SRCC GBO 2011 Question Paper

• Locate the Subject (someone), Verb (locked), and Object (classroom).
• The object of the active sentence becomes the subject of the passive sentence.
• The tense of the "be" verb must match the tense of the active sentence. (past perfect- had locked)
• The original subject (someone) is introduced with "by" (or dropped if unknown/unimportant), dropped here, since there is unknown person (someone)
  
  Using these rules, " The classroom had been locked." is the right choice.

• Locate the Subject (we), Verb (rebuked), and Object (him).
• The object of the active sentence becomes the Subject of the passive sentence. 
• The tense of the "be" verb must match the tense of the active sentence. (simple past- justly rebuked)
• The original Subject(we) is introduced with "by" (or dropped if unknown/unimportant)
  
  Using these rules, "He was justly rebuked by us for acting so selfishly" is the correct choice.

• Locate the Subject (remark), Verb (wounded), and Object (feelings).
• The object of the active sentence becomes the subject of the passive sentence.
• The tense of the "be" verb must match the tense of the active sentence. (present perfect- has wounded)
• The original subject (remark) is introduced with "by" (or dropped if unknown/unimportant)
  
  Using these rules, " My feelings have been wounded by that remark" is the right choice.

The series can be broken as: $$\ \frac{\ 1}{2}+\ \ \frac{\ 3}{4}+\ \frac{\ 7}{8}+.......\ \ \frac{\ 1023}{1024}$$ or $$\left(1-\ \frac{\ 1}{2}\right)+\left(\ 1-\ \frac{\ 1}{4}\right)+......$$

This becomes: 1+1+1+1..... n times - { $$\ \frac{\ 1}{2}+\ \frac{\ 1}{2^2}+.......\ \ \frac{\ 1}{2^n}$$ }

Here, n = 10

Hence, the sum becomes 10 - (Sum of GP upto 10 terms with a common ratio of 0.5)

This gives the answer as: 10 - $$\ \frac{\ 1}{2}\times\ \ \frac{\ \ 1-\frac{\ 1}{2^{10}}}{1-\ \frac{\ 1}{2}}$$ or approximately 9 (slightly greater than 9)

In binomial expansion of $$\left(1+x\right)^n$$, if x is really close to 0, it could be written as: $$(1 + x)^n \approx 1 + nx$$.
So, $$1.001001001^{\frac{1}{3}}$$ can be written as $$1+\frac{1}{3}\times\ 0.001001001$$ and,
$$1.001001^{\frac{1}{3}}$$ can be written as $$1+\frac{1}{3}\times\ 0.001001$$.
Their difference comes out to be: $$\frac{1}{3}\left(0.001001001-0.001001\right)=\frac{0.000000001}{3}=\frac{1}{3}\times\ 10^{-9}=0.33\times\ 10^{-9}=3.3\times10^{-10}$$.

$$a^3-1$$ could be written as (a-1)($$a^2+ab+b^2$$) and $$a^3+1$$ could be written as (a+1)($$a^2-ab+b^2$$).
So, the expression could be written as: $$\frac{\left(2-1\right)\left(3-1\right)\left(4-1\right)...}{\left(2+1\right)\left(3+1\right)\left(4+1\right)...}\times\ \frac{\left(2^2+2+1^2\right)\left(3^2+3+1^2\right)...}{\left(2^2-2+1^2\right)\left(3^2-3+1^2\right)...}$$.
The first fraction will be: $$\frac{1\cdot2\cdot3...99}{3\cdot4\cdot5...101}=\frac{1\cdot2}{100\cdot101}=\frac{1}{50\cdot101}$$.
For the 2nd part, $$a^2+a+1^2=\left(a+1\right)^2-\left(a+1\right)+1^2$$. This means that $$2^2+2+1^2=\left(3\right)^2-\left(3\right)+1^2$$.
So, the 2nd fraction could be written as: $$\frac{7\cdot13\cdot21...}{3\cdot7\cdot13...}=\frac{100^2+100+1^2}{3}=\frac{10101}{3}$$.
The product of the 2 fractions will be : $$\frac{1}{50\cdot101}\times\ \frac{10101}{3}=0.667\ \approx\ \frac{2}{3}$$

Since the square root of a number is between 6 and 7.

So, number itself will lie between $$6^2 = 36$$ and $$7^2 = 49$$

Cube root of 36 = 3.3 (Approx)

Cube root of 49= 3.66 (Approx)

Hence, the cube root of the number will be between 3 and 4. 

$$3^{210} = (3^2)^{105} = 9^{105} < 17^{105} => 3^{210} < 17^{105}$$

$$7^{140} = (7^4)^{35} = 2401^{35}$$

$$17^{105} = (17^3)^{35} = 4913^{35} => 7^{140} < 17^{105}$$

$$17^{105} = (17^5)^{21} = 1419857‬^{21}$$

$$31^{84} = (31^4)^{21} = 923521^{21} => 31^{84} < 17^{105}$$

Hence, largest of the number = $$17^{105}$$

Let 18 consecutive natural number = a, a + 1, a + 2, a + 3, ........., a + 17.

Sum = $$\frac{n}{2}[2a + (n - 1)d]$$

= $$\frac{18}{2}[2a + (18 - 1)1]$$

= 9 (2a + 17)

= 18a + 153

Put a = 1, Sum = 18 + 153 = 171

Put a = 2, Sum = 36 + 153 = 189

Put a = 3, Sum =54 + 153 = 207

Put a = 4, Sum =72 + 153 = 225 [Required answer]

$$\frac{1}{1 + 1^2 + 1^4} + \frac{2}{1 + 2^2 + 2^4} + \frac{3}{1 + 3^2 + 3^4}+ ....... + \frac{99}{1 + 99^2 + 99^4}$$

= $$\frac{1}{3} + \frac{2}{21} + \frac{3}{91} + ....... +$$

= $$\frac{1}{1 \cdot 3} + \frac{2}{3 \cdot 7} + \frac{3}{7 \cdot 13} + ....... +$$

$$\frac{1}{1 \cdot 3}$$ this can be written as $$\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)$$

$$ \frac{2}{3 \cdot 7}$$ this can be written as $$\frac{1}{2}\left(\frac{1}{3}-\frac{1}{7}\right)$$

$$\frac{3}{7 \cdot 13}$$ this can be written as $$\frac{1}{2}\left(\frac{1}{7}-\frac{1}{13}\right)$$

The whole expression can be written as

=$$\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)$$+ $$\frac{1}{2}\left(\frac{1}{3}-\frac{1}{7}\right)$$+$$\frac{1}{2}\left(\frac{1}{7}-\frac{1}{13}\right)$$ +....

