# CAT 2023 QA Slot 1 | Previous Year CAT Paper

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**1. CAT 2023 QA Slot 1 | Modern Math - Permutation & Combination**

The number of all natural numbers up to 1000 with non-repeating digits is:

- A.
648

- B.
585

- C.
738

- D.
504

Answer: Option C

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**Explanation** :

**Single-digit such numbers** = 9

**2-digit such numbers: __ __**

Ten’s digit can be filled in 9 ways (i.e., 1, 2, 3, …, 9)

Unit’s digit can be filled in 9 ways (i.e., any of the 10 single digits except the one at ten’s place)

⇒ Total such 2-digit numbers = 9 × 9 = 81

**3-digit such numbers: __ __ __**

Hundred’s digit can be filled in 9 ways (i.e., 1, 2, 3, …, 9)

Ten’s digit can be filled in 9 ways (i.e., any of the 10 single digits except the one at hundred’s place)

Unit’s digit can be filled in 8 ways (i.e., any of the 10 single digits except the ones at hundred’s and ten’s place)

⇒ Total such 3-digit numbers = 9 × 9 × 8 = 648

∴ Total such numbers = 9 + 81 + 648 = 738.

Hence, option (c).

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**2. CAT 2023 QA Slot 1 | Geometry - Triangles**

In a right-angled triangle ∆ABC, the altitude AB is 5 cm, and the base BC is 12 cm. P and Q are two points on BC such that the areas of ∆ABP, ∆ABQ and ∆ABC are in arithmetic progression. If the area of ∆ABC is 1.5 times the area of ∆ABP, the length of PQ, in cm, is

Answer: 2

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**Explanation** :

Given, AB = 5 and BC = 12

Area of ∆ABP = ½ × AB × BP = 2.5 × BP …(1)

Area of ∆ABQ = ½ × AB × BQ = 2.5 × BQ …(2)

Area of ∆ABC = ½ × AB × BC = 2.5 × BC …(3)

Given, (3) = 1.5 × (1)

⇒ 2.5 BC = 1.5 × 2.5 × BP

⇒ BC = 1.5 × BP

⇒ 12 = 1.5 × BP

⇒ BP = 8

Also, Areas of ∆ABP, ∆ABQ and ABC are in arithmetic progression.

⇒ 2 × (2) = (1) + (3)

⇒ 2 × 2.5 × BQ = 2.5 × BP + 2.5 × BC

⇒ 2BQ = BP + BC

⇒ 2BQ = 8 + 12

⇒ BQ = 10

∴ PQ = BQ – BP = 10 – 8 = 2

Hence, 2.

**Concept**:

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**3. CAT 2023 QA Slot 1 | Arithmetic - Mixture, Alligation, Removal & Replacement**

A mixture P is formed by removing a certain amount of coffee from a coffee jar and replacing the same amount with cocoa powder. The same amount is again removed from mixture P and replaced with same amount of cocoa powder to form a new mixture Q. If the ratio of coffee and cocoa in the mixture Q is 16 : 9, then the ratio of cocoa in mixture P to that in mixture Q is

- A.
4 : 9

- B.
1 : 3

- C.
1 : 2

- D.
5 : 9

Answer: Option D

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**Explanation** :

Let the initial quantity of coffee in the jar be 100 kg and r kg is replaced each time.

Since r kg out of 100 kg is removed, fraction of coffee removed = r/100

∴ fraction of coffee remaining = $\left(1-\frac{\mathrm{r}}{100}\right)$

⇒ Quantity of coffee remaining after first replacement = 100 × $\left(1-\frac{\mathrm{r}}{100}\right)$

And quantity of cocoa after first replacement = r kgs.

⇒ Similarly, quantity of coffee remaining after second replacement = 100 × ${\left(1-\frac{\mathrm{r}}{100}\right)}^{2}$

Now, after 2nd replacement coffee and cocoa are in the ratio of 16 : 9

⇒ Quantity of coffee left after 2nd replacement = $\frac{16}{16+9}$ × 100 = 64 kg

⇒ 100 × ${\left(1-\frac{\mathrm{r}}{100}\right)}^{2}$ = 64

⇒ ${\left(1-\frac{\mathrm{r}}{100}\right)}^{2}$= $\frac{64}{100}$

⇒ $\left(1-\frac{\mathrm{r}}{100}\right)$ = $\frac{8}{10}$ = $\frac{4}{5}$

⇒ r = 20 kg

∴ 20 kg of cocoa is added after 1st replacement.

