Grandmother
- (e)None of these

**Hint**

T is R's grandmother.

**Solution:**

The given information can be expressed through the family tree as:

From the family tree, it can be concluded that:

P is the great-grandson of T.

P and Q are siblings.

R is the mother of P and Q.

S is R's father and the grandfather of P and Q.

T is S's mother, R's grandmother and great-grandmother of P and Q.

Hence, P is the great-grandson of T.

- (a)Sister
- (b)Sister-in-law
- (c)Niece
- (d)Wife
- (e)None of these

**Hint**

A and D are brother, such that, A, B, and D are siblings.

**Solution:**

A - B's brother

C - D's father

E - B's mother

This can be expressed through the family tree as:

Since A and B are siblings, E is the mother of A also.

A and D are brothers. So A, B, D are siblings.

C is D's father. So, C - father of A, B and D.

E - mother of B and A so, D's as well.

Since, E - mother and C - father

Therefore, E is the wife of C.

- (a)Daughter-in-law
- (b)Daughter
- (c)Aunt
- (d)Nephew
- (e)None of these

**Hint**

A, B and the mother of C are siblings.

**Solution:**

It is given that:

B= brother of A.

C's mother = only sister of A.

This can be expressed through the family tree as:

Since, B= brother of A and C's mother =sister of A. It can be concluded that B is also brother of C's mother.

It means A, B and C's mother are siblings.

Now, D is the maternal grandmother of C. In other words D is mother of C's mother.

As concluded above- A, B and C's mother are siblings.

D is the mother of A, B and C's mother.

However, from the given statements A can be male or female. Hence, A can be daughter/son of D.

From the given options, 'Daughter' is the right option.

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- (a)ORMLAN
- (b)ONRMLA
- (c)ONMRLA
- (d)ONMRAL

**Hint**

Rearrange the letters of the word.

**Solution:**

Try to determine the relationship between the first pair of words. Both words are related to each other in some way.

The positions of the first and second letters of the word are interchanged.

ME TA PH OR = EM AT HP RO

NO RM AL = ON MR LA

Hence, the correct answer is option (C).

- (a)Palpate
- (b)Placate
- (c)Perjure
- (d)Placebo

**Hint**

First word is the synonym of the second.

**Solution:**

Try to determine the relationship between the first pair of words. Both words are related to each other in some way.

Chide is a synonym of scold which means to speak out in angry or displeased rebuke. Similarly, placate is a synonym of soothe which means make (someone) less angry or hostile.

Hence, the correct answer is placate.

- (a)$0.00025$
- (b)$0.0025$
- (c)$0.025$
- (d)$0.25$

**Hint**

In a pair, the second number is the square of the first number.

**Solution:**

According to the given analogy, the second number of the pair denotes the square of the first number.

${\left(0.01\right)}^{2}:0.0001$

Similarly,

${\left(0.05\right)}^{2}:0.0025$

Hence, the correct answer is $0.0025.$

- (a)Ear, Nose, Eyesight: Vision
- (b)Plus, Minus, Multiple: Division
- (c)Winter, Spring, Summer: Seasons
- (d)Humid, Hot, Tundra: Region

**Hint**

Here, the given words belong to the same category.

**Solution:**

Try to determine the relationship between the first pair of words. Both words are related to each other in some way.

Hinduism, Islam, and Christianity are different religions. Similarly, winter, spring, and summer are different seasons.

Hence, the correct answer is option 'C'.

- (a)BROW
- (b)DRAW
- (c)CRAW
- (d)SLAW

**Hint**

Rearrange the letters of the word.

**Solution:**

First, we have to found the relation between the WOLF: FLOW.

Writing the word WOLF from Backward, we get FLOW.

Applying the same relation in the same way

Writing the word WARD from Backward, we get DRAW.

So, the answer is DRAW.

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- (a)Withdrawal
- (b)Deduction
- (c)Deposit
- (d)Debit

**Hint**

Majority of the words refer to taking money out of an account.

**Solution:**

Given words:

Withdrawal, Deduction, Deposit, Debit.

All the words are showing a deduction except deposit which means submitting a amount.

Therefore, deposit does not belong to the group from which other option belong.

- (a)$62-37$
- (b)$74-40$
- (c)$85-60$
- (d)$103-78$

**Hint**

Check the difference between the numbers of the given pairs.

