CAT 2018 Question Paper Quant Slot 1

Givne that: $$2^{x}=3^{\log_{5}{2}}$$

$$\Rightarrow$$ $$2^{x}=2^{\log_{5}{3}}$$

$$\Rightarrow$$ $$x=\log_{5}{3}$$

$$\Rightarrow$$ $$x=\log_{5}{\dfrac{3*5}{5}}$$

$$\Rightarrow$$ $$x=\log_{5}{5}+\log_{5}{\dfrac{3}{5}}$$

$$\Rightarrow$$ $$x=1+\log_{5}{\dfrac{3}{5}}$$. Hence, option D is the correct answer. 

Given that two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm.

In the diagram we can see that AB = 6 cm, CD = 4 cm and MN = 1 cm.

We can see that M and N are the mid points of AB and CD respectively. 

AM = 3 cm and CD = 2 cm. Let 'OM' be x cm. 

In right angle triangle AMO,

$$AO^2 = AM^2 + OM^2$$ 

$$\Rightarrow$$ $$AO^2 = 3^2 + x^2$$  ... (1)

In right angle triangle CNO,

$$CO^2 = CN^2 + ON^2$$ 

$$\Rightarrow$$ $$CO^2 = 2^2 + (OM+MN)^2$$ 

$$\Rightarrow$$ $$CO^2 = 2^2 + (x+1)^2$$  ... (2)

We know that both AO and CO are the radius of the circle. Hence $$AO = CO$$

Therefore, we can equate equation (1) and (2)

$$3^2+x^2$$=$$2^2+(x+1)^2$$

$$\Rightarrow$$ x = 2 cm

Therefore, the radius of the circle 

$$AO = \sqrt{AM^2 + OM^2}$$ 

$$\Rightarrow$$ $$AO=\sqrt{3^2+2^2}=\sqrt{13}$$. Hence, option A is the correct answer. 

Let the efficiency of humans be 'h' and the efficiency of robots be 'r'.
In the first case, 
Total work = (15h + 5r) * 30......(i)
In the second case,
Total work = (5h + 15r) * 60......(ii)
On equating (i) and (ii), we get
(15h + 5r) * 30 = (5h + 15r) * 60
Or, 15h + 5r = 10h + 30r
Or, 5h = 25r
Or, h = 5r
Total work = (15h + 5r) * 30 = (15h + h) * 30 = 480h
Time taken by 15 humans = $$\dfrac{\text{480h}}{\text{15h}}$$ days= 32 days.
Hence, option C is the correct answer.

Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD?

We know that AC is the diameter and $$\angle$$ ABC = 90°. AC = 2*13 = 26 cm

In right angle triangle ABC,

$$AC^2 = AB^2 + BC^2$$

$$\Rightarrow$$ $$AB^2+BC^2=26^2$$

$$\Rightarrow$$ $$AB^2+BC^2=676$$

Let us check with the options.

Option (A): $$24^2+10^2 = 676$$. Hence, this is a possible answer. 

Option (B): $$25^2+9^2 = 706 \neq 676$$.  Hence, this is an incorrect pair.

Option (C): $$25^2+10^2 = 725 \neq 676$$.  Hence, this is an incorrect pair.

Option (D): $$24^2+12^2 = 720 \neq 676$$. Hence, this is an incorrect pair.

Therefore, we can say that option A is the correct answer.

It is given that EFGH is also a square whose area is 62.5% of that of ABCD. Let us assume that E divides AB in x : 1. Because of symmetry we can se that points F, G and H divide BC, CD and DA in x : 1.

Let us assume that 'x+1' is the length of side of square ABCD. 

Area of square ABCD = $$(x+1)^2$$ sq. units.

Therefore, area of square EFGH = $$\dfrac{62.5}{100}*(x+1)^2$$ = $$\dfrac{5(x+1)^2}{8}$$ ... (1)

In right angle triangle EBF,

$$EF^2 = EB^2 + BF^2$$

$$\Rightarrow$$ $$EF = \sqrt{1^2+x^2}$$

Therefore, the area of square EFGH = $$EF^2$$ = $$x^2+1$$ ... (2)

By equating (1) and (2),

$$x^2+1 = \dfrac{5(x+1)^2}{8}$$

$$\Rightarrow$$ $$8x^2+8 = 5x^2+10x+5$$

$$\Rightarrow$$ $$3x^2-10x+3 = 0$$

$$\Rightarrow$$ $$(x - 3)(3x - 1) = 0$$

$$\Rightarrow$$ $$x = 3$$ or $$1/3$$

The ratio of length of EB to that of CG = 1 : x 

EB : CG = 1 : 3 or 3 : 1. Hence, option D is the correct answer. 

