CAT 2020 Quant Question: Averages
A batsman played n + 2 innings and got out on all occasions. His average score in these n + 2 innings was 29 runs and he scored 38 and 15 runs in the last two innings. The batsman scored less than 38 runs in each of the first n innings. In these n innings, his average score was 30 runs and lowest score was x runs. The smallest possible value of x is
A. 1
B. 3
C. 2
D. 4
Answer:
The average of the first n innings = 30
The average of all the n+2 innings = 29
29(n + 2) = n(30) + 2(26.5)
n = 5 (You may choos eto use the allegation method too)
The average of the first 5 innings = 30 and in none of the innings did the batsman score 38 or above.
Let’s assume the scores of the 5 innings of the batsman in ascending order to be 30, 30, 30, 30, 30. (This complies with the avberage being 30)
The maximum score that he can score is 37 runs, Let’s distribute the score of the first innings among the rest 4, as 7 per innings so that the score of the first innings is minimized.
Now the scores will be 2, 37, 37, 37, 37
Now we can’t distribute the score of the first innings anymore, because, that makes the score of atleast one of the remainin 4 overs 38 or greater.
Hence the least score of any inning is 2.
The answer is, “2”
CAT 2020 Quant Question: Mixtures
Two alcohol solutions, A and B, are mixed in the proportion 1:3 by volume. The volume of the mixture is then doubled by adding solution A such that the resulting mixture has 72% alcohol. If solution A has 60% alcohol, then the percentage of alcohol in solution B is
A. 94%
B. 92%
C. 90%
D. 89%
Answer:
We are told that the A and B are mixed in the ratio 1:3
That means, one unit of A is mixed with 3 units of B
Observe that the total Volume is 1 + 3 = 4 units.
The volume is doubled by adding A, this implies that a quantity of 4 units of A is added to the existing 4units of the mixture.
The total volume will now be 4 + 4 = 8 units.
Of which three units are B and the rest 5 units are A. Hence the ratio of A to B in the final mixture is 5 : 3
We are told that A has a concentration of 60% alcohol and the final mixture has a concentration of 72% alcohol
300 + 3 × Concentration of B = 576
3 × Concentration of B = 576 - 300 = 276
3 × Concentration of B = 276 = 270 + 6
3 × Concentration of B = 270 + 6
Concentration of B =
Concentration of B = 90 + 2 = 92.
The answer is, “92%”
CAT 2020 Quant Question: Co-ordinate Geometry
The vertices of a triangle are (0,0), (4,0) and (3,9). The area of the circle passing through these three points is
Answer:
A circle is circumscribed on the points (0,0) (4,0) and (3,9)
We are asked to find the area of this circle
We know that the area of a triangle with sides a, b and c, inscribed in a circle of radius R is given by
With the help of the co-ordinates of the triangle it is easy to find the height of the triangle, which is QS = 9 units.
Hence the area of the triangle = A = 
× base × height = × PR × QS = × 4 × 9 = 18 sq. units
a = PR = 4 units
Now that we know the radius of the circle the area of the circle 
Hence, the answer is, 
CAT 2020 Quant Question: Simple interest & Compound interest
A person invested a certain amount of money at 10% annual interest, compounded half-yearly. After one and a half years, the interest and principal together became Rs 18522. The amount, in rupees, that the person had invested is
Answer:
A 10% annual interest is a 5% bi-annual interest.
Hence interst for half an year is 5%.
The Principal, P is invested for 1.5 years or 3 terms of half years.
P = 2 × 20 × 20 × 20 = 16000.
The answer is, “16000”
CAT 2020 Quant Question: Speed, time & distance
A and B are two railway stations 90 km apart. A train leaves A at 9:00 am, heading towards B at a speed of 40 km/hr. Another train leaves B at 10:30 am, heading towards A at a speed of 20 km/hr. The trains meet each other at
A. 11 : 20 am
B. 11 : 00 am
C. 10 : 45 am
D. 11 : 45 am
Answer:
Let’s represent the positions of the trains at 9:00 AM
The positions of the two trains at 10:30 AM will be
The answer is, “11 : 00 am”
CAT 2020 Quant Question: Speed, time & distance
Vimla starts for office every day at 9 am and reaches exactly on time if she drives at her usual speed of 40 km/hr. She is late by 6 minutes if she drives at 35 km/hr. One day, she covers two-thirds of her distance to office in one-thirds of her usual time to reach office, and then stops for 8 minutes. The speed, in km/hr, at which she should drive the remaining distance to reach office exactly on time is
A. 27
B. 28
C. 29
D. 26
Answer:
Vimala’s speed has decreased from 40kmph to 35 kmph. That means her current speed is 
of the original Speed. That means, her current speed is 
of the original speed. Therefore, the time taken will be 
of the original time taken.
