# XAT 2018 QADI | Previous Year XAT Paper

XAT 2018 Data Interpretation and Quantitative Ability

**1. XAT 2018 QADI | Algebra - Number Theory**

Find the value of the expression: 10 + 10^{3} + 10^{6} + 10^{9}

- A.
1010101010

- B.
1001000010

- C.
1001000110

- D.
1001001010

- E.
100010001010

Answer: Option D

**Explanation** :

10 + 10^{3} + 10^{6} + 10^{9} =

10 +

1000 +

1000000 +

1000000000 +

= 1001001010

Hence, option (d).

Workspace:

**2. XAT 2018 QADI | Arithmetic - Time & Work**

Abdul, Bimal, Charlie and Dilbar can finish a task in 10, 12, 15 and 18 days respectively. They can either choose to work or remain absent on a particular day. If 50 percent of the total work gets completed after 3 days, then, which of the following options is possible?

- A.
Each of them worked for exactly 2 days.

- B.
Bimal and Dilbar worked for 1 day each, Charlie worked for 2 days and Abdul worked for all 3 days.

- C.
Abdul and Charlie worked for 2 days each, Dilbar worked for 1 day and Bimal worked for all 3 days.

- D.
Abdul and Dilbar worked for 2 days each, Charlie worked for 1 day and Bimal worked for all 3 days.

- E.
Abdul and Charlie worked for 1 day each, Bimal worked for 2 days and Dilbar worked for all 3 days.

Answer: Option E

**Explanation** :

Let the total units of work = LCM of (10, 12, 15 and 18) = 180

Therefore, efficiencies of Abdul, Bimal, Charlie and Dilbar individually in one day will be 18 units, 15 units, 12 units and 10 units per day respectively.

Let us check options now:

Option (a): Total work done = 2 × (18 + 15 + 12 + 10)

= 110 units

Option (b): Total work done = 1 × (15 + 10) + 2 × 12 + 3 × 18

= 25 + 24 + 54 = 103 units

Option (c): Total work done = 2 × (18 + 2) + 1 × 10 + 3 × 15

= 40 + 10 + 45 = 95 units

Option (d): Total work done = 2 × (18 + 10) + 1 × 12 + 3 × 15

= 56 + 12 + 45 = 113 units

Option (e): Total work done = 1 × (18 + 12) + 2 × 15 + 3 × 10

= 30 + 30 + 30 = 90 units

Hence, option (e).

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**3. XAT 2018 QADI | Geometry - Quadrilaterals & Polygons**

If the diagonals of a rhombus of side 15 cm are in the ratio 3:4, find the area of the rhombus.

- A.
54 sq. cm.

- B.
108 sq. cm.

- C.
144 sq. cm.

- D.
200 sq. cm.

- E.
None of the above

Answer: Option E

**Explanation** :

We know that the diagonals of a rhombus bisect each other at right angles.

So, let’s assume that the two diagonals have lengths 6x and 8x.

⇒ (3x)^{2} + (4x)^{2} = 152

⇒ 25x^{2} = 225

⇒ x^{2} = 9

⇒ x = 3

∴ The length of the diagonals are 18 and 24.

∴ Area of the rhombus = ½ × product of the diagonals = ½ × 18 × 24 = 216

Hence, option (e).

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**4. XAT 2018 QADI | Arithmetic - Percentage**

The price of a product is P. A shopkeeper raises its price by X% and then offers a discount of Y% on the raised price. The discounted price again becomes P. If Y is the difference between X and Y, then find X.

- A.
20

- B.
25

- C.
50

- D.
100

- E.
None of the above

Answer: Option D

**Explanation** :

Given, $P\left(1+\frac{X}{100}\right)\left(1-\frac{Y}{100}\right)$ = P

⇒ $\left(1+\frac{X}{100}\right)\left(1-\frac{Y}{100}\right)$ = 1 …(1)

⇒ X > Y

It is also given that X – Y = Y.

