Name: QUANTITATIVE APTITUDE AND REASONING
Author: Shyam Saraf,Abhilasha Swarup

1.

Which amongst the following is a prime number?

(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.

2.

Zero

(0)

is a _____.

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

3.

If '

n

' is even, which of the following is odd?

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

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)

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

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

$= 4 + 2 + 4$

$= 10$

Here, $10$ is an even number

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

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

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

$= 2 (1) + 0$

$= 2$

Here, $2$ is an even number

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

$= n^{3} + 2 n$

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

$= 8 + 4$

$= 12$

Here, $12$ is even number.

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

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

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

$= 11$

Here, $11$ is odd number.

Hence, option (d) is correct answer.

4.

If '

n

' is even, which of the following may not be even?

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

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

Hence, option D is correct.

5.

134565

is divisible by _____.

(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.

1.

Preeti went to the market and purchased some mangoes, out of which

10 %

were bad. She gives

33.33 %

of the remaining to the poor and then

180

had left. How many mangoes did Preeti have initially?

(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 $60 % = 180$
$\Rightarrow 1 % = 3 \Rightarrow 1 \times 100 % = 3 \times 100$

$\Rightarrow 100 % = 300$

Hence, the correct answer is $300$ mangoes.

2.

If the price of chocolates is raised by

20 %

, then find out by what percentage a child must reduce his consumption so that there is no change in his expenditure.

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

Hint

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

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 $= [\frac{R}{100 + R} \times 100] %$

Here, $R = 20 %$
$∴ [\frac{20}{100 + 20} \times 100] % = \frac{20}{120} \times 100 = \frac{50}{3} %$
$\Rightarrow 16.66 %$
Hence, option A is correct.

3.

If

2

litres of water is evaporated from

10

litres of a solution containing

12 %

salt, then find the percentage of salt in the remaining solution.

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

Hint

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

Solution:

If $2$ litres of water evaporated from $10$ liters of solution
Remaining solution $= 10 - 2 = 8$ litres
$∴ 10$ litres of $12 %$ $= 10 \times \frac{12}{100} = 1.2$ (as only water evaporate from solution)

$∴$ Solution in $8$ litres is $= \frac{1 .2}{8} \times 100$

$= 0 .6 \times 25 = 15 %$

Salt in the remaining solution is $15 %$

4.

If the salary of a person increases by

40 %

and then reduces by

20 %

, find the change in salary.

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

Hint

Percentage Change $= (\frac{N e w V a l u e - o l d V a l u e}{O l d V a l u e} \times 100)$

Solution:

Let the original salary be $₹ x$
Salary after increase

$= 40 % x = x + \frac{40}{100} x = \frac{140 x}{100}$

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

Salary after decrease

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

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

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

$= \frac{35 x - 7 x}{25} = \frac{28 x}{25}$

Absolute increase

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

$= \frac{28 x - 25 x}{25} = \frac{3 x}{25}$

Percentage increase

$= \frac{3 x}{25} \times \frac{100}{x} = 12$

So, salary increases by $12 %$ .

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

5.

A builder owns a house worth

40

lakhs. He sells it to Mr. Rahul at a profit of

20 %

. After

3

months, Mr. Rahul gets a promotion and has to go to Hyderabad. So, he decides to sell the house back at a loss of

20 %

to the builder. In this transaction, what does the builder make?

(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; $P r o f i t = S . P - C . P .$

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.

1.

Shilpa purchased a computer table for

₹ 1, 260

, but due to some scratches on its top, she had to sell it for

₹ 1, 197

. Find her loss percentage.

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

Hint

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

Solution:

According to the question,

Cost price of laptop (CP) $= ₹ 1, 260$

Selling price of laptop (SP) $= ₹ 1, 197$

We know that, loss $loss = CP - SP$
$₹ 1, 260 - ₹ 1, 197 = ₹ 63$

$loss %$ $= \frac{Loss}{CP} \times 100$

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

$loss % = 5 %$

Hence, the correct answer is $5 %$ .

2.

A false dealer promises to sell at cost price but does a fraud of

10 %

while purchasing and

20 %

while selling. Find his profit percentage.

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

Hint

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

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 $Rs . 100$ be the cost price.
According to question he gains a profit of $10 %$ while purchasing i.e., $10 %$ of $Rs . 100$ is $Rs . 10$ . So, the cost price is $Rs . 90$ .

While selling, he gains a profit of $20 %$ on cost price i.e., $\frac{100 \times 20}{100} = Rs . 20$

Dealer gains total profit $= Rs . (10 + 20) = Rs . 30$
Profit percentage $= \frac{30}{90} \times 100 = 33 .33 %$ .

Hence, dealer gets a profit of $33.33 %$

3.

A vendor bought

40

mangoes at

Rs . 60

and sold them at

Rs . 20

per dozen. Find his gain percent.

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

Hint

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

Solution:

According to the given question,

Cost of $40$ mangoes $= Rs . 60$

SP of $12$ mangoes $= Rs . 20$

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

Using formula,

$Profit = SP - CP$

$= 66.66 - 60 = Rs . 6.66$

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

4.

A trader wants to make a profit of

40 %

by selling

360

books, each with equal cost. However, he sold

160

less books at a price which is two-thirds the previous S.P. Find his profit percentage.

