CAT 2023 QA Slot 3 | Previous Year CAT Paper

Answer: Option C

Video Explanation

Explanation :

A system of linear equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂ has infinite solutions if

$\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}$ = $\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}$ = $\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}$

For, x + y = 4 and (a + 5)x + (b² -15)y = 8b to have infinite solution

⇒ $\frac{1}{\mathrm{a}+5}$ = $\frac{1}{{\mathrm{b}}^2-15}$ = $\frac{4}{8\mathrm{b}}$   ...(1)

⇒ 8b = 4(b² -15)

⇒ 4b² - 8b - 60 = 0

⇒ b² - 2b - 15 = 0

⇒ (b - 5)(b + 3) = 0

⇒ b = 5 or -3

From (1): ⇒ a + 5 = b² - 15
⇒ a = b² - 20

∴ a = 5 when b = 5 ⇒ ab = 25
or, a = -11 when b = -3 ⇒ ab = 33

∴ Maximum value of ab = -11 × -3 = 33

Hence, option (c).

Answer: Option C

Video Explanation

Explanation :

8^m = 2^3m and 8ⁿ = 2³ⁿ 

Now, 2^3m < 2^x < 2³ⁿ 

We have to find least possible value of (m + n) such that there are 41 possible values of x.

Least possible of m = 1, hence we get, 2³ < 2^x < 2³ⁿ 

Now, x can be any interger from 4 till 44 (41 values).

∴ Least possible value of 3n = 45, hence n = 15.

∴ Least possible value of m + n = 1 + 15 = 16.

Hence, option (c).

Answer: Option A

Video Explanation

Explanation :

Given, x⁸ + ${\left(\frac{1}{\mathrm{x}}\right)}^8$ = 47

⇒ x⁸ + ${\left(\frac{1}{\mathrm{x}}\right)}^8$ + 2 = 49

⇒ (x⁴)² + ${\left(\frac{1}{{\mathrm{x}}^4}\right)}^2$ + $2\times {\mathrm{x}}^4\times \frac{1}{{\mathrm{x}}^4}$ = 49

⇒ ${\left({\mathrm{x}}^4+\frac{1}{{\mathrm{x}}^4}\right)}^2$ = 49

⇒ ${\mathrm{x}}^4+\frac{1}{{\mathrm{x}}^4}$ = 7 [-7 will be rejected as LHS should be positive]

Now, ${\mathrm{x}}^4+\frac{1}{{\mathrm{x}}^4}$ + 2 = 9

⇒ ${\left({\mathrm{x}}^2\right)}^2+{\left(\frac{1}{{\mathrm{x}}^2}\right)}^2$ + $2\times {\mathrm{x}}^2\times \frac{1}{{\mathrm{x}}^2}$ = 9

⇒ ${\left({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}\right)}^2$ = 9

⇒ ${\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}$ = 3

Similarly, $\mathrm{x}+\frac{1}{\mathrm{x}}$ = $\sqrt{5}$

Cubing both sides, we get

⇒ x³ + $\frac{1}{{\mathrm{x}}^3}$ + $3\times \mathrm{x}\times \frac{1}{\mathrm{x}}\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)$ = 5√5

⇒ x³ + $\frac{1}{{\mathrm{x}}^3}$ = 2√5

Again cubing both sides, we get

⇒ x⁹ + $\frac{1}{{\mathrm{x}}^9}$ + $3\times {\mathrm{x}}^3\times \frac{1}{{\mathrm{x}}^3}\left({\mathrm{x}}^3+\frac{1}{{\mathrm{x}}^3}\right)$ = 40√5

⇒ x⁹ + $\frac{1}{{\mathrm{x}}^9}$ = 34√5

Hence, option (a).