Also, quantity of cocoa after 2nd replacement = 100 – 64 = 36 kgs

⇒ Required ratio = 20 : 36 = 5 : 9.

Hence, option (d).

**Concept**:

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**4. CAT 2023 QA Slot 1 | Arithmetic - Time, Speed & Distance**

Brishti went on an 8-hour trip in a car. Before the trip, the car had travelled a total of x km till then, where x is a whole number and is palindromic, i.e., x remains unchanged when its digits are reversed. At the end of the trip, the car had travelled a total of 26862 kms till then, this number again being palindromic. If Brishti never drove at more than 110 kmph, then the greatest possible average speed at which she drove during the rip, in kmph was?

- A.
110

- B.
80

- C.
90

- D.
100

Answer: Option D

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**Explanation** :

Total distance travelled at the end of the trip = 26,862.

This includes the distance which car had travelled before the trip and the distance it travels in the 8 hours of the trip.

To maximize speed in 8 hours, we need to maximize distance travelled in 8 hours.

∴ We need to minimize the distance travelled by car before the trip starts, hence we need to minimize x.

Since Brishti drove at less than 110 kmph, hence she must have travelled less than 110 × 8 = 880 kms in the last 8 hours.

∴ She must have travelled more than 26862 – 880 = 25,982 kms before the last 8 hours.

⇒ x > 25,982 and it has to be a palindrome.

Least palindrome greater than 25,982 is 26062.

∴ Car travelled at least 26062 kms before the trip, hence maximum distance travelled during the trip = 26862 – 26062 = 800 kms.

⇒ Maximum average speed for the trip = 800/8 = 100 kmph.

Hence, option (d).

**Concept**:

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**5. CAT 2023 QA Slot 1 | Algebra - Number Theory**

If x and y are real numbers such that x^{2} + (x – 2y - 1)^{2} = -4y(x + y), then the value of x - 2y is?

- A.
1

- B.
-1

- C.
2

- D.
0

Answer: Option A

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**Explanation** :

Given, x^{2} + (x – 2y - 1)^{2} = -4y(x + y)

⇒ x^{2} + (x – 2y - 1)^{2} = -4yx - 4y^{2}

⇒ x^{2} + 4yx + 4y^{2} + (x – 2y - 1)^{2} = 0

⇒ (x^{2} + 2 × x × 2y + (2y)^{2}) + (x – 2y - 1)^{2} = 0

⇒ (x + 2y)^{2} + (x – 2y - 1)^{2} = 0

Sum of squares of two number can be 0 only when both the numbers are 0.

∴ x + 2y = 0 and x – 2y – 1 = 0

⇒ x – 2y = 1

Hence, option (a).

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**6. CAT 2023 QA Slot 1 | Algebra - Quadratic Equations | Algebra - Inequalities & Modulus**

The number of integral solutions of equation 2|x|(x^{2} + 1) = 5x^{2} is?

Answer: 3

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**Explanation** :

**Case 1**: x ≥ 0 ⇒ |x| = x

∴ 2 × x × (x^{2} + 1) = 5x^{2}

⇒ 2x(x^{2} + 1) = 5x^{2}

⇒ 2x(x^{2} + 1) - 5x^{2} = 0

⇒ x[2(x^{2} + 1) - 5x] = 0

⇒ x(2x^{2} – 5x + 2) = 0

⇒ x(2x^{2} – 4x - x + 2) = 0

⇒ x[2x(x – 2) - (x - 2)] = 0

⇒ x(2x - 1)(x - 2) = 0

⇒ x = 0 or ½ or 2.

We need only integral solutions hence acceptable answers are 0 and 2.

**Case 2**: x < 0 ⇒ |x| = -x

∴ 2 × -x × (x^{2} + 1) = 5x^{2}

⇒ -2x(x^{2} + 1) = 5x^{2}

⇒ 2x(x^{2} + 1) + 5x^{2} = 0

⇒ x[2(x^{2} + 1) + 5x] = 0

⇒ x(2x^{2} + 5x + 2) = 0

⇒ x(2x^{2} + 4x + x + 2) = 0

⇒ x[(2x(x + 2) + (x + 2)] = 0

⇒ x(2x + 1)(x + 2) = 0

⇒ x = 0 or -1/2 or -2

We need only integral solutions hence acceptable answers are 0 and -2.