The terms get cancelled out and we will be remained with the first from the first term and the second from the last term.

The second term in each of these terms form a pattern.

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

For the first term n=1, $$(n+1)^2-(n+1)+1$$ = 3

For the second term n=2, $$(n+1)^2-(n+1)+1$$ = 7

For the second term n=3, $$(n+1)^2-(n+1)+1$$ = 13

For the 99th term n=99, $$(n+1)^2-(n+1)+1$$ = 9901.

The final value of the expression is $$\frac{1}{2}-\frac{1}{9901}$$

This will be in between 0.49 and 0.5

Going by the traditional method since the numbers are easy to calculate, the value of $$71^2 - 37^2 - 51$$ is 3621.

The prime factorisation of 3621 is $$\ 1\times\ 3\times\ 17\times\ 71$$

Hence, there are three prime numbers that divide the expression.

$$x^2 - x - 1 = 0$$

$$x^2 = 1 + x$$ Or $$x^2 - x = 1$$

$$= x^3 - 2x + 1 = x.x^2 - 2x + 1$$

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

$$=x + x^2 - 2x + 1$$

$$=x^2 - x + 1$$

$$=1 + 1$$

= 2

$$120 = 2 * 2 * 2 * 3 * 5 = 2^3 * 3^1 * 5^1$$

Sum of factors = $$(2^0 + 2^1 + 2^2 + 2^3) * (3^0 * 3^1) * (5^0 * 5^1) = (1 + 2 + 4 + 8) * (1 + 3) * (1 + 5) = 15 * 4 * 6 = 360$$

Ssum of the reciprocals factors = $$\frac{360}{120} = 3$$

x% of y is 1% of z

$$\frac{x}{100} * y = \frac{1}{100} * z$$

xy = z ....... (1)

y% of z is 1% of x 

$$\frac{y}{100} * z = \frac{1}{100} * x$$

yz = x ....... (2)

z% of x is 1% of y

$$\frac{z}{100} * x = \frac{1}{100} * y$$

zx = y ....... (3)

Now,

xy + yz + zx = z + x + y

Put value of z from equation (1) in equaiton (2):

y * xy = x

$$y^2 = 1$$

y = 1 Similarly, 

x = z = 1

Hence, x + y + z = 1 + 1 + 1 = 3

The given expression is $$(10^{4n^2 + 8} + 1)^2$$.

The expression will be of the form $$(10000....1)^2$$

Any number of the form 1000(x times)1 will be of the form 1000(x times)20000(x times)1.

For example $$1001^2$$=1002001

So, the sum of the digits will always we 1+2+1=4

$$=x^4y + xy^4$$

$$=xy(x^3 + y^3)$$

$$=xy(x + y)(x^2 + y^2 - xy)$$

$$=xy(x^2 + y^2 - xy)$$                                  [Since, x + y = 1]

$$=xy[(x + y)^2 - 2xy - xy)]$$

$$=xy[1 - 3xy]$$

Ley xy = a

$$S=a[1 - 3a]$$

$$S=a - 3a^2$$

$$\frac{\text{d}S}{\text{d}a} = 1 - 6a$$

For maximum; 

$$\frac{\text{d}S}{\text{d}a} = 1 - 6a = 0$$

$$a = xy = \frac{1}{6}$$

After putting $$xy = \frac{1}{6}$$ :

$$=xy[1 - 3xy]$$

$$=\frac{1}{6}(1 - 3 * \frac{1}{6})$$

$$=\frac{1}{12}$$

The average of the 15 values is given as 13.

Therefore, the total values will be $$15\times\ 13=195$$

As we want the maximum value of second largest number

So we will take the minimum value of first thirteen (15-2=13) natural numbers. 

As it is given in the question, we need to take different natural numbers, so the smallest 13 different natural number will be 

1, 2, 3, 4...13 Now when we sum the first 13 natural numbers, we get 

The formula for the summation of first n natural number is $$\dfrac{n\left(n+1\right)}{2}$$

For $$n=13$$, the sum will be equal to $$\dfrac{13\left(13+1\right)}{2}=91$$.

Therefore we will subtract 91 from the total value so 195-91=104

Now we need to find the maximum second-to-last number, so 104/2=52.

Now the two numbers will be 52-1=51 and 52+1=53, as the numbers should be different.

Therefore, the maximum value of the second largest number will be 51.

Hence option B is the correct answer.

Let first term $$a_1 = a$$ and common difference = d

$$a_2 = a + d; a_3 = a + 2d$$

Now, 

$$a_1 + a_8 + a_{12} + a_{19} = 224$$

a + a + 7d + a + 11d + a + 18d = 224

4a + 36d = 224

a + 9d = 56 ....... (1)

= $$a_1 + a_2 + a_3 + ........ + a_{19}$$

= a + a + d + a + 2d + .......... + a + 18d

= 19a + 171d

= 19(a + 9d)

= 19 * 56

= 1064

$$V \propto \frac{1}{\sqrt{m}}$$

$$m \propto \frac{1}{t^2}$$

Now, 

$$V \propto \frac{1}{\sqrt{\frac{1}{t^2}}}$$

$$V \propto t$$

Let side of square = 'a'

Diameter of circle = Diagonal of square = $$a\sqrt{2}$$

Radius of circle = $$\frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$$

Let side of triangle = s

Length of iInradius of triangle = $$\frac{s}{2\sqrt{3}} = \frac{a}{\sqrt{2}}$$

$$s = a\sqrt{6}$$

Area of triangle = $$\frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4}(a\sqrt{6})^2 = \frac{3\sqrt{3}a^2}{2}$$

