# XAT 2021 DM | Previous Year XAT Paper

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**Read the following scenario and answer the three questions that follow.**

A quick survey at the end of a purchase at buyagain.com asks the following three questions to each shopper:

1. Are you shopping at the website for the first time? (YES or NO)

2. Specify your gender: (MALE or FEMALE)

3. How satisfied are you? (HAPPY, NEUTRAL or UNHAPPY)

240 shoppers answer the survey, among whom 65 are first time shoppers. Furthermore:

i. The ratio of the numbers of male to female shoppers is 1 : 2 while the ratio of the numbers of unhappy, happy and neutral shoppers is 3 : 4 : 5

ii. The ratio of the numbers of happy first-time male shoppers, happy returning male shoppers, unhappy female shoppers, neutral male shoppers, neutral female shoppers and happy female shoppers is 1 : 1 : 4 : 4 : 6 : 6

iii. Among the first-time shoppers, the ratio of the numbers of happy male, neutral male, unhappy female and the remaining female shoppers is 1 : 1 : 1 : 2, while the number of happy first-time female shoppers is equal to the number of unhappy first-time male shoppers

**1. XAT 2021 DM | Algebra - Simple Equations**

Which among the following cannot be determined uniquely?

- A.
The number of first-time happy male shoppers

- B.
The number of returning male shoppers

- C.
All the numbers can be determined uniquely

- D.
The number of returning unhappy female shoppers

- E.
The number of first-time neutral male shoppers

Answer: Option C

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**Explanation** :

From the given data the following table can be created:

Hence the value of x=10

All the values can be uniquely determined

Workspace:

**2. XAT 2021 DM | LR - Cubes**

The six faces of a wooden cube of side 6 cm are labelled A, B, C, D, E and F respectively. Three of these faces A, B, and C are each adjacent to the other two, and are painted red. The other three faces are not painted. Then, the wooden cube is neatly cut into 216 little cubes of equal size. How many of the little cubes have no sides painted?

- A.
125

- B.
135

- C.
91

- D.
108

- E.
100

Answer: Option A

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**Explanation** :

Since A, B and C are adjacent faces. If we remove them, the resultant solid will also be a cube with side 5.

Hence total number of cubes unpainted = 5^{3} = 125

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**3. XAT 2021 DM | Geometry - Basics**

ABC is a triangle with integer-valued sides AB = 1, BC >1, and CA >1. If D is the mid-point of AB, then, which of the following options is the closest to the maximum possible value of the angle ACD (in degrees)?

- A.
15

- B.
30

- C.
45

- D.
75

- E.
60

Answer: Option A

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**Explanation** :

Given, AB = 1, AC > 1 and BC > 1

If ABC was an equilateral triangle i.e if AB = AC = BC = 1, ACD would have been 30°.

As C moves away from A and B, the angle decreases, hence the angle should be less than 30 and among the given options only 15 is less than 30.

Workspace:

**4. XAT 2021 DM | Algebra - Simple Equations**

Find z, if it is known that:

a: -y^{2} + x^{2} = 20

b: y^{3} - 2x^{2} - 4z ≥ -12 and

c: x, y and z are all positive integers

- A.
Any integer greater than 0 and less than 24

- B.
24

- C.
We need one more equation to find z

- D.
6

- E.
1

Answer: Option E

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**Explanation** :

Since x^{2} - y^{2} = 20 and x, y, z are positive integers, the only values possible are x = 6 and y = 4.

y^{3} - 2x^{2} - 4z 12 ≥ -12

Keeping the values of x and y, we get z ≤ 1

⇒ z = 1

Workspace:

**5. XAT 2021 DM | Algebra - Simple Equations**

An encryption system operates as follows:

Step 1. Fix a number k (k ≤ 26).

Step 2. For each word, swap the first k letters from the front with the last k letters from the end in reverse order. If a word contains less than 2k letters, write the entire word in reverse order.