Answer: Option B

Video Explanation

Explanation :

Given, $\frac{1}{2}$, $\frac{{\log }_3(2^x-9)}{{\log }_{3}4}$ and $\frac{{\log }_5(2^x+{\displaystyle \frac{17}{2}})}{{\log }_{5}4}$ are in arithmetic progression,

⇒ 1/2, log₄ (2^x - 9) and log₄ (2^x + 17/2) are in AP

⇒ 2 × log₄ (2^x - 9) = 1/2 + log₄ (2^x + 17/2)

⇒ log₄ (2^x - 9)² = log₄ 2 + log₄ (2^x + 17/2)

⇒ log₄ (2^x - 9)² = log₄ 2 + log₄ (2^x + 17/2)

⇒ log₄ (2^x - 9)² = log₄ 2 × (2^x + 17/2)

⇒ (2^x - 9)² = 2 × (2^x + 17/2)

⇒ (a - 9)² = 2a + 17    [Assuming 2^x = a]

⇒ a² - 18a + 81 = 2a + 17

⇒ a² - 20a + 64 = 0

⇒ (a - 16)(a - 4) = 0

⇒ a = 2^x = 16   [4 is rejected as (a - 9) cannot be negative]

Now the first term of the AP = 1/2 and

second term of the AP = log₄ (2^x - 9) = log₄ (16 - 9) = log₄ 7

Common difference = log₄ 7 - 1/2 = log₄ 7 - log₄ 2 = log₄ 7/2

Hence, option (a).

Answer: 468

Video Explanation

Explanation :

A number having 15 factors can be of the form:

a¹⁴ or a⁴ × b² 

Case 1: a¹⁴ 
The smallest possible such number = 2¹⁴ = 1024 × 16 > 16000

Case 2: a⁴ × b² 
The smallest possible such number = 2⁴ × 3²  = 16 × 9 = 144

The second smalled possible such number = 3⁴ × 2²  = 81 × 4 = 324

∴ The required sum = 144 + 324 = 468.

Hence, 468.

Answer: 9

Video Explanation

Explanation :

Let the roots of the given quandratic equation be p and q.

⇒ $\frac{1}{\mathrm{p}}-\frac{1}{\mathrm{q}}$ = $\frac{1}{3}$   ...(1)

⇒ $\frac{1}{{\mathrm{p}}^2}+\frac{1}{{\mathrm{q}}^2}$ = $\frac{5}{9}$   ...(2)

Squaring (1) we get

⇒  $\frac{1}{{\mathrm{p}}^2}+\frac{1}{{\mathrm{q}}^2}-\frac{2}{\mathrm{pq}}$ = $\frac{1}{9}$

⇒  $\frac{5}{9}-\frac{2}{\mathrm{pq}}$ = $\frac{1}{9}$

⇒ pq = 9/2 = c

From (2) we get

⇒ 9(p² + q²) = 5p²q² 

⇒ 9(p² + q²) = 5(pq)²  

⇒ 9((p + q)² - 2pq) = 5(9/2)²  

⇒ (p + q)² - 9 = 45/4

⇒ (p + q)² = 81/4

⇒ (p + q) = ± 9/2 = - b

largest possible value of b = 9/2

∴ Largest possible value of a + b = 9/2 + 9/2 = 9.

Hence, 9.

Answer: 5

Video Explanation

Explanation :

Given, 5^n-1 < 3ⁿ⁺¹ 

Putting values of n = 1, 2 and so on we see that the above inequality is true for n = 1, 2, 3, 4 and 5 only.

Now, 3ⁿ⁺¹ < 2^n+m is true of all values of n.

Taking n = 5, we get 
3⁶ < 2^5+m 
⇒ 729 < 2^5+m 

The least power of 2 greater than 729 is 1024 (2¹⁰)

∴ 2¹⁰  =  2^5+m 
⇒ 5 + m = 10
⇒ m = 5
∴ Least value of m = 5.

If we check for other values of n, we may get smaller value of m, but those values will not suffice when n = 5.

Hence, 5.

Answer: Option C

Video Explanation

Explanation :

The merchange earns profit due to increase in price and due to cheating (quantity) while buying and selling.

multiplication factor for price: 110/100 × 0.95

multiplication factor for quantity: while buying = 105/100
                                                    while selling = 100/95
overall multiplication factor for quantity: 105/100 × 100/95 = 105/95

⇒ overall multiplication factor (including price and quantity) = 110/100 × 0.95 × 105/95 = 11/10 × 105/100 = 1.155

∴ overal % profit = (1.155 - 1) × 100% = 15.5%

Hence, option (c).