Area of square = $$a^2$$

Required ratio = $$\frac{3\sqrt{3}a^2}{2}: a^2 = 3\sqrt{3}: 2$$

Radius of larger circle = PC + Circum Radius of $$\triangle{ABC}$$            [PC = Radius of smaller circle = 1 cm]

$$\triangle{ABC}$$ is equilateral triangle with side = 2 cm

Circum of an equilateral triangle = $$\frac{a}{\sqrt{3}} = \frac{2}{\sqrt{3}}$$

Radius of large circle = $$1 + \frac{2}{\sqrt{3}} = \frac{\sqrt{3} + 2}{\sqrt{3}} = \frac{3 + 2\sqrt{3}}{3}$$

a + b < c + d ..... (1)

b + c < d + e ...... (2)

c + d < e + a ....... (3)

d + e < a + b ....... (4)

From (1) and (4):

d + e < a + b < c + d => e < c

From (1) and (3):

a + b < c + d < e + a => b < e

From (2) and (4):

b + c < d + e < a + b => c < a

Now, 

b < e < c < a

Let b = 1, e = 2, c = 3 and a = 4

From (1) and (3): 

a + b < c + d => 4 + 1 < 3 + d => d > 2 => d > e

c + d < e + a => 3 + d < 2 + 4 => d < 3 => d < c

Hence, 

b < e < d < c < a

Largest number is a and smallest number is b.

$$2^{x + y} = 10$$ ..... (1)

$$2^{y + z} = 20$$ ....... (2)

$$2^{z + x} = 30$$........ (3)

From (1), (2) and (3):

$$2^{2(x + y + z)} = 10 * 20 * 30 = 6000 = 400 * 15$$

$$2^{x + y + z} = 20\sqrt{15}$$ .... (4)

From (3) and (4):

$$\frac{2^{x + y + z}}{2^{y + z}} = \frac{20\sqrt{15}}{20}$$

$$2^x = \sqrt{15}$$

Since 'x' and 'y' are positive integers. Let x = y = 1

x + y = 1 + 1 = 2

$$\sqrt{3xy} = \sqrt{3 * 1 * 1} = \sqrt{3} = 1.732$$

$$\frac{x^2 + y^2}{x + y} = \frac{1^2 + 1^2}{1 + 1} = 1$$

$$\frac{x^4 + x^2y^2 + y^4}{x^3 + y^3} = \frac{1^4 + 1^2 * 1^2 + 1^4}{1^3 + 1^3} = 1.5$$

Hence, x + y is largest. 

The equation shall be: 25a+16b+20c = 213 and a+b+c = 12

Since the final total is 213, the only possibility of a sum ending with 3 is when the resultant of 16 ends with 8 (48 or 128) and the resultant of 25 ends with 5, thereby giving 3 as the unit digit of the sum.

Case 1: When 16b is 48

This means b = 3

Hence, 25a + 20c = 165

Now, a+c = 9 (as b = 3)

These two equations are inconsistent since one of the values is coming out a negative integer. 

Case 2: When 16b = 128

This means b = 8

Hence, 25a + 20c = 85

And a+c=4 (as b=8)

Hence, a=1 and c=3

By this, we know that the number of months with 25 problems is 1.

Let y $$=\sin^6 x + \cos^6 x$$

y $$=(\sin^2 x)^3 + (\cos^2 x)^3$$

y $$=(\sin^2 x + \cos^2 x)^3 - 3\sin^2 x \cos^2 x(\sin^2 x + \cos^2 x)$$

y $$=1 - 3\sin^2 x \cos^2 x$$

y $$=1 - 3(\sin x \cos x)^2$$

y $$=1 - \frac{3}{4}(2\sin x \cos x)^2$$

y $$=1 - \frac{3}{4}(sin2 x)^2$$

When sin2x is maximum, y will be minumum and maximum value of sin2x is 1. 

y (min) = $$1 - \frac{3}{4} * 1 = \frac{1}{4}$$

When sin2x is minimum, y will be maximum and minimum value of sin2x is 0.

y (max) = $$1 - \frac{3}{4} * 0 = 1$$

Required ratio = 1: $$\frac{1}{4}$$ = 4: 1

The best way to proceed with this question is to assume m as a very small integer. For this question, let us say that m=2

Hence, a = 11 and b = 105

Hence,  $$\sqrt{ab + 1}$$ is $$\sqrt{1156}$$, which is 34.

Clearly, the number of 3s is m-1, aa m-1 = 1.

This is clearly in line with option A.

Simplifying the expression: $$\ \frac{\ \log\ x}{\log\ 3}\times\ \ \frac{\ \log\ 2x}{\log\ x}\times\ \ \frac{\ \log\ y}{\log\ 2x}$$, which is $$\ \log_3y$$

Now, the RHS of the equation is equal to 2. 

Hence, $$\ \log_3y$$ = 2, or y is 

Let the length of each small rectangular strips is 'l' and its breadth is 'b'.

Breadth of piece of paper = Length of each small strips = l

Length of piece of paper = Sum of breadth of 'n' small strips = nb

Now, 

$$\frac{l}{b} = \frac{nb}{l}$$

$$l^{2} = nb^{2}$$

$$l = \sqrt{n}b$$

$$l: b = \sqrt{n}: 1$$

Since, $$\triangle{PQR}$$ and $$\triangle{QRS}$$ are similar equilateral triangles. 

Side of $$\triangle{PQR}$$ = Diameter of any circle = 4 cm

$$Area of \triangle{PQR} = \frac{\sqrt{3}}{4}PR^{2} = \frac{1}{2} * QR * \frac{PS}{2}$$

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

$$PS = 4\sqrt{3}$$

Lenght of BC = PS + 2 + 2 = $$4\sqrt{3} + 4 = (4{\sqrt{3} + 1}) cm$$

Let length, breadth and height of cuboid is 'l', 'b' and 'h' respectively.

