# CAT 2019 QA Slot 2 | Previous Year Questions

**1. CAT 2019 QA Slot 2 | Algebra - Quadratic Equations | Algebra - Logarithms**

The real root of the equation 2^{6x} + 2^{3x+2} - 21 = 0 is

- A.
$\frac{{\mathrm{log}}_{2}\left(3\right)}{3}$

- B.
log

_{2}9 - C.
$\frac{{\mathrm{log}}_{2}\left(7\right)}{3}$

- D.
log

_{2}27

Answer: Option A

**Explanation** :

Given: 2^{6x} + 2^{3x+2} - 21 = 0

Replace 2^{3x} with y

So, 2^{6x} = y^{2}

Now, 2^{6x} + 2^{3x+2} - 21 = 0 can be rewritten as y^{2} + 2^{2} × 2^{3x} - 21 = 0

y^{2} + 4y - 21 = 0

Solving the above quadratic equation,

(y + 7) (y - 3) = 0

So, y = -7 or +3

y = - 7 is rejected (since, y = 2^{3x} which should always be positive)

⇒ 2^{3x} = 3

Taking log on both sides,

log_{2}3 = 3x

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

Hence, option (a).

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**2. CAT 2019 QA Slot 2 | Arithmetic - Average**

The average of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average of these 20 integers?

- A.
4

- B.
5

- C.
4.5

- D.
3.5

Answer: Option C

**Explanation** :

Exactly 20 of the 30 integers do not exceed 5 that means 10 of the 30 integers are greater than 5.

Sum of all 30 integers = 30 × 5 = 150.

To keep the average of the 20 integers as high as possible, we need to keep the average of the 10 integers above 5 as low as possible. Since we are dealing with integers, the least value that the 10 integers above 5 can take is 6.

So, the sum of the 10 integers = 10 × 6 = 60.

Hence, the sum of the remaining 20 integers = 150 - 60 = 90

∴ The average of the remaining 20 = 90/20 = 4.5

Hence, option (c).

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**3. CAT 2019 QA Slot 2 | Algebra - Number Theory**

Let a, b, x, y be real numbers such that a^{2} + b^{2} = 25 , x^{2} + y^{2} = 169 and ax + by = 65. If k = ay - bx, then

- A.
k = 0

- B.
k > 513

- C.
k = 513

- D.
0 < k ≤ 513

Answer: Option A

**Explanation** :

Given: a^{2} + b^{2} = 25 and x^{2} + y^{2} = 169

We know 5^{2} = 25 and 13^{2} = 169

Multiply both equations to get (a^{2} + b^{2}) (x^{2} + y^{2}) = 25 × 169

(a^{2} + b^{2}) × (x^{2} + y^{2}) = 4225

We know, 4225 = 65^{2}

We also know that ax + by = 65

So, numerically (Not algebraically),

(a^{2} + b^{2}) × (x^{2} + y^{2}) = (ax + by)^{2}

Expanding the equation,

⇒ (ax)^{2} + (ay)^{2} + (bx)^{2} + (by)^{2} = (ax)^{2} + (by)^{2} + 2axby

⇒ (ay)^{2} + (bx)^{2} = 2axby

⇒ (ay)^{2} + (bx)^{2} - 2axby = 0

This is of the form, (p - q)^{2}

(ay - bx)^{2} = 0

⇒ ay - bx = 0 = k

Hence, option (a).

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**4. CAT 2019 QA Slot 2 | Geometry - Triangles**

In a triangle ABC, medians AD and BE are perpendicular to each other, and have lengths 12 cm and 9 cm, respectively. Then, the area of triangle ABC, in sq cm, is

- A.
80

- B.
68

- C.
72

- D.
78

Answer: Option C

**Explanation** :

As AD and BE are medians, so G is the centroid of ∆ABC.

So, AG : GD = BG : GE = 2 : 1.

∴ AG = 8 cm, GD = 4 cm, BG = 6 cm and GE = 3 cm.

If we join centroid with all 3 vertices, we get 3 triangles of equal areas.

∴ Area of ∆ABC = 3 × Area of ∆ABG

⇒ Area of ∆ABC = 3 × (1/2 × 8 × 6) = 72 sq. cm.

Hence, option (a).