Step 3. Replace each letter by a letter k spaces ahead in the alphabet. If you cross Z in the process to move k steps ahead, start again from A.

Example: k = 2: zebra → arbez → ctdgb.

If the word “flight” becomes “znmorl” after encryption, then the value of k:

- A.
5

- B.
4

- C.
7

- D.
Cannot be determined uniquely from the given information

- E.
6

Answer: Option E

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**Explanation** :

Flight become znmorl

Let's assume k > 3

So flight will become thgilf → znmrol. Hence the value of k will be 6

Workspace:

**Read the following scenario and answer the three questions that follow.**

The following plot describes the height (in cm), weight (in kg), age (in years) and gender (F for female, M for male) of 20 patients visiting a hospital.

A person’s body mass index (BMI) is calculated as weight (in kg) divided by squared height (measured in square metres). For example, a person weighing 100 kg and of height 100 cm (1m) will have a BMI of 100. A person with BMI less than or equal to 18.5 is considered as underweight, above 18.5 but less than or equal to 25 as normal weight, above 25 but less than or equal to 30 as overweight, and above 30 as obese.

**6. XAT 2021 DM | DI - Tables & Graphs**

The average age of the female patients who weigh 50 kg or above is approximately

- A.
62

- B.
65

- C.
68

- D.
70

- E.
Cannot be determined from the given data

Answer: Option A

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**Explanation** :

There are 5 ladies whose weights are 50 or above

There ages are 50,50, 70,60 and 80

k > 3

Average = 310/5 = 62

Workspace:

**Read the following scenario and answer the three questions that follow.**

The following plot describes the height (in cm), weight (in kg), age (in years) and gender (F for female, M for male) of 20 patients visiting a hospital.

A person’s body mass index (BMI) is calculated as weight (in kg) divided by squared height (measured in square metres). For example, a person weighing 100 kg and of height 100 cm (1m) will have a BMI of 100. A person with BMI less than or equal to 18.5 is considered as underweight, above 18.5 but less than or equal to 25 as normal weight, above 25 but less than or equal to 30 as overweight, and above 30 as obese.

**7. XAT 2021 DM | Algebra - Simple Equations**

The highest BMI among all patients is approximately

- A.
20

- B.
33

- C.
30

- D.
27

- E.
23

Answer: Option D

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**Explanation** :

For the highest BMI, weight should be as high as possible and height as little as possible.

Hence it is possible with the person with a weight of 69 kg and a height of 1.6m

His BMI will be (${\left(\frac{69}{1.6}\right)}^{2}$) = 27

Workspace:

**Read the following scenario and answer the three questions that follow.**

The following plot describes the height (in cm), weight (in kg), age (in years) and gender (F for female, M for male) of 20 patients visiting a hospital.

A person’s body mass index (BMI) is calculated as weight (in kg) divided by squared height (measured in square metres). For example, a person weighing 100 kg and of height 100 cm (1m) will have a BMI of 100. A person with BMI less than or equal to 18.5 is considered as underweight, above 18.5 but less than or equal to 25 as normal weight, above 25 but less than or equal to 30 as overweight, and above 30 as obese.

**8. XAT 2021 DM | Algebra - Simple Equations**

The BMI of the oldest person considered as normal weight is approximately

- A.
20

- B.
25

- C.
22

- D.
24

- E.
19

Answer: Option A

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**Explanation** :

The BMI of 1st oldest person = ($\left(\frac{40}{1.5}\right)$)^{2} = 17.77

The BMI of next oldest person = ($\left(\frac{61}{1.75}\right)$) = 19.9

Workspace:

**9. XAT 2021 DM | Arithmetic - Ratio, Proportion & Variation**

The topmost point of a perfectly vertical pole is marked A. The pole stands on a flat ground at point D. The points B and C are somewhere between A and D on the pole. From a point E, located on the ground at a certain distance from D, the points A, B and C are at angles of 60, 45 and 30 degrees respectively. What is AB : BC : CD?

