# CAT 2022 QA Slot 2 | Previous Year CAT Paper

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**1. CAT 2022 QA Slot 2 | Algebra - Logarithms**

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

Answer: 47

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**Explanation** :

Given, $\left(\frac{4-{\mathrm{log}}_{2}n}{3-{\mathrm{log}}_{4}n}\right)$ < 0

**Case 1:** 4 – log_{2}n < 0 and 3 – log_{4}n > 0

⇒ log_{2}n > 4 and log_{4}n < 3

⇒ n > 16 and n < 64

∴ integral values of n can be 17, 18, …, 63 i.e., 47 values.

**Case 2:** 4 – log_{2}n > 0 and 3 – log_{4}n < 0

⇒ log_{2}n < 4 and log_{4}n > 3

⇒ n < 16 and n > 64

∴ No integral values of n is possible.

Hence, 47.

Workspace:

**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

Answer: Option C

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**Explanation** :

**Case 1:** Number of 4-digit numbers

__ __ __ __

Here,

unit’s digit can be either 2, 3, 4 or 5 i.e., 4 ways,

ten’s digit can be chosen from remaining 5 numbers in 5 ways,

hundred’s digit can be chosen from remaining 4 numbers in 4 ways.

thousand’s digit can be chosen from remaining 3 numbers in 3 ways.

⇒ Total such 4-digit numbers = 4 × 5 × 4 × 3 = 240

**Case 2:** Number of 5-digit numbers

__ __ __ __ __

Here,

unit’s digit can be either 1, 2, 3, 4 or 5 i.e., 5 ways,

ten’s digit can be chosen from remaining 5 numbers in 5 ways,

hundred’s digit can be chosen from remaining 4 numbers in 4 ways.

thousand’s digit can be chosen from remaining 3 numbers in 3 ways.

ten thousand’s digit can be chosen from remaining 2 numbers in 2 ways.

⇒ Total such 5-digit numbers = 5 × 5 × 4 × 3 × 2 = 600

**Case 2:** Number of 6-digit numbers

__ __ __ __ __

Here,

unit’s digit can be either 1, 2, 3, 4 or 5 i.e., 5 ways,

ten’s digit can be chosen from remaining 5 numbers in 5 ways,

hundred’s digit can be chosen from remaining 4 numbers in 4 ways.

thousand’s digit can be chosen from remaining 3 numbers in 3 ways.

ten thousand’s digit can be chosen from remaining 1 number in 1 way.

One lakh’s digit

⇒ Total such 6-digit numbers = 5 × 5 × 4 × 3 × 2 × 1 = 600

∴ Total required numbers = 240 + 600 + 600 = 1440.

Hence, option (c).

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**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) = x^{2}. If f(x^{2} - x) = 5, then the value of the sum f(g(5)) + g(f(5)) is

Answer: 12

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**Explanation** :

f(x^{2} – x) = 5

put x = 0 ⇒ f(0) = 5

Given f(x) + f(x – 1) – 1 = 0

⇒ f(x) = 1 – f(x - 1) …(1)

put x = 1 in (1)

⇒ f(1) = 1 – f(0)

⇒ f(1) = 1 – 5 = -4

Put x = 2 in (1)

⇒ f(2) = 1 – f(1) = 1 – (-4) = 5

∴ f(odd value of x) = -4 &

f(even value of x) = 5

Now, we have f(g(5)) + g(f(5))

⇒ f(g(5)) + g(f(5)) = f(5^{2}) + g(-4)

⇒ f(g(5)) + g(f(5)) = f(25) + 16

⇒ f(g(5)) + g(f(5)) = -4 + 16

⇒ f(g(5)) + g(f(5)) = 12

Hence, 12.

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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

Answer: Option A

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**Explanation** :

Let container A initially have 100 liters of sugar while container B have 100 liters of milk.

Now the second container has sugar and milk in the ratio of 1 : 2.

When half i.e., 75 liters of it is transferred, 25 liters of sugar and 50 liters of milk will be transferred.

Now the first container has sugar and milk in the ratio of 3 : 2.