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**5. CAT 2019 QA Slot 2 | Algebra - Progressions**

Let a_{1} , a_{2} be integers such that a_{1} - a_{2} + a_{3} - a_{4} + ........ + (-1)^{n-1} a_{n} = n , for n ≥ 1. Then a_{51} + a_{52} + ........ + a_{1023} equals

- A.
-1

- B.
1

- C.
0

- D.
10

Answer: Option B

**Explanation** :

Given that a_{1} - a_{2} + a_{3} - a_{4} + ........ + (-1)^{n-1} a_{n} = n

Put n = 1 ⇒ a_{1} = 1

Put n = 2 ⇒ a_{1} - a_{2} = 2 ⇒ a_{2} = 1 - 2 = -1

Put n = 3 ⇒ a_{1} - a_{2} + a_{3} = 3 ⇒ a_{3} = 3 – 1 -1 = 1

Hence, the series proceeds as 1, -1, 1, -1, ...

i.e. odd term of the series = +1

& even terms of the series = -1

Then a_{51} + a_{52} + ........ + a_{1023} = 1 + (-1) + .... + 1

⇒ a_{51} + a_{52} + ........ + a_{1023} = 1

Hence, option (b).

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**6. CAT 2019 QA Slot 2 | Algebra - Number Theory**

How many factors of 2^{4} × 3^{5} × 10^{4} are perfect squares which are greater than 1?

Answer: 44

**Explanation** :

N = 2^{4} × 3^{5} × 10^{4 }= 2^{4} × 3^{5} × 2^{4} × 5^{4} = 2^{8} × 3^{5} × 5^{4}

Now, for any number to be a perfect square, it must have even powers.

So, if we consider 2^{8}, only powers of 0, 2, 4, 6, 8 can lead to perfect squares i.e. 5 ways

Now,

If we consider 3^{5}, only powers of 0, 2, 4 can lead to perfect squares i.e. 3 ways

If we consider 5^{4}, only powers of 0, 2, 4 can lead to perfect squares i.e. 3 ways

So, total number of possibilities = 5 × 3 × 3 ways = 45 ways.

Since we need to find the number of factors greater than 1,

Required number of ways = 45 -1 = 44 ways

Hence, 44.

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**7. CAT 2019 QA Slot 2 | Geometry - Circles**

Two circles, each of radius 4 cm, touch externally. Each of these two circles is touched externally by a third circle. If these three circles have a common tangent, then the radius of the third circle, in cm, is

- A.
π/3

- B.
1

- C.
1/√2

- D.
√2

Answer: Option B

**Explanation** :

Let the radius of the third circle be ‘x’ cm.

AF = DE = 4 cm ⇒ AK = DC = (4 – x) cm

In ∆ADC, AC^{2} = AD^{2} + DC^{2}.

(4 + x)^{2} = 4^{2} + (4 – x)^{2}

∴ x = 1.

Hence, option (a).

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**8. CAT 2019 QA Slot 2 | Algebra - Number Theory**

What is the largest positive integer such that $\frac{({n}^{2}+7n+12)}{({n}^{2}-n-12)}$ is also positive integer?

- A.
6

- B.
8

- C.
16

- D.
12

Answer: Option D

**Explanation** :

Given : $\frac{({n}^{2}+7n+12)}{({n}^{2}-n-12)}$

$\frac{({n}^{2}+7n+12)}{({n}^{2}-n-12)}$ = $\frac{({n}^{2}+4n+3n+12)}{({n}^{2}-4n+3n-12)}$ = $\frac{(n+4)(n+3)}{(n-4)(n+3)}$

Now, n cannot be equal to -3, since denominator cannot be 0

⇒ $\frac{({n}^{2}+7n+12)}{({n}^{2}-n-12)}$ = $\frac{(n+4)}{(n-4)}$ = $\frac{(n-4+8)}{(n-4)}$ = $1+\frac{8}{(n-4)}$

For $\frac{({n}^{2}+7n+12)}{({n}^{2}-n-12)}$ to be an integer, 8/(n-4) should also be an integer.

Largest value of n – 4 = 8

∴ largest possible value of n = 12.

Hence, option (d).

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**9. CAT 2019 QA Slot 2 | Arithmetic - Percentage**

In 2010, a library contained a total of 11500 books in two categories - fiction and non-fiction. In 2015, the library contained a total of 12760 books in these two categories. During this period, there was 10% increase in the fiction category while there was 12% increase in the non-fiction category. How many fiction books were in the library in 2015?

