CAT 2023 QA Slot 2 | Previous Year CAT Paper

Answer: Option B

Explanation :

For any natural numbers m, n and k, such that k divides both m + 2n and 3m + 4n, k must be a common divisor of

k divides m + 2n ⇒ m + 2n = a × k   ...(1)
k divides 3m + 4n ⇒ 3m + 4n = b × k   ...(2)

(1) × 3 - (2)
⇒ 3m + 6n - (3m + 4n) = 3ak - bk
⇒ 2n = (3a - b)k
∴ 2n is divisible by k

(2) - (1) × 2
⇒ 3m + 4n - 2(m + 2n) = bk - 2ak
⇒ m = (b - 2a)k
∴ m is divisible by k

∴ 2n and m are divisible by k
⇒ k is common divisor of 2n and m.

Hence, option (b).

Answer: Option D

Explanation :

Given, ${2^{4x}}^2$ - $2^{2x^2+x+16}$ + $2^{2x+30}$ = 0

⇒ ${\left({2^{2x}}^2\right)}^2$ - $2\times 2^{2x^2}\times 2^{x+15}$ + ${\left(2^{x+15}\right)}^2$ = 0

⇒ ${\left({2^{2x}}^2-2^{x+15}\right)}^2$ = 0

⇒ ${2^{2x}}^2$ = $2^{x+15}$

⇒ 2x² = x + 15

⇒ 2x² - x - 15 = 0

⇒ (2x + 5)(x - 3) = 0

⇒ x = -5/2 or 3.

Sum of all possible values of x = 3 + (-5/2) = 1/2

Hence, option (d).

Concept:

Indices

Algebraic Identities

Answer: Option D

Explanation :

Given, $\frac{\mathrm{x}}{\mathrm{y}}$ < $\frac{\mathrm{x}+3}{\mathrm{y}-3}$ 

⇒ $\frac{\mathrm{x}}{\mathrm{y}}$ - $\frac{\mathrm{x}+3}{\mathrm{y}-3}$ < 0

⇒ $\frac{\mathrm{x}(\mathrm{y}-3)-\mathrm{y}(\mathrm{x}+3)}{\mathrm{y}(\mathrm{y}-3)}$ < 0

⇒ $\frac{\mathrm{xy}-3x-\mathrm{yx}-3y}{\mathrm{y}(\mathrm{y}-3)}$ < 0

⇒ $\frac{-3(x+y)}{\mathrm{y}(\mathrm{y}-3)}$ < 0

⇒ $\frac{(x+y)}{\mathrm{y}(\mathrm{y}-3)}$ > 0   ...(1)

Now, checking options.

Option (a): If y > 10, y(y - 3) > 0
From (1): ⇒ (x + y) > 0
⇒ -x < y
Hence, option (b) is incorrect.

Options (b) and (c) are slightly difficult to check. We can come back to them if required.

Option (d): If y < 0, y(y - 3) > 0
From (1): ⇒ (x + y) > 0
⇒ -x < y
Hence, option (d) is correct.

Hence, option (d).

Answer: Option A

Explanation :

a^mbⁿ = 144¹⁴⁵ 

⇒ a^mbⁿ = (12²)¹⁴⁵ 

⇒ a^mbⁿ = (2⁴ × 3²)¹⁴⁵ 

⇒ a^mbⁿ = 2⁵⁸⁰ × 3²⁹⁰  

To maximise n - m, we need to maximise n and minimise m.
∴ Let 3²⁹⁰ = X

⇒ a^mbⁿ = X × 2⁵⁸⁰  

∴ a = X, m = 1, b = 2 and n = 580

⇒ n - m = 580 - 1 = 579

Hence, option (a). 

Concept:

Indices

Answer: 6

Explanation :

The given equation can be written as: x² - 2x + 1 + 2kx + 11 = 0
⇒ x² + (2k - 2)x + 12 = 0

The given quadratic has no real roots, hence discriminant is less than 0.
⇒ (2k - 2)² - 4 × 1 × 12 < 0
⇒ 4(k - 1)² - 48 < 0
⇒ (k - 1)² < 12

Largets integral value of k satisying above inequality is 4.

