CAT 2021 Question Paper Slot -1 | QA
CAT Quantitative Aptitude (QA) | CAT Previous Year Paper
The CAT 2021 Question Paper Slot 1 QA (Quantitative Ability) section was primarily focused on Arithmetic, followed by Algebra. Within Arithmetic, topics like Speed-Time-Distance, Mixture, and Alligations were heavily represented. However, this year brought a surprise with fewer questions from Geometry compared to previous years. There were 8 TITA questions in total. Overall, the difficulty level of this section was moderate.
CAT 2021 Slot 1 - Quantitative Aptitude
CAT 2021 Slot 1 – QA
2. The natural numbers are divided into groups as (1), (2, 3, 4), (5, 6, 7, 8, 9), ….. and so on. Then, the sum of the numbers in the 15th group is equal to
A. 6119
B. 7471
C. 6090
D. 4941
The answer is Option A
x2+x2 + 4xy+ 4y2 =0
x2 + 4xy+ 4y2 + (x-2y-1)2 = 0
(x + y)2 + (x – 2y-1)2 = 0
We know that,
If a2 + b2 = 0, then a = b = 0
x + 2y = 0 and x-2y-1 = 0
Hence, x – 2y = 1
CAT 2021 Slot 1 – QA
4. Suppose the length of each side of the regular hexagon ABCDEF is 2 cm. If T is the mid point of CD, then the length of AT, in cm, is
A. √14
B. √13
C. √15
D. √12
CAT 2021 Slot 1 – QA
5.
CAT 2021 Slot 1 – QA
6. Amal purchases some pens at 8 each. To sell these, he hires an employee at a fixed wage. He sells 100 of these pens at 12 each. If the remaining pens are sold at 11 each, then he makes a net profit of 300, while he makes a net loss of 300 if the remaining pens are sold at 9 each. The wage of the employee, in INR, is
CAT 2021 Slot 1 – QA
7. How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order
CAT 2021 Slot 1 - Quantitative Aptitude
CAT 2021 Slot 1 – QA
8. Suppose hospital A admitted 21 less Covid infected patients than hospital B, and all eventually recovered. The sum of recovery days for patients in hospitals A and B were 200 and 152, respectively. If the average recovery days for patients admitted in hospital A was 3 more than the average in hospital B then the number admitted in hospital A was
CAT 2021 Slot 1 – QA
9. The number of integers n that satisfy the inequalities | n – 60 | < | n – 100 |< | n – 20 | is
A. 18
B. 20
C. 21
D. 19
CAT 2021 Slot 1 – QA
10.
CAT 2021 Slot 1 – QA
11. Anu, Vinu and Manu can complete a work alone in 15 days, 12 days and 20 days, respectively. Vinu works everyday. Anu works only on alternate days starting from the first day while Manu works only on alternate days starting from the second day. Then, the number of days needed to complete the work is
A. 6
B. 5
C. 8
D. 7
CAT 2021 Slot 1 – QA
12. Onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg, respectively. A family spends a fixed amount of money on onion for each of the first three months, and then spends half that amount on onion for each of the next two months. The average expense for onion, in rupees per kg, for the family over these 5 months is closest to
A. 18
B. 16
C. 20
D. 26
CAT 2021 Slot 1 – QA
13. If the area of regular hexagon is equal to the area of an equilateral triangle of side 12 cm, then the length, in cm, of each the hexagon is
A. 6√6
B. √6
C. 4√6
D. 2√6
CAT 2021 Slot 1 - Quantitative Aptitude
CAT 2021 Slot 1 – QA
15. The amount Neeta and Geeta together earn in a day equals what Sita alone earns in 6 days. The amount Sita and Neeta together earn in a day equals what Geeta alone earns in 2 days. The ratio of the daily earnings of the one who earns the most to that of the one who earns the least is
A. 7 : 3
B. 11 : 3
C. 11 : 7
D. 3 : 2
CAT 2021 Slot 1 – QA
17. Amar, Akbar and Anthony are working on a project. Working together Amar and Akbar can complete the project in 1 year, Akbar and Anthony can complete in 16 months, Anthony and Amar can complete in 2 years. If the person who is neither the fastest nor the slowest works alone, the time in months he will take to complete the project is
CAT 2021 Slot 1 - Quantitative Aptitude
CAT 2021 Slot 1 – QA
19. The strength of an indigo solution in percentage is equal to the amount of indigo in grams per 100 cc of water. Two 800 cc bottles are filled with indigo solutions of strengths 33% and 17%, respectively. A part of the solution from the first bottle is thrown away and replaced by an equal volume of the solution from the second bottle. If the strength of the indigo solution in the first bottle has now changed to 21% then the volume, in cc, of the solution left in the second bottle is
CAT 2021 Slot 1 – QA
20. A circle of diameter 8 inches is inscribed in a triangle ABC where angle ABC = 90°. If BC = 10 inches then the area of the triangle in square inches is
CAT 2021 Slot 1 – QA
21.
The answer option is A.
If 1/(√y+√z) is the arithmetic mean of 1/√x +1/√z & 1/(√x+√y)
Then = 2/(√y+√x) = 1/(√x+√z) + 1/(√x+√y)
Rationalizing,
(2 √y-√z)/(y-z) = (√x-√y)/(x-z) + (√x-√y)/(x-y) ———- (1)
So, starting with options
If y, x & z are in AP,
⇒x-y=z-x=d & z-y=2d
⇒from (1),
RHS,
= (√x-√z)/(-d) + (√x-√y)/d=(√x-√y+√z-√x)/d
= (√z-√y)/d
LHS =(2(√x-√(y)))/(-2d)=(√z-√y)/d= RHS.
Hence, LHS = RHS
So, ⇒(b)is true.
Using the examples of the circle L
x^2+y^2+4x-6y-3=0, we can find the point of center of the circle and the radius (x^2+4x+4)+(y^2-6y+9)-4-9-3=0
(〖x+2)]^2+(〖y-3)〗^2=16
So, radius (r) = 4, Centre = (-2, 3)
DIAGRAM
OP being the radius of bigger circle (L) with center at O.
We can find the measurement of OP using trigonometry.
Sin 30^0 =OQ/OP=1/2=4/OP
⇒OP=8=radius of bigger circle (L)
Therefore, Example of the bigger circle (L) with centre (-2, 3)
and radius 8 is 〖(x+2)〗^2+〖(y-3)〗^2=8^2
Substituting x=6,in the above circle equation we can find.
〖(6+2)〗^2+〖(y-3)〗^2=64
8^2+〖(y-3)〗^2=64
(y-3)^2=0
⇒ y = 3.
So, the solution of this question is (a) (6, 3).
CAT 2021 Slot 1 - Quantitative Aptitude