∴ Acceptable integral solutions are -2, 0 and 2, i.e., 3 integral solutions.

Hence, 3.

**Concept**:

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**7. CAT 2023 QA Slot 1 | Arithmetic - Profit & Loss**

Gita sells two objects A and B at the same price such that she makes a profit of 20% on object A and a loss of 10% on object B. If she increases the selling price such that objects A and B are still sold at an equal price and a profit of 10% is made on object B, then the profit made on object A will be nearest to

- A.
49%

- B.
42%

- C.
45%

- D.
47%

Answer: Option D

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**Explanation** :

Let the selling price be Rs. x

Cost price of item sold to A = x/1.2

Cost price of item sold to B = x/0.9

To make calculations easier assume x = 108 (LCM of 12 and 9)

∴ Cost price of item sold to A = x/1.2 = 90

∴ Cost price of item sold to B = x/0.9 = 120

Now, selling price is increased such that a profit of 10% is made for item B.

⇒ New selling price = 120 × 1.1 = Rs. 132

∴ For A, cost price = Rs. 90 and selling price = Rs. 132.

⇒ % profit for A = $\frac{(132-90)}{90}$ × 100% = 46.66% ≈ 47%.

Hence, option (d).

**Concept**:

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**8. CAT 2023 QA Slot 1 | LR - Clocks**

The minor angle between the hour hand and minute hand of a clock was observed at 8:48 am. The minimum deviation (in min) after 8:48 am when the angle increases by 50% is?

- A.
36/11

- B.
24/11

- C.
2

- D.
4

Answer: Option B

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**Explanation** :

The angle between the two hands at h hours and m minutes = $\left|30\mathrm{h}-\frac{11}{2}\mathrm{m}\right|$

∴ Angle at 8:48 am = $\left|30\times 8-\frac{11}{2}\times 48\right|$ = |240 - 264| = 24°.

Now, the angle between the two hands should increase by 50% i.e., 12°.

Relative speed of the two hands = 6 – ½ = 11/2°/min.

∴ Time taken for angle to increase 12° = $\frac{12}{11/2}$ = $\frac{24}{11}$ minutes.

Hence, option (b).

**Concept**:

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**9. CAT 2023 QA Slot 1 | Algebra - Quadratic Equations**

Let α and β be two distinct root of the equation 2x^{2} – 6x + k = 0, such that (α + β) and αβ are the two roots of the equation x^{2} + px + p = 0. Then the value of 8(k - p)?

Answer: 6

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**Explanation** :

α and β be two distinct root of 2x^{2} – 6x + k = 0,

∴ α + β = -(-6)/2 = 3

And, α × β = k/2 = k/2

Now, 3 and k are the roots of the equation x^{2} + px + p = 0.

∴ Sum of the roots = 3 + k/2 = -(p)/1 = -p …(1)

∴ Product of the roots = 3 × k/2 = (p)/1 = p …(2)

(1) + (2)

⇒ 3 + k/2 + 3k/2 = p – p = 0

⇒ k = -3/2

⇒ p = -9/4 [from (2)]

Now, we need to find 8(k - p)

= $8\left(\frac{-3}{2}-\frac{-9}{4}\right)$

= $8\left(\frac{3}{4}\right)$

= 6

Hence, 6.

**Concept**:

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**10. CAT 2023 QA Slot 1 | Arithmetic - Simple & Compound Interest**

Anil invests Rs. 22000 for 6 years in a certain scheme with 4% interest per annum, compounded half-yearly. Sunil invests in the same scheme for 5 years, and then reinvests the entire amount received at the end of 5 years for one year at 10% simple interest. If the amounts received by both at the end of 6 years are same, then the initial investment made by Sunil, in rupees, is

Answer: 20808

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**Explanation** :

Half yearly interest = 4/2 = 2%.

Amount Anil gets after 6 years = 22000 × ${\left(1+\frac{2}{100}\right)}^{6\times 2}$

Let the amount which Sunil invests = S.

Amount Sunil gets after 5 years = S × ${\left(1+\frac{2}{100}\right)}^{5\times 2}$

Sunil invests this amount for 6^{th} year at 10% p.a.

