CAT 2022 QA Slot 2 questions and solutions

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1. CAT 2022 QA Slot 2 | Algebra - Logarithms

The number of distinct integer values of n satisfying $\frac{4 - \log_{2} n}{3 - \log_{4} n}$ < 0, is

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Answer: 47

Video Explanation

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Explanation :

Given, $(\frac{4 - \log_{2} n}{3 - \log_{4} n})$ < 0

Case 1: 4 – log₂n < 0 and 3 – log₄n > 0
⇒ log₂n > 4 and log₄n < 3
⇒ n > 16 and n < 64
∴ integral values of n can be 17, 18, …, 63 i.e., 47 values.

Case 2: 4 – log₂n > 0 and 3 – log₄n < 0
⇒ log₂n < 4 and log₄n > 3
⇒ n < 16 and n > 64
∴ No integral values of n is possible.

Hence, 47.

2. CAT 2022 QA Slot 2 | Modern Math - Permutation & Combination

The number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once, is

A.
1200
B.
1420
C.
1440
D.
1480

3. CAT 2022 QA Slot 2 | Algebra - Functions & Graphs

Suppose for all integers x, there are two functions f and g such that f(x) + f(x - 1) - 1 = 0 and g(x) = x². If f(x² - x) = 5, then the value of the sum f(g(5)) + g(f(5)) is

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4. CAT 2022 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement

There are two containers of the same volume, first container half-filled with sugar syrup and the second container half-filled with milk. Half the content of the first container is transferred to the second container, and then the half of this mixture is transferred back to the first container. Next, half the content of the first container is transferred back to the second container. Then the ratio of sugar syrup and milk in the second container is

A.
5 : 6
B.
4 : 5
C.
5 : 4
D.
6 : 5

5. CAT 2022 QA Slot 2 | Algebra - Progressions

Consider the arithmetic progressions 3, 7, 11, ... and let An dentoe the sum of the first n terms of this progression. Then the value of $\frac{1}{25} {\sum A_{n}}_{n = 1}^{25}$

A.
415
B.
404
C.
455
D.
442

Answer: Option C

Video Explanation

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Explanation :

A_n = 3 + 7 + 11 + ...

⇒ A_n = $\frac{n}{2}$ [2 × 3 + (n - 1) × 4] = $\frac{n}{2}$ [4n + 2] = 2n² + n

Now, $\frac{1}{25}$ $\sum_{n = 1}^{25}$ [2n² + n]

= $\frac{1}{25}$ [2 × $(\frac{25 \times 26 \times 51}{6})$ + $(\frac{25 \times 26}{2})$ ]

= $\frac{1}{25}$ [25 × 26 × 17 + 25 × 13]

= 26 × 17 + 13 = 455

Hence, option (c).

6. CAT 2022 QA Slot 2 | Algebra - Number Theory

If a and b are non-negative real numbers such that a + 2b = 6, then the average of the maximum and minimum values of (a + b) is:

A.
4
B.
4.5
C.
3.5
D.
3

Answer: Option B

Video Explanation

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Explanation :

Given, a + 2b = 6.
⇒ a + b = 6 - b

∴ (a + b) will be maximum when b is least. Least value of b can be 0, since b cannot be negative.
⇒ (a + b)_max = 6

∴ (a + b) will be minimum when b is highest. Highest value of b can be 3, since a cannot be negative.
⇒ (a + b)_min = 3

⇒ Average of highest and lowest values of (a + b) = $\frac{3 + 6}{2}$ = 4.5

Hence, option (b).

7. CAT 2022 QA Slot 2 | Arithmetic - Simple & Compound Interest

Mr. Pinto invests one-fifth of his capital at 6%, one-third at 10% and the remaining at 1%, each rate being simple interest per annum. Then, the minimum number of years required for the cumulative interest income from these investments to equal or exceed his initial capital is

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Answer: 20

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Explanation :

Let the total investment be Rs. 1500.

⇒ Rs. 300 is invested at 6% ⇒ Interest/year = Rs. 18
⇒ Rs. 500 is invested at 10% ⇒ Interest/year = Rs. 50
⇒ Rs. 700 is invested at 1% ⇒ Interest/year = Rs. 7

∴ Total interest received/year = 18 + 50 + 7 = Rs. 75

⇒ Time required to receive Rs. 1500 as interest = $\frac{1500}{75}$ = 20 years.

Hence, 20.