⇒ X = 2Y

Substituting this in (1)

⇒ $\left(1+\frac{2Y}{100}\right)\left(1-\frac{Y}{100}\right)$ = 1

⇒ 1 - ($\frac{2{Y}^{2}}{10000}$ + $\frac{Y}{100}$ = 1

⇒ Y = 50

∴ X = 100

Hence, option (d).

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**5. XAT 2018 QADI | Arithmetic - Mixture, Alligation, Removal & Replacement**

A mixture comprises water and liquids A and B. The volume of water is 1/3rd of the total mixture and the volume of liquids A and B are in the ratio 5:3. To remove the water, the mixture is passed through a porous medium which completely absorbs the water and partially absorbs liquid A. Altogether this porous medium absorbs 200 ml of the initial mixture. If the ratio of volume of liquids A and B in the residual concentrated mixture becomes 7:9 then find the volume of water absorbed by the porous medium.

- A.
60 ml

- B.
200/3 ml

- C.
80 ml

- D.
100 ml

- E.
120 ml

Answer: Option E

**Explanation** :

Let the volume of liquid A and B in the initial mixture be 5x and 3x ml.

∴ Total volume of A and B = 8x ml, this is 2/3^{rd} of the total mixture.

⇒ Total volume of the mixture = 3/2 × 8x = 12x ml and volume of water = 1/3^{rd} of 12x i.e., 4x ml.

∴ Volume of water, liquid A and liquid B in the mixture will be 4x, 5x and 3x respectively.

When this mixture passes through the porous material whole of water and part of liquid A is absorbed. B is not absorbed by the porous material.

Now, since the ratio of volume of liquids A and B in the residual concentrated mixture becomes 7:9

⇒ (Remaining volume of A) : (3x) = 7 : 9

⇒ Remaining volume of A = 7x/3 ml.

∴ Volume of A absorbed = 5x – 7x/3 = 8x/3

Volume of water absorbed = 4x ml

⇒ 4x + 8x/3 = 200

⇒ x = 30.

∴ Volume of water absorbed = 4x = 120 ml.

Hence, option (d).

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**6. XAT 2018 QADI | Modern Math - Probability**

A coin of radius 3 cm is randomly dropped on a square floor full of square shaped tiles of side 10 cm each. What is the probability that the coin will land completely within a tile? In other words, the coin should not cross the edge of any tile.

- A.
0.91

- B.
0.5

- C.
0.49

- D.
0.36

- E.
0.16

Answer: Option E

**Explanation** :

Let’s consider one of the tiles with side 10 cm. For the coin to land completely within the tile, its centre should fall anywhere inside the square of side 4 cm.

∴ Favourable area = 4 × 4 = 16

Total area available = 10 × 10 = 100

∴ Required probability = 0.16

Hence, option (e).

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**7. XAT 2018 QADI | Geometry - Mensuration**

It takes 2 liters to paint the surface of a solid sphere. If this solid sphere is sliced into 4 identical pieces, how many liters will be required to paint all the surfaces of these 4 pieces.

- A.
2.2 liters

- B.
2.5 liters

- C.
3 liters

- D.
4 liters

- E.
None of the above

Answer: Option D

**Explanation** :

Total outer surface area of a sphere of radius r = 4πr^{2}

When the sphere is cut into 4 identical pieces, each piece will expose πr^{2}/2 + πr^{2}/2 = πr^{2} area more.

∴ Now, 4πr^{2} additional area needs to be painted which will require 2 more liters of paint.

∴ 4 litres of paint will be needed.

Hence, option (d).

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**8. XAT 2018 QADI | Arithmetic - Time, Speed & Distance**

Every day a person walks at a constant speed, V_{1} for 30 minutes. On a particular day, after walking for 10 minutes at V_{1}, he rested for 5 minutes. He finished the remaining distance of his regular walk at a constant speed, V_{2}, in another 30 minutes. On that day, find the ratio of V_{2} and his average speed (i.e., total distance covered /total time taken including resting time).