Since she is 6 minutes late when travelling at 35kmph, 
of the original time taken is 6 minutes.
Therefore the total time taken originally is 7 × 6 = 42 minutes.
of the original time taken = 
= 14 minutes.
of the total distance is generally covered in 14 minutes.
Currently the same distance is to be covered in (42 - 14 - 8 = 20) minutes.
So, to reach the Destination on time, Vimala, must cover the same of the Distance in 20 minutes instead of 14 minutes.
That means she has to 
of the initial time.
Which is 
of the original time
This implies that, in order to reach at the same point of time. The speed had to 
or 70% of the original speed.
So, she has to travel at 70% of 40kmph = 28kmph.
The answer is, “28”
CAT 2020 Quant Question: Linear & quadratic equations
Let m and n be positive integers, If 
+ mx + 2n = 0 and + 2nx + m = 0 have real roots, then the smallest possible value of m + n is
A. 8
B. 6
C. 5
D. 7
Answer:
The roots of a Quadratic Equation of the form, 
+ bx + c = 0
are given by x = 
In order for these roots to be real, the portion under the square root should not be negative.
In other words 
- 4ac’, determines whether the roots are real or imaginary.
Hence, rightly, it is called the Determinant(D).
If the Determinant, D is greater than or equal to 0, then the equation has real roots.
For D to be greater than or equal to 0, ≥ 4ac.
We are told that 
+ mx + 2n = 0 and + 2nx + m = 0 have real roots,
That means, 
≥ 4(2n) and 
≥ 4m.
Instead of trying to solve through equations, the two inequalities ≥ 8n and 
≥ m.
Let’s try to solve inputting specific numbers.
If n = 1; ≥ 8; m ≥ 3
If m ≥ 3 at n = 1, ≥ m stands invalid.
If n = 2; ≥ 16; m ≥ 4
If m ≥ 4 at n = 2, 
≥ m stands valid, if m = 4 and n = 2
Hence 4, 2 is the smallest(and the only) pair that m, n can take.
So, the minimum sum of m and n is 4+2 = 6.
Hence, the answer is, “6”
CAT 2020 Quant Question: Geometry
In a trepezium ABCD, AB is parallel to DC, BC is perpendicular to DC and ∠BAD = 45°. If DC = 5 cm, BC = 4 cm, the area of the trepezium in sq. cm is
Answer:
Let’s draw the Trapezium.
Now drop a perpendicular from D on to AB touching AB at E.
In ΔAED, ∠AED = 90°, ∠DAE = 45°, therefore ∠EDA = 45°.
Clearly ΔAED is an iscoceles right triangle and Quadrilateral BCDE is a rectangle.
Therefore, BC = DE = 4cm.
Since ΔAED is an iscoceles triangle, AE = ED = 4cm.
Area of the Trepezium = Area of ΔAED + Area of Rectangle BCDE
Area of the Trepezium = 
× AE × ED + DC × BC
Area of the Trepezium = × 4 × 4 + 5 × 4
Area of the Trepezium = 8 + 20
Area of the Trepezium = 28 
Hence, the answer is, “28”
CAT 2020 Quant Question: Co-ordinate Geometry
The points (2 , 1) and (-3 , -4) are opposite vertices of a parellelogram. If the other two vertices lie on the line x + 9y + c = 0, then c is
A. 15
B. 13
C. 14
D. 12
Answer:
The line x + 9y + c = 0, forms one of the diagonal of the parallelogram.
The other diagonal connects the points (2, 1) and (-3, -4).
The diagonals of a parellogram bisect each other, hence the line x + 9y + c = 0 should pass through the midpoint of the line joining (2, 1) and (-3, -4).
The midpoint of the line joining (2, 1) and (-3, -4) = 
= (-0.5 , -1.5)
Therefore, the line x + 9y + c = 0 passes through (-0.5 , -1.5),
Hence, (-0.5) + 9(-1.5) + c = 0
(-0.5) + 27(-0.5) + c = 0
28 (-0.5) + c = 0
-14 + c = 0
c = 14.
Hence, the answer is, “14”
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CAT 2020 Quant Question: Digits
How many pairs (a,b) of positive integers are there such that a ≤ b and ab = 
A. 2019
B. 2018
C. 2020
D. 2017