Now, we have k/4y + 9y
= 4/4y + 9y
= 1/y + 9y

We know AM ≥ GM
⇒ (1/y + 9y)/2 ≥ $\sqrt{\frac{1}{y}\times 9y}$
⇒ (1/y + 9y)/2 ≥ 3
⇒ 1/y + 9y ≥ 6

∴ Least possible value of 1/y + 9y = 6.

Hence, 6.

Concept:

No Real Roots of a Quadratic

Answer: 15

Explanation :

Positive integers having exactly two distinct factors other than 1 and itself will be of the form a × b or a³, where a and b are prime numbers.

Case 1: a × b < 50. Possible combinations are:
If one of them is 2, the other prime number can be 3, 5, 7, 11, 13, 17, 19, 23 i.e., 8 possibilities.
If one of them is 3, the other prime number can be 5, 7, 11, 13 i.e., 4 possibilities.
If one of them is 5, the other prime number can be 7 i.e., 1 possibility.
∴ 13 cases.

Case 2: a³. Possible values of a are 2 and 3 i.e., numbers are 8 and 27.
∴ 2 cases.

∴ Total 13 + 2 = 15 such numbers are possible.

Hence, 15.

Concept:

Number of Factors

Answer: 7

Explanation :

Given, ${\log }_{\sqrt{3}} \left(x\right)$ + $\frac{{\log }_x 25}{{\log }_x (0.008)}$ = $\frac{16}{3}$

⇒ ${\log }_{3^{1/2}} \left(x\right)$ + $\frac{{\log }_x 5^2}{{\log }_x {\left(5\right)}^{-3}}$ = $\frac{16}{3}$

⇒ $2\times {\log }_3 \left(x\right)$ + $\frac{2\times {\log }_x 5}{-3\times {\log }_x \left(5\right)}$ = $\frac{16}{3}$

⇒ $2\times {\log }_3 \left(x\right)$ - $\frac{2}{3}$ = $\frac{16}{3}$

⇒ $2\times {\log }_3 \left(x\right)$ = $\frac{16}{3}$ + $\frac{2}{3}$ = 6

⇒ log₃ x = 3

⇒ x = 27

Now, ${\log }_3 \left(3x^2\right)$ = ${\log }_3 (3\times {27}^2)$ = ${\log }_3 \left(3^7\right)$ = 7

Hence, 7.

Concept:

Logarithm Properties

Answer: Option B

Explanation :

Let time taken by A and C alone to fill the tank is A and C hours respectively.
∴ Time taken B alone = (A - 1) hours.

Case 1: All three can fill the tank in 2 hours.
⇒ $\frac{1}{\mathrm{A}}$ - $\frac{1}{\mathrm{A}-1}$ + $\frac{1}{\mathrm{C}}$ = $\frac{1}{2}$   ...(1)

Case 2: B works for 1 hour and C works for 5/4 hours.
⇒ - $\frac{1}{\mathrm{A}-1}\times 1$ + $\frac{1}{\mathrm{C}}\times \frac{9}{4}$ = 1   ...(2)

(1) × 9/4 - (2)
⇒ $\frac{9}{4\mathrm{A}}$ - $\frac{9}{4(\mathrm{A}-1)}$ + $\frac{1}{\mathrm{A}-1}\times 1$ = $\frac{9}{8}$ - 1

⇒ $\frac{9}{4\mathrm{A}}$ - $\frac{5}{4(\mathrm{A}-1)}$ = $\frac{1}{8}$

⇒ $\frac{18}{\mathrm{A}}$ - $\frac{10}{(\mathrm{A}-1)}$ = 1

⇒ A(A - 1) = 18(A - 1) - 10A

⇒ A² - A = 8A - 18

⇒ A² - 9A + 18 = 0

⇒ A = 3 or 6 hours.

Since A should be less than 5, hence we accept only A = 3 hours.

Now, from (1), we have
⇒ $\frac{1}{\mathrm{A}}$ - $\frac{1}{\mathrm{A}-1}$ + $\frac{1}{\mathrm{C}}$ = $\frac{1}{2}$

⇒ $\frac{1}{3}$ - $\frac{1}{2}$ + $\frac{1}{\mathrm{C}}$ = $\frac{1}{2}$

⇒ C = 3/2 hours = 90 minutes.