- A.
(3 + √3) : (1 + √3) : 1

- B.
(3 - √3) : 1 : (√3 - 1)

- C.
1 : 1 : 1

- D.
(3 - √3) : (√3 - 1) : 1

- E.
(√3 - 1) : 1 : (3 - √3)

Answer: Option D

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**Explanation** :

Let ED = √3x

In triangle CDE, tan 30 = $\frac{CD}{\sqrt{3}x}$ ⇒ CD = x

In triangle BDE, tan 45 = $\frac{BD}{\sqrt{3}x}$ ⇒ BD = $\sqrt{3}$x ⇒ BC = $\sqrt{3}$x - x

In triangle ÖDE, tan 60 = $\frac{AD}{\sqrt{3}x}$ ⇒ AD = 3x ⇒ AB = 3x - $\sqrt{3}$x

AB : BC : CD = (3 - √3) : (√3 - 1) : 1

Workspace:

**10. XAT 2021 DM | Geometry - Circles**

Two circles P and Q, each of radius 2 cm, pass through each other’s centres. They intersect at points A and B. A circle R is drawn with diameter AB. What is the area of overlap (in square cm) between the circles R and P?

- A.
$\frac{8\pi}{3}-2\sqrt{3}$

- B.
$\frac{8\pi}{3}$

- C.
$\frac{13\pi}{3}-\sqrt{3}$

- D.
$\frac{17\pi}{6}-2\sqrt{3}$

- E.
$\frac{17\pi}{6}-\sqrt{3}$

Workspace:

**11. XAT 2021 DM | Arithmetic - Time, Speed & Distance**

Four friends, Ashish, Brian, Chaitra, and Dorothy, decide to jog for 30 minutes inside a stadium with a circular running track that is 200 metres long. The friends run at different speeds. Ashish completes a lap exactly every 60 seconds. Likewise, Brian, Chaitra and Dorothy complete a lap exactly every 1 minute 30 seconds, 40 seconds and 1 minute 20 seconds respectively. The friends begin together at the start line exactly at 4 p.m. What is the total of the numbers of laps the friends would have completed when they next cross the start line together ?

- A.
43

- B.
36

- C.
They will never be at the start line together again before 4:30 p.m.

- D.
47

- E.
28

Answer: Option D

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**Explanation** :

All the four friends will meet at the starting point after LCM(60,90,40,80) = 720 seconds.

Number of laps by A in 720 seconds = 12

Number of laps by B in 720 seconds = 8

Number of laps by C in 720 seconds = 18

Number of laps by D in 720 seconds = 9

Together they complete = 47 laps

Workspace:

**12. XAT 2021 DM | Modern Math - Probability**

Zahir and Raman are at the entrance of a dark cave. To enter this cave, they need to open a number lock. Raman sees a note on a rock: “ ... chest of pure diamonds kept for the smart one ... number has six digits ... second last digit is 2, third last is 4 ... divisible by all prime numbers less than 15 ...”. Excited, Zahir and Raman seek your help: which of these can be the first digit of the six digit number that will help them open the lock?

- A.
5

- B.
3

- C.
9

- D.
1

- E.
4

Answer: Option E

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**Explanation** :

Let the 6 digit number be _ _ _ 42_

It is divisible by 2,3,5,7,11,13

Hence, the number is a multiple of 2 × 3 × 5 × 7 × 11 × 13 = 30030

Now, 30030 × k = _ _ _ 4 2 _

Last digit would definitely be 0.

Second last digits which is 2 should be the last digit of 3 × k. Hence, last digit of k should be 4.

∴ k = 4 or 14 or 24 and so on.

The only possible value of k from the one's given above satisying the conditions given is 14.

∴ 30030 × 14 = 420420

∴ The first digit is 4.

Hence, option (e).

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