CAT 2020 Question Paper Slot 2

Let the distance from A to B be "x", then the distance from B to C will be 3x. Now the speed of Train 2 is double of Train 1. Let the speed of Train 1 be "v", then the speed of Train 2 will be "2v" while travelling from A to B.

Time taken by Train 1 = (x/v)

Time taken by Train 2 = (x/2v)

Now from B to C distance is "3x" and the speed of Train 2 is (v) and the speed of Train 1 is (2v).

Time taken by Train 1 = 3x/2v

Time taken by Train 2 = 3x/v

Total time taken by Train 1 = x/v(1+(3/2)) = (5/2)(x/v)

Total time taken by Train 2 = x/v(3+(1/2))= (7/2)(x/v)

Ratio of time taken = $$\frac{5}{\frac{2}{\frac{7}{2}}}=\frac{5}{7}$$

Let Jack take "t" days to complete the work, then John will take "2t" days to complete the work. So work done by Jack in one day is (1/t) and John is (1/2t) . 

Now let Jim take "m" days to complete the work. According to question, $$\frac{1}{t}+\frac{1}{m}=\frac{3}{2t}\ or\ \frac{1}{m}=\frac{1}{2t\ }or\ m=2t$$ Hence Jim takes "2t" time to complete the work.

Now let the three of them complete the work in "p" days. Hence John takes "p+3" days to complete the work.

$$\frac{1}{2t}\left(m+3\right)=\left(\frac{4}{2t}\right)m$$

$$\frac{1}{2t}\left(m+3\right)=\left(\frac{4}{2t}\right)m$$

or m=1. Hence JIm will take (1+3)=4 days to complete the work. Similarly John will also take 4 days to complete the work

$$f\left(x\right)=\ x^2+ax+b$$ 

$$f\left(x+1\right)=x^2+2x+1+ax+a+b$$

$$f\left(x-1\right)=x^2-2x+1+ax-a+b$$

$$ g(x)=f(x+1)-f(x-1)= 4x+2a$$

Now $$g(20) = 72$$ from this we get $$a = -4$$ ; $$f\left(x\right)=x^2-4x\ +b$$

For this expression to be greater than zero it has to be a perfect square which is possible for $$b\ge\ 4$$

Hence the smallest value of 'b' is 4.

For two years the compound interest is $$\frac{PR(1)}{100}+\frac{PR(1)}{100}\left(1+\frac{PR(1)}{100}\right)$$

For three years the simple interest is $$\frac{9PR}{100}$$

Now R(1)= 5% and R=3%

Hence $$\frac{5P}{100}+\frac{5P}{100}\left(1.05\right)-\frac{9P}{100}=1125$$

$$\frac{-4P}{100}+\frac{5.25P}{100}=1125$$

$$\frac{1.25P}{100}=1125$$

Solving we get P= 90000

We can form the following figure based on the given information:

Since OA = 4 m and OB=3 m; AB = 5 m. OR bisects the chord into PC and QC. 

Since AB = 5 m, we have $$a+b = 5      ...(i)$$  Also, $$4^2\ -k^2=a^2...\left(ii\right)$$ and $$3^2\ -k^2=b^2...\left(iii\right)$$

Subtracting (iii) from (ii), we get: $$a^2\ -b^2=7...\left(iv\right)$$

Substituting (i) in (iv), we get $$a - b = 1.4      ...(v)$$; $$\left[\left(a+b\right)\left(a\ -b\right)=7;\ \therefore\ \left(a-b\right)=\frac{7}{5}\right]$$ 

Solving (i) and (v), we obtain the value of $$a=3.2$$ and $$b=1.8$$

Hence, $$k^2\ =\ 5.76$$

Moving on to the larger triangle $$\triangle\ POC$$, we have $$5^2-k^2=\left(x+a\right)^2$$; 

Substituting the previous values, we get: $$(25-5.76)=\left(x+3.2\right)^2$$ 

$$\sqrt{19.24}=\left(x+3.2\right)$$ or $$x = 1.19 m$$

Similarly, solving for y using $$\triangle\ QOC$$, we get $$y=2.59 m$$

Therefore, $$PQ = 5+2.59+1.19 = 8.78 \approx\ 8.8 m$$

Hence, Option A is the correct answer.

Now car A beats car B by 45km. Let the speed of car A be v(a) and speed of car B be v(b).

$$\frac{v\left(a\right)}{v\left(b\right)}=\frac{m}{m-45}$$ .....(1)where '"m" is the entire distance of the race track.

Moreover $$\frac{v\left(b\right)}{v\left(c\right)}=\frac{m}{m-50}$$.......(2)

and finally $$\frac{v\left(a\right)}{v\left(c\right)}=\frac{m}{m-90}$$......(3)

Multiplying (1) and (2) we get (3). $$\frac{m}{m-90}=\frac{m}{m-45}\left(\frac{m}{m-50}\right)$$

Solving we get m=450 which is the length of the entire race track

Here there are two cases possible

Case 1: When 7 is at the left extreme

In that case 3 can occupy any of the three remaining places and the remaining two places can be taken by (0,1,2,4,5,6,8,9)

So total ways 3(8)(7)= 168

Case 2: When 7 is not at the extremes

Here there are 3 cases possible. And the remaining two places can be filled in 7(7) ways.(Remember 0 can't come on the extreme left)

Hence in total 3(7)(7)=147 ways

Total ways 168+147=315 ways

Let the sides of the rectangle be "a" and "3a" m. Hence the perimeter of the rectangle is 8a.

Let the side of the equilateral triangle be "m" cm. Hence the perimeter of the equilateral triangle is "3m" cm. Now we know that 8a+3m=90......(1)

Moreover area of the equilateral triangle is $$\frac{\sqrt{\ 3}}{4}m^2$$ and area of the rectangle is $$3a^2$$

According to the relation given $$\left(\frac{\sqrt{\ 3}}{4}m^2\right)^{^2}=\ 3a^2$$

$$\frac{3}{16}m^4=\ 3a^2\ or\ a^2=\frac{m^4}{16}$$

$$a=\frac{m^2}{4}$$  

Substituting this in (1) we get $$2m^2+3m-90\ =0$$ solving this we get m=6 (ignoring the negative value since side can't be negative)

Hence a=9 and the longer side of the rectangle will be 3a=27cm

Let the amount of money with Amal and Sunil be 6x and 4x. Now the amount of money with Mita be 5x. Difference between the largest and smallest amount is ₹400 i.e. 6x-4x=400 or 2x=400 or x=200 . Amount of money with Sunil is 200(4)=₹800

On expanding the expression we get $$1-\log_ab+1-\log_ba$$

$$or\ 2-\left(\log_ab+\frac{1}{\log_ba}\right)$$

Now applying the property of AM>=GM, we get that  $$\frac{\left(\log_ab+\frac{1}{\log_ba}\right)}{2}\ge1\ or\ \left(\log_ab+\frac{1}{\log_ba}\right)\ge2$$ Hence from here we can conclude that the expression will always be equal to 0 or less than 0. Hence any positive value is not possible. So 1 is not possible.