Hence, option (b).

Concept:

Time & Work

Answer: Option D

Explanation :

Let the amount paid at the end of 2nd year be Rs. x.

∴ 2,00,000 × ${\left(1+\frac{8/2}{100}\right)}^{2\times 3}$ = 10,320 × ${\left(1+\frac{8/2}{100}\right)}^{2\times 2}$ + x

⇒ 200000 × (1.04)⁶ = 10320 × (1.04)⁴ + x

⇒ 253063 = 12072 + x

⇒ x = 240991

∴ Total interest paid = (10320 + 240991) - 200000 = 51311

Hence, option (d).

Concept:

Compounding Half-Yearly

Installments

Answer: Option C

Explanation :

Speed of
Ravi = 40 × 5/18 = 100/9 m/s and
Ashok = 50 × 5/18 = 250/18 m/s

Time taken for them to meet = $\frac{225}{{\displaystyle \frac{100}{9}}+{\displaystyle \frac{250}{18}}}$ = $\frac{225\times 18}{{\displaystyle 450}}$ = 9 secs.

⇒ Vijay and Ravi should also meet in 9 secs.
∴ 9 = $\frac{54}{V-{\displaystyle \frac{100}{9}}}$

⇒ 9V - 100 = 54

⇒ 9V = 154

⇒ V = 154/9 m/s = 154/9 × 18/5 kmph = 61.6 kmph

Hence, option (c).

Concept:

Relative Speed

Answer: Option B

Explanation :

Minu purchases a pair of sunglasses at Rs.1000 and sells to Kanu at 20% profit. Then, Kanu sells it back to Minu at 20% loss. Finally, Minu sells the same pair of sunglasses to Tanu. If the total profit made by Minu from all her transactions is Rs.500, then the percentage of profit made by Minu when she sold the pair of sunglasses to Tanu is

Initially cost price of Minu = Rs. 1000

Minu sells it to Kanu at 20% profit i.e., at Rs. 1000 + 200 = Rs. 1200
Profit earned by Minu = Rs. 200

Now, Kanu sells it back to Minu at 20% loss on 1200 i.e., at 1200 + 240 = Rs. 960

Now, let Minu sells it to Tanu at a profit of x.
∴ 200 + x = 500
⇒ x = 300

∴ Minu earns 300 profit by selling the sunglasses to Tanu, hence her profit % = 300/960 × 100% = 31.25%

Hence, option (b).

Answer: Option B

Explanation :

Price (P) ∝ (weight)² 
⇒ P = k × w² 

Total value after breaking of the stone will be highest if one of the pieces is as heavy as possible and others are as light as possible i.e., the weights of the pieces are 1, 2, 3 and 12 units.
∴ Total value in this case = k × (1)² + k × (2)² + k × (3)² + k × (12)² = 159k

Total value after breaking of the stone will be least if weights of the pieces are as close to each other as possible i.e., the weights of the pieces are 3, 4, 5, and 6 units.
∴ Total value in this case = k × (3)² + k × (4)² + k × (5)² + k × (6)² = 86k

⇒ 158k – 86k = 288000
⇒ 72k = 288000
⇒ k = 4000

∴ Weight of original stone = 4000 × (18)² = 1296000.

Hence, option (b).

Answer: Option A

Explanation :

Let the total number of employees be 100 and their average salary be s.

⇒ There are 20 manufacturing employees and 80 non-manufacturing
Let the average salary of manufacturing employees be 'x' and that of non-manufacturing employees be 'y'.

Total Salary of manufacturing employees is one-sixth that of total employees:
∴ 20 × x = 1/6 × 100 × s
⇒ x = 5s/6

Total Salary of nonmanufacturing employees is five-sixth that of total employees:
∴ 80 × x = 5/6 × 100 × s
⇒ x = 25s/24

The ratio of the average salary obtained by the manufacturing employees to the average salary obtained by the non-manufacturing employee = 5s/6 : 25s/24 = 4 : 5.

