CAT 2019 Question Paper | CAT QA
CAT Question Paper | CAT Previous Year Paper

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 2.

Let the amount invested in the ratio 2 : 1 be 2x and x.

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

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

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

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

(I) + (II) + (III) = 76000.

Solving we get; x = 2L.

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

Answer: 9

Let the number of members who play exactly one game, two games and all three games be a, b and c respectively.

So, a + 2b + 3c = 144 + 123 + 132 = 399 ...(I)

Total number of members = a + b + c = 256 ....(II)

Subtracting (II) from (I), we get;

b + 2c = 143 ....(III)

Also, b + 3c = 58 + 25 + 63 = 146 ....(IV)

Solving (III) and (IV), we get; c = 3 and b = 137.

Using (II), a = 116.

Number of members who can play only tennis = 123 − (58 + 25 − 3) = 43.

Hence option 2.

log₅ (x + y) + log₅ (x − y) = 3.

∴ log₅ [(x + y)(x − y)] = 3.

∴ (x + y)(x − y) = 5³ = 125

So, x² − y² = 125  ....(I)

log₂ y − log₂ x = 1 − log₂ 3. 

∴ log₂ (3y/x) = 1

∴ (3y/x) = 2¹ = 2.

So, 3y = 2x .... (II)

Solving (I) and (II), we get;

x = 15 and y =10.

∴ xy = 15 × 10 = 150. 

Hence option 2.

Let AQ be x, so AP = 2x.

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 4.

∣x² − x − 6∣ = x + 2

∴ âˆ£(x − 3)(x + 2)∣ = x + 2 

Case 1: x < −2. 

(−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.

(−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) = x + 2 

∴ x = 4.

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

Hence option 1.  

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   ...(I)   

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

Solving (I) and (II), 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 3.

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 1.

−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 4.

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 2.

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%.

Answer: 20.

f(1) = 2

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.

Answer: 3.

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    ...(I)

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    ...(II)  

Solving (I) and (II), we get; b = 4/5.

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

Hence option 4.

Population of the town at the beginning of the;

First year = p.

Second year = 3 + 2p.

Third year = 3 + 2(3 + 2p) = 9 + 4p = 3(1 + 2) + 4p

Fourth year = 3 + 2(9 + 4p) = 21 + 8p = 3(1 + 2 + 4) + 8p

So, we can deduce that population at the beginining of the nth year = 3(1 + 2¹ + 2² + ...2⁽ⁿ⁻²⁾) + 2⁽ⁿ⁻¹⁾ p

= 3[2⁽ⁿ⁻¹⁾ − 1] + 2⁽ⁿ⁻¹⁾ p

= 2⁽ⁿ⁻¹⁾ [3 + p] − 3

If we consider 2019 as 1st year, then 2034 is 16th year. (2034 − 2019 + 1 = 16)

So, population at the beginining of 2034 (n = 16) =  2⁽¹⁶⁻¹⁾ [3 + 1000] − 3

= [2¹⁵ × 1003] − 3

Hence option 3.

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)

Solving this equation, we get; x = 880.

Answer: 880.

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.  

Answer: 10.

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

∴ a = (6/5)b = (4/5)k.

∴ (k/b) = 3/2 ....(I)

Kamala's new income = (96/100)k = 0.96k and Bimla's new income = (11/10)b = 1.1b.   

Using (I), required percentage = [{(0.96k)/(1.1b)} − 1] × 100 = 30.9 ≈ 31%.

Hence option 1.

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 2.

Let the work done by one man and one machine 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.

Answer: 13.

For x > 0: 6x² + 1 = 5x ⇒  6x² − 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: 6x² + 1 = −5x ⇒  6x² + 5x + 1 = 0. The two roots of this equation are −1/2 and −1/3.

∴ Number of roots = 2 + 1 + 2 = 5.

Answer: 5.

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 3.

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 3.

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 − 2¹¹) = 3 × 2¹¹ (2 − 1) = 6144.

Answer: 6144. 

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 ...(I)   

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 ...(II)

Solving (I) and (II), we get; a = 2. 

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

Hence option 4.

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

Let total distance travelled be 3s.

∴ 3s = 10t + 20t + 30t = 60t.

∴ s = 20t.

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

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

Hence option 1.

Car starting at 10 am: Let the speed and time taken be a and t respectively.

Car starting at 11 am: Let the speed be b. Time taken = t − 1.

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

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.

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

Henc option 4.

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.

Diameter = √(15² + 9²) = √306.

Radius = (√306)/2 = 8.74 ≈ 9 units.

Answer: 9

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.

Henc option 1.

|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 âˆ†ABC + Area âˆ†DEF

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

Required area = 1 + 1 = 2 square units.

Answer: 2.

Weight/Volume for liquid 1 = 1000 g/L.

Weight/Volume for liquid 2 = 800 g/L.

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

So, Weight/Volume for the mixture = 960 g/L.

Using the alligation cross;

(Liquid 1 in the mixture)/(Liquid 2 in the mixture) = (960 − 800)/(1000 − 960) = 160/40 = 4/1.

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

Hence option 3. 

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

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

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

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

(19/2) + 4 + 3n = 4m + (3/2) ⇒ 6n + 24 = 8m ...(I)

4 + 2m = n  ...(II)

Solving (I) and (II), we get;

m = −12.

Hence option 2.

Let the brick have length, breadth, height as x, y and z respectively. Also let the lengths of the three different diagonals be 3k, (2√3)k and (√15)k.  

∴ x² + y² = (3k)² = 9k² ...(I)

y² + z² = [(2√3)k]² = 12k² ...(II)

z² + x² = [(√15)k]² = 15k² ...(II)

Adding (I), (II) and (III), we get;

x² + y² + z² = 18k² ...(IV)

Using (IV) along with any of (I), (II) and (III), we get;

x = k√6 , y = k√3 and z = 3k, 

Required ratio = (k√3)/3k = 1/√3.

Hence option 3.

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)]

Consider (1,1) to (4,6): âˆ†x = 4 − 1 = 3 and âˆ†y = 6 − 1 = 5. So we need to go 3 units horizontally and 5 units vertically. So a total of 8 units.

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

Consider (4,6) to (8,10): âˆ†x = 8 − 4 = 4 and âˆ†y = 10 − 6 = 4. So we need to go 4 units horizontally and 4 units vertically. So 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.

Answer: 3920.

Resulting hexagon will be regular only if the hexagon is symmetrical wrt the original triangle. That is only possible when the corners cut are all equilateral and the side length of the hexagon is equal to side length of the equilateral corner (x say). 

H = 6 × [(√3)/4] x² (∵ A regular hexagon consists of six equilateral triangles of side length equal to the side length of the hexagon)

T = [(√3)/4] (3x)² (∵ Side length of the triiangle = x + x + x = 3x)

∴ H/T = 6/9 = 2/3.

Hence option 4.