- A.
6600

- B.
6160

- C.
6000

- D.
5500

Answer: Option A

**Explanation** :

In 2010 : Total books = 11500

Let the number of fiction books be f, non-fiction books will be (11500 - f)

In 2015 : Total books = 12760

Increase in total number of books = 12760 - 11500 = 1260

Fiction books increase by 10%, non-fiction books increase by 12%.

Hence, f × $\frac{10}{100}$ + (11500 - f) × $\frac{12}{100}$ = 1260

10f + 12 × 11500 – 12f = 126000

x = 6000 books

So, Fiction books in 2015, 6000 + 600 = 6600 books

Hence, option (a).

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**10. CAT 2019 QA Slot 2 | Algebra - Functions & Graphs**

Let f be a function such that f(mn) = f(m) × f(n) for every positive integers m and n. If f(1), f(2) and f(3) are positive integers, f(1) < f(2), and f(24) = 54, then f(18) equals

Answer: 12

**Explanation** :

f(mn) = f(m) × f(n)

f(1), f(2) & f(3) are positive integers.

Now we know, f(24) = 54

So, f(2) × f(3) × f(4) = f(2) × f(3) × f(2)^{2} = 54

f(3) × f(2)^{3} = 54

Also, we know, 54 = 2 × 3^{3}

Therefore, f(2) = 3, f(3) = 2

Now, f(18) = f(3)^{2} × f(2)

f(18) = 2^{2} × 3 = 12

Hence, 12.

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**11. CAT 2019 QA Slot 2 | Geometry - Quadrilaterals & Polygons**

Let A and B be two regular polygons having a and b sides, respectively. If b = 2a and each interior angle of B is 3/2 times each interior angle of A, then each interior angle, in degrees, of a regular polygon with a + b sides is

Answer: 150

**Explanation** :

Sum of the interior angles of a n-sided Polygon = (n - 2) × 180

Measure of each interior angle = $\frac{(n-2)}{n}$× 180°

So, Interior angle of the polygon with side a = $\frac{(a-2)}{a}$ × 180°

Interior angle of the polygon with side b = $\frac{(2a-2)}{2a}$ × 180°

It is given that interior angle of side b is 3/2 times to that of the polygon with side A

⇒ $\frac{(2a-2)}{2a}$ × 180° = $\frac{3}{2}\left[\frac{(a-2)}{a}\times 180\xb0\right]$

⇒ (2a - 2) × 180 = (3a - 6) × 180

⇒ 2a - 2 = 3a - 6

⇒ a = 4. So, b = 8

Now, (a + b) sides = 8 + 4 = 12 sides

Interior angle =$\frac{12-2}{12}$× 180° = 150°

Hence, 150.

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**12. CAT 2019 QA Slot 2 | Arithmetic - Time, Speed & Distance**

A cyclist leaves A at 10 am and reaches B at 11 am. Starting from 10:01 am, every minute a motorcycle leaves A and moves towards B. Forty-five such motorcycles reach B by 11 am. All motorcycles have the same speed. If the cyclist had doubled his speed, how many motorcycles would have reached B by the time the cyclist reached B?

- A.
22

- B.
20

- C.
15

- D.
23

Answer: Option C

**Explanation** :

It is given that the cyclist starts at 10:00 am from A and reaches B at 11:00 am

Now, Motorcyclists start every minute from 10:01 am, and 45 such motorcyclists reach B before 11:00 am

If they leave one by one every minute, the 45^{th} motorcyclist would have left by 10:45 am to reach B at 11:00 am.

Thus, time taken by one motorcyclist to reach B from A = 15 minutes.

Now, the cyclist doubles his speed. This means, he reaches B at 10:30 am

So, the last motorcyclist should have left A by 10:15 am

Thus, 15 motorcyclists would have reached B by the time the cyclist reaches B

Hence, option (c).

