CAT 2019 QA Slot 1 | Previous Year CAT Paper

Answer: Option D

Video Explanation

Explanation :

The hexagon will be regular only if the hexagon is symmetrical with resepect to the original triangle.

This is only possible when the corners cut are all equilateral (of let's say side x) and the side length of the hexagon is equal to side length of the equilateral corner (again x).

H = 6 × $\frac{\sqrt{3}}{4}$x² (∵ A regular hexagon consists of six equilateral triangles of side length equal to the side length of the hexagon)

T = $\frac{\sqrt{3}}{4}$(3x)² = $\frac{9\sqrt{3}}{4}$x² (∵ Side length of the triangle = x + x + x = 3x)

∴ $\frac{\mathrm{H}}{\mathrm{T}}$ = $\frac{6}{9}$ = $\frac{2}{3}$.

Hence, option (d).  

Answer: 3920

Video Explanation

Explanation :

We have to go from (1, 1) to (8, 10) via (4, 6)

Number of paths from (1,1) to (8,10) = [Number of paths from (1,1) to (4,6)] × [Number of paths from (4,6) to (8,10)]

Path from (1,1) to (4,6):

Number of horizontal displacements (∆x) = 4 − 1 = 3 units and

Number of vertical displacements (∆y) = 6 − 1 = 5 units.

Hence, a total of 8 units.

∴ Number of paths from (1,1) to (4,6) = ⁸C₃ × ⁵C₅ = 56.
 
Path from (4,6) to (8,10):

Number of horizontal displacements (∆x) = 8 − 4 = 4 units and

Number of vertical displacements (∆y) = 10 − 6 = 4 units.

Hence, a total of 8 units.

∴ Number of paths from (4,6) to (8,10) = ⁸C₄ × ⁴C₄ = 70.

Total required number of paths = 56 × 70 = 3920.

Hence, 3920.

Answer: Option C

Video Explanation

Explanation :

Ratio of the three diagonals is 3 : 2√3 : √15

Let the legths of the three diagonals be 3k, (2√3)k and (√15)k.

And, the brick have length, breadth, height as x, y and z respectively. 
 
∴ x² + y² = (3k)² = 9k² ...(1)

y² + z² = [(2√3)k]² = 12k² ...(2)

z² + x² = [(√15)k]² = 15k² ...(3)

Adding (1), (2) and (3), we get;
 
x² + y² + z² = 18k² ...(4)

Using (4) along with any of (1), (2) and (3), we get;

x = k√6 , y = k√3 and z = 3k, 

Required ratio = (k√3)/3k = 1/√3.

Hence, option (c).

Answer: Option B

Video Explanation

Explanation :

(√2)¹⁹ 3⁴ 4² 9^m 8ⁿ = 3ⁿ 16^m (64)^1/4

∴ 2^19/2 3⁴ 2⁴ 3^2m 2³ⁿ = 3ⁿ 2^4m 2^6/4

∴ 2^19/2 3⁴ 2⁴ 3^2m 2³ⁿ = 3ⁿ 2^4m 2^3/2

∴ 2^{[(19/2) + 4 + 3n]} 3^{(4 + 2m)} = 2^{[4m + (3/2)]} 3ⁿ

Equating the exponents of 2 and 3 on both sides, we get;

Powers of 2 : (19/2) + 4 + 3n = 4m + (3/2) 

⇒ 6n + 24 = 8m ...(1)

Powers of 3 : 4 + 2m = n  ...(2)

Solving (1) and (2), we get;

m = −12.

Hence, option (b).

Answer: Option C

Video Explanation

Explanation :

Density of liquid 1 = 1000 g/L.

Density of liquid 2 = 800 g/L.

Half litre of the mixture weighs 480 gm, so 1 L of the mixture weighs 960 gm.

So, density of the mixture = 960 g/L.  

Using the alligation cross;

$\frac{Liquid 1}{Liquid 2}=\frac{(960-800)}{(1000-960)}$ = $\frac{4}{1}$.

Percentage of liquid 1 in the mixture = (4/5) × 100 = 80%.

Hence, option (c). 

