CAT 2021 Quant Question:
The sides AB and CD of a trapezium ABCD are parallel, with AB being the smaller side. P is the midpoint of CD and ABPD is a parallelogram. If the difference between the areas of the parallelogram ABPD and the triangle BPC is 10 sq cm, then the area, in sq cm, of the trapezium ABCD is
A. 20
B. 25
C. 40
D. 30
Answer:
We can construct a Quadrilateral ABCD from the given information:
Point P divides the line CD into two equal halves,
Further we know that the area of the parallelogram ABPD is twice the area of the triangle BCP, and hence the heights of both will also be the same.
Now, from here we can say that the area of the parallelogram will be twice the area of the triangle.
Let the area of the triangle be ‘A’. Then the area of the parallelogram BCP will be 2A.
So, this difference will be 2A - A = 10
Hence A = 10
Then the area of the parallelogram would be 2A + A = 30 sq cm
The answer is 30
Choice D is the correct answer.
CAT 2021 Quant Question:
Raj invested ₹ 10000 in a fund. At the end of first year, he incurred a loss but his balance was more than ₹ 5000. This balance, when invested for another year, grew and the percentage of growth in the second year was five times the percentage of loss in the first year. If the gain of Raj from the initial investment over the two year period is 35%, then the percentage of loss in the first year is
A. 15
B. 5
C. 10
D. 70
Answer:
Let the total investment be P, then at the end of the first year, after a loss of x percentage the value of the investment becomes
P(1 - x)
Then in the next year the value of the investment increases by 5x
So, the overall value of the investment becomes
P(1 - x)(1 + 5x)
This is an increase of 35% from the initial investment amount.
Hence,
1.35 P = P (1 - x)(1 + 5x)
Going by the options we can see that the equation is satisfied when x = 10
Hence 10 is the answer
The answer is 10
Choice C is the correct answer.
CAT 2021 Quant Question:
Suppose one of the roots of the equation a 
where a, b and c are rational numbers and a ≠ 0. If 
A. 2
B. 3
C. 4
D. 1
Answer:
For a quadratic equation, 
the roots are given by 
Since the roots, a, b, c are all rational and one of the roots, 2 + √ 3 is irrational, the other root of the equation will also be irrational and a conjugate of 2 + √ 3.
That is, the other root is, 2 - √ 3.
Sum of roots is given by 
= (2 + √ 3) + (2 - √ 3) = 4.
And Product of roots is given by 
But here the quadratic equation is
Therefore,
a = ∓2
|a| = 2
Hence, the answer is '2’
Choice A is the correct answer.
CAT 2021 Quant Question:
For a sequence of real numbers 
for all natural numbers n, then the 
A. 2
B. -2
C. 200
D. -200
Answer:
We are given a generic formula in ‘n’ for the sum of first n terms, with alternating signs.
(Observe that the even terms have a negative sign, while the odd terms have a positive sign)
Alternate Solution:
Observe that 
is an increasing function for all natural numbers.
That is, if x > y; 
+ 2x > 
+ 2y.
So adding the 
term to the sequence increases the sequence by 2n + 1
The even terms of the sequence are subtracted, but then the sum of the terms in the sequence still increases, this means that the term is negative if n is even and term is positive if n is odd, but the magnitude of the term is is always (2n +1)
The answer is '-2’
Choice B is the correct answer.
CAT 2021 Quant Question:
Two pipes A and B are attached to an empty water tank. Pipe A fills the tank while pipe B drains it. If pipe A is opened at 2 pm and pipe B is opened at 3 pm, then the tank becomes full at 10 pm. Instead, if pipe A is opened at 2 pm and pipe B is opened at 4 pm, then the tank becomes full at 6 pm. If pipe B is not opened at all, then the time, in minutes, taken to fill the tank is
A. 140
B. 120
C. 144
D. 264
Answer:
Let the filling rate of pipe A be ‘a’ and the rate of emptying the tank of pipe B be ‘b’.