- A.
1 : 1

- B.
1 : 2

- C.
2 : 3

- D.
2 : 1

- E.
None of the above

Answer: Option A

**Explanation** :

The person usually walks at V_{1} for 30 minutes.

∴ Distance travelled = 30V_{1}.

Now the person travels for 10 minutes. Distance remaining after this = 30V_{1} – 10V_{1} = 20V_{1}

This remaining distance is covered by him at V_{2} in 30 minutes.

⇒ V_{2} = 20V_{1}/30 = 2V_{1}/3

∴ His average speed for the entire journey = 30V_{1}/45 = 2V_{1}/3

Now,

V_{2} : Average speed = 2V_{1}/3 : 2V_{1}/3 = 1 : 1

Hence, option (a).

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**9. XAT 2018 QADI | Geometry - Trigonometry**

A boat, stationed at the North of a lighthouse, is making an angle of 30° with the top of the lighthouse. Simultaneously, another boat, stationed at the East of the same lighthouse, is making an angle of 45° with the top of the lighthouse. What will be the shortest distance between these two boats? The height of the lighthouse is 300 feet. Assume both the boats are of negligible dimensions.

- A.
300 feet

- B.
600√3 feet

- C.
300√3 feet

- D.
600 feet

- E.
None of the above

Answer: Option D

**Explanation** :

Let LM be the lighthouse and B_{1} and B_{2} be the positions of the two boats.

In ∆LMB_{1},

Tan 30° = LM/MB_{1}

⇒ MB_{1} = LM√3 = 300√3

Also, in ∆LMB_{2},

Tan 45° = LM/MB_{2}

⇒ MB_{2} = LM = 300

In ∆MB_{1}B_{2},

(B_{1}B_{2})^{2} = (MB_{1})^{2} + (MB_{2})^{2}

⇒ (B_{1}B_{2})^{2} = (300√3)^{2} + (300)^{2}

⇒ (B_{1}B_{2})^{2} = 300^{2} × [(√3)^{2} + (1)^{2}]

⇒ (B_{1}B_{2})^{2} = 300^{2} × 4

⇒ B_{1}B_{2} = 300 × 2 = 600

Hence, option (d).

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**10. XAT 2018 QADI | Algebra - Quadratic Equations**

Two different quadratic equations have a common root. Let the three unique roots of the two equations be A, B and C - all of them are positive integers. If (A + B + C) = 41 and the product of the roots of one of the equations is 35, which of the following options is definitely correct?

- A.
The common root is 29.

- B.
The smallest among the roots is 1.

- C.
One of the roots is 5.

- D.
Product of the roots of the other equation is 5.

- E.
All of the above are possible, but none are definitely correct.

Answer: Option C

**Explanation** :

A, B and C are positive integers.

Let the common root is A.

∴ A × B = 35

This is possible when (A, B) is (1, 35) or (35, 1) or (5, 7) or (7, 5)

C = 41 – A - B

For each of these 4 possibilities value of C = 5 or 5 or 29 or 29

In each of these 4 possibilities, one of the roots is definitely 5.

Hence, option (c).

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**11. XAT 2018 QADI | Algebra - Progressions**

An antique store has a collection of eight clocks. At a particular moment, the displayed times on seven of the eight clocks were as follows: 1:55 pm, 2:03 pm, 2:11 pm, 2:24 pm, 2:45 pm, 3:19 pm and 4:14 pm. If the displayed times of all eight clocks form a mathematical series, then what was the displayed time on the remaining clock?

- A.
1:53 pm

- B.
1:58 pm

- C.
2:18 pm

- D.
3:08 pm

- E.
5:08 pm

Answer: Option B

**Explanation** :

Observe the following times.

Thus, we see that “5” is missing in the first difference. Therefore, the missing time is 5 minutes before 2.03, that is, 1.58.

Hence, option (b).

