# XAT 2012 QA | Previous Year XAT Paper

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**1. XAT 2012 QA | Algebra - Number Theory**

Three Vice Presidents (VP) regularly visit the plant on different days. Due to labour unrest, VP (HR) regularly visits the plant after a gap of 2 days. VP (Operations) regularly visits the plant after a gap of 3 days. VP (Sales) regularly visits the plant after a gap of 5 days. The VPs do not deviate from their individual schedules. CEO of the company meets the VPs when all the three VPs come to the plant together. CEO is one leave from January 5th to January 28th, 2012. Last time CEO met the VPs on January 3, 2012. When is the next time CEO will meet all the VPs?

- A.
February 6, 2012

- B.
February 7, 2012

- C.
February 8, 2012

- D.
February 9, 2012

- E.
None of the above

Answer: Option C

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**Explanation** :

VP (HR) visits on every third day, VP (Operations) visits on every fourth day and VP (Sales) visits on every sixth day.

Hence, all of them will visit together on every twelfth day.

Now, all VPs visited together on January 3, 2012.

Hence, they will visit on, 15th January, 27th January, 8th February and so on.

Hence, option (c).

Workspace:

**2. XAT 2012 QA | Arithmetic - Average**

Ramesh analysed the monthly salary figures of five vice presidents of his company. All the salary figures are integers. The mean and the median salary figures are Rs. 5 lakh, and the only mode is Rs. 8 lakh. Which of the options below is the sum (in Rs. lakh) of the highest and the lowest salaries?

- A.
9

- B.
10

- C.
11

- D.
12

- E.
None of the above.

Answer: Option A

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**Explanation** :

Mean of the salaries of the five vice presidents is Rs. 5 lakhs. Hence, sum of the salaries of the five vice presidents = 25 lakhs.

Now, median of the salaries is Rs. 5 lakhs and 8 is the only mode.

Hence, the highest salary and second highest salaries are both 8 lakhs.

Hence, sum of two lowest salaries = 25 – (5 + 8 + 8) = 4 Lakhs.

As 8 is the only mode hence, the only combination of the lowest salaries is 1 lakhs and 3 lakhs.

Hence, lowest salary = Rs. 1 lakh.

Hence, the required sum = 8 + 1 = 9 lakhs.

Hence, option (a).

Workspace:

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

The following graphs shows the revenue (in $ million) of three companies in their initial six years of operations, in an economy which is characterized by a persistent inflation.

**3. XAT 2012 QA | DI - Tables & Graphs**

In 2010, which could be a valid statement about the revenues (adjusted for inflation) of these three companies?

- A.
Revenues of all three companies were equal.

- B.
Revenues of all three companies could be equal.

- C.
Revenue of Yahoo was definitely less than Facebook which was definitely less than Google.

- D.
Total of Yahoo and Facebook was definitely higher than that of Google.

- E.
None of the above.

Answer: Option E

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**Explanation** :

No data has been given in the question as to which year amongst the given years in the graph is 2010.

Hence this question is incorrect.

Note: By trial and error method, let us assume all the years one by one to be 2010.

Revenues of all the three companies are not equal in any of the 6 years and hence option A and B are incorrect.

Revenues of Google were less than that of Facebook and Yahoo in the first three years mentioned in the graph and hence option C is also incorrect.

Total of Yahoo and Facebook was less than Google in the fifth year shown in the graph and so option D is also wrong.

If this question needs to be compulsorily answered, then option E, none of the above would be the correct one.

However this doesn’t refute the fact that this question is an incorrect one as data about the regarding which year represents 2010 is not given.

Hence, option (e).

Workspace:

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

The following graphs shows the revenue (in $ million) of three companies in their initial six years of operations, in an economy which is characterized by a persistent inflation.

**4. XAT 2012 QA | DI - Tables & Graphs**

The difference in the average percentage increase in revenues, from 4^{th} to 6^{th} year, of Yahoo and Facebook is:

- A.
35%

- B.
40%

- C.
45%

- D.
50%

- E.
55%

Answer: Option A

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**Explanation** :

Revenue of Yahoo in 2006 = 1200

Revenue of Yahoo in 2004 = 250

So, percentage increase = (950/250) × 100 = 380

Average percentage increase = 380/2 = 190

Revenue of Facebook in 2006 = 2000

Revenue of Facebook in 2004 = 350

So, percentage increase = (1650/350) × 100 = 235.71

So, percentage difference in both the values =

[(235.71 – 190)/190] × 100 = 24.05

None of the options are less than 35%,so the question is incorrect.

