# XAT 2011 QA | Previous Year XAT Paper

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

The following graphs give annual data of Assets, Sales (as percentage of Assets) and Spending on Corporate Social Responsibility (CSR) (as percentage of Sales), of a company for the period 2004 – 2009.

**1. XAT 2011 QA | Algebra - Simple Equations**

In which year was the increase in spending on CSR, vis-à-vis the previous year, the maximum?

- A.
2005

- B.
2006

- C.
2007

- D.
2008

- E.
2009

Answer: Option B

**Explanation** :

The data can be tabulated as follows:

Hence, the increase in spending on CSR was the maximum in 2006.

Hence, option (b).

Workspace:

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

The following graphs give annual data of Assets, Sales (as percentage of Assets) and Spending on Corporate Social Responsibility (CSR) (as percentage of Sales), of a company for the period 2004 – 2009.

**2. XAT 2011 QA | Algebra - Simple Equations**

Of the years indicated below, in which year was the ratio of CSR/Assets the maximum?

- A.
2004

- B.
2005

- C.
2006

- D.
2007

- E.
2008

Answer: Option E

**Explanation** :

The ratio of CSR/Assets was the maximum in 2008 in the years given in the options.

Hence, option (e).

Workspace:

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

The following graphs give annual data of Assets, Sales (as percentage of Assets) and Spending on Corporate Social Responsibility (CSR) (as percentage of Sales), of a company for the period 2004 – 2009.

**3. XAT 2011 QA | Algebra - Simple Equations**

What was the maximum value of spending on CSR activities in the period 2004-2009?

- A.
Rs. 0.5 Crore

- B.
Rs 1.0 Crore

- C.
Rs. 2.0 Crore

- D.
Rs. 3.0 Crore

- E.
Rs. 4.0 Crore

Answer: Option D

**Explanation** :

Maximum spending on CSR was approximately 3 crore.

Hence, option (d).

Workspace:

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

**4. XAT 2011 QA | Algebra - Simple Equations**

In which year, did the spending on CSR (measured in Rs) decline, as compared to previous year?

- A.
2006

- B.
2007

- C.
2008

- D.
2009

- E.
None of above

Answer: Option E

**Explanation** :

It is clear that CSR Does not decreases over the specified period.

Hence, option (e).

Workspace:

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

Five years ago Maxam Glass Co. had estimated its staff requirements in the five levels in their organization as: Level-1: 55; Level-2: 65; Level-3: 225; Level-4: 255 & Level-5: 300. Over the years the company had recruited people based on ad-hoc requirements, in the process also selecting ex-defence service men and ex-policemen. The following graph shows actual staff strength at various levels as on date.

**5. XAT 2011 QA | Algebra - Simple Equations**

The level in which the Ex-Defence Servicemen are highest in percentage terms is:

- A.
1

- B.
2

- C.
3

- D.
4

- E.
5

Answer: Option D

**Explanation** :

The data can be tabulated as follows

From the table, the percentage of ex-defence servicemen is highest in level 4.

Hence, option (d).

Alternatively,

It is quite clear that the percentage of Ex-defence serviceman comes closer to 20 in level 4 and 5 only, while in other levels it is less than 15.

Now, as 25 × 330 > 60 × 130, the percentage of Ex-defence servicemen is highest in level 4.

Hence, option (d).

Workspace:

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

Five years ago Maxam Glass Co. had estimated its staff requirements in the five levels in their organization as: Level-1: 55; Level-2: 65; Level-3: 225; Level-4: 255 & Level-5: 300. Over the years the company had recruited people based on ad-hoc requirements, in the process also selecting ex-defence service men and ex-policemen. The following graph shows actual staff strength at various levels as on date.

**6. XAT 2011 QA | Algebra - Simple Equations**

If the company decides to abolish all vacant posts at all levels, which level would incur the highest reduction in percentage terms?

- A.
1

- B.
2

- C.
3

- D.
4

- E.
5

Answer: Option D

**Explanation** :

It is quite clear that level 4 will incur the highest reduction.

Hence, option (d).

Workspace:

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

Five years ago Maxam Glass Co. had estimated its staff requirements in the five levels in their organization as: Level-1: 55; Level-2: 65; Level-3: 225; Level-4: 255 & Level-5: 300. Over the years the company had recruited people based on ad-hoc requirements, in the process also selecting ex-defence service men and ex-policemen. The following graph shows actual staff strength at various levels as on date.

**7. XAT 2011 QA | Algebra - Simple Equations**

Among all levels, which level has the lowest representation of Ex-policemen?

- A.
1

- B.
2

- C.
3

- D.
4

- E.
5

Answer: Option C

**Explanation** :

It is quite clear that Level three represents the lowest number of Ex-policeman.