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**12. XAT 2018 QADI | Arithmetic - Ratio, Proportion & Variation**

The number of boys in a school was 30 more than the number of girls. Subsequently, a few more girls joined the same school. Consequently, the ratio of boys and girls became 3:5. Find the minimum number of girls, who joined subsequently.

- A.
31

- B.
51

- C.
52

- D.
55

- E.
Solution not possible

Answer: Option C

**Explanation** :

Let the number of girls in the school initially be ‘g’.

∴ Number of boys = g + 30.

If ‘x’ new girls who joined the school.

⇒ (g + 30)/(g + x) = 3/5

∴ 5g + 150 = 3g + 3x

∴ x = 2g/3 + 50

For x to be minimum g should be minimum. Also, g should be a multiple of 3 for x to be an integer.

⇒ Minimum value of g = 3

∴ The minimum value of x = 52.

Hence, option (c).

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**13. XAT 2018 QADI | Arithmetic - Time, Speed & Distance**

A girl travels along a straight line, from point A to B at a constant speed, V_{1} meters/sec for T seconds. Next, she travels from point B to C along a straight line, at a constant speed of V_{2} meters/sec for another T seconds. BC makes an angle 105° with AB. If CA makes an angle 30° with BC, how much time will she take to travel back from point C to A at a constant speed of V_{2} meters/sec, if she travels along a straight line from C to A?

- A.
0.5(√3 - 1)T

- B.
T

- C.
0.5(√3 + 1)T

- D.
√3T

- E.
None of the above

Answer: Option C

**Explanation** :

Refer to the diagram below.

Distance AB = V_{1}T and distance BC = V_{2}T

Also, ∠CAB = 180 – 30 – 105 = 45°

Let BD be drawn perpendicular to AC.

In ΔBDA,

BD = AB sin 45° = V_{1}T/√2

Also, AD = AB cos 45° = V_{1}T/√2

In ΔBDC,

BD = BC sin 30° = V_{2}T/2

Also, DC = BC cos 30° = √3V_{2}T/2

Now, V_{1}T/√2 = V_{2}T/2

⇒ V_{1} = V_{2}/√2

AC = AD + DC = V_{1}T/√2 + √3V_{2}T/2

= V_{2}T/2 + √3V_{2}T/2

= V_{2}T(1 + √3)/2

Time taken to travel AC at speed V_{2} = [V_{2}T(1 + √3)/2] ÷ V_{2} = T(1 + √3)/2 = 0.5(1 + √3)T

Hence, option (c).

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**14. XAT 2018 QADI | Miscellaneous**

In the final semester, an engineering college offers three elective courses and one mandatory course. A student has to register for exactly three courses: two electives and the mandatory course. The registration in three of the four courses is: 45, 55 and 70. What will be the number of students in the elective with the lowest registration?

- A.
35

- B.
40

- C.
42

- D.
45

- E.
Either B or D

Answer: Option E

**Explanation** :

There are two cases to be considered.

Since there is only one mandatory course, all students will register for this course.

Since each student registers for exactly 3 courses, it means that sum of registrations for all four courses will be = 3 × (number of students).

**Case 1**: The mandatory course is one of the courses with registrations 45, 55 or 70.

The only possibility is that the mandatory course has 70 registrations, i.e., total number of students is 70.

∴ The total number of students = 70 and the number of student-course combinations = 3 × 70 = 210

∴ 70 + 55 + 45 + X = 210

∴ X = 40

Therefore, the minimum registrations for a course = 40.

**Case 2**: The mandatory course is not one with registrations 45, 55 or 70.

If the total number of students is X, the mandatory course will have X registrations.

Now, X will definitely be greater than or equal to 70.

∴ Minimum number of registrations in this case = 45

Hence, option (e).

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**15. XAT 2018 QADI | Algebra - Number Theory**

X and Y are the digits at the unit's place of the numbers (408X) and (789Y) where X ≠ Y. However, the digits at the unit's place of the numbers (408X)^{63} and (789Y)^{85} are the same. What will be the possible value(s) of (X + Y)?

