# XAT 2020 QADI | Previous Year XAT Paper

Previous year paper questions for XAT 2020 QA

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**1. XAT 2020 QADI | Geometry - Triangles**

Two lighthouses, located at points A and B on the earth, are 60 feet and 40 feet tall respectively. Each lighthouse is perfectly vertical and the land connecting A and B is perfectly flat. The topmost point of the lighthouse at A is A’ and of the lighthouse at B is B’. Draw line segments A’B and B’A, and let them intersect at point C’. Drop a perpendicular from C’ to touch the earth at point C. What is the length of CC’ in feet?

- A.
20

- B.
30

- C.
24

- D.
The distance between A and B is also needed to solve this

- E.
25

Answer: Option C

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**Explanation** :

∆BAA’ ~ ∆BCC’

∴ $\frac{BC}{BA}=\frac{CC\text{'}}{AA\text{'}}$ …(1)

Also, ∆ABB’ ~ ∆ACC’

∴ $\frac{AC}{AB}=\frac{CC\text{'}}{BB\text{'}}$ …(2)

Adding (1) and (2) we get

$\frac{AC+BC}{AB}=\frac{CC\text{'}}{AA\text{'}}+\frac{CC\text{'}}{BB\text{'}}$

⇒ 1 = $\frac{CC\text{'}}{AA\text{'}}+\frac{CC\text{'}}{BB\text{'}}$

⇒ $\frac{1}{CC\text{'}}=\frac{1}{AA\text{'}}+\frac{1}{BB\text{'}}$

⇒ $\frac{1}{CC\text{'}}=\frac{1}{60\text{'}}+\frac{1}{40}$ = $\frac{1}{24}$

∴ CC’ = 24.

Hence, option (c).

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**2. XAT 2020 QADI | Miscellaneous**

If A ʘ B = (A + B) × B, then what is (5 ʘ 2) ʘ 5 ?

- A.
95

- B.
275

- C.
125

- D.
74

- E.
200

Answer: Option A

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**Explanation** :

Given, A ʘ B = (A + B) × B

∴ (5 ʘ 2) = (5 + 2) × 2 = 14

Now, (5 ʘ 2) ʘ 5 = 14 ʘ 5

⇒ 14 ʘ 5 = (14 + 5) × 5 = 19 × 5 = 95

Hence, option (a).

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**3. XAT 2020 QADI | Arithmetic - Percentage**

Nalini has received a total of 600 WhatsApp messages from four friends Anita, Bina, Chaitra and Divya. Bina and Divya have respectively sent 30% and 20% of these messages, while Anita has sent an equal number of messages as Chaitra. Moreover, Nalini finds that of Anita’s, Bina’s, Chaitra’s and Divya’s messages, 60%, 40%, 80% and 50% respectively are jokes. What percentage of the jokes, received by Nalini, have been sent neither by Divya nor by Bina?

- A.
65.12

- B.
61.4

- C.
57

- D.
38.6

- E.
34.88

Answer: Option B

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**Explanation** :

Number of messages send by Bina = 30% of 600 = 180

Number of messages send by Divya = 20% of 600 = 120

Remaining messages (300) are sent by Anita and Chitra equally i.e., Anita and Chitra sent 150 messages each.

Jokes sent by Anita = 60% of 150 = 90

Jokes sent by Bina = 40% of 180 = 72

Jokes sent by Chitra = 80% of 150 = 120

Jokes sent by Divya = 50% of 120 = 60

Total Jokes sent = 90 + 72 + 120 + 60 = 342

∴ Percentage of jokes that were neither sent by Bina or Divya = 210/342 × 100 = 61.4%

Hence, option (b).

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**4. XAT 2020 QADI | Algebra - Progressions**

A man is laying stones, from start to end, along the two sides of a 200-meter-walkway. The stones are to be laid 5 meters apart from each other. When he begins, all the stones are present at the start of the walkway. He places the first stone on each side at the walkway’s start. For all the other stones, the man lays the stones first along one of the walkway’s sides, then along the other side in an exactly similar fashion. However, he can carry only one stone at a time. To lay each stone, the man walks to the spot, lays the stone, and then walks back to pick another. After laying all the stones, the man walks back to the start, which marks the end of his work. What is the total distance that the man walks in executing this work? Assume that the width of the walkway is negligible.