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

Hint

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

Solution:

Let us consider the cost price of a book be $Rs . 1$ .
Then, $360$ books each with equal cost will be $Rs . 360$
With $40 %$ of profit, the selling price of books will be

$SP = Profit + CP$

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

$= 144 + 360 = 504$

According to the question, the trader sells $160$ books less, Which means he sells only $200$ books.
So, the CP of $200$ books is $Rs . 200$
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,

$Profit = SP - CP$ $= 336 - 200 = 136$
$Profit %$ $= \frac{136}{200} \times 100 %$

$= 68 %$

Hence, the $Profit %$ of the trader is $68 %$ .

5.

If the C.P. of

20

pens is equal to the S.P. of

16

pens, find the gain or loss percentage.

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

Hint

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

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
$Gain$ $= 20 - 16 = 4$

$Gain %$ $= \frac{profit}{CP} \times 100$

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

Hence, the $Gain %$ is $25 %$ .

1.

A sum of money lent out at S.I. doubled itself in

10

years. In how many years will it become three times itself ?

(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 $2 P$ in $10$ years.

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

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

$R = 10$

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

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

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

$N = 20$ years.

2.

In what time a sum kept at S.I., will become four times itself at

12 %

?

(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 $4 x$ 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 $4 P$ .

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

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

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

$T = 25$ years

Hence, The correct answer is option (C).

3.

Shilpi invests a sum of money at S.I. and it becomes eight times itself in

12

years. Find in how many years it will become

15

times at the same rate.

(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 $= 8 P$

S.I. $= 8 P - P$

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

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

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

S.I. $= 15 P - P$

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

$T = 24$ years.

4.

Two friends A and B invested equal sums of money for

2

years at

10 %

p.a. A invested at C.I. whereas B invested at S.I. A gets

R s . 82

more than what B gets. How much money did each invest?

(a) $R s . 8, 200$
(b) $R s . 6, 500$
(c) $R s . 7, 000$
(d) $R s . 8,500$

Hint

Using Formula:

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

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 $R s . 82$ more than B

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

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

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

$\Rightarrow P = R s . 8, 200$

Hence the correct answer is $R s . 8, 200$ .

5.

If the annual increase of male population in a town is

10 %

and the present population is

17715610

, what was the male population

3

years ago?

(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., $P \times {(1 + \frac{r}{100})}^{t}$ .

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 $\times {(1 + \frac{10}{10})}^{3}$

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

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

$= 13310 \times 1000$

$= 1331 \times 10^{4}$

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

1.

Two numbers are in the ratio

3 : 4

. If

10

is added to both the numbers, the ratio becomes

4 : 5

. Find the sum of both the numbers.

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

Hint

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

Solution:

According to the question,

Two numbers are in the ratio $3 : 4$

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

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

i.e.

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

or,

$15 x + 50 = 16 x + 40$

or,

$x = 10$

Therefore, the numbers are

$3 x = 3 \times 10 = 30$

and

$4 x = 4 \times 10 = 40$

The sum of the numbers

$= 30 + 40 = 70$ .

2.

If

a : b = 2 : 3

, then the value of

(3 a + b) : (7 a - b)

is _____.

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

Hint

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

Solution:

According to the question,

$a : b = 2 : 3$

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

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

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

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

Therefore,

$3 a + b : 7 a - b = 9 : 11 .$

3.

The ratio of the rate of flow of water in pipes varies inversely as the square of the radius of the pipes. What is the ratio of the rates of flow in two pipes of radius

4 cm

and

8 cm

, respectively?

(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 $4$ $c m$ and $8$ $c m$ ,

so their squares would be $16$ $c m$ and $64$ $c m$ 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 $4 : 1$ .

4.

Two bags have a certain number of tennis balls. First, half the tennis balls from Bag

1

are transferred to Bag

2

, and 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. The ratio of tennis balls in Bag

1

to that in Bag

2

, now, is

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

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

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

Balls in Bag $2$

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

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

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

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

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

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

$= 2 : 1$

5.

If

3, x,

and

27

are in continued proportion, find

x

.

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

Hint

If $a, b & c$ 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 $x = 9 .$

1.

In three numbers, the first is twice the second and thrice the third. If the average of these three numbers is

44,

then the first number is _____.

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

Hint

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

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

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.,

$x = 2 y = 3 z$

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

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

or, $11 x = 44 \times 3 \times 6$

or, $x = 72$

Thus, the first number is $72 .$

2.

The average of

10

numbers is

40.2 .

Two numbers

13

and

18

are removed and

31

is added instead. Find the new average.

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

Hint

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

Solution:

According to the question,

Let the sum of those ten numbers be $x .$

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

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

$x = 402$

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

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

3.

The average of

11

numbers is

10.9 .

If the average of the first

6

numbers is

10.5

and that of the last six digits is

11.4,

then the middle number is _____.

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

Hint

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

Solution:

According to the question,

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

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

$x = 119.9$

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

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

$y = 63$

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

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

$z = 68.4$

Then the middle number is

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

$= 63 + 68.4 - 119.9$

$= 11.5 .$

4.

In a class of

100

students, there are

70

boys whose average marks in a subject are

75 .

If the marks of the complete class are

72,

then what are average marks of the girls?