However, “None of the above” is not one of the given options and hence we will have to take the option closest to 24.05 as the correct answer.

Hence, option (a).

Workspace:

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

The following graphs shows the revenue (in $ million) of three companies in their initial six years of operations, in an economy which is characterized by a persistent inflation.

**5. XAT 2012 QA | DI - Tables & Graphs**

What would have been Facebook’s revenue (in $ million) in its sixth year of operation if the company had matched Google’s percentage growth in revenues from the fifth to the sixth year?

Choose the option that is nearest to the answer.

- A.
1600

- B.
1700

- C.
1900

- D.
2100

- E.
None of the above.

Answer: Option A

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**Explanation** :

Percentage increase in Google’s growth between 5th and 6th year = (3200 – 1500)/1500 × 100 = 113.33

So, required value of Facebook revenue after 6th year = (750 × 213.33)/100 ≈ 1600

Hence, option (a).

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**6. XAT 2012 QA | Arithmetic - Percentage**

Tina, Mina, Gina, Lina and Bina are 5 sisters, aged in that order, with Tina being the eldest. Each of them had to carry a bucket of water from a well to their house. Their buckets’ capacities were proportional to their ages. While returning, equal amount of water got splashed out of their buckets. Who lost maximum amount of water as a percentage of the bucket capacity?

- A.
Tina

- B.
Mina

- C.
Gina

- D.
Lina

- E.
Bina

Answer: Option E

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**Explanation** :

Let T, M, G, L and B be the capacities of Tina’s, Mina’s, Gina’s, Lina’s and Bina’s bucket.

Hence, T > M > G > L > B

Assume that they spill x litres of water.

Hence, the percentages of the water spilled by them are;

(x/T) × 100, (x/M) × 100, (x/G) × 100, (x/L) × 100 and (x/B) × 100 respectively.

As, , T > M > G > L > B, this implies that x/ T < x/M < x/G < x/L < x/B

Hence, percentage of water spilled is highest for Bina.

Hence, option (e).

Workspace:

**7. XAT 2012 QA | Geometry - Mensuration**

Ram, a farmer, managed to grow shaped-watermelons inside glass cases of different shapes. The shapes he used were: a perfect cube, hemi-spherical, cuboid, cylindrical along with the normal spherical shaped watermelons. Thickness of the skin was same for all the shapes. Each of the glass cases was so designed that the total volume and the weight of the all the water-melons would be equal irrespective of the shape.

A customer wants to but water-melons for making juice, for which the skin of the water-melon has to be peeled off, and therefore is a waste. Which shape should the customer buy?

- A.
Cube

- B.
Hemi-sphere

- C.
Cuboid

- D.
Cylinder

- E.
Normal spherical

Answer: Option E

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**Explanation** :

Let V be the volume of water-melon, S be the total surface area and t be the thickness of the skin, then volume useable for Juice is, V – St

Hence, if total surface area is minimum, then useable volume of the water-melon will be highest.

Now, for equal volume, sphere has the least surface area.

Hence, option (e).

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**8. XAT 2012 QA | Arithmetic - Simple & Compound Interest**

A man borrows Rs. 6000 at 5% interest, on reducing balance, at the start of the year. If he repays Rs. 1200 at the end of each year, find the amount of loan outstanding, in Rs., at the beginning of the third year.

- A.
3162.75

- B.
4125.00

- C.
4155.00

- D.
5100.00

- E.
5355.00

Answer: Option C

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**Explanation** :

After one year, amount due = 6000 × 1.05 – 1200 = 5100

Hence, amount due after two years = 5100 × 1.05 – 1200 = 4155

Hence, option (c).

Workspace:

**9. XAT 2012 QA | Geometry - Mensuration**

A spherical metal of radius 10 cm is molten and made into 1000 smaller spheres of equal sizes. In this process the surface area of the metal is increased by:

- A.
1000 times

- B.
100 times

- C.
10 times

- D.
No change

- E.
None of the above

Answer: Option E

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**Explanation** :

Let *r* be the radius of the smaller sphere.

Now, the volume of the big sphere and the 1000 small spheres is same.

Hence, we have,

Hence, *r*^{3} = 1

Hence, *r* = 1

Now, total surface area of big sphere = 4 × π × 10^{2} = 400π

Total surface area of 1000 new sphere = 1000 × 4π × 1^{2} = 4000π

Hence, total surface area increases by, (4000π – 400π) / 400π = 9 times.

Hence, option (e).

