CAT 2019 Question Paper Slot 2

When A and B met for the first time at 10:00 AM, A covered 60% of the track.

So B must have covered 40% of the track.

It is given that A returns to P at 10:12 AM i.e A covers 40% of the track in 12 minutes

60% of the track in 18 minutes

B covers 40% of track when A covers 60% of the track.

B covers 40% of the track in 18 minutes.

B will cover the rest 60% in 27 minutes, hence it will return to B at 10:27 AM

$$a_1 - a_2 + a_3 - a_4 + .... + (-1)^{n - 1} a_n = n$$

It is clear from the above equation that when n is odd, the co-efficient of a is positive otherwise negative.

$$a_1 - a_2 = 2$$

$$a_1 = a_2 + 2$$   

$$a_1 - a_2 + a_3 = 3$$

On substituting the value of $$a_1$$ in the above equation, we get

$$a_3$$ = 1

$$a_1 - a_2 + a_3 - a_4 = 4$$

On substituting the values of $$a_1, a_3$$ in the above equation, we get

$$a_4$$ = -1

$$a_1 - a_2 + a_3 - a_4 +a_5 = 5$$

On substituting the values of $$a_1, a_3, a_4$$ in the above equation, we get

$$a_5$$ = 1

So we can conclude that $$a_3, a_5, a_7....a_{n+1}$$ = 1 and $$a_2, a_4, a_6....a_{2n}$$ = -1

Now we have to find the value of $$a_{51} + a_{52} + .... + a_{1023}$$

Number of terms = 1023=51+(n-1)1

n=973

There will be 486 even and 487 odd terms, so the value of $$a_{51} + a_{52} + .... + a_{1023}$$ = 486*-1+487*1=1

Let 'h' be the height of the triangle ABC, semiperimeter(S) $$= \frac{4+4+r+4+4+r}{2} = 8+r$$, 
$$a=4+r, b=4+r, c=8$$

Area of triangle ABC $$=\ \ \sqrt{\ s\cdot\left(s-a\right)\left(s-b\right)\left(s-c\right)}=$$
$$= \sqrt{\left(\ 8+r\right)\times\ 4\times\ 4\times\ r}$$ = $$\ \frac{\ 1}{2}\times\ \left(4+4\right)\times\ height$$

Height (h) = $$\sqrt{\ \left(8+r\right)r}$$

Now, $$ h + r = 4 \longrightarrow  \sqrt{\ \left(8+r\right)r} + r = 4$$ (Considering the height of the triangle)

$$\sqrt{\ \left(8+r\right)r}$$=4-r

16r=16

r=1


Alternatively,

$$\text{AE}^@+\text{EC}^2=\text{AC}^2 \longrightarrow 4^2+\left(4-r\right)^2 = \left(4+r\right)^2 \longrightarrow\ \longrightarrow\ \longrightarrow\ r=1$$

The roots of $$x^2 - 4x - log_{2}{A} = 0$$ will be real and distinct if and only if the discriminant is greater than zero

16+4*$$log_{2}{A}$$ > 0

$$log_{2}{A}$$ > -4

A> 1/16

Given,

The quadratic equation $$x^2 + bx + c = 0$$ has two roots 4a and 3a

7a=-b

12$$a^2$$ = c

We have to find the value of  $$b^2 + c$$ = 49$$a^2$$+ 12$$a^2$$=61$$a^2$$

Now lets verify the options 

61$$a^2$$ = 3721 ==> a= 7.8 which is not an integer

61$$a^2$$ = 361 ==> a= 2.42 which is not an integer

61$$a^2$$ = 427 ==> a= 2.64 which is not an integer

61$$a^2$$ = 549 ==> a= 3 which is an integer

It is given that the base of the pyramid is square and each of the four sides are equilateral triangles.

Length of each side of the equilateral triangle = 20cm

Since the side of the triangle will be common to the square as well, the side of the square = 20cm

Let h be the vertical height of the pyramid ie OA

OB = 10 since it is half the side of the square 

AB is the height of the equilateral triangle i.e 10$$\sqrt{\ 3}$$

AOB is a right angle, so applying the Pythagorean formula, we get

$$OA^2+ OB^2= AB^2$$

$$h^2$$ + 100= 300

h=10$$\sqrt{\ 2}$$

Let p be the length of AP.

It is given that $$\angle\ BAC\ =90\ $$ and $$\angle\ APC\ =90\ $$

Let $$\angle\ ABC\ =\theta\ $$, then $$\angle\ BAP\ =90-\theta\ $$ and $$\angle\ BCA\ =90-\theta\ $$

So $$\angle\ PAC\ =\theta\ $$

Triangles BPA and APC are similar

$$p^2=x\left(20-x\right)$$

We have to maximize the value of p, which will be maximum when x=20-x

x=10

$$\sqrt{\log_{e}{\frac{4x - x^2}{3}}}$$ will be real if $$\log_e\ \frac{\ 4x-x^2}{3}\ \ge\ 0$$

$$\frac{\ 4x-x^2}{3}\ >=\ 1$$

$$\ 4x-x^2-3\ >=\ 0$$

$$\ x^2-4x+3\ =<\ 0$$

1=< x=< 3

$$5^x - 3^y = 13438$$ and $$5^{x - 1} + 3^{y + 1} = 9686$$

$$5^{x} + 3^{y}*15 = 9686*5$$

$$5^{x} + 3^{y}*15 = 48430$$

16*$$3^y$$=34992

$$3^y$$ = 2187

y = 7

$$5^x$$=13438+2187=15625

x=6

x+y = 13

Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively.

The amount of salt in vessels A, B, C = 50 ml, 110 ml, 160 ml respectively.

The amount of water in vessels A, B, C = 450 ml, 390 ml, 340 ml respectively.

In 100 ml solution in vessel A, there will be 10ml of salt and 90 ml of water

Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A

i.e after the first transfer, the amount of salt in vessels A, B, C = 40, 120, 160 ml respectively.

after the second transfer, the amount of salt in vessels A, B, C =40, 100, 180 ml respectively.

After the third transfer, the amount of salt in vessels A, B, C = 70, 100, 150 respectively.

Each transfer can be captured through the following table.

Percentage of salt in vessel A =$$\ \frac{\ 70}{500}\times\ 100$$

=14%