XAT 2012 Question Paper - QUANT

VP (HR) visits the plant after a gap of 2 days i.e every third day. Similarly, VP (Operations) visits the plant every fourth day and VP (Sales) visits the plant every sixth day. To find the number of days after which all the three will come together is the lowest common multiple of 3, 4 and 6 which is 12. So, every 12th day, all three VPs would come together. If their meeting was held on 3rd January, then their subsequent meetings would be held on 15th January, 27th January, 8th February, 20th February and so on. However, the CEO is not available till 28th January. So, their next meeting would be held on 8th February.
hence, option C is the correct answer.

Median = 5 , Mode = 8

Mean = 5, => Sum of 5 salaries = $$25$$

As mode is 8, it will occur 2 times but not 3 ($$\because$$ sum is 25)

Also, median is 5, the third salary is 5 and the first two are less than 5

=> Sum of third , fourth and fifth salary = 5 + 8 + 8 = 21

Sum of first two = 25 - 21 = 4

First and second salaries cannot be same ($$\because$$ mode is 8)

=> First and second salary = 1 and 3

$$\therefore$$ Sum (in lakh) of the highest and the lowest salaries = 1 + 8 = 9

We are not aware that which year, out of the given year is 2010. Hence, we can say that the data is insufficient. Therefore, option E is the best suitable answer. 

The revenue of Yahoo in 4th year = 250 millions.

The revenue of Yahoo in 6th year = 1200 millions.

Therefore, average percentage increase in Yahoo's revenue in a year = $$[(\dfrac{1200}{250})^{0.5}-1]*100$$ = 119%

The revenue of Facebook in year 4 = 300 millions

The revenue of Facebook in year 6 = 2000 millions

Therefore, average percentage increase in Facebook's revenue in a year = $$[(\dfrac{2000}{300})^{0.5}-1]*100$$ = 159%

Therefore, the required difference = 159 - 119 = 40%. Hence, option B is the correct answer.

Google’s percentage growth in revenues from the fifth to the sixth year = $$\dfrac{3200-1500}{1500}\times 100$$ = 113.33 percent. 

Therefore, Facebook's revenue in the 6th year = $$\dfrac{100+113.33}{100}\times 750$$ = 1600 millions. 

Therefore, option A is the correct answer.

Let the capacities of bucket of water carried by Tina, Mina, Gina, Lina and Bina respectively be $$W_T, W_M, W_G, W_L, W_B$$

It is given that : $$W_T > W_M > W_G > W_L > W_B$$

Let they spill $$x$$ litres of water from the bucket.

Thus, %age of water spilled by them respectively be

= $$\frac{x}{W_T} \times 100 $$, $$\frac{x}{W_M} \times 100$$, $$\frac{x}{W_G} \times 100$$, $$\frac{x}{W_L} \times 100$$, $$\frac{x}{W_B} \times 100$$

$$\because W_T > W_M > W_G > W_L > W_B$$

$$\therefore \frac{x}{W_T} \times 100$$ < $$ \frac{x}{W_M} \times 100$$ < $$\frac{x}{W_G} \times 100$$ < $$\frac{x}{W_L} \times 100$$ < $$\frac{x}{W_B} \times 100$$

Thus, Bina lost maximum amount of water as a percentage of the bucket capacity.

Let the volume of watermelon = $$V$$

Total surface area = $$S$$

Thickness of the skin = $$t$$

=> Volume usable for juice = $$V - S t$$

Hence, if total surface area is minimum then usable volume of the watermelon will be max.

For equal volume, sphere has the least surface area.

Ans - (E)

Amount man gets after 1 year

= $$6000 + (\frac{6000 \times 5 \times 1}{100}) - 1200$$

= $$6000 + 300 - 1200 = 5100$$

$$\therefore$$ Amount at the beginning of third year, i.e. after 2 years

= $$5100 + (\frac{5100 \times 5 \times 1}{100}) - 1200$$

= $$5100 + 255 - 1200 = 4155$$

Radius of larger sphere = $$R = 10$$ cm

Let radius of each of the smaller spheres = $$r$$ cm

=> $$\frac{4}{3} \pi R^3 = 1000 \times \frac{4}{3} \pi r^3$$

=> $$10^3 = 1000 r^3$$

=> $$r = \sqrt[3]{1} = 1$$ cm

Initial surface area of sphere = $$4 \pi R^2 = 4 \pi \times 100 = 400 \pi$$

Final surface area of 1000 spheres = $$1000 \times 4 \pi r^2 = 1000 \times 4 \pi = 4000 \pi$$

$$\therefore$$ Increase in surface area = $$4000 \pi - 400 \pi = 3600 \pi$$

=> $$\frac{3600 \pi}{400 \pi} = 9$$ times

Let volume of jug = $$v$$ litre

After first replacement, volume of orange juice = $$(10 - v)$$ litre

Volume of pineapple juice = $$v$$ litre

After second replacement, volume of orange juice remaining 

= $$(10 - v) - (\frac{10 - v}{10} v) = \frac{(10 - v)^2}{10}$$

Volume of pineapple juice remaining = $$v - \frac{v^2}{10} = \frac{v (10 - v)}{10}$$