Then from the given information we can say that, when pipe A is kept open from 2PM to 10PM, i.e. for 8 hours and pipe B drains from 3PM till 10PM (7 hours), the tank gets filled up.
So,
8a - 7b = 1 [1]
Also, from the second statement,
A is kept open from 2 PM to 6 PM (4 hours), and B is kept open for 2 hours (4 PM to 6PM), the tank gets filled up.
4a - 2b = 1 [2]
Multiplying equation [2] and subtracting [1] from it we get
8a - 4b - (8a - 7b) = 2 - 1
7b - 4b = 1
b = 1/3
From here we can compute the value of a
Putting the value of b in eq [2]
4a - 2(1/3) = 1
4a - 2/3 = 1
4a = 1 + 2/3
4a = 5/3
a = 5/12
Hence the rate of filling is 5/12.
If only pipe a is kept open, the tank will get filled in, say ‘n’ hours
n a = 1
n (5/12) = 1
So, n = 12/5
= 2.4 hours
= 144 minutes
The answer is 144
Choice C is the correct answer.
CAT 2021 Quant Question:
A person buys tea of three different qualities at ₹ 800, ₹ 500, and ₹ 300 per kg, respectively, and the amounts bought are in the proportion 2 : 3 : 5. She mixes all the tea and sells one-sixth of the mixture at ₹ 700 per kg. The price, in INR per kg, at which she should sell the remaining tea, to make an overall profit of 50%, is
A. 692
B. 688
C. 653
D. 675
Answer:
Tea of three different qualities costing ₹ 800, ₹ 500, and ₹ 300 per kg, respectively, is bought in amounts which are in the proportion 2 : 3 : 5.
Let us assume that the amounts in which the Tea is bought is 6 : 9 : 15
(This is because 2 + 3 + 5 = 10, which is not a multiple of 6, further reading into the question, we find that we need to deal with one-sixth of the quantity, so having the total quantity as a multiple of 6 is beneficial to us.)
So, Total Cost Price = ₹ 800 * 6 + ₹ 500 * 9 + ₹ 300 * 15 = ₹ 13,800
Profit % = 50%
Profit = 50% of Cost Price = 50% of ₹ 13,800 = ₹ 6,900
Therefore, Selling Price = Cost Price + Profit = ₹ 13,800 + ₹ 6,900 = ₹ 20,700
Since one-sixth that 
The Selling Price of these 5kgs is 5 * 700 = ₹ 3,500
The remaining ₹ 20,700 - ₹ 3,500 = ₹ 17,200 must be generated from selling the remaining 30 - 5 = 25 kgs.
So, the selling price of the remaining tea per kg must be 
= 172 * 4 = ₹ 688 per kg
The answer is 688
Choice B is the correct answer.
CAT 2021 Quant Question:
If a certain weight of an alloy of silver and copper is mixed with 3 kg of pure silver, the resulting alloy will have 90% silver by weight. If the same weight of the initial alloy is mixed with 2 kg of another alloy which has 90% silver by weight, the resulting alloy will have 84% silver by weight. Then, the weight of the initial alloy, in kg, is
A. 3
B. 2.5
C. 4
D. 3.5
Answer:
Let the weight of Silver in the initial Alloy be Ag kgs
and the weight of Copper in the initial Alloy be Cu kgs
So, the total weight of the initial Alloy = (Ag + Cu) kgs
This Alloy is mixed with 3kgs of Pure Silver to form a new mixture.
Total weight of the mixture = (Ag + Cu) + 3 kgs
Weight of Silver in the mixture = (Ag + 3) kgs
Since this mixture contains 90% Silver,
The initial Alloy is mixed with 2kgs of another alloy having 90% Silver by weight.
Total weight of the mixture = (Ag + Cu) + 2 kgs
Weight of Silver in the mixture = (Ag + 90% of 2) kgs = (Ag + 1.8) kgs
Since this mixture contains 84% Silver,
10(Ag + 3) = 9(Ag + Cu + 3)
Ag + 3 = 9 Cu
Ag = 9 Cu - 3
Simplifying (1)
4(9 Cu - 3) = 21 Cu - 3
36 Cu - 12 = 21 Cu - 3
15 Cu = 9
Ag = 9 Cu - 3 = 9(0.6) - 3 = 5.4 - 3 = 2.4
The weight of the initial Alloy = Ag + Cu = 2.4 + 0.6 = 3 kgs
The answer is '3’
Choice A is the correct answer.