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

Hint

$T o t a l m a r k s = A v e r a g e m a r k s \times N u m b e r o f s t u d e n t s .$

Average marks of girls $= \frac{T o t a l m r a k s o f t h e c l a s s - T o t a l m a r k s o f b o y s}{N u m b e r o f g i r l s i n t h e c l a s s}$ .

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 $x .$

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

$x = 5250$

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

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

$y = 7200$

Then the average marks of the girls will be,

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

Hence, the correct answer is $65$ .

5.

The average of

x

and

y

is

40 .

If

z

= 10

then what is the average of

x, y

and

z ?

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

Hint

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

Solution:

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

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

$x + y = 80$

If $z = 10,$ then the average of $x, y$ and $z$ will be,

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

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

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

1.

In what ratio should

80 %

milk solution be mixed with a

40 %

milk solution in order to get a

45 %

milk solution?

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

Hint

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

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,

$80 % o f x + 40 % o f y = 45 % o f (x + y) \frac{80}{100} \times x + \frac{40}{100} \times y = \frac{45}{100} \times (x + y) \frac{1}{100} (80 x + 40 y) = \frac{1}{100} (45 x + 45 y) 80 x + 40 y = 45 x + 45 y 80 x - 45 x = 45 y - 40 y 35 x = 5 y \frac{x}{y} = \frac{5}{35} ∴ x : y = 1 : 7$

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

2.

In a class,

70 %

of the boys passed the examination, while only

58 %

girls passed the examination. If the overall pass percentage of boys and girls is

64 %

, and the number of boys is

36

, then find the number of girls in the class.

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

Hint

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

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 $= 36$

3.

$40 litres$ of Juice A costing $₹ 20 / litre$ is mixed with $20$ $litres$ of Juice B costing $₹ x / litre$ to produce a mixture of costing $₹ 25 / litre$ . What is the per litre price of Juice B?

(a) $₹ 33$
(b) $₹ 34$
(c) $₹ 35$
(d)None of these
(e) $₹ 40$

Hint

If two ingredients are mixed, then $\frac{Q u a n t i t y o f c h e a p e r}{Q u a n t i t y o f d e a r e r} = \frac{C . P . o f d e a r e r - M e a n P r i c e}{M e a n p r i c e - C . P . o f c h e a p e r} .$

Solution:

According to the question,

Cheaper price: Juice A $= ₹ 20 / litre$

Mean price: Mixture of Juice A and Juice B $= ₹ 25 / litre$

Let the Dearer price be $' x'$ .

For ratio:

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

$= (x - 25) : (25 - 20) = (x - 25) : 5$

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:

$(x - 25) : 5 = 2 : 1$

$x - 25 = 10$

$x = 35$

Therefore, the dearer price or the price of Juice B per litre is $₹ 35$ .

4.

The $x$ litres of Juice A costing $₹ 25 / litre$ are mixed with $y$ litres of Juice B costing $₹ 42 / litre$ to produce a mixture that is sold for $₹ 36.30 / litre$ , making $10 %$ profit. Find the ratio of $x & y$ .

(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:

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

Solution:

According to the question,

Price of Juice A is $₹ 25 / litre$ .

Price of Juice B is $₹ 42 / litre$ .

Profit $= 10 %$

Selling price $= ₹ 36.30 / litre$

Selling price of the mixture is $₹ 36.30$ , which includes $10 %$ profit.

Cost price of the mixture $= \frac{36.30}{110} \times 100 = \frac{3630}{110} = ₹ 33$

Now,

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

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

5.

A container has milk and water in the ratio

5 : 4

. Now,

20

litres of water is added due to which the ratio is reversed. What is the initial quantity of liquid available in the container?

(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: $\frac{Quantity of milk}{quantity of water + 20 litres of water added} = 4 : 5$ .

Solution:

According to the question,

Ratio of milk and water is $5 : 4$

Then, let the quanity of milk be $5 x$ and water be $4 x$ .

Hence, the total quantity of solution will be

$5 x + 4 x = 9 x$ .

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

$(5 x) : (4 x + 20) = 4 : 5$ (reversed)

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

$25 x = 16 x + 80$

$25 x - 16 x = 80$

$9 x = 80$

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

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

$= 80$ litres.

1.

A and B together finish a work in

20

days. They worked together for

10

days and then B left. After another

20

days, A finished the remaining work. In how many days can A alone finish the entire job?

(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 $' n'$ days, then A's one day work $= \frac{1}{n}$

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 $(\frac{10}{20}) = 0.5$ 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.

2.

A and B together can do a piece of work in

40

days. A worked for

10

days and then left. Then, B finished the remaining work alone in

60

days. In how many days did B finish the entire work alone?

(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,

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

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 $x$ days and $y$ days, respectively.

Then,
$\frac{1}{x} + \frac{1}{y} = \frac{1}{40}$

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

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

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

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

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

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

3.

Harshita and Peter together can do a piece ofwork in

16

days. Harshita alone can do the same work in

24

days. In how many days can Peter alone finish the job?

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

Hint

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

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 $=$ $\frac{1}{Number of days to complete the work}$

(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.

4.

Chand can complete a work in

10

days and Eshan in

20

days. They start the work together but after

5

days, Chand leaves. The remaining work is completed by Eshan alone. How many days will Eshan take to finish the remaining work?

