Given that CD is  the diameter = 22 cm.
CE = 7, hence ED = 15 cm

In the circle, we know that, AExEB = CExED
Thus xy = 7×15 = 105 and x + y = 20.5
Now (x – y)² = (x + y)² – 4xy = 20.5² – 4×105
or (x – y)² = 0.25
Thus difference of AE and EB = 0.5

Let us first calculate the number of paths from a point \((x_1, y_1)\) to \((x_2, y_2)\). Suppose A is (0, 0) and B is (4, 3). From A to B, there are 7 steps, 4 horizontal and 3 vertical. To reach from A to B, a total of 7 steps have to be taken, out of which 4 are horizontal and 3 are vertical steps. Among 4 horizontal steps and among 3 vertical steps order is not important. If we select any 4 horizontal steps out of total 7 steps, remaining 3 steps are automatically selected. Hence the number of ways = \(^{\rm{7}}{{\rm{C}}_{\rm{4}}}{\rm{  = }}{{\rm{ }}^{\rm{7}}}{{\rm{C}}_{\rm{3}}}\) = 35.

In general number of paths from the point \((x_1, y_1)\) to \((x_2, y_2)\) is \(^{({x_2} - {x_1} + {y_2} - {y_1})}{{\rm{C}}_{({x_2} - {x_1})}}\)

In this question suppose A =  (1,1), B = (8,10) and C = (4,6)

Number of horizontal and vertical steps from A to C are 3 and 5, thus the number of paths from A to C = \(^{\rm{8}}{{\rm{C}}_{\rm{3}}}{\rm{ }}\) = 56.

Number of horizontal and vertical steps from C to B are 4 and 4, thus the number of paths from C to B = \(^{\rm{8}}{{\rm{C}}_{\rm{4}}}{\rm{ }}\) = 70.

Therefore number of paths from A to B via C = 56×70 = 3920

We know that if  \({a^x} = b\), then \(a = {b^{1/x}}\), here

\({5.55^x} = 1000 \Rightarrow 5.55 = {1000^{1/x}}\)

\({0.555^y} = 1000 \Rightarrow 0.555 = {1000^{1/y}}\)

Dividing the first equation by second, we have

\(\frac{{5.55}}{{0.555}} = {1000^{\frac{1}{x} - \frac{1}{y}}}\)

\( \Rightarrow 10 = {1000^{\frac{1}{x} - \frac{1}{y}}}\)

Hence \(\cfrac{1}{x} - \cfrac{1}{y} = \frac{1}{3}\).

Suppose number of players of Football, Tennis and Cricket are represented by n(F), n(T) and n(C), then
n(F) = 144, n(T) = 123 and n(C) = 132
Also n(FnT) = 58, n(FnC) = 63 and n(TnC) = 25

Now n(FUTUC) = n(F) + n(T) + n(C) – n(FnT) – n(FnC) – n(TnC) + n(FnTnC)
or 256 = 144 + 123 + 132 – 58 – 63 – 25 + n(FnTnC)
or n(FnTnC) = 3
Thus number of members who can play only tennis = 123 – 55 – 3 – 22 = 43

Suppose total marks in the examination = T and pass marks = x, then
40% of T + 50% of 40% of T = x – 35
=> 0.60T = x – 35 .......(1)
Again marks are increased by 20%, i.e. 0.60T + 20% of 0.60T = 0.72T
=> 0.72T = x + 7 .........(2)
From equations (1) and (2), 0.12T = 42 or T = 350.
From the equation (1), x = 0.60T + 35 = 60% of T + 35
Note that 35 is 10% of T, thus x = 60% of T + 10% of T = 70% of T

We know that each exterior angle of any regular hexagon = 60°. Thus \(\angle {\rm{UPA}} = \angle {\rm{PUA}} = \angle {\rm{PAU}} = {60^ \circ }\) so AP = PU = UA

Similarly it can be shown easily that angle CTS and angle QRB are also equilateral. Area of the hexagon, H = \(6 \times \frac{{\sqrt 3 }}{4}{x^2}\)
Area of the equilateral triangle ABC, T = \(\frac{{\sqrt 3 }}{4}{(3x)^2}\)
Ratio H : T = 6 : 9 = 2: 3

The line 3x + 5y = 45 cuts the x axis at A (15, 0) and y axis at B(0, 9). Since the triangle AOB is a right angled, so AB is the diameter of its circum-circle. 

