# XAT 2016 QADI | Previous Year Questions

Previous year paper questions for XAT 2016 QA

**1. XAT 2016 QADI | Algebra - Progressions**

Consider the set of numbers {1, 3, 3^{2}, 3^{3},…...,3^{100}}. The ratio of the last number and the sum of the remaining numbers is closest to:

- A.
1

- B.
2

- C.
3

- D.
50

- E.
99

Answer: Option B

**Explanation** :

The last number is 3^{100}.

Sum of the remaining numbers = S = 1 + 3 + 3^{2} + … + 3^{99} …(1)

This is a GP with a = 1; r = 3 and n = 100

∴ S = $\frac{1\times ({3}^{100}-1)}{3-1}$

∴ Required ratio = $\hspace{0.17em}\frac{{3}^{100}}{\frac{1\times ({3}^{100}-1)}{2}}$ = $\hspace{0.17em}\frac{2}{1-{\displaystyle \frac{1}{{3}^{100}}}}$

∵ 1/3^{100} is approximately zero, hence it can be ignored.

∴ Ratio = 2

Hence, option (b).

Workspace:

**2. XAT 2016 QADI | Data Sufficiency**

Anita, Biplove, Cheryl, Danish, Emily and Feroze compared their marks among themselves. Anita scored the highest marks, Biplove scored more than Danish. Cheryl scored more than at least two others and Emily had not scored the lowest.

**Statement I**: Exactly two members scored less than Cheryl.

**Statement II**: Emily and Feroze scored the same marks.

Which of the following statements would be sufficient to identify the one with the lowest marks?

- A.
Statement I only.

- B.
Statement II only.

- C.
Both Statement I and Statement II are required together.

- D.
Neither Statement I nor Statement II is sufficient.

- E.
Either Statement I or Statement II is sufficient.

Answer: Option B

**Explanation** :

Anita scored the highest marks, hence Anita cannot have the lowest marks.

Biplove scored more than Danish, hence Biploye cannot have the lowest marks.

Cheryl scored more than at least two others, hence Cheryl cannot have the lowest marks.

Emily had not scored the lowest.

∴ Either Danish or Feroze will have the lowest marks.

From Statement I: Cheryl’s has the 4th highest marks

Here, we cannot uniquely identify the person with least marks with only this information.

From Statement II: Emily and Feroze scored same marks.

∵ Emily doesn’t have lowest marks, Feroze also cannot have the lowest marks.

∴ Only Danish can have the lowest marks.

⇒ Statement II is sufficient to identify the person with the lowest marks

Hence, option (b).

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**3. XAT 2016 QADI | Arithmetic - Profit & Loss**

Rani bought more apples than oranges. She sells apples at Rs. 23 apiece and makes 15% profit. She sells oranges at Rs. 10 apiece and marks 25% profit. If she gets Rs. 653 after selling all the apples and oranges, find her profit percentage.

- A.
16.8%

- B.
17.4%

- C.
17.9%

- D.
18.5%

- E.
19.1%

Answer: Option B

**Explanation** :

Let the number of apples sold be ‘a’ and the number of oranges sold be ‘o’.

∴ Total Selling Price = 23a + 10b = 653

In R.H.S., there is a 3 in the unit’s place

∴ 23a + 10b should end with 3.

Now, unit’s digit of 10b will be ‘0’ hence units digit of 13a should be 3.

For this the values possible values of ‘a’ are 1, 11, 21, …

When a = 11, b = 40 [not possible since a should be more than b.]

When, a = 21, b = 17 which is in line with the condition of a > b

When a ≥ 31, b is negative, hence we will not consider those values.

∴ a = 21 and b = 17.

Now, the profit per apple is 15% and profit per orange is 25%

Cost price of each apple = 23/1.15 = Rs. 20

Cost price of each orange = 10/1.25 = Rs. 8

∴ Total Cost price = 20a + 8b = Rs.556

∴ Profit percent = ((653 - 556)/ 556) × 100 = 17.4%

Hence, option (b).

Workspace:

**4. XAT 2016 QADI | Venn Diagram**

In an amusement park along with the entry pass a visitor gets two of the three available rides (A, B and C) free. On a particular day 77 opted for ride A, 55 opted for B and 50 opted for C; 25 visitors opted for both A and C, 22 opted for both A and B, while no visitor opted for both B and C. 40 visitors did not opt for ride A and B, or both. How many visited with the entry pass on that day?