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**10. XAT 2012 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

Gopal sells fruit juice mixture using orange juice and pineapple juice. Gopal prepares this mixture by drawing out a jug of orange juice from a 10 litre container filled with orange juice, and replacing it with pineapple juice. If Gopal draws out another jug of the resultant mixture and replaces it with pineapple juice, the container will have equal volumes of orange juice and pineapple juice. The volume of the jug in litres, is

- A.
2

- B.
>2 and ≤2.5

- C.
2.5

- D.
>2.5 and ≤3

- E.
≥3

Answer: Option D

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**Explanation** :

Assume that the volume of the jug is l liters.

Hence, after first replacement, the juice mixture contains l liters of pineapple juice.

When the juice mixture is drawn out for the second time using the jug, the amount of pineapple juice in the jug = l × ( l / 10)

This is replaced by l litres of pineapple juice.

Hence, amount of pineapple juice after two replacements = l + ( l – l × ( l / 10)) = 5

Hence, we get,

l^{2} – 20l + 50 = 0

Solving the above quadratic we get,

l = 5(2 - √2) or l = 5(2 + √2)

As, 5(2 + √2) > 10, l = 5(2 - √2) ≈ 2.92

Hence, option (d).

Workspace:

**11. XAT 2012 QA | Algebra - Simple Equations**

Nikhil’s mother asks him to buy 100 pieces of sweets worth Rs. 100/-. The sweet shop has 3 kinds of sweets, kajubarfi, gulabjamun and sandesh. Kajubarfi costs Rs. 10/- per piece, gulabjamun costs Rs. 3/- per piece and sandesh costs 50 paise per piece. If Nikhil decides to buy at least one sweet of each type, how many gulabjamuns should he buy?

- A.
1

- B.
2

- C.
3

- D.
4

- E.
5

Answer: Option A

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**Explanation** :

Let Nikhil buy a, b and c pieces of kajubarfi, gulabjamun and sandesh respectively.

Hence, we have,

a + b + c = 100, and … (i)

10a + 3b + 0.5c = 100 … (ii)

By, 2 × ii – i, we get,

19a + 5b = 100

Hence, a = (100 – 5b)/19

Now, (100 – 5b)/19 will be positive integer only if b = 1.

In that case, a = 5.

Hence, Nikhil must buy 1 gulabjamun.

Hence, option (a).

Workspace:

**12. XAT 2012 QA | Algebra - Simple Equations**

A potter asked his two sons to sell some pots in the market. The amount received for each pot was same as the number of pots sold. The two brothers spent the entire amount on some packets of potato chips and one packet of banana chips. One brother had the packet of banana chips along with some packets of potato chips, while the other brother just had potato chips. Each packet of potato chips costs Rs. 10/- and the packet of banana chips costs less than Rs. 10/-. The packets of chips were divided between the two brothers so each brother received equal number of packets. How much money should one brother give to the other to make the division financially equitable?

- A.
1

- B.
2

- C.
4

- D.
5

- E.
7

Answer: Option B

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**Explanation** :

Let the potter’s sons have *x* pots.

Hence, they received Rs. *x*^{2} after selling these pots.

As the price of one banana wafer packet is less than Rs. 10 hence, *x*^{2} will not be a multiple of 100.

Assume that they bought *n* packets of potato wafers.

Hence, total number of wafers packet = *n* + 1

Hence, each son gets (*n* + 1)/2 packets.

Hence, *n* is odd.

Let *b* be the price of banana wafers.

Hence, they have Rs. (*n* × 10 + *b*)

As *n* is odd, tens place of (*n* × 10 + *b*) is odd.

Now, each brother can have equal money if total amount earned by them is even.

Hence, *b* must be even.

Hence, we have the following condition,

*x*^{2} = 10 × *n* + *b* such that *b* is even and *n* is odd.

Hence, *x* is an even number.

Now, if unit’s digit is of *x* is 2 or 8, then tens place of *x*^{2} will be even.

This is violates our condition that tens digit of *x*^{2} is odd and hence it is not possible.

Hence, unit’s place digit of *x* is 4 or 6.

In either case, unit digit of *x*^{2} is 6.

Hence, *b* = 6.

Hence, the son having banana wafers owes Rs. ((*n* – 1) × 10/2 + 6) and the other son owes ((*n* + 1) × 10/2)

Hence, one of the son has Rs. ((*n* + 1) × 10/2) – ((*n* – 1) × 10/2 + 6) = Rs. 4 more than the other.

