CAT 2018 Question Paper Slot 2

Let 't' pm be the time when the tank is emptied everyday. Let 'a' and 'b' be the liters/hr filled by pump A and pump B respectively.

On Monday, A alone completed filling the tank at 8 pm. Therefore, we can say that pump A worked for (8 - t) hours. Hence, the volume of the tank = a*(8 - t) liters. 

Similarly, on Tuesday, B alone completed filling the tank at 6 pm. Therefore, we can say that pump B worked for (6 - t) hours. Hence, the volume of the tank = b*(6 - t) liters. 

On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. Therefore, we can say that pump A worked for (5 - t) hours  and pump B worked for 2 hours. Hence, the volume of the tank = a*(5 - t)+2b liters. 

We can say that  a*(8 - t) =  b*(6 - t) =  a*(5 - t) + 2b

a*(8 - t) = a*(5 - t) + 2b

$$\Rightarrow$$ 3a = 2b   ... (1)

a*(8 - t) =  b*(6 - t)

Using equation (1), we can say that 

$$a*(8-t)=\dfrac{3a}{2}*(6-t)$$

$$t = 2$$

Therefore, we can say that the tank gets emptied at 2 pm daily. We can see that A takes 6 hours and pump B takes 4 hours alone. 

Hence, working together both can fill the tank in = \dfrac{6*4}{6+4} = 2.4 hours or 2 hours and 24 minutes. 

The pumps started filling the tank at 2:00 pm. Hence, the tank will be filled by 4:24 pm.  

Given that the arithmetic mean of x, y and z is 80.

$$\Rightarrow$$ $$\dfrac{x+y+z}{3} = 80$$

$$\Rightarrow$$ $$x+y+z = 240$$  ... (1)

Also,  $$\dfrac{x+y+z+v+u}{5} = 75$$

$$\Rightarrow$$ $$\dfrac{x+y+z+v+u}{5} = 75$$

$$\Rightarrow$$ $$x+y+z+v+u = 375$$

Substituting values from equation (1),

$$\Rightarrow$$ $$v+u = 135$$

It is given that u=(x+y)/2 and v=(y+z)/2.

$$\Rightarrow$$ $$(x+y)/2+(y+z)/2 = 135$$

$$\Rightarrow$$ $$x+2y+z = 270$$

$$\Rightarrow$$ $$y = 30$$   (Since $$x+y+z = 240$$) 

Therefore, we can say that $$x+z = 240 - y = 210$$. We are also given that x ≥ z, 

Hence, $$x_{min}$$ = 210/2 = 105. 

Time taken to cover first 50 km at 100 km/hr = $$\frac{1}{2}$$ hr.
Time taken to cover second 50 km at 50 km/hr = 1 hr.
Time taken to cover last 50 km at 25 km/hr = 2 hr.
When car 2 starts, car 1 has already covered 20 km.
So, time taken by car 1 to reach B after car 1 starts = total time - time required to travel first 20 km
= 3 hr 30 min - 12 min = 3 hr 18 min
Distance travelled by car 2 = (50 + 50 + 45) = 145 km
Distance from B = (150 - 145) km = 5 km

Hence, 5 is the correct answer.

Let 'a' and 'b' are those two numbers. 

$$\Rightarrow$$ $$a^2+b^2 = 97$$

$$\Rightarrow$$ $$a^2+b^2-2ab = 97-2ab$$

$$\Rightarrow$$ $$(a-b)^2 = 97-2ab$$

We know that $$(a-b)^2$$ $$\geq$$ 0

$$\Rightarrow$$ 97-2ab $$\geq$$ 0

$$\Rightarrow$$ ab $$\leq$$ 48.5

Hence, ab $$\neq$$ 64. Therefore, option D is the correct answer.

Let 'a' and 'b' be the length of sides of the rectangle. (a > b)

Area of the rectangle = a*b

Perimeter of the rectangle = 2*(a+b)

$$\Rightarrow$$ $$\dfrac{a*b}{(2*(a+b))^2}=\dfrac{1}{25}$$

$$\Rightarrow$$ $$25ab=4(a+b)^2$$

$$\Rightarrow$$ $$4a^2-17ab+4b^2=0$$

$$\Rightarrow$$ $$(4a-b)(a-4b)=0$$

$$\Rightarrow$$ $$a = 4b$$ or $$\dfrac{b}{4}$$

We initially assumed that a > b, therefore a $$\neq$$ $$\dfrac{b}{4}$$.

Hence, a = 4b

$$\Rightarrow$$ b : a = 1 : 4

Let us draw the diagram according to the available information.

We can see that triangle BPC and BQC are inscribed inside a semicircle. Hence, we can say that

$$\angle$$ BPC = $$\angle$$ BQC = 90°

Therefore, we can say that BQ  $$\perp$$ AC and CP $$\perp$$ AB.

In triangle ABC,

Area of triangle = (1/2)*Base*Height = (1/2)*AB*CP = (1/2)*AC*BQ

$$\Rightarrow$$ BQ = $$\dfrac{AB*CP}{AC}$$ = $$\dfrac{30*20}{25}$$ = 24 cm.

We know that area of the triangle = 32 sq. units, BC = 8 units

Therefore, the height of the perpendicular drawn from point A to BC = 2*32/8 = 8 units.

Let us draw a possible diagram of the given triangle. 

We can see that if A coincide with (-4, 0) then the distance between A and (0, 0) = 4 units.

If we move the triangle up or down keeping the base BC on x = 4, then point A will move away from origin as vertical distance will come into factor whereas horizontal distance will remain as 4 units.

Hence, we can say that minimum distance between A and origin (0, 0) = 4 units. 

Area of the semicircle with AB as a diameter = $$\dfrac{1}{2}*\pi*(\dfrac{AB^2}{4})$$

$$\Rightarrow$$ $$\dfrac{1}{2}*\pi*(\dfrac{AB^2}{4})$$ = $$72*\pi$$

$$\Rightarrow$$ $$AB = 24 cm$$

Given that area of the rectangle ABCD = 768 sq.cm

$$\Rightarrow$$ AB*BC = 768

$$\Rightarrow$$ BC = 32 cm

We can see that the perimeter of the remaining shape = AD + DC + BC + Arc(AB)

$$\Rightarrow$$ 32+24+32+$$\pi*24/2$$

$$\Rightarrow$$ $$88+12\pi$$

If we carefully observe set A, then we find that $$6^{2n} -35n - 1$$ is divisible by 35. So, set A contains multiples of 35. However, not all the multiples of 35 are there in set A, for different values of $$n$$.
For $$n = 1$$, the value is 0, for $$n = 2$$, the value is 1225 which is the 35th multiple of 3.
If we observe set B, it consists of all the multiples of 35 including 0.
So, we can say that every member of set A will be in B while every member of set B will not necessarily be in set A.
Hence, option A is the correct answer.

$$4^{n} > 17^{19}$$

$$\Rightarrow$$ $$16^{n/2} > 17^{19}$$

Therefore, we can say that n/2 > 19

n > 38

Hence, option D is the correct answer.