SNAP 2009 Question Paper

There are three possibilities for sum of uppermost faces of dice is equal to 15. They are

5, 5, 5 - only 1 arrangement

4, 5, 6 - 3! = 6 arrangements

3, 6, 6 - $$\frac{3!}{2}$$ = 3 arrangements

Out of these 10 arrangements, there are 2 cases in which 4 occurs in first roll ( 4,5,6 or 4,6,5)

Therefore, probability = $$\frac{2}{10}$$ = $$\frac{1}{5}$$

Answer is option B.

Let speed of boat be V and speed of stream be U 
so as per case 1 
Downstream i.e V+U = 30/2 =15
And Upstream i.e V-U = 30/6 = 5 
Now adding 
we get 2V =2
V =10

For a number to be divisible by 4, last 2 digits must be divisible by 4

Last 2 digits can be 12, 24, 32, 52

_, _, _, _, _

Last 2 digits have 4 possibilities and remaining digits can be arranged in 3! ways

Number of 5 digit numbers divisible by 4 are 3! $$\times\ $$ 4 = 24

Total number of 5 digit numbers that can be formed = 5!

Therefore, probability = $$\frac{24}{5!}\ $$ = $$\frac{1}{5}$$

Answer is option A.

Sum of observation is given as = 70

Number of observation = 20

Therefore, Mean = 70/20=3.5

But, here observation starts from 23.

Therefore, Mean = 23 + 3.5 = 26.5

Gold in 50g of alloy

$$=80\times\frac{50}{100}=40g$$

Let x g gold must be added.

Now, according to the question,

$$40+x=\frac{90}{100}\left(50+x\right)$$

$$x=50g$$

Thus, 50 g of gold must be added to it to get 90%.

Thus, the correct option is A.

Let the incomes of two persons are 7x and 3x
Expenditure of first person = 7x - 300
Expenditure of second person = 3x - 300
According to the question,
7x−300/3x−300=5/2
=> 14x-600 = 15x-1500
x = 900
Income of first person
= 7x = 7×900=RS.6300

Old price of rice = Rs.24/k
New price of rice = Rs.27 per/kg
Therefore, Increase in price of rice = New price - Old price = 27 - 24 = Rs.3/kg
Therefore, Increased percentage in price
= 3/24×100=12.5 %
Now, reduction in consumption, so as to keep the expenditure fixed.
According to the formula,
$$\frac{a}{100+a}\times\ 100$$
= $$\frac{12.5}{100+12.5}\times\ 100$$
=11.1 %

The number of different journey tickets that are required by the authorities = The number of selecting 2 stations × \times\ × 2 (For every pair stations say A and B, there are two tickets required A to B and B to A)
= $$^{10}C_2\times\ 2$$ = 90

We know that Circumference is dependent linearly on radius 
so if circumference is increased by 5% , we can conclude radius is increased by 5%
Now Area varies directly with square of radius 
Now if r is changed to 1.05 r 
so r^2 will change to (1.05r)(1.05r) = 1.1025 r^2
So we can conclude area will change by 10.25%

ABACUS 
Vowels = AAU 
Remaining BCS
Now AAU will be considered 1 and B,C,S separate 
so these can be arranged in 4! ways 
Vowels AAU can be arranged in 3!/2! ways 
so total arrangements = $$\frac{4!\times\ 3!}{2!}$$