Hence, option (c).

Workspace:

**8. XAT 2011 QA | Algebra - Simple Equations**

In a locality, there are ten houses in a row. On a particular night a thief planned to steal from three houses of the locality. In how many ways can he plan such that no two of them are next to each other?

- A.
56

- B.
73

- C.
80

- D.
120

- E.
None of the above

Answer: Option A

**Explanation** :

Out of the 10 houses, we first place the 7 houses in which the thief will not steal. There are 8 spaces in between and at either ends of these 7 houses where we can place the three houses in which he plans to steal. Thus, the three houses can be selected in ^{8}C_{3} = 56 ways.

Hence, option (a).

Workspace:

**9. XAT 2011 QA | Algebra - Simple Equations**

If x = (9 + 4√5)^{48} = [x] + f

where [x] is defined as integral part of *x* and *f* is a faction, then *x(1 – f)* equals

- A.
1

- B.
Less than 1

- C.
More than 1

- D.
Between 1 and 2

- E.
None of the above

Answer: Option A

**Explanation** :

x = (9 + 4√5)^{48}

Let

y = (9 - 4√5)^{48}

Now,

(9 + 4√5)^{48} × (9 - 4√5)^{48} = (81 - 80)^{48} = 1 ...(i)

Also, (9 + 4√5)^{48} + (9 - 4√5)^{48}

= [^{48}C_{0} 9^{48} + ^{48}C_{1}9^{47} (4√5) + ^{48}C_{2}9^{46} (4√5)^{2} + .... + ^{48}C_{47} (9)(4√5)^{47} + ^{48}C_{48} (4√5)^{48}]

+ [ ^{48}C_{0} 9^{48} - ^{48}C_{1} 9^{47} (4√5) + ^{48}C_{2} 9^{46} (4√5)_{2} - ... - ^{48}C_{47} (9)(4√5)^{47} + ^{48}C_{48} (4√5)^{48}]

= 2[^{48}C_{0} 9^{48} + ^{48}C_{2} 9^{46} (4√5)^{2} + ... ^{48}C_{48} (4√5)^{48}]

∴ x + y = 2(k) = even

Now,

0 < 9 - 4√5 < 1

∴ 0 < (9 - 4√5)^{48} < 1

∴ 0 < y < 1 ...(ii)

Also, x = [x] + f, 0 < f < 1 … (iii)

∴ [x] + f + y is even

As [x] is an integer, f + y is an integer.

From (ii) and (iii)

0 < f + y < 2

∴ f + y = 1

Now, x (1 – f) = xy

But from (i), xy = 1

∴ x(1 – f) = 1

Hence, option (a).

Workspace:

**10. XAT 2011 QA | Algebra - Simple Equations**

Let *a _{n}* = 1 1 1 1 1 1 1….. 1, where 1 occurs

*n*number of times. Then,

i. *a*_{741} is not a prime.

ii. *a*_{534} is not a prime.

iii. *a*_{123} is not a prime.

iv.* a*_{77} is not a prime.

- A.
(i) is correct.

- B.
(i) and (ii) are correct.

- C.
(ii) and (iii) are correct.

- D.
All of them are correct.

- E.
None of them is correct.

Answer: Option D

**Explanation** :

*a*_{741} has 1 written 741 times.

∴ The sum of digits of *a*_{741} is divisible by 3.

∴ *a*_{741} is divisible by 3.

∴ *a*_{534} and *a*_{123} are also not prime by the same logic.

∴ (i), (ii), and (iii) are correct.

Option (4) can be safely assumed to be the correct option.

Hence, option (d).

Workspace:

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

Total income tax payable is obtained by adding two additional surcharges on calculated income tax.

- Education Cess: An additional surcharge called ‘Education Cess’ is levied at the late of 2% on the amount of income tax.
- Secondary and Higher Education Cess: An additional surcharge called ‘Secondary and Higher Education Cess` is levied at the rate of 1% on the amount of income tax.

**11. XAT 2011 QA | Algebra - Simple Equations**

Sangeeta is a young working lady. Towards the end of the financial year 2009-10, she found her total annual income to be Rs. 3, 37, 425/-. What % of her income is payable as income tax?

- A.
5.64

- B.
6.82

- C.
7.38

- D.
8.10

- E.
None of the above

Answer: Option A

**Explanation** :

Income = 3,37,425

Tax up to Rs. 1,90,000 = 0

Tax for the next Rs. 1,10,000 = 10% of 1,10,000

= Rs. 11000

Tax for the next Rs. 37425 = 20% of 37,425

= Rs. 7485

∴ Income tax = Rs. 18,485

Education cess = 18485 × 0.02

Secondary and higher education cess = 18485 × 0.01.