Example: If M = 3 then the digit at unit's place of the number (2M) is 3 (as the number is 23) and the digit at unit's place of the number (2M)^{2 }is 9 (as 23^{2} is 529).

- A.
9

- B.
10

- C.
11

- D.
12

- E.
None of the above

Answer: Option B

**Explanation** :

For various powers the units digit of a number always a cycle of 4 terms.

We need unit’s digits of (X)63 and (Y)85 to be same

Unit’s digit of X63 will be same as unit’s digit of X3

Similarly, unit’s digit of Y85 is same as unit’s digit of Y1.

∴ unit’s digits of (X)3 should be same as units digit of Y.

This is possible (from the table) when (X, Y) = (2, 8) or (8, 2) or (7, 3) or (3, 7).

In any of these cases X + Y = 10.

Hence, option (b).

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**16. XAT 2018 QADI | Algebra - Inequalities & Modulus**

If 2 ≤ |x – 1|×|y + 3| ≤ 5 and both x and y are negative integers, find the number of possible combinations of x and y.

- A.
4

- B.
5

- C.
6

- D.
8

- E.
10

Answer: Option E

**Explanation** :

Since 2 ≤ |x – 1| × |y + 3| ≤ 5, the value of |x – 1| × |y + 3| can be 2, 3, 4 or 5. Thus, we have the following cases.

**Case 1**: |x – 1| × |y + 3| = 2

⇒ |x – 1| = 2 and |y + 3| = 1

⇒ x = -1 and y = -4 or -2

Note: |x - 1| cannot be equal to 1 since x has to be a negative number.

∴ 2 possibilities.

**Case 2**: |x – 1| × |y + 3| = 3

⇒ |x – 1| = 3 and |y + 3| = 1

⇒ x = -2 and y = -4 or -2

∴ 2 possibilities.

**Case 3**: |x – 1| × |y + 3| = 4

⇒ |x – 1| = 4 and |y + 3| = 1

⇒ x = -3 and y = -4 or -2

⇒ |x – 1| = 2 and |y + 3| = 2

⇒ x = -1 and y = -1 or -5

∴ 4 possibilities.

**Case 3**: |x – 1| × |y + 3| = 5

⇒ |x – 1| = 5 and |y + 3| = 1

⇒ x = -4 and y = -4 or -2

∴ 2 possibilities.

∴ Total 2 + 2 + 4 + 2 = 10 possibilities.

Hence, option (e).

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**17. XAT 2018 QADI | Algebra - Number Theory**

David has an interesting habit of spending money. He spends exactly £X on the X^{th} day of a month. For example, he spends exactly £5 on the 5^{th} of any month. On a few days in a year, David noticed that his cumulative spending during the last 'four consecutive days' can be expressed as 2^{N} where N is a natural number. What can be the possible value(s) of N?

- A.
5

- B.
6

- C.
7

- D.
8

- E.
N can have more than one value

Answer: Option B

**Explanation** :

Sum of any four consecutive numbers = x + x + 1 + x + 2 + x + 3 = 4x + 3.

This sum is never divisible by 4. But 2^{N} is always divisible by 4 (for N ≥ 2).

∴ The four consecutive days do not fall in the same month. Let’s look at the various possibilities.

Now, there are 4 possibilities for number of days in a month.

The only possibility for cumulative spending during the last 'four consecutive days' to be expressed as 2^{N} = 64 = 2^{6}.

∴ N = 6

Hence, option (b).

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**18. XAT 2018 QADI | Geometry - Mensuration**

A cone of radius 4 cm with a slant height of 12 cm was sliced horizontally, resulting into a smaller cone (upper portion) and a frustum (lower portion). If the ratio of the curved surface area of the upper smaller cone and the lower frustum is 1:2, what will be the slant height of the frustum?