- A.
16,200 meters

- B.
8,200 meters

- C.
8,050 meters

- D.
16,400 meters

- E.
4,100 meters

Answer: Option D

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**Explanation** :

On one side:

To place 1^{st} rock, he had to travel 10 m.

For 2nd rock he had to travel 20 m…,

Similarly, till last rock he had to travel 400 m.

Total sum would be 10 + 20 + 30 + ... + 400 = 40/2 × 410 = 8200 m.

Similarly, on the other side he will travel 8200 m.

Total 8200 + 8200 = 16400.

Hence, option (d).

Workspace:

**5. XAT 2020 QADI | Arithmetic - Time, Speed & Distance**

Two friends, Ram and Shyam, start at the same point, at the same time. Ram travels straight north at a speed of 10km/hr, while Shyam travels straight east at twice the speed of Ram. After 15 minutes, Shyam messages Ram that he is just passing by a large telephone tower and after another 15 minutes Ram messages Shyam that he is just passing by an old banyan tree. After some more time has elapsed, Ram and Shyam stop. They stop at the same point of time. If the straight-line distance between Ram and Shyam now is 50 km, how far is Shyam from the banyan tree (in km)? (Assume that Ram and Shyam travel on a flat surface.)

- A.
20√5 + 5

- B.
20√5 - 5

- C.
5√21

- D.
45

- E.
115/3

Answer: Option D

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**Explanation** :

Distance travelled by Shyam till Telephone tower = 20 × 1/4 = 5 kms

Distance travelled by Ram till Banyan tree = 10 × ½ = 5 kms

Let the two of them stop after t hours from starting.

∴ (50)^{2} = (10t)^{2} + (20t)^{2}

⇒ 2500 = 100t^{2} + 400t^{2}

⇒ 5 = t^{2}

Now, (BS)^{2} = 52 + (20t)^{2}

⇒ (BS)^{2} = 25 + 400t^{2}

⇒ (BS)^{2} = 25 + 400 × 5 = 2025

⇒ BS = √2025 = 45

Hence, option (d).

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**6. XAT 2020 QADI | Geometry - Mensuration**

A rectangular swimming pool is 50 meters long and 25 meters wide. Its depth is always the same along its width but linearly increases along its length from 1 meter at one end to 4 meters at the other end. How much water (in cubic meters) is needed to completely fill the pool?

- A.
3750

- B.
2500

- C.
1250

- D.
1874

- E.
3125

Answer: Option E

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**Explanation** :

Volume of the tank = Volume of the cuboid + Volume of the triangular prism

The volume of cuboid will be 50 × 25 × 1 = 1250

The volume of triangular prism = Area of triangle × width = 1/2 × 3 × 50 × 25 = 1875

∴ Total volume = 1875 + 1250 = 3125.

Hence, option (e).

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**7. XAT 2020 QADI | Geometry - Circles | Geometry - Quadrilaterals & Polygons**

A rectangular field is 40 meters long and 30 meters wide. Draw diagonals on this field and then draw circles of radius 1.25 meters, with centers only on the diagonals. Each circle must fall completely within the field. Any two circles can touch each other but should not overlap.

What is the maximum number of such circles that can be drawn in the field?

- A.
39

- B.
40

- C.
37

- D.
36

- E.
38

Answer: Option C

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**Explanation** :

Length of the diagonal = √(30^{2}+40^{2} ) = 50 m.

Each circle on the end of the diagonal will touch sides of the rectangular field

Using Pythagoras' theorem, the distance between the vertex of the rectangle and centre of the first circle drawn on the diagonal (OC) = 1.25√2

Distance between the vertex of the rectangle and circumference of the first circle drawn on the diagonal (OD) = 1.25√2 - 1.25 = 0.51 meters

Space that cannot be used to draw circle otherwise they will go outside rectangle on every diagonal = 0.51 × 2 = 1.02 meters

Space that can be used to draw circles = length of diagonal - unused space = 50 - 1.02 = 48.98 meters

On every diagonal, maximum number of such circles = usable length/diameter of each circle = 48.98/2.5 = 19.6

∴ Maximum 19 circles can be drawn on a diagonal.