Total volume of pineapple juice = $$\frac{v (10 - v)}{10} + v = \frac{20v - v^2}{10}$$

It is given that container has equal volumes of both juices.

=> $$\frac{(10 - v)^2}{10} = \frac{20v - v^2}{10}$$

$$100 + v^2 - 20v = 20v -v^2$$

$$2v^2 -40v +100=0$$

=> $$v^2 - 20v + 50 = 0$$

=> $$v = 17.07 , 2.93$$

$$\because$$ Container is 10 litres, => $$v \neq 17.07$$

$$\therefore v = 2.93$$ litres

Let Nikhil buy $$x, y$$ and $$z$$ pieces of kajubarfi, gulabjamun and sandesh respectively.     $$(x,y,z \geq 1)$$

=> $$x + y + z = 100$$ --------------Eqn(I)

Also, $$10x + 3y + \frac{1}{2} z = 100$$

=> $$20x + 6y + z = 200$$ ----------Eqn(II)

Subtracting eqn(I) from (II), we get :

=> $$19x + 5y = 100$$

=> $$y = \frac{100 - 19x}{5}$$

If $$x = 1$$, $$y$$ will not be natural. The only value of $$x$$ for natural $$y$$ is $$x = 5$$

=> $$y = \frac{100 - 95}{5} = 1$$

$$\therefore$$ Nikhil must buy 1 gulabjamun.

Let 'n' be the number of potato chips bought by the brothers. Also let 'x' be the cost price of a banana chips. ( x < 10) 

Total number of chips purchases = (n + 1). It is given that each brother has equal number of chips packets i.e. (n + 1) is an even number or we can say that 'n' is odd. 

Total amount spend by the brother on these chips packets = 10n + x. It is given that the amount received for each pot was same as the number of pots sold.

Hence, we can say that 10n + x is a perfect square. We can see that the tens place digit is an odd number. 

Perfect squares ending with an odd digit in the tens place = 16, 36, 196, 256 and so on {All (10a $$\pm$$ 4)$$^2$$ type numbers}

We can see that unit place is 6 in all cases and that will be the same as cost price of a banana chips packet. 

The difference between the amount with the two friends = Cost price of 1 potato chips packet - Cost price of 1 banana chips packet = 10 - 6 = 4

Hence,we can say that the brother, who has only chips packets with him, should given Rs.2 to the other brother so that they have the same amount with them.

Therefore, option B is the correct answer. 

To find : $$(BD)^2 + (BE)^2 + (BF)^2 = ?$$

AC = 80 m

AD = DE = EF = FC = 20

Let $$AB = a$$ and $$BC = b$$

In $$\triangle$$ ABC

$$(a)^2 + (b)^2 = (80)^2$$

Also, $$(BE) = 1/2 (AC) = 40 $$

=> $$BE = 40$$

Using Apollonius theorem in $$\triangle$$ ABE, as AD = DE

=> $$(AB)^2 + (BE)^2 = 2 [(BD)^2 + (AD)^2]$$

=> $$(BD)^2 + 20^2 = \frac{1}{2} (a^2 + 40^2)$$ --------Eqn(I)

Similarly, for $$\triangle$$ BEC, as EF = FC

=> $$(BE)^2 + (BC)^2 = 2 [(BF)^2 + (FC)^2]$$

=> $$(BF)^2 + 20^2 = \frac{1}{2} (b^2 + 40^2)$$ --------Eqn(II)

Adding eqns (I) & (II), we get :

=> $$(BD)^2 + (BF)^2 + 20^2 + 20^2$$ $$= \frac{1}{2} (a^2 + 40^2 + b^2 + 40^2)$$

=> $$(BD)^2 + (BF)^2 + 20^2 + 20^2$$ $$= \frac{1}{2} (80^2 + 40^2 + 40^2)$$

 => $$(BD)^2 + (BF)^2 = 4800 - 800 = 4000$$

$$\therefore$$ $$(BD)^2 + (BE)^2 + (BF)^2 = 4000 + 40^2$$

= $$4000 + 1600 = 5600$$

At the start, Ramya has 10 litres petrol. She purchased 20 litres, 15 litres and 10 litres.