Let the distance between A and B be 4x. 
Length of BP is thrice the length of AP.
=> AP = x and BP = 3x

Let the speed of car 1 be s and the speed of car 2 be 0.5s. 
Car 2 reaches P one hour (60 minutes) after Car 1 reaches P.

=> x/s + 60 = 3x/0.5s
x/s + 60 = 6x/s
5x/s = 60
x/s = 12

Time taken by car 1 in reaching P from A = x/s = 12 minutes.
Therefore, 12 is the correct answer.  

f(x) = min ($${2x^{2},52-5x}$$)
The maximum possible value of this function will be attained at the point in which $$2x^2$$ is equal to $$52-5x$$.

$$2x^2 = 52-5x$$
$$2x^2+5x-52=0$$
$$(2x+13)(x-4)=0$$
=> $$x=\frac{-13}{2}$$ or $$x = 4$$
It has been given that $$x$$ is a positive real number. Therefore, we can eliminate the case $$x=\frac{-13}{2}$$.
$$x=4$$ is the point at which the function attains the maximum value. $$4$$ is not the maximum value of the function. 
Substituting $$x=4$$ in the original function, we get, $$2x^2 = 2*4^2= 32$$.
f(x) = $$32$$.
Therefore, 32 is the right answer.

We know that one of the 3 numbers is 37.
Let the product of the other 2 numbers be x.
It has been given that 73x-37x = 720
36x = 720
x = 20

Product of 2 real numbers is 20. 
We have to find the minimum possible value of the sum of the squares of the 2 numbers. 
Let x=a*b
It has been given that a*b=20

The least possible sum for a given product is obtained when the numbers are as close to each other as possible. 
Therefore, when a=b, the value of a and b will be $$\sqrt{20}$$.

Sum of the squares of the 2 numbers = 20 + 20 = 40.

Therefore, 40 is the correct answer. 

Train T starts at 3 PM and train S starts at 4 PM. 
Let the speed of train T be t.
=> Speed of train S = 0.75t.

When the trains meet, train t would have traveled for one more hour than train S. 
Let us assume that the 2 trains meet x hours after 3 PM. Trains S would have traveled for x-1 hours. 

Distance traveled by train T = xt
Distance traveled by train S = (x-1)*0.75t = 0.75xt-0.75t

We know that train T has traveled three fifths of the distance. Therefore, train S should have traveled two-fifths the distance between the 2 cities. 

=> (xt)/(0.75xt-0.75t) = 3/2
2xt = 2.25xt-2.25t
0.25x = 2.25
x = 9 hours. 

Train T takes 9 hours to cover three-fifths the distance. Therefore, to cover the entire distance, train T will take 9*(5/3) = 15 hours. 
Therefore, 15 is the correct answer.

The selling price of the mixture is Rs.40/kg. 
Let a be the price of 1 kg of tea A in the mixture and b be the price per kg of tea B. 
It has been given that the profit is 10% if the 2 varieties are mixed in the ratio 3:2
Let the cost price of the mixture be x. 
It has been given that 1.1x = 40
x = 40/1.1
Price per kg of the mixture in ratio 3:2 = $$\frac{3a+2b}{5} $$
$$\frac{3a+2b}{5} = \frac{40}{1.1}$$
$$3.3a+2.2b=200$$ --------(1)

The profit is 5% if the 2 varieties are mixed in the ratio 2:3.
Price per kg of the mixture in ratio 2:3 = $$\frac{2a+3b}{5}$$
$$\frac{2a+3b}{5} = \frac{40}{1.05}$$
$$2.1a+3.15b=200$$ ------(2)

Equating (1) and (2), we get, 
$$3.3a+2.2b = 2.1a+3.15b$$
$$1.2a=0.95b$$
$$\frac{a}{b} = \frac{0.95}{1.2}$$
$$\frac{a}{b} = \frac{19}{24}$$

Therefore, option C is the right answer. 