- A.
102

- B.
115

- C.
130

- D.
135

- E.
150

Answer: Option E

**Explanation** :

Let the Venn diagram be as shown in the figure,

No one can take all three rides, hence g = 0.

22 people take rides A and B,

∴ d = 22

25 people take rides A and C,

∴ f = 25

50 people take ride C,

∴ c = 50 – 25 = 25.

40 people don’t take A or B or both,

∴ 40 = c + h

⇒ h = 40 – 25 = 15

∴ Total number of people visiting the park = (77 + 55 + 50 – 25 – 22) + 15 = 150.

Hence, option (e).

Workspace:

**5. XAT 2016 QADI | Geometry - Quadrilaterals & Polygons**

∆ABC and ∆XYZ are equilateral triangles of 54 cm sides. All smaller triangles like ∆ANM, ∆OCP, ∆QPX etc. are also equilateral triangles. Find the area of the shape MNOPQRM.

- A.
243√3 sq. cm.

- B.
486√3 sq. cm.

- C.
729√3 sq. cm.

- D.
4374√3 sq. cm.

- E.
None of the above

Answer: Option A

**Explanation** :

Since all triangles are equilateral triangles

This is possible when YM = MN = NZ = 1/3 of YZ = 54/3 = 18 cm.

∴ Side of the regular hexagon MNOPQRM is 18 cm.

∴ Area of the hexagon = 6 × √3/4 × (18)^{2} = 486√3.

Hence, option (b).

Workspace:

**6. XAT 2016 QADI | Geometry - Triangles**

In the figure below, AB = AC = CD. If ∠ADB = 20°, what is the value of ∠BAD?

- A.
40°

- B.
60°

- C.
70°

- D.
120°

- E.
140°

Answer: Option D

**Explanation** :

In ∆ ACD, AC = CD

⇒ ∠CAD = ∠CDA = 20°

In ∆ ABC

∠ACB = ∠CAD + ∠CDA = (20 + 20)° = 40° …(Exterior angle theorem)

Also, AC = AB

⇒ m∠ABC = ∠ACB = 40°

∴ m∠BAD = 180 – 40 – 20 = 120°

Hence, option (d).

Workspace:

**7. XAT 2016 QADI | Geometry - Mensuration**

Akhtar plans to cover a rectangular floor of dimensions 9.5 meters and 11.5 meters using tiles. Two types of square shaped tiles are available in the market. A tile with side 1 meter costs Rs. 100 and a tile with side 0.5 meters costs Rs. 30. The tiles can be cut if required. What will be the minimum cost of covering the entire floor with tiles?

- A.
10930

- B.
10900

- C.
11000

- D.
10950

- E.
10430

Answer: Option A

**Explanation** :

Consider the diagram below:

Area of a 1 × 1 tile = 1 cm^{2}

Total area of A = 9 × 11 = 99 cm^{2}

∴ Number of 1 × 1 tiles required to cover A = 99/1 = 99 tiles.

Total Area of B + C = 9 × 0.5 + 11 × 0.5 = 20 × 0.5 = 10 cm^{2}

∴ Number of 1 × 1 tiles required to cover B and C = 10/1 = 10 tiles.

Now the remaining area i.e., D = 0.5 × 0.5 = 0.25 cm^{2}

It would be better to but a 0.5 × 0.5 tile to cover D.

∴ The minimum cost of covering entire floor with tiles

= (99 + 10) × 100 + 1 × 30 = 10930

Hence, option (a).

Workspace:

**8. XAT 2016 QADI | Arithmetic - Time, Speed & Distance**

Pradeep could either walk or drive to office. The time taken to walk to the office is 8 times the driving time. One day, his wife took the car making him walk to office. After walking 1 km, he reached a temple when his wife call to say that he can now take the car. Pradeep figure that continuing to walk to the office will take as long as walking back home and then driving to the office. Calculate the distance between the temple and the office.

- A.
1

- B.
7/3

- C.
9/7

- D.
16/7

- E.
16/9

Answer: Option C

**Explanation** :

Time taken to walk to the office is 8 times the driving time. It means that speed while driving is 8 times the speed while walking.

Let the walking speed be ‘s’ kmph

∴ Driving speed = 8s kmph

Let the distance between home and office be D kms.

Now, time taken to (D – 1) kms is same as time taken to walk 1 km and then drive D kms.