When half i.e., 62.5 liters of it is transferred, 37.5 liters of sugar and 25 liters of milk will be transferred.

∴ Ratio of sugar and milk in 2nd container = 62.5 : 75 = 625 : 750 = 25 : 30 = 5 : 6

Hence, option (a).

Workspace:

**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 $\frac{1}{25}\underset{\mathrm{n}=1}{\overset{25}{\sum {\mathrm{A}}_{\mathrm{n}}}}$

- A.
415

- B.
404

- C.
455

- D.
442

Answer: Option C

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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] = 2n^{2} + n

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

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

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

= 26 × 17 + 13 = 455

Hence, option (c).

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**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

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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).

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**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

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.

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**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:

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 = $\frac{544x-520x}{4x}$ = 6 hours.

Hence, 6.

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**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

Answer: Option C

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**Explanation** :

For (15,000)! to be completely divisible by (n!)!, n! ≤ 15,000

Now we know,

6! = 720

7! = 5,040

8! = 40,320

∵ n ≤ 15,000, highest value n can take is 7

Hence, option (c).

Workspace:

**10. CAT 2022 QA Slot 2 | Arithmetic - Average**

The average of a non-decreasing sequence of N number a_{1}, a_{2}, ..., a_{N} is 300. If a1 is replaced by 6a_{1}, the new average becomes 400. Then, the number of possible values of a_{1} is

Answer: 14

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**Explanation** :

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

⇒ a_{1} + a_{2} + ... + a_{N} = 300N ...(1)

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

⇒ 6a_{1} + a_{2} + ... + a_{N} = 400N ...(2)

(2) - (1)

⇒ 5a_{1} = 100N

⇒ a_{1} = 20N

Since, a_{1} is the least of the given numbers it cannot be more than the average, hence a_{1} ≤ 300.

⇒ N ≤ 15

If N = 1, a_{1} = 20 and average cannot be equal to 300.

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

Hence, 14.

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**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

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**Explanation** :

Let f(x) = ax^{2} + 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).

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

Let r and c be real numbers. If r and -r are roots of 5x^{3} + cx^{2} - 10x + 9 = 0, then c equals

- A.
4

- B.
-9/2

- C.
-4

- D.
9/2

Answer: Option B

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**Explanation** :

In a polynomial ax^{3} + bx^{2} + cx + d = 0, whose roots are α, β and γ.

⇒ α + β + γ = -b/a

⇒ αβ + βγ + γα = c/a

⇒ αβγ = -d/a

If roots of 5x^{3} + cx^{2} – 10x + 9 = 0, r, -r and γ

⇒ r – r + γ = -c/5

⇒ γ = -c/5 …(1)

⇒ r × -r + r × γ – r × γ = (-10)/5 = -2

⇒ r^{2} = 2 …(2)

⇒ r × - r × γ = -(9/5)

⇒ r^{2} × γ = 9/5

⇒ γ = 9/10 …(3)

From (1) and (3)

⇒ -c/5 = 9/10

⇒ c = -9/2

Hence, option (b).

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**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

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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).

Workspace:

**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

Answer: Option B

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**Explanation** :

Total marks of all 5 students = 5 × 38 = 190

**Case 1:** Least marks of Amit.

To minimize Amit’s marks we should maximize the marks of other 4 students.

⇒ 3 students get more than 32, hence their maximum marks can be 48, 49 and 50. [Marks are distinct]

2 students (including Amit) scored less than or equal to 32.

⇒ Maximum marks of 2nd student can be 32

∴ Sum of marks of these 4 students = 32 + 48 + 49 + 50 = 179

⇒ Least marks of Amit = 190 – 179 = 11

**Case 2:** Maximum marks of Amit.

Since Amit and one other student scored less than or equal to 32 marks, maximum marks Amit can score is 31. [Amit scored least and all scores are distinct.]

∴ Difference between highest and least score of Amit = 31 – 11 = 20

Hence, option (b).