8. CAT 2022 QA Slot 2 | Arithmetic - Time & Work

Working alone, the times taken by Anu, Tanu and Manu to complete any job are in the ratio 5 : 8 : 10. They accept a job which they can finish in 4 days if they all work together for 8 hours per day. However, Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day. Then, the number of hours that Manu will take to complete the remaining job working alone is:

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Answer: 6

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Explanation :

Ratio of time taken by Anu, Tanu and Manu is 5 : 8 : 10.
⇒ Ratio of efficiencies of Anu, Tanu and Manu = $\frac{1}{5}$ : $\frac{1}{8}$ : $\frac{1}{10}$ = 8 : 5 : 4
Let their efficiencies be 8x, 5x and 4x respectively per hour.

Total work done by them in 4 days = (8x + 5x + 4x) × 4 × 8 = 17x × 32 = 544x

Now, Anu and Tanu worked by 6 days working 6 hours 40 minute i.e., 6 $\frac{2}{3}$ hours daily.
∴ Worked completed by Anu and Tanu = 13x × 6 × $\frac{20}{3}$ = 520x

⇒ Time taken by Manu to complete the remaining work = $\frac{544 x - 520 x}{4 x}$ = 6 hours.

Hence, 6.

9. CAT 2022 QA Slot 2 | Algebra - Number Theory

For some natural number n, assume that (15,000)! is divisible by (n!)!. The largest possible value of n is:

A.
6
B.
5
C.
7
D.
4

10. CAT 2022 QA Slot 2 | Arithmetic - Average

The average of a non-decreasing sequence of N number a₁, a₂, ..., a_N is 300. If a1 is replaced by 6a₁, the new average becomes 400. Then, the number of possible values of a₁ is

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Answer: 14

Video Explanation

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Explanation :

Given, $\frac{a_{1} + a_{2} + . . . + a_{N}}{N}$ = 300

⇒ a₁ + a₂ + ... + a_N = 300N ...(1)

Also, $\frac{6 a_{1} + a_{2} + . . . + a_{N}}{N}$ = 400

⇒ 6a₁ + a₂ + ... + a_N = 400N ...(2)

(2) - (1)

⇒ 5a₁ = 100N

⇒ a₁ = 20N

Since, a₁ is the least of the given numbers it cannot be more than the average, hence a₁ ≤ 300.

⇒ N ≤ 15

If N = 1, a₁ = 20 and average cannot be equal to 300.

Hence, N can take all values from 2 till 15, i.e., 14 values.

Hence, 14.

11. CAT 2022 QA Slot 2 | Algebra - Quadratic Equations

Let f(x) be a quadratic ploynomial in x such that f(x) ≥ 0 for all real numbers x. If f(2) = 0 and f(4) = 6, then f(-2) is equal to

A.
24
B.
6
C.
36
D.
12

Answer: Option A

Video Explanation

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Explanation :

Let f(x) = ax² + bx + c
Since, f(x) ≥ 0, and f(2) = 0, it means the graph of the quadratic lies above x-axis and touches x-axis at x = 2.

Since x = 2 is the only root of the equation (i.e., graph is symmetric about x = 2),
⇒ f(2 + a) = f(2 – a)
⇒ f(2 + 2) = f(2 - 2)
⇒ f(4) = f(0)
⇒ f(0) = 6
⇒ c = 6

f(2) = 0
⇒ 4a + 2b + c = 0
⇒ 4a + 2b = -6 …(2)

f(4) = 6
⇒ 16a + 4b + c = 6
⇒ 16a + 4b = 0 …(3)

Solving (2) and (4), we get
a = -3/2 and b = -6

⇒ f(-2) = $\frac{3}{2}$ (-2)² – 6 × - 2 + 6 = 6 + 12 + 6 = 24

Hence, option (a).

12. CAT 2022 QA Slot 2 | Algebra - Quadratic Equations

Let r and c be real numbers. If r and -r are roots of 5x³ + cx² - 10x + 9 = 0, then c equals

A.
4
B.
-9/2
C.
-4
D.
9/2

13. CAT 2022 QA Slot 2 | Algebra - Simple Equations

Manu earns ₹4000 per month and wants to save an average of ₹550 per month in a year. In the first nine months, his monthly expense was ₹3500, and he foresees that, tenth month onward, his monthly expense will increase to ₹3700. In order to meet his yearly savings target, his monthly earnings, in rupees, from the tenth month onward should be:

A.
4400
B.
4300
C.
4200
D.
4350

Answer: Option A

Video Explanation

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Explanation :