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

Hint

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

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 $= L . C . M . (10, 20) = 20$ 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.

5.

Pipes A, B, and C can fill a tank alone in

10

hours,

15

hours, and

20

hours, respectively. If all the three pipes are simultaneously opened, in how much time will the tank be full?

(a) $4 h$
(b) $4 \frac{7}{13} h$
(c) $4 \frac{8}{13} h$
(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.,

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

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

1.

Arun and Varun are moving towards each other at the speed of

10 m / s

and

12 m / s

, respectively. They are

880 m

apart. Find the time taken by them to meet each other.

(a) $10 s$
(b) $30 s$
(c) $40 s$
(d) $60 s$

Hint

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

Solution:

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

Time taken by them to meet each other

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

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

2.

The ratio of the speeds of Kittu and Bittu is

3 : 5

. If Bittu runs

120 m

then, Sanu runs

90 m

in the same time. If Kittu can run

1 km

in

10 \sec

, what time will Sanu take to run the same distance?

(a) $10.5 s$
(b) $8 s$
(c) $20 s$
(d) $15 s$

Solution:

Let the speeds of Kittu, Bittu and Sanu be $S_{k}, S_{b},$ and $S_{s}$ respectively.

Then according the data given,

$\frac{S_{k}}{S_{b}} = \frac{3}{5}$

$\frac{S_{b}}{S_{s}} = \frac{120}{90} = \frac{4}{3}$

$\frac{S_{k}}{S_{b}} \times \frac{S_{b}}{S_{s}} = \frac{3}{5} \times \frac{4}{3}$

$\frac{S_{k}}{S_{s}} = \frac{4}{5}$

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

$S_{s} = \frac{5}{4} \times S_{k} = 125 m / s$

Thus, time taken by Sanu to cover

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

Hence, the correct answer is $8 s$ .

3.

Sunil walks to school at the rate of

4 kmph

and reaches

10 \min

late. Next day, he increases his speed by

2 kmph

and then reaches school

16 \min

early. Find the distance between his school and home.

(a) $10 km$
(b) $5.2 km$
(c) $9.2 km$
(d) $5.8 km$

Hint

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

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 $=$ $26 \times 2$ $=$ $52 \min$

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

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

Distance between Sunil's school and home is $5.2 km$ .

Hence, option B is correct.

4.

A car is moving along the four sides of a square at the speeds of

25 kmph

,

50 kmph

,

75 kmph

, and

100 kmph

, respectively. Find the average speed of the car around the square field.

(a) $45 kmph$
(b) $48 kmph$
(c) $50 kmph$
(d)None of these

Hint

Square has four equal sides, resulting into equal distance,

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

Solution:

Let the side of the car speed be $x km .$

Let the average speed of the car be $y$ $kmph$ .

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

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:

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

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

Therefore, average speed is $48 kmph .$

5.

Bittu travelled from his home to his office at the speed of

40 kmph

, and returned from office to his home at

60 kmph

. If the whole journey took

2 h 20 \min

, then find the distance of the office from his home.

(a) $55 km$
(b) $56 km$
(c) $65 km$
(d) $66 km$

Hint

$A v e r a g e s p e e d$ $= \frac{2 xy}{x + y}$ .

Solution:

Bittu's speed from his home to office,

$x = 40 kmph$ .

Bittu's speed from office to home,

$y = 60 kmph$ .

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

Distance travelled in $2$ hours $20$ minutes. i.e, $(\frac{7}{3} h r s)$

$D i s \tan c e = Speed \times Time = 48 \times \frac{7}{3} = 112 km .$

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

Hence, the correct answer is $56 k m$ .

1.

The age of Ron’s father is

5

times the Ron’s age. Seven years ago, the sum of their ages was

40

years. Find their present ages (in years).

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

Solution:

We assume that;

The present age of Ron $= x$

Then, the present age of Ron’s father $= 5 x$

According to the question, we can write as :

$(x - 7) + (5 x - 7) = 40$

$x + 5 x - 14 = 40$

$x + 5 x = 40 + 14$

$6 x = 54$

$x = 9$

Hence, the present age of Ron’s father

$= 5 x = 45$

Therefore, option B is correct.

2.

In a cinema theatre, there are two types of tickets; one is sold at

Rs . 80

per ticket and the other at

Rs . 60

per ticket. A total of

100

tickets were sold on a particular day and the total revenue on that day was

Rs . 7000

. Find the number of tickets sold at

Rs . 60

.

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

Solution:

We assume that;

The no. of tickets sold at $Rs . 60$ $= x$

The no. of tickets sold at $Rs . 80$ $= y$

According to the question,

$x + y = 100 \dots \dots . . 1$

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

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

$- 2 y = - 100$

$y = 50$

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

$x = 50$

Thus, the number of tickets sold at

$Rs . 60 = 50$

Therefore, option A is correct.

3.

Saumya, on her birthday, has a certain number of toffees to divide equally among

20

students in her class. If the number of toffees and the number of students, both, were increased by five, each student would receive four toffees less. How many toffees does Saumya have?

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

Solution:

We assume that :

Each student receives the number of toffees $= x$

Then, Saumya has total number of toffees

$= 20 x$

According to the question,

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

Further, we can solve as :

$20 x + 5 = 25 x - 100$

$105 = 5 x$

$x = 21$

Therefore, total number of toffees

$= 20 x = 20 \times 21 = 420$

Hence, the correct answer is $420$ .