Answer: 2

Video Explanation

Explanation :

|x| ≥ 1 ⇒ x  ≥ 1 and x ≤ −1. This is represented as the shaded region in the figure below.

|x| + |y| ≤ 2 : This will have four subcases depending on which quadrant the point is.

1st quadrant (x > 0; y > 0) : y ≤ 2 − x
2nd quadrant (x < 0; y > 0) : y ≤ 2 + x
3rd quadrant (x < 0; y < 0) : y ≥ −2 − x
4th quadrant (x > 0; y < 0) : y ≥ x − 2

Combining these four, the graph for |x| + |y| ≤ 2 is as shown below.

The intersection of |x| + |y| ≤ 2 and |x| ≥ 1 is as shown as the shaded area in the following image.

We need to find Area ∆ABC + Area ∆DEF

Area ∆ABC = Area ∆DEF = (1/2) × BC × AM = (1/2) × 2 × 1 = 1 square unit.

Required area = 1 + 1 = 2 square units.

Hence, 2.

Answer: Option A

Video Explanation

Explanation :

The circle can be drawn as shown below. Diameter of the circle = 2 × 11 = 22 cm.

CE = 7, so ED = CD − CE = 22 − 7 = 15.

Let AE be x units, so EB = AB − AE = 20.5 − x.

Using the intersecting chords theorem; AE × EB = CE × ED.

∴ x(20.5 − x) = 7 × 15

Solving this equation, we get x = 10.5, 10.

So, if (AE, BE) = (10.5, 10) or (10,10.5)

Required difference = |BE − AE| = |10.5− 10| = 0.5.

Hence, option (a).

Answer: 9

Video Explanation

Explanation :

3x + 5y − 45 = 0 cuts the coordinate axes at C(15,0) and A(0,9) as shown in the image below. 

∠ABC = 90°, so we can deduce that AC is the diameter of the circumcircle.
[In a right triangle, hypotenuese is the diameter of circumcircle.]

Diameter = √(15² + 9²) = √306.

Radius = (√306)/2 = 8.74 ≈ 9 units.

Hence, 9.

Answer: Option D

Video Explanation

Explanation :

First car starts at 10 am : Let the speed and time taken be a and t respectively.

Second car starts at 11 am : Let the speed be b. Time taken = t − 1.

Required percentage = [(b − a)/a] × 100 = [(b/a) − 1] × 100

Required percentage will be highest when b/a is highest.

Now, since distances covered by both cars are same, so D = at = b(t − 1)

∴ b/a = t/(t − 1) = 1/[1 − (1/t)]

(b/a) is maximum when t is minimum.

t_min = 6 (given)

∴ b/a = 6/5.

Maximum required percentage = [(6/5) − 1] × 100 = 20%.

Hence, option (d).

Answer: Option A

Video Explanation

Explanation :

Let time taken by Amal be 3t, so time taken with each speed = 3t/3 = t.  

Let total distance travelled be 3d. Hence, Bimal travels d at each of the 3 different speeds.

For Amal : 3d = 10t + 20t + 30t = 60t.

∴ d = 20t.

Time taken by Bimal = (d/10) + (d/20) + (d/30) = (11d/60) = 11t/3.

Required percentage = [{(11t/3)/3t} − 1] × 100 = (2/9) × 100 = 22.22 ≈ 22%.

Hence, option (a).

Answer: Option D

Video Explanation

Explanation :

Let total work be the LCM of 12 and 9 = 36 units.

Let the efficiency of A and B be a and b respectively.

Work done per day when A and B are working together = 36/12 = 3 units.

∴ a + b = 3 ...(1)   

Work done per day when A is working at half efficieny and B is working at thrice efficiency = 36/9 = 4 units.

∴ (a/2) + 3b = 4 ...(2)

Solving (1) and (2), we get; a = 2. 

Time taken by A alone to complete the work = 36/2 = 18 days.

Hence, option (d).

Answer: 6144

Video Explanation

Explanation :

Eleventh term of the series (a₁₁) = S₁₁ − S₁₀  (where S_n represents the sum of n terms of the series)

S₁₁ = 3(2¹² − 2) and S₁₀ = 3(2¹¹ − 2).