∴ Amount Sunil gets at the end of 6 years = S × ${\left(1+\frac{2}{100}\right)}^{5\times 2}$ × 1.1

⇒ S × ${\left(1+\frac{2}{100}\right)}^{5\times 2}$ × 1.1 = 22000 × ${\left(1+\frac{2}{100}\right)}^{6\times 2}$

⇒ S × 1.1 = 22000 × ${\left(1+\frac{2}{100}\right)}^{2}$

⇒ S = 20000 × (1.02)^{2}

⇒ S = 20000 × 1.0404 = 20,808.

Hence, 20808.

**Concept**:

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**11. CAT 2023 QA Slot 1 | Arithmetic - Time, Speed & Distance**

Arvind travels from town A to town B, and Surbhi from town B to town A, both starting at the same time along the same route. After meeting each other, Arvind takes 6 hours to reach town B while Surbhi takes 24 hours to reach town A. If Arvind travelled at a speed of 54 km/h, then the distance, in km, between town A and town B is

Answer: 972

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**Explanation** :

After meeting they take 6 and 24 hours to reach opposite end.

∴ Time taken to meet = $\sqrt{6\times 24}$ = √144 = 12 hours.

⇒ X takes a total of 12 + 6 = 18 hours to reach from A to B.

∴ Distance between A and B = distance travelled by X in 18 hours = 18 × 54 = 972 kms.

Hence, 972.

**Concept**:

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**12. CAT 2023 QA Slot 1 | Algebra - Logarithms**

If x and y are positive real numbers such that log_{x} (x^{2} + 12) = 4 and 3log_{y} x = 1, then x + y equals?

- A.
11

- B.
68

- C.
20

- D.
10

Answer: Option D

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**Explanation** :

Given, log_{x} (x^{2} + 12) = 4

⇒ x^{2} + 12 = x^{4}

⇒ a + 12 = a^{2} [Take x^{2} = a]

⇒ a^{2} – a – 12 = 0

⇒ (a – 4)(a + 3) = 0

⇒ a = -3 or 4. [a = x^{2} cannot be negative hence -3 is rejected]

⇒ x^{2} = 4

⇒ x = ± 2 [x = -2 is rejected as log is not defined for negative numbers]

⇒ x = 2

Also, 3log_{y} x = 1

⇒ 3log_{y} 2 = 1

⇒ log_{y} 2^{3} = 1

⇒ 2^{3} = y^{1}

⇒ y = 8

∴ x + y = 2 + 8 = 10

Hence, option (d).

**Concept**:

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**13. CAT 2023 QA Slot 1 | Algebra - Progressions | Algebra - Surds & Indices**

A lab experiment measures the number of organisms at 8 am every day. Starting with 2 organisms on the first day, the number of organisms on any day is equal to 3 more than twice the number on the previous day. If the number of organisms on the n^{th} day exceeds one million, then the lowest possible value of n is

Answer: 19

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**Explanation** :

Number of micro-organisms on day 1 = 2

Number of micro-organisms on day 2 = 2 × 2 + 3

= 2^{2} + 3

Number of micro-organisms on day 3 = 2 × (2^{2} + 3) + 3

= 2^{3} + 3 × (2 + 1)

Number of micro-organisms on day 4 = 2 × (2^{3} + 3 × (2 + 1)) + 3

= 2^{4} + 3 × (2^{2} + 2 + 1)

Number of micro-organisms on day 5 = 2 × (2^{4} + 3 × (2^{2} + 2 + 1)) + 3

= 2^{5} + 3 × (2^{3} + 2^{2} + 2 + 1)

∴ Number of micro-organisms at the end of day n = 2^{n} + 3 × (2^{n-2} + … + 2^{2} + 2 + 1)

= 2 × 2^{n-1} + 3 × (2^{n-1} - 1)

= 5 × 2^{n-1} - 3

Now, 5 × 2^{n-1} - 3 ≥ 10,00,000

⇒ 2^{n-1} ≥ 2,00,000 + 3/5

The least value of (n - 1) satisfying above inequality is 18.

⇒ n - 1 = 18

⇒ n = 19

Hence, 19.