**Solution:**

According to the given options, the odd pair is $74-40$ as the difference between them is $34$ while the difference between the numbers of the other pair is $25$.

For example, $62-37=25$

Hence, the correct answer is $74-40$.

- (a)ABZY
- (b)BCYX
- (c)CDVW
- (d)DEVU

**Hint**

For all the other groups except one, the last two letters are consecutive(in the alphabet series) in the reverse order.

**Solution:**

For all the other groups except CDVW, the last two letters are consecutive (in the alphabet series) in the reverse order.

In option (C), the last two words 'VW' is in sequence.

So, the correct answer is option (C).

- (a)Pond-Lake
- (b)Pistol-Gun
- (c)Car-Bus
- (d)Church-Monument

**Hint**

Both the words of the pair are specific terms.

**Solution:**

The odd pair is Church-Monument as the relationship between them is that the second variable is a generic term while the first is a specific term (All Churches are monuments). For other pairs, both the words of the pair are specific terms.

- (a)Insurance
- (b)Provident Fund
- (c)Salary
- (d)Shares

**Hint**

All the words denote types of investments except one.

**Solution:**

Given words:

Insurance, Provident Fund, Salary, Shares.

The odd word is salary as all the other words denote types of investments while salary is the amount paid to an employee for his work.

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- (a)Valid
- (b)Invalid
- (c)False
- (d)Can not determine.

**Hint**

If the pairs are logically consistent for the statement, then the statement is valid.

**Solution:**

The condition is as follows:

Here P: Harry is Cute; Q: Harry is Smart.

Condition Valid Invalid False P OR Q NOT P$\to $Q

NOT Q$\to $P

P$\to $Q

P $\to $NOT Q

Q $\to $P

Q$\to $ NOT P

NOT P $\to $NOT Q

NOT Q$\to $NOT P

The logically consistent pairs for the statement, 'Harry is cute or smart' are

Harry is not cute $\to $ Harry is smart.

So, this statement is valid.

- (a)Valid
- (b)Invalid
- (c)False
- (d)Can not determine

**Hint**

If the pairs are logically consistent for the statement, then the statement is valid.

**Solution:**

The condition is as follows:

Here P: Ron is Not Cute; Q: Ron is Intelligent.

Condition Valid Invalid False P OR Q NOT P$\to $Q,

NOT Q$\to $P

P$\to $Q,

P $\to $NOT Q,

Q $\to $P,

Q$\to $NOT P

NOT P $\to $NOT Q,

NOT Q$\to $NOT P

The logically consistent pairs for the statement, 'Ron is not cute or intelligent' are

Ron is (not) not cute $\to $ Ron is intelligent.

So, this statement is valid.

- (a)Valid
- (b)Invalid
- (c)False
- (d)Can not determine

**Hint**

Hermoine is either smart or sweet.

**Solution:**

The condition are as follows:

Here P: Hermione is Sweet; Q: Hermione is Smart.

Condition Valid Invalid False P OR Q NOT P→Q,

NOT Q→P

P→Q,

P →NOT Q,

Q →P,

Q→NOT P

NOT P →NOT Q,

NOT Q→NOT P

The logically consistent pairs for the statement, 'Hermione is sweet or smart' are

She is smart $\to $ She is sweet. (P$\to $Q)

So, this statement is invalid.

- (a)Valid
- (b)Invalid
- (c)False
- (d)Can not determine

**Hint**

It is based on Not P$\to $ Q.

**Solution:**

The condition are as follows:

Here P: Ginny is redhead; Q: Ginny is Blonde.

Condition Valid Invalid False P OR Q NOT P→Q

NOT Q→P

P→Q

P →NOT Q

Q →P

Q→ NOT P

NOT P →NOT Q

NOT Q→NOT P

The logically consistent pairs for the statement, 'Ginny is a redhead or a blonde' are

She is not a redhead $\to $ She is a blonde.

(NOT P$\to $ Q)

So, this statement id valid.

- (a)Valid
- (b)Invalid
- (c)False
- (d)Can not determine

**Hint**

Hagrid does not have a merman if he has a dinosaur.

**Solution:**

The condition are as follows:

Here P: Hagrid has a dinosaur; Q: Hagrid has a merman.

