SNAP 2007 Question Paper

Let total number of days=20x

Number of rainy days = 2x

Number of rainbow days = x

Number of non rainbow days= 19x

$$\therefore\ $$ Percentage of non rainbow days = $$\frac{19x}{20x}\ \cdot\ 100\ =95$$

Considering statement 1 alone,

Let the number of balls in bag III be x.

The number of balls in bag I becomes x+5.

And the rest, which is 10- (x+x+5)= 5-2x balls are with C.

We cannot find the value of x, so Statement 1 alone is not sufficient.

Considering statement 2 alone, 

Let the number of balls in bag I be 2a

Balls in bag II will be a and the rest, i.e 10-3a will be in bag 3.

Again, we cannot find the value of a in this case, and hence, Statement 2 alone is not sufficient.

Considering both the statements and using the number of balls with each one of them as found using statement 1, we get

$$5+a=2\left(5-2a\right)$$

=>5+a= 10-4a

=> 5a=5, or a=1.

So, balls in Bag I= 5+a=6

balls in Bag II= 5-2a=3

balls in Bag III= a= 5.

Therefore, both statement I and II are required.

Let the side of the square lake be of length x kms.

Using pythagorus theorem, we can find the length of the diagonal as $$\sqrt{x^2+x^2}=\sqrt{2x^2}= x\sqrt{2}$$

'.' Area of the square= 50 sq. kms,

$$x^2=50$$ or x= $$5\sqrt{\ 2}$$

Therefore length of the diagonal = $$5\sqrt{\ 2}\cdot\sqrt{\ 2}=5\cdot2=10$$ km

Let's assume $$x$$ to be product of $$a, a+1, a+2,$$ and $$a+3$$

Hence, $$x=a(a+1)(a+2)(a+3)$$

$$n=a(a+1)(a+2)(a+3)+1$$

$$n=a(a+3)(a+1)(a+2)+1$$

$$n=(a^2+3a)(a^2+3a+2)+1$$

Let's assume$$a^2+3=p$$

$$n=p(p+2)+1$$

$$n=p^2+2p+1$$

$$n=(p+1)^2$$

Hence, we can say that n is a perfect square. 

Hence, 3 is true and 2 is automatically false as perfect squares can't be prime. 

Product of any 4 consecutive integers is always even, as one of the numbers among the 4 consecutive integers will be even and any number multiplied by an even number is even. x is even, hence x+1 is odd. thus n is odd. 

Hence both 1 and 3 are true.

Number of vowels in word EXAMINATION =6

Number of consonants in word EXAMINATION =5

Number of ways in which 6 vowels can be arranged in 6 odd places=(6!)/2!*2!   (1)

Number of ways in which 5 consonants can be arranged in 5 even places = 5!/2!        (2)
Now total arrangements = (1)*(2)
=10,800

The 6th and the 16th child divide the circle into two halves as it is given that the 6th child is diametrically opposite the 16th child. 

When we consider one of the two halves, it will contain students numbered from 7 to 15, i.e a total of 9 children. 

Similarly, the other side must also have 9 children.

Now, including the two pairs of 9 children at each half along with child numbered 6 and 16, we have a total of 9+9+2= 20 children.

Let us assume the number of apples purchased by the man to be x.

Then, the number of oranges will be (40-x).

Let the price of each apple be Rs. y and that of an orange be Rs. z

It is given that, xy+ (40-x)z= 17.

=> xy+40z-xz=17 ---------------(1)

Now, if the number of apples and oranges are interchanged, the apples will be (40-x) and the oranges will be x.

Given, the new total cost will be Rs. 15

So, (40-x)y + xz=15

=>40y-xy +xz= 15 ---------------(2)

Adding eqns 1 and 2, we get,

xy+40z-xz+40y-xy+xz=15+17

=> 40(y+z)=32

=>y+z= $$\ \frac{\ 32}{40}=\ \frac{\ 4}{5}$$

Or, the cost of one apple and one orange, x+y= Rs.$$\frac{\ 4}{5}=\ 80\ paise$$

By looking into the option, we can see that we only need to check the divisibility of the given number by 3 and 11.

Divisibility test by 3 : The given number, 311311311311311311311 has seven 3s and fourteen 1s.

So, the sum of the digits in the number= (7*3)+(14*1)= 21+14=35.

Since, 35 is not divisible by 3, the given number is not divisible by 3.

Divisibility test by 11: The sum of digits at odd places= (1+3+1+1+3+1+1+3+1+1+3)= 19

The sum of digits at even places= Sum of all the digits- Sum of digits at odd places= 35-19= 16.

Since, the difference between the the sum of digits at odd places and the sum of digits at even places, i.e. 19-16 is not a multiple of 11.

So, the given number is also not divisible by 11.

Hence, option D

Let us assume that the investment made at bank A at 6% S.I. is Rs. x

Since, total investment= Rs. 9000, Investment made at bank B at 8% S.I.= Rs. (9000-x)

From bank A, interest earned= $$\ \frac{\ P.r.t}{100}$$, where P= Principal amount, r= rate of Interest and t= time period of investment in years.

So, Interest from bank A= $$\ \frac{\ x.6.3}{100}$$= $$\ \frac{\ 18x}{100}$$

From Bank B, Interest earned for 3 years= $$\ \frac{\ \left(9000-x\right).8.3}{100}=\ \frac{\ 216000-24x}{100}$$

Equating the sum of interests earned from both the banks to the value given i.e. 1800, we find,

1800=$$\ \frac{\ 216000-24x}{100}+\ \frac{\ 18x}{100}=\ \frac{\ 216000-6x}{100}$$

=>$$\ 1800\cdot100=\ \ 216000-6x$$

=> $$\ 6x=\ \ 216000-180000$$

=>x=$$\ x=\ \ \ \frac{\ 36000}{6}$$

.'. x= Rs. 6000- Option B

Let us assume that the length of each bogy is 1 unit, and the speed of both A and B be s units.

Initially, both A and B had 12 bogies, and ran in opposite direction.

So, using the concept of relative speed, the effective speed of the two trains taken together, when travelling in opposite direction adds up.

Effective distance covered = 12+12 units=24.

Effective speed= 2s units.

Now, using time= $$\ \frac{\ Dis\tan ce}{Speed}$$, we get 2 mins= $$\ \frac{\ 24}{2s}$$

=>120=$$\ \frac{\ 12}{s}$$

.'.s= $$\ \frac{\ 1}{10}$$

In the new scenario, length of B becomes 16 units.

So, Effective distance covered = 12+16 units=28 units.

Effective speed= 2s units.

New time= $$\ \frac{\ 28}{2s}=\ \frac{\ 14}{s}$$=140.

So, increase in time in the new case= 140-120 seconds= 20 Seconds- Option D.