- A.
12 - √3

- B.
12 - 2√3

- C.
12 - 3√3

- D.
12 - 4√3

- E.
None of the above

Answer: Option D

**Explanation** :

Curved surface area of the original larger cone = π × Radius × Slant height = 48π

Therefore, the curved surface of the smaller cone = 1/3 × 48π = 16π

Now, the radius and slant of height of the smaller cone would be reduced in equal proportions from the larger cone.

Therefore, slant height of the frustum would be 1/√3 time the corresponding values of the larger cone.

⇒ Slant height of the smaller cone = 1/√3 × 12 = 4√3

∴ Slant height of the frustum = 12 - 4√3

Hence, option (d).

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**19. XAT 2018 QADI | Geometry - Circles**

Two circles with radius 2R and √2R intersect each other at points A and B. The centers of both the circles are on the same side of AB. O is the center of the bigger circle and ∠AOB is 60°. Find the area of the common region between two circles.

- A.
(√3 – π – 1)R

^{2} - B.
(√3 – π)R

^{2} - C.
(13π/6 + 1 - √3)R

^{2} - D.
(13π/6 + √3)R

^{2} - E.
None of the above

Answer: Option C

**Explanation** :

Let O is the center of the bigger circle. So, OA = 2R. Let P be the center of the smaller circle. So, PA = R√2.

In ∆OAM, ∠AOM = 30° and OA = 2R, so AM =OA × Sin 30 = 2R/2 = R. Also, OM = R√3

In ∆PAM, let ∠APM = θ and PA = R√2 and AM = R, so Sin θ = AM/PA = R/ R√2 = 1/√2, so θ = 45°

Therefore ∠APB = 2∠APM = 2 × 45 = 90°. Also, PM = PA × Cos 45 = R√2 × Cos 45 = R

Common area between the two circles is the area of the smaller circle minus the area of the shaded region.

Area of the shaded region = Area of quadrant APB + Area ∆OPA + Area ∆OPB – Area of sector AOB

Area of quadrant APB = π × (R√2)^{2}/4

Area ∆OPA + Area ∆OPB = Area ∆AOB – Area ∆PAB = [(1/2) × OM × AB) – [(1/2) × PM × AB]

OM = R√3 and AB = AM × 2 = R × 2 = 2R

So, Area ∆OPA + Area ∆OPB = [(1/2) × R√3 × 2R) – [(1/2) × R × 2R] = (√3 – 1)R^{2}

Area of sector AOB = (π/6)(2R)^{2}

Hence, Area of the shaded region = [π × (R√2)^{2}/4] + [(√3 – 1)R^{2}] +[(π/6)(2R)^{2}]

= [(√3 – 1)R^{2}] – [πR^{2}/6]

Common area between the two circles = Area of the smaller circle – Area of the shaded region

= π(R√2)^{2} – {[(√3 – 1)R^{2}] – [πR^{2}/6] = (13π/6) + 1 – √3) R^{2}

Hence, option (c).

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**20. XAT 2018 QADI | Data Sufficiency**

These statements provide data that may help answer the respective questions. Read the questions and the statements and determine if the data provided by the statements is sufficient or insufficient, on their own or together, to answer the questions. Accordingly, choose the appropriate option given below the questions.

A group of six friends noticed that the sum of their ages is the square of a prime number. What is the average age of the group?

**Statement I**: All members are between 50 and 85 years of age.

**Statement II**: The standard deviation of their ages is 4.6.

- A.
Statement I alone is sufficient to answer.

- B.
Statement II alone is sufficient to answer.

- C.
Either of the statement is sufficient to answer.

- D.
Both statements are required to answer.

- E.
Additional information is required.

Answer: Option A

**Explanation** :

**From statement I: **

Sum of their ages lies between (6 × 50) and (6 × 85), i.e., 300 and 510.

The only square of a prime number in this range is 19^{2} = 361.

∴ The average of their ages = 361/6.