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**13. CAT 2019 QA Slot 2 | Algebra - Quadratic Equations | Algebra - Logarithms**

Let A be a real number. Then the roots of the equation x^{2} - 4x - log_{2}A = 0 are real and distinct if and only if

- A.
A < 1/16

- B.
A < 1/8

- C.
A > 1/8

- D.
A > 1/16

Answer: Option D

**Explanation** :

For roots of a quadratic equation to be real and distinct, Discriminant > 0

So, for x^{2} − 4x – log_{2}A = 0,

D = (-4)^{2} - (4 × 1 × (-log_{2}A)) > 0

⇒ 16 + 4 × log_{2}A > 0

⇒ log_{2}A > -4

⇒ A > 2^{-4}

⇒ A > 1/16

Hence, option (d).

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**14. CAT 2019 QA Slot 2 | Arithmetic - Time, Speed & Distance**

John jogs on track A at 6 kmph and Mary jogs on track B at 7.5 kmph. The total length of tracks A and B is 325 metres. While John makes 9 rounds of track A, Mary makes 5 rounds of track B. In how many seconds will Mary make one round of track A?

Answer: 48

**Explanation** :

Let the track length for John be 'a' and for Mary be 'b'

So, Distance travelled by John = 9a

Distance travelled by Mary = 5b

Now, Time taken by John = 9a/6 hours

Time taken by Mary = 5b/7.5 hours

We know that Time taken by John = Time taken by Mary

⇒ 9a/6 = 5b/7.5

⇒ a = 4b/9

Total track length = 325 meters

So, 4b/9 + b = 325 meters

⇒ 13b/9 = 325 meters

⇒ b = 225 meters

⇒ a = 100 meters

Mary jogs at 7.5 Kmph = 7.5 × 5/18

So, time taken = $\frac{100}{(7.5\times {\displaystyle \frac{5}{18}})}$ = 48 seconds.

Hence, 48.

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**15. CAT 2019 QA Slot 2 | Arithmetic - Time & Work**

Anil alone can do a job in 20 days while Sunil alone can do it in 40 days. Anil starts the job, and after 3 days, Sunil joins him. Again, after a few more days, Bimal joins them and they together finish the job. If Bimal has done 10% of the job, then in how many days was the job done?

- A.
13

- B.
12

- C.
15

- D.
14

Answer: Option A

**Explanation** :

Anil in one day can do 1/20^{th} of the work

Sunil in one day can do 1/40^{th} of the work

Anil starts the job and Sunil joins him after three days.

So, Anil would have done 3/20^{th} of the work by the time Sunil joins

After Sunil joins, they both would be doing 3/40^{th} of work everyday

Now, it is known that Bimal joins them after some days and finishes 10% of the work (i.e. 1/10^{th }of the work).

Now, Anil alone had done 3/20^{th} of the work in first 3 days and Bimal completes 1/10^{th} of the work

So, in total they would have done 3/20 + 1/10 = 1/4^{th} of the work.

Remaining work = 3/4^{th}, which would be done by Anil and Sunil together.

Anil and Sunil together comple 1/20^{th} + 1/40^{th} = 3/40^{th }work in 1 day.

Hence, time taken by them to comple 3/4^{th} of the work = 10 days.

Combining everything,

Total number of days = 3 + 10 = 13 days.

Hence, option (a).

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**16. CAT 2019 QA Slot 2 | Arithmetic - Ratio, Proportion & Variation**

In an examination, Rama's score was one-twelfth of the sum of the scores of Mohan and Anjali. After a review, the score of each of them increased by 6. The revised scores of Anjali, Mohan, and Rama were in the ratio 11 : 10 : 3. Then Anjali's score exceeded Rama's score by

- A.
26

- B.
32

- C.
24

- D.
35

Answer: Option B

**Explanation** :

It is given that the scores of Anjali, Mohan and Rama after review were in the ratio 11 : 10 : 3

So, let their values be 11x, 10x and 3x respectively.

It is known that their score increased by 6 after review.

So, scores before review = 11x - 6, 10x - 6 and 3x - 6 respectively

Now, from the data given : (11x - 6 + 10x - 6) × 1/12 = 3x - 6

⇒ 21x - 12 = 36x - 72

⇒ 60 = 15x

⇒ x = 4

So, marks after revision are 44, 40 and 12 respectively.

Therefore, Anjali's score exceeded Rama's by 44 - 12 = 32 marks

Hence, option (b).