At the end, she has left 5 litres of petrol.

Petrol used = $$(10 + 20 + 15 + 10 - 5) = 50$$ litres

Distance travelled = $$800 - 400 = 400$$ km

$$\therefore$$ Mileage of Ramya's car = $$\frac{400}{50} = 8$$ km/litre

Ramya's car has 5 litres in the tank. She can fill a maximum of 30 litres more as tank capacity is 35 litres.

The cost of petrol in Rampur is Rs. 45/litre. As the cost of petrol is lower at all the succeeding petrol pumps and hence to minimise the cost, she will fill enough petrol to reach first petrol pump i.e. 150 km.

Initially she can travel = $$8 \times 5 = 40$$ km using 5 litres of petrol

Hence to travel 110 km, she will need = $$\frac{110}{8} = 13.75$$ litre at the rate of Rs. 45/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 first pump and the distance is 50 km.

=> She needs = $$\frac{50}{8} = 6.25$$ litres at the rate of Rs. 40/litre

For the rest of the journey (200 km), she will need = $$\frac{200}{8} = 25$$ litres at the rate of Rs. 35/litre

$$\therefore$$ Total cost = $$(13.75 \times 45) + (6.25 \times 40) + (25 \times 35)$$

= $$618.75 + 250 + 875 = 1743.75 \approx Rs. 1744$$

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

Hence, 10 ml of potency 1 solution is equivalent to

= $$\frac{10}{50} = \frac{1}{5}$$

Similarly, 15 ml of potency 2 solution corresponds to = $$\frac{15}{50} \times 2 = \frac{3}{5}$$

and 30 ml of potency 4 solution corresponds to = $$\frac{30}{50} \times 4 = \frac{12}{5}$$

$$\therefore$$ Required dosage 

= $$\frac{1}{5} + \frac{3}{5} + \frac{12}{5}$$

= $$\frac{16}{5} = 3.2$$ tablets

Let Ram replaces $$x$$ litres of 12 % sol. with 39 % solution

Now, quality of 12 % solution in 27 litre = $$\frac{12}{100} \times 27$$

=> After replacing we have volume of 12 % solution

= $$(\frac{12}{100} \times 27) - (\frac{12 x}{100}) + (\frac{39 x}{100})$$

= $$\frac{324 + 27 x}{100}$$

This is equal to 27 litre of 21 % solution.

=> $$\frac{324 + 27 x}{100} = \frac{21}{100} \times 27$$

=> $$27x = 567 - 324 = 243$$

=> $$x = \frac{243}{27} = 9$$

Alternate Solution:

The mixture of two solutions 12% alcohol and 39% alcohol should yield 21% alcohol solution. 

Using allegation:  

The ratio = 18/9 =2:1

So the amount of 39% solution required = 27*1/(2+1) =9

Let the distance between two termini = $$P$$ km
Let the speed of the bus started from Terminus A be x and that of the bus started from terminus B be y.
The two buses met at a distance of 7km from Terminus A
Since the time of travel for both buses is the same.
$$\frac{7}{x}$$=$$\frac{P-7}{y}$$
$$\frac{x}{y}$$=$$\frac{7}{P-7}$$  ----------- Eq (1)
They met again at a distance of 4 km from terminus B.
Distance travelled by bus which started from Terminus A = P+4
Distance travelled by bus which started from Terminus A = 2P-4
So $$\frac{x}{y}$$ = $$\frac{P+4}{2P-4}$$   --Eg (2)
On solving Eq 1& 2, we get P=17 km
Each bus covers a distance of 17*2=34 km on a round trip.
Each bus makes 5 round trips in a day =34*5=170 km
Cost of running one bus = $$170 \times 20 = 3400$$
$$\therefore$$ Cost of running both buses = $$3400 \times 2 = Rs. 6,800$$

Shortest distance between A and B = $d = \sqrt{(AM)^2 + (BM)^2}$ ----------Eqn(1)

In $$\triangle$$ OPN

=> $$sin 45 = \frac{PN}{PO}$$

=> $$\frac{1}{\sqrt{2}} = \frac{PN}{50}$$

=> $$PN = \frac{50}{\sqrt{2}} = 25 \sqrt{2}$$

=> $$BM = PN = 25 \sqrt{2}$$

Again, $$tan 45 = \frac{PN}{ON}$$

=> $$ON = 25 \sqrt{2}$$

=> $$AM=AO\ +\ ON\ +\ NM\ =\ 50\ +\ 25\sqrt{2}\ +\ 50\ =\ 100\ +\ 25\sqrt{2}$$

Using eqn(I), we get :