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**14. CAT 2023 QA Slot 1 | Algebra - Number Theory**

Let n be the least positive integer such that 168 is a factor of 1134^{n}. If m is the least positive integer such that 1134^{n} is a factor of 168^{m}, then m + n equals

- A.
24

- B.
12

- C.
15

- D.
9

Answer: Option C

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**Explanation** :

168 = 2^{3} × 21 = 2^{3} × 3 × 7

1134 = 2 × 567 = 2 × 3^{4} × 7

Now (1134)^{n} = 2^{n} × 34^{n} × 7^{n}

Since 168 (2^{3} × 3 × 7) completely divides 1134_{n} (2^{n} × 34^{n} × 7^{n})

⇒ Power of 2 in 1134_{n} ≥ Power of 2 in 168

⇒ n ≥ 3

Similarly, we can check for power of 3 and power of 7 and we get the least value of n = 3.

Now (168)^{m} = 2^{3m} × 3^{m} × 7^{m}

Since 1134^{n} (2^{n} × 3^{4n} × 7^{n} = 2^{3} × 3^{12} × 7^{3}) completely divides 168m (2^{3m} × 3^{m} × 7^{m})

⇒ Power of 3 in 168^{m} ≥ Power of 3 in 1134^{n}

⇒ m ≥ 12

Similarly, we can check for power of 2 and power of 7 and we get the least value of m = 12.

∴ n + m = 3 + 12 = 15

Hence, option (c).

**Concept**:

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**15. CAT 2023 QA Slot 1 | Algebra - Quadratic Equations**

The equation x^{3} + (2r + 1)x^{2} + (4r - 1)x + 2 = 0 has -2 as one of the roots. If the other roots are real, then the minimum possible non-negative integer value of r is?

Answer: 2

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**Explanation** :

Let p and q be the other two real roots of the given cubic equation.

Product of the three roots of the given cubic equation = - $\frac{2}{1}$ = -2 × p × q

⇒ q = 1/p

∴ The three roots are -2, p and 1/p

Sum of the three roots of the given cubic equation = - $\frac{2\mathrm{r}+1}{1}$ = -2 + p + $\frac{1}{\mathrm{p}}$

⇒ p + $\frac{1}{\mathrm{p}}$ = -2r + 1

We know than sum of a number and its reciprocal is either less than or equal to - 2 or greater than or equal to 2.

⇒ 2 ≤ p + $\frac{1}{\mathrm{p}}$ ≤ -2

⇒ 2 ≤ -2r + 1 ≤ -2

⇒ 1 ≤ -2r ≤ -3

⇒ -1/2 ≥ r ≥ 3/2

∴ Least non-negative integral value of r is 2.

Hence, 2.

**Concept**:

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**16. CAT 2023 QA Slot 1 | Arithmetic - Percentage**

The salaries of three friends Sita, Gita and Mita are initially in the ratio 5 : 6 : 7, respectively. In the first year, they get salary hikes of 20%, 25% and 20%, respectively. In the second year, Sita and Mita get salary hikes of 40% and 25%, respectively, and the salary of Gita becomes equal to the mean salary of the three friends. The salary hike of Gita in the second year is

- A.
26%

- B.
25%

- C.
30%

- D.
28%

Answer: Option A

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**Explanation** :

Let the initial salaries of A, B and C be 5x, 6x and 7x respectively.

Salary of A increases by 20% and then by 40%.

∴ Salary of A after 2 years = 5x × 1.2 × 1.4 = 8.4x

Salary of C increases by 20% and then by 25%.

∴ Salary of C after 2 years = 7x × 1.2 × 1.25 = 10.5x

At the end of 2 years, B's salary is average of all three, hence B's salary will also be average of salaries of A and C.

⇒ Salary of B after 2 years = (8.4x + 10.5x)/2 = 9.45

Let the % increase in B's salary be P% in 2^{nd} year. Increase in first year is 25%.

⇒ 9.45x = 6x × 1.25 × (1 + P/100)

⇒ (1 + P/100) = 9.45/7.5 = 1.26

⇒ P = 26%

Hence, option (a).

**Concept: **Percentage Change

**Concept**:

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**17. CAT 2023 QA Slot 1 | Arithmetic - Average**

In an examination, the average marks of 4 girls and 6 boys is 24. Each of the girls has the same marks while each of the boys has the same marks. If the marks of any girl is at most double the marks of any boy, but not less than the marks of any boy, then the number of possible distinct integer values of the total marks of 2 girls and 6 boys is

- A.
20

- B.
19

- C.
21

- D.
22

Answer: Option C

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**Explanation** :

Let the average marks of a boy and girl be b and g respectively.