Diameter of the circle, AB = \(\sqrt{15^2+9^2}=\sqrt{306}\) we know that \(\sqrt{306}\) lies between 17 and 18, so the radius lies between 8.5 to 9. The value of radius closest to the nearest integer = 9

There are two possible cases:
Case (i) \(m\) is even, then \(m + 1\) will be odd and
\(f(m) = m(m + 1) f(m + 1) = m + 1 + 3 = m + 4\)
Using the relation \(8f(m + 1) - f(m) = 2\), we have
\(8(m + 4) – m(m + 1) = 2\)
=> \(m^2 – 7m – 30 = 0\)  or \((m – 10)(m + 3) = 0\), or \(m = 10\) as \(m\) is positive. 
Case (ii) \(m\) is odd, then \(m + 1\) will be even and \(f(m) = m + 3\)\(f(m + 1) = (m + 1)(m + 2) = {m^2} + 3m + 2\).
Using the relation \(8f(m + 1) - f(m) = 2\), we have
\(8({m^2} + 3m + 2) - (m + 3) = 2\)
\( \Rightarrow 8{m^2} + 23m + 14 = 0\)
Above equation will not have positive roots as all the signs are positive. Thus value of \(m\) is unacceptable. Hence m can have only one value i.e.  \(m = 10\)

Given that initial population (in 2019) = 1000
Population after 1 year = 2×1000 + 3
Population after 2 years = 2×(2×1000 + 3) + 3 = 2²×1000 + 9
Population after 3 years = 2×(2²×1000 + 9) + 3 = 2³×1000 + 21
It is clear that population after 15 years (in 2034) will have a factor of 2¹⁵, so choice (3) can be correct. Let us write the population again in the format of the number given in the choice (3).
Population after 1 year = 2×1000 + 3 = 2×(1003 – 3) + 3 = 2×1003 – 3
Population after 2 years = 2²×1000 + 9 = 2²×(1003 – 3) + 9 = 2²×1003 – 3
Population after 3 years = 2³×1000 + 21 = 2³×(1003 – 3) + 21 = 2³×1003 – 3
clearly choice (3) is correct.

Suppose the amount invested in two other deposits = 2x and x (in lakhs), so the amount invested in fixed deposit = (15 – 3x) lakhs.
Total interest earned is (15 – 3x)×100000×0.06 + (2x)×100000×0.04 + (x)×100000×0.03 = 76000
=> 90 – 18x + 8x + 3x = 76 or x = 2 lakhs.
Hence amount invested in fixed deposit = 15 – 2x – x = 15 – 6 = 9 lakhs

Suppose the numbers are x and y, then
\(\cfrac{{{x^3} - {y^3}}}{{{{(x - y)}^3}}} = \cfrac{{157}}{3}\) => \(\cfrac{{(x - y)({x^2} + {y^2} + xy)}}{{{{(x - y)}^3}}} = \cfrac{{157}}{3}\) 
\(\cfrac{{({x^2} + {y^2} + xy)}}{{{{(x - y)}^2}}} = \cfrac{{157}}{3}\)
Lets change the expression in the form of (x + y), we have \(\cfrac{{{{(x + y)}^2} - xy}}{{{{(x + y)}^2} - 4xy}} = \cfrac{{157}}{3}\)
Put (x + y) = k,
 \( \Rightarrow \cfrac{{{k^2} - xy}}{{{k^2} - 4xy}} = \cfrac{{157}}{3}\) or \(154{k^2} = 625xy = 625 \times 616\) Thus k² = 625×4 or k = 50.

On selling a pen at 5% loss and a book at 15% gain, Karim gains Rs. 7. If he sells the pen at 5% gain and the book at 10% gain, he gains Rs. 13. What is the cost price of the book in Rupees?

Suppose the cost price of pen is 100p and that of a book is 100b, then

Suppose the time taken in the journey by the first and second car are (t + 1) and t.
Let the speeds of the cars are v1 and v2, and the length of the journey is d, then
\(\frac{d}{{{v_1}}} = t + 1\) and \(\frac{d}{{{v_2}}} = t\)
Dividing first equation by second, we have \(\frac{{{v_2}}}{{{v_1}}} = \frac{{t + 1}}{t} = 1 + \frac{1}{t}\)
Given that first car travelled for at least 6 hours, so t + 1 ≥ 6 or t ≥ 5.
When t is minimum, then the ratio \(\frac{{{v_2}}}{{{v_1}}}\) will be the largest, thus \(\frac{{{v_2}}}{{{v_1}}}\) = 1.20