According to the question: 

l + b + h = 19 ........ (1)

Diagonal of cuboid = $$\sqrt{l^{2} + b^{2} + h^{2}} = 5\sqrt{5}$$

$$l^{2} + b^{2} + h^{2} = 125$$ ...... (2)

We know that: 

$$(l + b + h)^{2} = l^{2} + b^{2} + h^{2} + 2(lb + bh + hl)$$

From eqaution (1) and (2):

$$19^{2} = 125 + 2(lb + bh + hl)$$

$$2(lb + bh + hl) = 361 - 125 = 236$$

Surface area of cuboid = $$2(lb + bh + hl) = 236 cm^{2}$$

Base radius of cone = AQ = BQ = 5 cm

AB = 2AQ = 10 cm

Height of cone = PQ = 12 cm

Let QY = x and OQ = 2x

MN = XY = 2x

Now, 

$$\frac{MN}{AB} = \frac{PO}{PQ}$$

$$\frac{2x}{10} = \frac{12 - 2x}{12}$$

6x = 30 - 5x

$$x = \frac{30}{11}$$

Ddiameter of the cylinder = XY = 2x = $$\frac{60}{11} cm$$

We will assume the Selling Price = 100x. This gives his profit = 25% of SP i.e. 25x as he makes 25% profit on the Selling Price.
Now, the Cost Price for him = SP - Profit = 75x.
This Selling Price is achieved when he gave 20% discount on the MP.
This means that 0.8*MP = SP ==> MP = 125x.
Also, the CP is 25% off of the List Price. It means 0.75*LP = CP ==> LP = 100x.
MP is 125x and LP is 100x which means that MP is 125% of LP.

ABC is sector that is cut from a circle of diameter 12 cm. 

So, AB = AC = $$\frac{12}{2} = 6 cm$$

Let AO = R = Radius of circle that circumscribed the sector. 

OM is perpendicular to AM. AM = $$\frac{AB}{2} = \frac{6}{2} = 3 cm$$

Since, $$\angle{BAC} = \theta, then \angle{BAO} = \frac{\theta}{2}$$

Now, 

In $$\triangle{AMO}:$$

$$sec\frac{\theta}{2} = \frac{AO}{AM}$$

$$sec\frac{\theta}{2} = \frac{R}{3}$$

$$R = 3sec\frac{\theta}{2}$$

River is flowing from A to C, that means A to C is downstream and C to A is upstream. 

Since, Time taken to go from A to C in downstream = 7 hours

Time taken to go from A to B in downstream = $$\frac{7}{2} = 3.5 hours$$

Since, time taken to go from A to B and come back to the same point = 5 hours 15 minutes = $$5 + \frac{15}{60} = 5.25 hours$$

Now, time taken to go from B to A = 5.25 - 3.5 = 1.75 hours

Hence, time taken to go from C to A = 1.75 * 2 = 3.5 hours = $$3\frac{1}{2} hours$$

In A, 40% of students scored above 60% marks, the number of them being equal to 1160. Hence, the total number of students is 2900.

In B, 30% of students scored above 60% marks, the number of them being equal to 1170. Hence, the total number of students is 3900.

In C, 50% of students scored above 60% marks, the number of them being equal to 1170. Hence, the total number of students is 2340.

In D, 15% of students scored above 60% marks, the number of them being equal to 1200. Hence, the total number of students is 8000.

In E, 40% of students scored above 60% marks, the number of them being equal to 1600. Hence, the total number of students is 4000.

Hence, D has the largest number of students.

In A, 40% of students scored above 60% marks, the number of them being equal to 1160. Hence, the total number of students is 2900.

In B, 30% of students scored above 60% marks, the number of them being equal to 1170. Hence, the total number of students is 3900.

In C, 50% of students scored above 60% marks, the number of them being equal to 1170. Hence, the total number of students is 2340.

In D, 15% of students scored above 60% marks, the number of them being equal to 1200. Hence, the total number of students is 8000.

In E, 40% of students scored above 60% marks, the number of them being equal to 1600. Hence, the total number of students is 4000.

The number of students scored less than or equal to 60% from A is 1740.

The number of students scored less than or equal to 60% from B is 2730.

The number of students scored less than or equal to 60% from C is 1170.

The number of students scored less than or equal to 60% from D is 6800.

The number of students scored less than or equal to 60% from A is 2400.

The sum of these numbers is 14840.

The average will be 14840/5=2986.

So, option B is the answer.

In A, 40% of students scored above 60% marks, the number of them being equal to 1160. Hence, the total number of students is 2900.

In B, 30% of students scored above 60% marks, the number of them being equal to 1170. Hence, the total number of students is 3900.

In C, 50% of students scored above 60% marks, the number of them being equal to 1170. Hence, the total number of students is 2340.

In D, 15% of students scored above 60% marks, the number of them being equal to 1200. Hence, the total number of students is 8000.

In E, 40% of students scored above 60% marks, the number of them being equal to 1600. Hence, the total number of students is 4000.

The number of students of schools A and E together = 2900+4000=6900.

The number of students of school D is more than the number of students of schools A and E together = 8000-6900=1100.

The percentage more is by $$\frac{1100}{6900}\cdot100=15.9$$

So, option A is the answer.

Number of students of school B scoring marks less than or equal to 60% = $$1170 * \frac{70}{30} = 2730$$

Number of students of school C scoring marks less than or equal to 60% = $$1170 * \frac{50}{50} = 1170$$

Number of students of schools A and D together scoring marks above 60% = 1160 + 1200 = 2360

Required ratio = (2730 + 1170): 2360 = 3900: 2360 = 

Production of scooters in 2002 = 140

Production of scooters in 2003 = 80

Drop in number of production of scooters in 2003 from 2002 = 140 - 80 = 60

Required percent = (60/140) * 100 = 42.9% (Approx)

Production of motorbikes in 2003 = 130

Pproduction of scooters in 2003 = 80

Required percent = (130/80) * 100 = 162.5%

Average production of motorbikes per year during the period 2000 to 2004 = $$\frac{140 + 120 + 100 + 130 + 110}{5} = 120$$

Hence, in the year 2001 production of motorbikes was equal to the average production of motorbikes per year during the period 2000 to 2004.

Number of scooters produced during the period 2000 to 2003 = 120 + 100 + 140 + 80 = 440

Number of motorbikes produced during the period 2000 to 2003 = 140 + 120 + 100 + 130 = 490

Required ratio = 440: 490 = 44: 49

Although it is given that two unfortunate incidents happened with Mr.A, we cannot say with full conviction that misfortune 'never' comes alone. The word never makes it an over-stretched statement. Hence, the conclusion is a long drawn one. 