Hence, the question can be answered using statement I alone.

**From statement II:**

Standard Deviation = 4.6 = $\frac{\sqrt{\sum ({x}_{i}-\overline{x})}}{6}$

Since there are two unknowns here, the question cannot be answered using this statement alone.

Hence, option (a).

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**21. XAT 2018 QADI | Data Sufficiency**

These statements provide data that may help answer the respective questions. Read the questions and the statements and determine if the data provided by the statements is sufficient or insufficient, on their own or together, to answer the questions. Accordingly, choose the appropriate option given below the questions.

Harry and Sunny have randomly picked 5 cards each from a pack of 10 cards, numbered from 1 to 10. Who has randomly picked the card with number 2 written on it?

**Statement I**: Sum of the numbers on the cards picked by Harry is 5 more than that of Sunny.

**Statement II**: One has exactly four even numbered cards while the other has exactly four odd numbered cards.

- A.
Statement I alone is sufficient to answer.

- B.
Statement II alone is sufficient to answer.

- C.
Either of the statement is sufficient to answer.

- D.
Both statements are required to answer.

- E.
Additional information is required.

Answer: Option D

**Explanation** :

**From statement I:**

The sum of all the 10 cards = 55.

The sum of the numbers picked by Harry is 30 and that picked by Sunny is 25.

A sum of 30 can be obtained if the cards picked are 2, 4, 6, 8 and 10.

A sum of 25 can be obtained if the cards picked are 2, 3, 4, 7 and 9.

∴ We cannot determine who picked card number 2.

**From statement II:**

We cannot ascertain who picked card number 2.

From statements I and II together:

Sum of 4 odd cards and 1 even card will always be even

∴ Harry should get 4 odd cards and 1 even card such that his sum is 30.

Case 1:

Harry picks – 1, 3, 10, 7, 9 (Sum is 30)

Sunny picks – 2, 4, 6, 8, 5 (Sum is 25)

In this case Sunny picks card 2.

Case 2:

Harry picks – 1, 8, 5, 7, 9 (Sum is 30)

Sunny picks – 2, 4, 6, 3, 10 (Sum is 25)

In this case also Sunny picks card 2.

Case 3:

Harry picks – 6, 3, 5, 7, 9 (Sum is 30)

Sunny picks – 2, 4, 1, 8, 10 (Sum is 25)

In this case also Sunny picks card 2.

There will be no other possibility.

∴ Both statements together are required to answer the question.

Hence, option (d).

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**Answer the next 3 questions based on the information given below.**

Six teams are playing in a hockey tournament where each team is playing against every other team exactly once. At an intermediate stage, the status is as follows;

Note:

- The team that scores more goals than it concedes wins the match, while if both the teams score the same no. of goals, the match is declared drawn.
- In a match played between Team X and Team Y, if team X scores 1 and concedes none, then the score line would read. Team X-Team Y (1-0).

**22. XAT 2018 QADI | DI - Games & Tournaments**

Which of the following matches are yet to be played?

- A.
Team A – Team B and Team C – Team D

- B.
Team C – Team D and Team E – Team F

- C.
Team E – Team F and Team B – Team D

- D.
Team C – Team D and Team A – Team E

- E.
Team A – Team B and Team E – Team F

Answer: Option E

**Explanation** :

Team A wins both its matches and the only teams to lose a match are E and F.

∴ A played and won against E and F.

D drew both its matches. D would draw its matches against B (1 - 1) and C (0 - 0).

Since team B drew 1-1 with team D, team B must have won the other game 4-0.

If B won (4 - 0) against E, then E would’ve won its other match (1 – 0). But E has to lose both its matches.

∴ B won (4 - 0) against F and F lost its match against A (0 - 3).

Now, since A won against F (3 - 0) hence it won its match against E by (2 - 1)

Now since E lost its match against A by (1 - 2), E would’ve lost its other match by (0 - 2) against C.

[TBP = To be played]

Hence, option (e).

