# CAT 2019 QA Slot 1 | Previous Year CAT Paper

**1. CAT 2019 QA Slot 1 | Geometry - Triangles**

Corners are cut off from an equilateral triangle T to produce a regular hexagon H. Then, the ratio of the area of H to the area of T is

- A.
5 : 6

- B.
4 : 5

- C.
3 : 4

- D.
2 : 3

Answer: Option D

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

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

Hence, option (d).

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**2. CAT 2019 QA Slot 1 | Modern Math - Permutation & Combination**

With rectangular axes of coordinates, the number of paths from (1, 1) to (8, 10) via (4, 6), where each step from any point (x, y) is either to (x, y + 1) or to (x + 1, y), is

Answer: 3920

**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) = ^{8}C_{3} × ^{5}C_{5} = 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) = ^{8}C_{4} × ^{4}C_{4} = 70.

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

Hence, 3920.

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**3. CAT 2019 QA Slot 1 | Geometry - Mensuration**

If the rectangular faces of a brick have their diagonals in the ratio 3 : 2√3 : √15, then the ratio of the length of the shortest edge of the brick to that of its longest edge is?

- A.
√3 : 2

- B.
2 : √5

- C.
1 : √3

- D.
√2 : √3

Answer: Option C

**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.

∴ x^{2} + y^{2} = (3k)^{2} = 9k^{2} ...(1)

y^{2} + z^{2} = [(2√3)k]^{2} = 12k^{2} ...(2)

z^{2} + x^{2} = [(√15)k]^{2} = 15k^{2} ...(3)

Adding (1), (2) and (3), we get;

x^{2} + y^{2} + z^{2} = 18k^{2} ...(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).

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**4. CAT 2019 QA Slot 1 | Algebra - Surds & Indices**

If m and n are integers such that (√2)^{19} 3^{4} 4^{2} 9^{m} 8^{n} = 3^{n} 16^{m} (64)^{1/4}, then m is

- A.
-16

- B.
-12

- C.
-24

- D.
-20

Answer: Option B

**Explanation** :

(√2)^{19} 3^{4} 4^{2} 9* ^{m}* 8

*= 3*

^{n}^{n}16

^{m}(64)

^{1/4}

∴ 2^{19/2} 3^{4} 2^{4} 3^{2}* ^{m}* 2

^{3}

*= 3*

^{n}^{n}2

^{4m}2

^{6/4}

∴ 2^{19/2} 3^{4} 2^{4} 3^{2}* ^{m}* 2

^{3}

*= 3*

^{n}^{n}2

^{4m}2

^{3/2}

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

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

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

⇒ 6*n* + 24 = 8*m* ...(1)

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

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

m = −12.

Hence, option (b).

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**5. CAT 2019 QA Slot 1 | Arithmetic - Mixture, Alligation, Removal & Replacement**

A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm. If half litre of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, in terms of volume, is

- A.
70

- B.
85

- C.
80

- D.
75

Answer: Option C

**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;

$\frac{Liquid1}{Liquid2}=\frac{(960-800)}{(1000-960)}$ = $\frac{4}{1}$.

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

Hence, option (c).

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**6. CAT 2019 QA Slot 1 | Geometry - Coordinate Geometry**

Let S be the set of all points (x, y) in the x-y plane such that |x| + |y| ≤ 2 and |x| ≥ 1. Then, the area, in square units, of the region represented by S equals

Answer: 2

**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.

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**7. CAT 2019 QA Slot 1 | Geometry - Circles**

In a circle of radius 11 cm, CD is a diameter and AB is a chord of length 20.5 cm. If AB and CD intersect at a point E inside the circle and CE has length 7 cm, then the difference of the lengths of BE and AE, in cm, is

- A.
0.5

- B.
2.5

- C.
3.5

- D.
1.5

Answer: Option A

**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).

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**8. CAT 2019 QA Slot 1 | Geometry - Coordinate Geometry**

Let T be the triangle formed by the straight line 3x + 5y − 45 = 0 and the coordinate axes. Let the circumcircle of T have radius of length L, measured in the same unit as the coordinate axes. Then, the integer closest to L is

Answer: 9

**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 = √(15^{2} + 9^{2}) = √306.

Radius = (√306)/2 = 8.74 ≈ 9 units.

Hence, 9.

