CAT 2021 Quant Question:
In a triangle ABC , ∠ BCA =50°. D and E are points on AB and AC, respectively, such that AD = DE. If F is a point on BC such that BD = DF, then ∠ FDE, in degrees, is equal to
A. 100
B. 80
C. 96
D. 72
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Answer:
We draw the diagram carefully with the given information.
From the triangle ABC,
∠A + ∠B + ∠C = 180°
∠A + ∠B + 500 = 180°
∠A + ∠B = 130°
In the quadrilateral CFDE,
∠C + ∠F + ∠D + ∠E = 360°
50° + 180° - ∠A + ∠x + 180° - ∠B = 360°
50° + ∠x = ∠A + ∠B
50° + ∠x = 130°
∠x = 80°
∠FDE = 80°
Hence, the answer is '80’
Choice B is the correct answer.
CAT 2021 Quant Question:
For a real number a, if 
= 4 then a must lie in the range
A. 4 < a < 5
B. 3 < a < 4
C. a > 5
D. 2 < a < 3
Answer:
Hence, the answer is '4 < a < 5’
Choice A is the correct answer.
CAT 2021 Quant Question:
The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is
Answer:
The arithmetic mean of the scores of 25 students = 50
Sum of scores of these students = 25 × 50 = 1250
For the scores of the top 5 students to be as high as possible, the score of the bottom 20 students should be as low as possible.
The minimum score is 30, and the scores of the bottom 20 students are distinct integers.
In order for the bottom 20 scores to be as low as possible, they must be,
30, 31, 32, … 49
Sum of the bottom 20 scores
= 30 + 31 + 32 + … + 49
= (30 + 0) + (30 + 1) + (30 + 2) + … + (30 + 19)
= 20 × 30 + (0 + 1 + 2 + 3 + … + 19)
= 600 + 
= 600 + 190
= 790
Therefore, the maximum sum of the scores of the top 5 students = 1250 - 790 = 460
The maximum score of the five toppers, since they all scored the same marks = 
The answer is 92
CAT 2021 Quant Question:
One part of a hostel’s monthly expenses is fixed, and the other part is proportional to the number of its boarders. The hostel collects ₹ 1600 per month from each boarder. When the number of boarders is 50, the profit of the hostel is ₹ 200 per boarder, and when the number of boarders is 75, the profit of the hostel is ₹ 250 per boarder. When the number of boarders is 80, the total profit of the hostel, in INR, will be
A. 20800
B. 20200
C. 20500
D. 20000
Answer:
Let the fixed cost be ₹ F and the variable cost be ₹ V.
Since the profit per border is ₹200 when there are 50 borders
The expenses of the Hostel is,
F + 50(V) = 50 (1600 - 200)
F + 50(V) = 50 (1400) — (1)
Since the profit per border is ₹250 when there are 75 borders
The expenses of the Hostel is,
F + 75(V) = 75 (1600 - 250)
F + 75(V) = 75 (1350) — (2)
(2) - (1)
25(V) = 75(1350) - 50(1400)
25(V) = 25( 3(1350) - 2(1400) )
V = 3(1350) - 2(1400)
V = 4050 - 2800
V = 1250
F + 75(V) = 75 (1350)
F + 75(1250) = 75 (1350)
F = 75(100) = 7500
The Expenditure for 80 borders will be,
= F + 80(V)
= 7500 + 80(1250)
The revenue collected from 80 students is,
= 80(1600)
Hence, the profit is,
= 80(1600) - (7500 + 80(1250))
= 80(1600 - 1250) - 7500
= 80(350) - 7500
= 100(8×35 - 75)
= 20500
Hence the total profit when there are 80 borders is ₹20500.
The answer is 20500
Choice C is the correct answer.
CAT 2021 Quant Question:
One day, Rahul started a work at 9 AM and Gautam joined him two hours later. They then worked together and completed the work at 5 PM the same day. If both had started at 9 AM and worked together, the work would have been completed 30 minutes earlier. Working alone, the time Rahul would have taken, in hours, to complete the work is
A. 12
B. 11.5
C. 12.5
D. 10
Answer:
Let R be the fraction of work done by Rahul in 1 hour.
and G be the fraction of work done by Gautam in 1 hour.
Initially Rahul works from 9AM to 5PM (8 hours) and Gautam works for 2 hours less.
8R + 6G = 1 whole unit of work
If they start together they finish 30 minutes earlier or if they start at 9 AM, they finish at 4:30PM (7.5 hours)
7.5R + 7.5G = 1 whole unit of work
8R + 6G = 7.5R + 7.5G
0.5R = 1.5G
R = 3G
This means Rahul is thrice as efficient as Gautam.
8R + 6G = 1 whole unit of work
8R + 2R = 1 whole unit of work
10R = 1 whole unit of work
R is the fraction of work done by Rahul in 1 hour.
Since 10R = 1
R alone takes 10 hours to finish the job.
The answer is '10’
Choice D is the correct answer.