Workspace:

**23. XAT 2018 QADI | DI - Games & Tournaments**

Which of the following score line is a possible outcome in the tournament?

- A.
Team A – Team D (1-0)

- B.
Team A – Team E (2-1)

- C.
Team B – Team D (1-0)

- D.
Team C – Team F (2-0)

- E.
None of the above

Answer: Option B

**Explanation** :

Consider the solution to first question of this set.

Team A won against E by (2 - 1).

Hence, option (b).

Workspace:

**24. XAT 2018 QADI | DI - Games & Tournaments**

Which of the following score line is not a possible outcome in the tournament?

- A.
Team A – Team F (4-0)

- B.
Team B – Team F (4-0)

- C.
Team C – Team D (0-0)

- D.
Team C – Team E (2-0)

- E.
All of the above options are possible

Answer: Option A

**Explanation** :

Consider the solution to first question of this set.

Team A won against F by (3 - 0).

Hence, option (a).

Workspace:

**Answer the next 3 questions based on the information given below.**

The graphs below represent the performance of four professors, across years, measured on four variables: Percentage of time spent on teaching, percentage of time spent on research, feedback (on a scale of 10, right-hand side) and number of publications (right-hand side). Assume that the cumulative time spent per year on research and teaching activities are same for all four professors and each of them taught only one course of 90 classroom hours.

**25. XAT 2018 QADI | DI - Tables & Graphs**

Which of the following, shows the maximum year to year percentage growth in feedback?

- A.
Professor Artihmetic during 2013-2014

- B.
Professor Algebra during 2015-2016

- C.
Professor Calculus during 2012-2013

- D.
Professor Calculus during 2014-2015

- E.
None of the above

Answer: Option C

**Explanation** :

Option (a): Growth is from 4 to 6.5, i.e., 62.5%.

Option (b): Growth is from 5 to 8, i.e., 60%.

Option (c): Growth is from 4 to 7, i.e., 75%.

Option (d): Growth is from 5.5 to 9, i.e., 63.6%.

Clearly, maximum growth is from 4 to 7 i.e., 75%.

Hence, option (c).

Workspace:

**26. XAT 2018 QADI | DI - Tables & Graphs**

Count the number of instances in which "annual decreasing efforts in research" is accompanied with "annual increase in feedback"?

- A.
5

- B.
7

- C.
9

- D.
11

- E.
None of the above

Answer: Option C

**Explanation** :

Prof. Arithmetic: 2 instances in years 2014 and 2016.

Prof. Algebra: 3 instances in years 2012, 2015 and 2016.

Prof. Geometry: 2 instances in years 2011 and 2014.

Prof. Calculus: 2 instances in years 2011 and 2015.

Therefore, there are 9 instances.

Hence, option (c).

Workspace:

**27. XAT 2018 QADI | DI - Tables & Graphs**

Research efficiency is the ratio of cumulative number of publication for a period of 3 years to the cumulative number of hours spent on research activity in those 3 years. Which of the following professors is the least efficient researcher for the period 2015 to 2017?

- A.
Professor Arithmetic

- B.
Professor Algebra

- C.
Professor Geometry

- D.
Professor Calculus

- E.
Cannot be determined as research is expressed in percentage

Answer: Option A

**Explanation** :

Since cumulative time spent per year on research and teaching activities is the same for all professors, we’ll directly take the percentages for calculations.

Research efficiency of Prof. Arithmetic = (2 + 3 + 3)/(70 + 50 + 60) = 2/45

Research efficiency of Prof. Algebra = (3 + 2 + 3)/(35 + 25 + 40) = 2/25

Research efficiency of Prof. Geometry = (3 + 1 + 4)/(75 + 25 + 20) = 2/30

Research efficiency of Prof. Calculus = (3 + 3 + 2)/(55 + 55 + 40) = 2/37.5

∴ The least efficient is Prof. Arithmetic.

Hence, option (a).

Workspace:

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