Answer: Option C

Video Explanation

Explanation :

Let the speeds of river, faster and slower boats be r, f and s km/hr respectively and distance between A and B be 12 kms.

For Faster boat:
⇒ f + r = 12/2 = 6   ...(1)
⇒ f - r = 12/3 = 4   ...(2)

(1) - (2)
⇒ 2r = 6 - 4 = 2
⇒ r = 1

For Slower boat:
⇒ $\frac{12}{\mathrm{s}-1}$ + $\frac{12}{\mathrm{s}+1}$ = 6

⇒ $\frac{12\times 2\mathrm{s}}{{\mathrm{s}}^2-{\left(1\right)}^2}$ = 6

⇒ s² - 1 = 4s

⇒ s² - 4s - 1 = 0

∴ s = $\frac{4\pm \sqrt{16+4}}{2}$ = 2 ± √5 = 2 + √5 [negative value of s is rejected]

Now, time taken by the slower boat to go from A to B

= $\frac{12}{(2+\sqrt{5})+1}$ = $\frac{12}{(3+\sqrt{5})}$ = $\frac{12(3-\sqrt{5})}{9-5}$ = 3(3 - √5).

Hence, option (c).

Concept: Boats & Streams

Answer: Option A

Video Explanation

Explanation :

Let the total work to be done = LCM(7, 15, 6) = 630 units.

⇒ Combined efficiency of Rahul, Rakshita and Gurmeet < 630/7 = 90 units/day

All 3 worked for 6 days and then Rakshita worked for 3 days. Calculating total work done
∴ 630 = (combined efficiency of all 3) × 6 + (efficiency of Rakshita) × 3
⇒ (efficiency of Rakshita) × 3 = 630 - (combined efficiency of all 3) × 6
⇒ (efficiency of Rakshita) = 210 - (combined efficiency of all 3) × 2
⇒ (efficiency of Rakshita) > 210 - 90 × 2
⇒ (efficiency of Rakshita) > 30

∴ Rakshita's efficiency is greater than 30 units/day.

⇒ Time taken by Rakshita < 630/30 = 21 days.

∴ Rakshita will take less than 21 days to finish the job. Hence, she cannot take 21 days to finish the job.

Hence, option (a).

Answer: Option A

Video Explanation

Explanation :

Population at the end of 2021 = 1,00,000 × (1 - x/100)
Population at the end of 2022 = 1,00,000 × (1 - y/100) × (1 + x/100)   ...(1)

∴ Overall % change = -y + x + (-y × x)/100   [using formula = a + b + ab/100]

Since population in 2022 is greater than that in 2020
⇒ x > y.
∴ x = y + 10

Hence, the population decreases by y% and then increases by (y + 10)%
∴ Overall % change = -y + (y + 10) + (-y × (y + 10))/100 = 10 - y(y + 10)/100   [using formula = a + b + ab/100]

Overall change should be positive
∴ Overall % change = 10 - y(y + 10)/100 > 0
⇒ y(y + 10) < 1000

Higheset value of y satisfying the above inequality is 27.

∴ The population will decrease by a maximum of 27%.
⇒ Least population in 2021 = 1,00,000 × (1 - 27%) = 73,000.

Hence, option (a).

Concept: Successive Percentage Change Formula

Answer: Option C

Video Explanation

Explanation :

Let the old average be 'a'.
∴ Total weight of the three students = 3a

Let the weight of D and E be D kgs and E kgs respectively.

Because of D the average reduces by x kg.
⇒ a - x = (3a + D)/4
⇒ 4a - 4x = 3a + D
⇒ D = a - 4x   ...(1)

Because of E the average increases by 2x kg.
⇒ a + 2x = (3a + E)/4
⇒ 4a + 8x = 3a + E
⇒ E = a + 8x   ...(2)

We know, E - D = 12
⇒ (a + 8x) - (a - 4x) = 12
⇒ 12x = 12
⇒ x = 1

Hence, option (c).

Answer: Option B

Video Explanation

Explanation :

Let 20 kgs and 30 kgs of A and B are mixed.
∴ Amount of sugar in C = 2/5 × 20 + 3/10 × 30 = 17 kgs

Now we have 50 kgs of C mixed with 50 kgs of milk i.e., 100 kgs of final solution.