It is given that magazines 'often' have reader's column, and a group of friends will launch a magazine. Hence, they may or may not have a reader's column. Hence, we don't have enough data to substantiate the column. 

Since both statements show that certain unachievable/seemingly impossible things were done by people having strong faith, we can necessarily deduce that faith has great power. This is because the seemingly impossible tasks clearly had the support of immense faith. 

The first statement is not something that can be proven. Also, the conclusion is again a superstitious statement. Hence, we cannot be so sure of ba conclusion which is superstitious and cannot be deduced from a logical statement. This makes the conclusion a long drawn one. 

The first statement talks about negligence playing a role in 'most' of the accidents. The second statement describes how a cobbler was hit by a bus. However, we cannot deduce that the cobbler was negligent in this case, as not all accidents happen due to negligence. Hence, this is a long-drawn one. 

Hair is on the head, and similarly, teeth are inside the mouth. 

Fore and hind are totally opposite in nature (antonyms). Similarly, north and south are antonyms. 

The relationship in the pair Rogue : Rascal is one of synonyms, as both words describe a dishonest or unprincipled person. We must find the option that shares this synonym relationship:

Option A: Murderer : Cruelty — This is a Characteristic relationship; cruelty is a trait of a murderer, but they are not synonyms.

Option B: Polite : Harsh — This is an Antonym relationship; the words have opposite meanings.

Option C: Spendthrift : Extravagant — This is a Synonym relationship; both words describe someone who spends money in a reckless or wasteful manner.

Option D: Notorious : Famous — While both mean well-known, they have opposite connotations (negative vs. positive), so they are not true synonyms in this context.
Hence the correct answer is Option C.

The similar relationship is in Option C i.e. Pink : Blue.

As a watch is used to measure time, the barometer is used to measure pressure. Option B is close yet incorrect, as lactometer is used to measure milk's density. 

Statement 1: R is taller than P. This statement alone is not sufficient to answer the question.

Statement 2: Q is taller than P. This statement alone is not sufficient to answer the question.

Combining the statements, we get both Q and R to be taller than P. Hence, P is the shortest.

So, the correct choice is Option (D).

Suppose A's salary is 4x and B's salary is 3x.

Statement 1: B's salary is 75% of that of A's salary. This statement is the same as the original statement since 75% equals 3/4. So, we dont get any valuable iformation from this statement.

Statement 2: B's salary is 4500 Rs.

Combining with original statement, 3x = 4500

Therefore, x = 1500

And 4x = 6000.

Hence, we get A's salary as Rs 6000. 

So, the second statement is sufficient to get A's salary. Hence Option (B).

Statement 1: Total weight of the books on the table is 130 kg. 

This does not tell us anything on the number of books. 

Statement 2: Total weight of table is 130 kg.

Again, this statement does not tell us anything on the number of books.

Combining the 2 statements, also does not give us any information on the number of books.

Hence, Option C.

lee pee tin - Always Keep Smiling.

Statement 1: tin lut lee - Always Keep left

Combing the code with the given statement, tin and lee stand for Always and keep, hence, pee should stand for Smiling. Thus, this statement alone is sufficient to get the code for Smiling.

Statement 2: dee pee - Rose smiling

Combining the code with the given statement, dee stands for Rose and pee stands for smiling. Thus, this statement also alone is sufficient to get the code for Smiling.

Hence Option (A).

If we know that X is 22 years old during September, 1993, we can calculate his present age by adding up the number of years from 1993 to present. Hence statement 2 is sufficient to get X's present age. Statement 1 mentions the month but not the year, so that statement a;on is not sufficient to give us X's age. 

The statement says: A further confirmation by the Finance Minister that the prices of essential items would not be raised was welcome feature for the public.

Implication 1: He proposes to raise prices of luxury items. The statement says that the prices of essential items would not be raised, but it does not mention anything about the price of the luxury items. Hence it is not an implicit implication.

Implication 2: Some people had doubted the Finance Minister's announcement regarding the stability of prices of essential items. The statement does not say anything about the people's reaction to the Finance Minister's announcement on the stability of price of essential items. So, we are not sure if they doubted the announcement. Hence this is also not an implicit implication.

Hence, option (C).

The statement says: Prayers are sometimes answered when there is no chance or hope.

Implication 1:Some times, we do get helped miraculously. This is an implicit implication because, it is a direct implication of the statement that the prayers are answered miraculously when there is no chance or hope.

Implication 2:God shows favour only when we are in absolute despair. This is not an implicit implication because the statement describes the scenario when there is no chance or hope, but it never says that prayers are not answered otherwise. Prayers may be answered even when we are not in absolute despair or hopeless.

Hence, option (A).

The statement says: Virtually all bosses are problem bosses in one way or the other.

Implication 1: No employee other than boss creates problems. This statement says only about bosses. It does not mention anything about employees other than bosses. So, this is not an implicit implication.  

Implication 2: Boss and employees never have coordination. This statement is not related to the given statement in any way. So, this is also not an implicit implication.

Hence, option(C).

The statement says: First defeat often becomes a key to victory for some players.

Implication 1: After first defeat they come to know where they stand and what they have to achieve. This gives us a reason why defeat helps some players to be victorious for the future. So, this is an implicit implication.

Implication 2: After first defeat, every player is bound to win. This statement does not provide any reason as to why defeat becomes a key to victory for some players. So this is not an implicit implication.

Hence, Option (A).

The statement says: Road and other areas around hospitals are marked as silence zone.

Implication 1: Hospitals do not have sound-proof doors. If hospitals did have sound-proof doors, the areas did not need to be marked as silence zone. So this implication is implicit.

Implication 2: Noisy vehicles disturb patients. Again, if noisy vehicles did not disturb patients, they don't need to be marked as silence zones. So this implication is implicit as well.

Hence, Option (D).

Considering the block of logic, and substituting words by their position in the statement, i.e Original statement (You do things to help others in difficulty) -> 12345678

Step 1: 21645387

Step 2: 26154837

Step 3: 26518437

Step 4: 62581473

...

For getting step 7: We impose step 3 transformation on the step-4 transformed word 62581473.

Let's suppose the Step-4 transformed word is A, an the Step-7 transformed word is B.