We can see that T$$_{2}$$ is formed by using the mid points of T$$_{1}$$. Hence, we can say that area of triangle of T$$_{2}$$ will be (1/4)th of the area of triangle T$$_{1}$$.

Area of triangle T$$_{1}$$ = $$\dfrac{\sqrt{3}}{4}*(24)^2$$ = $$144\sqrt{3}$$ sq. cm

Area of triangle T$$_{2}$$ = $$\dfrac{144\sqrt{3}}{4}$$ = $$36\sqrt{3}$$ sq. cm 

Sum of the area of all triangles =  T$$_{1}$$ + T$$_{2}$$ + T$$_{3}$$ + ... 

$$\Rightarrow$$ T$$_{1}$$ + T$$_{1}$$/$$4$$ + T$$_{1}$$/$$4^2$$ + ... 

$$\Rightarrow$$ $$\dfrac{T_{1}}{1 - 0.25}$$

$$\Rightarrow$$ $$\dfrac{4}{3}*T_{1}$$

$$\Rightarrow$$ $$\dfrac{4}{3}*144\sqrt{3}$$

$$\Rightarrow$$ $$192\sqrt{3}$$

Hence, option D is the correct answer. 

Given that: $$\log_{12}{81}=p$$

$$\Rightarrow$$ $$\log_{81}{12}=\dfrac{1}{p}$$

$$\Rightarrow$$ $$\log_{3}{3*4}=\dfrac{4}{p}$$

$$\Rightarrow$$ $$1+\log_{3}{4}=\dfrac{4}{p}$$

Using Componendo and Dividendo, 

$$\Rightarrow$$ $$\dfrac{1+\log_{3}{4}-1}{1+\log_{3}{4}+1}=\dfrac{4-p}{4+p}$$

$$\Rightarrow$$ $$\dfrac{\log_{3}{4}}{2+\log_{3}{4}}=\dfrac{4-p}{4+p}$$

$$\Rightarrow$$ $$\dfrac{\log_{3}{4}}{\log_{3}{9}+\log_{3}{4}}=\dfrac{4-p}{4+p}$$

$$\Rightarrow$$ $$\dfrac{\log_{3}{4}}{\log_{3}{36}}=\dfrac{4-p}{4+p}$$

$$\Rightarrow$$ $$3*\dfrac{4-p}{4+p}=\dfrac{3\log_{3}{4}}{\log_{3}{36}}$$

$$\Rightarrow$$ $$3*\dfrac{4-p}{4+p}=\dfrac{\log_{3}{64}}{\log_{3}{36}}$$

$$\Rightarrow$$ $$3*\dfrac{4-p}{4+p}=\log_{36}{64}$$

$$\Rightarrow$$ $$3*\dfrac{4-p}{4+p}=\log_{6^2}{8^2}=\log_{6}{8}$$. Hence, option D is the correct answer.

We have to equate the installments and the amount due either at the time of borrowing or at the time when the entire loan is repaid. Let us bring all values to the time frame in which all the dues get settled, i.e, by the end of 2 years. 

John borrowed Rs. 2,10,000 from the bank at 10% per annum. This loan will amount to 2,10,000*1.1*1.1 = Rs.2,54,100 by the end of 2 years. 
Let the amount paid as installment every year be Rs.x.

John would pay the first installment by the end of the first year. Therefore, we have to calculate the interest on this amount from the end of the first year to the end of the second year. The loan will get settled the moment the second installment is paid. 

=> 1.1x + x = 2,54,100
2.1x = 2,54,100
=> x = Rs. 1,21,000.

Therefore, 121000 is the correct answer. 

Let us assume that A can complete 'a' units of work in a day and B can complete 'b' units of work in a day. 
A works alone till half the work is completed. 
A and B work together for 4 days and B works alone to complete the last 5% of the work. 
=> A and B in 4 days can complete 45% of the work.

Let us assume the total amount of work to be done to be 100 units.
4a + 4b = 45 ---------(1)
B needs 25% more time than A to finish a job. 
=> 1.25*b = a ----------(2)

Substituting (2) in (1), we get, 
5b+4b = 45
9b = 45
b = 5 units/day

B alone can finish the job in 100/5 = 20 days. 
Therefore, option A is the right answer.