Now, on every diagonal, one circle will be at the centre (intersection of diagonals) and 9 circles will be on each half of the diagonal

⇒ The circle in centre will be common for both diagonals.

∴ Total circles = 19 + 19 – 1 = 37.

Hence, option (c).

Workspace:

**8. XAT 2020 QADI | Algebra - Number Theory**

What is the remainder if 19^{20} – 20^{19} is divided by 7?

- A.
5

- B.
1

- C.
6

- D.
0

- E.
3

Answer: Option A

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**Explanation** :

$R\left[\frac{{19}^{20}}{7}\right]$ = $R\left[\frac{{5}^{20}}{7}\right]$ = $R\left[\frac{{25}^{10}}{7}\right]$ = $R\left[\frac{{4}^{10}}{7}\right]$ = $R\left[\frac{{16}^{5}}{7}\right]$ = $R\left[\frac{{2}^{5}}{7}\right]$ = $R\left[\frac{32}{7}\right]$ = 4

$R\left[\frac{{20}^{19}}{7}\right]$ = $R\left[\frac{{(-1)}^{19}}{7}\right]$ = -1

$R\left[\frac{{19}^{20}-{20}^{19}}{7}\right]$ = 4 - (-1) = 5

Hence, option (a).

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**9. XAT 2020 QADI | Geometry - Mensuration**

Six drums are used to store water. Five drums are of equal capacity, while the sixth drum has double the capacity of each of these five drums. On one morning, three drums are found half full, two are found two-thirds full and one is found completely full. It is attempted to transfer all the water to the smaller drums. How many smaller drums are adequate to store the water?

- A.
Four but not three

- B.
Three or four, depending on which drum had how much water initially

- C.
Five but not four

- D.
Five may be inadequate, depending on which drum had how much water initially

- E.
Three but not two

Answer: Option C

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**Explanation** :

Let the volume of 5 smaller drums be V

∴ Volume of the bigger drum = 2V

On one morning, three drums are found half full, two are found two-thirds full and one is found completely full.

Case 1: We will get the maximum water when the bigger drum is completely full

∴ Amount of water = 3 × V/2 + 2 × 2V/3 + 1 × 2V = 29V/6 = 4.83V

∴ Number of smaller drums required to store 29V/6 water = 5

Case 2: We will get the minimum water when the bigger drum is only half full

∴ Amount of water = 1 × 2V/2 + 2 × V/2 + 2 × 2V/3 + 1 × V = 13V/3 = 4.133V

∴ Number of smaller drums required to store 13V/3 water = 5

∴ In both cases we need at least 5 smaller drums to store the water.

Hence, option (c).

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**10. XAT 2020 QADI | Algebra - Number Theory**

When expressed in a decimal form, which of the following numbers will be nonterminating as well as non-repeating?

- A.
$\left(\frac{\pi}{2}\right)\left[\frac{1}{\pi}+1\right]-\frac{\pi}{2}$

- B.
sin

^{2}1° + sin^{2}2° + … + sin^{2}89° - C.
$\frac{\sqrt[3]{729}}{3}+\frac{22}{7}$

- D.
$(4-\pi )\left[1+\frac{\pi}{4}+{\left(\frac{\pi}{4}\right)}^{2}+{\left(\frac{\pi}{4}\right)}^{3}+...infiniteterms\right]$

- E.
$\sqrt{2}\left(3\sqrt{2}-\frac{4}{\sqrt{2}}\right)+\sqrt{3}$

Answer: Option E

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**Explanation** :

We need to check options.