Suppose work done per day by A and B are a and b, then 12(a + b) = total work = W
when A works half as efficiently as she usually does, and B works thrice as efficiently as he usually does, the work done in 9 days is 9(0.5a + 3b) = total work = W
As the total wok remains the same, hence 12a + 12 b = 4.5a + 27b => a = 2b and W = 36b = 18a
Time taken by A alone to finish the work = (18a)/a = 18 days

11th term can also be written as: \(({a_1} + {a_2} + {a_3} + ..... + {a_{11}}) - ({a_1} + {a_2} + {a_3} + ..... + {a_{10}})\) \( = 3({2^{12}} - 2) - 3({2^{11}} - 2)\) 

\( = 3({2^{12}} - {2^{11}}) = 3 \times {2^{11}}(2 - 1)\) = 6144

We know that for any two positive numbers, arithmetic mean ≥ Geometric Mean
\(\frac{{{2^x} + {2^{ - x}}}}{2} \ge \sqrt {{2^x}{2^{ - x}}} \)
\( \Rightarrow {2^x} + {2^{ - x}} \ge 2\)
The value 2 is obtained at x = 0 only. But the value of 2cos (x(x + 1) will always be less than or equal to 2. So the two values will be equal at x = 0 only.

Suppose length of the racecourse is x. In the same time first, second and third horses are covering the distances x, (x – 11) and (x – 90) m. Ratio of the speeds of the second and third horses = (x – 11) : (x – 90) Also when second horse covers x metres, in the same time the third covers (x – 80) m, so ratio of their speeds = x : (x – 80)

\(\frac{{x - 11}}{{x - 90}} = \frac{x}{{x - 80}}\) \( \Rightarrow {x^2} - 91x + 880 = {x^2} - 90x\)

Hence x = 880 m

We know that angle subtended by diameter at any point of the circumference is 90°, so the angles P and Q are 90°.

Now AP² = 10² – BP² = 10² – 6² = 64 or AP = 8 and AQ = 8/2 = 4

Now BQ² = 10² – 4² = 84 or BQ = √84 = 9.16 closest answer among the 4 choices is 9.1

Suppose total time taken by Amal is 3t, and as Amal took each mode of transport 1/3 of his total journey time, so the total journey is 10t + 20t + 30t = 60t
As Bimal took each mode of transport 1/3 of the total distance, total time taken by Bimal = (20t/10) + (20t/20) + (20t/30) = 3t + (2t/3)
Required percentage = \(\frac{{3t + \frac{{2t}}{3} - 3t}}{{3t}} \times 100 = 22.22\% \)
Choice (1) is the correct answer.

Interest earned is directly proportional to rate of interest × amount invested Thus ratio of the interest earned = (3×6) : (4×5) : (5×4) = 18 : 20 : 20
Suppose their interest incomes are 18x, 20x and 20x
Difference in the income Bina and Amla = 20x – 18x = 2x = 250
=> x = 125
So the total income = 18x + 20x + 20x = 58x = 58×125 = 7250.

Arranging powers of 2 and 3, we have
\({2^{\frac{{19}}{2}}} \cdot {3^4} \cdot {2^4} \cdot {3^{2m}} \cdot {2^{3n}} = {3^n} \cdot {2^{4m}} \cdot {2^{\frac{6}{4}}}\)
\( \Rightarrow {2^{\frac{{19}}{2} + 4 + 3n}} \cdot {3^{2m + 4}} = {3^n} \cdot {2^{\frac{6}{4} + 4m}}\)
Equating powers of 2 and 3, we have, \(n = 4 + 2m\) \(4m = 12 + 3n\)
Solving the above equations, m = – 12

Weight of liquid 1 per litre = 1000 g
Weight of liquid 2 per litre = 800 g
Weight of the mixture 1 per litre = 2×480 = 960 g
Using Alligation rule,

The ratio of liquid 1 and liquid 2 in the mixture = 4 : 1, so the liquid 1 is 80% of the mixture.

The first equaltion log₅(x + y) + log₅(x – y) = 3 can be written as log ₅(x + y)(x – y) = 3
=> x² – y² = 5³ = 125 ............(1)
and the second equation can be written as log ₂y – log₂x = log₂2 – log₂3
=> log₂(y/x) = log₂(2/3) or y/x = 2/3
Let x = 3k and y = 2k, from equation (1), 9k ² – 4k² = 125 or k² = 25
Now xy = (3k)(2k) = 6k² = 150

Suppose the lengths of the three diagonals are 3k, 2√3k and √15k, and the three sides are a, b and c, then a ² + b² = 9k²

b² +c² = 12k²

c² + a² = 15k² 

Adding the three equations, 2(a² + b² + c² ) = 36k²

=> a² + b² + c² = 18k²

Thus a² = 6k², b² = 3k² and c² = 9k²

Ratio of the shortest and the longest sides = 1 : √3

|x| + |y| = 2 represents 4 lines, x + y = 1, x – y = 1, – x – y = 1 and – x + y = 1. These four lines represent a square. |x| = 1 represents two straight lines x = 1 and x = – 1. The line x = 1 cuts the graph |x| +|y| = 2 at two points (1, 1) and (1, – 1), similarly the line x = –1 cuts the graph |x| +|y| = 2 at two points (–1, 1) and (–1, – 1) The region S is shown by the shaded part.