∴ Total income Tax payable = 18485 + 18485 × 0.02 + 18485 × 0.01 = Rs. 19039.55

∴ Required percentage = 19039.55 / 337425 × 100 ≈ 5.64%

Hence, option (a).

Workspace:

**12. XAT 2011 QA | Algebra - Simple Equations**

Mr. Madan observed his tax deduction at source, done by his employer, as Rs. 3,17,910/-. What was his total income (in Rs.) if he neither has to pay any additional tax nor is eligible for any refund?

- A.
13,48,835/-

- B.
14,45,522/-

- C.
14,47,169/-

- D.
15,70,718/-

- E.
None of the above

Answer: Option A

**Explanation** :

Let Madan’s income be (x + 5) lakhs.

Then his income tax = 0 × 160000 + 0.1 × 140000 + 0.2 × 200000 + 0.3 × x = (54000 + 0.3x)

Total tax payable including surcharges = 1.03 (54000 + 0.3x)

∴ 317910 = 1.03 (54000 + 0.3x)

∴ x ≈ 848835

∴ Madam’s income = 848833 + 500000 = Rs. 1348835

Hence, option (a).

Workspace:

**13. XAT 2011 QA | Algebra - Simple Equations**

A straight line through point P of a triangle PQR intersects the side QR at the point S and the circumcircle of the triangle PQR at the point T. lf S is not the centre of the circumcircle, then which of the following is true?

- A.
$\left(\frac{1}{PS}\right)$ + $\left(\frac{1}{ST}\right)$ < $\frac{2}{\sqrt{\left(QS\right)\left(QR\right)}}$

- B.
$\left(\frac{1}{PS}\right)$ + $\left(\frac{1}{ST}\right)$ < $\frac{4}{QR}$

- C.
$\left(\frac{1}{PS}\right)$ + $\left(\frac{1}{ST}\right)$ > $1\sqrt{\left(QS\right)\left(QR\right)}$

- D.
$\left(\frac{1}{PS}\right)$ + $\left(\frac{1}{ST}\right)$ > $\frac{4}{QR}$

- E.
None of the above

Answer: Option D

**Explanation** :

As S is not the circumcentre, PS ≠ ST and QS ≠ SR

∵ PT and QR are chords of the circle intersecting at S, PS × ST = QS × SR … (i)

We know that Arithmetic mean ≥ Geometric mean

$\therefore \frac{PS+ST}{2}$ ≥ $\sqrt{PS\times ST}$

But as PS ≠ ST,

$\frac{PS+ST}{2}$ > $\sqrt{PS\times ST}$

$\therefore \frac{PS+ST}{2}$ > $\sqrt{QS\times SR}$

$\therefore \frac{PS+ST}{2}$ > $2\sqrt{QS\times SR}$

$\therefore \frac{PS+ST}{PS\times ST}$ > $\frac{2\sqrt{QS+SR}}{QS\times SR}$

$\therefore \frac{1}{PS}$ + $\frac{1}{ST}$ > $\frac{2}{\sqrt{QS\times SR}}$ ...(i)

∴ Option 1 is false.

Also,

$\frac{QS+SR}{2}$ > $\sqrt{QS\times SR}$

$\therefore \frac{2}{QR}$ < $\frac{1}{\sqrt{QS\times SR}}$

$\therefore \frac{4}{QR}$ < $\frac{2}{\sqrt{QS\times SR}}$

$\therefore \frac{1}{PS}$ + $\frac{1}{ST}$ > $\frac{2}{\sqrt{QS\times SR}}$ > $\frac{4}{QR}$ ...From (i)

Hence, option (d).

**Note: **As the result is a general one, we can, without loss of generality, consider an equilateral triangle PQR with point S being the mid-point of QR and verify all options using numbers.

Workspace:

**14. XAT 2011 QA | Algebra - Simple Equations**

What is the maximum possible value of (21 Sin X + 72 Cos X)?

- A.
21

- B.
57

- C.
63

- D.
75

- E.
None of the above

Answer: Option D

**Explanation** :

Let f (X) = 21 sin X + 72 cos X

∴ f ’(X) = 21 cos X – 72 sin X

At f ’(X) = 0

21 cos X = 72 sin X

∴ tan X = $\frac{21}{72}$

Drawing the corresponding right triangle we have the following:

∴ f ‘’(X) = –21 sin X – 72 cos X

= - 21 × $\frac{7}{25}$ - 72 × $\frac{24}{25}$ < 0

∴ f (X) has a maximum at f ’(X) = 0

∴ Maximum value of f(x) = 21 × $\frac{21}{75}$ + 72 × $\frac{72}{75}$ = 75

Hence, option (d).