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**17. CAT 2019 QA Slot 2 | Arithmetic - Percentage**

In an examination, the score of A was 10% less than that of B, the score of B was 25% more than that of C, and the score of C was 20% less than that of D. If A scored 72, then the score of D was

Answer: 80

**Explanation** :

Given: A scored 72

A's score was 10% less than B

So, Score of B = 72/0.9 = 80

We know that B was 25% more than C

So, C × 5/4 = 80

⇒ C = 64

Now, we know that C scored 20% less than D.

So, C = 4/5 × D

⇒ 64 = 4/5 × D

⇒ D = 80 marks

Hence, 80.

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**18. CAT 2019 QA Slot 2 | Geometry - Mensuration**

The base of a regular pyramid is a square and each of the other four sides is an equilateral triangle, length of each side being 20 cm. The vertical height of the pyramid, in cm, is

- A.
10√2

- B.
8√3

- C.
12

- D.
5√5

Answer: Option A

**Explanation** :

In the above image, the side length of the square is equal to the side length of the equilateral triangle.

So, AB = BC = OC = 20 cm.

∆AOM is a right angled triangle with ∠AMO = 90°, AO = 20 cm. OM is the height of the pyramid.

AC is the diagonal of the square base, so AM = AC/2 = (20√2)/2 = 10√2.

So using pythagoras theorem; OM^{2} = AO^{2} − AM^{2} = 20^{2} − (10√2)^{2} = 400 − 200 = 200.

∴ OM = 10√2.

Hence, option (b)

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**19. CAT 2019 QA Slot 2 | Algebra - Logarithms | Algebra - Inequalities & Modulus**

If x is a real number, then $\sqrt{{\mathrm{log}}_{e}\left(\frac{4x-{x}^{2}}{3}\right)}$ is a real number if and only if

- A.
-3 ≤ x ≤ 3

- B.
1 ≤ x ≤ 2

- C.
1 ≤ x ≤ 3

- D.
-1 ≤ x ≤ 3

Answer: Option C

**Explanation** :

It is given that, $\sqrt{{\mathrm{log}}_{e}\left(\frac{4x-{x}^{2}}{3}\right)}$ is a real number

Therefore, ${\mathrm{log}}_{e}\left(\frac{4x-{x}^{2}}{3}\right)$ ≥ 0

⇒ $\frac{4x-{x}^{2}}{3}$ ≥ 1

⇒ 4x – x^{2} ≥ 3

⇒ x^{2} - 4x + 3 ≤ 0

⇒ (x - 1)(x - 3) ≤ 0

⇒ x ∈ [1, 3]

Hence, option (c).

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**20. CAT 2019 QA Slot 2 | Geometry - Triangles**

Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then the maximum possible length of AP, in cm, is

- A.
10

- B.
8√2

- C.
6√2

- D.
5

Answer: Option A

**Explanation** :

∠BAC = 90°, so BC can be considered diameter of a circle with center anywhere on BC.

AP is maximum when it is the perpendicular bisector from the vertex to the hypotenuse. So, BP = PC = 20/2 = 10. This implies that P is the center of the circle.

∴ AP = BP = PC = 10 cm.

Hence, option (b).

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**21. CAT 2019 QA Slot 2 | Arithmetic - Time, Speed & Distance**

Two ants A and B start from a point P on a circle at the same time, with A moving clock-wise and B moving anti-clockwise. They meet for the first time at 10:00 am when A has covered 60% of the track. If A returns to P at 10:12 am, then B returns to P at

- A.
10 : 27 am

- B.
10 : 25 am

- C.
10 : 45 am

- D.
10 : 18 am

Answer: Option A

**Explanation** :

We are told that, by the time A and B meet for the first time, A covers 60% of the distance, while B covers 40% of the distance.

So, the speeds of A and B are in the ratio 60:40 or 3:2

Hence, the time they take to cover a particular distance will be in the ratio 2:3

We know that A covers 60% of the distance at 10:00 AM and covers 100% of the distance at 10:12 AM.

That means A takes 12 minutes to cover 40% of the track. So to cover the entire track he must have taken 12 + 12 + 6 = 30 minutes. (because 40% + 40% + 20% = 100%)

Since the time taken by A and B to complete the track are in the ratio 2:3, the time taken by B to complete the track will be 45 minutes.

At 10:00 AM, B has covered 40% of the track. If we can find out what time does B take to complete the remaining 60% of the track, we can find the finish time of B.

Time required to complete 60% of the track = 60% of 45 = 27 minutes.