∴ $\frac{D-1}{s}$ = $\frac{1}{s}$ + $\frac{D}{8s}$

⇒ 8(D – 1) = 8 + D

⇒ 7D = 16

⇒ D = 16/7

∴ Distance between temple and office = D – 1 = 16/7 - 1 = 9/7

Hence, option (c).

Workspace:

**9. XAT 2016 QADI | Miscellaneous**

Each day on Planet M is 10 hours, each hour 60 minutes and each minute 40 seconds. The inhabitants of Planet M use 10 hour analog clock with an hour hand, a minute hand and a second hand. If one such clock shows 3 hours 42 minutes and 20 seconds in a mirror what will be the time in Planet M exactly after 5 minutes?

- A.
6 hours 18 minutes 20 seconds

- B.
6 hours 22 minutes 20 seconds

- C.
6 hours 23 minutes 20 seconds

- D.
7 hours 17 minutes 20 seconds

- E.
7 hours 23 minutes 20 seconds

Answer: Option B

**Explanation** :

For hours there are 10 division and 10 and 5 are opposite to each other.

When the mirror shows 3 hours 42 minutes and 20 seconds, the hour hand is between 3 and 4.

∴ The hour hand must be between 6 and 7

For minutes there are 60 division hence 60 and 30 minute mark will be opposite each other.

In the mirror the minute hand is between 42 minute mark and 43 minute mark.

∴ The minute hand must be between 17 minute mark and 18 minute mark

For seconds there are 40 divisions hence 40 and 20 second mark will be opposite each other.

In the mirror the second hand is at 20.

∴ The second hand will be at 20.

So, the actual time must be 6 hours 17 minutes and 20 seconds

So, after five minutes, the time will be 6 hours 22 minutes and 20 seconds.

**Alternately,**

The sum of the time visible in the mirror and the actual time should be equal to 10:00:00 hours.

10:00:00 – 03:42:20 = 06:17:20 hours

Thus, actual time is 6 hours 17 minutes and 20 seconds.

So, after 5 minutes the time will be 6 hours 22 minutes and 20 seconds

Hence, option (b).

Workspace:

**10. XAT 2016 QADI | Geometry - Mensuration**

A square piece of paper is folded three times along its diagonal to get an isosceles triangle whose equal sides are 10 cm. What is the area of the unfolder original piece of paper?

- A.
400 sq.cm.

- B.
800 sq.cm.

- C.
800√2 sq.cm.

- D.
1600 sq.cm.

- E.
Insufficient data to answer

Answer: Option A

**Explanation** :

When a square piece of paper is folded along its diagonal we will get a right isosceles triangle whose area becomes half.

Hence, after 3 such folds we get a right isosceles triangle, whose area is 1/8th the area of the original square.

Area of the folded triangle = ½ × 10 × 10 = 50.

∴ Area of the original square = 8 × 50 = 400 cm^{2}.

**Alternately,**

If the side of the isosceles triangle is 10 cm then we get the side of the square as 20 cm on unfolding it.

Hence, the area of square = 400 cm^{2}.

Hence, option (a).

Workspace:

**11. XAT 2016 QADI | Geometry - Triangles**

The difference between the area of the circumscribed circle and the area of the inscribed circle of an equilateral triangle is 2156 sq. cm. What is the area of the equilateral triangle?

- A.
686√3

- B.
1000

- C.
961√2

- D.
650√3

- E.
None of the above

Answer: Option A

**Explanation** :

Let ‘a’ be the side of the triangle.

Circum-radius of an equilateral triangle = a/√3

∴ Area of the circum-circle = π × (a/√3)^{2} = πa^{2}/3

In-radius of an equilateral triangle = a/2√3

∴ Area of the in-circle = π × (a/2√3)^{2} = πa^{2}/12

According to the question,

πa^{2}/3 - πa^{2}/12 = 2156

⇒ πa^{2}/4 = 2156

⇒ a^{2} = 2156 × 4 × 7/22 = 2744

∴ Area of the equilateral triangle = √3/4 × a^{2} = √3/4 × 2744 = 686√3.

Hence, option (a).

Workspace:

**12. XAT 2016 QADI | Geometry - Trigonometry | Data Sufficiency**

A person standing on the ground at point A saw an object at point B on the ground at a distance of 600 meters. The object started flying towards him at an angle of 30° with the ground. The person saw the object for the second time at point C flying at 30° angle with him. At point C, the object changed direction and continued flying upwards. The person saw the object for the third time when the object was directly above him. The object was flying at a constant speed of 10 kmph.