**Option (a)**: $\frac{\mathrm{\pi}}{2}\left[\frac{1}{\mathrm{\pi}}+1\right]-\frac{\mathrm{\pi}}{2}$

= $\frac{1}{2}+\frac{\mathrm{\pi}}{2}-\frac{\mathrm{\pi}}{2}$ = $\frac{1}{2}$ = 0.5 (terminating)

**Option (b)**: sin21° + sin22° + … + sin288° + sin289°

= sin21° + sin22° + … + sin245° + … + cos22° + cos21°

= 44 + ½ = 44.5 (terminating)

**Option (c)**: $\frac{\sqrt[3]{729}}{3}+\frac{22}{7}$

= $\frac{9}{3}+\frac{22}{7}$ = 3 + 22/7 (terminating)

**Option (d)**: $\left(4-\mathrm{\pi}\right)[1+\frac{\mathrm{\pi}}{4}+{\left(\frac{\mathrm{\pi}}{4}\right)}^{2}+{\left(\frac{\mathrm{\pi}}{4}\right)}^{3}+...inifiniteterms]$

= $\left(4-\mathrm{\pi}\right)\left[\frac{1}{1-{\displaystyle \frac{\mathrm{\pi}}{4}}}\right]$ = 4 (non-terminating)

**Option (e)**: $\sqrt{2}\left(3\sqrt{2}-\frac{4}{\sqrt{2}}\right)+\sqrt{3}$

= 6 - 4 + √3 = 2 + √3 (non-terminating and non-repeating)

Hence, option (e).

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**11. XAT 2020 QADI | Arithmetic - Time & Work**

X, Y, and Z are three software experts, who work on upgrading the software in a number of identical systems. X takes a day off after every 3 days of work, Y takes a day off after every 4 days of work and Z takes a day off after every 5 days of work.

Starting afresh after a common day off,

i) X and Y working together can complete one new upgrade job in 6 days

ii) Z and X working together can complete two new upgrade jobs in 8 days

iii) Y and Z working together can complete three new upgrade jobs in 12 days

If X, Y and Z together start afresh on a new upgrade job (after a common day off), exactly how many days will be required to complete this job?

- A.
2 days

- B.
3.5 days

- C.
2.5 days

- D.
4 days

- E.
5 days

Answer: Option C

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**Explanation** :

Let efficiency per day of X, Y and Z be x, y and z respectively.

Let the amount of work required for an upgrade = T units.

(i) X and Y working together can complete one new upgrade job in 6 days. In 6 days X will work for 5 days and Y will work for 5 days.

∴ 5x + 5y = T …(1)

(ii) Z and X working together can complete two new upgrade jobs in 8 days. X will work for 6 days and Z will work for 7 days.

∴ 6x + 7z = 2T …(2)

(iii) Y and Z working together can complete three new upgrade jobs in 12 days. Y will work for 10 days and Z will work for 10 days.

∴ 10y + 10z = 3T …(3)

Solving (1) , (2) and (3) we get

x = T/10, y = T/10 and z = T/5

If X, Y and Z work together their combined efficiency = T/10 + T/10 + T/5 = 2T/5

∴ Time required by three of them together to complete 1 upgrade = T/(2T/5) = 2.5 days.

Hence, option (c).

Workspace:

**12. XAT 2020 QADI | Modern Math - Probability**

Ashok has a bag containing 40 cards, numbered with the integers from 1 to 40. No two cards are numbered with the same integer. Likewise, his sister Shilpa has another bag containing only five cards that are numbered with the integers from 1 to 5, with no integer repeating. Their mother, Latha, randomly draws one card each from Ashok’s and Shilpa’s bags and notes down their respective numbers. If Latha divides the number obtained from Ashok’s bag by the number obtained from Shilpa’s, what is the probability that the remainder will not be greater than 2?

- A.
0.91

- B.
0.87

- C.
0.94

- D.
0.73

- E.
0.8

Answer: Option B

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**Explanation** :

The number of ways of selecting one card from Ashok's bag and other from Shilpa’s bag = ^{40}C_{1} × ^{5}C_{1} = 200 ways

Favourable cases:

**Cards 1 or 2 or 3**: If card 1 or 2 or 3 is drawn from Shilpa’s bad remainder in each case will be less than or equal to 2.

∴ Favourable cases = ^{40}C_{1} × ^{3}C_{1} = 120 ways

**Card 4**: If card 4 is drawn from Shilpa’s bag we need to eliminate those cases when the remainder will be 3.

⇒ We need to eliminate those cards from Ashok’s bag which have number of the form 4x + 3.

∴ Cards of the form 4x + 3 are 3, 7, 11, …., 39 i.e., 10 cards.

