# XAT 2023 QADI | Previous Year XAT Paper

XAT 2023 QADI

**1. XAT 2023 QADI | Algebra - Simple Equations**

Raju and Sarita play a number game. First, each one of them chooses a positive integer independently. Separately, they both multiply their chosen integers by 2, and then subtract 20 from their resultant numbers. Now, each of them has a new number. Then, they divide their respective new numbers by 5. Finally, they added their results and found that the sum is 16. What can be the maximum possible difference between the positive integers chosen by Raju and Sarita?

- A.
67

- B.
58

- C.
49

- D.
40

- E.
None of the above

Answer: Option B

**Explanation** :

Let the numbers choosen by Raju and Sarita be r and s respectively.

Multiplying both the numbers with 2, we get the numbers as 2r and 2s respectively.

Subtracting 20 from both the numbers, we get the numbers as 2r - 20 and 2s - 20 respectively.

Dividing both the numbers by 5, we get the numbers as (2r - 20)/5 and (2s - 20)/5 respectively.

∴ (2r - 20)/5 + (2s - 20)/5 = 16

⇒ (2r - 20) + (2s - 20) = 80

⇒ 2r + 2s = 120

⇒ r + s = 60

For maximum difference one number should be least possible and the other maximum possible.

Since r and s both are positive integers, the least value one of them can take is 1 hence the maximum value of other will be (60 - 1 = 59)

⇒ Maximum difference between them = 59 - 1 = 58.

Hence, option (b).

Workspace:

**2. XAT 2023 QADI | Geometry - Coordinate Geometry**

ABC is a triangle and the coordinates of A, B and C are (a, b - 2c), (a, b + 4c) and (-2a, 3c) respectively where a, b and c are positive numbers. The area of the triangle ABC is:

- A.
6abc

- B.
9abc

- C.
6bc

- D.
9ac

- E.
None of the above

Answer: Option D

**Explanation** :

The length of AC = (b + 4c) - (b - 2c) = 6c

Altitude from C to AB = a - (-2a) = 3a

∴ Area of ABC = 1/2 × b × h = 1/2 × 3a × 6c = 9ac

Hence, option (d).

Workspace:

**3. XAT 2023 QADI | Arithmetic - Simple & Compound Interest**

Jose borrowed some money from his friend at simple interest rate of 10% and invested the entire amount in stocks. At the end of the first year, he repaid 1/5th of the principal amount. At the end of the second year, he repaid half of the remaining principal amount. At the end of third year, he repaid the entire remaining principal amount. At the end of the fourth year, he paid the last three years’ interest amount. As there was no principal amount left, his friend did not charge any interest in the fourth year. At the end of fourth year, he sold out all his stocks. Later, he calculated that he gained Rs. 97500 after paying principal and interest amounts to his friend. If his invested amount in the stocks became double at the end of the fourth year, how much money did he borrow from his friend?

- A.
250000

- B.
200000

- C.
150000

- D.
125000

- E.
None of the above

Answer: Option D

**Explanation** :

Let total amount borrowed by Jose = Rs. 5x.

Principal paid back is x at the end of 1^{st} year, 2x at the end of 2^{nd} year and 2x at the end of 3^{rd} year.

Total interest accumulated = x × 1 × 0.1 + 2x × 2 × 0.1 + 2x × 3 × 0.1 = 1.1x

∴ Total amount Jose needs to pay = 5x + 1.1x = 6.1x

Value of stocks doubled i.e., 10x

Now, Jose's profit = 97500

⇒ 97500 = 10x - 6.1x

⇒ 3.9x = 97500

⇒ x = 25,000

∴ Money borrowed by Jose = 5x = 1,25,000

Hence, option (d).

Workspace:

**4. XAT 2023 QADI | Geometry - Coordinate Geometry**

ABC is a triangle with BC = 5. D is thefoot of the perpendicular from A on BC. E is a point on CD such that BE = 3.

**The value of AB ^{2} - AE^{2} + 6CD is:**

- A.
5

- B.
10

- C.
14

- D.
18

- E.
21

Answer: Option E

**Explanation** :

Given, BC = 5 and BE = 3

In △ABD ⇒ AB^{2} = AD^{2} + BD^{2} ...(1)

In △AED ⇒ AE^{2} = AD^{2} + DE^{2} ...(2)

(1) - (2)

⇒ AB^{2} - AE^{2} = BD^{2} - DE^{2}

⇒ AB^{2} - AE^{2} = x^{2} - (3 - x)^{2}

⇒ AB^{2} - AE^{2} = x^{2} - 9 - x^{2} + 6x

⇒ AB^{2} - AE^{2} = - 9 + 6x

We need to find AB^{2} - AE^{2} + 6CD

= (-9 + 6x) + 6 × (5 - x)

= -9 + 6x + 30 - 6x

= 21

Hence, option (e).