CAT 2021 Quant Question:
The total of male and female populations in a city increased by 25% from 1970 to 1980. During the same period, the male population increased by 40% while the female population increased by 20%. From 1980 to 1990, the female population increased by 25%. In 1990, if the female population is twice the male population, then the percentage increase in the total of male and female populations in the city from 1970 to 1990 is
A. 68.75
B. 68.50
C. 68.25
D. 69.25
Answer:
From 1970 to 1980,
the male population increased by 40%
the female population increased by 20%
the overall population increased by 25%
1.4M + 1.2F = 1.25(M + F)
1.4M + 1.2F = 1.25M + 1.25F
1.4M - 1.25M = 1.25F - 1.2F
0.15M = 0.05F
F = 3M
From 1980 to 1990,
Female population increased by 25%
Therefore, the female population in 1990 = 1.25 × 1.2F = 1.5F
Since, the female population in 1990 is twice the male population,
Male population in 1990 = 1.5F ÷ 2 = 0.75F
Since F = 3M,
Male population in 1990 = 2.25M
Total population in 1970 = M + F = M + 3M = 4M
Total population in 1990 = 2.25M + 1.5F = 2.25M + 4.5M = 6.75M
The percentage increase in population = 
The answer is '68.75’
Choice A is the correct answer.
CAT 2021 Quant Question:
Let ABCD be a parallelogram. The lengths of the side AD and the diagonal AC are 10 cm and 20 cm, respectively. If the angle ∠ADC is equal to 30° then the area of the parallelogram, in sq. cm, is
Answer:
Revising the Cosine rule and the area of the triangle using the Sine rule…
We draw the described parallelogram ABCD.
Applying the cosine rule in triangle ADC,
DC is the length of one of the sides of the parallelogram, hence it can’t be negative.
Area of the parallelogram ABCD = 5 (DC)
Choice D is the correct answer.
CAT 2021 Quant Question:
A park is shaped like a rhombus and has area 96 sq m. If 40 m of fencing is needed to enclose the park, the cost, in INR, of laying electric wires along its two diagonals, at the rate of ₹125 per m, is
Answer:
The are of the rhombus is given by,
Also,
The cost of laying electric wires along the diagonals at the rate of ₹125 per meter
= 28 × 125
= ₹3500
The answer is 3500
CAT 2021 Quant Question:
If 3x+2|y|+y=73x+2|y|+y=7 and x+|x|+3y=1x+|x|+3y=1, then x+2yx+2y is
Answer:
We are given two equations: 3x + 2|y| + y = 7 and x + |x| + 3y = 1
|x| or the modulus of x, is a function of x, that gives the magnitude of x.
|x| = -(x); if x is negative, and
|x| = x; if x is non-negative.
Therefore, depending on whether ‘x’ and ‘y’ are positive or negative, we assume the following cases for the two equations given.
Case (i): ‘x’ and ‘y’ are both positive.
|x| = x and |y| = y
3x + 2|y| + y = 7
x + |x| + 3y = 1
3x + 2y + y = 7
x + x + 3y = 1
3x + 3y = 7
2x + 3y = 1
Solving the two equations, we get x = 6 and y = -11/3.
Since this is contradictory to the assumption that y is positive, we discard this case.
Case (ii): ‘x’ and ‘y’ are both negative.
|x| = -x and |y| = -y
3x + 2|y| + y = 7
x + |x| + 3y = 1
3x - 2y + y = 7
x - x + 3y = 1
3x - y = 7
3y = 1
Solving the two equations, we get x = 22/9 and y = 1/3.
Since this is contradictory to the assumption that both x and y are both negative, we discard this case.
Case (iii): ‘x’ is negative and ‘y’ is positive.
|x| = -x and |y| = y
3x + 2|y| + y = 7
x + |x| + 3y = 1
3x + 2y + y = 7
x - x + 3y = 1
3x + 3y = 7
3y = 1
Solving the two equations, we get x = 2 and y = 1/3.
Since this is contradictory to the assumption that both x is negative, we discard this case.
Case (iv): ‘x’ is positive and ‘y’ is negative.
|x| = x and |y| = -y
3x + 2|y| + y = 7
x + |x| + 3y = 1
3x - 2y + y = 7
x + x + 3y = 1
3x - y = 7
2x + 3y = 1
Solving the two equations, we get x = 2 and y = -1.
This satisfies the assumption that x is positive and y is negative.
Hence x = 2 and y = -1
Therefore, x + 2y = 2 + 2(-1) = 0
Hence, x + 2y = 0.
Hence, the answer is '0’
Choice A is the correct answer.
CAT 2021 Quant Question:
In a tournament, a team has played 40 matches so far and won 30% of them. If they win 60% of the remaining matches, their overall win percentage will be 50%. Suppose they win 90% of the remaining matches, then the total number of matches won by the team in the tournament will be
A. 86
B. 84
C. 78
D. 80
Answer:
Let the number of matches to be played be ‘x’
We are given that 40 matches are already played and 30% of them are won.
If 60% of the remaining matches are won, then the overall win percentage is 50%.
30% of 40 + 60% of x = 50% of (40 + x)
0.3 (40) + 0.6 (x) = 0.5 (40 + x)
12 + 0.6 (x) = 20 + 0.5 (x)
0.1 (x) = 8
x = 80
The number of matches to be played is 80.
If the team wins 90% of the remaining matches, it would win 90% of 80 = 72 matches.
Total matches won by the team
= 30% of 40 + 72
= 12 + 72 = 84
The answer is '84’
Choice B is the correct answer.