Hence, option (a).

Answer: 7

Explanation :

4 liters out of 40 liters is remvoed, hence 4/40 = 1/10^th is removed every time.
∴ 1 - 1/10 = 9/10^th remains every time.

Quantity of milk will become less than that of water as soon as quantity of milk goes below 20 liters.

∴ Quantity of milk remaining after n replacements = 40 × (9/10)ⁿ < 20
⇒ (0.9)ⁿ < 0.5

Least value of n satisfying above inequality is 7.

Hence, 7.

Answer: 16

Explanation :

n people get average Rs. 352.
∴ Total amount distributed = 352n.

Now, 2 people get a total of 506 + 506 = Rs. 1012
⇒ Remaining amount = 352n - 1012

The average amount received by other is less than or equal to 330
∴ $\frac{352\mathrm{n}-1012}{\mathrm{n}-2}$ ≤ 330
⇒ 352n - 1012 ≤ 330n - 660
⇒ 22n ≤ 352
⇒ n ≤ 352/22 = 16
⇒ n ≤ 16

∴ n can take maximum value of 16.

Hence, 16.

Answer: 407

Explanation :

Let the number of white and blue shirts bought is 'w' and 'b' respectively.

Total cost price = 1000w + 1125b

⇒ Average cost price/shirt = $\frac{1000\mathrm{w}+1125\mathrm{b}}{\mathrm{w}+\mathrm{b}}$

⇒ Average marked price/shirt = $\frac{1000\mathrm{w}+1125\mathrm{b}}{\mathrm{w}+\mathrm{b}}\times 1.25$

⇒ Average sellingprice/shirt = $\frac{1000\mathrm{w}+1125\mathrm{b}}{\mathrm{w}+\mathrm{b}}\times 1.25\times 0.9$

⇒ Total selling price = (1000w + 1125b) × 1.25 × 0.9

∴ Profit = 51000 = (1000w + 1125b) × 1.25 × 0.9 - (1000w + 1125b)
⇒ 51000 = (1000w + 1125b) × (1.25 × 0.9 - 1)
⇒ 51000 = (1000w + 1125b) × (0.125)
⇒ 1000w + 1125b = 408000
⇒ 40w + 45b = 16320
⇒ 8w + 9b = 3264

w + b will be maximum when we maximum the variable with least coefficiency and minimise the variable with highest coefficient.
Least possible value of of b cannot be 0 as at least one shirt of each type is bought. Hence, the next least possible value of b possible is 8.

∴ b = 8 and w = 399

∴ Highest possible value of w + b = 399 + 8 = 407

Hence, 407.

Answer: Option A

Explanation :

​​​​​​​

Concept: Diameter of a circle makes an angle of 90° on the circle.

∴ (2r)² = a² + b² 
⇒ 4r² = a² + b² 

Now, Area of the triangle = 1/2 × base × height
= 1/2 × a × b
= $\frac{\mathrm{ab}({\mathrm{a}}^2+{\mathrm{b}}^2)}{2({\mathrm{a}}^2+{\mathrm{b}}^2)}$
= $\frac{\mathrm{ab}\times 4{\mathrm{r}}^2}{2({\mathrm{a}}^2+{\mathrm{b}}^2)}$
= $\frac{2{\mathrm{abr}}^2}{{\mathrm{a}}^2+{\mathrm{b}}^2}$

Hence, option (a).​​​​​​​$$

Answer: Option A

Explanation :

​​​​​​​

Area of △ABP, △APQ and AQCD are in GP.
Let their areas be A, Ar and Ar² 
Now, Area of AQCD = 4 × Area of △ABP
⇒ Ar² = 4 × A
⇒ r = 2

∴ Area of △ABP : Area of △APQ : Area of AQCD = 1 : 2 : 4

From the figure given above
Area of △ABP = 1/2 × BP × AB = 1/2 × BP × 9 = 4.5 × BP
Area of △APQ = 1/2 × PQ × AB = 1/2 × PQ × 9 = 4.5 × PQ
Area of AQCD = 1/2 × (QC + AD) × AB = 1/2 × (QC + 6) × 9 = 4.5 × (QC + 6)