∴ a₁₁ = 3(2¹² − 2) - 3(2¹¹ - 2) = 3(2¹² − 2¹¹) = 3 × 2¹¹ (2 − 1) = 6144.

Hence, 6144. 

Answer: Option C

Video Explanation

Explanation :

Let the two numbers be x and y.

So, xy = 616 and (x³ − y³)/(x − y)³ = 157/3. 

3(x³ − y³) = 157(x − y)³

∴ 3(x³ − y³) =157 [x³ − y³ − 3xy(x − y)]

∴ 154(x³ − y³) = 3 × 157 × xy(x − y)

∴ 154(x³ − y³) = 3 × 157 × 616 × (x − y)  [Using xy = 616]

∴ (x − y)(x² + y² + xy) = 1884(x − y)

∴ (x² + y² + xy) = 1884 (If x ≠ y)

Adding xy both sides, we get;

(x²+ y² + 2xy) = 1884 + xy = 1884 + 616 = 2500.

∴ x + y = 50.

Hence, option (c).

Answer: Option C

Video Explanation

Explanation :

Let the total marks be 100x.

Meena's score = 40x.

Meena's score after review = 40x + [(40x)/2] = 60x.

Passing marks = 60x + 35.

Post review score × (6/5) = 7 + Passing marks

∴ 60x × (6/5) = 60x + 42.

Solving this equation we get; x = 3.5.

So, passing marks = 60x + 35 = (60 × 3.5) + 35 = 245 and total marks = 100x = 100 × 3.5 = 350.

Percentage score needed to pass the examination = (Passing marks/Total marks) × 100 = (245/350) × 100 = 70%.

Hence, option (c).

Answer: 5

Video Explanation

Explanation :

Since we have |x|, we will have to consider cases where, x < 0 or x = 0 or x > 0.

For x > 0: 6$x^2$ + 1 = 5x ⇒  6$x^2$ − 5x + 1 = 0. The two roots of this equation are 1/2 and 1/3.

For x = 0, LHS = RHS, ∴ x = 0 is a root of the equation.

For x < 0: 6$x^2$ + 1 = −5x ⇒  6$x^2$ + 5x + 1 = 0. The two roots of this equation are −1/2 and −1/3.

∴ Number of roots = 2 + 1 + 2 = 5.

Answer: 5.

Answer: 13

Video Explanation

Explanation :

Let the work done by one man and one machine per day be x and y respectively.

Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job.

Since efficiency is inversely proportional to the time taken, so the efficiency of 3 men and 8 machines is twice that of 8 men and 3 machines.

∴ (3x + 8y) = 2(8x + 3y)

∴ 13x = 2y.

So, work done by 13 men in a day = work done by 2 machines in a day.

∴ If two machines can finish the job in 13 days, same work will be done by 13 men in 13 days.

Hence, 13.

Answer: Option B

Video Explanation

Explanation :

Let marks of Gautam be G.

∴ G + (62 × 21) = T ...(I) (where T is the total marks of all 22 students)

82.5 + (21 × x) = T ...(II) (where x is the average marks of 21 students other than Ramesh) 

The average score of all the 22 students is one more than the average score of the 21 students other than Ramesh.

∴ (T/22) = 1 + x ...(III)

Solving (I), (II) and (III), we get; x = 60.5, T = 1353 and G = 51.

Hence, option (b).

Answer: Option A

Video Explanation

Explanation :

Let incomes of Amala, Bimala and Kamala be a, b and k respectively.

∴ a = $\frac{6}{5}$ × b = $\frac{4}{5}$ × k.

∴ $\frac{\mathrm{k}}{\mathrm{b}}$ = $\frac{3}{2}$

Let Kamala's income be 300 and Bimala's income be 200.

∴ Kamala's new income = 300 × 0.96 = 288 and Bimla's new income = 200 × 1.1 = 220.   

∴ Required percentage = $\left(\frac{288-220}{220}\right)$ × 100 = $\left(\frac{68}{220}\right)$  × 100 = 30.9 ≈ 31%.

Hence, option (a).

Answer: 10

Video Explanation

Explanation :

Case I: m is odd.

So, (m + 1) is even.  