Workspace:

**15. XAT 2011 QA | Algebra - Simple Equations**

The scheduling officer for a local police department is trying to schedule additional patrol units in each of two neighbourhoods – southern and northern. She knows that on any given day, the probabilities of major crimes and minor crimes being committed in the northern neighbourhood were 0.418 and 0.612, respectively, and that the corresponding probabilities in the southern neighbourhood were 0.355 and 0.520. Assuming that all crime occur independent of each other and likewise that crime in the two neighbourhoods are independent of each other, what is the probability that no crime of either type is committed in either neighbourhood on any given day?

- A.
0.069

- B.
0.225

- C.
0.69

- D.
0.775

- E.
None of the above

Answer: Option A

**Explanation** :

Required probability = (1 – 0.418) (1 – 0.612) (1 – 0.355) (1 – 0.520) ≈ 0.069

Hence, option (a).

Workspace:

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

A man standing on a boat south of a light house observes his shadow to be 24 meters long, as measured at the sea level. On sailing 300 meters eastwards, he finds his shadow as 30 meters long, measured in a similar manner. The height of the man is 6 meters above sea level.

**16. XAT 2011 QA | Algebra - Simple Equations**

The height of the light house above the sea level is:

- A.
90 meters

- B.
94 meters

- C.
96 meters

- D.
100 meters

- E.
l06 meters

Answer: Option E

**Explanation** :

Let LM denote the light house of height h above the sea level.

Let KN denote the man and MN denote the south direction.

NS is the shadow of the man.

Then, KN = 6, NS = 24

Also, ∠ KNS = 90° and ∠ LMS = 90°.

By similarity of ΔLMS and ΔKNS,

$\frac{LM}{MS}$ = $\frac{6}{24}$ = $\frac{1}{4}$

∴ If LM = h, MS = 4h and MN = 4h – 24

The boat moves from N to P along the east.

∴ NP = 300

The man’s new position is AP.

∴ AP = 6, PB = 30

Δ APB ~ Δ LMB

$\therefore \frac{LM}{MB}$ = $\frac{6}{30}$ = $\frac{1}{5}$

∴ MP = 5*h* – 30

But MN^{2} + 300^{2} = MP^{2}

∴ 16(*h* – 6)^{2} + 300^{2} = 25(*h* – 6)^{2}

∴ 300^{2} = 25(*h* – 6)^{2} – 16(*h* – 6)^{2}

∴ 90000 = 9(*h* – 6)^{2}

∴ *h* = 106

Hence, option (e).

Workspace:

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

A man standing on a boat south of a light house observes his shadow to be 24 meters long, as measured at the sea level. On sailing 300 meters eastwards, he finds his shadow as 30 meters long, measured in a similar manner. The height of the man is 6 meters above sea level.

**17. XAT 2011 QA | Algebra - Simple Equations**

What is the horizontal distance of the man from the lighthouse in the second position?

- A.
300 meters

- B.
400 meters

- C.
500 meters

- D.
600 meters

- E.
None of the above

Answer: Option C

**Explanation** :

The horizontal distance of the man from the light house in the second position = 5h – 30 = 500 m

Hence, option (c).

Workspace:

**18. XAT 2011 QA | Algebra - Simple Equations**

A 25 ft long ladder is placed against the wall with its base 7 ft from the wall. The base of the ladder is drawn out so that the top comes down by half the distance that the base is drawn out. This distance is in the range:

- A.
(2, 7)

- B.
(5, 8)

- C.
(9, 10)

- D.
(3, 7)

- E.
None of the above

Answer: Option E

**Explanation** :

Let the base is drawn by x unit.

∴ Height of top of the ladder will decrease by $\frac{x}{2}$.

∴ By Pythagoras theorem,

$\therefore {\left(24-\frac{x}{2}\right)}^{2}$ + ${(7+x)}^{2}$ = 625

$\therefore \frac{{x}^{2}}{4}$ + ${x}^{2}$ - 24x + 14x = 0

$\therefore \frac{5{x}^{2}}{4}$ = 10x

∴ x = 8

Hence, option (e).

Workspace:

**19. XAT 2011 QA | Algebra - Simple Equations**

The domain of the function

*f*(*x*) = log_{7}{ log_{3}(log_{5}(20*x* – *x*^{2} – 91 ))} is:

- A.
(7, 13)

- B.
(8, 12)

- C.
(7, 12)

- D.
(12, 13)

- E.
None of the above

Answer: Option B

**Explanation** :

*f*(*x*) = log_{7}{ log_{3}(log_{5}(20*x* – *x*^{2} – 91 ))}

As log of negative numbers is not defined,

log_{3} (log_{5} (20x - x^{2} - 91)) > 0

∴ log_{5} (20x - x^{2} - 91)) > 3^{0}

∴ log_{5} (20x - x^{2} - 91) > 1

∴ 20x - x^{2} - 91 > 5

∴ 20x - x^{2} - 96 > 0

∴ x^{2} - 20x + 96 < 0

∴ (x - 12) (x - 8) < 0

∴ 8 < x < 12

Hence, option (b).