∴ Favourable cases = (40 – 10) × 1 = 30 ways

**Card 5**: If card 5 is drawn from Shilpa’s bag we need to eliminate those cases when the remainder will be 3 or 4.

⇒ We need to eliminate those cards from Ashok’s bag which have number of the form 5x + 3 or 5x + 4.

∴ Cards of the form 5x + 3 are 3, 8, 13, …., 38 i.e., 8 cards.

∴ Cards of the form 5x + 4 are 4, 9, 14, …., 39 i.e., 8 cards.

∴ Favourable cases = (40 - 16) × 1 = 24 ways

∴ Total Favourable cases = 120 + 30 + 24 = 174 ways

⇒ Probability of remainder not greater than 2 = 174/200 = 0.87

Hence, option (b).

Workspace:

**13. XAT 2020 QADI | Geometry - Circles**

In the figure given below, the circle has a chord AB of length 12 cm, which makes an angle of 60° at the center of the circle, O. ABCD, as shown in the diagram, is a rectangle. OQ is the perpendicular bisector of AB, intersecting the chord AB at P, the arc AB at M and CD at Q. OM = MQ. The area of the region enclosed by the line segments AQ and QB, and the arc BMA, is closest to (in cm^{2}):

- A.
137

- B.
63

- C.
35

- D.
69

- E.
215

Answer: Option D

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**Explanation** :

In a circle a chord which subtends an angle of 60° at the center is equal to the radius of the circle.

∴ OA = OB = AB = 12

∆OAB is an equilateral triangle

OP (height of an equilateral triangle) = √3/2 × 12 = 6√3

∴ PM = OM = OP = 12 - 6√3

Area of AQBMA = Area of Triangle ABQ – Area of segment AMBP

Area of AQBMA = Area of Triangle ABQ – (Area of minor arc AMBO - Area of ∆OAB)

Now, PQ = MQ + PM = 12 + (12 - 6√3) = 24 - 6√3

Area of triangle ABQ = 1/2 × 12 × (24 - 6√3) = 6(24 - 6√3) = 81.64

Also, Area of minor arc AMB – Area of OAB

= 60/360 × π × 144 - 1/2 × 12× 6√3 = 24π - 36√3 = 13.07

∴ Area of AQBMA = 68.57 ≈ 69.

Hence, option (d).

Workspace:

**14. XAT 2020 QADI | Arithmetic - Ratio, Proportion & Variation**

A box contains 6 cricket balls, 5 tennis balls and 4 rubber balls. Of these, some balls are defective. The proportion of defective cricket balls is more than the proportion of defective tennis balls but less than the proportion of defective rubber balls. Moreover, the overall proportion of defective balls is twice the proportion of defective tennis balls. What BEST can be said about the number of defective rubber balls in the box?

- A.
It is exactly 3

- B.
It is either 3 or 4

- C.
It is exactly 2

- D.
It is either 2 or 3

- E.
It is either 0 or 1

Answer: Option A

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**Explanation** :

Let the number of defective cricket, tennis and rubber balls = c, t and r respectively.

Given, r/4 > c/6 > t/5 ...(1)

and (c + t + r)/15 = 2 × t/5

⇒ c + r = 5t

Also, maximum value of c + r can be 5 + 4 = 9

⇒ The value of t can only be 1.

⇒ c + r = 5 …(2)

The only possible value to satisfy (1) and (2) is r = 3 and c = 2.

Hence, option (a).

Workspace:

**15. XAT 2020 QADI | Geometry - Circles | Geometry - Quadrilaterals & Polygons**

Mohanlal, a prosperous farmer, has a square land of side 2 km. For the current season, he decides to have some fun. He marks two distinct points on one of the diagonals of the land. Using these points as centers, he constructs two circles. Each of these circles falls completely within the land, and touches at least two sides of the land. To his surprise, the radii of both the circles are exactly equal to 2/3 km. Mohanlal plants potatoes on the overlapping portion of these circles.