Workspace:

**5. XAT 2023 QADI | Arithmetic - Average**

Amit has forgotten his 4-digit locker key. He remembers that all the digits are positive integers and are different from each other. Moreover, the fourth digit is the smallest and the maximum value of the first digit is 3. Also, he recalls that if he divides the second digit by the third digit, he gets the first digit. How many different combinations does Amit have to try for unlocking the locker?

- A.
2

- B.
1

- C.
4

- D.
5

- E.
3

Answer: Option E

**Explanation** :

Let the number be 'abcd' where all digits are positive single-digit distinct integers.

Now, d is the smallest of all the 4 digits while a = b/c.

Also, a ≤ 3.

**Case 1**: a = 1

Not possible since d is the smallest of all the digits.

**Case 2**: a = 2 ⇒ b = 2c

Also, since d is the smallest of the 4 digits, d = 1

Now, possible values of (b, c) are (2, 1), (4, 2), (6, 3) and (8, 4)

But c cannot be the least of the 4 digits and all digits must be distinct.

∴ Accepted values of (b, c) are (6, 3) and (8, 4)

⇒ (a, b, c, d) can be (2, 6, 3, 1) or (2, 8, 4, 1)

**Case 3**: a = 3 ⇒ b = 3c

Also, since d is the smallest of the 4 digits, d = 1 or 2

Now, possible values of (b, c) are (3, 1), (6, 2), (9, 3)

But c cannot be the least of the 4 digits and all digits must be distinct.

∴ Accepted values of (b, c) are (6, 2)

⇒ (a, b, c, d) can be (3, 6, 2, 1)

∴ Total acceptable values of (a, b, c, d) are (2, 6, 3, 1) or (2, 8, 4, 1) or (3, 6, 2, 1) i.e., 3 values.

Hence, option (e).

Workspace:

**6. XAT 2023 QADI | Algebra - Progressions**

Suppose Haruka has a special key ∆ in her calculator called delta key:

Rule 1: If the display shows a one-digit number, pressing delta key ∆ replaces the displayed number with twice its value.

Rule 2: If the display shows a two-digit number, pressing delta key ∆ replaces the displayed number with the sum of the two digits.

Suppose Haruka enters the value 1 and then presses delta key ∆ repeatedly.

After pressing the ∆ key for 68 times, what will be the displayed number?

- A.
7

- B.
4

- C.
10

- D.
2

- E.
8

Answer: Option A

**Explanation** :

Initially the number = 1

Output when ∆ is pressed 1 time = 2 [Single digit number 1 is doubled]

Output when ∆ is pressed 2 times = 4 [Single digit number 2 is doubled]

Output when ∆ is pressed 3 times = 8 [Single digit number 4 is doubled]

Output when ∆ is pressed 4 times = 16 [Single digit number 8 is doubled]

Output when ∆ is pressed 5 times = 7 [Digits of 2-digit number 16 are added]

Output when ∆ is pressed 6 times = 14 [Single digit number 7 is doubled]

Output when ∆ is pressed 7 times = 5 [Digits of 2-digit number 14 are added]

Output when ∆ is pressed 8 times = 10 [Single digit number 5 is doubled]

Output when ∆ is pressed 9 times = 1 [Digits of 2-digit number 10 are added]

After 9 steps we get back the same input 1.

Hence, when the input is 1, we will get the same output at the end of 9 or 18 or ... or 63 steps.

∴ Output at the end of 68 steps will be same as the output at the end of 5 steps i.e., 7.

Hence, option (a).

Workspace:

**7. XAT 2023 QADI | Algebra - Simple Equations**

There are three sections in a question paper and each section has 10 questions. First section only has multiple-choice questions, and 2 marks will be awarded for each correct answer. For each wrong answer, 0.5 marks will be deducted. Any unattempted question in this section will be treated as a wrong answer. Each question in the second section carries 3 marks, whereas each question in the third section carries 5 marks. For any wrong answer or un-attempted question in the second and third sections, no marks will be deducted. A student’s score is the addition of marks obtained in all the three sections. What is the sixth highest possible score?