Now, Area of △APQ = 2 × Area of △ABP
⇒ 4.5 × PQ = 2 × 4.5 × BP
⇒ PQ = 2BP

∴ QC = 6 - BP - PQ = 6 - 3BP

Now, Area of AQCD = 4 × Area of △ABP
⇒ 4.5 × (QC + 6) = 4 × 4.5 × BP
⇒ 4.5 × (6 - 3BP + 6) = 4 × 4.5 × BP
⇒ 6 - 3BP + 6 = 4BP
⇒ BP = 12/7

∴ PQ = 24/7 and

∴ QC = 6 - 3 × 12/7 = 6/7

⇒ BP : PQ : QC = 12/7 : 24/7 : 6/7 = 2 : 4 : 1

Hence, option (a).

Answer: 45

Explanation :

​​​​​​​

The area of given figure ABCD = 1/2 × (14 + 4) × 5 = 45.

Hence, 45.

Answer: Option B

Explanation :

If p² + q² - 29 = 2pq - 20 = 52 - 2pq, then the difference between the maximum and minimum possible value of (p³ - q³) is

Given, p² + q² - 29 = 2pq - 20
⇒ p² + q² - 2pq = 29 - 20
⇒ (p - q)² = 9
⇒ p - q = ± 3

Also given, p² + q² - 29 = 52 - 2pq
⇒ p² + q² + 2pq = 81
⇒ (p + q)² = 81
⇒ p + q = ± 9

Case 1: p - q = + 3 and p + q = + 9
Solving these 2 equations we get, p = 6 and q = 3
∴ p³ - q³ = 216 - 27 = 189

Case 2: p - q = - 3 and p + q = + 9
Solving these 2 equations we get, p = 3 and q = 6
∴ p³ - q³ = 27 - 216 = - 189

Case 3: p - q = + 3 and p + q = - 9
Solving these 2 equations we get, p = - 3 and q = - 6
∴ p³ - q³ = (-27) - (-216) = 189

Case 4: p - q = - 3 and p + q = - 9
Solving these 2 equations we get, p = - 6 and q =  - 3
∴ p³ - q³ = - 216 - (-27) = - 189

∴ Highest possible value of p³ - q³ = 189 least possible value of p³ - q³ = - 189.

∴ Required difference = 189 - (-189) = 378

Hence, option (b).

Answer: Option C

Explanation :

Let both the series a₁, a₂, a₃, ... and b₁, b₂, b₃, ... be in arithmetic progression such that the common differences of both the series are prime numbers. If a₅ = b₉, a₁₉ = b₁₉ and b₂ = 0, then a₁₁ equal?

Let the common difference of a_n and b_n be p and q respectively, where both p and q are prime numbers.

Given,
a₅ = b₉,   ...(1)
a₁₉ = b₁₉   ...(2)

(2) - (1)
⇒ a₁₉ - a₅ = b₁₉ - b₉ 
⇒ 14p = 10q
⇒ p/q = 5/7

Since p and q are prime numbers, they must be equal to 5 and 7 respectively.

Now, b₂ = 0
⇒ b₉ = b₂ + 7q = 0 + 49 = 49

From (1):
a₅ = b₉ = 49

Now, a₁₁ = a₅ + 6p = 49 + 30 = 79

Hence, option (c).

Concept:

Arithmetic Progression

Answer: 967

Explanation :

The given series are APs

Series a_n is: 13, 19, 25, 31, 37, 43, 49, ...
Series b_n is: 15, 22, 29, 36, 43, 50, ...

For 2 APs, their common terms are also in AP, with common difference as LCM of common difference of the orignal 2 APs.
The first common term of the two series is 43 and the common difference of the two series is LCM (6, 7) = 42
∴ The series comprising of common terms is 43, 85, 127, ...

Now, nth term of this series = 43 + 42(n - 1) = 42n + 1

⇒ 42n + 1 < 1000
⇒ n < 999/42 = 23.78
∴ Highest possible value of n = 23

⇒ Highest three-digit term common to both the original series = 42 × 23 + 1 = 967.

Hence, 967.