Condition Valid Invalid False P OR Q NOT P→Q

NOT Q→P

P→Q

P →NOT Q

Q →P

Q→ NOT P

NOT P →NOT Q

NOT Q→NOT P

The logically consistent pairs for the statement, 'Hagrid has a dinosaur or he has a merman' are

Hagrid does not have a merman $\to $ he has a dinosaur.

(NOT Q $\to $ P)

So, this statement is valid.

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

**Hint**

There are some literates who may not be scientists.

**Solution:**

Given statements,

a. All scientists are fools.

b. All fools are literates.

From the given statements, the relationship between all of above are as follows,

Now,

Conclusion I. All literates are scientists.

From the figure, it is clear that

All literates are not scientists, only some literates are scientists.

Therefore,

Conclusion I is not true.

Conclusion II. All scientists are literates.

From the figure, it is clear that

All scientists are literates.

Therefore,

Conclusion II is true.

Hence, only conclusion II follows.

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

**Hint**

No clear relation between cat and pens.

**Solution:**

Given statements,

a. All pens are elephants.

b. Some elephants are cats.

By the given statements the relationship between all of above are as follow,

Now,

Conclusion I. Some pens are cats.

From the figure, it is clear that

Some pens are not cats as there is no direct relation between the two.

Therefore,

Conclusion I is not true.

Conclusion II. All pens are cats.

From the figure, it is clear that

All pens are not cats as there is no direct relation between the two.

Therefore,

Conclusion II is not true.

Hence, neither I nor II follows.

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

**Hint**

No apple is orange and all orange is banana, so all bananas are apple.

**Solution:**

Given statements,

a. No apple is an orange.

b. All bananas are oranges.

By the given statements the relationship between all of above are as follow,

Now,

Conclusion I. No apple is a banana.

From the figure, it is clear that

No apple is banana, as no apple is orange.

Therefore,

Conclusion I is true.

Conclusion II. Some oranges are bananas.

From the figure, it is clear that

Some oranges are bananas, as all bananas are oranges.

Therefore,

Conclusion II is true.

Hence, both conclusion I and II follow.

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

**Hint**

No direct relation between parrot and cuckoo is given.

**Solution:**

Given statements,

a. No ducks are parrots.

b. No ducks are cuckoos.

By the given statements the relationship between all of above are as follow,

Now,

Conclusion I. Some parrots are cuckoos.

In the given statement, the relation between parrots and cuckoos are not given.

Conclusion II. Some parrots are not cuckoos.

In the given statement, the relation between parrots and cuckoos are not given.

By observing both the conclusion, it is clear that "Some parrots may or may not be cuckoos".

Hence, either conclusion I or II follows.

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

**Hint**

No direct relation between girl and children is given.

**Solution:**

Given statements,

a. No boys are girls.

b. Some boys are children.

By the given statements the relationship between all of above are as follow,

Now,

Conclusion I. Some girls are not children.

From the figure, it is clear that

Some girls are not children, as no direct relation between the two is given.

Therefore,

Conclusion I is not true.

Conclusion II. No girls are children.

In the given statements, the relation between girls and children are not given.

Therefore,

Conclusion II is not true.

Hence, neither conclusion I nor II follows.

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- (a)If only assumption I is implicit
- (b)If only assumption II is implicit
- (c)If either I or II is implicit
- (d)If neither I nor II is implicit
- (e)If both I and II are implicit

**Hint**

Read both the statements carfully and see if one or both are implicit in the given statement..

**Solution:**

Clearly assumption I is implicit because the bank expects to work efficiently after the reduction of workforce as it would lead to easier management.

Assumption II is not implicit. It may or may not be an assumption.

- (a)If only assumption I is implicit
- (b)If only assumption II is implicit
- (c)If either I or II is implicit
- (d)If neither I nor II is implicit
- (e)If both I and II are implicit

**Hint**

Rs. 1,999 per month is very affordable by most people for a car.

**Solution:**

Since nothing has been mentioned about used cars in the given statement, we cannot assume anything about used cars. Hence, I is not implicit. However, it has been assumed that people can afford an amount of Rs. 1999 per month to buy a new car.

- (a)If only assumption I is implicit
- (b)If only assumption II is implicit
- (c)If either I or II is implicit
- (d)If neither I nor II is implicit
- (e)If both I and II are implicit

**Hint**

The scheme would encourage those owning an old fridge to go for a new one at a reasonable price whereas an advertisement highlights that which appeals to the masses.