4.

If the sum of the two numbers is

20

, the difference of their squares is

240

. Find the larger number.

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

Solution:

We assume that the numbers are $x & y$ .

According to the question;

$x + y = 20 \dots \dots . . 1$

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

From equation $2,$

$(x + y) (x - y) = 240$

Put the value from eq. $1,$

$20 (x - y) = 240$

$x - y = 12 \dots \dots 3$

Now, from $1$ and $3,$

$2 x = 32$

$x = 16$

$y = 4$

Therefore, the bigger number is $16 .$

Hence, the correct answer is $16$ .

5.

Pawan has

₹ 840

and he divides this amount among his brothers Ram, Shyam and Krish such that Ram has thrice the amount received by Shyam and Krish has twenty more than twice the amount received by Ram. Find the amount received by Krish.

(a) $₹ 500$
(b) $₹ 166$
(c) $₹ 246$
(d) $₹ 512$

Solution:

We assume that;

Shyam received the amount $= x$

Ram received the amount $= 3 x$

Krish received the amount $= 6 x + 20$

According to the question, we can write as :

$x + 3 x + 6 x + 20 = 840$

$10 x = 820$

$x = 82$

Thus, Krish received the amount

$= (82 \times 6) + 20 = ₹ 512$

Therefore, $₹ 512$ is correct answer.

1.

If

\log_{4} x

=

- 3

, find the value of

x

.

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

Hint

If $\log_{b} a = m$ then $a = b^{m}$ .

Solution:

We know that:

If $\log_{b} a = m$ then $a = b^{m}$ .

According to the question,

$\log_{4} x = - 3$

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

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

2.

The value of

\log_{4} \log_{25} 625

is

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

Hint

$loga^{a} = 1$ .

Solution:

According to the question,

$\log_{4} \log_{25} 625 = \log_{4} \log_{25} {(25)}^{2}$

Now,

$\log_{25} {(25)}^{2} = 2 \log_{25} 25 = 2 \times 1 = 2$

$\Rightarrow \log_{4} \log_{25} {(25)}^{2} = \log_{4} 2$

$\log_{4} 2 = 2 \log_{2^{2}} 2 = \frac{1}{2} \log_{2} 2 = \frac{1}{2}$

3.

Find the value of

\log_{2 \sqrt{2}} 16

.

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

Hint

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

Solution:

According to the question,

$\log_{2 \sqrt{2}} 16 = \log_{{(\sqrt{2})}^{3}} {(\sqrt{2})}^{8}$

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

Therefore,

$= \frac{8}{3} \log_{\sqrt{2}} \sqrt{2}$

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

4.

Find the value of

\log_{120} 2 + \log_{120} 3 + \log_{120} 4 + \log_{120} 5 .

(a) $\log_{120} 60$
(b) $\log 1$
(c) $1$
(d) $0$

Hint

It is related to, $loga xy = loga x + loga y .$

Solution:

According to the question,

$\log_{120} 2 + \log_{120} 3 + \log_{120} 4 + \log_{120} 5$

We know that, (

$\log ab = \log a + \log b$ )

$\Rightarrow \log_{120} (2 \times 3 \times 4 \times 5) .$

$\Rightarrow \log_{120} 120 = 1$

$(as \log_{a} a = 1)$

5.

Find the value of

\log_{10} 50,

given that

\log_{10} 2 = 0.3010 .

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

Hint

$\log_{a} \frac{x}{y} = \log_{a} x - \log_{a} y .$

Solution:

According to the question,

$\log_{10} 50 = \log_{10} (\frac{100}{2})$

$= \log_{10} 100 - \log_{10} 2$

$= \log_{10} 10^{2} - \log_{10} 2$

$= 2 \log_{10} 10 - \log_{10} 2$

$= 2 \times 1 - 0.3010$

$= 1.699 .$

1.

How many triangles can be formed by joining

12

points,

6

of which are collinear?

(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

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

2.

In how many ways can the letters of the words NATURE be arranged so that the vowels are always together?

(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 $144$ .

3.

In a meeting room, eight chairs are numbered from

1

to

8

. Two women first choose two chairs from those marked

1

to

4

and four men select four chairs from the remaining. Find the number of possible arrangements:

(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 $= {}^{4}P_{2} \times {}^{6}P_{4}$

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

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

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

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

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

4.

Ram has six coins of different values. In how many ways can he put all the coins in his two pockets?

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

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.

5.

In an exam consisting of

12

questions divided into two sections of

6

questions each, a student has to answer only

7

. A maximum of four questions can be attempted from each group. In how many ways can he select the seven questions in all?

(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.

1.

Three dice are thrown. What is the probability that the resultant will have at least one result as

6

?

(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'.

2.

The probabilities of two events are

0.35

and

0.60

. The probability of both occurring together is

0.25

. What is the probability of none of the events occurring?

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

Hint

First, we calculate the value of $P (X or Y) = P (X) + P (Y) - P (X and Y)$ , then calculate the value of $P (none)$ .

Solution:

Let the two events be $X$ and $Y$ .