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

=> $$d^2 = 10000 + 5000 \sqrt{2} + 1250 + 1250 $$ = $$12500 + 5000 \sqrt{2}$$

=> $$d^2 = 2500 (5 + 2 \sqrt{2}) = 2500 (5 + \sqrt{8})$$ -----------Eqn(II)

Also, it is given that : $$d = a\sqrt{b+\sqrt{c}}$$

=> $$d^2 = a^2 (b + \sqrt{c})$$ -----------Eqn(III)

Comparing, eqn(II) & (III), we get :

=> $$a^2 (b + \sqrt{c}) = 2500 (5 + \sqrt{8})$$

=> $$a = 50 , b = 5 , c = 8$$

$$\therefore a + b + c = 50 + 5 + 8 = 63$$

Let cost of 1 roti, 1 tadka and 1 tea be Rs. $$x, y, z$$ respectively.

Acc to ques,

=> $$10x + 4y + z = 80$$ <-- Equation 1

and $$7x + 3y + z = 60$$ <-- Equation 2

We have three equations and two unknowns. So let us try to find the value of $$x$$ and $$y$$ in terms of $$z$$.

Multiplying Equation 1 with 3 and subtracting 4 times (Equation 2) from it gives,

$$3*(10x+4y+z)-4*(7x+3y+z)=3*80-4*60$$

Or, $$2x-z =0$$

=> $$x = \frac{z}{2}$$

Substituting it in Equation 1, we get $$6z+4y=80$$ or $$y = 20 - \frac{3}{2} z$$

To find : $$5x + 5y + 5z = ?$$

= $$5 [\frac{z}{2} + (20 - \frac{3 z}{2}) + z]$$

= $$5 \times 20 = 100$$

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

Given : $$c = 4$$

To find : $$a = ?$$

Solution : $$a + 2b + 3c = 15 + 12 + 8$$

=> $$a + 2b + 3c = 35$$ -------------Eqn(I)

Also, $$b + 3c = 6 + 7 + 4$$

=> $$b + (3 \times 4) = 17$$

=> $$b = 17 - 12 = 5$$

Using eqn(I),

=> $$a + (2 \times 5) + (3 \times 4) = 35$$

=> $$a + 10 + 12 = 35$$

=> $$a = 35 - 22 = 13$$

Height = 7 cm and Radius = 0.25 cm

Volume of cone = $$\frac{1}{3} \pi r^2 h$$

= $$\frac{1}{3} \times \frac{22}{7} \times (0.25)^2 \times 7 = 0.458 cm^3$$

$$\because$$ $$0.458 cm^3$$ can write $$330$$ words

=> $$1 cm^3$$ can write = $$\frac{330}{0.458} = 720.05$$ words

Now, $$1$$ litre = $$1000 cm^3$$

=> $$\frac{3}{5}$$ litre = $$\frac{3}{5} \times 1000 = 600 cm^3$$

$$\therefore$$ $$600 cm^3$$ can write = $$600 \times 720.05$$

$$\approx 4,32,000$$ words

Total runs scored in 4's = $$\dfrac{43.16}{100}\times 306$$ = 132 

Therefore, total number of 4's scored by the batsman = $$\dfrac{132}{4}$$ = 33

Total runs scored in 6's = $$\dfrac{3.92}{100}\times 306$$ = 12 

Therefore, total number of 6's scored by the batsman = $$\dfrac{12}{6}$$ = 2

Hence, we can say that option E is the correct answer. 

Net increment in the number of runs due to 5 the dot balls resulting in 4s and two 3s resulting in two 1s = 4*5 - 2*2 = 16.

Therefore, the total number of runs scored by the batsman = 16 + 306 = 322 runs.

The number of runs scored by 4's = 132 + 20 = 152        {We know that batsman scored 132 runs from 4s alone}

The angle subtended by 4's = $$\dfrac{152}{322}\times 360$$ = 169.93 $$\approx$$ 170.

It is given that the 3rd disk also has the maximum diameter. This is possible only when all 3 discs touch each other externally. 

Also the sum of the diameter of the two disks is 10+20 = 30 feets. Hence, we can say that centres of these two disk lie on a diameter of original plywood.The figure can be drawn as shown below,

Here, in the diagram O is the centre of original disk. A and C are the centres of disks having radius 10 feet and 5 feet respectively. D is the touching point of circles with centre A and C.

Let 'x' be the radius of the disk that is to be cut out from the remaining part.

We can say that AO = OD = CD = 5 feet. Also, AB = 10+x, BC = 5+X, OB = 15-x and AC = 15 feet. 

