SNAP 2006 Question Paper

The identical terms will be of the form k*LCM(d1,d2) +smallest common number .

Where d1 , d2 are common differences in two series and smallest common term in two series = 6

So N= k*LCM ( 2,3) + 6

N= 6k+6   (where k is a whole number)

Since the last number in 1st series is 200 and last number in 2nd series is 300, it is sufficient to check multiples of till 200

Therefore total number of multiples of 6 below 200 = 33

Let x,y and z be the number of cows,  horses and chickens respectively.

Since he has Rs 10,000.

$$1000x+300y+50z$$ = 10,000 ....(I)

$$ x+y+z$$ = 100 ....(II)

300*eqn(II)- eqn(I) gives

$$250z - 700x = 20,000$$

$$z\ =\ \ \frac{\ 20,000+700x}{250}$$

since $$z$$ is an integer then $$Remainder\left(\frac{\ 20,000+700x}{250}\right)$$ = 0

$$Remainder\left(\frac{\ 20,000+700x}{250}\right)$$ = $$Remainder\left(\frac{200x}{250}\right)$$

$$Remainder\left(\frac{200x}{250}\right)$$ = 0 is possible when x takes values 5 ,10, 15 and so on

When x = 5, then z = 94 and y = 1

When x = 10 then y = 108 ( rejected as total number of animals has to be 100 )

Thus he has to buy 94 chickens

Option A : $$1\ +\ 2^{3\ }+\ 3^{3\ }\ -1\ -4^3=-29$$

Option B: $$1\ +\ 2^{3\ }+\ 3^{3\ }\ +1\ -4^3=-27$$

Option C: $$1\ -\ 2^{3\ }-3^{3\ }\ +1\ +4^3=31$$

Option D: $$1\ -\ 2^{3\ }-3^{3\ }\ -1\ +4^3=29$$

Total taking = number  of participants * cost per participant

 38950 =  $$2\times\ 5^2\times\ 19\times\ 41$$

95 is the only factor of 38950 which lies between 45 to 100.

Hence 95 is number of participants.

Cost per participant = 38950/95 = 410

T(n) be $$n^{th}$$ term

We see that T(n)= T(n-1) * T(n-2)

Hence next term is 36*216 = 7776

Let the amount Tanya paid be 6x

Amount paid by Veena is 3x

Then Amita paid by 2x

Required fraction $$\frac{3x}{6x+3x+2x}$$ = $$\frac{3}{11}$$

In Round 1 : 64 matches are played among 128 members

In Round 2: 32 matches are played among 64 members

In Round 3 : 16 matches are played among 32 members

and so on till 1 match is played among final 2 Players.

Total matches = 1+2+4+8+16+32+64 = 127

Let speed of crew= 'a' mph

Speed of current = 'b' mph

Down stream ,the speeds will be added 

Upstream,the speeds will be subtracted

a+b = $$\frac{\left(10\right)}{\frac{5}{6}}$$

a-b = $$\frac{\left(12\right)}{\frac{3}{2}}$$

Solving the above two equations  , a=10 and b=2

Probability = $$\frac{\left(Number\ of\ favourable\ cases\right)}{Total\ number\ of\ cases}$$

Here the possible outcomes are GG or GB

So probability of two girls = 1/2

Assume the distance to be $$x$$ km.

When he travels at the speed of 2.5 km/hr he is 6 minutes late.

When he travels at the speed of 3.5 km/hr he is 6 minutes early

$$\frac{\ x}{2.5}$$ - $$\frac{\ x}{3.5}$$ = $$\frac{\ 12}{60}$$

$$\frac{\ 2x}{5}-\frac{\ 2x}{7}\ =\frac{\ 1}{5}$$

$$\frac{\ 4x}{35}=\frac{\ 1}{5}$$

$$\ x=\frac{\ 7}{4}=\ 1\frac{\ 3}{4}$$ km