Hence, he must give Rs. 2 to the other to have financially equitable division.

Hence, option (b).

Workspace:

**13. XAT 2012 QA | Geometry - Triangles**

A city has a park shaped as a right angled triangle. The length of the longest side of this park is 80 m. The Mayor of the city wants to construct three paths from the corner point opposite to the longest side such that these three paths divide the longest side into four equal segments. Determine the sum of the squares of the lengths of the three paths.

- A.
4000 m

- B.
4800 m

- C.
5600 m

- D.
6400 m

- E.
7200 m

Answer: Option C

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**Explanation** :

Here D is midpoint of AC, E is the midpoint of AD and F is the midpoint of CD.

Hence, AE = ED = DF = CF = 20

Let AB = *c* and BC = *a*

Applying Apollonius theorem in ∆ABC, we get,

BD^{2} + 40^{2} = 1/2 × (*c*^{2} + *a*^{2}) = 1/2 × 80^{2}

BD^{2} = 40^{2} … (i)

Now, applying Apollonius theorem in ∆ABD, we get,

BE^{2} + 20^{2} = 1/2 × (*c*^{2} + BD^{2}) … (ii)

Similarly, applying Apollonius in ∆CDB, we get,

BF^{2} + 20^{2} = 1/2 × (*a*^{2} + BD^{2}) … (iii)

Adding (ii) and (iii), and substituting value from (i), we get,

BE^{2} + BF^{2} + 2 × 20^{2} = 1/2 (*a*^{2} + *c*^{2} + 2 × BD^{2}) = 1/2 (80^{2} + 2 × 40^{2}) = 3 × 40^{2}

Hence, BE^{2} + BF^{2} + BD^{2} = 3 × 40^{2} + 40^{2} – 2 × 20^{2} = 5600

Hence, option (c).

Workspace:

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

Ramya, based in Shanpur, took her car for a 400 km trip to Rampur. She maintained a log of the odometer readings and the amount of petrol she purchased at different petrol pumps at different prices (given below). Her car already had 10 litres of petrol at the start of the journey, and she first purchased petrol at the start of the journey, as given in table below, and she had 5 litres remaining at the end of the journey.

**14. XAT 2012 QA | LR - Mathematical Reasoning**

What has been the mileage (in kilometers per litre) of her car over the entire trip?

- A.
8.00

- B.
8.50

- C.
9.00

- D.
9.50

- E.
None of the above

Answer: Option A

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**Explanation** :

Ramya has 10 litres petrol in her car to start with.

She purchased 20 litres, 15 litres and 10 litres on three occasions.

She was left with 5 litres in her car at the end of the journey.

Hence, she utilized (10 + 20 + 15 + 10 – 5) = 50 litres.

She travelled (800 – 400) = 400 km.

Hence, the mileage of Ramya’s car = 400/50 = 8 km/litre

Hence, option (a).

Workspace:

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

Ramya, based in Shanpur, took her car for a 400 km trip to Rampur. She maintained a log of the odometer readings and the amount of petrol she purchased at different petrol pumps at different prices (given below). Her car already had 10 litres of petrol at the start of the journey, and she first purchased petrol at the start of the journey, as given in table below, and she had 5 litres remaining at the end of the journey.

**15. XAT 2012 QA | LR - Mathematical Reasoning**

Her car’s tank-capacity is 35 litres. Petrol costs Rs. 45/- litre in Rampur. What is the minimum amount of money she would need for purchasing petrol for the return trip from Rampur to Shanpur, using the same route? Assume that the mileage of the car remains unchanged throughout the route, and she did not use her car to travel around in Rampur.

- A.
1714

- B.
1724

- C.
1734

- D.
1744

- E.
Data insufficient to answer.

Answer: Option D

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**Explanation** :

Ramya’s car already has 5 litres in the tank.

Her car’s tank capacity is 35 litres.

She can fill a maximum of 30 litres more.

The petrol cost in Rampur is Rs. 45/litre

As the cost of petrol is lower at all the succeeding petrol pumps that come on the way and hence, to minimize the cost, she will fill enough petrol to reach the first pump i.e 150 km.

She already has 5 litres using which she can travel 40 km.

Hence, to travel 110 km, she will need 110/8 = 13.75 litres at the rate of 45 per litre.

On reaching the first petrol pump in the reverse journey, she will fill up enough petrol to reach the second petrol pump as the cost of petrol in the second pump is less than the cost of the first pump.

This distance is 50 km.

Ramya needs 50/8 = 6.25 litres at the rate of Rs. 40/litre.