Also, let 'y' be the length of BD.

By applying apollonius theorem in triangle ABD,

$$AB^2+BD^2=2(OB^2+OD^2)$$

$$(10+x)^2+y^2=2((15-x)^2+5^2)$$

$$x^2+20x+100+y^2=500-60x+2x^2$$

$$y^2 - x^2+80x=400$$ ... (1) 

Similarly, by applying apollonius theorem in triangle OBC,

$$OB^2 + BC^2 = 2(BD^2+DC^2)$$

$$(15-x)^2 + (5+x)^2 = 2(y^2+5^2)$$

$$2x^2-20x+250=2y^2+50$$

$$2x^2-2y^2-20x+200=0$$ ... (2) 

By equation (1) and (2) we can say that, 

$$160x-20x+200=800$$

$$x=600/140 = 30/7$$

Therefore, the diameter of the largest disk that can be cut out from the remaining portion of the plywood piece = 2*30/7 = 8.57 feet. Hence, option C is the correct answer.

Statement iii and iv are redundant. Thus, we can cancel out first three options.

From (i) , (ii) , (v), we have : 

Let A,B,C be the ages of Jose's sons in ascending order

and $$A + B + C = 13$$

and $$A + B = 4$$

=> $$C = 13 - 4 = 9$$

Now, from (vi) we get $$A = B$$

=> $$A = B = \frac{4}{2} = 2$$

$$\therefore$$ i, ii, v and vi are required to answer the question.

Ratio of profits earned by Ram : Shyam = 4000 : 6000

= 2 : 3

If profit < 2,00,000

% of profit earned by Shyam = $$\frac{3}{5} \times$$ 100 = 60%

If 2,00,000 < profit < 4,00,000, he gets 20 % and 60 % of the remaining profit.

% of profit earned by Shyam = 20% + .80 $$\times$$ 60% = 68%

If profit > 4,00,000

% of profit earned by Shyam = 35 % + .65 $$\times$$ 60% = 74%

Now, for first 2,00,000 profit earned by Shyam = $$\frac{60}{100} \times$$ 2,00,000 = Rs. 1,20,000

For second 2,00,000 profit earned by Shyam = $$\frac{68}{100} \times$$ 2,00,000 = Rs. 1,36,000

Let total profit earned by them = Rs. (4,00,000 + $$x$$)

=> From $$Rs. x$$ profit, Shyam received = 3,67,000 - 1,20,000 - 1,36,000 = Rs. 1,11,000

=> $$\frac{74}{100} \times x$$ = 1,11,000

=> $$x$$ = 1,11,000 $$\times \frac{100}{74}$$ = 1,50,000

$$\therefore$$ Total profit = 4,00,000 + 1,50,000 = Rs. 5,50,000

Let length = $$l$$ ft and breadth = $$b$$ ft

ABCD is original plot. First square AEFD is sold of side = DF = $$b$$, => FC = $$(l - b)$$

After that square EBHG is sold of side GH = $$(l - b)$$

=> HC = $$b - (l - b) = (2b - l)$$

Perimeter of remaining land GHCF = $$2 \times [(l - b) + (2b - l)]$$

= $$2b$$

Perimeter of original land ABCD = $$2 (l + b)$$

Acc to ques

=> $$\frac{2b}{2 (l + b)} = \frac{3}{8}$$

=> $$8b = 3l + 3b$$

=> $$5b = 3l$$

Only one combination is possible, i.e. $$l = 5$$ and $$b = 3$$     ($$\because$$ It is given that : $$l < 2b$$) 

=> Area of land = $$5 \times 3 = 15$$ sq. ft

=> Cost of land = $$15 \times 1000 = Rs. 15,000$$

For overall profit to be 10 %, S.P. = $$15,000 \times \frac{110}{100}$$

= $$Rs. 16,500$$

Side of AEFD = $$b = 3$$

=> S.P. of AEFD = $$3^2 \times 1200 = Rs. 10,800$$

Side of EBHG = $$(l - b) = 5 - 3 = 2$$

S.P. of EBHG = $$2^2 \times 1150 = Rs. 4,600$$

Length of remaining rectangular land GHCF = $$(l - b) = 5 - 3 = 2$$

Breadth = $$(2b - l) = 6 - 5 = 1$$

Let selling price per sq. ft of GHCF = $$Rs. x$$

S.P. of GHCF = $$2 \times 1 \times = Rs. 2x$$

$$\therefore$$ Total S.P.

=> $$10,800 + 4,600 + 2x = 16,500$$

=> $$2x = 16,500 - 15,400$$

=> $$x = \frac{1100}{2} = Rs. 550$$