Given, (4g + 6b)/10 = 24

⇒ 4g + 6b = 240

⇒ 2g + 3b = 120 ...(1)

Also, b ≤ g ≤ 2b

⇒ 2b ≤ 2g ≤ 4b

⇒ 5b ≤ 2g + 3b ≤ 7b ...(2)

From (1) & (2), we get

5b ≤ 120 ≤ 7b

⇒ b ≤ 24 and b ≤ 17(1/7) ...(3)

Now we need to find integral values of 2g + 6b

= 2g + 3b + 3b

= 120 + 3b

120 + 3 × 120/7 ≤ 120 + 3b ≤ 120 + 3 × 24 ...from (3)

⇒ 171.42 ≤ 120 + 3b ≤ 192

∴ Integral possible values of 2g + 6b are from 172 till 192 i.e., 21 possible integral values.

Hence, option (c).

Workspace:

**18. CAT 2023 QA Slot 1 | Arithmetic - Time & Work**

The amount of job that Amal, Sunil and Kamal can individually do in a day, are in harmonic progression. Kamal takes twice as much time as Amal to do the same amount of job. If Amal and Sunil work for 4 days and 9 days, respectively, Kamal needs to work for 16 days to finish the remaining job. Then the number of days Sunil will take to finish the job working alone, is

Answer: 27

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**Explanation** :

Let the time taken by A and K to complete a work is x and 2x days respectively.

Work done in a day is the efficiency of a person.

Hence, if efficiencies of A, S and K are in Harmonic Progression, time taken by them to finish a work will be in Arithmetic Progression.

∴ Time taken by S alone is arithmetic mean of time taken by A and K alone.

⇒ Time taken by S alone = (x + 2x)/2 = 1.5x

Using unitary method

⇒ 1 = $\frac{1}{\mathrm{x}}\times 4$ + $\frac{1}{1.5\mathrm{x}}\times 9$ + $\frac{1}{2\mathrm{x}}\times 16$

⇒ x = 4 + 6 + 8

⇒ x = 18

∴ S will take 1.5 × 18 = 27 days to finish the job alone.

Hence, 27.

**Concept**:

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**19. CAT 2023 QA Slot 1 | Algebra - Surds & Indices**

If $\sqrt{5x+9}$ + $\sqrt{5x-9}$ = 3(2 + √2), then find the value of $\sqrt{10x+9}$?

- A.
3√7

- B.
3√5

- C.
4√3

- D.
7√3

Answer: Option A

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**Explanation** :

Given, $\sqrt{5x+9}$ + $\sqrt{5x-9}$ = 6 + 3√2

⇒ $\sqrt{5x+9}$ + $\sqrt{5x-9}$ = √36 + √18

∴ $\sqrt{5x+9}$= √36 and $\sqrt{5x-9}$= √18

⇒ 5x + 9 = 36 and 5x - 9 = 18

⇒ 5x = 27

⇒ 10x = 54

⇒ 10x + 9 = 63

⇒ $\sqrt{10x+9}$ = √63

⇒ $\sqrt{10x+9}$ = 3√7

Hence, option (a).

Workspace:

**20. CAT 2023 QA Slot 1 | Algebra - Progressions | Algebra - Surds & Indices**

For some positive and distinct real numbers x, y and z if $\frac{1}{\sqrt{\mathrm{y}}+\sqrt{\mathrm{z}}}$ is the arithmetic mean of $\frac{1}{\sqrt{\mathrm{x}}+\sqrt{\mathrm{z}}}$ and $\frac{1}{\sqrt{\mathrm{x}}+\sqrt{y}}$, then the relationship which will always hold true, is?

- A.
x, y and z are in Arithmetic Progression

- B.
y, x and z are in Arithmetic Progression

- C.
√x, √y and √z are in Arithmetic Progression

- D.
√y, √x and √z are in Arithmetic Progression

Answer: Option B

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**Explanation** :

Given, $\frac{2}{\sqrt{\mathrm{y}}+\sqrt{\mathrm{z}}}$ = $\frac{1}{\sqrt{\mathrm{x}}+\sqrt{\mathrm{z}}}$ + $\frac{1}{\sqrt{\mathrm{x}}+\sqrt{y}}$

⇒ $\frac{2}{\sqrt{\mathrm{y}}+\sqrt{\mathrm{z}}}$ = $\frac{\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}+\sqrt{\mathrm{x}}+\sqrt{\mathrm{z}}}{(\sqrt{\mathrm{x}}+\sqrt{\mathrm{z}})(\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}})}$

⇒ 2(√x + √z)(√x + √y) = (2√x + √y + √z)(√y + √z)

⇒ 2x + 2√xy + 2√zx + 2√zy = 2√xy + 2√xz + y + √yz + zy + z

⇒ 2x = y + z

∴ y, x and z are in Arithmetic Progression

Hence, option (b).