For the rest of the journey (200 km) she will need 200/8 = 25 litres at the rate of Rs. 35/litre.

Hence the total cost is (13.75 × 45 + 6.25 × 40 + 25 × 35) = 1743.75

Hence, option (d).

Workspace:

**16. XAT 2012 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

A medical practitioner has created different potencies of a commonly used medicine by dissolving tables in water and using the resultant solution.

*Potency 1 solution: When 1 tablet is dissolved in 50 ml, the entire 50 ml is equivalent to one dose.Potency 2 solution: When 2 tablets are dissolved in 50 ml, the entire 50 ml of this solution is equivalent to 2 doses,… and so on.*

This way he can give fractions of tablets based on the intensity of infection and the age of the patient.

For particular patient, he administers 10 ml of potency 1, 15 ml of potency 2 and 30 ml of potency 4. The dosage administered to the patient is equivalent to

- A.
> 2 and ≤ 3 tablets

- B.
> 3 and ≤ 3.25 tablets

- C.
> 3.25 and ≤ 3.5 tablets

- D.
> 3.5 and ≤ 3.75 tablets

- E.
> 3.75 and ≤ 4 tablets

Answer: Option B

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**Explanation** :

50 ml of potency 1 solution is equivalent to 1 tablet; 50 ml of potency 2 solution is equivalent to 2 tablets and so on.

Hence, 10 ml of potency 1 solution is equivalent to 10/50 = 1/5 tablet.

Similarly, 15 ml of potency 2 and 30 ml of potency 4 corresponds to 15/50 × 2 and 30/50 × 4 tablets respectively.

Hence, the dosage administered is equivalent to

1/5 + 3/5 + 12/5 = 16/5 = 3.2 tablets

Hence, option (b).

Workspace:

**17. XAT 2012 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

Ram prepares solutions of alcohol in water according to customers’ needs. This morning Ram has prepared 27 litres of a 12% alcohol solution and kept it ready in a 27 litre delivery container to be shipped to the customer. Just before delivery, he finds out that the customer had asked for 27 litres of 21% alcohol solution. To prepare what the customer wants, Ram replaces a portion of 12% solution by 39% solution. How many litres of 12% solution are replaced?

- A.
5

- B.
9

- C.
10

- D.
12

- E.
15

Answer: Option B

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**Explanation** :

Let Ram replace x litres of 12% solution with 39% solution.

Hence, amount of alcohol in new solution = (27 – x) × 0.12 + x × 0.39 = 27 × 0.12 + x × 0.27

Now, new concentration of the solution is 21%, hence, volume of alcohol = 27 × .21

Hence, 27 × 0.12 + x × 0.27 = 27 × 0.21

Hence, 0.12 + x/100 = 0.21

Hence, x = 9.

Hence, option (b).

Workspace:

**18. XAT 2012 QA | Arithmetic - Time, Speed & Distance**

City Bus Corporation runs two buses from terminus A to terminus B, each bus making 5 round trips in a day. There are no stops in between. These buses ply back and forth on the same route at different but uniform speeds. Each morning the buses start at 7 AM from the respective terminuses. They meet for the first time at a distance of 7 km from terminus A. Their next meeting is at a distance of 4 km from terminus B, while travelling in opposite directions. Assuming that the time taken by the buses at the terminuses is negligibly small, and the cost of running a bus is Rs. 20 per km, find the daily cost of running the buses (in Rs.).

- A.
3200

- B.
4000

- C.
6400

- D.
6800

- E.
None of the above

Answer: Option D

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**Explanation** :

Let the distance between the two terminuses be x km.

Now, relative distance travel by the two buses before they meet for the first time = x km

Similarly, relative distance travelled by the two buses after first meet and before second meet = 2x

Now, bus originating from terminus A travels 7 km before the first meet.

Hence, it should travel 2 × 7 = 14 km after first meet and before second meet.

Hence, total distance travelled by that bus before second meet = 7 + 14 = 21 km

Now, second meet occurs at 4 km from the terminus B.

Hence, total distance travelled by the bus starting from terminus A (from the beginning till they meet for the second time) = x + 4 km

Hence, x + 4 = 21

Hence, x = 17 km

Hence, one bus travels 34 × 5 = 170 km a day.

Hence, cost of running one bus = 170 × 20 = Rs. 3400

Hence, cost of running two buses = 3400 × 2 = Rs. 6800

Hence, option (d).