∴ 8[(m + 1)(m + 2)] − (m + 3) = 2

∴ 8m² + 23m + 11 = 0.

Both roots of this equation are negative as sum of the roots (−23/8) is negative and the product (11/8) is positive. But it is given that m is a positive integer. Hence this case is discarded.

Case II: m is even.

So, (m + 1) is odd.

∴ 8(m + 3 + 1) − m(m + 1) = 2.

∴ m² − 7m − 30 = 0

Solving this equation, we get; m = 10 or −3.

Since m is positive, m = 10.  

Hence, 10.

Answer: 880

Video Explanation

Explanation :

Let x be the length of the racecourse. 

The first horse beat the second by 11 metres and the third by 90 metres.

∴ Distances travelled by the first, second and third horse are x, x − 11 and x − 90 respectively. 

The second horse beat the third by 80 metres.

Distances travelled by the second and third horse are x and x − 80 respectively.

Ratio of speeds of second and third horse is constant which is equal to the ratio of the distances travelled by the second and third horse.

∴ (x − 11)/(x − 90) = x/(x − 80)

⇒ x² - 91x + 880 = x² - 90x

⇒ x = 880

Hence, 880.

Concept:

Races

Answer: Option C

Video Explanation

Explanation :

From the given data, populaiton in 

2019 : p

after 1 year i.e. in 2020 : 2p + 3

after 2 years i.e. in 2021 : 2(2p + 3) + 3 = 2²p + 2×3 + 3

after 3 years i.e. in 2022 : 2(2²p + 2×3 + 3) = 2³p + 2²×3 + 2×3 + 3

...

after n years i.e. : = 2ⁿp + 2^n-1×3 + 2^n-2×3 + ... + 3

Hence, population in 2034 i.e. after 15 years = 2¹⁵p + 2¹⁴×3 + 2¹³×3 + ... + 3

= 2¹⁵p + 3(2^n-1 + 2^n-2 + ... + 1)

= 2¹⁵p + 3(2¹⁵ - 1)/(2 - 1)

= 2¹⁵ × 1000 + 3(2¹⁵ - 1)

= 2¹⁵ × 1003 - 3

Hence, option (c).

Answer: Option D

Video Explanation

Explanation :

Let the cost price of one pen and one book be 100p and 100b respectively.

On selling a pen at 5% loss and a book at 15% gain, Karim gains Rs. 7.

∴ 95p + 115b = 7 + (100p + 100b) ⇒ 15b − 5p = 7    ...(1)

On selling the pen at 5% gain and the book at 10% gain, he gains Rs. 13.

∴ 105p + 110b = 13 + (100p + 100b)  ⇒ 10b + 5p = 13    ...(2)  

Solving (1) and (2), we get; b = 4/5.

So, cost price of one book = 100b = 100 × (4/5) = Rs. 80.

Hence, option (d).

Answer: 20

Video Explanation

Explanation :

Let the total number of students be 100x.

So, number of girls and boys are 60x and 40x respectively.

There are 30 more girls than boys ∴ 60x = 40x + 30. 

∴ x = 3/2.

Number of students who passed = 68x = 68 × (3/2) = 102 out of which 30 boys passed.  

So, number of girls who passed = 102 − 30 = 72  

Number of girls who did not pass = 60x − 72 = 90 − 72 = 18. 

Required percentage = [18/90] × 100 = 20%.

Hence, 20.

Answer: Option B

Video Explanation

Explanation :

The best approach to solving such questions in exams is to put values and then cross checking the options.

Let n = 2, so we will have three terms in AP (a₁, a₂ and a₃). Let a₁ = a₂ = a₃ = 1.

1/(√a₁ + √a₂) = 1/2.

1/(√a₂ + √a₃) = 1/2.

∴ [1/(√a₁ + √a₂)] + [1/(√a₂ + √a₃)] = (1/2) + (1/2) = 1.

Put n = 2 in;

Option 1: 2/0 = not defined. So this option is incorrect.

Option 2: 2/(1 + 1) = 2/2 = 1. So this option is correct.

Option 3: (2 − 1)/(1 + 1) = 1/2. So this option is incorrect.