**Concept**:

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**21. CAT 2023 QA Slot 1 | Geometry - Coordinate Geometry**

Let C be the circle x^{2} + y^{2} + 4x - 6y - 3 = 0 and L be the locus of the point of intersection of a pair of tangents to C with the angle between the two tangents equal to 60 degree. Then, the point at which L touches the line x = 6 is?

- A.
(6, 6)

- B.
(6, 8)

- C.
(6, 3)

- D.
(6, 4)

Answer: Option C

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**Explanation** :

Let C be the circle x^{2} + y^{2} + 4x - 6y - 3 = 0 and L be the locus of the point of intersection of a pair of tangents to C with the angle between the two tangents equal to 60 degree. Then, the point at which L touches the line x = 6 is?

Given, x^{2} + y^{2} + 4x - 6y - 3 = 0

⇒ x^{2} + 4x + 4 + y^{2} - 6y + 9 - 3 - 4 - 9 = 0 [Adding and Subtracting 4 and 9]

⇒ (x + 2)^{2} + (y - 3)^{2} = 16

⇒ (x + 2)^{2} + (y - 3)^{2} = 4^{2}

This represents the equation of a circle with radius = 4 units and center (O) at (-2, 3).

Point L lies on line x = 6, hence x-coordinate of L is 6. Let y-coordinate of L be 'y'.

Let P be the point where tangent from L touches the given circle.

∆LOP is a 30-60-90° triangle.

⇒ OL = 2 × OP = 8

∴ OL = 8 = $\sqrt{(6-{(-2))}^{2}+{(3-\mathrm{y})}^{2}}$

⇒ 64 = (8)^{2} + (3 - y)^{2}

⇒ y = 3

∴ Coordinate of L = (6, 3)

Hence, option (c).

**Note**: A locus of points is a set of points that all satisfy some given condition or property.

Here L is the locus of the point of intersection of a pair of tangents to C with the angle between the two tangents equal to 60 degree. Hence, L will be a circle bigger than the original circle as shown. Tangents drawn from any point on this circle to the smaller circle will make an angle of 60 degree.

**Concept**:

Workspace:

**22. CAT 2023 QA Slot 1 | Geometry - Quadrilaterals & Polygons**

A quadrilateral ABCD is inscribed in a circle such that AB : CD = 2 : 1 and BC : AD = 5 : 4. If AC and BD intersect at the point E, then AE : CE equals

- A.
2 :1

- B.
5 : 8

- C.
1 : 2

- D.
8 : 5

Answer: Option D

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**Explanation** :

∠DAC = ∠DBC [angle subtended in same segment by same chord are equal]

Similarly,

∠ADB = ∠ACB

∠CDB = ∠CAB

∠ACD = ∠ABD

∆AED is similar to ∆BEC

[∠DAC = ∠DBC and ∠ADB = ∠ACB]

∴ $\frac{\mathrm{ED}}{\mathrm{EC}}$ = $\frac{\mathrm{AD}}{\mathrm{BC}}$ = $\frac{4}{5}$ ...(1)

∆AEB is similar to ∆DEC

[∠CDB = ∠CAB and ∠ACD = ∠ABD]

∴ $\frac{\mathrm{AE}}{\mathrm{ED}}$ = $\frac{\mathrm{AB}}{\mathrm{CD}}$ = $\frac{2}{1}$ ...(2)

(1) × (2)

⇒$\frac{\mathrm{ED}}{\mathrm{EC}}\times \frac{\mathrm{AE}}{\mathrm{ED}}$ = $\frac{4}{5}\times \frac{2}{1}$ = $\frac{8}{5}$

Hence, option (d).

**Concept**:

Angle subtended in same segment of a circle by same chord are equal

Workspace:

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