Find the angle at which the object was flying after the person saw it for the second time. You may use additional statement(s) if required.

Statement I: After changing direction the object took 3 more minutes than it had taken before.

Statement II: After changing direction the object travelled an additional 200√3 meters.

Which of the following is the correct option?

- A.
Statement I alone is sufficient to find the angle but statement II is not.

- B.
Statement II alone is sufficient to find the angle but statement I is not.

- C.
Statement I and Statement II are consistent with each other.

- D.
Statement I and Statement II are inconsistent with each other.

- E.
Neither Statement I nor Statement II is sufficient to find the angle.

Answer: Option D

**Explanation** :

From the given data,

m∠CAB = 30°; m∠CBA = 30° and AB = 600 m

∴ BC = AC = 200√3 m … [By sine rule]

Statement I: After changing the direction the object took 3 more minutes than it had taken before.

The object travels 200√3 m from B to C at 10 km/hr

Thus, in 3 minutes it can travel 500 m. Hence, the object travels a total of 500 + 200√3m from C.

Thus, we know the hypotenuse CD by which we can find out the angle.

Statement II: After changing directions, the object travels 200√3 m.

Since, the object travels the same distance as before, this can only happen if the object stays on the course as before without changing any direction.

Thus, we can clearly see that the two angles from the statements are inconsistent with each other.

Hence, option (d).

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**13. XAT 2016 QADI | Algebra - Functions & Graphs**

f is a function for which f(1) = 1 and f(x) = 2x + f(x – 1ؘ) for each natural number x ≥ 2. Find f(31).

- A.
869

- B.
929

- C.
951

- D.
991

- E.
None of the above

Answer: Option D

**Explanation** :

Given function can be rearranged as,

f(x) – f(x – 1) = 2x

⇒ f(2) – f(1) = 2 × 2

⇒ f(3) – f(2) = 2 × 3

⇒ f(4) – f(3) = 2 × 4

…

⇒ f(31) – f(30) = 2 × 31

Adding all these equations we get

f(31) – f(1) = 2 × (2 + 3 + 4 + … + 31)

f(31) – 1 = 2 × ((31×32)/2-1) = 990

⇒ f(31) = 991

**Alternately,**

From the given function ;

f(2) = 4 + f(1) = 5 = 22 + 1

f(3) = 6 + f(2) = 6 + 5 = 11 = 32 +2

f(4) = 8 + f(3) = 8 + 11 = 19 = 42 + 3

Hence we can see that f(x) = x2 + (x-1)

Hence f(31) = 312 + 30 = 991.

Hence, option (d).

Workspace:

**14. XAT 2016 QADI | Algebra - Number System**

Two number in the base system B are 2061_{B} and 601_{B}. The sum of these two numbers in decimal system is 432. Find the value of 1010B in decimal system.

- A.
110

- B.
120

- C.
130

- D.
140

- E.
150

Answer: Option C

**Explanation** :

(2061)_{B} = (1 + 6B + 2B3)_{10}

(601)_{B} = (1 + 6B2)_{10}

∴ 2B^{3} + 6B^{2} + 6B + 2 = 432

⇒ B^{3} + 3B^{2} + 3B + 1 = 216

⇒ (B + 1)^{3} = 216

⇒ B = 5

∴ 1010_{5} = (1 × 53 + 1 × 5)_{10} = 13010

Hence, option (c).

Workspace:

**15. XAT 2016 QADI | Arithmetic - Time & Work**

A water tank has M inlet pipes and N outlet pipes. An inlet pipe can fill the tank in 8 hours while an outlet pipe can empty the full tank in 12 hours. If all pipes are left open simultaneously, it takes 6 hours to fill the empty tank. What is the relationship between M and N?

- A.
M : N = 1 : 1

- B.
M : N = 2 : 1

- C.
M : N = 2 : 3

- D.
M : N = 3 : 2

- E.
None of the above

Answer: Option E

**Explanation** :

Let the total capacity of the tank be 24 units.

∴ The work done by an inlet pipe = 24/8 = 3 units/hour and

Work done by an outlet pipe = 24/12 = 2 units/hour

Now, work done by M inlet and N outlet pipes in an hour = 3M – 2N

It taken 6 hours for these pipes to fill the tank completely.