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**9. CAT 2019 QA Slot 1 | Arithmetic - Time, Speed & Distance**

Two cars travel the same distance starting at 10:00 am and 11:00 am, respectively, on the same day. They reach their common destination at the same point of time. If the first car travelled for at least 6 hours, then the highest possible value of the percentage by which the speed of the second car could exceed that of the first car is

- A.
25

- B.
10

- C.
30

- D.
20

Answer: Option D

**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).

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**10. CAT 2019 QA Slot 1 | Arithmetic - Time, Speed & Distance**

One can use three different transports which move at 10, 20, and 30 kmph, respectively. To reach from A to B, Amal took each mode of transport for 1/3 of his total journey time, while Bimal took each mode of transport for 1/3 of the total distance. The percentage by which Bimal’s travel time exceeds Amal’s travel time is nearest to

- A.
22

- B.
21

- C.
20

- D.
19

Answer: Option A

**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).

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**11. CAT 2019 QA Slot 1 | Arithmetic - Time & Work**

At their usual efficiency levels, A and B together finish a task in 12 days. If A had worked half as efficiently as she usually does, and B had worked thrice as efficiently as he usually does, the task would have been completed in 9 days. How many days would A take to finish the task if she works alone at her usual efficiency?

- A.
36

- B.
24

- C.
12

- D.
18

Answer: Option D

**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).

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**12. CAT 2019 QA Slot 1 | Algebra - Progressions**

If a_{1} + a_{2} + a_{3} + … + a_{n} = 3(2^{n+1} – 2), for every n ≥ 1, then a_{11} equals

Answer: 6144

**Explanation** :

Eleventh term of the series (a_{11}) = S_{11} − S_{10} (where S_{n} represents the sum of n terms of the series)

S_{11 }= 3(2^{12} − 2) and S_{10} = 3(2^{11} − 2).

∴ a_{11} = 3(2^{12} − 2) - 3(2^{11} - 2) = 3(2^{12} − 2^{11}) = 3 × 2^{11} (2 − 1) = 6144.

Hence, 6144.

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**13. CAT 2019 QA Slot 1 | Algebra - Number Theory**

The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157 : 3, then the sum of the two numbers is

- A.
58

- B.
85

- C.
50

- D.
95

Answer: Option C

**Explanation** :

Let the two numbers be x and y.

So, xy = 616 and (x^{3} − y^{3})/(x − y)^{3} = 157/3.

3(x^{3} − y^{3}) = 157(x − y)^{3}

∴ 3(x^{3} − y^{3}) =157 [x^{3} − y^{3} − 3xy(x − y)]

∴ 154(x^{3} − y^{3}) = 3 × 157 × xy(x − y)

∴ 154(x^{3} − y^{3}) = 3 × 157 × 616 × (x − y) [Using xy = 616]

∴ (x − y)(x^{2} + y^{2} + xy) = 1884(x − y)

∴ (x^{2} + y^{2} + xy) = 1884 (If x ≠ y)

Adding xy both sides, we get;

(x^{2}+ y^{2} + 2xy) = 1884 + xy = 1884 + 616 = 2500.

∴ x + y = 50.

Hence, option (c).

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**14. CAT 2019 QA Slot 1 | Arithmetic - Percentage**

Meena scores 40% in an examination and after review, even though her score is increased by 50%, she fails by 35 marks. If her post-review score is increased by 20%, she will have 7 marks more than the passing score. The percentage score needed for passing the examination is

- A.
60

- B.
80

- C.
70

- D.
75

Answer: Option C

**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).

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**15. CAT 2019 QA Slot 1 | Algebra - Quadratic Equations**

The number of solutions to the equation |x|(6${x}^{2}$ + 1) = 5${x}^{2}$ is

Answer: 5

**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.

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**16. CAT 2019 QA Slot 1 | Arithmetic - Time & Work**

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. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?

Answer: 13

**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.

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**17. CAT 2019 QA Slot 1 | Arithmetic - Average**

Ramesh and Gautam are among 22 students who write an examination. Ramesh scores 82.5. The average score of the 21 students other than Gautam is 62. The average score of all the 22 students is one more than the average score of the 21 students other than Ramesh. The score of Gautam is

- A.
53

- B.
51

- C.
49

- D.
48

Answer: Option B

**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).

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**18. CAT 2019 QA Slot 1 | Arithmetic - Percentage**

The income of Amala is 20% more than that of Bimala and 20% less than that of Kamala. If Kamala's income goes down by 4% and Bimala's goes up by 10%, then the percentage by which Kamala's income would exceed Bimala's is nearest to

- A.
31

- B.
29

- C.
28

- D.
32

Answer: Option A

**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).

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**19. CAT 2019 QA Slot 1 | Algebra - Functions & Graphs**

For any positive integer n, let f(n) = n(n + 1) if n is even, and f(n) = n + 3 if n is odd. If m is a positive integer such that 8 f(m + 1) − f(m) = 2, then m equals

Answer: 10

**Explanation** :

Case I: m is odd.

So, (m + 1) is even.

∴ 8[(m + 1)(m + 2)] − (m + 3) = 2

∴ 8m^{2} + 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.

∴ m^{2} − 7m − 30 = 0

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

Since m is positive, m = 10.

