Quant For CAT 2020 (Slot - 3) PDF Download

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 ,

⇒ 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 , 

⇒ x = 92%

Given,

Given,

⇒ 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, 

∴ 

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,

⇒ n = 180.

∴ Extra 180 - 140 = 40 workers are needed. 

x₁​ = −1
x₁​ = x₂ + 2 ⇒ x₂ = x₁ - 2 = -3
Similarly, 
x₃ = x₁ - 5 = -6
x₄ = -10
.
.
The series is -1, -3, -6, -10, -15......
When the differences are in AP, then the nth term is 

and finally log_a 2 = 3
Substituting this in (1) we get log_a 5+log_a 3 = A − 3
Now we have two equations in two variables (1) and (2) . On solving we get

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, 
⇒ 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 = 

= 28  km/hr

Equation of circle x² + y² + 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 = -2g = −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

Therefore, Area = π r²

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

Let 14^a =36^b =84^c = k

Given, 
Rate of interest = 10%
Since it is compounded half-yearly, R = 5%
n = 3
We know, 

18522 = P(1 + 0.05)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 ⇒ SP₁​ = 5⋅100⋅1.2 = Rs 600
Next 15 kgs are sold after giving 10% discount ⇒ SP₂ ​= 15⋅100⋅1.2⋅0.9 = 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 ≃ 150
Hence, the seller should further mark up by 

= 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) = 

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 a/d ≠ b/e.
Therefore k/4 ≠ 1/k  ⇒ k²  ≠ 4
⇒ k  ≠  |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 = 42⁰¹⁷ = 2⁴⁰³⁴ 
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 ≤ b and ab = 4²⁰¹⁷
= 2018. 

Anil and Sunil will meet at a first point after LCM (3/15, 3/10) = 3/5 hr
In the mean time, distance travelled by ravi = 8 x 3/5 = 4.8 km

Given, BC = DE = 4
CD = BE = 5
In triangle ADE, EAD=45^{0}$$
tan 45 = 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 = 1/2 (sum of the opposite sides) . height = 1/2(4 + 6) . 2
= 10

The given function is equivalent to f(x) = a^x 
Given, f(5) = 4

⇒ 

⇒ 

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