**Solution:**

Clearly, the scheme would encourage those owning an old fridge to go for a new one at a reasonable price without the hassles of disposing of the old one. So, I is implicit. Besides, an advertisement highlights that which appeals to masses and which customers crave for. So, II is also implicit.

- (a)If only assumption I is implicit
- (b)If only assumption II is implicit
- (c)If either I or II is implicit
- (d)If neither I nor II is implicit
- (e)If both I and II are implicit

**Hint**

The school has planned to charge fees based on incomes of parents so the school will assume that parents will tell the truth about their incomes and that they will also agree to pay such fees.

**Solution:**

In reality, due to disparity in the fee structure of various educational institutions, merit is often looked down upon and weightage in given to the financial condition.

Today, degrees and certificates can be availed by money. The goverment's initiative of linking the school fee with the parents incomes will actually motivate parents to furnish authentic information regarding their incomes to the school authorities besides there will be no delay in the fee payments protocol, as it can be easily done by the parents.

- (a)If only assumption I is implicit
- (b)If only assumption II is implicit
- (c)If either I or II is implicit
- (d)If neither I nor II is implicit
- (e)If both I and II are implicit

**Hint**

People affected by cyclone can be rehabilitated in many other ways.

**Solution:**

Engaging the army for the rehabilitation work by the government implies that the government assumes that the army possesses the required ability to rehabilitate the affected people rapidly. Hence, II is implicit but I is not implicit because of the word 'only'. There may be other agencies available for the service.

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

**Hint**

Both irregular and regular students can fail.

**Solution:**

The given statement clearly states that all irregular and some regular students fail in the examinations.

This, in turn, approaches that each one of the students who is successful is regular.

However, now not all regular students are successful. So, neither I nor II follows.

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

**Hint**

Catchy slogans attracts people.

**Solution:**

The slogan given in the statement is genuinely a catchy one which suggests that catchy slogans do attract people. So, I does not follow. Nothing about people's choice for colours may be deduced from the statement. Thus, II additionally does not follow.

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

**Hint**

Most people continue to be controlled by oppressive governments by force, and not by choice.

**Solution:**

It is mentioned in the statement that the majority of the people in the world are compelled to be administrated by governments which refuse them personal liberty and the right to dissent.

The citizens are not detached to these rights and have a desire for them. But the governments deny them these freedoms. So, only conclusion II follows.

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

**Hint**

The one's who protested are released, not the one's who murdered.

**Solution:**

The statement says that the political prisoners can be divided into two parts - The one who had been released and those who had been put in prison for political protests. However, no individuals who were released are involved in murder. This means that no political prisoner had committed murder. So, I follow. Clearly, II isn't directly associated with the given statement and does not follow.

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

**Hint**

Earlier, because of Fewer options, man had less choices to make for his destiny.

**Solution:**

Since there were fewer options available earlier but now the modem man influences his destiny. So, I directly follow While for conclusion II, no information is given in the statement.

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- (a)if only the course of action I follows
- (b)if only course of action II follows
- (c)if either course of action I or II follows
- (d)if neither course of action I nor II follows
- (e)if both courses of action I and II follow

**Hint**

Only those actions can be undertaken which have not been implemented yet, and are not risky.

**Solution:**

In general, the police already patrols at night.

Releasing the three persons is not advisable, as it involves risk and may lead to many problems.

Hence, neither courses of action are valid.

- (a)if only the course of action I follows
- (b)if only course of action II follows
- (c)if either course of action I or II follows
- (d)if neither course of action I nor II follows
- (e)if both courses of action I and II follow

**Hint**

The first course of action is a preventive measure, whereas the second course of action is a corrective measure.

**Solution:**

Here, both the measures are correct. I is advisable because by doing so, the number of cases of the disease will be restricted.

II is also advisable because proper facilities in the hospitals will be helpful in the treatment of the patients.

- (a)if only the course of action I follows
- (b)if only course of action II follows
- (c)if either course of action I or II follows
- (d)if neither course of action I nor II follows
- (e)if both courses of action I and II follow

**Hint**

The first course of action is a preventive measure, whereas the second course of action is a corrective measure.

**Solution:**

Here, both the courses of action are advisable as building stronger protection walls can restrict further cases of such accidents, and arresting the bus driver would be helpful in understanding the errors that causes the accident.