Hence, 10.

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**20. CAT 2019 QA Slot 1 | Arithmetic - Time, Speed & Distance**

In a race of three horses, the first beat the second by 11 metres and the third by 90 metres. If the second beat the third by 80 metres, what was the length, in metres, of the racecourse?

Answer: 880

**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)

⇒ x^{2} - 91x + 880 = x^{2} - 90x

⇒ x = 880

Hence, 880.

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**21. CAT 2019 QA Slot 1 | Algebra - Progressions**

If the population of a town is p in the beginning of any year then it becomes 3 + 2p in the beginning of the next year. If the population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be

- A.
(1003)

^{15}+ 6 - B.
(997)

^{15}- 3 - C.
(1003)2

^{15}- 3 - D.
(997)2

^{14}+ 3

Answer: Option C

**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 = 2^{2}p + 2×3 + 3

after 3 years i.e. in 2022 : 2(2^{2}p + 2×3 + 3) = 2^{3}p + 2^{2}×3 + 2×3 + 3

...

after n years i.e. : = 2^{n}p + 2^{n-1}×3 + 2^{n-2}×3 + ... + 3

Hence, population in 2034 i.e. after 15 years = 2^{15}p + 2^{14}×3 + 2^{13}×3 + ... + 3

= 2^{15}p + 3(2^{n-1} + 2^{n-2} + ... + 1)

= 2^{15}p + 3(2^{15} - 1)/(2 - 1)

= 2^{15 }× 1000 + 3(2^{15} - 1)

= 2^{15 }× 1003 - 3

Hence, option (c).

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**22. CAT 2019 QA Slot 1 | Arithmetic - Percentage**

On selling a pen at 5% loss and a book at 15% gain, Karim gains Rs. 7. If he sells the pen at 5% gain and the book at 10% gain, he gains Rs. 13. What is the cost price of the book in Rupees?

- A.
95

- B.
85

- C.
100

- D.
80

Answer: Option D

**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).

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**23. CAT 2019 QA Slot 1 | Arithmetic - Percentage**

In a class, 60% of the students are girls and the rest are boys. There are 30 more girls than boys. If 68% of the students, including 30 boys, pass an examination, the percentage of the girls who do not pass is

Answer: 20

**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.

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**24. CAT 2019 QA Slot 1 | Algebra - Progressions**

If a_{1}, a_{2}, ... are in A.P., then, $\frac{1}{\sqrt{{a}_{1}}+\sqrt{{a}_{2}}}$ + $\frac{1}{\sqrt{{a}_{2}}+\sqrt{{a}_{3}}}$ + ... + $\frac{1}{\sqrt{{a}_{n}}+\sqrt{{a}_{n+1}}}$ is equal to

- A.
$\frac{n}{\sqrt{{a}_{1}}-\sqrt{{a}_{n+1}}}$

- B.
$\frac{n}{\sqrt{{a}_{1}}+\sqrt{{a}_{n+1}}}$

- C.
$\frac{n-1}{\sqrt{{a}_{1}}+\sqrt{{a}_{n-1}}}$

- D.
$\frac{n-1}{\sqrt{{a}_{1}}+\sqrt{{a}_{n}}}$

Answer: Option B

**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 (a_{1}, a_{2} and a_{3}). Let a_{1} = a_{2} = a_{3} = 1.

1/(√a_{1} + √a_{2}) = 1/2.

1/(√a_{2} + √a_{3}) = 1/2.

∴ [1/(√a_{1} + √a_{2})] + [1/(√a_{2} + √a_{3})] = (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).

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**25. CAT 2019 QA Slot 1 | Arithmetic - Time, Speed & Distance**

The wheels of bicycles A and B have radii 30 cm and 40 cm, respectively. While traveling a certain distance, each wheel of A required 5000 more revolutions than each wheel of B. If bicycle B traveled this distance in 45 minutes, then its speed, in km per hour, was

- A.
14π

- B.
18π

- C.
16π

- D.
12π

Answer: Option C

**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).

Workspace:

**26. CAT 2019 QA Slot 1 | Algebra - Quadratic Equations**

The product of the distinct roots of ∣x^{2} − x − 6∣ = x + 2 is

- A.
-16

- B.
-4

- C.
-8

- D.
-24

Answer: Option A

**Explanation** :

∣x^{2} − 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).

Workspace:

**27. CAT 2019 QA Slot 1 | Geometry - Circles**

AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to

- A.
7.8

- B.
8.5

- C.
9.3

- D.
9.1

Answer: Option D

**Explanation** :

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

So AB^{2} = AP^{2} + PB^{2}

⇒ 10^{2} = (2x)^{2} + 6^{2}

∴ x = 4.

