CAT 2020 Question Paper Slot 3

Initially let's consider A and B as one component

The volume of the mixture is doubled by adding A(60% alcohol) i.e they are mixed in 1:1 ratio and the resultant mixture has 72% alcohol.

Let the percentage of alcohol in component 1 be 'x'.

Using allegations , $$\frac{\left(72-60\right)}{x-72}=\frac{1}{1}$$ => x= 84

Percentage of alcohol in A = 60% => Let's percentage of alcohol in B = x%

 The resultant mixture has 84% alcohol. ratio = 1:3

Using allegations , $$\frac{\left(x-84\right)}{84-60}=\frac{1}{3}$$

=> x= 92%

Given, $$\frac{\text{sum of scores in n matches+38+15}}{n+2}=29$$

Given, $$\frac{\text{sum of scores in n matches}}{n}=30$$

=> 30n + 53 = 29(n+2) => n=5

Sum of the scores in 5 matches = 29*7 - 38-15 = 150

Since the batsmen scored less than 38, in each of the first 5 innings. The value of x will be minimum when remaining four values are highest

=> 37+37+37+37 + x = 150

=> x = 2

To have real roots the discriminant should be greater than or equal to 0.

So, $$m^2-8n\ge0\ \&\ 4n^2-4m\ge0$$

=> $$m^2\ge8n\ \&\ n^2\ge m$$

Since m,n are positive integers the value of m+n will be minimum when m=4 and n=2.

.'. m+n=6.

Let the desired efficiency of each worker '6x' per day.

140*6x*200= 6 km ...(i)

In 60 days 60/200*6=1.8 km of work is to be done but actually 1.5km is only done.

Actual efficiency 'y'= 1.5/1.8 *6x =5x.

Now, left over work = 4.5km which is to be done in 140 days with 'n' workers whose efficiency is 'y'.

=> n*5x*140=4.5 ...(ii)

(i)/(ii) gives,

$$\frac{\left(140\cdot6x\cdot200\right)}{\left(n\cdot5x\cdot140\right)}=\frac{6}{4.5}$$

=> n=180.

.'. Extra 180-140 =40 workers are needed. 

$$x_1=-1$$

$$x_1=x_2+2$$ => $$x_2=x_1-2$$ = -3

Similarly, 

$$x_3=x_1-5=-6$$

$$x_4=-10$$

.

.

The series is -1, -3, -6, -10, -15......

When the differences are in AP, then the nth term is $$-\frac{n\left(n+1\right)}{2}$$

$$x_{100}=-\frac{100\left(100+1\right)}{2}=-5050$$

$$\log_a30=A\ or\ \log_a5+\log_a2+\log_a3=A$$...........(1)

$$\log_a\left(\frac{5}{3}\right)=-B\ or\ \log_a3-\log_a5=B$$.............(2)

and finally $$\log_a2=3$$

Substituting this in (1) we get $$\log_a5+\log_a3=A-3$$

Now we have two equations in two variables (1) and (2) . On solving we get

$$\log_a3=\frac{\left(A+B-3\right)}{2\ }or\ \log_3a=\frac{2}{A+B-3}$$

Let tom's age = x

=> Dick=3x

=>harry = 6x

Given, 

3x+1 = (x+3x+6x)/3 

=> x= 3

Hence, Harry's age = 18 years

Let distance = d

Given, $$\frac{d}{35}-\frac{d}{40}=\frac{6}{60}$$

=> d = 28km

The actual time  taken to travel 28km = 28/40 = 7/10 hours = 42 min.

Given time taken to travel 58/3 km = 1/3 *42 = 14 min.

Then a break of 8 min.

To reach on time, he should cover remaining 28/3 km in 20 min => Speed = $$\frac{\left(\frac{28}{3}\right)}{\frac{20}{60}}=28\ $$ km/hr

$$0.2<\frac{n}{11}<0.5$$ 

=> 2.2<n<5.5

Since n is an even natural number, the value of n = 4

$$0.2<\frac{m}{20}<0.5$$  => 4< m<10. Possible values of m = 5,6,7,8,9

Since $$0.2<\frac{n}{m}<0.5$$, the only possible value of m is 9

Hence m-2n = 9-8 = 1

Total number of numbers from 100 to 999 = 900

The number of three digits numbers with unique digits:

_ _ _

The hundredth's place can be filled in 9 ways ( Number 0 cannot be selected)

Ten's place can be filled in 9 ways

One's place can be filled in 8 ways

Total number of numbers = 9*9*8 = 648

Number of integers in the set {100, 101, 102, ..., 999} have at least one digit repeated = 900 - 648 = 252