Therefore applying Step-3 transformation on A will give us B.

A = 62581473

B = 26518437 transformation on A = [2 index of A][6 INDEX OF A][5 index of A][1 index OF A][8 index of A][4 index of A][3 index of A][7 index of A] = 24163857.

Substituting the words in the given statement 'why are you not in contact with me' to Step-2 word -> 26154837

2 why

6  are

1 you

5  not

4  in

8 contact

3  with

7 me.

Now 7th Step word = 24163857 = Why in you are with contact not me.

Hence, Option C.

Considering the block of logic, and substituting words by their position in the statement, i.e Original statement (You do things to help others in difficulty) -> 12345678

Step 1: 21645387

Step 2: 26154837

Step 3: 26518437

Step 4: 62581473

...

For getting step 6: We impose step 2 transformation on the step-4 transformed word 62581473.

Let's suppose the Step-4 transformed word is A, an the Step-6 transformed word is B.

Therefore applying Step-2 transformation on A will give us B.

A = 62581473

B = 26154837 transformation on A = [2 index of A][6 INDEX OF A][1 index of A][5 index OF A][4 index of A][8 index of A][3 index of A][7 index of A] = 24618357.

Therefore the word after 6th step = B = Above is the message to you from me = 24618357.

2 Above

4 is 

6 the

1 message

8 to 

3 you

5 from

7 me

Word after 3rd step = 26518437 = Above the from message to is you me.

Hence, Option B.

Considering the block of logic, and substituting words by their position in the statement, i.e Original statement (You do things to help others in difficulty) -> 12345678

Step 1: 21645387

Step 2: 26154837

Step 3: 26518437

Step 4: 62581473

...

Consider the statement given in the question, I am talking about your Delhi Patna Agra.

Assigning index to each word in their order in the statement:

I - 1

am - 2

talking - 3  

about - 4 

your - 5

Delhi - 6

Patna - 7

Agra - 8

The second statement is About am I talking Delhi Agra Patna you. Writing in the form of indices of the words in the original statement, it becomes 42136875.

Step 8 arrangement gives us the same.

For getting step 8: We impose step 4 transformation on the step-4 transformed word 62581473.

Let's suppose the Step-4 transformed word is A, an the Step-8 transformed word is B.

Therefore applying Step-4 transformation on A will give us B.

A = 62581473

B = 62581473 transformation on A = [6 index of A][2 INDEX OF A][5 index of A][8 index OF A][1 index of A][4 index of A][7 index of A][2 index of A] = 42136875.

Hence the right answer is step VIII.

Considering the block of logic, and substituting words by their position in the statement, i.e Original statement (You do things to help others in difficulty) -> 12345678

Step 1: 21645387

Step 2: 26154837

Step 3: 26518437

Step 4: 62581473

...

For getting step 7: We impose step 3 transformation on the step-4 transformed word 62581473.

Let's suppose the Step-4 transformed word is A, an the Step-7 transformed word is B.

Therefore applying Step-3 transformation on A will give us B.

A = 62581473

B = 26518437 transformation on A = [2 index of A][6 INDEX OF A][5 index of A][1 index OF A][8 index of A][4 index of A][3 index of A][7 index of A] = 24163857.

Therefore the word after 7th step = B = Serious I the am to about you conversation

Considering the block of logic, and substituting words by their position in the statement, i.e Original statement (You do things to help others in difficulty) -> 12345678

Step 1: 21645387

Step 2: 26154837

Step 3: 26518437

Step 4: 62581473

...

For getting step 8: We impose step 4 transformation on the step-4 transformed word 62581473.

Let's suppose the Step-4 transformed word is A, an the Step-8 transformed word is B.

Therefore applying Step-4 transformation on A will give us B. 

A = 62581473

B = 62581473 transformation on A = [6 index of A][2 INDEX OF A][5 index of A][8 index OF A][1 index of A][4 index of A][7 index of A][2 index of A] = 42136875.

Therefore the word after 8th step = B =  For am I waiting reply future in your

Statement 1. A$$\le$$K

Statement 2. N<A

Statement 3. M=N

Statement 4. R>M

Hence R>M=N<A$$\le$$K

Conclusion 1. K>N True

Conclusion 2. N<R True

Hence, option D.

Statement 1. A > T

Statement 2. R$$\ge$$A

Statement 3. D=R

Statement 4. S<D

Therefore: S<D=R$$\ge$$A>T

Conclusions:

1. D=A May/may not be true (Can't be definitely termed as true)

2. D>A May/may not be true (Can't be definitely termed as true)

Hence, Option C

E + E means E is equal to E.
H $$\times\ $$ F means H is smaller than F.
O - E means O is greater than E and,
H = G means H is smaller than or equal to G.

Conclusion 1: O - H means O > H. With the given statements, we can not be certain of this conclusion.
Conclusion 2: G x F means G is smaller than F. H is smaller than both F and G but we can not be certain of the values of F and G so this conclusion also does not follow.
Thus, the answer is Option C.

R x S -> R$$<$$S

D - E -> D$$>$$E

G $$\div$$ S -> G$$\ge$$S

E + G -> E$$=$$G

Hence, D$$>$$E$$=$$G$$\ge$$S$$>$$R

Conclusion: 1. R$$<$$D True

2. D$$>$$G True

Hence Option D

M = W -> M $$\le\ $$ W

B + A -> B = A

M + A -> M = A

G - W -> G $$>$$ W

Hence, A = B = M $$\le$$ W $$<$$ G

Conclusions:

1. B+M -> B=M -> True

2. M-B -> M$$>$$B -> False

Hence, Option A

Lets use the following symbols for male and female: Male - ' and Female - ".

For the order of intelligence, lets rank them from 1 to 7, where 1 denotes most intelligent and 7 denotes the least intelligent.

From hint (3), we get G is a female and the most intelligent. Therefore, G"(1).

Using hint (4) and (7), we get that B is the father of E, and F is the grandfather of E, and E is male. Also, e get E as the least intelligent and F as the second most intelligent. So, we can map them as

Using hint (10), we get that D is another son of F. Adding it to our mapping, wet get,

Using hint (2) which says there are 2 married couple, and hint (11) that says there are no widow or widower, we get to know the married couples are F and his wife and B and his wife.