Workspace:

**20. XAT 2011 QA | Algebra - Simple Equations**

There are four machines in a factory. At exactly 8 pm, when the mechanic is about to leave the factory, he is informed that two of the four machines are not working properly. The mechanic is in a hurry, and decides that he will identify the two faulty machines before going home, and repair them next morning. It takes him twenty minutes to walk to the bus stop. The last bus leaves at 8:32 pm. If it takes six minutes to identify whether a machine is defective or not, and if he decides to check the machines at random, what is the probability that the mechanic will be able to catch the last bus?

- A.
0

- B.
$\frac{1}{6}$

- C.
$\frac{1}{4}$

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

- E.
1

Answer: Option D

**Explanation** :

If the mechanic wants to catch the bus, he will have 12 minutes to inspect the machines. As inspecting one machine takes 6 minutes, he will be able to identify the faulty machines if the first two machines he inspects are both faulty or both working properly.

Suppose A, B, C and D are the machines, and A and B are faulty. He can inspect these machines in ^{4}P_{4} = 24 ways. If he inspects A and B first, he will be able to catch the bus. If he inspects C and D first, he will know that A and B are faulty and still he will be able to catch the bus.

There are 4 ways in which he can inspect A and B first and 4 ways in which he can inspect C and D first.

∴ Probability that he will be able to catch the bus

= $\frac{8}{24}$ = $\frac{1}{3}$

Hence, option (d).

Workspace:

**21. XAT 2011 QA | Algebra - Simple Equations**

The football league of a certain country is played according to the following rules:

- Each team plays exactly one game against each of the other teams.
- The winning team of each game is awarded l point and the losing team gets 0 point.
- If a-match ends in a draw, both the teams get 1/2 point.

After the league was over, the teams were ranked according to the points that they earned at the end of the tournament. Analysis of the points table revealed the following:

- Exactly half of the points earned by each team were earned in games against the ten teams which finished at the bottom of the table.
- Each of the bottom ten teams earned half of their total points against the other nine teams in the bottom ten.
- How many teams participated in the league?

- A.
16

- B.
18

- C.
19

- D.
25

- E.
30

Answer: Option D

**Explanation** :

Let there be 10 + n teams.

One point is awarded for each match.

The bottom 10 teams will play ^{10}C_{2} = 45 matches against each other and 10 × n matches against the top n teams.

Similarly top n teams will play ^{n}C_{2} matches against each other & 10n matches against the bottom 10 teams.

Now, each of the bottom ten teams earned half of their total points against each other.

Hence they earned 45 points against the top n teams.

Hence, top n teams earned 10n – 45 points against the bottom 10 teams.

Now, half of the points earned by each team were earned in games against the ten teams which finished at the bottom 10.

Hence, top n team earned half of their points against the bottom 10 teams.

Hence, ^{n}C_{2} = 10n – 45

$\therefore \frac{n(n+1)}{2}$ = 10n - 45

∴ ${n}^{2}$ - 21n + 90 = 0

∴ (n - 6) (n - 15) = 0

∴ n = 6 or n = 15

Now, If n = 6 then top 6 teams will earn ^{6}C_{2} + 10n – 45 = 30 points, with an average of 5 points.

The bottom 10 will earn an average of 9 points.

But this is not possible, as the average score of the top n should be greater than the bottom 10.

Hence, n = 15

∴ Total number of teams = 10 + n = 10 + 15 = 25

Hence, option (d).

Workspace:

**22. XAT 2011 QA | Algebra - Simple Equations**

In a city, there is a circular park. There are four points of entry into the park, namely - P, Q, R and S. Three paths were constructed which connected the points PQ, RS, and PS. The length of the path PQ is 10 units, and the length of the path RS is 7 units. Later, the municipal corporation extended the paths PQ and RS past Q and R respectively, and they meet at a point T on the main road outside the park. The path from Q to T measures 8 units, and it was found that the angle PTS is 60 . Find the area (in square units) enclosed by the paths PT, TS, and PS.

- A.
36√3

- B.
54√3

- C.
72√3

- D.
90√3

- E.
None of the above

Answer: Option C

**Explanation** :

The following diagram represents the given scenario.

Let, TR = x

∴ 8 × 18 = x (7 + x)

∴ x^{2} + 7x – 144 = 0

∴ x = 9 or x = – 16

But x cannot be negative.

Hence, x = 9

A(∆TPS) = $\frac{1}{2}$ × 18 × 16 × sin 60°

= 72√3 sq.units

Hence, option (c).

