XAT 2012

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 :

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

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

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

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

= $$4000 + 1600 = 5600$$