Option 4: (2 − 1)/(1 + 1) = 1/2. So this option is incorrect.

Hence, option (b).

Answer: Option C

Video Explanation

Explanation :

Let the radius of A and B be a and b respectively. (a = 30 and b = 40)  

Distance travelled by any wheel D = (Circumference C) × (Number of revolutions N)

∴ D ∝ R × N (∵ C ∝ Radius R)

∴ N ∝ 1/R

∴ N_a/N_b = b/a = 40/30 = 4/3   ...(1)   

It is given that N_a = N_b + 5000 ...(2)

Solving (1) and (2), we get; N_a = 20000 and N_b = 15000.

D = C_a × N_a = 2πa × N_a = v_b × (3/4)   [where v_b is the velocity of B]

∴ 2π × (30 × 10^-5) × 20000 = v_b × (3/4)

∴ v_b = 16π.

Hence, option (c).

Answer: Option A

Video Explanation

Explanation :

∣x² − x − 6∣ = x + 2

∴ ∣(x − 3)(x + 2)∣ = x + 2 

Case 1: x < −2. i.e. (x − 3)(x + 2) > 0

(x - 3)(x + 2) = x + 2  

∴ x = 4. (which is rejected since  4 is not less than −2)

Case 2: x = −2.

This is a real root of this equation.

Case 3: −2 < x < 3. i.e. (x − 3)(x + 2) < 0

- (x - 3)(x + 2) = x + 2  

∴ x = 2.

Case 4: x = 3.

This does not satisfy the equation so x = 3 is not a root of this equation. 

Case 5: x > 3. (x − 3)(x + 2) > 0

(x − 3)(x + 2) = x + 2 

∴ x = 4.

Required product = (−2) × 2 × 4 = −16.

Hence, option (a).

Answer: Option D

Video Explanation

Explanation :

Since diameter subtends an angle of 90° at any point on the circumference, ∆APB is a right angled triangle.

So AB² = AP² + PB²

⇒ 10² = (2x)² + 6²

∴ x = 4.

∆AQB is also a right angled triangle, so AB² = AQ² + QB²

∴ 10² = x² + QB²

Put x = 4 to get QB = √84 = 9.165 ≈ 9.1 cm.

Hence, option (d).

Answer: Option B

Video Explanation

Explanation :

Let the amount invested by Amala, Bina and Gouri be 3x, 4x and 5x respectively.  

Also, let the respective rates be 6r, 5r and 4r.

So, the respective ratio of simple interest is (3x × 6r) : (4x × 5r) : (5x × 4r) = 18xr : 20xr : 20xr.

Bina's interest income exceeds Amala's by Rs 250.

∴ 20xr − 18xr = 250.

∴ xr = 125.

Total interest income = 18xr + 20xr + 20xr = 58xr = 58 × 125 = Rs. 7250.

Hence, option (b).

Answer: 9

Video Explanation

Explanation :

Since the amount invested is in the ratio 2 : 1, we can assum the amout to be 2x and x respectively.

∴ Amount invested in fixed deposit = 15L − 3x   (where L is lakhs)

Simple interest earned on the fixed deposit = [(15L − 3x) × (6/100) × 1]  ...(1)

Simple interest earned on 2x principle = 2x × (4/100) × 1 ...(2)

Simple interest earned on x principle = x × (3/100) × 1 ...(3)

(1) + (2) + (3) = 76000.

Solving we get; x = 2L.

So, amount invested in fixed deposit = 15L − 3x = 15L − 6L = 9L.

Hence, 9.

Answer: Option B

Video Explanation

Explanation :

From observing the data given, we find that it is a closed 3 set Venn diagram.

Let the three sports be F, T and C for Football, Tennis and Cricket respectively

n(FUTUC) = 256, 

n(F) = 144, n(T) = 123, n(C) = 132, 

n(F∩T) = 58, n(C∩T) = 25, n(F∩C) = 63

We know that (AUBUC) = n(A) + n(B) +n(C) - n(A∩B) - n(B∩C) - n(C∩A) + n(A∩B∩C)

So, 256 = 144 + 123 + 132 - 58 - 25 - 63 + n(F∩T∩C)

n(F∩T∩C) = 256 - 144 + 123 +132 - 146

n (F∩T∩C) = 256 - 253 = 3

Now, n(Students who play only Tennis) = 123 - (55 + 3 + 22) = 123 - 80

n(Students who play only Tennis) = 43 students.