Workspace:

**19. XAT 2012 QA | LR - Directions**

Shyam, a fertilizer salesman, sells directly to farmers. He visits two villages A and B. Shyam starts from A, and travels 50 meters to the East, then 50 meters North-East at exactly 45° to his earlier direction, and then another 50 meters East to reach village B. If the shortest distance between villages A and B is in the form of $a\sqrt{b+\sqrt{c}}$ meters, find the value of a + b + c.

- A.
52

- B.
54

- C.
58

- D.
59

- E.
None of the above

Answer: Option E

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**Explanation** :

Total distance travelled by Shyam is;

$50+50+\frac{50}{\sqrt{2}}$ km towards east, and $\frac{50}{\sqrt{2}}$ km towards north

Hence, smallest distance, say d, between village A and Village B is;

$d=\sqrt{({(100+25\sqrt{2})}^{2}+{\left(25\sqrt{2}\right)}^{2})}$

Hence, ${d}^{2}={(100+25\sqrt{2})}^{2}+{\left(25\sqrt{2}\right)}^{2}={a}^{2}(b+\sqrt{c})$

But, ${(100+25\sqrt{2})}^{2}+{\left(25\sqrt{2}\right)}^{2}=2500(5+2\sqrt{2})$

$=2500(5+\sqrt{8})$

Hence, *a*^{2} = 2500, *b* = 5 and *c* = 8

Hence, *a* + *b* + *c* = 50 + 5 + 8 = 63

Hence, option (e).

Workspace:

**20. XAT 2012 QA | Algebra - Simple Equations**

Three truck drivers, Amar, Akbar and Anthony stop at a road side eating joint. Amar orders 10 rotis, 4 plates of tadka, and a cup of tea. Akbar orders 7 rotis, 3 plates of tadka, and a cup of tea. Amar pays Rs. 80 for the meal and Akbar pays Rs. 60. Meanwhile, Anthony orders 5 rotis, 5 plates of tadka and 5 cups of tea. How much (in Rs.) will Anthony pay?

- A.
75

- B.
80

- C.
95

- D.
100

- E.
None of the above

Answer: Option D

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**Explanation** :

Let cost of a Roti, a plate of Tadka and a cup of tea be Rs. a, b and c respectively.

Hence, we have,

10a + 4b + c = 80, and … (I)

7a + 3b + c = 60 … (II)

Now, by II – I,

3a + b = 20 … (III)

Hence, by, (I) – 3(III), we have,

a + b + c = 20

Hence, 5a + 5b + 5c = 100

Hence, Anthony will pay Rs. 100.

Hence, option (d).

Workspace:

**21. XAT 2012 QA | Algebra - Simple Equations**

A computer program was tested 300 times before its release. The testing was done in three stages of 100 tests each. The software failed 15 times in Stage I, 12 times in Stage II, 8 times in Stage III, 6 times in both Stage I and Stage II, 7 times in both Stage II and Stage III, 4 times in both Stage I and Stage III, and 4 times in all the three stages. How many times the software failed in a single stage only?

- A.
10

- B.
13

- C.
15

- D.
17

- E.
35

Answer: Option B

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**Explanation** :

Assume that the software fails a, b and c times in a single stage, in two stages and in all stages respectively.

Hence, b + 3c = 6 + 7 + 4 = 17

But c = 4, hence, b = 5

Similarly, we have,

a + 2b + 3c = 15 + 12 + 8

Hence, a = 13

Hence, option (b).

Workspace:

**22. XAT 2012 QA | Geometry - Mensuration**

Suresh, who runs a bakery, uses a conical shaped equipment to write decorative labels (e.g., Happy Birthday etc.) using cream. The height of this equipment is 7 cm and the diameter of the base is 5 mm. A full charge of the equipment will write 330 words on an average. How many words can be written using three fifth of a litre of cream?

- A.
45090

- B.
45100

- C.
46000

- D.
43200

- E.
None of the above

Answer: Option E

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**Explanation** :

Volume of the equipment is;

1/3 × π × *r*^{2} × *h* = 1/3 × 22/7 × 7 × (0.25)^{2} = 11/6 cm^{3}

Now, 11/24 cm^{3} can write 330 words.

Hence, 1 cm^{3} can write,

330 × 24/11 = 720

We know that 1 cm^{3} = 1 ml

3/5^{th} of a litre is 600 ml which equals 600 cm^{3}

Hence, 600 cm^{3} will write, 720 × 600 = 432000 words

Hence, option (e).

Workspace:

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

The following pie chart shows the percentage distribution of runs scored by a batsman in a test innings.