Workspace:

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

From a group of 545 contenders, a party has to select a leader. Even after holding a series of meetings, the politicians and the general body failed to reach a consensus. It was then proposed that all 545 contenders be given a number from 1 to 545. Then they will be asked to stand on a podium in a circular arrangement, and counting would start from the contender numbered 1. The counting would be done in a clockwise fashion. The rule is that every alternate contender would be asked to step down as the counting continued, with the circle getting smaller and smaller, till only one person remains standing. Therefore the first person to be eliminated would be the contender numbered 2.

**23. XAT 2011 QA | Algebra - Simple Equations**

Which position should a contender choose if he has to be the leader?

- A.
3

- B.
67

- C.
195

- D.
323

- E.
451

Answer: Option B

**Explanation** :

Let, *f*(*n*)represent the position of winner when *n* persons are standing in a circle.

*f *(*n*) = 2*l* + 1

where, *n* = 2* ^{m}* +

*l*and 0 ≤

*l*< 2

^{m}Now, *n *= 545

∴ *n* = 512 + 33

∴ *n* = 2^{9} + 33

* l* = 33

∴ *f *(545) = 2 × 33 + 1 = 67

Hence, option (b).

Workspace:

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

From a group of 545 contenders, a party has to select a leader. Even after holding a series of meetings, the politicians and the general body failed to reach a consensus. It was then proposed that all 545 contenders be given a number from 1 to 545. Then they will be asked to stand on a podium in a circular arrangement, and counting would start from the contender numbered 1. The counting would be done in a clockwise fashion. The rule is that every alternate contender would be asked to step down as the counting continued, with the circle getting smaller and smaller, till only one person remains standing. Therefore the first person to be eliminated would be the contender numbered 2.

**24. XAT 2011 QA | Algebra - Simple Equations**

One of the contending politicians, Mr. Chanaya, was quite proficient in calculations and could correctly figure out the exact position. He was the last person remaining in the circle. Sensing foul play the politicians decided to repeat the game. However, this time, instead of removing every alternate person, they agreed on removing every 300th person from the circle. All other rules were kept intact. Mr. Chanaya did some quick calculations and found that for a group of 542 people the right position to become a leader would be 437. What is the right position for the whole group of 545 as per the modified rule?

- A.
3

- B.
194

- C.
249

- D.
437

- E.
543

Answer: Option C

**Explanation** :

Let *f *(*n, k*) represent the position of a winner when there are *n *people out of which every *k*^{th} person is eliminated.

We have,

*f *(*n, k*) = (*f *(*n* – 1*, k*) + *k*)mod* n*

Now *f *(542, 300) = 437

Hence,

*f* (543, 300) = (437 + 300) mod 543 = 194

*f *(544, 300) = (194 + 300) mod 544 = 494

*f *(545, 300) = (494 + 300) mod 545 = 249

∴ A contender at 249^{th} position will win the election.

Hence, option (c).

Workspace:

**25. XAT 2011 QA | Algebra - Simple Equations**

Little Pika who is five and half years old has just learnt addition. However, he does not know how to carry. For example, he can add 14 and 5, but he does not know how to add 14 and 7. How many pairs of consecutive integers between 1000 and 2000 (both 1000 and 2000 included) can Little Pika add?

- A.
150

- B.
155

- C.
156

- D.
258

- E.
None of the above

Answer: Option C

**Explanation** :

Let a, b and c be three digits belonging to the set {0, 1, 2, 3, 4}

The first number of the two consecutive numbers can be of following types

Hence total number of pairs = 125 + 25 + 5 + 1 = 156

Hence, option (c).

Workspace:

**26. XAT 2011 QA | Algebra - Simple Equations**

In the country of Twenty, there are exactly twenty cities, and there is exactly one direct road between any two cities. No two direct roads have an overlapping road segment. After the election dates are announced, candidates from their respective cities start visiting the other cities. Following are the rules that the election commission has laid down for the candidates:

- Each candidate must visit each of the other cities exactly once.
- Each candidate must use only the direct roads between two cities for going from one city to another.
- The candidate must return to his own city at the end of the campaign.
- No direct road between two cities would be used by more than one candidate.

The maximum possible number of candidates is

- A.
5

- B.
6

- C.
7

- D.
8

- E.
9

Answer: Option E

**Explanation** :

There are 20 cities connected to each other by roads.

Hence there will be $\frac{20\times 19}{2}$ = 190 roads.

Let the maximum number of candidates be n.

Hence the number of roads used by these n candidates will be 20n

Now,

20n ≤ 190

∴ greatest n = 9

Hence, option (e).