- A.
5( 𝜋 + 4)/27

- B.
2(2 𝜋 − 3 √3 )/27

- C.
4(𝜋 − 3 √3 )/27

- D.
𝟐(𝝅 − 2)/9

- E.
(𝝅 − 2)/9

Answer: Option D

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**Explanation** :

Length of the diagonal AC = 2√2

⇒ AX = ½ of diagonal = √2

Length of AC_{1} = 2/3 × √2 = 2√2/3

⇒ C_{1}X = AX – AC_{1} = √2 - 2√2/3 = √2/3

Using Pythagoras theorem:

XM = $\sqrt{{\left(\frac{2}{3}\right)}^{2}-{\left(\frac{\sqrt{2}}{3}\right)}^{2}}$ = $\frac{\sqrt{2}}{3}$

In ∆C_{1}XM,

C_{1}X = XM = √2/3 and ∠C_{1}XM = 90°

⇒ ∠XC_{1}M = 45°

And ∠LC_{1}M = 90°

∴ Area of ∆LC_{1}M = ½ × 2/3 × 2/3 = 2/9

Area of minor arc LM = Area of sector LC_{1}M - Area of ∆LC_{1}M

= $\frac{90}{360}\times \mathrm{\pi}\times {\left(\frac{2}{3}\right)}^{2}-\frac{2}{9}$ = $\frac{\mathrm{\pi}-2}{9}$

∴ Area of overlapping region = $2\left(\frac{\mathrm{\pi}-2}{9}\right)$

Hence, option (d).

Workspace:

**16. XAT 2020 QADI | Arithmetic - Time, Speed & Distance**

A hare and a tortoise run between points O and P located exactly 6 km from each other on a straight line. They start together at O, go straight to P and then return to O along the same line. They run at constant speeds of 12 km/hr and 1 km/hr respectively. Since the tortoise is slower than the hare, the hare shuttles between O and P until the tortoise goes once to P and returns to O. During the run, how many times are the hare and the tortoise separated by an exact distance of 1 km from each other?

- A.
40

- B.
24

- C.
48

- D.
42

- E.
22

Answer: Option A

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**Explanation** :

Tortoise will take 12 hours to complete 1 round. During this, hare will make 12 rounds of OP.

In the first round, both started from point O. After some time, distance between them will be 1 km.

After some more time, when hare is returning from P to O, before and after crossing tortoise, hare will be two more times 1 km apart from tortoise. So, in first round, there are three such occurrences.

In the second round, when hare starts from point O, while going and returning, there will be four occurrences when before and after crossing the tortoise, hare will be exactly 1 km apart. But the first occurrence of round 2 is already counted in round 1. So, in second round as well, there will be total 3 occurrences.

In each of the third, fourth and fifth rounds, there will be 4 such occurrences.

In the sixth round, because tortoise will be at point P, there will be only 2 cases.

Now, till round 6 there are 20 such occurrences.

And from round 7 to 12, it will exactly same but in reverse order.

∴ Total such occurrences = 20 × 2 = 40.

Hence, option (a).

Workspace:

**17. XAT 2020 QADI | Arithmetic - Ratio, Proportion & Variation**

X, Y and Z start a web-based venture together. X invests Rs. 2.5 lakhs, Y invests Rs. 3.5 lakhs, and Z invests Rs. 4 lakhs. In the first year, the venture makes a profit of Rs. 2 lakhs. A part of the profit is shared between Y and Z in the ratio of 2 : 3, and the remaining profit is divided among X, Y and Z in the ratio of their initial investments.

The amount that Z receives is four times the amount that X receives. How much amount does Y receive?

- A.
Rs. 80,200

- B.
Rs. 75,000

- C.
Rs. 93,750

- D.
Rs. 74,250

- E.
Rs. 1,02,500

Answer: Option B

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**Explanation** :

Let Rs. 5P is divided between Y and Z in the ratio of 2 : 3.

∴ Y received 2P and Z received 3P.

Remaining profit = 2,00,000 – 5P

This will be dived amongst X, Y and Z in the ratio 2.5 : 3.5 : 4 = 5 : 7 : 8.