CAT 2021 Quant Question:
Anil can paint a house in 12 days while Barun can paint it in 16 days. Anil, Barun, and Chandu undertake to paint the house for ₹ 24000 and the three of them together complete the painting in 6 days. If Chandu is paid in proportion to the work done by him, then the amount in INR received by him is
Answer:
Anil can complete the job in 12 days,
This means, in one day, he completes 
th fraction of the job.
Barun can complete the job in 16 days,
This means, in one day, he completes 
th fraction of the job.
Anil, Barun and Chandu together finish the job in 6 days.
So, the fraction of work done by Arun = 
and, the fraction of work done by Barun = 
The remaining work is done by Chandu.
So, the fraction of work done by Chandu = 
Since Chandu does ⅛ th of the job, he should receive ⅛ th of the Money.
Money received by Chandu =
× ₹ 24000 = ₹ 3000
The answer is '3000’
CAT 2021 Quant Question:
Mira and Amal walk along a circular track, starting from the same point at the same time. If they walk in the same direction, then in 45 minutes, Amal completes exactly 3 more rounds than Mira. If they walk in opposite directions, then they meet for the first time exactly after 3 minutes. The number of rounds Mira walks in one hour is
Answer:
Let ‘a’ be the number of minutes required by Amal to complete one round,
and ‘m’ be the number of minutes required by Mira to complete one round.
In 45, minutes, number of rounds completed by Amal = 
In 45, minutes, number of rounds completed by Mira = 
We know that, in 45 minutes, Amal completed 3 more rounds than Mira.
Therefore, 
In 3, minutes, number of rounds completed by Amal = 
In 3, minutes, number of rounds completed by Mira = 
We know that, in 3 minutes, Amal and Mira together complete one round.
Therefore, 
So, Mera needs 
minutes to complete 1 round.
She needs 15 minutes to complete 2 rounds and
60 minutes to complete 8 rounds.
So Meera covers 8 rounds in 1 hour.
The answer is 8
CAT 2021 Quant Question:
A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is
Answer:
We will select the 4 digits first and arrange them later.
Out of the 4 digits, one of them should be 2 and one of them should be 3.
2, 3, , .
So, we just need to select the other two digits…
The two digits could be (1,1), (2, 2), (3, 3), (1, 2), (1, 3), or (2, 3).
So, the selection of numbers could be…
2, 3, 1, 1
2, 3, 2, 2
2, 3, 3, 3
2, 3, 1, 2
2, 3, 1, 3
2, 3, 2, 3
Each of these selections could be re-arranged in a number of ways.
So total number of possibilities = (12 + 4 + 4 + 12 + 12 + 6) = 50 ways.
Alternate method:
(Arrangements with at least one 2 and one 3) = (All possible arrangements) - (Arrangements with either 1 or 2) - (Arrangements with either 1 or 3) + (Arrangements with only 1)
Think why we need to add (Arrangements with only 1)!
_, _, _, _
(All possible arrangements) = 
Each blank could be any one of 1, 2 or 3.
(Arrangements with either 1 or 2) = 
Each blank could be any one of 1 or 2.
(Arrangements with either 1 or 3) = 
Each blank could be any one of 1 or 3.
(Arrangements with only 1) = 1
Each blank is filled with 1.
(Arrangements with at least one 2 and one 3) = (All possible arrangements) - (Arrangements with either 1 or 2) - (Arrangements with either 1 or 3) + (Arrangements with only 1)
(Arrangements with at least one 2 and one 3) = 
(Arrangements with at least one 2 and one 3) = 81 - 16 - 16 + 1
(Arrangements with at least one 2 and one 3) = 82 - 32 = 50
Hence, the answer is 50