Hence, B complete one single round at 10:27 AM.

Hence, option (a).

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**22. CAT 2019 QA Slot 2 | Algebra - Number Theory**

How many pairs (m,n) of positive integers satisfy the equation m^{2} + 105 = n^{2}?

Answer: 4

**Explanation** :

We know that m^{2} + 105 = n^{2}

105 = n^{2} - m^{2}

105 = (n + m) × (n - m)

105 can be written as 3 × 5 × 7, which can be written as product of two numbers in 4 ways i.e. (105 × 1), (35 × 3), (21 × 5), (15 × 7)

Solving each of these 4 cases will give corresponding values of m and n.

Hence, 4.

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**23. CAT 2019 QA Slot 2 | Arithmetic - Percentage**

The salaries of Ramesh, Ganesh and Rajesh were in the ratio 6 : 5 : 7 in 2010, and in the ratio 3 : 4 : 3 in 2015. If Ramesh's salary increased by 25% during 2010-2015, then the percentage increase in Rajesh's salary during this period is closest to

- A.
7

- B.
8

- C.
9

- D.
10

Answer: Option A

**Explanation** :

The salaries of Ramesh, Ganesh and Rajesh were in the ratio 6 : 5 : 7 in 2010

In 2010, let salary of Ramesh = 6x, Ganesh = 5x & Rajesh = 7x

In 2015, their salaries are in the ratio 3 : 4 : 3 respectively.

In 2015, let salary of Ramesh = 3y, Ganesh = 4y & Rajesh = 3y

Now, Ramesh’s salary increased by 25%

Hence, 3y = 5/4 × 6x

⇒ y = 2.5x

∴ Rajesh’s salary in 2015 = 3y = 3 × 2.5x = 7.5x

Percentage increase in Rajesh's salary during 2010 - 2015 = $\frac{7.5-7}{7}$ × 100% ≈ 7.

Hence, option (a).

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**24. CAT 2019 QA Slot 2 | Geometry - Mensuration**

A man makes complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius 3 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is

- A.
1044(4 + π)

- B.
8464π

- C.
928π

- D.
1026(1 + π)

Answer: Option D

**Explanation** :

As the total number of cylinders to be kept at a minimum, the volume of each cylinder must be maximum possible.

HCF(405, 783, 351) = 27 ⇒ Volume of each cylinder = 27 cc.

∴ πr^{2}h = 27 (where r and h represent the radius and height of the cylinder)

Given that r = 3, so πh = 3.

Number of cylinders of;

Iron = 405/27 = 15.

Aluminium = 783/27 = 29.

Copper = 351/27 = 13.

∴ Total number of cylinders = 15 + 29 + 13 = 57.

Total surface area of all the cylinders = 57 × [2πr^{2} + 2πrh]

Put r = 3 and πh = 3 to get;

Required surface area = 1026(1 + π) sq. cm.

Hence, option (d).

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**25. CAT 2019 QA Slot 2 | Algebra - Quadratic Equations**

The quadratic equation x^{2} + bx + c = 0 has two roots 4a and 3a, where a is an integer. Which of the following is a possible value of b^{2} + c?

- A.
3721

- B.
549

- C.
361

- D.
427

Answer: Option B

**Explanation** :

Given quadratic equation is x^{2} + bx + c = 0

Sum of roots = -b

⇒ -b = 4a + 3a

⇒ -b = 7a

Product of the roots = c

⇒ c = 4a × 3a

⇒ c = 12a^{2}

Now, b^{2} = 49a^{2} and c =12a^{2}

Hence, b^{2} + c = 49a^{2} + 12a^{2}

⇒ b^{2} + c = 61a^{2}

Final answer must be a multiple of 61 & the multiple should be a perfect square.

Going through the options,

3721 = 61 × 61 (multiple is not a perfect square)

361 (not a multiple of 61)

427 = 61 × 7 (multiple is not a perfect square)

549 = 61 × 9 = 61 × 3^{2}

Thus, possible value of b^{2} + c = 549.

Hence, option (b).