- A.
92.5

- B.
94

- C.
95.5

- D.
95

- E.
None of the above

Answer: Option B

**Explanation** :

Marking Scheme:

Section A: Correct Answer = +2; Wrong/Unattempted Question = -0.5

Section B: Correct Answer = +3; Wrong/Question = -0.5

Section C: Correct Answer = +5; Wrong/Question = -0.5

Each section has 10 questions.

∴ Maximum marks = 10 × 2 + 10 × 3 + 10 × 5 = 100.

For each unattempted/wrong question in section A, marks obtained will go down by 2.5.

For each unattempted/wrong question in section B, marks obtained will go down by 3.

For each unattempted/wrong question in section C, marks obtained will go down by 5.

⇒ 2^{nd} highest marks (when 1 Q is wrong in section A) = 100 - 2.5 = 97.5

⇒ 3^{rd} highest marks (when 1 Q is wrong in section B) = 100 - 3 = 97

⇒ 4^{th} highest marks (when 1 Q is wrong in section C) = 100 - 5 = 95

⇒ 5^{th} highest marks (when 1 Q is wrong in section A and B) = 100 - 5.5 = 94.5

⇒ 6^{th} highest marks (when 2 Qs are wrong in section B) = 100 - 6 = 94

Hence, option (b).

Workspace:

**8. XAT 2023 QADI | Geometry - Quadrilaterals & Polygons**

ABCD is a trapezoid where BC is parallel to AD and perpendicular to AB. Kindly note that BC < AD. P is a point on AD such that CPD is an equilateral triangle. Q is a point on BC such that AQ is parallel to PC. If the area of the triangle CPD is 4√3, find the area of the triangle ABQ.

- A.
2√3

- B.
4√3

- C.
4

- D.
4√3

- E.
None of the above

Answer: Option A

**Explanation** :

Given that, CPD is an equilateral triangle

∠CPD = ∠PDC = ∠DCP = 60°

AQ || PC ⇒ ∠CPD = ∠QAP = 60°

BC || AD ⇒ ∠QAP = ∠AQB = 60°

⇒ AQCP is a parallelogram, hence AQ = PC = a

∆AQB is a 30°-60°-90° triangle.

⇒ BQ = 1/2 × AQ = a/2

Comapring ∆AQB and ∆CPD

Base AQ = 1/2 × PD while the heights of both triangles is same.

⇒ Area of ∆AQB = 1/2 × Area of ∆CPD = 2√3

Hence, option (a).

Workspace:

**9. XAT 2023 QADI | Arithmetic - Profit & Loss**

Rajnish bought an item at 25% discount on the printed price. He sold it at 10% discount on the printed price. What is his profit in percentage?

- A.
10

- B.
15

- C.
17.5

- D.
20

- E.
None of the above

Answer: Option D

**Explanation** :

Let the printed price be Rs. 100.

Cost price for Rajnish after 25% discount = Rs. 75

Rajnish sold it at 10% discount in printed price i.e., at Rs. 90

∴ Required profit % = $\frac{90-75}{75}\times 100$ = $\frac{15}{75}\times 100$ = 20%

Hence, option (d).

Workspace:

**10. XAT 2023 QADI | Geometry - Trigonometry**

Find the value of:

$\frac{{\mathrm{sin}}^{6}15\xb0+{\mathrm{sin}}^{6}75\xb0+6{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}{{\mathrm{sin}}^{4}15\xb0+{\mathrm{sin}}^{4}75\xb0+5{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}$

- A.
sin 15° + sin 75°

- B.
6/5

- C.
1

- D.
sin 15° cos 15°

- E.
None of the above

Answer: Option C

**Explanation** :

$\frac{{\mathrm{sin}}^{6}15\xb0+{\mathrm{sin}}^{6}75\xb0+6{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}{{\mathrm{sin}}^{4}15\xb0+{\mathrm{sin}}^{4}75\xb0+5{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}$

Let sin^{2}15° = a and sin^{2}75° = b

∴ $\frac{{\mathrm{sin}}^{6}15\xb0+{\mathrm{sin}}^{6}75\xb0+6{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}{{\mathrm{sin}}^{4}15\xb0+{\mathrm{sin}}^{4}75\xb0+5{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}$ = $\frac{{\mathrm{a}}^{3}+{\mathrm{b}}^{3}+6\mathrm{ab}}{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}+5\mathrm{ab}}$