Monthly savings of Manu = Rs. 550
⇒ Yearly savings of Manu = 12 × 550 = Rs. 6,600

Total expense of Manu for the first 9 months = 9 × 3500 = Rs. 31,500
Total expense of Manu for the last 3 months = 3 × 3700 = 11,100
⇒ Yearly expense of Manu = 11,100 + 31,500 = Rs. 42,600

Monthly income of Manu = Rs. 4,000
⇒ Income of Manu for first 9 months = 9 × 4000 = Rs. 36,000

⇒ Manu’s income for last 3 months = 42,600 + 6,600 – 36,000 = Rs. 13,200
⇒ Monthly income of Manu for last 3 months = $\frac{13200}{3}$ = Rs. 4,400

Hence, option (a).

14. CAT 2022 QA Slot 2 | Algebra - Simple Equations

Five students, including Amit, appear for an examination in which possible marks are integers between 0 and 50, both inclusive. The average marks for all the students is 38 and exactly three students got more than 32. If no two students got the same marks and Amit got the least marks among the five students, then the difference between the highest and lowest possible marks of Amit is

A.
24
B.
20
C.
21
D.
22

15. CAT 2022 QA Slot 2 | Algebra - Quadratic Equations

The number of integral solutions of the equation ${(x^{2} - 10)}^{(x^{2} - 3 x - 10)}$ = 1 is

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Answer: 4

Video Explanation

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Explanation :

Given, (x² - $10)^{x^{2} - 3 x - 10}$

For this to be equal to 1

Case 1: x² - 10 = 1

⇒ x = $\sqrt{11}$
Rejected as x is not an integer.

Case 2: x² – 3x – 10 = 0
⇒ (x – 5)(x + 2) = 0
⇒ x = 5 or -2 i.e., 2 integral values of x.

Case 3: x² – 10 = -1 and x² – 3x – 10 = even
If x² – 10 = -1 ⇒ x = ±3
For both x = +3 and -3 x² – 3x – 10 is even, hence, 2 integral values of x.

⇒ Total 4 integral values of x are possible.

Hence, 4.

16. CAT 2022 QA Slot 2 | Arithmetic - Percentage

In an election, there were four candidates and 80% of the registered voters casted their votes. One of the candidates received 30% of the casted votes while the other three candidates received the remaining casted votes in the proportion 1 : 2 : 3. If the winner of the election received 2512 votes more than the candidate with the second highest votes, then the number of registered voters was:

A.
60288
B.
50240
C.
40192
D.
62800

Answer: Option D

Video Explanation

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Explanation :

Let the total number of registered voters be 100x.
Number of votes casted = 80x

Votes for Candidate 1 = 30% of 80x = 24x
∴ Remaining three candidates will recieve = 80x - 24x = 56x votes.

Remaining 3 candidates get votes in the ratio of 1 : 2 : 3 of the remaining 56x votes.
⇒ Votes of Candidate 2 = $\frac{1}{6}$ × 56x = $\frac{28 x}{3}$
⇒ Votes of Candidate 3 = $\frac{2}{6}$ × 56x = $\frac{56 x}{3}$
⇒ Votes of Candidate 4 = $\frac{3}{6}$ × 56x = 28x

Highest number of votes is received by Candidate 4 while second highest is by Candidate 1.
⇒ 28x – 24x = 2512
⇒ x = 2512/4 = 628

∴ total number of registered votes = 100x = 62800

Hence, option (d).

17. CAT 2022 QA Slot 2 | Algebra - Simple Equations

In an examination, there were 75 questions. 3 marks were awarded for each correct answer, 1 mark was deducted for each wrong answer and 1 mark was awarded for each unattempted question. Rayan scored a total of 97 marks in the examination. If the number of unattempted questions was higher than the number of attempted questions, then the maximum number of correct answers that Rayan could have given in the examination is:

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18. CAT 2022 QA Slot 2 | Geometry - Triangles

In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If ∠BAC = 45° and ∠ABC = θ, then AD/BE equals

A.
1
B.
√2 cosθ
C.
√2 sinθ
D.
(sinθ + cosθ)/√2

Answer: Option C

Video Explanation

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Explanation :

Area of ∆ABC = $\frac{1}{2}$ × AB × AC × sin 45° = $\frac{A B \times A C}{2 \sqrt{2}}$ ...(1)

Area of ∆ABC = $\frac{1}{2}$ × BA × BC × sin θ° = $\frac{B A \times B C \times \sin θ}{2}$ ...(2)

(1) = (2)

⇒ $\frac{A B \times A C}{2 \sqrt{2}}$ = $\frac{B A \times B C \times \sin θ}{2}$

⇒ $\frac{A C}{B C}$ = √2 × sin⁡𝜃 ...(3)

Area of ∆ABC = $\frac{1}{2}$ × AD × BC = $\frac{1}{2}$ × AC × BE

⇒ $\frac{A D}{B E}$ = $\frac{A C}{B C}$

⇒ $\frac{A D}{B E}$ = √2 × sin⁡𝜃

Hence, option (c).