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**26. CAT 2019 QA Slot 2 | Algebra - Simple Equations**

In a six-digit number, the sixth, that is, the rightmost, digit is the sum of the first three digits, the fifth digit is the sum of first two digits, the third digit is equal to the first digit, the second digit is twice the first digit and the fourth digit is the sum of fifth and sixth digits. Then, the largest possible value of the fourth digit is

Answer: 7

**Explanation** :

Let the first digit be = a

Second digit = Twice of first digit = 2a

Third digit = First digit = a

Fifth digit = Sum of first two digits = a + 2a = 3a

Sixth digit = Sum of first three digits = a + 2a + a = 4a

Fourth digit = Sum of fifth and sixth digit = 3a + 4a = 7a

Largest possible value of fourth digit = 7 (it has to be a single digit integer)

Hence, 7.

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**27. CAT 2019 QA Slot 2 | Arithmetic - Profit & Loss**

Mukesh purchased 10 bicycles in 2017, all at the same price. He sold six of these at a profit of 25% and the remaining four at a loss of 25%. If he made a total profit of Rs. 2000, then his purchase price of a bicycle, in Rupees, was

- A.
2000

- B.
6000

- C.
8000

- D.
4000

Answer: Option D

**Explanation** :

Let the cost price of one bicycle = Rs. x

Total cost price = Rs. 10x

He made a total profit of 25% on 6 cycles and 25% loss on 4 cycles and made a profit of Rs. 2000

So, 2000 = 6 × x/4 - 4 × x/4

2000 = x/2

x = Rs. 4000

Hence, option (d).

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**28. CAT 2019 QA Slot 2 | Algebra - Progressions**

The number of common terms in the two sequences: 15, 19, 23, 27, ...... , 415 and 14, 19, 24, 29, ...... , 464 is

- A.
20

- B.
18

- C.
21

- D.
19

Answer: Option A

**Explanation** :

First series - 15, 19, 23, 27, ......, 415 ⇒ Common difference = 4

Second series - 14, 19, 24, 29, ......, 464 ⇒ Common difference = 5

Common terms in both sequences = 19, 39, 59, ...., (Common difference = LCM (4,5) = 20)

Now :

19, 39, 59, ...., = (20 - 1), (40 -1), (60 -1), ...., (400-1) (There is no room for 419, as the first series ends at 415)

399 = 400 - 1 = 20 × 20 - 1

Hence, the number of common terms in the two sequences = 20

Hence, option (a).

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**29. CAT 2019 QA Slot 2 | Algebra - Progressions**

If (2n+1) + (2n+3) + (2n+5) + ... + (2n+47) = 5280 , then what is the value of 1 + 2 + 3 + ... + n?

Answer: 4851

**Explanation** :

Given: (2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = 5280

⇒ (2n + 2n +.... + 2n) + (1 +3 + 5 +.... + 47) = 5280

Odd numbers from 1 to 47 are added in the above series.

n^{th} odd natural number = 2n - 1

∴ Number of terms from 1 to 47 = (47 + 1)/2 = 24 terms

∴ Given equation becomes 2n × 24 + (1 + 3 + 5 + … + 47) = 5280

Sum of first n odd natural numbers = n^{2}.

⇒ 2n × 24 + 24^{2} = 5280

⇒ 2n × 24 = 5280 - 24^{2}

⇒ 2n = 220 - 24

⇒ n = 98

Hence, 1 + 2 + 3 + ... + 98 = $\frac{98\times 99}{2}$ = 4851.

Hence, 4851.

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**30. CAT 2019 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement**

The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively. Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A. The strength, in percentage, of the resulting solution in vessel A is

- A.
15

- B.
12

- C.
13

- D.
14

Answer: Option D

**Explanation** :

Vessels A, B and C contains salt solution of strengths 10%, 22% and 32% respectively

It is also given that the amount of salt solution = 500 ml

So, Vessels A, B and C contains salt of 50 grams, 110 grams and 160 grams respectively

100 ml of Solution is transferred from A to B:

A would have 400 ml, B would have 600 ml of solution

Amount salt from A which is transferred to B = (Initial salt amount)/5 = 10 grams

So, Total salt in B = 110 + 10 = 120 grams (After first transfer)

Total salt in A = 40 grams (After first transfer)

Now, 100 ml from Vessel B is transferred to Vessel C

So similarly, 1/6th of salt would transfer from B to C

Total Salt in B = 120 - 20 = 100 grams (After second transfer)

Total Salt in C = 160 + 20 = 180 grams (After second transfer)

Now, 100 ml from Vessel C is transferred to Vessel A

So similarly, 1/6^{th} of salt would transfer from C to A

Total Salt in C = 160 - 30 = 130 grams (After third transfer)

Total Salt in A = 40 + 30 = 70 grams (After third transfer)

So, Vessel A contains 70 grams Salt in 500 ml solution

Strength of Salt Solution in Vessel A = 14%

Hence, option (d).