**23. XAT 2012 QA | DI - Tables & Graphs**

If the batsman has scored a total of 306 runs, how many 4s and 6s did he hit?

- A.
31 and 3 respectively

- B.
32 and 2 respectively

- C.
32 and 3 respectively

- D.
33 and 1 respectively

- E.
33 and 2 respectively

Answer: Option E

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**Explanation** :

Total runs scored by the batsman = 306.

43.14% of 306 = (43.14/100) × 306 = 132 runs.

This is equal to 33 fours.

3.94% of 306 is equivalent to 4% of 300 runs which is equal to 12 runs.

This is equal to 2 sixes.

Hence, option (e).

Workspace:

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

The following pie chart shows the percentage distribution of runs scored by a batsman in a test innings.

**24. XAT 2012 QA | DI - Tables & Graphs**

If 5 of the dot balls had been hit for 4s, and if two of the shots for which the batsman scored 3 runs each had fetched him one run instead, what would have been the central angle of the sector corresponding to the percentage of runs scored in 4s?

- A.
160

- B.
163

- C.
165

- D.
167

- E.
170

Answer: Option E

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**Explanation** :

Considering the changes mentioned in the question, total runs = 306 + 20 – 4 = 322.

As calculated in the previous question, runs scored in fours = 132.

So, runs scored in fours = 132 + 20 = 152.

So, central angle = (152/322) × 360 = 169.93

So, as per the options, the answer should be 170.

Hence, option (e).

Workspace:

**25. XAT 2012 QA | Geometry - Mensuration**

Carpenter Rajesh has a circular piece of plywood of diameter 30 feet. He has cut out two disks of diameter 20 feet and 10 feet. What is the diameter of the largest disk that can be cut out from the remaining portion of the plywood piece?

- A.
>8.00 feet and ≤ 8.20 feet

- B.
> 8.21 feet and ≤ 8.40 feet

- C.
> 8.41 feet and ≤ 8.60 feet

- D.
> 8.61 feet and ≤ 8.80 feet

- E.
> 8.81 feet and ≤ 9.00 feet

Answer: Option C

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**Explanation** :

Consider the given diagram,

Here, O is the center of the largest circle, A and B are the centers of the circle having radius 10 and 5 feet respectively.

Let C be the center of the largest circle that can be cut from the remaining portion.

The circles having radius 10 and 5 cm touch each other at point D.

Let radius of the largest circle that can be cut from the remaining portion be *r*.

Now, one can easily observe that AO = OD = BD = 5 cm.

Now, AC = 10 + *r*, and BC = 5 + *r*, and OC = 15 – *r*

Let DC = *a*.

Now, applying Apollonius in triangle ADC, we have,

(10 + *r*)^{2} + *a*^{2} = 2((15 – *r*)^{2} + 5^{2})

i.e., *a*^{2} – *r*^{2} + 80*r* = 400 … (I)

Similarly, applying Apollonius theorem in triangle OCB, we get,

(15 – *r*)^{2} + (5 + *r*)^{2} = 2(*a*^{2} + 5^{2})

i.e. 2*a*^{2} – 2*r*^{2} + 20*r* = 200 … (II)

By, 2 × (I) – (II), we get,

140*r* = 600

Hence, *r* = 30/7

Hence, diameter of the required circle = 60/7 ≈ 8.57

Hence,** **option 3.

Workspace:

**26. XAT 2012 QA | LR - Selection & Distribution**

Lionel and Ronaldo had a discussion on the ages of Jose’s sons. Ronaldo made following statements about Jose’s sons:

- Jose has three sons.
- The sum of the ages of Jose’s sons is 13.
- The product of the ages of the sons is the same as the age of Lionel.
- Jose’s eldest son, Zizou weighs 32 kilos.
- The sum of the ages of the younger sons of Jose is 4.
- Jose has fathered a twin.
- Jose is not the father of a triplet.
- The LCM of the ages of Jose’s sons is more than the sum of their ages.

Which of the following combination gives information sufficient to determine the ages of Jose’s sons?

- A.
i, ii, iii and iv

- B.
i, ii, iv and vi

- C.
i, ii, iii and v

- D.
i, ii, v and vii

- E.
i, ii, v and vi

Answer: Option E

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**Explanation** :

Consider each option separately,

Consider option A:

In this case, we don’t know the Lionel’s age.

Hence, the only conclusion can be derived is;

Jose has 3 sons and sum of their age is 13.