∴ 6(3M – 2N) = 24

∴ 3M – 2N = 4

Now when N = 1, M = 2.

Also when N = 4, M = 4

Similarly we get infinite values of N for which we get different values of M and the ratio M : N cannot be determined uniquely.

Hence, option (e).

Workspace:

**16. XAT 2016 QADI | Arithmetic - Profit & Loss**

Company ABC starts an educational program in collaboration with Institute XYZ. As per the agreement, ABC and XYZ will share profit in 60 : 40 ratio. The initial investment of Rs. 100,000 on infrastructure is borne entirely by ABC whereas the running cost of Rs. 400 per student is borne by XYZ. If each student pays Rs. 2000 for the program find the minimum number of students required to make the program profitable, assuming ABC wants to recover its investment in the very first year and the program has no seat limits.

- A.
63

- B.
84

- C.
105

- D.
157

- E.
167

Answer: Option C

**Explanation** :

Initial investment by ABC = 1,00,000

Let the total number of students be x.

Net profit from a student = 2000 – 400 = Rs. 1600

Net profit from ‘x’ students = 1600x

This will be divided between ABC and XYZ in the ratio of 60 : 40 i.e., 3 : 2.

∴ ABC will receive 1600x × 3/5

Hence, 1600x × 3/5 ≥ 1,00,000

⇒ x ≥ 104.2

∴ Minimum number of students should be 105.

Hence, option (c).

Workspace:

**17. XAT 2016 QADI | Algebra - Number Theory**

For two positive integers a and b, if (a + b)^{(a+b)} is divisible by 500, then the least possible value of a × b is:

- A.
^{8} - B.
9

- C.
10

- D.
12

- E.
None of the above

Answer: Option B

**Explanation** :

500 = 2^{2} × 5^{3}

For (a + b)^{(a+b)} to be divisible by 500, it should have power of 2 as well as 5 in it.

∴ a + b should have power of both 2 as well as 5.

The least such number is 10^{10}.

∴ a + b = 10.

∴ a × b is minimum when a = 9 and b = 1 (or vice-versa)

Hence, the least possible value is 9.

Hence, option (b).

Workspace:

**18. XAT 2016 QADI | Arithmetic - Time, Speed & Distance**

Study the figure below and answer the question:

Four persons walk from Point A to Point D following different routes. The one following ABCD takes 70 minutes. Another person takes 45 minutes following ABD. The third person takes 30 minutes following route ACD. The last person takes 65 minutes following route ACBD. IF all were to walk at the same speed, how long will it take to go from point B to point C?

- A.
10 mins

- B.
20 mins

- C.
30 mins

- D.
40 mins

- E.
Cannot be answered as the angles are unknown.

Answer: Option C

**Explanation** :

Let AB = a , BC = b , BD = c , AC = d and CD = e and let the speed of the persons be V.

From given information :

a + b + e = 70V …(1)

a +c = 45V … (2)

d + e = 30V … (3)

d + b + c = 65V … (4)

Adding (1) and (4) we get (a + c) + (d + e) + 2b = 135V

From (2) and (3) ; 45V + 30V + 2b = 135V

b = 30V.

b/V = 30.

Hence , it takes 30 min to go from B to C

Hence, option (c).

Workspace:

**19. XAT 2016 QADI | Algebra - Inequalities & Modulus**

a, b, c are integers, |a| ≠ |b| ≠ |c| and –10 ≤ a, b, c ≤ 10. What will be the maximum possible value of [abc – (a + b + c)]?

- A.
524

- B.
693

- C.
731

- D.
970

- E.
None of the above

Answer: Option C

**Explanation** :

abc – (a + b + c) = a(bc – 1) – b – c .

It is clear that the expression would be maximum when b and c are negative and their modulus are maximum possible.

∴ b = –10 , c = –9.

Now a has to be positive and maximum possible. It has to be 8.

Hence the expression becomes equal to 731.

Hence, option (c).

Workspace:

**20. XAT 2016 QADI | Geometry - Quadrilaterals & Polygons**

ABCD is a quadrilateral such that AD = 9 cm, BC = 13 cm and ⎿DAB = ⎿BCD = 90°. P and Q are two points on AB and CD respectively, such that DQ : BP = 1 : 2 and DQ is an integer. How many values can DQ take, for which the maximum possible area of the quadrilateral PBQD is 150 sq.cm?