∆AQB is also a right angled triangle, so AB^{2} = AQ^{2} + QB^{2}

∴ 10^{2} = x^{2} + QB^{2}

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

Hence, option (d).

Workspace:

**28. CAT 2019 QA Slot 1 | Arithmetic - Simple & Compound Interest**

Amala, Bina, and Gouri invest money in the ratio 3 : 4 : 5 in fixed deposits having respective annual interest rates in the ratio 6 : 5 : 4. What is their total interest income (in Rs) after a year, if Bina's interest income exceeds Amala's by Rs 250?

- A.
7000

- B.
7250

- C.
6350

- D.
6000

Answer: Option B

**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).

Workspace:

**29. CAT 2019 QA Slot 1 | Arithmetic - Simple & Compound Interest**

A person invested a total amount of Rs 15 lakh. A part of it was invested in a fixed deposit earning 6% annual interest, and the remaining amount was invested in two other deposits in the ratio 2 : 1, earning annual interest at the rates of 4% and 3%, respectively. If the total annual interest income is Rs 76,000 then the amount (in Rs lakh) invested in the fixed deposit was

Answer: 9

**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.

Workspace:

**30. CAT 2019 QA Slot 1 | Venn Diagram**

A club has 256 members of whom 144 can play football, 123 can play tennis, and 132 can play cricket. Moreover, 58 members can play both football and tennis, 25 can play both cricket and tennis, while 63 can play both football and cricket. If every member can play at least one game, then the number of members who can play only tennis is

- A.
45

- B.
43

- C.
32

- D.
38

Answer: Option B

**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).

Workspace:

**31. CAT 2019 QA Slot 1 | Algebra - Logarithms**

Let x and y be positive real numbers such that log_{5}(x + y) + log_{5}(x - y) = 3, and log_{2}y - log_{2}x = 1 - log_{2}3. Then xy equals

- A.
250

- B.
150

- C.
100

- D.
25

Answer: Option B

**Explanation** :

log_{5}(x + y) + log_{5}(x - y) = 3

∴ log_{5}(x + y)(x - y) = 3.

∴ (x + y)(x − y) = 5^{3} = 125

So, x^{2} - y^{2} = 125 ....(1)

log_{2}y - log_{2}x = 1 - log_{2}3

∴ ${\mathrm{log}}_{2}\left(\frac{3y}{x}\right)$ = 1

∴ (3y/x) = 2^{1} = 2.

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

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

x = 15 and y =10.

∴ xy = 15 × 10 = 150.

Hence, option (b).

Workspace:

**32. CAT 2019 QA Slot 1 | Algebra - Surds & Indices**

If 5.55^{x} = 0.555^{y} = 1000, then the value of (1/x) − (1/y) is

- A.
1/3

- B.
2/3

- C.
3

- D.
1

Answer: Option A

**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} = 10^{3} = 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 10^{3} = 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).

Workspace:

**33. CAT 2019 QA Slot 1 | Algebra - Functions & Graphs**

Consider a function f satisfying f(x + y) = f(x) f(y) where x, y are positive integers and f(1) = 2. If f(a + 1) + f(a + 2) +…+ f(a + n) = 16(2^{n} – 1) then a is equal to

Answer: 3

**Explanation** :

Given, f(1) = 2

Now,

f(2) = f(1 + 1) = f(1) × f(1) = 2 × 2 = 4 = 2^{2}.

f(3) = f(1 + 2) = f(1) × f(2) = 2 × 2^{2} = 2^{3}.

f(3) = f(1 + 3) = f(1) × f(3) = 2 × 2^{3} = 2^{4}.

So, f(n) = 2^{n}.

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

∴ 2^{(a + 1)} + 2^{(a + 2)} + ... + 2^{(a + n)} = 2^{4}(2^{n} – 1)

∴ 2^{(a + 1)}[1 + 2 + ... 2^{(n−1)}] = 2^{4}(2^{n} – 1)

∴ 2^{(a + 1)}(2^{n} – 1) = 2^{4}(2^{n} – 1)

∴ 2^{(a + 1)} = 2^{4}

So, a + 1 = 4

∴ a = 3.

Hence, 3.

Workspace:

**34. CAT 2019 QA Slot 1 | Geometry - Trigonometry | Algebra - Inequalities & Modulus**

The number of the real roots of the equation 2cos(x(x + 1)) = 2^{x} + 2^{–x} is

- A.
0

- B.
2

- C.
infinite

- D.
1

Answer: Option D

**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 ≤ (a^{2} + 1)/a ≤ 2

∴ −2a ≤ (a^{2} + 1) ≤ 2a

∴ (a^{2} + 1 + 2a) ≥ 0 ⇒ (a + 1)^{2} ≥ 0, so a ∈ R.

Also, a^{2} + 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).

Workspace:

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