Workspace:

**27. XAT 2011 QA | Algebra - Simple Equations**

The micromanometer in a certain factory can measure the pressure inside the gas chamber from 1 unit to 999999 units. Lately this instrument has not been working properly. The problem with the instrument is that it always skips the digit 5 and moves directly from 4 to 6. What is the actual pressure inside the gas chamber if the micromanometer displays 0030l6?

- A.
2201

- B.
2202

- C.
2600

- D.
2960

- E.
None of the above options

Answer: Option A

**Explanation** :

Skipping the digit 5 converts the counting base system to 9 from 10.

Actual values of digits greater than 5 will change i.e. face value of 6 will be 5, 7 will be 6, and so on.

Hence, 3016 displayed by micromanometer is actually 3015 in base 9

Now, (3015)_{9 }= 3 × 9^{3 }+ 0 × 9^{2 }+ 1 × 9 + 5 = 2201

Hence, option (a).

Workspace:

**28. XAT 2011 QA | Algebra - Simple Equations**

Consider a square ABCD of side 60 cm. lt contains arcs BD and AC drawn with centres at A and D respectively. A circle is drawn such that it is tangent to side AB, and the arcs BD and AC. What is the radius of the circle?

- A.
9 cm

- B.
10 cm

- C.
12 cm

- D.
15 cm

- E.
None of the above

Answer: Option B

**Explanation** :

Let O be the centre of the smaller circle.

Let the small circle touch AB at P.* *OQ* *is perpendicular to AD*.*

Now,* *AO = 60 – *r* and &* *OP = *r*

Now,

QO^{2} = AP^{2} = (60 – *r*)^{2} – *r*^{2} … (i)

Now, in ΔDQO*,*

QO^{2} = DO^{2} – DQ^{2}

Now, DO = 60 + *r* and DQ = 60 – *r*

DQ* = *60* – r*

∴ OQ^{2} = (60 + *r*)^{2} – (60 – *r*)^{2} … (ii)

∴ (60 – *r*)^{2} +3600 – 120*r* = (60 + *r*)^{2}

From (i) and (ii),

∴ *r* = 10

Hence, option (b).

Workspace:

**29. XAT 2011 QA | Algebra - Simple Equations**

There are 240 second year students in a B-School. The Finance area offers 3 electives in the second year. These are Financial Derivatives, Behavioural Finance, and Security Analysis. Four students have taken all the three electives, and 48 students have taken Financial Derivatives. There are twice as many students who study Financial Derivatives and Security Analysis but not Behavioural Finance, as those who study both Financial Derivatives and Behavioural Finance but not Security Analysis, and 4 times as many who study all the three. 124 students study Security Analysis. There are 59 students who could not muster courage to take up any of these subjects. The group of students who study both Financial Derivatives and Security Analysis but not Behavioural Finance, is exactly the same as the group made up of students who study both Behavioural Finance and Security Analysis. How many students study Behavioural Finance only?

- A.
29

- B.
30

- C.
32

- D.
35

- E.
None of the above

Answer: Option A

**Explanation** :

There are 240 students in all, out of which 59 do not study any subject out of the given three.

∴ 181 study one or more of Financial Derivatives (FD), Behavioral Finance (BF) and Security Analysis (SA).

No. of students who study both FD and SA but not BF

= 4 × no. of students who study all three = 4 × 4 = 16

= 2 × no. of students who study both FD and BF but not SA

Also, no. of students who study both FD and SA but not BF = no. of students who study both BF and SA = 16

∴ We have the following:

∴ No. of students who study only BF

= 181 – 124 – 20 – 8 = 29

Hence, option (a).

Workspace:

**30. XAT 2011 QA | Algebra - Simple Equations**

ln a plane rectangular coordinate system, points L, M, N and O are represented by the coordinates (–5, 0), (1, –1), (0, 5), and (–1, 5) respectively. Consider a variable point P in the same plane. The minimum value of PL + PM + PN + PO is

- A.
1 + √37

- B.
5√2 + 2√10

- C.
√41 + √37

- D.
√41 + 1

- E.
None of the above

Answer: Option B

**Explanation** :

PL + PM will be minimum if P lies on LN. PM + PO will be minimum if P lies on OM.

∴ PL + PM + PN + PO will be minimum, only if P is the point of intersection of diagonals of quadrilateral LMNO.

Now, LN = $\sqrt{{(-5-0)}^{2}+{(0-5)}^{2}}$ = $5\sqrt{2}$

MO = $\sqrt{(1-{(-1)}^{2}+{(-1-5)}^{2}}$ = 2$\sqrt{10}$

∴ LN + MO = $5\sqrt{2}$ + 2$\sqrt{10}$

∴ PL + PM + PN + PO = 5$\sqrt{2}$ + 2$\sqrt{10}$

Hence, option (b).