- A.
14

- B.
12

- C.
10

- D.
9

- E.
8

Answer: Option D

**Explanation** :

Let DQ = x and PB = 2x.

From the diagram A(△BDP) = ½ × AD × PB = ½ × 9 × 2x.

Also A(△BDQ) = ½ × BC × DQ = ½ × 13 × x

A(ꐎBPDQ) = A(△BDQ) + A(△BDP) = ½ × 31 × x = 150

∴ x = 300 / 31

∴ x can have integer values from 1 to 9.

Hence , there are 9 values of DQ.

Hence, option (d).

Workspace:

**21. XAT 2016 QADI | Algebra - Number Theory**

If a, b and c are 3 consecutive integers between –10 to +10 (both inclusive), how many integer values are possible for the expression $\frac{{a}^{3}+{b}^{3}+{c}^{3}+3abc}{{(a+b+c)}^{2}}$?

- A.
0

- B.
1

- C.
2

- D.
3

- E.
4

Answer: Option C

**Explanation** :

If a, b and c are three consecutive numbers

⇒ a = b – 1 and c = b + 1

Substituting this in the expression we get the expression we get,

= $\frac{{(b-1)}^{3}+{b}^{3}+{(b+1)}^{3}+3(b-1)b(b+1)}{{(b-1+b+b+1)}^{2}}$

= $\frac{6{b}^{3}+3b}{{\left(3b\right)}^{2}}$

= $\frac{2{b}^{2}+1}{3b}$

Only for b = ±1 is the above expression an integer.

Hence, option (c).

Workspace:

**22. XAT 2016 QADI | Geometry - Circles**

In the figure below, two circular curves create 60° and 90° angles with their respective centres. If the length of the bottom curve Y is 10, find the length of the other curve.

- A.
15π/√2

- B.
20π√2/3

- C.
60π/√2

- D.
20π/3

- E.
15π

Answer: Option A

**Explanation** :

From the diagram, we can say that Y makes an angle of 60° and X an angle of 90°.

For Y ; ⅙ × 2πr = 10π.( r is radius of circle of which Y is a part).

∴ r = 30.

From the diagram ;

the radius of other circle = 15√2.

∴ Length of X = ¼× 2× π ×15√2 = 15π/√2.

Hence, option (a).

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**Answer the next 4 questions based on the information given below.**

This graph depicts the last eight years annual salaries (in Rs. lacs.) offered to student during campus placement. Every year 100 students go through placement process. However, at least one of them fails to get placed. The salaries of all unplaced students are marked zero and represented in the graph.

The bold line in the graph presents Mean salaries at various years.

**23. XAT 2016 QADI | DI - Tables & Graphs**

In which year were a maximum number of students offered salaries between Rs. 20 to Rs. 30 lacs (both inclusive)?

- A.
2008

- B.
2009

- C.
2010

- D.
2012

- E.
Cannot be determined

Answer: Option E

**Explanation** :

For a salary range we can get the highest and the lowest number of people in that salary range but cannot get the actual distribution of students.

In the year 2008, for 20-30 lac salary range, we have a complete salary range (76-100) inside it and some students of the range 51-75.

Hence, there will be at least 26 students and at max 49 students in 20-30 lac salary range.

In the year 2009 some students of the range 51-75 and some students of the range 75-100.

Hence, there will be minimum 1 student and at max 49 students in 20-30 lac salary range.

In the year 2010 some students of the range 26-50 range, all students from 51-75 range and some students of the range 76-100.

Hence, there will be minimum 27 students and at max 73 students in 20-30 lac salary range.

In the year 2012 we have a complete salary range 51-75 inside it and some students of the range (76-100).

Hence, there will be minimum 26 students and at max 49 students in 20-30 lac salary range.

Since we cannot determine the exact students and cannot infer a particular year. Hence the answer cannot be determined.

Hence, option (e).

Workspace:

**24. XAT 2016 QADI | DI - Tables & Graphs**

Identify the years in which the annual median salary is higher by at least 60% than the average salary of the preceding year?

- A.
2009, 2010

- B.
2012, 2014

- C.
2009, 2010, 2012

- D.
2009, 2012, 2014

- E.
2009, 2010, 2012, 2014

Answer: Option B

**Explanation** :

In the given chart the median salary is the junction of the second and third category from the bottom.

For 2009 the median salary is approximately 24 lakhs.

