CAT 2020 Question Paper Slot 1

$$2^{Y^2({\log_{3}{5})}}=5^{Y^2(\log_3 2)}$$

Given, $$5^{Y^2\left(\log_32\right)}=5^{\left(\log_23\right)}$$

=> $$Y^2\left(\log_32\right)=\left(\log_23\right)=>Y^2=\left(\log_23\right)^2$$

=>$$Y=\left(-\log_23\right)^{\ }or\ \left(\log_23\right)$$

since Y is a negative number, Y=$$\left(-\log_23\right)=\left(\log_2\frac{1}{3}\right)$$

Let the initial number of chocolates be 64x.

First child gets 32x+1 and 32x-1 are left.

2nd child gets 16x+1/2 and 16x-3/2 are left

3rd child gets 8x+1/4 and 8x-7/4 are left

4th child gets 4x+1/8 and 4x-15/8 are left

5th child gets 2x+1/16 and 2x-31/16 are left.

Given, 2x-31/16=0=> 2x=31/16 => x=31/32.

.'. Initially the Gentleman has 64x i.e. 64*31/32 =62 chocolates.

Let the number be 'abc'. Then, $$2<a\times\ b\times\ c<7$$. The product can be 3,4,5,6.

We can obtain each of these as products with the combination 1,1, x where x = 3,4,5,6. Each number can be arranged in 3 ways, and we have 4 such numbers: hence, a total of 12 numbers fulfilling the criteria.

We can factories 4 as 2*2 and the combination 2,2,1 can be used to form 3 more distinct numbers. 

We can factorize 6 as 2*3 and the combination 1,2,3 can be used to form 6 additional distinct numbers. 

Thus a total of 12 + 3 + 6 = 21 such numbers can be formed.

The graphs of $$2^{x}+2^{-x} and 2-(x-2)^{2}$$ never intersect. So, number of solutions=0.

Alternate method:

We notice that the minimum value of the term in the LHS will be greater than or equal to 2 {at x=0; LHS = 2}. However, the term in the RHS is less than or equal to 2 {at x=2; RHS = 2}. The values of x at which both the sides become 2 are distinct; hence, there are zero real-valued solutions to the above equation.

$$(x^{2}-7x+11)^{(x^{2}-13x+42)}=1$$

if $$(x^{2}-13x+42)$$=0 or $$(x^{2}-7x+11)$$=1 or $$(x^{2}-7x+11)$$=-1 and $$(x^{2}-13x+42)$$ is even number

For x=6,7 the value $$(x^{2}-13x+42)$$=0

$$(x^{2}-7x+11)$$=1 for x=5,2.

$$(x^{2}-7x+11)$$=-1 for x=3,4 and for X=3 or 4, $$(x^{2}-13x+42)$$ is even number.

.'. {2,3,4,5,6,7} is the solution set of x.

.'. x can take six values. 

Let the base radius be 3r.

Height of upper cone is 9 so, by symmetry radius of upper cone will be r.

Volume of frustum=$$\frac{\pi}{3}\left(9r^2\cdot27-r^2.9\right)$$

Volume of upper cone = $$\frac{\pi}{3}.r^2.9$$

Difference= $$\frac{\pi}{3}\cdot9\cdot r^2\cdot25=225$$ => $$\frac{\pi}{3}\cdot r^2=1$$

Volume of larger cone = $$\frac{\pi}{3}\cdot9r^2\cdot27=243$$

The area of the region contained by the lines $$\mid x\mid-y\leq1,y\geq0$$ and $$y\leq1$$ is the white region.

Total area = Area of rectangle + 2 * Area of triangle = $$2+\left(\frac{1}{2}\times\ 2\times\ 1\right)\ =3$$

Hence, 3 is the correct answer.

Let ABCD be the rectangle with length 2l and breadth 2b respectively.

Area of the circle i.e. area of painted region = $$\pi\ b^2$$.

Given, 4lb-$$\pi\ b^2$$=(2/3)$$\pi\ b^2$$.

=> 4lb=(5/3)$$\pi\ b^2$$.

=>l=$$\frac{5\pi}{12}b$$.

Given, 4lb=135 => 4*$$\frac{5\pi}{12}b^2$$=135 => b= $$\frac{9}{\sqrt{\ \pi\ }}$$

=> l=$$\frac{15}{4}\sqrt{\ \pi\ }$$
Perimeter of rectangle =4(l+b)=4($$\frac{15}{4}\sqrt{\ \pi\ }$$+$$\frac{9}{\sqrt{\ \pi\ }}$$ )=$$3\sqrt{\pi}(5+\frac{12}{\pi})$$.

Hence option A is correct. 

Let the length of radius be 'r'.

From the above diagram,

$$x^2+r^2=6^2\ $$....(i)

$$\left(10-x\right)^2+r^2=8^2\ $$----(ii)

Subtracting (i) from (ii), we get: 

x=3.6 => $$r^2=36-\left(3.6\right)^2$$ ==> $$r^2=36-\left(3.6\right)^2\ =23.04$$.

Area of circle = $$\pi\ r^2=23.04\pi\ $$

Area of rhombus= 1/2*d1*d2=1/2*12*16=96.

.'. Ratio of areas = 23.04$$\pi\ $$/96=$$\frac{6\pi}{25}$$

$$x_0=max(x_1,x_2,....,x_{12})$$

$$x_0$$ will be minimum if x1,x2...x12 are close to each other

100/12=8.33

.'. max$$(x_1,x_2,....,x_{12})$$ will be minimum if $$(x_1,x_2,....,x_{12})$$=(9,9,9,9,8,8,8,8,8,8,8,8,)

.'. Option B is correct.