SNAP 2010 Question Paper

Let A, B, C, D have a, b, c, d chocolates respectively initially.

D gets thrice as many as he already has from A which makes a total of 4d chocolates with D.

He gives one-eigth of that to B which leaves D with 3.5d Chocolates.

Now 3.5 d = 35 => d = 10

Now, A has given b, 2c, 3d chocolates to B, C, D respectively at first which left him with (a-b-2c-3d) chocolates.

Then he gave 10% of the remaining chocolates to C and 20% to B.

He finally has 0.7(a-b-2c-3d) chocolates.

0.7(a-b-2c-3d) = 35

=>(a-b-2c-3d) = 50 = k (say)

C got 2c chocolates at first and then again 10% of 'k' from A.

C finally has 3c + 0.1k = 35

=> c = 10

B got b chocolates at first and then again 20% of 'k' from A and 0.5d from D.
B finally has 2b + 0.5d + 0.2k = 35

=> b = 10

For A, 0.7(a-b-2c-3d) = 35

=> a = 110

102+9+90 = 201

203+70+7 =280

In the 3 given terms, the 1st letters have 1 letter between them -> TUV , VWX

the 2nd letters have 2 letters between them -> BCDE , EFGH

Similarly, the 3rd letters have 3 letters between them and the 4th letters have 4 letters between them.

So, the next term of the series will be ZKXS.

Let us name the statements-

S1: I got the top score.

S2: Ricky was second.

B1: I got the top score.

B2: Sachin was second.

R1: I got the top score.

R2: Sachin was third.

Case 1:

Let S1 be true. That means B1 and R1 are false because only one of them can be 1st.

If S1 is true, S2 has to be false; meaning Ricky was 3rd.

So, the order we get as of now is Sachin, Brian, Ricky.

R2 says Sachin was 3rd but both R1 and R2 cannot be false. So, this case is rejected.

Case 2:

Let R1 be true. Then R2 is false meaning Sachin was second. So the order we get as of now is Ricky, Sachin, Brian.

If R1 is true, S1 and B1 are false. If S1 is false, S2 has to be true. But Ricky cannot be second.

So, this case is also rejected.

Case 3:

Let B1 be true. Then B2 is false meaning Sachin was third. So the order we get as of now is Brian, Ricky, Sachin.

Both R1 and S1 are false as B1 is true. R2 and S2 are true and support the order we got.

Let number of people who prefer tea =3k

those who prefer = 2k

those who like both =k

5k = 60

=> k=12

The clock strikes thrice at 3.

There are two intervals between 3 strikes. For nine strikes there will be 8 intervals.

2 intervals take 6 seconds. => 8 intervals will take 24 seconds.

E3 cannot start last as his name has number 3 and E1 cannot finish 1st as his name has number 1. So the one starting last and finishing 1st will be E2. Then E3 will start 1st as E1 cannot.

E3 cannot start last as his name has number 3 and E1 cannot finish 1st

as his name has number 1. So the one starting last and finishing 1st

will be E2. Then E3 will start 1st as E1 cannot. E3 cannot finish 3rd, so E1 will.

From the given statements, let us rank students from brightest to dullest from left to right. Let the Doon school student be K.

From the first condition, we get K,A,Q,R in order.

From the second condition, we get P,K,C,A,Q,R.

So P is the brightest.

From the given statements, let us rank students from brightest to dullest from left to right. Let the Doon school student be K.

From the first condition, we get K,A,Q,R in order.

From the second condition, we get P,K,C,A,Q,R.

So R is the dullest.