SNAP 2006 Question Paper

I have won X number of chess games and my brother has won 7 games.

For every game I win I get 1 chocolate and and I give 1 chocolate if my brother wins. Let A be number of chocolates with me initially.

Then , A +X - 7 = A +20

X= 27

Total games played = Games won by me + Games won by my brother = 27+7 = 34

$$9^{th}$$ term = D

$$16^{th}$$ term = J

$$24^{th}$$ term = Y

$$27^{th}$$ term = S

No meaningful word can be made from the terms hence X is the answer

$$5^{\frac{1}{4}}$$ * $$125^{0.25}$$

= $$5^{0.25}$$ * $$5^{3*0.25}$$

= $$5^{0.25+0.75}$$

=$$5^1$$

=5

Initially, A has  10 L wine and B has 10L water

Process 1 : When 1 L of water is poured from B to A

A now has 11L ( 10 L wine and 1L water) and B has 9L water

Process 2:  1L from A is put into B

1L of liquid from A will have the composition of 10/11L Wine and 1/11L Water

Composition of A(10L)  = $$\frac{100}{11}$$ L wine and $$\frac{10}{11}$$ L water

Composition of B(10L) = $$\frac{10}{11}$$ L wine and $$\frac{100}{11}$$ L water

Let a,b,c and d represent the money with Anand, Binoy, Chetna and Dharma respectively.

It is given that a+b+c+d = 47 ......(I)

a+b = 27 ........(II)

a+c = 25 ........(III)

a+d = 23 ........(IV)

Adding eqn (II),(III) and (IV) and subtracting (I)

(a+b)+(a+c)+(a+d) - (a+b+c+d) = (23+25+27-47)

2a = 28 or a = 14

from eqn (II) b = 13

Let them meet at time "t"

Floor of David at time t = 11+57t

Floor of Albert at time t = 51- 63t

When they meet their floor will be same. 11+57t = 51 - 63t

t = $$\dfrac{1}{3}$$ minutes.

Floor at which they meet = 11 + 57 *$$\dfrac{1}{3}$$ = 11+19 = 30

Average of 5 consecutive number = $$\frac{\left(n\right)+\left(n+1\right)+\left(n+2\right)+\left(n+3\right)+\left(n+4\right)}{5}$$ = n+2

If 2 more consecutive number are added then its average = $$\frac{\left(n\right)+\left(n+1\right)+\left(n+2\right)+\left(n+3\right)+\left(n+4\right)\ +\left(n+5\right)+\left(n+6\right)}{7}$$ = n+3

Hence average is increased by 1

Here T(n) represent the $$n^{th}$$ term of series

T(1) - T(2) = 1

T(2) - T(3) = 3

T(3) -T(4) = 5

T(4)- T(5) = 9.. is incorrect since we notice that difference between consecutive terms are consecutive odd numbers.

T(5) = 34 is incorrect 

Given P =6Q

Required answer is

$$\ \frac{\ P-Q}{P}\times\ 100$$%

= $$\ \frac{\ 5Q}{6Q}\times\ 100$$% (since P=6Q)

= $$83\frac{\ 1}{3}$$ %

Family of  numbers when  divided by a,b,c and leaves remainders x,y,z respectively such that (a-x)=(b-y)=(c-z)=v are of the form N = k *LCM ( a,b,c) - v

Therefore  N = k*LCM(32,40,72) - 22 => N=1440k-22

Smallest such number is obtained when k=1 which is 1418