Now, given: $P (X) = 0.35$

$P (Y) = 0.60$

$P (X and Y) = 0.25$

We know that,

$P (X or Y) = P (X) + P (Y) - P (X and Y)$

Substituting the given values we get:

$P (X or Y) = 0.35 + 0.60 - 0.25$

$\Rightarrow$ $0.70$

$P (none) = 1 - P (X or Y)$

$\Rightarrow 1 - 0.70 = 0.30$

Therefore, the correct answer is option 'B'.

3.

Four friends are playing with coins. They tossed up their coins simultaneously. What is the probability of getting exactly two tails?

(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'.

4.

In a toss of

20

coins one after the other, find the probability of getting exactly two heads on consecutive throws.

(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 $(2^{20})$ 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'.

5.

In a shooting competition, the probabilities of hitting the target for three shooters

P, Q,

and

R

are

\frac{1}{4}, \frac{1}{3}

and

\frac{2}{5}

respectively. If all of them fire independently, calculate the probability of exactly one of them hitting the target.

(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,

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

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

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

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

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

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

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

$= P (P) \times P (Q') \times P (R') + P (Q) \times P (P') \times P (R') + P (R) \times P (P') \times P (Q')$

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

$= \frac{9}{20}$

1.

The length and breadth of a rectangular room are

24

m

and

18

m

respectively. Find the area of the floor of the room (in

sq . m .

).

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

Solution:

We know that,

Area of the rectangular room

$=$ Lenght $\times$ Spred

Area of the rectangular room

$= 24 \times 18 = 432 sq . m .$

Therefore, $432 sq . m .$ is the correct answer.

2.

The length and breadth of a rectangular room are

24 m

and

18 m

, respectively. Find the maximum possible length of a rod that can be put on the floor of the room.

(a) $24 m$
(b) $30 m$
(c) $32 m$
(d)None of these
(e) $35 m$

Hint

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

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 $= \sqrt{{length}^{2} + {breadth}^{2}}$

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

Therefore, option B is correct.

3.

The length of a rectangle is three times its breadth. The diagonal is

100 cm

. Find the length of the rectangle.

(a) $86 cm$
(b) $94.8 cm$
(c) $31.6 cm$
(d)None of these
(e) $80 cm$

Solution:

Let the breadth of rectangle be $x cm$

According to given question,

Length of rectangle $= 3 x cm$

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

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

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

$x = 10 \sqrt{10}$

$x = 31.6$

Hence, length of rectangle

$= 3 \times 31.6 = 94.8 cm$

Therefore, option B is correct.

4.

The length of a wall is

\frac{2}{3}

times its height. If the area of the wall is

54 m^{2}

, find the sum of the length and the height of the wall.

(a) $15 m$
(b) $20 m$
(c) $9 m$
(d) $6 m$
(e)None of these

Solution:

Let the height of wall be $x m$

According to given question

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

Area $= length \times height$

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

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

$x^{2} = 81$

$x = 9$

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

Therefore, sum of length and height $= 6 + 9 = 15 m$

Therefore, option A is correct.

5.

How many tiles, of dimension

15 cm

by

20 cm,

will be required to pave the floor of a meeting hall

12 m

long and

10 m

wide?

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

Hint

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

Solution:

We know that, length of hall $= 12 m = 1200 cm$

Breadth $= 10 m = 1000 cm$

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

Therefore, option A is correct.

1.

What is the angle (in degrees) between the two hands of a clock at

4 : 20

?

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

Hint

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

Solution:

Given:

Time $= 4 : 20$

According to question,

We know that,

$θ = |\frac{11}{2} m - 30 h|$

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

$θ = |\frac{11}{2} (20) - 30 (4)|$

$θ = 10^{o}$

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

2.

What is the angle (in degrees) between the two hands of a clock at

3 : 40

?

(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 $(30 h + \frac{m}{2}) - 6 m = 30 h - (\frac{11}{2}) m$

Solution:

Given:

Time $= 3 : 40$

According to question,

$θ = |\frac{11}{2} m - 30 h|$

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

$θ = |\frac{11}{2} (40) - 30 (3)|$

$θ = 130^{o}$ .

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

3.

What is the angle (in degrees) between the two hands of a clock at

7 : 46

?

(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 $(30 h + \frac{m}{2}) - 6 m = 30 h - (\frac{11}{2}) m$ .

Solution:

Given:

Time $= 7 : 46$

According to question,

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

$θ = |\frac{11}{2} m - 30 h|$

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

$θ = |\frac{11}{2} (46) - 30 (7)|$

$θ = 317^{o}$

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

4.

Between

6

and

7

o'clock, when are the two hands of a clock perpendicular to each other?

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

Hint

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

Solution:

We know that,

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

Here, $H = 6$ and $θ = 90^{o}$

Plugging the values, we get

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

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

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

Therefore, two hands are perpendicular at

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

5.

Between

8

and

9

o'clock, when are the two hands of a clock opposite to each other?

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

Hint

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

Solution:

We know that,

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

Here, $H = 8$ and $θ = 180^{o}$

Plugging the values, we get:

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

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

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

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

1.

In the following question, a term of the letter series is missing as shown by a question mark(?). Choose the missing term out of the given alternatives. U, O, I, ?, A

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

2.

In the following question, various terms of a letter series are given with terms missing as shown by a question mark(

?

). Choose the missing term out of the given alternatives. Y,W,U,S,Q,?,?