- (a)If only I follows.
- (b)If only II follows.
- (c)Neither I nor II follows.
- (d)Both follow.

**Hint**

The action to be taken depends on the transmissibility of the virus.

**Solution:**

Course of action I is not advisable because such actions must be taken only in extreme and exceptional conditions. Otherwise, this would cause inconvenience to several people.

Course of action II is advisable because it will be helpful in tackling the situation and prevent further spread of the disease. Therefore, only course of action II follows.

- (a)if only the course of action I follows
- (b)if only course of action II follows
- (c)if either course of action I or II follows
- (d)if neither course of action I nor II follows
- (e)if both courses of action I and II follow

**Hint**

The government must estimate the volume of foodgrains and the number of clothes required, as well as predict the number of people affected.

**Solution:**

According to the situation, both courses of action are valid, as identifying & providing relief as well as collecting foodgrains & clothes will lead to the proper implementation of the relief programme.

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

**Hint**

The increased stamp duty would have increased the real estate prices, whereas the statement says the prices have decreased.

**Solution:**

There isn't much evidence to show relation between the stamp duty and the real estate prices. Thus, both the statements are effects of independent causes.

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

**Hint**

Large exports lead to big profits.

**Solution:**

Steel manufacturing companies in India made profit because they exported huge quantities of steel to many Asian countries. Therefore, (B) is the cause and (A) is the effect.

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

**Hint**

Number of students aspiring for government colleges is more than the seats available, while it is the complete opposite in private colleges.

**Solution:**

Both the statement are in contradiction to each other. There were more students in government engineering colleges than the number of seats but in private engineering colleges the seats were still vacant as there weren't enough students. Hence, both the statements (A) and (B) are the effects of independent causes.

- (a)if statement (A) is the cause and statement (B) is its effect.
- (b)if statement (B) is the cause and statement (A) is the effect.
- (c)if both the statements (A) and (B) are independent causes.
- (d)if both the statements (A) and (B) are effects of independent causes.
- (e)if both the statements (A) and (B) are effects of some common cause.

**Hint**

The sector that performs better gets more favour.

**Solution:**

Both (A) and (B) are effects of independent causes because the banks decided to give advances to priority sector on par with corporate sector although the priority sector is the better sector in terms of bad loans, it should have got some favour.

- (a)if statement (A) is the cause and statement (B) is its effect.
- (b)if statement (B) is the cause and statement (A) is the effect.
- (c)if both the statements (A) and (B) are independent causes.
- (d)if both the statements (A) and (B) are effects of independent causes.
- (e)if both the statements (A) and (B) are effects of some common cause.

**Hint**

More international flights will bring more tourists to India.

**Solution:**

To substantiate the increase of the flow of foreign tourists to India, private airline companies are allowed to operate on specified international routes. Hence, (B) is the cause and (A) is its effect.

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### Importance of QUANTITATIVE APTITUDE AND REASONING

Practice Now#### Question-wise Difficulty Level Summary

Additionally, we tag all questions of any book to difficulty, ranging from 1 to 10. This tagging helps create personalised learning paths based on the performance of the student.

The table given below shows the difficulty tagged to the questions of this textbook -

Difficulty | No. of Questions |
---|---|

Easy | 2074 |

Medium | 606 |

Hard | 1 |

## Why is the QUANTITATIVE APTITUDE AND REASONING Important?

Exam Name |
---|

Madhya Pradesh Patwari |

IBPS RRB Office Assistant Mains |

IBPS Clerk Mains |

SBI PO Mains |

Andhra Pradesh Police SI Prelims |

Rajasthan Patwari |

IBPS RRB Officer Scale-I Prelims |

SBI Clerk Mains |

Bihar Cooperative Bank Assistant Manager Prelims |

Rajasthan RSMSSB Gram Vikas Adhikari VDO |

NABARD Development Assistant Prelims |

SBI PO Prelims |

IIHM eCHAT |

IBPS RRB Officer Scale-I Mains |

SBI Clerk Prelims |

IMU CET |

IBPS PO Mains |

IBPS Clerk Prelims |

Bihar Cooperative Bank Assistant Prelims |

RBI Assistant Prelims |

IBPS RRB Office Assistant Prelims |

LIC AAO Prelims |

Andhra Pradesh Police Constable Prelims |

BCA Entrance |

IBPS PO Prelims |