It has been given that the digits in the number should appear in the ascending order. Therefore, there is only 1 possible arrangement of the digits once they are selected to form a number.
There are 9 numbers (1,2,3,4,5,6,7,8,9) in total.
2 digit numbers can be formed in $$9C_2$$ ways.
3 digit numbers can be formed in $$9C_3$$ ways.
..................................................
..9 digit number can be formed in 9C9 ways. 

We know that $$nC_0+nC_1+nC_2+.........+nC_n =2^n$$
=> $$9C_0 + 9C_1+9C_2+....9C_9 = 2^9$$
$$9C_0 + 9C_1 + ...9C_9 = 512$$

We have to subtract $$9C_0$$ and $$9C_1$$ from both the sides of the equations since we cannot form single digit numbers. 
=> $$9C_2 + 9C_3+...+9C_9=512-1-9$$
$$9C_2+9C_3+...+9C_9=502$$

Therefore, $$502$$ is the right answer. 

We are given that diameter of base = 8 ft. Therefore, the radius of circular base = 8/2 = 4 ft 

In triangle OAB and OCD

$$\dfrac{OA}{AB} = \dfrac{OC}{CD}$$

$$\Rightarrow$$ AB = $$\dfrac{3*4}{12}$$ = 1 ft.

Therefore, the volume of remaining part = Volume of entire cone - Volume of smaller cone

$$\Rightarrow$$ $$\dfrac{1}{3}*\pi*4^2*12-\dfrac{1}{3}*\pi*1^2*3$$

$$\Rightarrow$$ $$\dfrac{1}{3}*\pi*189$$

$$\Rightarrow$$ $$\dfrac{22}{7*3}*189$$

$$\Rightarrow$$ $$198$$ cubic ft

Let the number of marbles with Raju and Lalitha initially be 4x and 9x. 
Let the number of marbles that Lalitha gave to Raju be a.

It has been given that (4x+a)/(9x-a) = 5/6
24x + 6a = 45x - 5a
11a = 21x
a/x = 21/11

Fraction of original marbles given to Raju by Lalitha = a/9x (Since Lalitha had 9x marbles initially).
a/9x = 21/99
 = 7/33.

Therefore, option D is the right answer. 

Let us draw a Venn diagram using the information present in the question. 

It is given that the number of students studying H equals that studying E. 

Let 'x' be the total number of students who studied H, and H and P but mot E.We can also say that the same will be the number of students who studied E, and E and P but not H.Therefore,

x + 20 + 10 + x = 74 

x = 22

Hence, the number of students studying H = 22 + 10+ 20 = 52

Let the price of peanuts be Rs. 100x per kg
Then, the price of walnuts = Rs. 300x per kg
Cost price of peanuts for the shopkeeper = Rs. 110x per kg
Cost price of walnuts for the shopkeeper = Rs. 360x per kg
Total cost incurred to the shopkeeper while buying = Rs.(8 * 110x + 16 * 360x) = Rs. 6640x
Since, 5kg walnut and 3kg peanuts are lost in transit, the shopkeeper will be remained with (16-5)+(8-3)=16kgs of nuts
Total selling price that the shopkeeper got = Rs. (166 * 16) = Rs. 2656
Profit = 25%
So, cost price = Rs. 2124.80
Therefore, 6640x = 2124.80
On solving, we get x = 0.32
Therefore, price of walnuts = Rs. (300 * 0.32) = Rs. 96 per kg.
Hence, option A is the correct answer

It has been given that among 200 students, 105 like pizza and 134 like burger.
The question asks us to find out the number of students who can be liking only burgers among the given values. 

The least number of students who like only burger will be obtained when everyone who likes pizza likes burger too. 
In this case, 105 students will like pizza and burger and 134-105 = 29 students will like only burger. Therefore, the number of students who like only burger cannot be less than 29.

The maximum number of students who like only burger will be obtained when we try to separate the 2 sets as much as possible. 
There are 200 students in total. 105 of them like pizza. Therefore, the remaining 95 students can like only burger and 134-95 = 39 students can like both pizza and burger. As we can see, the number of students who like burger cannot exceed 95. 