⇒ Concentration of sugar in final solution = 17/(50 + 50) × 100% = 17%

Hence, option (b).

Answer: 340

Video Explanation

Explanation :

Let total number of fruits at the beginning of the day = 5x
∴ Number of mangoes = 40% of 5x = 2x

Let the number of apples be '5a', hence the number of bananas = 5x - 5a.

⇒ He sells 50% of 2x mangoes, 96 bananas and 40% of 5a apples i.e., x + 96 + 2a fruits.

Also, he sold 50% of his fruits
⇒ x + 96 + 2a = 50% of 5x = 2.5x
⇒ 1.5x = 96 + 2a
⇒ x = 64 + 4a/3
⇒ 5x = 320 + 20a/3

To minimise 5x, we need to minimise 2aa/3.

Minimum integral value of 20a/3 will be 20 when a = 3.

∴ 5x = 320 + 20 = 340

Hence, 340.

Answer: 42

Video Explanation

Explanation :

Let the number of coins collected each week by A and B be 3x and 4x respectively.

For A: Coins collected by A in 5 weeks is a multiple of 7
∴ 3x × 5 is a multiple of 7
⇒ x is a multiple of 7.

For B: Coins collected by B in 3 weeks is a multiple of 24
∴ 4x × 3 is a multiple of 24
⇒ x is a multiple of 2.

∴ x is a multiple of 7 and 2, hence least possible value of x = 14.

⇒ Minimum possible number of coins collected by A in one week is = 3x = 42.

Hence, 42.

Answer: 36

Video Explanation

Explanation :

Let work done per day (efficiency) of Gautam and Suhani are 'g' and 's' units.

A shortfall of 40% for Gautam is compensated by 50% extra work done by Suhani.
⇒​​​​​​​ 0.4 × g = 0.5 × s
⇒​ g/s = 5/4
⇒​ Let g = 5x and s = 4x

Together they can complete the work in 20 days.
⇒​ Total work to be done = 20 × (5x + 4x) = 180x units

The faster among the two is Gautam whose efficiency is 5x.

∴ Time required by Gautam alone = 180x / 5x = 36 days.

Hence, 36.

Answer: Option A

Video Explanation

Explanation :

In △OBD, ∠OBD = 90 - 75 = 15° 
In △EBC, ∠ECB = 90 - 15 = 75° 
Similarly, ∠B = 75° and ∠A = 180 - 75 - 75 = 30°

Area of △ABC = $\frac{1}{2}\times \mathrm{AD}\times \mathrm{BC}$ = $\frac{1}{2}\times \mathrm{BE}\times \mathrm{AC}$

⇒ $\frac{\mathrm{AD}}{\mathrm{BE}}$ = $\frac{\mathrm{AC}}{\mathrm{BC}}$   ...(2)

Area of △ABC = $\frac{1}{2}\times \mathrm{AB}\times \mathrm{AC}\times \mathrm{Sin} 30°$ = $\frac{1}{2}\times A\mathrm{B}\times \mathrm{BC}\times \mathrm{Sin} 75°$

⇒ $\frac{\mathrm{AC}}{\mathrm{BC}}$ = $\frac{Sin 75°}{Sin 30°}$ = 2 × Sin 75° = 2 × Cos (90 - 15)° = 2 × Cos 15°

⇒ $\frac{\mathrm{AD}}{\mathrm{BE}}$ = $\frac{\mathrm{AC}}{\mathrm{BC}}$ = 2 × Cos 15°

Hence, option (a).

Answer: Option A

Video Explanation

Explanation :

Let the longer side of the rectangle is 2a and the shorter side is b as shown in the figure.

Using pythagoras theorem, we get
a² + b² = 2² = 4   ...(1)

Also, AM ≥ GM
⇒ $\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2}$ ≥ $\sqrt{{\mathrm{a}}^2{\mathrm{b}}^2}$

⇒ a² + b² ≥ 2ab

⇒ 4 ≥ 2ab

⇒ 2 ≥ ab

∴ Highest value of ab = 2, when a = b.

⇒ Ratio of longer side to shorter side = 2a : b = 2 : 1.

Hence, option (a).

Answer: 54

Video Explanation

Explanation :

Let the exterior angle be x°, hence interior angle is (x + 120)°.