Using hint (5), we get that C has one son and one daughter, so C can not be F's wife, because he has 2 sons, so, C is the wife of B. Also, B and C have a daughter. Mapping into the existing solution, we get:

Now, we have 2 people remaining, both are females. One of them is the daughter of B and C, and the other one is the wife of F. Using hint (9) that none of he married topped the IQ test, and using hint (3) that G is the topper of the IQ Test, we get G as the daughter of B and C, and the remaining person A is the wife of F. Mapping into the figure, we get,

From hint (10), D is more intelligent than B, but less intelligent than C. Also, using (8), B is more intelligent than A. Using hint (5), C is more intelligent than B.

We know that G is ranked 1, F is ranked 2, E is ranked 7. Using the remaining ranks and applying them to the deduction from the previous line, we get, C as 3, D as 4, B as 5 and A as 6. Hence the figure comes out as:

So, this is our final figure, depicting all relationships and the intelligence test ranks.

From the fig., A is married to F.

Hence, option (A).

Lets use the following symbols for male and female: Male - ' and Female - ".

For the order of intelligence, lets rank them from 1 to 7, where 1 denotes most intelligent and 7 denotes the least intelligent.

From hint (3), we get G is a female and the most intelligent. Therefore, G"(1).

Using hint (4) and (7), we get that B is the father of E, and F is the grandfather of E, and E is male. Also, e get E as the least intelligent and F as the second most intelligent. So, we can map them as

Using hint (10), we get that D is another son of F. Adding it to our mapping, wet get,

Using hint (2) which says there are 2 married couple, and hint (11) that says there are no widow or widower, we get to know the married couples are F and his wife and B and his wife.

Using hint (5), we get that C has one son and one daughter, so C can not be F's wife, because he has 2 sons, so, C is the wife of B. Also, B and C have a daughter. Mapping into the existing solution, we get:

Now, we have 2 people remaining, both are females. One of them is the daughter of B and C, and the other one is the wife of F. Using hint (9) that none of he married topped the IQ test, and using hint (3) that G is the topper of the IQ Test, we get G as the daughter of B and C, and the remaining person A is the wife of F. Mapping into the figure, we get,

From hint (10), D is more intelligent than B, but less intelligent than C. Also, using (8), B is more intelligent than A. Using hint (5), C is more intelligent than B.

We know that G is ranked 1, F is ranked 2, E is ranked 7. Using the remaining ranks and applying them to the deduction from the previous line, we get, C as 3, D as 4, B as 5 and A as 6. Hence the figure comes out as:

So, this is our final figure, depicting all relationships and the intelligence test ranks.

From the figure, G is the daughter of D's brother, hence niece.

Option (D). 

Lets use the following symbols for male and female: Male - ' and Female - ".

For the order of intelligence, lets rank them from 1 to 7, where 1 denotes most intelligent and 7 denotes the least intelligent.

From hint (3), we get G is a female and the most intelligent. Therefore, G"(1).

Using hint (4) and (7), we get that B is the father of E, and F is the grandfather of E, and E is male. Also, e get E as the least intelligent and F as the second most intelligent. So, we can map them as

Using hint (10), we get that D is another son of F. Adding it to our mapping, wet get,

Using hint (2) which says there are 2 married couple, and hint (11) that says there are no widow or widower, we get to know the married couples are F and his wife and B and his wife.

Using hint (5), we get that C has one son and one daughter, so C can not be F's wife, because he has 2 sons, so, C is the wife of B. Also, B and C have a daughter. Mapping into the existing solution, we get:

Now, we have 2 people remaining, both are females. One of them is the daughter of B and C, and the other one is the wife of F. Using hint (9) that none of he married topped the IQ test, and using hint (3) that G is the topper of the IQ Test, we get G as the daughter of B and C, and the remaining person A is the wife of F. Mapping into the figure, we get,

From hint (10), D is more intelligent than B, but less intelligent than C. Also, using (8), B is more intelligent than A. Using hint (5), C is more intelligent than B.

We know that G is ranked 1, F is ranked 2, E is ranked 7. Using the remaining ranks and applying them to the deduction from the previous line, we get, C as 3, D as 4, B as 5 and A as 6. Hence the figure comes out as:

So, this is our final figure, depicting all relationships and the intelligence test ranks.

From the diagram, we can deduce that C is the third most intelligent.

Hence Option (A).

Lets use the following symbols for male and female: Male - ' and Female - ".

For the order of intelligence, lets rank them from 1 to 7, where 1 denotes most intelligent and 7 denotes the least intelligent.

From hint (3), we get G is a female and the most intelligent. Therefore, G"(1).

Using hint (4) and (7), we get that B is the father of E, and F is the grandfather of E, and E is male. Also, e get E as the least intelligent and F as the second most intelligent. So, we can map them as

Using hint (10), we get that D is another son of F. Adding it to our mapping, wet get,

Using hint (2) which says there are 2 married couple, and hint (11) that says there are no widow or widower, we get to know the married couples are F and his wife and B and his wife.

Using hint (5), we get that C has one son and one daughter, so C can not be F's wife, because he has 2 sons, so, C is the wife of B. Also, B and C have a daughter. Mapping into the existing solution, we get:

Now, we have 2 people remaining, both are females. One of them is the daughter of B and C, and the other one is the wife of F. Using hint (9) that none of he married topped the IQ test, and using hint (3) that G is the topper of the IQ Test, we get G as the daughter of B and C, and the remaining person A is the wife of F. Mapping into the figure, we get,

From hint (10), D is more intelligent than B, but less intelligent than C. Also, using (8), B is more intelligent than A. Using hint (5), C is more intelligent than B.

We know that G is ranked 1, F is ranked 2, E is ranked 7. Using the remaining ranks and applying them to the deduction from the previous line, we get, C as 3, D as 4, B as 5 and A as 6. Hence the figure comes out as:

So, this is our final figure, depicting all relationships and the intelligence test ranks.

C is A's Daughter-In-Law, whereas A is C's mother-in-law.

Hence Option (A).

Lets use the following symbols for male and female: Male - ' and Female - ".