The number of students who like only burger should lie between 29 and 95 (both the values are included). 
93 is the only value among the given options that satisfies this condition and hence, option D is the right answer. 

$$f(x + 2) = f(x) + f(x + 1)$$
As we can see, the value of a term is the sum of the 2 terms preceding it. 

It has been given that $$f(11) = 91$$ and $$f(15) = 617$$.
We have to find the value of $$f(10)$$.

Let $$f(10)$$ = b
$$f(12)$$ = b + 91
$$f(13)$$ = 91 + b + 91 = 182 + b
$$f(14)$$ = 182+b+91+b = 273+2b
$$f(15)$$ = 273+2b+182+b = 455+3b
It has been given that 455+3b = 617
3b = 162
=> b = 54

Therefore, 54 is the correct answer. 

The possible average age of people whose ages are below 51 years will be maximum if the average age of the number of people aged 51 years and above is minimum. Hence, we can say that that there are 30 people having same age 51 years. 

Let 'x' be the maximum average age of people whose ages are below 51. 

Then we can say that, 

$$\dfrac{51*30+39*x}{30+39} = 38$$

$$\Rightarrow$$ $$1530+39x = 2622$$

$$\Rightarrow$$ $$x = 1092/39 = 28$$

Hence, we can say that option D is the correct answer. 

Let the price of paint B be x. 
Price of paint A = x+8

We know that the amount of paint B in the mixture does not exceed the amount of paint A. Therefore, paint B can at the maximum compose 50% of the mixture. 

The seller sells 10 litres of paint at Rs.264 earning a profit of 10%. 
=> The cost price of 10 litres of the paint mixture = Rs. 240

Therefore, the cost of 1 litre of the mixture = Rs.24

We have to find the highest possible cost of paint B. 
When we increase the cost of paint B, the cost of paint A will increase too. If the cost price of the mixture is closer to the cost of paint B, then the amount of paint B present in the mixture should be greater than the amount of paint A present in the mixture.
The highest possible cost of paint B will be obtained when the volumes of paint A and paint B in the mixture are equal. 

=> (x+x+8)/2 = 24
2x = 40
x = Rs. 20

Therefore, option C is the right answer. 

In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP is perpendicular to CD. Then the area, in sq cm, of triangle APD is

Given that area of parallelogram = 72 sq cm

Area of triangle ABC = $$\dfrac{1}{2}$$*area of parallelogram

(1/2)*AB*BC*sinABC = $$\dfrac{1}{2}$$*72

sinABC = $$\dfrac{1}{2}$$

$$\angle$$ ABC = 30°

Let us draw a perpendicular CQ from C to AB. 

$$\angle\ B\ =\ \angle\ D\ =\ 30^{\circ\ }$$ ( opposite angles in a parallelogram)

$$\angle\ QCB\ =\ \angle\ DAP\ =\ 60^{\circ\ }$$  ( Sum of angles in a triangle = 180)

In triangles APD and CQB, AP= CQ , AD=CB ( opposite sides of a parallelogram) , $$\angle\ QCB\ =\ \angle\ DAP\ =\ 60^{\circ\ }$$

Hence APD and CQB are congruent triangles using SAS property.

Therefore, we can say that area of triangle APD = area of triangle CQB

In right angle triangle CQB,

QB = CBcos30° = $$16*\dfrac{\sqrt{3}}{2}$$ = $$8\sqrt{3}$$ cm

CQ = CBsin30° = $$16*\dfrac{1}{2}$$ = $$8$$ cm

Therefore, area of triangle CQB = 1/2*CQ*QB = $$1/2*8*8\sqrt{3}$$ = $$32\sqrt{3}$$

Hence, we can say that area of triangle APD = $$32\sqrt{3}$$.