19. CAT 2022 QA Slot 2 | Geometry - Quadrilaterals & Polygons

Regular polygons A and B have number of sides in the ratio 1 : 2 and interior angles in the ratio 3 : 4. Then the number of sides of B equals

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Answer: 10

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Explanation :

Interior angle of a n-sided regular polygon = $\frac{(n - 2) \times 180 °}{n}$

Let the number of sides of polygon A and B be n and 2n respectively.

⇒ $\frac{\frac{(n - 2) \times 180 °}{n}}{\frac{(2 n - 2) \times 180 °}{2 n}}$ = $\frac{3}{4}$

⇒ $\frac{n - 2}{n - 1}$ = $\frac{3}{4}$

⇒ 4n – 8 = 3n – 3

⇒ n = 5

∴ Number of sides of polygon B = 2n= 10.

Hence, 10.

20. CAT 2022 QA Slot 2 | Algebra - Progressions

On day one, there are 100 particles in a laboratory experiment. On day n, where n greater than or 2, one out of every n particles produces another particle. If the total number of particles in the laboratory experiment increases to 1000 on day m, then m equals.

A.
16
B.
17
C.
19
D.
18

Answer: Option C

Video Explanation

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Explanation :

On nth day ‘n’ particles produce 1 extra particle.
⇒ For every n particles on previous day, their will be (n + 1) particles next day.

∴ On nth day, the number of particles will become $\frac{n + 1}{n}$ times the number of particles of previous day.

⇒ Number of particles after day 2 = 100 × $(\frac{3}{2})$
⇒ Number of particles after day 3 = 100 × $(\frac{3}{2})$ × $(\frac{4}{3})$
⇒ Number of particles after day 4 = 100 × $(\frac{3}{2})$ × $(\frac{4}{3})$ × $(\frac{5}{4})$

...

⇒ Number of particles after day m = 100 × $(\frac{3}{2})$ × $(\frac{4}{3})$ × $(\frac{5}{4})$ × ... × $(\frac{m + 1}{m})$ = 100 × $(\frac{m + 1}{2})$

⇒ 100 × $(\frac{m + 1}{2})$ = 1000

⇒ $(\frac{m + 1}{2})$ = 10

⇒ m = 19

Hence, option (c).

21. CAT 2022 QA Slot 2 | Arithmetic - Time, Speed & Distance

Two ships meet mid-ocean, and then, one ship goes south and the other ship goes west, both travelling at constant speeds. Two hours later, they are 60 km apart. If the speed of one of the ships is 6 km per hour more than the other one, then the speed, in km per hour, of the slower ship is

A.
18
B.
24
C.
12
D.
20

22. CAT 2022 QA Slot 2 | Geometry - Triangles

The length of each side of an equilateral triangle ABC is 3 cm. Let D be a point on BC such that the area of triangle ADC is half the area of triangle ABD. Then the length of AD, in cm, is

A.
√7
B.
√6
C.
√5
D.
√8

Answer: Option A

Video Explanation

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Explanation :

Area of ∆ACD is half of area of ∆ABD.
Since their height is same, ratio of their areas will be same as the ratio of their bases.
⇒ BD = 2CD
⇒ BD = 2 cm and CD = 1 cm.

Taking E as the midpoint of BC, BE = 3/2 = 1.5
⇒ ED = 2 – 1.5 = 0.5 cm.

Also, AE = height of the equilateral triangle = $\frac{\sqrt{3}}{2}$ × 3 = $\frac{3 \sqrt{3}}{2}$

In ∆AED
⇒ AD² = AE² + ED²
⇒ AD² = ${(\frac{3 \sqrt{3}}{2})}^{2}$ + (0.5)²

⇒ AD² = $\frac{27}{4}$ + $\frac{1}{4}$ = 7

⇒ AD = $\sqrt{7}$

Hence, option (a).

CAT 2022 QA Slot 2 | Previous Year CAT Paper

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