CAT 2020 Question Paper Quant 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

Let the total marks be 100x

Marks obtained by Bishnu = 52x

Marks obtained by Asha = 64x

Marks obtained by Ramesh = 52x+23

Marks obtained by Ramesh = 64x-34

=> 52x+23 = 64x-34

=> x = $$\frac{19}{4}$$

Marks obtained by Geeta =84x = 84*19/4 = 399

Let  $$14^a=36^b=84^c$$ = k

=> a = $$\log_{14}k$$ , b = $$\log_{36}k$$, c=$$\log_{84}k$$

$$6b(\frac{1}{c}-\frac{1}{a})$$ = $$6\cdot\frac{1}{2}\log_6k\left(\log_k84-\log_k14\right)$$ = 3

Given, 

Rate of interest = 10%

Since it is compounded half-yearly, R=5%

n=3

We know, A = $$P\left(1+\frac{R}{100}\right)^{^n}$$

18522 = $$P\left(1+0.05\right)^3$$

=> P = 16000

Let the cost price of 1kg of sugar = Rs 100

The total cost price of 35 kg = Rs3500

Marked up price per kg = Rs 120

GIven, the final profit is 15% => Final SP of 35 kg = 3500 *1.15 = Rs 4025

First 5 kg's are sold at 20% marked up price => $$SP_1=5\cdot100\cdot1.2$$ = Rs 600

Next 15 kgs are sold after giving 10% discount => $$SP_2=15\cdot100\cdot1.2\cdot0.9\ =\ 1620$$

3kgs of sugar got wasted

=> 23 kg of sugar was sold at Rs (600 +1620) = Rs  2220

Remaining 12kg should be sold at Rs 4025 - 2220 = Rs1805

=> SP of 1kg = 1805/12 $$\simeq\ 150$$

Hence, the seller should further mark up by $$\frac{\left(150-120\right)}{120}\cdot100\ =\ 25\%$$

The midpoints of two diagonals of a parallelogram are the same

Hence the midpoint of (2,1) and (-3,-4) lie on $$x+9y+c=0$$

midpoint of (2,1) and (-3,-4) = ($$\frac{2-3}{2},\frac{1-4}{2}$$) = (-1/2 , -3/2)

Keeping this cordinates in the above line equation, we get c = 14

The distance travelled by A between 9:00 Am and 10:30 Am is 3/2*40 =60 km.

Now they are separated by 30 km

Let the time taken to meet =t 

Distance travelled by A in time t + Distance travelled by B in time t = 30

40t + 20t =30 => t=1/2 hour

Hence they meet at 11:00 AM

Possible values of x = 3,4,5,6,7,8,9

When x = 3, there is no possible value of y

When x = 4, the possible values of y = 22

When x = 5, the possible values of y=21,22

When x = 6, the possible values of y = 20.21,22

When x = 7, the possible values of y = 19,20,21,22

When x = 8, the possible values of y=18,19,20,21,22

When x = 9, the possible values of y=17,18,19,20,21,22

The unique values of N = 26,27,28,29,30,31

Two linear equations ax+by= c and dx+ ey = f have a unique solution if $$\frac{a}{d}\ne\ \frac{b}{e}$$

Therefore, $$\frac{k}{4}\ne\ \frac{1}{k}$$ => $$k^2\ne\ 4$$

=> k $$\ne\ $$ |2|

The number of multiples of 2 between 1 and 120 = 60

The number of multiples of 5 between 1 and 120 which are not multiples of 2 = 12

The number of multiples of 7 between 1 and 120 which are not multiples of 2 and 5 = 7

Hence, number of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7 = 120 - 60 - 12 - 7 = 41

$$ab\ =\ 4^{2017}=2^{4034}$$

The total number of factors = 4035.

out of these 4035 factors, we can choose two numbers a,b such that a<b in [4035/2] = 2017.

And since the given number is a perfect square we have one set of two equal factors.

.'. many pairs(a, b) of positive integers are there such that $$a\leq b$$ and $$ab=4^{2017}$$ = 2018. 

Anil and Sunil will meet at a first point after LCM ( $$\frac{3}{15},\frac{3}{10}$$) = 3/5 hr

In the mean time, distance travelled by ravi = 8 * 3/5 = 4.8 km

Given, BC = DE = 4

CD = BE = 5

In triangle ADE, EAD=45^{0}$$

$$\tan\ 45\ =\ \frac{DE}{AE}$$ => AE = 4

Area of trapezium = Area of rectangle BCDE + Area of triangle AED

= 20 + 8 = 28

The required figure is a trapezium with vertices A(0,0), B(2,0), C(2,4) and D(0,6)

AB = 2 BC = 4 and AD = 6

Area of trapezium = $$\frac{1}{2}\left(sum\ of\ the\ opposite\ sides\right)\cdot height$$ = $$\frac{1}{2}\left(4+6\right)\cdot2\ =\ 10$$

The given function is equivalent to f(x) = $$a^x$$

Given, f(5) = 4

=> $$a^5=4=>\ a=4^{\frac{1}{5}}$$

=> f(x) = $$4^{\frac{x}{5}}$$

f(10) - f(-10) = 16 - 1/16 = 15.9375

$$\frac{\left(2\cdot4\cdot8\cdot16\right)}{\left(\log_22^2\right)^2\cdot\left(\log_{2^2}2^3\right)^3\cdot\left(\log_{2^3}2^4\right)^4}\cdot$$

= $$\frac{2^{10}}{4\cdot\left(\frac{3}{2}\right)^3\cdot\left(\frac{4}{3}\right)^4}=24$$

Equation of circle $$x^2+y^2+2gx+2fy+c=0$$

It passes through (0,0), (4,0) and (3,9). Substitute each point in the above equation:

=> On substituting the value (0,0) in the above equation, we obtain: $$c=0$$

=> On substituting the value (4,0) in the above equation, we obtain:  $$16+0+8g+0 = 0$$ ; $$g=-2$$

=> On substituting the value (3,9) in the above equation, we obtain: $$9+81-12+18f = 0$$ ; $$f= -13/3$$

Radius of the circle r = $$\sqrt{\ g^2+f^2-c}$$ => $$r^2=\frac{205}{9}$$

Therefore, Area = $$\pi\ r^2=\frac{205\pi\ }{9}$$