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

3.

In the following question, a term is missing from the letter series as shown by a question mark (

?

). Choose the missing term out of the given alternatives.

A, B, D, G, ?

(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.

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

Therefore the missing term is $K$ .

4.

In the following question, a term is missing from the letter series as shown by a question mark (

?

). Choose the missing term out of the given alternatives.

Z, U, Q, ?, L

(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.

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

Hence the missing term is $N$ .

5.

In the following question, a term from the letter series is missing as shown by a question mark (

?

). Choose the missing term out of the given alternatives.

A, C, F, H, ?, M

(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:

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

Therefore the missing term is $K$ .

1.

Sukhinder remembers that his father's birthday is between 13 and 16 of June, whereas his sister remembers that their father's birthday is between 14 and 18 of June. On which day is their father's birthday?

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

2.

If

5

^th January

1991

was a Saturday. What day of the week was

3

^rd March

1992

?

(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 $1992 = 5 + 1 + 3 = 9$ days i.e. $2$ odd days.

Hence, $3$ ^rd March $1992$ will be $2$ days after Sunday, i.e. Tuesday.

3.

Kamini ranks

16

^th from the top and

15

^thfrom the bottom in a certain examination. How many students are there in her class?

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

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

$= 15 + 14 + 1 = 30$

4.

How many

7

are there in the following number sequence, which is immediately preceded by

5

, but not immediately followed by

3

?

375745739785778971576574375738

(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:

$37 (57) 4573978 (57) 78971 (57) 6 (57) 4375738$

Thus, there are four such $7$ .

5.

Reaching the place of meeting on Tuesday,

15

min before

8 : 30

hrs, Anuj found himself half an hour earlier than the man who was

40

min late. What was the scheduled time of the meeting?

(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.

1.

Aman started walking towards the north. After walking

30 m

, he turned towards the left and walked

40 m

. He then turned left and walked

30 m

. He again turned left and walked

50 m

. How far is he from his original position?

(a) $40 m$
(b) $10 m$
(c) $50 m$
(d) $30 m$
(e)None of these

Solution:

He walks from $A$ to $B 30 m$ towards the north.

$B$ to $C = 40 m$

$C$ to $D = 30 m$

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

Hence, $10 m$ is the correct answer.

2.

Ram went

15 k m

to the west from my house, then turned left and walked

20 k m

. He then turned east and walked

25 k m

and, finally, turning left, covered

20 k m

. How far is he now from his house?

(a) $10 k m$
(b) $25 k m$
(c) $20 k m$
(d) $15 k m$
(e)None of these

Solution:

Starts from point A

Distance covers from A to B $= 15 k m$

B to C $= 20 k m$

C to D $= 25 k m$

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

At point E, he is facing the north direction.

Hence, he is $10 k m$ far from his house.

3.

Shyam walks

6 km

to the east and then turns to the south walking

2 km

. Again, he turns to the east and walks

2 km

. Next, he turns towards the north and walks

8 km

. Find the distance between the initial and terminating points.

(a) $12 km$
(b) $10 km$
(c) $8 km$
(d) $16 km$
(e)None of these

Hint

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

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

Solution:

Draw the diagram, according to the given conditions.

In the above diagram $BCDE$ is a square

$BE = 2 km :$ $DE = 2 km$

$AE = 8 km :$ $FE = 6 km$

Hence $AF$ :

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

$64 + 36 = x^{2}$

$100 = x^{2}$

$x = 10$

Thus, Shyam's distance from the starting point

$= 10 km$ .

4.

Kamu walks

5 km

straight from her house towards the west, then turns right and walks

3 km

. Thereafter, she takes a left tum and walks

2 km

. Further, she turns left and walks

3 km

. Finally, she turns right and walks

3 km

. In what direction is she now from her house?

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

Solution:

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

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

Hence, the required answer is the west direction.

5.

Hari travelled

17 km

to the east, turned left and went

15 km

: he again turned left and went

17 km

. How far is he from the starting point?

(a) $17 km$
(b) $2 km$
(c) $15 km$
(d) $32 km$
(e)None of these

Solution:

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

The distance from his starting point to the final point is $15 km$ .

Hence, $15 km$ is the answer.

1.

It is given that: 'M

Δ

N' means M is the mother of N.
'M

•

N' means M is the sister of N.
'M

*

N' means M is the father of N.
'M

β

N' means M is the brother of N. Then, which of the following means that L is the paternal grandfather of O?

(a) $L * R • M β K β O$
(b) $R * L Δ P β K • O$
(c) $L * M Δ R * β O$
(d) $L * R β M * K β O$

Solution:

According to the question, we can clearly see that:

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

Similarly,

$• and β \to$ brother - sister (so, there is no generation gap)

Now checking by the option, here we will replace ' $Δ$ and $*$ ' by ' $+ 1$ ' and ' $•$ and $β$ ' by ' $0$ '.

In option (a), we have:

$L (+ 1) R (0) M (0) K (0) O \Rightarrow + 1 + 0 + 0 + 0 = + 1$

Similarly, in option (c), we have:

$L (+ 1) M (+ 1) R (+ l) K (0) O \Rightarrow + 1 + 1 + 1 + 0 = + 3$

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.

2.