Sum of interior and exterior angles is 180°.
⇒ x + (x + 120) = 180
⇒ x = 30°

∴ Interior angle of this regular polygon = 30 + 120 = 150°

Each interior angle of a regular polygon of n sides = (n - 2)/n × 180° = 150°
⇒ (n - 2)/n = 5/6
⇒ n = 12

Now, number of diagonals in a n-sided polygon = n × (n - 3)/2 = 12 × 9/2 = 54

Hence, 54.

Answer: Option C

Video Explanation

Explanation :

n can take any value from 1 till 99.

∴ a_n = 54, 62, 70, 78, 86, 94, 102, ..., 838 and
b_n = 102, 106, 110, ..., 494

These two are arithmetic progressions. Common terms of 2 APs are also in AP whose common difference is LCM of the common difference of original APs.
Common difference of a_n and b_n is 8 and 4 respectively.
∴ The common difference of the common terms = LCM(8, 4) = 8

⇒ The common terms are 102, 110, 118, ...

Now, n^th term of this series = 102 + 8(n - 1)
This should be less than or equal to 494
⇒ 102 + 8(n - 1) ≤ 494
⇒ n ≤ 50

Sum of all these 50 terms = 50/2 × (102 + 494) = 14900

Hence, 14900.

Answer: Option C

Video Explanation

Explanation :

Given, 1 + $\left(1+\frac{1}{3}\right)\frac{1}{4}$ + $\left(1+\frac{1}{3}+\frac{1}{9}\right)\frac{1}{16}$ + $\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)\frac{1}{64}$ + ..., 

= 1 + $\left(\frac{1}{4}+\frac{1}{3}\times \frac{1}{4}\right)$ + $\left(\frac{1}{16}+\frac{1}{3}\times \frac{1}{16}+\frac{1}{9}\times \frac{1}{16}\right)$ + $\left(\frac{1}{64}+\frac{1}{3}\times \frac{1}{64}+\frac{1}{9}\times \frac{1}{64}+\frac{1}{27}\times \frac{1}{64}\right)$ + ...

= 1 + $\left(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...\right)$ + $\frac{1}{3}\left(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...\right)$ + $\frac{1}{9}\left(\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+...\right)$ + $\frac{1}{27}\left(\frac{1}{64}+\frac{1}{256}+\frac{1}{1024}...\right)$ + ...

= 1 + $\left(\frac{1/4}{1-1/4}\right)$ + $\frac{1}{3}\left(\frac{1/4}{1-1/4}\right)$ + $\frac{1}{9}\left(\frac{1/16}{1-1/4}\right)$ + $\frac{1}{27}\left(\frac{1/64}{1-1/4}\right)$ + ...

= 1 + $\frac{1}{3}$ + $\frac{1}{9}$ + $\frac{1}{9}\times \frac{1}{12}$ + $\frac{1}{9}\times \frac{1}{144}$ + ...

= 1 + $\frac{1}{3}$ + $\frac{1}{9}\left(1+\frac{1}{12}+\frac{1}{144}+...\right)$

= 1 + $\frac{1}{3}$ + $\frac{1}{9}\times \frac{1}{1-1/12}$

= 1 + $\frac{1}{3}$ + $\frac{4}{33}$

= $\frac{33+11+4}{33}$

= $\frac{48}{33}$

= $\frac{16}{11}$

Hence, option (c).

Answer: 3

Video Explanation

Explanation :

Suppose f(x, y) is a real-valued function such that f(3x + 2y, 2x - 5y) = 19x, for all real numbers x and y. The value of x for which f(x, 2x) = 27, is

Let
3x + 2y = a   ...(1) and
2x - 5y = b   ...(2)

Solving (1) and (2), we get
x = (5a + 2b)/19 and y = (2a - 3b)/19

∴ f(3x + 2y, 2x - 5y) = 19x can be rewritten as

f(a, b) = 5a + 2b

Now, substituting a = x and b = 2x, we get
​​​​​​​⇒ f(x, 2x) = 5x + 4x = 9x
​​​​​​​⇒ 27 = 5x + 4x = 9x
​​​​​​​⇒ x = 3

Hence, 3.