SNAP 2015 Question Paper

The total number of multiples of 4 is 16. (04, 08, 12, ...., 64)

The numbers, which are multiples of 4, have their unit digit 0/2/4/6/8

The multiples of 4, with their unit digit as 8, are (08,28,48), and the multiples of 4, with their unit digit as 6, are (16, 36, 56).

The multiples of 4, with their unit digit as 4, are (04, 24, 44, 64), and the multiples of 4, with their unit digit as 2, are (12, 32, 52), the multiples of 4, with their unit digit as 0, are (20, 40, 60).

It is given that the number present in the units places and tens places is swapped, post which they are written in ascending order, then which of the following numbers will be at 10th place from the last => (16+1-10) = 7th from the beginning.

In ascending order (after swapping the digits), the least value will be those numbers that had '0' as their unit place. After that, the numbers had '2' as their unit place, and so on.

7th from the beginning = (3 numbers that had 0 as their unit place) + (3 numbers that had 2 as their unit place) + (1st number that had '4' as their unit place)

Hence, the first number that had 4 as their unit place now becomes (04 => 40). Hence, 40 is the correct answer.

Probability of Ajit = 5/8
Probability of Ravi = 3/5
Probability of Hari = ½
Probability that the target is hit by at least 2 persons = 1 - [Prob of no hit + Prob of 1 hit].
=$$1-\left(\frac{3}{8}\times\ \frac{2}{5}\times\ \frac{1}{2}+\frac{5}{8}\times\ \frac{2}{5}\times\ \frac{1}{2}+\frac{3}{8}\times\ \frac{3}{5}\times\ \frac{1}{2}+\frac{3}{8}\times\ \frac{2}{5}\times\ \frac{1}{2}\right)$$
= $$1-\frac{31}{80}=\frac{49}{80}$$

mlog(a) + nlog(b) = log($$a^{\frac{1}{m}}\cdot b^{\frac{1}{n}}$$)

Therefore the given LHS reduces to log(2$$\sqrt{\ xy}$$) which is equal to log(x+y)

Remove log on both sides

2$$\sqrt{\ xy}$$=x+y

$$\left(\sqrt{\ x}\right)^{2\ }+\ \left(\sqrt{\ y}\right)^2\ -2\sqrt{\ x\cdot y}=0$$

$$\left(\sqrt{\ x}-\sqrt{\ y}\right)^2=0$$

=> x=y

$$\log\left(a^m\right)\ =\ m\log\left(a\right)\ and\ \ \log_aa$$ = 1

$$\log_55^2\ +\ \log_2\left(\log_33^4\right)$$

2 + $$\ \log_24$$

2+ $$\ \log_22^2$$

4

Let AB represent the rock cliff with its height be h

 D represents the initial point and C is the point at which he observers the boat after Ten minutes less than half of an hour (20 minutes).

BD = $$\frac{h}{\tan\ \left(30\right)}\ =\ \sqrt{\ 3}h$$

BC = $$\frac{h}{\tan\ \left(60\right)}\ =\ \frac{h}{\sqrt{\ 3}}$$

In 20 minutes boat covered CD length which is = $$\sqrt{\ 3}h\ -\ \frac{h}{\sqrt{\ 3}}=\ \frac{2h}{\sqrt{\ 3}}$$

Hence remaining lenght $$\frac{h}{\sqrt{\ 3}}$$(BC) will be covered in 10 minutes

Exactly 1 subject = a+b+c ---> Represented by X

Exactly 2 subjects=d+e+f -----> Represented by Y

Exactly 3 subjects= g -----> Represented by Z

So X + Y+ Z + none = total --------------> (I)

German + French + Spanish = (a+b+c) +2(d+e+f) + 3(g) = X+ 2Y + 3Z -----------> (II)

So X+2Y+3Z= 30+28+32=90 

Given X = 54

So 2Y +3Z = 36--------> (1)

Given ,

French and German = 6 => d+g = 6

German and Spanish = e+g = 8

French and Spanish = f + g = 10

adding all the three (d+e+f) + 3g = 24

Y + 3Z = 24 ------> (2)

solving 1 and 2 you get Y=12 and Z=4

Therefore Total = X+Y+Z+None = 54+12+4=70

P(atleast 1 solved) = 1 - P(no one solved)

= 1-(1-0.7)x(1-0.6)

= 1 - (0.3)x(0.4)

= 1 - 0.12

= 0.88

Let the rate of interest be x% and Principal amount be P

Simple interest in 1st case = $$\frac{\left(P\cdot3\cdot X\right)}{100}$$

Simple interest in 2nd case = $$\frac{\left(P\cdot3\cdot\left(X+2\right)\right)}{100}$$

Given the difference is 360 

$$\frac{\left(3\cdot P\right)}{100}\left(X+2-X\right)$$ = 360

So on solving we get P=6000.

Let  the total amount invested be 9x

So the interest for 2 years on 4x amount = $$\frac{\left(4x*2*6\right)}{100}$$

     the interest for 2 years on 4x amount = $$\frac{\left(5x*2*7\right)}{100}$$

Given , Sum of these = 348

1.18x = 348 => x$$\simeq\ $$295

So total amount invested = 2655

Probability = Number of favourable cases/ total number of cases

favourable cases= $$7_{c_2}$$ 

total cases = $$9_{c_2}$$

Probability = $$\frac{\left(7c_2\right)}{\left(9_{c_c}\right)}$$ =7/12