XAT 2012 Question Paper

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 = $$3400 \times 2 = Rs. 6,800$$

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.