SNAP 2014

Let smaller number be $$x$$

The bigger number will be $$6x+15$$

Their difference = $$5x+15 = 1365$$

$$x = 270$$ is required answer

Since AB is the diameter $$\angle\ ADB\ =\ 90^0$$

Which gives $$\angle\ DAB\ =\ 50^0$$

ABCD is cyclic quadrilateral so $$\angle\ A+\angle\ C\ =\ 180^0$$

Thus $$\angle\ DCB\ =\ 130^0$$

In$$\bigtriangleup\ $$ BCD, 2x+130 = 150

$$x\ =\ 25^0$$

Probability of alternate balls when first ball is black = $$\frac{5}{8}\times\ \frac{3}{7}\times\ \frac{4}{6}\times\ \frac{2}{5}=\frac{1}{14}$$

Probability of alternate balls when first ball is white = $$\frac{3}{8}\times\ \frac{5}{7}\times\ \frac{2}{6}\times\ \frac{4}{5}=\frac{1}{14}$$

Overall Probability = $$\frac{1}{14}+\ \frac{1}{14}=\frac{1}{7}$$

The given equation can be written as 

$$\log\left(10\right)^{x\ }\ +\ \log\left(1+2^x\right)=\log\left(5\right)^x+\log6$$

$$\log\left(10\right)^{x\ }\left(1+2^x\right)=\log\left(5\right)^x\cdot6$$    (  since logA + logB=logAB)

$$\log\ \frac{\left(2^x\cdot5^x\right)\left(1+2^x\right)}{5^x\cdot6}=0$$    ( since logA - logB=logA/B)

$$\frac{\left(2^x\ +2^{2x\ }\right)}{6}=10^0$$  ($$Since\ \log_aN\ =x\ \ =>N=a^x$$)

$$2^{^x}+2^{2x}=6$$

The above  equation is satisfied only when x=1

For odd function f(x) = -f(-x)

for option A : f(x) = $$2^{-x.x}$$, f(-x) = $$2^{-(-x).(-x)}$$ = $$2^{-x.x}$$ = f(x). Hence Option A is not correct

for Option B : f(x) = $$2^{x-x.x.x.x}$$, f(-x) = $$2^{(-x)-(-x).(-x).(-x).(-x)}$$ = $$2^{(-x)-x.x.x.x)}$$ $$\ne\ -f\left(-x\right)$$

Neither A or B are correct

Total surface area of cube = $$6\times\ 8^2\ =384$$

Amount of paint required = $$\ \frac{\ 384}{16}=24\ kgs$$

Cost of 24kgs paint = $$24\times\ 36.5\ =876$$

Total distance covered by horse = Circumference of the circle.

$$2.5 \times 66 = 2\pi \times $$ (radius)

$$2.5 \times 66 = 2 \times \frac{22}{7}\times $$ (radius)

radius $$=\ 26.25$$

Let the percentage of salt in the solution be $$x$$%. The amount of salt will be constant. Upon equating

$$5\%of\ 6l\ =\ x\%\ of\ 5l$$

It gives $$x=6$$

While finding the last 2 digits, we can ignore the hundredth digit of each number and the last two digit will be the same as last two digit in case of $$22\times\ 23\times\ 25\times\ 27\times\ 29$$.

25*22= 550, which is an important result. 

The rest numbers, 23, 27 and 29 when multiplied by 550, will result in the same last 2 digits as odd number multiplied by 5 will always give 5 as the last digit and we already have 0 at the end of it. So, odd number* 50= X50 always.

So, the last 2 digits will always be 50.

Let the original fraction be $$\ \frac{\ a}{b}$$.

Given that Numerator, a is increased by 200% or it becomes 3 times. So, new numerator= 3a.

The denominator increases by 200%, and it also becomes 3 times. So, new denominator= 3b.

So, the new fraction= $$\ \frac{\ 3a}{3b}$$, which is same as $$\ \frac{\ a}{b}$$, the old fraction. 

The value of the fraction is given to be $$\ 2\frac{\ 4}{5}=\ \ \frac{\ 14}{5}$$, which is also the old fraction.

Since, none of the first three options match. Option D is the correct choice.