Given that $$U^{2}+(U-2V-1)^{2}$$= −$$4V(U+V)$$ 

$$\Rightarrow$$ $$U^{2}+(U-2V-1)(U-2V-1)$$= −$$4V(U+V)$$

$$\Rightarrow$$ $$U^{2}+(U^2-2UV-U-2UV+4V^2+2V-U+2V+1)$$ = −$$4V(U+V)$$ 

$$\Rightarrow$$ $$U^{2}+(U^2-4UV-2U+4V^2+4V+1)$$ = −$$4V(U+V)$$ 

$$\Rightarrow$$ $$2U^2-4UV-2U+4V^2+4V+1=−4UV-4V^2$$

$$\Rightarrow$$ $$2U^2-2U+8V^2+4V+1=0$$

$$\Rightarrow$$ $$2[U^2-U+\dfrac{1}{4}]+8[V^2+\dfrac{V}{2}+\dfrac{1}{16}]=0$$

$$\Rightarrow$$ $$2(U-\dfrac{1}{2})^2+8(V+\dfrac{1}{4})^2=0$$

Sum of two square terms is zero i.e. individual square term is equal to zero. 

$$U-\dfrac{1}{2}$$ = 0 and $$V+\dfrac{1}{4}$$ = 0

U = $$\dfrac{1}{2}$$ and V = $$-\dfrac{1}{4}$$

Therefore, $$U+3V$$ = $$\dfrac{1}{2}$$+$$\dfrac{-1*3}{4}$$ = $$\dfrac{-1}{4}$$. Hence, option C is the correct answer.

It is given that radius of the circle = 1 cm

Chord AB subtends an angle of 60° on the centre of the given circle. R be the region bounded by the radii OA, OB and the arc AB.

Therefore, R = $$\dfrac{60°}{360°}$$*Area of the circle = $$\dfrac{1}{6}$$*$$\pi*(1)^2$$ = $$\dfrac{\pi}{6}$$ sq. cm

It is given that OC = OD and area of triangle OCD is half that of R. Let OC = OD = x.

Area of triangle COD = $$\dfrac{1}{2}*OC*OD*sin60°$$

$$\dfrac{\pi}{6*2}$$ = $$\dfrac{1}{2}*x*x*\dfrac{\sqrt{3}}{2}$$

$$\Rightarrow$$ $$x^2 = \dfrac{\pi}{3\sqrt{3}}$$

$$\Rightarrow$$ $$x$$ = $$(\frac{\pi}{3\sqrt{3}})^{\frac{1}{2}}$$ cm. Hence, option C is the correct answer. 

Let the time taken by Partha to cover 60 km be x hours.
Narayan will cover 60 km in x-4 hours. 

Speed of Partha = $$\frac{60}{x}$$
Speed of Narayan = $$\frac{60}{x-4}$$

Partha reaches the mid-point of A and B two hours before Narayan reaches B.

=> $$\dfrac{30}{\frac{60}{x}} + 2 = \dfrac{60}{\frac{60}{(x-4)}}$$

$$\frac{x}{2} + 2 = x-4$$
$$\frac{x+4}{2}=x-4$$
$$x+4=2x-8$$
$$x=12$$

Partha will take 12 hours to cross 60 km.
=> Speed of Partha = $$\frac{60}{12}=5$$ Kmph.
Therefore, option D is the right answer. 

Let x = $$a$$, y = $$ar$$ and z = $$ar^2$$
It is given that, 5x, 16y and 12z are in AP.
so, 5x + 12z = 32y
On replacing the values of x, y and z, we get
$$5a + 12ar^2 = 32ar$$
or, $$12r^2 - 32r + 5$$ = 0
On solving, $$r$$ = $$\frac{5}{2}$$ or $$\frac{1}{6}$$

For $$r$$ = $$\frac{1}{6}$$, x < y < z is not satisfied.

So, $$r$$ = $$\frac{5}{2}$$

Hence, option C is the correct answer.

Given that $$x^{2018}y^{2017}=\frac{1}{2}$$  ... (1)

$$x^{2016}y^{2019}=8$$ ... (2)

Equation (2)/ Equation (1)

$$\dfrac{y^2}{x^2} = \dfrac{8}{1/2}$$

$$\dfrac{y}{x} = 4$$ or $$-4$$

Case 1: When $$\dfrac{y}{x} = 4$$

$$x^{2018}(4x)^{2017}=\dfrac{1}{2}$$

$$x^{2018+2017}(2)^{4034}=\dfrac{1}{2}$$

$$x^{4035}=\dfrac{1}{(2)^{4035}}$$

$$x=\dfrac{1}{2}$$

Since, $$\dfrac{y}{x} = 4$$, => y = 2

Therefore, $$x^{2}+y^{3}$$ = $$\dfrac{1}{4}+8$$ = $$\dfrac{33}{4}$$

Case 2: When $$\dfrac{y}{x} = -4$$

$$x^{2018}(-4x)^{2017}=\dfrac{1}{2}$$

$$x^{2018+2017}(2)^{4034}=\dfrac{-1}{2}$$

$$x^{4035}=\dfrac{1}{(-2)^{4035}}$$

$$x=\dfrac{-1}{2}$$

Since, $$\dfrac{y}{x} = -4$$, => y = 2

Therefore, $$x^{2}+y^{3}$$ = $$\dfrac{1}{4}+8$$ = $$\dfrac{33}{4}$$. Hence, option D is the correct answer. 