For the order of intelligence, lets rank them from 1 to 7, where 1 denotes most intelligent and 7 denotes the least intelligent.

From hint (3), we get G is a female and the most intelligent. Therefore, G"(1).

Using hint (4) and (7), we get that B is the father of E, and F is the grandfather of E, and E is male. Also, e get E as the least intelligent and F as the second most intelligent. So, we can map them as

Using hint (10), we get that D is another son of F. Adding it to our mapping, wet get,

Using hint (2) which says there are 2 married couple, and hint (11) that says there are no widow or widower, we get to know the married couples are F and his wife and B and his wife.

Using hint (5), we get that C has one son and one daughter, so C can not be F's wife, because he has 2 sons, so, C is the wife of B. Also, B and C have a daughter. Mapping into the existing solution, we get:

Now, we have 2 people remaining, both are females. One of them is the daughter of B and C, and the other one is the wife of F. Using hint (9) that none of he married topped the IQ test, and using hint (3) that G is the topper of the IQ Test, we get G as the daughter of B and C, and the remaining person A is the wife of F. Mapping into the figure, we get,

From hint (10), D is more intelligent than B, but less intelligent than C. Also, using (8), B is more intelligent than A. Using hint (5), C is more intelligent than B.

We know that G is ranked 1, F is ranked 2, E is ranked 7. Using the remaining ranks and applying them to the deduction from the previous line, we get, C as 3, D as 4, B as 5 and A as 6. Hence the figure comes out as:

So, this is our final figure, depicting all relationships and the intelligence test ranks.

Hence, A is of  different generation than B,C and D.

Hence, option (D).

Lets use the following symbols for male and female: Male - ' and Female - ".

For the order of intelligence, lets rank them from 1 to 7, where 1 denotes most intelligent and 7 denotes the least intelligent.

From hint (3), we get G is a female and the most intelligent. Therefore, G"(1).

Using hint (4) and (7), we get that B is the father of E, and F is the grandfather of E, and E is male. Also, e get E as the least intelligent and F as the second most intelligent. So, we can map them as

Using hint (10), we get that D is another son of F. Adding it to our mapping, wet get,

Using hint (2) which says there are 2 married couple, and hint (11) that says there are no widow or widower, we get to know the married couples are F and his wife and B and his wife.

Using hint (5), we get that C has one son and one daughter, so C can not be F's wife, because he has 2 sons, so, C is the wife of B. Also, B and C have a daughter. Mapping into the existing solution, we get:

Now, we have 2 people remaining, both are females. One of them is the daughter of B and C, and the other one is the wife of F. Using hint (9) that none of he married topped the IQ test, and using hint (3) that G is the topper of the IQ Test, we get G as the daughter of B and C, and the remaining person A is the wife of F. Mapping into the figure, we get,

From hint (10), D is more intelligent than B, but less intelligent than C. Also, using (8), B is more intelligent than A. Using hint (5), C is more intelligent than B.

We know that G is ranked 1, F is ranked 2, E is ranked 7. Using the remaining ranks and applying them to the deduction from the previous line, we get, C as 3, D as 4, B as 5 and A as 6. Hence the figure comes out as:

So, this is our final figure, depicting all relationships and the intelligence test ranks.

Hence, A whose rank is 6, is less intelligent than all but E whose rank is 7.

Option (A).

Lets use the following symbols for male and female: Male - ' and Female - ". 

For the order of intelligence, lets rank them from 1 to 7, where 1 denotes most intelligent and 7 denotes the least intelligent.

From hint (3), we get G is a female and the most intelligent. Therefore, G"(1).

Using hint (4) and (7), we get that B is the father of E, and F is the grandfather of E, and E is male. Also, e get E as the least intelligent and F as the second most intelligent. So, we can map them as 

Using hint (10), we get that D is another son of F. Adding it to our mapping, wet get,

Using hint (2) which says there are 2 married couple, and hint (11) that says there are no widow or widower, we get to know the married couples are F and his wife and B and his wife.

Using hint (5), we get that C has one son and one daughter, so C can not be F's wife, because he has 2 sons, so, C is the wife of B. Also, B and C have a daughter. Mapping into the existing solution, we get:

Now, we have 2 people remaining, both are females. One of them is the daughter of B and C, and the other one is the wife of F. Using hint (9) that none of he married topped the IQ test, and using hint (3) that G is the topper of the IQ Test, we get G as the daughter of B and C, and the remaining person A is the wife of F. Mapping into the figure, we get,

From hint (10), D is more intelligent than B, but less intelligent than C. Also, using (8), B is more intelligent than A. Using hint (5), C is more intelligent than B.

We know that G is ranked 1, F is ranked 2, E is ranked 7. Using the remaining ranks and applying them to the deduction from the previous line, we get, C as 3, D as 4, B as 5 and A as 6. Hence the figure comes out as:

So, this is our final figure, depicting all relationships and the intelligence test ranks.

Hence, B is married to C.

Option (C).

Starting the sequence with abc. For the next set of terms, we use the original set (i.e abc) after incluing an additional a.

So, the next set of terms becomes aabc.

Similarly for the next set of terms, we add b to the previous term in the right place. Hence it becomes aabbc.

In a similar way, the next set of terms become aabbcc.

So, the final sequence comes out as abcaabcaabbcaabbcca. 

We start the series with a,b,c,d as the first 4 terms.

The next 4 terms can be derived from the first 4 in the following way.

1. Place the last term(i.e d) as the first term of the new series.

2. The next terms of the new series are the next terms of the original series in a circular fashion (a,b,c,d,a,b,c,d....etc)

So the second set of terms is d,a,b,c.

In a similar way, the subsequent terms should be c,d,a,b and b,c,d,a.

The final series is a,b,c,d,d,a,b,c,c,d,a,b,b,c,d,a.

We start the series with a,b,c as the first 3 terms.

For the next 3 terms, we start with the last term i.e 'c' of the previous series(a,b,c in his case), and continue with the next 2 terms in circular fashion (a,b,c,a,b,c). So the second set of 3 terms comes out as c,a,b.

Continuing in a similar way, the next 3 terms are b,c,a, and the subsequent terms are a,b,c.

Hence, the final series turns out as a,b,c,c,a,b,b,c,a,a,b,c.