Hence, option (b).

Answer: Option B

Video Explanation

Explanation :

log₅(x + y) + log₅(x - y) = 3

∴ log₅(x + y)(x - y) = 3.

∴ (x + y)(x − y) = 5³ = 125

So, x² - y² = 125  ....(1)

log₂y - log₂x = 1 - log₂3

∴ ${\log }_2\left(\frac{3y}{x}\right)$ = 1

∴ (3y/x) = 2¹ = 2.

So, 3y = 2x .... (2)

Solving (1) and (2), we get;

x = 15 and y =10.

∴ xy = 15 × 10 = 150. 

Hence, option (b).

Answer: Option A

Video Explanation

Explanation :

Given, 5.55^x = 0.555^y = 1000 = k say.

⇒ 5.55^x = k ⇒ 5.55 = k^1/x   ...(1)
⇒ 0.555^y = k ⇒ 0.555 = k^1/y   ...(2)
⇒ 1000^z = 10³ = k ⇒ 10 = k^1/3   ...(3)

⇒ 5.55 × 10 = 0.555
∴ k^1/x × k^1/3 = k^1/y
⇒ k^{1/x + 1/3} = k^1/y
⇒ 1/x + 1/3 = 1/y
⇒ 1/x - 1/y = 1/3

Alternately,
5.55^x = 0.555^y = 1000  

Taking log to the base 10, we get; 

log(5.55^x) = log(0.555^y) = log(1000) = log 10³ = 3 

∴ x log 5.55 = y log 0.555 = 3 

So, (1/x) = (log 5.55)/3 

(1/y) = (log 0.555)/3  

Log 0.555 = log 5.55 × 10⁻¹ = log 5.55 + log 10⁻¹ =  (log 5.55) − 1. 

So, (1/y) = [(log 5.55) − 1]/3 

∴  (1/x) − (1/y) = [(log 5.55)/3] − [{(log 5.55) − 1}/3] = 1/3. 

Hence, option (a).

Answer: 3

Video Explanation

Explanation :

Given, f(1) = 2

Now,

f(2) = f(1 + 1) = f(1) × f(1) = 2 × 2 = 4 = 2². 

f(3) = f(1 + 2) = f(1) × f(2) = 2 × 2² = 2³. 

f(3) = f(1 + 3) = f(1) × f(3) = 2 × 2³ = 2⁴.   

So, f(n) = 2ⁿ. 

f(a + 1) + f(a + 2) +…+ f(a + n) = 16(2ⁿ – 1) 

∴ 2^{(a + 1)} + 2^{(a + 2)} + ... + 2^{(a + n)} = 2⁴(2ⁿ – 1) 

∴ 2^{(a + 1)}[1 + 2 + ... 2⁽ⁿ⁻¹⁾] = 2⁴(2ⁿ – 1) 

∴ 2^{(a + 1)}(2ⁿ – 1) = 2⁴(2ⁿ – 1) 

∴ 2^{(a + 1)} = 2⁴

So, a + 1 = 4

∴ a = 3.

Hence, 3.

Answer: Option D

Video Explanation

Explanation :

−2 ≤ 2cos(x(x + 1)) ≤ 2 

∴ −2 ≤ 2^x + 2^–x ≤ 2 

Let 2^x be a, so 2^–x is 1/a. 

So, −2 ≤ a + (1/a) ≤ 2     

∴ −2 ≤ (a² + 1)/a ≤ 2 

∴ −2a ≤ (a² + 1) ≤ 2a 

∴ (a² + 1 + 2a) ≥ 0 ⇒ (a + 1)² ≥ 0, so a ∈ R. 

Also, a² + 1 − 2a  ≤ 0  ⇒ (a − 1)² ≤ 0, so a = 1. 

Hence a = 1. 

So, 2^x = 1. 

∴ x = 0.  

So, there is only one real root. 

Hence, option (d).