Let the efficiency of filling pipes be 'x' and the efficiency of draining pipes be '-y'.
In the first case, 
Capacity of tank = (6x - 5y) * 6..........(i)
In the second case,
Capacity of tank = (5x - 6y) * 60.....(ii)
On equating (i) and (ii), we get
 (6x - 5y) * 6 =  (5x - 6y) * 60
or, 6x - 5y = 50x - 60y
or, 44x = 55y
or, 4x = 5y
or, x = 1.25y

Capacity of the tank = (6x - 5y) * 6 = (7.5y - 5y) * 6 = 15y
Net efficiency of 2 filling and 1 draining pipes = (2x - y) = (2.5y - y) = 1.5y
Time required = $$\dfrac{\text{15y}}{\text{1.5y}}$$hours = 10 hours.

Hence, 10 is the correct answer.

$$\log_{2}({5+\log_{3}{a}})=3$$
=>$$5 + \log_{3}{a}$$ = 8
=>$$ \log_{3}{a}$$ = 3
or $$a$$ = 27

$$\log_{5}({4a+12+\log_{2}{b}})=3$$
=>$$4a+12+\log_{2}{b}$$ = 125
Putting $$a$$ = 27, we get
$$\log_{2}{b}$$ = 5
or, $$b$$ = 32

So, $$a + b$$ = 27 + 32 = 59
Hence, option A is the correct answer.

Let the total number of tests be 'n' and the average by 'A'
Total score = n*A
When 1st 10 tests are excluded, decrease in total value of scores = (nA - 20 * 10) = (nA - 200)
Also, (n - 10)(A + 1) = (nA - 200)
On solving, we get 10A - n = 190..........(i)
When last 10 tests are excluded, decrease in total value of scores = (nA - 30 * 10) = (nA - 300)
Also, (n - 10)(A - 1) = (nA - 300)
On solving, we get 10A + n = 310..........(ii)
From (i) and (ii), we get n = 60
Hence, 60 is the correct answer. 

At $$x = 0, 2^x = 1$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 1 + 2 = 3 Which is divisible by 3. Hence, x = 0 is one possible solution. 

At $$x = 1, 2^x = 2$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 2 + 2 = 3 Which is divisible by 4. Hence, x = 1 is one possible solution. 

At $$x = 2, 2^x = 4$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 4 + 2 = 6 Which is divisible by 3. Hence, x = 2 is one possible solution. 

At $$x = 3, 2^x = 8$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 8 + 2 = 3 Which is not divisible by 3 or 4. Hence, x = 3 can't be a solution. 

At $$x = 4, 2^x = 16$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 16 + 2 = 18 Which is divisible by 3. Hence, x = 4 is one possible solution. 

At $$x = 5, 2^x = 32$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 32 + 2 = 34 Which is not divisible by 3 or 4. Hence, x = 5 can't be a solution. 

At $$x = 6, 2^x = 64$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 64 + 2 = 66 Which is divisible by 3. Hence, x = 6 is one possible solution. 

At $$x = 7, 2^x = 128$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 128 + 2 = 130 Which is not divisible by 3 or 4. Hence, x = 7 can't be a solution. 

At $$x = 8, 2^x = 256$$ which is not in the given range [0.25, 200]. Hence, x can't take any value greater than 7.

Therefore, all possible values of x = {0,1,2,4,6}. Hence, we can say that 'x' can take 5 different integer values. 

Total marks = N
Pass marks = 45% of N = 0.45N
Marks obtained = 36
It is given that, obtained marks is 68% less than that pass marks
=>the obtained marks is 32% of the pass marks.
So, 0.32 * 0.45N = 36
On solving, we get N = 250
Hence, option B is the correct answer.