Now, a^{3} + b^{3} = (a + b)^{3} - 3ab(a + b), and

a^{2} + b^{2} = (a + b)^{2} - 2ab

∴ $\frac{{\mathrm{sin}}^{6}15\xb0+{\mathrm{sin}}^{6}75\xb0+6{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}{{\mathrm{sin}}^{4}15\xb0+{\mathrm{sin}}^{4}75\xb0+5{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}$ = $\frac{{(\mathrm{a}+\mathrm{b})}^{3}-3\mathrm{ab}(\mathrm{a}+\mathrm{b})+6\mathrm{ab}}{{(\mathrm{a}+\mathrm{b})}^{2}-2\mathrm{ab}+5\mathrm{ab}}$

Now, a + b = sin^{2}15° + sin^{2}75° = sin^{2}15° + cos^{2}15° = 1 [∵ Sin𝜃 = Cos(90 - 𝜃)]

∴ $\frac{{\mathrm{sin}}^{6}15\xb0+{\mathrm{sin}}^{6}75\xb0+6{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}{{\mathrm{sin}}^{4}15\xb0+{\mathrm{sin}}^{4}75\xb0+5{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}$ = $\frac{1-3\mathrm{ab}+6\mathrm{ab}}{1-2\mathrm{ab}+5\mathrm{ab}}$ = $\frac{1+3\mathrm{ab}}{1+3\mathrm{ab}}$ = 1

Hence, option (c).

Workspace:

**11. XAT 2023 QADI | Modern Math - Probability**

A painter draws 64 equal squares of 1 square inch on a square canvas measuring 64 square inches. She chooses two squares (1 square inch each) randomly and then paints them. What is the probability that two painted squares have a common side?

- A.
$\frac{112}{2016}$

- B.
$\frac{1}{3}$

- C.
$\frac{512}{100034}$

- D.
$\frac{3}{97}$

- E.
$\frac{7}{108}$

Answer: Option A

**Explanation** :

Required probability = $\frac{\mathrm{No}.\mathrm{of}\mathrm{ways}\mathrm{of}\mathrm{selecting}2\mathrm{smaller}\mathrm{squares}\mathrm{with}\mathrm{having}\mathrm{common}\mathrm{side}}{\mathrm{No}.\mathrm{of}\mathrm{ways}\mathrm{of}\mathrm{selecting}2\mathrm{smaller}\mathrm{squares}}$

No. of ways of selecting 2 smaller squares = ^{64}C_{2} = 32 × 63

Now, to select 2 squares with common side, there are two cases possible.

**Case 1**: Horizontal pair is selected.

In the first row 2 squares can be selected in 7 ways i.e., (R_{1}C_{1}, R_{2}C_{2}), (R_{1}C_{2}, R_{2}C_{3}), ..., (R_{1}C_{7}, R_{2}C_{8})

Similarly 2 squares can be selected from each row in 7 ways.

∴ No. of ways of selecting 2 smaller horizontal pair of squares = 7 × 8 = 56.

**Case 2**: Vertical pair is selected.

In the first column 2 squares can be selected in 7 ways i.e., (C_{1}R_{1}, C_{1}R_{2}), (C_{1}R_{2}, C_{1}R_{3}), ..., (C_{1}R_{7}, C_{1}R_{8})

Similarly 2 squares can be selected from each column in 7 ways.

∴ No. of ways of selecting 2 smaller vertical pair of squares = 7 × 8 = 56.

⇒ Total no. of ways of selecting 2 smaller squares with common side = 56 + 56 = 112

∴ Required probability = $\frac{112}{32\times 63}$ = $\frac{112}{2016}$

Hence, option (a).

Workspace:

**12. XAT 2023 QADI | Algebra - Number Theory**

Let x and y be two positive integers and p be a prime number. If x(x – p) – y(y + p) = 7p, what will be the minimum value of x – y?

- A.
1

- B.
3

- C.
5

- D.
7

- E.
None of the above

Answer: Option E

**Explanation** :

Given, x(x – p) – y(y + p) = 7p

⇒ x^{2} - xp - y^{2} - yp = 7p

⇒ x^{2} - y^{2} - xp - yp = 7p

⇒ (x - y)(x + y) - p(x + y) = 7p

⇒ (x + y)(x - y - p) = 7p

Since, 7 and p are both prime numbers, we have

⇒ (x + y)(x - y - p) = 7 × p or 7p × 1

i.e., one (x + y) and (x - y - p) can be 7 and p or 7p and 1.

**Case 1**: one (x + y) and (x - y - p) can be 7 and p

⇒ (x + y) + (x - y - p) = 7 + p

⇒ 2x = 7 + 2p

⇒ x = 3.5 + 2p

This is not possible as x should be a integer.