Read the following information carefully and answer the questions that follow: 'P

\div

Q' means P is the father of Q.
'P + Q' means P is the mother of Q.
'P - Q' means P is the brother of Q.
'P x Q' means P is the sister of Q. lf A + B

\div

C - D, then A is D's:

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

Solution:

The given equation can be decoded as:

A+B= A mother of B

B $\div$ 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.

1.

Directions: Select the related letters/words/numbers from the given alternatives. METAPHOR: EMATHPRO:: NORMAL:?

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

3.

Select the related number from the given alternatives.

0.01 : 0.0001 : : 0.05 : ?

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

Solution:

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

${(0.01)}^{2} : 0.0001$

Similarly,

${(0.05)}^{2} : 0.0025$

Hence, the correct answer is $0.0025 .$

1.

In the following question, find the odd word/letters/number/number pair from the given alternatives.

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

2.

In the following question, find the odd number pair from the given alternatives.

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

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 $74 - 40$ .

1.

Read the following conditional propositions and identify if they are valid, invalid, or false. Harry is cute or smart. Harry is not cute. Harry is smart

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

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.

2.

Read the following conditional propositions and identify if they are valid, invalid, or false. Ron is not cute or intelligent. Ron is cute. Ron is intelligent.

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

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.

3.

Read the following conditional propositions and identify if they are valid, invalid, or false. Hermione is sweet or smart. She is smart. She is sweet.

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

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.

4.

Read the following conditional propositions and identify if they are valid, invalid, or false. Ginny is a redhead or a blonde. She is not a redhead. She is a blonde.

(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.

5.

Read the following conditional propositions and identify if they are valid, invalid, or false. Hagrid has a dinosaur or he has a merman. Hagrid does not have a merman and so he has a dinosaur.

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

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.

1.

In the following question, two statements are followed by two conclusions numbered I and II. Assume the given statements to be true even if they are at variance with commonly known facts. Then choose one of the answer choices which are given as: (1) If only conclusion I follows. (2) If only conclusion II follows. (3) If either I or II follows. (4) If both I and II follow. (5) If neither I nor II follows. Statements: a. All scientists are fools. b. All fools are literates. Conclusions: I. All literates are scientists. II. All scientists are literates.

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

1.

In the following question, a statement is given, followed by two assumptions numbered I and II. An assumption is something supposed or taken for granted. You have to consider the statement and the following assumptions and decide which of the assumptions is implicit in the statement. Statement: Central Bank, which is the largest bank in the country, has decided to reduce its workforce by 30 per cent so that its banks may work efficiently. Assumptions: I. The bank can perform all its activities after the reduction in the workforce. II. The surplus employees may be asked to adopt an early retirement scheme before leaving the bank.

(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

1.

In the following question is given a statement followed by two conclusions numbered I and II. You have to assume everything in the statement to be true, and then consider the two conclusions together and decide which of them logically follows beyond a reasonable doubt from the information given in the statement. Statement : Irregularity is a cause for failure in exams. Some regular students fail in the examinations. Conclusions : I. All failed students are regular. II. All successful students are not regular.

(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

1.

In the following question, a statement is given followed by two courses of action numbered I and II. A course of action is a step or administrative decision to be taken for improvement, follow-up, or further action in regard to the problem, policy, etc., on the basis of the information given in the statement. You have to assume everything in the statement to be true, then decide which of the two given suggested courses of action logically follows for pursuing. Statement: Three persons were caught with huge arms and ammunition in the city. Courses of action : I. The police should be instructed for night patrolling. II. The three persons should be set free and their movements should be carefully watched to nab the other criminals.

(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

1.

In the following question, the two statements (A) and (B) are given. These statements may be either independent causes or may be effects of independent causes. One of these statements may be the effect of the other statement. Read both the statements and decide which of the following answer choice correctly depicts the relationship between the two statements. Mark answer: (A) The state government has now decided to increase the stamp duty on house purchases with immediate effect. (B) The real estate prices have decreased considerably during the last few months

(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.

Difficulty	No. of Questions
Easy	2074
Medium	606
Hard	1

Exam Name
Madhya Pradesh Patwari
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QUANTITATIVE APTITUDE AND REASONING

Contents of QUANTITATIVE APTITUDE AND REASONING

27 Chapters

7 Topics

2 Topics

4 Topics

4 Topics

4 Topics

2 Topics

1 Topics

4 Topics

7 Topics

4 Topics

2 Topics

3 Topics

1 Topics

2 Topics

2 Topics

1 Topics

3 Topics

1 Topics

1 Topics

1 Topics

1 Topics

1 Topics

1 Topics

1 Topics

1 Topics

1 Topics

1 Topics

Experience Tests Tailored to Match the Real Exam

Practise Questions with Solutions from the QUANTITATIVE APTITUDE AND REASONING

Chapter 1

Number System

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Chapter 2

Percentages

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Chapter 3

Profit and Loss

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Chapter 4

Simple Interest and Compound Interest

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Chapter 5

Ratio, Proportion, and Variation

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Chapter 6

Averages

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Chapter 7

Mixture and Alligation

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Chapter 8

Work, Pipes, and Cisterns

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Chapter 9

Time, Speed, and Distance

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Chapter 10

Basic Algebra

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Chapter 11

Logarithms and Indices

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Chapter 12

Permutation and Combination

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