CAT 2020 Question Paper Quant 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.

Let the speed of Car 2 be 'x' kmph and the time taken by the two cars to meet be 't' hours.

In 't' hours, Car 1 travels $$\left(60\ \times\ t\right)\ km$$ while Car 2 travels $$\left(x\ \times\ t\right)\ km$$

It is given that the time taken by Car 1 to travel $$\left(x\ \times\ t\right)\ km$$ is 45 minutes or (3/4) hours. $$\therefore\ \frac{\left(x\ \times\ t\right)}{60}\ =\ \frac{3}{4}\ $$ or $$t=\frac{180}{4x}$$....(i)

Similarly, the time taken by Car 2 to travel $$\left(60\ \times\ t\right)\ km$$ is 20 minutes or (1/3) hours. $$\therefore\ \frac{\left(60\times\ t\right)}{x}=\frac{1}{3}$$ or $$\therefore\ t=\frac{x}{180}$$....(ii)

Equating the values in (i) and (ii), and solving for x:

$$\therefore\ \frac{180}{4x}=\frac{x}{180}\ \ \longrightarrow\ \ \ x\ =90\ kmph$$

Hence, Option B is the correct answer.

Let the price of desktop and laptop be x,y respectively.

Given,

x+y=50000...(i)

12.x+0.9y=50000(1.02)=51000...(ii)

(ii)-0.9(i) gives

0.3x=6000=> x=20000.

Initially the amount of Dye and Water are 16,24 respectively.

To make the ratio of Dye to Water to 2:5 the amount of water should be 40l for 16l of Dye=> 16l of water is added.

Now, the Dye and Water arr 16,40 respectively.

After removing 1/4th of solution the amount of Dye and Water will be 12,30l respectively.

To have Dye and Water in the ratio of 2:3, for 30l of water we need 20l of Dye => 8l of Dye should be added.

Hence , 8 is correct answer. 

$$x=2^{12\left(7+4\sqrt{\ 3}\right)}$$.

$$x^{\frac{7}{2}}=2^{42\left(7+4\sqrt{\ 3}\right)}$$

$$x^{2\sqrt{\ 3}}=2^{24\sqrt{\ 3}\left(7+4\sqrt{\ 3}\right)}$$

$$\frac{x^{\frac{7}{2}}}{x^{2{\sqrt{3}}}}$$ = $$2^{\left(7+4\sqrt{\ 3}\right)\left(42-24\sqrt{\ 3}\right)}=2^{\left(7+4\sqrt{\ 3}\right)\left(7-4\sqrt{\ 3}\right)6}$$ =$$2^6$$.

Hence C is correct answer.

Let the volume of Metals A,B,C we 3x, 4x, 7x

Ratio weights of given volume be 5y,2y,6y

.'. 15xy+8xy+42xy=130 => 65xy=130 => xy=2.

.'.`The weight, in kg. of the metal C is 42xy=84.

Let $$a=x+\frac{1}{x}$$
So, the given equation is $$a^2-3a+2=0$$
So, $$a$$ can be either 2 or 1.

If $$a=1$$, $$x+\frac{1}{x}=1$$ and it has no real roots. 
If $$a=2$$, $$x+\frac{1}{x}=2$$ and it has exactly one real root which is $$x=1$$

So, the total number of distinct real roots of the given equation is 1

$$\frac{\log\ 5}{2\log2}\ =\frac{\log\ y}{2\log2}\cdot\frac{\log\ 5}{2\log6}$$

$$\log\ 36\ =\ \log\ y;\ \therefore\ y\ =36$$

The difference in the time take to traverse the same distance $$'d'$$ at two different speeds is 35 minutes. Equating this: $$\frac{d}{8}-\frac{d}{15}\ =\ \frac{35}{60}$$

On solving, we obtain $$d = 10 kms$$. Let $$x kmph$$ be the speed at which Amal needs to travel to reach the office in 50 minutes; then

$$\frac{10}{x}=\frac{50}{60}\ or\ x\ =\ 12\ kmph$$.Hence, Option B is the correct answer.

Let the total distance be 'D' km and the speed of the train be 's' kmph. The time taken to cover D at speed d is 't' hours. Based on the information: on equating the distance, we get $$s\ \times\ t\ =\ \frac{s}{3}\times\ \left(t+\frac{1}{2}\right)$$ 
On solving we acquire the value of $$\ t\ =\frac{1}{4}$$ or 15 mins. We understand that during the return journey, the first 5 minutes are spent traveling at speed 's' {distance traveled in terms of s = $$\ \frac{s}{12}$$}. Remaining distance in terms of 's' = $$\ \frac{s}{4}-\frac{s}{12}\ =\frac{s}{6}$$

The rest 4 minutes of stoppage added to this initial 5 minutes amounts to a total of 9 minutes. The train has to complete the rest of the journey in $$15 - 9 = 6 mins$$ or {1/10 hours}. Thus, let 'x' kmph be the new value of speed. Based on the above, we get $$\frac{s}{\frac{6}{x}}\ =\frac{1}{10}\ or\ x\ =\frac{10s}{6}$$

Since the increase in speed needs to be calculated: $$\frac{\left(\frac{10s}{6}\ -s\right)}{s}\times\ 100\ =\frac{200}{3}\approx\ 67\%$$ increase.