**Case 2**: one (x + y) and (x - y - p) can be 7p and 1.

⇒ (x + y) + (x - y - p) = 7p + 1

⇒ 2x = 8p + 1

⇒ x = 4p + 0.5

This is not possible as x should be a integer.

∴ No value of integral value of x is possible.

Hence, option (e).

Workspace:

**13. XAT 2023 QADI | Arithmetic - Average**

Five students appeared for an examination. The average mark obtained by these five students is 40. The maximum mark of the examination is 100, and each of the five students scored more than 10 marks. However, none of them scored exactly 40 marks. Based on the information given, which of the following MUST BE true?

- A.
At least, three of them scored a maximum of 40 marks

- B.
At least, three of them scored more than 40 marks

- C.
At least, one of them scored exactly 41 marks

- D.
At most, two of them scored more than 40 marks

- E.
At least, one of them scored less than 40 marks

Answer: Option E

**Explanation** :

Average of 5 studens is 40, hence their total marks = 200

If the average is 40, there are two possibilities:

**Case 1**: All score exactly 40, but that is not possible.

**Case 2**: At least one scores more than 40 and at least one scores less than 40.

Hence, option (e).

Workspace:

**14. XAT 2023 QADI | Algebra - Progressions**

Consider a_{n + 1} = $\frac{1}{1+{\displaystyle \frac{1}{{a}_{n}}}}$ for n = 1, 2, ...., 2008, 2009 where a_{1} = 1. Find the value of a_{1} a_{2 }+ a_{2} a_{3 }+ a_{3} a_{4 }+ ... + a_{2008} a_{2009 }

- A.
_{$\frac{2009}{1000}$} - B.
_{$\frac{2009}{2008}$} - C.
_{$\frac{2008}{2009}$} - D.
_{$\frac{6000}{2009}$} - E.
_{$\frac{2008}{6000}$}

Answer: Option C

**Explanation** :

Given, a_{n + 1} = $\frac{1}{1+{\displaystyle \frac{1}{{a}_{n}}}}$

⇒ a_{1} = 1

⇒ a_{2} = $\frac{1}{1+{\displaystyle \frac{1}{{a}_{1}}}}$ = $\frac{1}{2}$

⇒ a_{3} = $\frac{1}{1+{\displaystyle \frac{1}{{a}_{2}}}}$ = $\frac{1}{3}$

...

⇒ a_{n} = $\frac{1}{n}$

Now: a_{1} a_{2 }+ a_{2} a_{3 }+ a_{3} a_{4 }+ ... + a_{2008} a_{2009} = $\frac{1}{1\times 2}$ + $\frac{1}{2\times 3}$ + $\frac{1}{3\times 4}$ + ... + $\frac{1}{2008\times 2009}$

= $\left(\frac{1}{1}-\frac{1}{2}\right)$ + $\left(\frac{1}{2}-\frac{1}{3}\right)$ + $\left(\frac{1}{3}-\frac{1}{4}\right)$ + ... + $\left(\frac{1}{2008}-\frac{1}{2009}\right)$

= $\frac{1}{1}-\frac{1}{2009}$ = $\frac{2008}{2009}$

Hence, option (c).

Workspace:

**15. XAT 2023 QADI | Arithmetic - Percentage**

The Guava club has won 40% of their football matches in the Apple Cup that they have played so far. If they play another n matches and win all of them, their winning percentage will improve to 50. Further, if they play 15 more matches and win all of them, their winning percentage will improve from 50 to 60. How many matches has the Guava club played in the Apple Cup so far? In the Apple Cup matches, there are only two possible outcomes, win or loss; draw is not possible.

- A.
50

- B.
40

- C.
30

- D.
Cannot be determined, as the value of n is not given

- E.
60

Answer: Option A

**Explanation** :

Let the total matches played so far = m.

Number of matches won = 0.4m

After playing n more matches and winning all the matches, win rate becomes 50%.

Total matches played = m + n

Total matches won = 0.4m + n

⇒ 0.4m + n = 50% of (m + n)

⇒ 0.4m + n = 0.5m + 0.5n

⇒ 0.5n = 0.1m

⇒ m : n = 5 : 1

⇒ m = 5x and n = x ...(1)

After playing 15 more matches and winning all the matches, win rate becomes 60%

Total matches played = m + n + 15

Total matches won = 0.4m + n + 15

⇒ 0.4m + n + 15 = 60% of (m + n + 15)

⇒ 0.4m + n + 15 = 0.6m + 0.6n + 9

⇒ 0.4n + 6 = 0.2m

⇒ 0.4 × x + 6 = 0.2 × 5x

⇒ 0.4x + 6 = x

⇒ x = 10

∴ Matches played by Guava club so far = m = 5x = 50.