Workspace:

**31. XAT 2011 QA | Algebra - Simple Equations**

ln a bank the account numbers are all 8 digit numbers, and they all start with the digit 2. So, an account number can be represented as 2*x*_{1}*x*_{2}*x*_{3}*x*_{4}*x*_{5}*x*_{6}*x*_{7}. An account number is considered to be a ‘magic’ number if *x*_{1}*x*_{2}*x*_{3} is exactly the same as *x*_{4}*x*_{5}*x*_{6}, or *x*_{5}*x*_{6}*x*_{7} or both. *x*_{i} can take values from 0 to 9, but 2 followed by seven 0s is not a valid account number. What is the maximum possible number of customers having a ‘magic’ account number?

- A.
9989

- B.
19980

- C.
19989

- D.
19999

- E.
19990

Answer: Option C

**Explanation** :

*x*_{1}*x*_{2}*x*_{3}*x*_{4}*x*_{5}*x*_{6}*x*_{7 }can be of the form *abcdabc* or *abcabcd*.

*abc* can be chosen in 10 × 10 × 10 = 1000 ways.

*d* can be chosen in 10 ways.

∴ *x*_{1}*x*_{2}*x*_{3}*x*_{4}*x*_{5}*x*_{6}*x*_{7 }can be chosen in 2 × 1000 × 10 = 20000 ways.

However, this includes the ways in which all of *a*, *b*, *c* and *d* are 0, 1, 2, 3, …, 9, twice.

We subtract these numbers to get 20000 – 10 = 19990

We also don’t want 2000000.

∴ The total number of ways = 19989.

Hence, option (c).

Workspace:

**32. XAT 2011 QA | Algebra - Simple Equations**

In a list of 7 integers, one integer, denoted as x is unknown. The other six integers are 20, 4, 10, 4, 8, and 4. If the mean, median, and mode of these seven integers are arranged in increasing order, they form an arithmetic progression. The sum of all possible values of x is

- A.
26

- B.
32

- C.
34

- D.
38

- E.
40

Answer: Option E

**Explanation** :

It can be observed that, irrespective of the value of x, mode of these numbers will be 4.

Now, the median of these numbers will depend on the value of x,

If x < 4 then the median of these seven numbers will be 4.

Now, as the mode is 4, the median cannot be 4.

(the question states that mean, median and mode are arranged in ascending order.)

Hence x cannot be less than 4.

Now,

If 4 < x ≤ 8

the median will be x & the mean will be,

$\frac{50+x}{7}$

Now $\frac{50+x}{7}$ > 7 and x ≤ 8

∴ 4, x and $\frac{50+x}{7}$ will form an AP only if $\frac{50+x}{7}$ > x

∴ $\frac{50+x}{7}$ - x = x - 4

∴ $\frac{50+x}{7}$ + 4 = 2x

∴ x = 6

Hence, x = 6 is a possible answer.

Now, if x > 8 then median will be 8 & mean will be

$\frac{50+x}{7}$

Now, if x > 8 then $\frac{50+x}{7}$ is greater than 8.

∴ Increasing order of mean, median and mode will be,

4, 8, $\frac{50+x}{7}$

Now, they are in A.P.

∴ 8 - 4 = $\frac{50+x}{7}$ - 8

∴ 12 = $\frac{50+x}{7}$

∴ x = 12 × 7 – 50

∴ x = 34

Hence, sum of all possible values of x = 6 + 34 = 40

Hence, option (e).

Workspace:

**33. XAT 2011 QA | Algebra - Simple Equations**

Prof. Bee noticed something peculiar while entering the quiz marks of his five students into a spreadsheet. The spreadsheet was programmed to calculate the average after each score was entered. Prof. Bee entered the marks in a random order and noticed that after each mark was entered, the average was always an integer. In ascending order, the marks of the students were 71, 76, 80, 82 and 91. What were the fourth and fifth marks that Prof. Bee entered?

- A.
71 and 82

- B.
71 and 76

- C.
71 and 80

- D.
76 and 80

- E.
91 and 80

Answer: Option C

**Explanation** :

It is convenient to solve this question by evaluating options.

Option 1:

If 71 and 82 are fourth & fifth numbers then, sum of first three numbers is,

76 + 80 + 91 = 247

But, 247 is not divisible by 3.

Hence, they cannot be fourth & fifth numbers.

∴ Option 1 is not possible.

Option 2:

Fourth and fifth numbers are 71 and 76.

Sum of first three numbers will be,

80 + 91 + 82 = 253

It is also not divisible by 3.

Hence, option (b) is also not possible.

Option 3:

Fourth and fifth numbers are 71 and 80.

Hence, sum of first three numbers will be

76 + 82 + 91 = 249, which is divisible by 3.

Hence, they can be fourth and fifth numbers.

Hence, option (c).

Workspace:

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