Hence, Option C is the correct answer.

The four digit even numbers will be of form:

1100, 1122, 1144 ... 1188, 2200, 2222, 2244 ... 9900, 9922, 9944, 9966, 9988

Their sum 'S' will be (1100+1100+22+1100+44+1100+66+1100+88)+(2200+2200+22+2200+44+...)....+(9900+9900+22+9900+44+9900+66+9900+88)

=> S=1100*5+(22+44+66+88)+2200*5+(22+44+66+88)....+9900*5+(22+44+66+88)

=> S=5*1100(1+2+3+...9)+9(22+44+66+88)

=>S=5*1100*9*10/2 + 9*11*20

Total number of numbers are 9*5=45

.'. Mean will be S/45 = 5*1100+44=5544.

Option D

Let the length of the train be $$ l kms$$ and speed be $$ s kmph$$. Base on the two scenarios presented, we obtain:

$$\frac{l}{s-2}=\frac{90}{3600}$$....(i) and $$\frac{l}{s-4}=\frac{100}{3600}$$...(ii)

On dividing (ii) by (i) and simplifying we acquire the value of $$s$$ as $$22 kmph$$. Substituting this value in (i), we have $$l=\frac{90}{3600}\times\ 20\ kms$$ {keeping it in km and hours for convenience}

Since we need to find $$\frac{l}{s}$$, let this be equal to $$x$$. Then, $$x\ =\ 90\times\frac{20}{22}\ =81.81\ \approx\ 82\ \sec onds\ $$

Hence, Option B is the correct choice.

Since $$c<9$$, we can have the following viable combinations for $$b\times\ c\ =96$$ (given our objective is to minimize the sum):

$$48\times\ 2$$ ; $$32\times3$$ ; $$24\times\ 4$$ ; $$16\times6$$ ; $$12\times8$$ 

Similarly, we can factorize $$a\times\ b\ = 432$$ into its factors. On close observation, we notice that $$18\times24\ and\ 24\ \times\ 4\ $$ corresponding to $$a\times b\ and\ b\times\ c\ $$ respectively together render us with the least value of the sum of $$a+b\ +\ c\ \ =\ 18+24+4\ =46$$

Hence, Option D is the correct answer.

Let 'x' be the strength of group G. Based on the information, $$0.65x$$ constitutes of literate people {the rest $$0.35x$$ = illiterate}
Of this $$0.65x$$, 75% are old people =(0.75)0.65x old literates. The total number of old people in group G is $$0.72x$$ {72% of the total}. Thus, the total number of old people who are illiterate = $$0.72x-0.4875x\ =\ 0.2325x$$. This is $$\frac{0.2325x}{0.35x}\times\ 100\ \approx\ \ 66\%$$ of the total number of illiterates. Hence, Option D is the correct answer.

Let their individual Amounts be equal after 't' years. Let their initial investments amount to $$A_V$$ and $$A_J$$ ; 

$$A_V\ =10,000\left(1+\frac{5t}{100}\right)$$ and $$A_J\ =8,000\left(1+\frac{10\left(t-2\right)}{100}\right)$$

Equating both: $$10,000\left(1+\frac{5t}{100}\right)\ =8,000\left(1+\frac{10\left(t-2\right)}{100}\right)$$

On simplifying both sides, we get: $$15t\ =\ 180\ ;\ t\ =\ 12$$

Let 'r' be the root of the function. It follows that f(r) = 0. We can represent this as $$f\left(r\right)=f\left\{5-\left(5-r\right)\right\}$$

Based on the relation: $$f\left(5-x\right)=f\left(5+x\right)$$; $$f\left(r\right)=f\left\{5-\left(5-r\right)\right\}=f\left\{5+\left(5-r\right)\right\}$$

$$\therefore\ f\left(r\right)=f\left(10-r\right)$$

Thus, every root 'r' is associated with another root '(10-r)' [these form a pair]. For even distinct roots, in this case four, let us assume the roots to be as follows: $$r_1,\ \left(10-r_1\right),\ r_2,\ \left(10-r_2\right)$$

The sum of these roots = $$r_1\ +\left(10-r_1\right)+\ r_2+\ \left(10-r_2\right)\ =\ 20$$

Hence, Option D is the correct answer.

Given

A+(B+C)/2=5 => 2A+B+C=10....(i)

(A+C)/2 +B=7 => A+2B+C=14.....(ii)

(i)-(ii)=> B-A=4 => B=4+A.

Given, A, B, C are positive integers

If A=1=>B=5 => C=3

If A=2=>B=6 => C=0 but this is invalid as C is positive.

Similarly if A>2 C will be negative and cases are not valid.

Hence, A+B=6.