Hence, option (a).

Workspace:

**16. XAT 2023 QADI | Arithmetic - Simple & Compound Interest**

Separately, Jack and Sristi invested the same amount of money in a stock market. Jack’s invested amount kept getting reduced by 50% every month. Sristi’s investment also reduced every month, but in an arithmetic progression with a common difference of Rs. 15000. They both withdrew their respective amounts at the end of the sixth month. They observed that if they had withdrawn their respective amounts at the end of the fourth month, the ratio of their amounts would have been the same as the ratio after the sixth month. What amount of money was invested by Jack in the stock market?

- A.
Rs. 100000

- B.
Rs. 120000

- C.
Rs. 150000

- D.
Rs. 180000

- E.
None of the above

Answer: Option A

**Explanation** :

Let their initial amounts be Rs. x

**After 4 months:**

Jack's investment after 4 months = x × ${\left(\frac{1}{2}\right)}^{4}$ = $\frac{\mathrm{x}}{16}$

Sristi's investment after 4 months = x - 15000 × 4 = x - 60000

**After 6 months:**

Jack's investment after 6 months = x × ${\left(\frac{1}{2}\right)}^{6}$ = $\frac{\mathrm{x}}{64}$

Sristi's investment after 6 months = x - 15000 × 6 = x - 90000

Ratio of their investments after 4 months = Ratio of their investments after 6 months.

⇒ $\frac{\mathrm{x}/16}{\mathrm{x}-60000}$ = $\frac{\mathrm{x}/64}{\mathrm{x}-90000}$

⇒ 4(x - 90,000) = x - 60,000

⇒ 3x = 3,00,000

⇒ x = 1,00,000

Hence, option (a).

Workspace:

**17. XAT 2023 QADI | Arithmetic - Time, Speed & Distance**

**The problem below consists of a question and two statements numbered 1 & 2. You have to decide whether the data provided in the statements are sufficient to answer the question.**

Rahim is riding upstream on a boat, from point A to B, at a constant speed. The distance from A to B is 30 km. One minute after Rahim leaves from point A, a speedboat starts from point A to go to point B. It crosses Rahim’s boat after 4 minutes. If the speed of the speedboat is constant from A to B, what is Rahim’s speed in still water?

1. The speed of the speedboat in still water is 30 km/hour.

2. Rahim takes three hours to reach point B from point A.

- A.
Statement 1 alone is sufficient to answer the question, but statement 2 alone is not sufficient

- B.
Statement 2 alone is sufficient to answer the question, but statement 1 alone is not sufficient

- C.
Each statement alone is sufficient

- D.
Both statements together are sufficient, but neither of them alone is sufficient

- E.
Statements 1 & 2 together are not sufficient

Answer: Option D

**Explanation** :

Let the speed of river be 'r' kmph.

Rahim and Speedboat start from A and meet at a point M (say).

Going upstream, Rahim takes 5 minutes to cover this distance while Speedboat takes 4 minutes to cover the same distance.

∴ $\frac{{\mathrm{S}}_{\mathrm{Rahim}}-\mathrm{r}}{{\mathrm{S}}_{\mathrm{Speedboat}}-\mathrm{r}}$ = $\frac{4}{5}$ ...(1)

We need to find S_{Rahim}

**Statement (1)**: S_{Speedboat} = 30

If we put S_{Speedboat} = 30 in (1), this alone would not be sufficient to calculate S_{Rahim}.

∴ Statement (1) alone is not sufficient to answer the question.

**Statement (2)**: Rahim takes 3 hours to reach B.

⇒ S_{Rahim} - r = 30/3 = 10 kmph.

Since we don't know the value of r, we cannot determine the S_{Rahim}.

∴ Statement (2) alone is not sufficient to answer the question.

Using both statements together we have, S_{Speedboat} = 30 kmph and S_{Rahim} - r = 10 kmph.

Putting these values in (1), we get

$\frac{10}{30-\mathrm{r}}$ = $\frac{4}{5}$

⇒ r = 17.5 kmph.

⇒ S_{Rahim} - 17.5 = 10

⇒ S_{Rahim} - 17.5 = 27.5 kmph

∴ Using both statements we can get the answer.

Hence, option (d).

Workspace:

**18. XAT 2023 QADI | Geometry - Mensuration**

A non-flying ant wants to travel from the bottom corner to the diagonally opposite top corner of a cubical room. The side of the room is 2 meters. What will be the minimum distance that the ant needs to travel?