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

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

Answer: 4

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**Explanation** :

Given, (x^{2} - $10{)}^{{x}^{2}-3x-10}$

For this to be equal to 1

**Case 1:** x^{2} - 10 = 1

⇒ x = $\sqrt{11}$

Rejected as x is not an integer.

**Case 2:** x^{2} – 3x – 10 = 0

⇒ (x – 5)(x + 2) = 0

⇒ x = 5 or -2 i.e., 2 integral values of x.

**Case 3:** x^{2} – 10 = -1 and x^{2} – 3x – 10 = even

If x^{2} – 10 = -1 ⇒ x = ±3

For both x = +3 and -3 x^{2} – 3x – 10 is even, hence, 2 integral values of x.

⇒ Total 4 integral values of x are possible.

Hence, 4.

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**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

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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 = $\frac{28x}{3}$

⇒ Votes of Candidate 3 = $\frac{2}{6}$ × 56x = $\frac{56x}{3}$

⇒ 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).

Workspace:

**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:

Answer: 24

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**Explanation** :

Let the number of correct, wrong and un-attempted questions be c, w and u respectively.

Since u > c + w

⇒ u > 75 - u

⇒ u > 37.5

∴ Least possible value of u is 38

⇒ c + w + u = 75 …(1)

⇒ 3c – w + u = 97 …(2)

(2) + (1)

⇒ 4c + 2u = 172

⇒ 2c + u = 86

⇒ 2c = 86 - u

c will be greatest when u is least i.e., u = 38

⇒ 2c = 86 - 38 = 48

⇒ c = 24

∴ Maximum value c can take is 24.

Hence, 24.

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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

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**Explanation** :

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

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

(1) = (2)

⇒ $\frac{AB\times AC}{2\sqrt{2}}$ = $\frac{BA\times BC\times \mathrm{sin}\theta}{2}$

⇒ $\frac{AC}{BC}$ = √2 × sin𝜃 ...(3)

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

⇒ $\frac{AD}{BE}$ = $\frac{AC}{BC}$

⇒ $\frac{AD}{BE}$ = √2 × sin𝜃

Hence, option (c).

Workspace:

**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

Answer: 10

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**Explanation** :

Interior angle of a n-sided regular polygon = $\frac{(n-2)\times 180\xb0}{n}$

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

⇒ $\frac{{\displaystyle \frac{(n-2)\times 180\xb0}{n}}}{{\displaystyle \frac{(2n-2)\times 180\xb0}{2n}}}$ = $\frac{3}{4}$

⇒ $\frac{n-2}{n-1}$ = $\frac{3}{4}$

⇒ 4n – 8 = 3n – 3

⇒ n = 5

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

Hence, 10.

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**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

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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 × $\left(\frac{3}{2}\right)$

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

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

...

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

⇒ 100 × $\left(\frac{m+1}{2}\right)$ = 1000

⇒ $\left(\frac{m+1}{2}\right)$ = 10

⇒ m = 19

Hence, option (c).

Workspace:

**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

Answer: Option A

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**Explanation** :

Let the speed of the slower ship be x km/hr.

Distance travelled by it in 2 hours = 2x kms

Speed of faster ship = (x + 6) km/hr.

Distance travelled by it in 2 hours = 2(x + 6) kms

⇒ (2x)^{2} + (2(x + 6))^{2} = (60)2

⇒ 4x^{2} + 4(x^{2} 12x + 36) = 3600

⇒ 8x^{2} + 48x + 144 = 3600

⇒ 8x^{2} + 48x - 3456 = 0

⇒ x^{2} + 6x - 432 = 0

⇒ (x + 24)(x - 18) = 0

⇒ x = 18 (-24 is rejected)

∴ Speed of slower person is 18 km/hr

Hence, option (a).

Workspace:

**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

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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

⇒ AD^{2} = AE^{2} + ED^{2}

⇒ AD^{2} = ${\left(\frac{3\sqrt{3}}{2}\right)}^{2}$ + (0.5)^{2}

⇒ AD^{2} = $\frac{27}{4}$ + $\frac{1}{4}$ = 7

⇒ AD = $\sqrt{7}$

Hence, option (a).

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