For 2008 the average salary is approximately 16 lakhs.

The median of 2009 is 50% more than the average of 2008.

Hence, 2009 does not satisfy the given criterion.

Therefore, all four options (a), (c), (d) and (e) are eliminated.

Hence, option (b).

Workspace:

**25. XAT 2016 QADI | DI - Tables & Graphs**

Identify the number of years in which the difference between the average salaries of the top 25% and the bottom 25% is more than Rs. 20 lacs:

- A.
0

- B.
1

- C.
2

- D.
3

- E.
None of the above

Answer: Option E

**Explanation** :

The number of years cannot be calculated with precision as the distribution pattern is unknown.

Hence, option (e).

Workspace:

**26. XAT 2016 QADI | DI - Tables & Graphs**

If the average salary is computed excluding students with no offers, in how many years will the new average salary be greater than the existing median salary? Refer the table below for number of students without offers.

- A.
3

- B.
4

- C.
5

- D.
6

- E.
Cannot be solved without additional information.

Answer: Option A

**Explanation** :

Mean salary of job offered students = $\frac{Averageofallstudents\times 100}{100-numberofstudentswithoutjoboffer}$

From the given chart and table, the following table is concluded

From the table it can be easily seen that the Mean salary after discarding the unemployed students is greater than the mean for years 2008, 2010 and 2013.

Hence, option (a).

Workspace:

**Answer the following questions based on the information given below.**

Study the data given in the table below and answer the question that follow:

All figures are in percentage

Based on survey of ‘shop types’ Kamath categorized Indian states into four geographical regions as shown in the table above. His boss felt that the categorization was inadequate since important labels were missing. Kamath argued that no further labels are required to interpret the data.

**27. XAT 2016 QADI | DI - Tables & Graphs**

A consultant observing the data made the following two inferences:

**Inference I**: The number of Grocers per-thousand-population is the highest in North India.

**Inference II**: The number of Cosmetic per-thousand-population is the highest in South India.

Which of following options is DEFINITELY correct?

- A.
Inference I alone is correct.

- B.
Inference II alone is correct.

- C.
Either of the inferences is correct

- D.
Neither of the inference is correct.

- E.
Inference I will be Correct only if inference II is correct.

Answer: Option D

**Explanation** :

Since there is no information available on the population of the different areas neither of the statements can be concluded from the given data.

Therefore, neither of the statements is correct.

Hence, option (d).

Workspace:

**28. XAT 2016 QADI | DI - Tables & Graphs**

The average size of Food Shops in East India was twice that of Food Shops in West India. Which of the following cannot be inferred from the above data?

- A.
As far as ‘Food Shops’ are concerned, customers in East India prefer spatial surroundings compared to customers in the West India.

- B.
As far as ‘Food Shops’ are concerned, Rentals are very high in West India compared to East India.

- C.
The ratio of customers buying from ‘Food Shops’ in East India to customers buying from ‘Food Shops’ in West India is 15.8 : 11.8.

- D.
There are 740 ‘Food Shops’ in West India.

- E.
There are 240 ‘Food Shops’ in South India.

Answer: Option C

**Explanation** :

There is no data available on the number or distribution of customers buying from “Food Shops” in East India or West India, while the statement C talks about the ratio of customers which cannot be inferred.

Hence, option (c).

Workspace:

**29. XAT 2016 QADI | DI - Tables & Graphs**

Bala collected the same data five years after Kamath, using the same categorization. His data is presented below:

Which of the following statements can DEFINITELY be concluded?

- A.
In the last four years the number of Electrical hardware shop types has increased in North India.

- B.
In the last four years the number of Grocers shop types has increased in South India.

- C.
For the last four years in All India the number of Chemists shop types has remained constant.

- D.
In the four years in East India the number of ‘other’ shop type has decreased.

- E.
As per the new survey conducted Pan Bidi shops in East India are next only to Grocers.

Answer: Option E

**Explanation** :

The two given tables provide information on the distribution percentage of the different types of shops in two different years.

However there is no information is available on the total number of shops in the two different years.

Hence, no comparison is not possible between the two years.

Options (a), (b), (c) and (d) talk about the increase or decrease in the number of shops with respect to each other.

Hence, they cannot be definitely concluded.

Statement E talks about the distribution in the same year as well as the same region.

Therefore, it can be concluded by comparing the percentage distribution.

Hence, option (e).

Workspace:

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