Consider option B:

In this case also, we only know that Jose has 3 sons and sum of their ages is 13 and two of them are twin.

But it still doesn’t provide adequate information to calculate age of each son.

Consider option C:

As, age of Lionel is not known, we only know that the sum of ages of Jose’s three son is 13 out of which sum of the age of the younger two brother is 4.

By this we can calculate age of the Jose’s eldest son, but we cannot age of the remaining two children.

Consider option D:

In this case, statement 7 is redundant, as by statement 2 only we can guess that Jose is not the father of a triplet.

Hence this statement doesn’t provide adequate information to guess the ages of all the three children.

Consider option E:

By, i, ii and iii, we can conclude that the age of the Jose’s eldest son is 9 and sum of the other two children is 4.

Now, by statement vi, Jose has fathered twins, we can easily conclude that the younger two children are twins.

Hence, their ages are 2 and 2 respectively.

Hence, E is sufficient to answer.

Hence, option (e).

Workspace:

**27. XAT 2012 QA | Arithmetic - Ratio, Proportion & Variation**

Ram and Shyam form a partnership (with Shyam as working partner) and start a business by investing Rs. 4000 and Rs. 6000 respectively. The conditions of partnership were as follows:

- In case of profits till Rs. 200,000 per annum, profits would be shared in the radio of the invested capital.
- Profits from Rs. 200,001 till Rs. 400,000 Shyam would take 20% out of the profit, before the division of remaining profits, which will then be based on ratio of invested capital.
- Profits in excess of Rs. 400,000, Shyam would take 35% out of the profits beyond Rs. 400,000, before the division of remaining profits, which will then be based on ratio of invested capital.

If Shyam’s share in a particular year was Rs. 367000, which option indicates the total business profit (in Rs.) for that year?

- A.
520,000

- B.
530,000

- C.
540,000

- D.
550,000

- E.
None of the above

Answer: Option D

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**Explanation** :

For first Rs. 200000, Shyam gets, 6000/(4000 + 6000) × 100 = 60% of the profit.

For next Rs. 200000, he gets 20% + plus 60% of the remaining profit.

i.e. 20% + 80 × 0.6% = 68%

Similarly, for a profit margin greater than Rs. 400000, he will get, 35% + 65 × 0.6% = 74% of the profits beyond Rs. 400000

Now, for a profit of first Rs. 400000, Shyam will receive 200000 × (68 + 60)/100 = 256000

But Shyam earns a total profit of 367000.

Let total profit earned by them be Rs. 400000 + x.

Hence, Shyam received 367000 – 256000 = Rs. 110000 from Rs. x profit.

i.e. Rs. 110000 is 74% of x.

Hence, x = 110000/0.74 = 150000

Hence, total profit earned by them = 400000 + 150000 = Rs. 550000

Hence, option (d).

Workspace:

**28. XAT 2012 QA | Geometry - Mensuration**

A property dealer bought a rectangular piece of land at Rs. 1000/sq. ft. The length of the plot is less than twice its breadth. Due to its size, there were no buyers for the full plot. Hence he decided to sell it in smaller sized pieces as given below.

The largest square from one end was sold at Rs. 1200/sq. ft. From the remaining rectangle the largest square was sold at Rs. 1150/sq. ft.

Due to crash in the property prices, the dealer found it difficult to make profit from the sale of the remaining part of the land. If the ratio of the perimeter of the remaining land to the perimeter of the original land is 3 : 8, at what price (in Rs.) the remaining part of the land is to be sold such that the dealer makes an overall profit of 10%?

- A.
500/sq. ft.

- B.
550/sq. ft.

- C.
600/sq. ft.

- D.
650/sq. ft.

- E.
None of the above.

Answer: Option B

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**Explanation** :

Consider the following diagram.

Here, a is the length of the plot and b is the height of the plot.

Hence, from the diagram, perimeter of the remaining portion is;

2 × (2b – a + a – b) = 2b

Perimeter of the original land = 2(a + b)

Hence, we have, 2b : 2(a + b) = 3 : 8

Hence, b : a = 3 : 5

Now, without loss of generality we can assume that a = 5 and b = 3

Hence, area of the land = 15 square unit.

Hence, cost of the land = 1000 × 15

Now, selling price of small and big squares are, 9 × 1200 and 4 × 1150 respectively.

Let he sells the remaining land at Rs. x/sq. ft.

Hence, we have,

9 × 1200 + 4 × 1150 + x × 2 = 1100 × 15

Hence, x = 550

Hence, option (b).

Workspace:

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