Area of the shaded part = 2×area of triangle
Area of one triangle = (1/2)×base×height = (1/2)×2×1 = 1
Area of the region S = 2

We know that |x| = x, when x > 0 and |x| = – x, when x < 0 x = 0 is a clear solution of the equation.

case 1: x > 0, |x| = x(6x² + 1) = 5x² => 6x² + 1 = 5x or (3x – 1)(2x – 1) = 0.  Thus x = 1/3 and x = 1/2

case 2: x < 0, |x| = – x(6x² + 1) = 5x² => 6x² + 1 = – 5x or (3x + 1)(2x + 1) = 0.  Thus x = – 1/3 and x = – 1/2 Hence there are 5 solutions.

Suppose one day work of a machine and a man are represented by M and m, then one day work by 3 men and 8 machines is double of 3 machines and 8 men (3m + 8M) = 2(3M + 8m)
=> M = (13/2) m
Work done by a machine in one day is 13/2 times that of a man.
Thus 2 machine can do the same work as 13 men do.

The expression |x² – x – 6| = |(x + 2)(x – 3)|
When x ≥ 3 or x ≤ – 2, |(x + 2)(x – 3)| = (x + 2)(x – 3)
So the equation |x² – x – 6| = x + 2 can be written as (x + 2)(x – 3) = (x + 2) => (x + 2)(x – 4) = 0
=> x = – 2 or x = 4
Both the solutions are in acceptable range.
When –2 ≤ x ≤ 3, |(x + 2)(x – 3)| = – (x + 2)(x – 3)
So the equation |x² – x – 6| = x + 2 can be written as – (x + 2)(x – 3) = (x + 2)
=> – (x + 2)(x – 2) = 0 => x = – 2 or x = 2
Both the solutions are in acceptable range.
Hence product of the roots = – 2×2×4 = – 16

Suppose each wheel of B requires n revolutions and each wheel of A requires n + 5000 revolutions, then 2π(0.4)×n = 2π(0.3)×(n + 5000)
=> n = 15000 Total distance = 2π(0.4)×15000 m = 12π km
This is covered in 45 m = 0.75 hours
Speed of B = \(\frac{{12\pi }}{{0.75}} = 16\pi \)

Given that f(1) = 2, thus f(2) = f(1 + 1) = f(1)f(1) = 2²

f(3) = f(1 + 2) = f(1)f(2) = 2×2² = 2³

f(4) = f(1 + 3) = f(1)f(3) = 2×2³ = 2⁴

So we can generalize as \(f(n) = 2n\)

Now we can calculate: f (a+1) = f(a)f(1) = 2f(a)

f (a+2) = f(a)f(2) = 2²f(a)

f (a+3) = f(a)f(2) = 2³f(a) and so on.

The sum f (a+1) + f (a+2) + ..... + f(a+n) will be

2f(a) + 2²f(a) + 2³f(a) + ......+2ⁿ f(a) = \(\frac{{2f(a)[{2^n} - 1]}}{{(2 - 1)}} = 2f(a)[{2^n} - 1]\)

But the sum is given as 16 (2ⁿ – 1) Hence 2f(a) = 16 or f(a) = 8 or a = 3.

Given that The average score of all the 22 students is 1 more than the average score of the 21 students other than Ramesh, it means average score increases by 1 when Ramesh is included. So score of Ramesh should be 22×1 more than the average of other 21.

Average of other 21 = 82.5 – 22 = 60.5

Average of all 22 is 60.5 + 1 = 61.5

The average score of the 21 students other than Gautam is 62, that mean average score decreases by 0.5 when Gautam is included, score of Gautam should be 22×0.5 less than the average of other 21 = 62 – 22×0.5 = 51. 

Alternate Method: Suppose score of Gautam is g and total score of 20 students other that Gautam and Ramesh is x, then 82.5 + x = 21×62 Solving these equations, g = 51.