- A.
6 meters

- B.
(2√2 + 2) meters

- C.
2√3 meters

- D.
2√6 meters

- E.
2√5 meters

Answer: Option E

**Explanation** :

In a 2D plane, minimum distance between 2 points is the straight line joining these points.

We can convert the cubical room in a 2D plane as shown in the figure.

∴ Shortest route from A to F = $\sqrt{{4}^{2}+{2}^{2}}$ = 2√5.

Hence, option (e).

Workspace:

**19. XAT 2023 QADI | Algebra - Inequalities & Modulus**

Given A = |x + 3| + |x - 2| - |2x - 8|. The maximum value of |A| is:

- A.
111

- B.
9

- C.
6

- D.
3

- E.
∞

Answer: Option B

**Explanation** :

A = |x + 3| + |x - 2| - |2x - 8|

The critical points are -3, 2 and 4.

**Case 1**: x ≥ 4

∴ A = x + 3 + x - 2 - (2x - 8)

⇒ A = 2x + 1 - 2x + 8

⇒ A = 9

**Case 2**: 2 ≤ x < 4

∴ A = x + 3 + x - 2 + (2x - 8)

⇒ A = 2x + 1 + 2x - 8

⇒ A = 4x - 7

∴ A ∈ [1, 9)

**Case 3**: -3 ≤ x < 2

∴ A = x + 3 - (x - 2) + (2x - 8)

⇒ A = x + 3 - x + 2 + 2x - 8

⇒ A = 2x - 3

∴ A ∈ [-9, 1)

**Case 4**: x < -3

∴ A = - (x + 3) - (x - 2) + (2x - 8)

⇒ A = - x - 3 - x + 2 + 2x - 8

⇒ A = - 9

From the above cases, The maximum value of |A| = 9

Hence, option (b).

Workspace:

**20. XAT 2023 QADI | Arithmetic - Mixture, Alligation, Removal & Replacement**

A small jar contained water, lime and sugar in the ratio of 90 : 7 : 3. A glass contained only water and sugar in it. Contents of both (small jar and glass) were mixed in a bigger jar and the ratio of contents in the bigger jar was 85 : 5 : 10 (water, lime and sugar respectively). Find the percentage of water in the bigger jar?

- A.
70

- B.
75

- C.
80

- D.
72.5

- E.
85

Answer: Option E

**Explanation** :

Let the quantity of water, line and sugar in the bigger jar be 85x, 5x and 10x respectively.

Now, percentage of water in bigger jar = $\frac{85\mathrm{x}}{85\mathrm{x}+5\mathrm{x}+10\mathrm{x}}\times 100\%$ = 85%

Hence, option (e).

Workspace:

**21. XAT 2023 QADI | Arithmetic - Time, Speed & Distance | Data Sufficiency**

**The problem below consists of a question and two statements numbered 1 & 2. You have to decide whether the data provided in the statements are sufficient to answer the question.**

In a cricket match, three slip fielders are positioned on a straight line. The distance between 1st slip and 2nd slip is the same as the distance between 2nd slip and the 3rd slip. The player X, who is not on the same line of slip fielders, throws a ball to the 3rd slip and the ball takes 5 seconds to reach the player at the 3rd slip. If he had thrown the ball at the same speed to the 1st slip or to the 2nd slip, it would have taken 3 seconds or 4 seconds, respectively. What is the distance between the 2nd slip and the player X?

1. The ball travels at a speed of 3.6 km/hour.

2. The distance between the 1st slip and the 3rd slip is 2 meters.

- A.
Statement 1 alone is sufficient to answer the question, but statement 2 alone is not sufficient

- B.
Statement 2 alone is sufficient to answer the question, but statement 1 alone is not sufficient

- C.
Each statement alone is sufficient

- D.
Both statements together are sufficient, but neither of them alone is sufficient

- E.
Statements 1 & 2 together are not sufficient

Answer: Option A

**Explanation** :

Discrepancy was found in this question and full marks were awarded to all candidates.

Workspace:

**Go through the information given below, and answer the THREE questions that follow. **

The three graphs below capture relationship between economic (and social) activities and subjective well-being. The first graph (Graph-1) captures relationship between GDP (percapita) and Satisfaction with life, across different countries and four islands: Gizo, Roviana, Niijhum Dwip, and Chittagong. The Graph-2 captures three different measures of subjective well-being (Satisfaction with life, Affect Balance and Momentary Affect) across the four islands, which have different levels monetization (Index). The Graph-3 captures levels of thirteen different socio-economic activities across four islands.