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**31. CAT 2019 QA Slot 2 | Algebra - Surds & Indices**

If 5^{x} - 3^{y} = 13438 and 5^{x-1} + 3^{y+1} = 9686, then x + y equals

Answer: 13

**Explanation** :

It is given that 5^{x} - 3^{y} = 13438 and 5^{x-1} + 3^{y+1} = 9686

Let 5^{x-1} = k and 3^{y} = m

So, 5k - m = 13538 …(1)

k + 3m = 9686 …(2)

Multiply (1) with 3,

15k - 3m = 40314 ----(3)

Adding (2) and (3) we get,

16k = 50000

k = 625 × 5

k = 5^{5}

We know, 5^{x-1} = k

So, x - 1 = 5

x = 6

We know, k + 3m = 9686

3125 + 3m = 9686

3m = 6561

m = 2187

m = 3^{7} = 3^{y}

y = 7

x + y = 6 + 7 = 13

Hence, 13.

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**32. CAT 2019 QA Slot 2 | Arithmetic - Simple & Compound Interest**

Amal invests Rs 12000 at 8% interest, compounded annually, and Rs 10000 at 6% interest, compounded semi-annually, both investments being for one year. Bimal invests his money at 7.5% simple interest for one year. If Amal and Bimal get the same amount of interest, then the amount, in Rupees, invested by Bimal is

Answer: 20920

**Explanation** :

The amount on the first investment of Anmol = 12,000 × (1.08) = 12,960

So the Interest on this investment is 12,960 - 12,000 = 960.

The amount on the second investment of Anmol = 10,000 × (1.03)^{2} = 10,609

So the Interest on this investment is 10,609 - 10,000 = 609.

So the total interest on these returns = 960 + 609 = 1,569.

Bimal has to get this as Simple Interest by investing X rupees at 7.5%

That means, X × 0.075 = 1,569

X = 20,920

So, Bimal has to invest 20,920 rupees.

Hence, 20920.

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**33. CAT 2019 QA Slot 2 | Arithmetic - Profit & Loss**

A shopkeeper sells two tables, each procured at cost price p, to Amal and Asim at a profit of 20% and at a loss of 20%, respectively. Amal sells his table to Bimal at a profit of 30%, while Asim sells his table to Barun at a loss of 30%. If the amounts paid by Bimal and Barun are x and y, respectively, then (x - y) / p equals

- A.
1

- B.
1.2

- C.
0.7

- D.
0.5

Answer: Option A

**Explanation** :

The Shopkeeper procures the table at price 'p'

He gains 20% on the transaction with Amal

So, Amal buys the table at '1.2p'

Amal sells athe table at 30% profit,

So, the Selling Price of Amal = 1.3 × 1.2p = 1.56p

⇒ x = 1.56p

The Shopkeeper loses 20% on the transaction with Asim

So, Asim buys the table at '0.8p'

Asim sells the table at 30% loss,

So, the Selling Price of Asim = 0.7 * 0.8p = 0.56p

⇒ y = 0.56p

(x - y)/p = (1.56p - 0.56p)/p = 1.

Hence, option (a).

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**34. CAT 2019 QA Slot 2 | Arithmetic - Time & Work**

John gets Rs 57 per hour of regular work and Rs 114 per hour of overtime work. He works altogether 172 hours and his income from overtime hours is 15% of his income from regular hours. Then, for how many hours did he work overtime?

Answer: 12

**Explanation** :

If John works the same number of regular and over-time hours say 'p'

The income would be 57p and 114p

Let's say that he works 'x' hours regular and 'y' hours overtime...

So, the income would be 57x and 114y

we are told that 114y is 15% of 57x

114y = 0.15 × 57x

y = 0.075x

we also know that x + y = 172

therefore, x + 0.075x = 1.075x = 172

x = 160

y = 172 - 160 = 12

Therefore, the number of hours he worked overtime is 12 hours.

Hence, 12.

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