CAT 2018 Question Paper Slot 1

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.