∴ X receives 5/20 × (2,00,000 – 5P) and Z receives 8/20 × (2,00,000 – 5P)

Total amount received by X = 5/20 × (2,00,000 – 5P)

Total amount received by Z = 3P + 8/20 × (2,00,000 – 5P)

According to the question:

3P + 8/20 × (2,00,000 – 5P) = 4(5/20 × (2,00,000 – 5P))

⇒ 3P + 80,000 – 2P = 2,00,000 – 5P

⇒ 6P = 1,20,000

⇒ P = 20,000

∴ Y receives = 2P + 7/20 × (2,00,000 – 5P) = 40,000 + 7/20 × 1,00,000 = 75,000

Hence, option (b).

Workspace:

**18. XAT 2020 QADI | Algebra - Simple Equations**

A shop sells bags in three sizes: small, medium and large. A large bag costs Rs.1000, a medium bag costs Rs.200, and a small bag costs Rs.50. Three buyers, Ashish, Banti and Chintu, independently buy some numbers of these types of bags. The respective amounts spent by Ashish, Banti and Chintu are equal. Put together, the shop sells 1 large bag, 15 small bags and some medium bags to these three buyers. What is the minimum number of medium bags that the shop sells to them?

- A.
7

- B.
5

- C.
9

- D.
4

- E.
10

Answer: Option A

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**Explanation** :

Let the number of medium bags sold = m

∴ Total amount spent by all three together = 1 × 1000 + 200 × m + 15 × 50 = 1750 + 200m.

Now since one of them buys a large bag (worth Rs. 1000) that person spends at least Rs. 1000.

∴ Each of them spends at least Rs. 1000 and hence total amount spent by them together is at least Rs. 3000.

⇒ 1750 + 200m ≥ 3000

⇒ m ≥ 6.25

∴ Minimum value of m can be 7.

If 7 medium bags are sold. Total amount spent = 1750 + 200 × 7 = 3150

⇒ Each of them spent 3150/3 = 1050

This is possible when

One of them buys a large bag and a small bag = 1000 + 50 = 1050

One other buys 5 medium bags and a small bag = 1000 + 50 = 1050

Third person buys 2 medium bags and 13 small bags = 400 + 650 = 1050

∴ Minimum number of medium size bags sold = 7.

Hence, option (a).

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**19. XAT 2020 QADI | Algebra - Number Theory**

Consider the four variables A, B, C and D and a function Z of these variables, Z = 15A^{2} − 3B^{4} + C + 0.5D. It is given that A, B, C and D must be non-negative integers and that all of the following relationships must hold:

i) 2A + B ≤ 2

ii) 4A + 2B + C ≤ 12

iii) 3A + 4B + D ≤ 15

If Z needs to be maximised, then what value must D take?

- A.
15

- B.
12

- C.
0

- D.
10

- E.
5

Answer: Option B

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**Explanation** :

To maximise Z, we need to maximise A, C and D and minimise B.

Given, 2A + B ≤ 2

A can be either 1 or 0.

**Case 1**: If A = 1, minimum value of B = 0

From (ii): Maximum value of C = 8

From (iii): Maximum value of D = 12

⇒ Z = 15A^{2} − 3B^{4} + C + 0.5D = 15 – 0 + 8 + 6 = 29

**Case 2**: If A = 0, minimum value of B = 0

From (ii): Maximum value of C = 12

From (iii): Maximum value of D = 15

⇒ Z = 15A^{2} − 3B^{4} + C + 0.5D = 0 – 0 + 12 + 7.5 = 19.5

We can see that maximum value of Z = 29, when D = 12.

Hence, option (b).

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**20. XAT 2020 QADI | Geometry - Circles**

XYZ is an equilateral triangle, inscribed in a circle. P is a point on the arc YZ such that X and P are on opposite sides of the chord YZ. Which of the following MUST always be true?

- A.
XZ + YP = XY + PZ

- B.
XP = YP + PZ

- C.
XP + PZ = XY + YP

- D.
XP = XY

- E.
XP = XY + YZ

Answer: Option B

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**Explanation** :

**Ptolemy’s theorem:**

Product of diagonal of a cyclic quadrilateral is equal to sum of product of opposite pair of sides.

∴ XP × YZ = PY × XZ + PZ × XY

⇒ XP × a = PY × a + PZ × a

⇒ XP = PY + PZ

Hence, option (b).

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