**22. XAT 2023 QADI | DI - Tables & Graphs**

Which of the following will BEST capture the relationship between GDP (x-axis) and Life Satisfaction (y-axis) of countries?

- A.
y = x

- B.
y = x

^{2} - C.
y = log(x)

- D.
y = $\frac{1}{{x}^{2}}$

- E.
y = e

^{x}

Answer: Option C

**Explanation** :

Graph for y = log x

Graph for y = x

Graph for y = x^{2}

Graph for y = 1/x^{2}

Graph for y = e^{x}

Hence, option (c).

Workspace:

**23. XAT 2023 QADI | DI - Tables & Graphs**

Which of the following, about the four islands, can be BEST inferred from the graphs?

- A.
Whenever affect balance increases, satisfaction with life decreases

- B.
Whenever Pleasant activities increase, satisfaction with life decreases

- C.
Whenever Religion increases, satisfaction with life decreases

- D.
Whenever satisfaction with life increases, family also increases

- E.
Index of monetization varies maximum in Niijhum Dwip

Answer: Option E

**Explanation** :

Workspace:

**24. XAT 2023 QADI | DI - Tables & Graphs**

Which of the following site has the highest fishing to economic ratio?

- A.
Gizo

- B.
Chittagong

- C.
Roviana

- D.
Niijhum Dwip

- E.
All four islands have equal dependence

Answer: Option C

**Explanation** :

Workspace:

**Go through the information given below, and answer the THREE questions that follow.**

The table captures Age and Gender distribution of Covid Positive Cases in a country. However, a part of data is missing, represented through unknown categories.

**25. XAT 2023 QADI | DI - Tables & Graphs**

In unknown age category, the ratio of males (unknown age category) and females (unknown age category) to total unknown cases (unknown age category) is same as the ratio of males (All) and females (All) to the total (total confirmed covid positive cases). How many females were in the unknown age category (rounded to nearest integer)?

- A.
120

- B.
140

- C.
110

- D.
125

- E.
130

Answer: Option D

**Explanation** :

Discrepancy was found in this question and full marks were awarded to all candidates.

Workspace:

**26. XAT 2023 QADI | DI - Tables & Graphs**

In which age category, the percentage of female covid patients is the HIGHEST?

- A.
61-70

- B.
31-40

- C.
41-50

- D.
51-60

- E.
81+

Answer: Option E

**Explanation** :

Percentage of females in category:

31-40 = 12693/24110 × 100% = 52.6%

41-50 = 7264/14120 × 100% = 51.4%

51-60 = 5567/11174 × 100% = 50%

61-70 = 3756/7300 × 100% = 51.4%

81+ = 1276/1996 × 100% = 63.9%

∴ Percentage of females is highest for 81+ age category.

Hence, option (e).

Workspace:

**27. XAT 2023 QADI | DI - Tables & Graphs**

Which of the following is true for “unknown gender Category”?

- Unknown age group patients are less likely (percentage term) to provide information about gender than any other age category
- Between 31 and 80, when age increases patients, in percentage terms, are less likely to provide information about gender
- Elderly (81+) category patients are more likely to give information about gender than 0-18 age group

- A.
3 only

- B.
2 only

- C.
1 and 2

- D.
1 and 3

- E.
2 and 3

Answer: Option D

**Explanation** :

Discrepancy was found in this question and full marks were awarded to all candidates.

Workspace:

**28. XAT 2023 QADI | DI - Tables & Graphs**

The addition of 7 distinct positive integers is 1740. What is the largest possible “greatest common

divisor” of these 7 distinct positive integers?

- A.
42

- B.
60

- C.
74

- D.
140

- E.
None of these

Answer: Option B

**Explanation** :

Let the greatest common divisor be m.

Now, the 7 numbers will be m × a, m × b, m × c, ..., m × g.

⇒ m × a + m × b + m × c + ... + m × g = 1740

⇒ m(a + b + c + ... + g) = 1740 ...(1)

For m to be highest possible, (a + b + c + ... + g) should be least possible.

Least possible value of (a + b + c + ... + g) = (1 + 2 + 3 + ... + 7) = 28

∴ (a + b + c + ... + g) ≥ 8

Now, (1) can be written as

m(a + b + c + ... + g) = 1740 = 2^{2} × 3 × 5 × 29

⇒ least possible value based on the given condition = 29

∴ m = 2^{2} × 3 × 5 = 60

Hence, option (b).

Workspace:

## Feedback

Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.