**DIRECTIONS: **Read the following given below and answer the questions that follow.

i) Eleven students Anwar, Baham, Chetan, Dayal, Eshwar, Farooq, Gajendra, Hariom, Inayat, Jatin and Kishore are sitting on a bench in a lecture room facing towards the teacher.

ii) Dayal is towards the left of Farooq and second to the right of Chetan.

iii) Eshwar is second to the left of Anwar and sitting on one end.

iv) Jatin is the neighbourer of Anwar and Baham and is third to the left of

Gajendra.

v) Hariom is next to left of Dayal and is third to the right of Inayat.

1) Which two students are sitting in two ends?

a) Kishore and Dayal b) Eshwar and Dayal

**c) Eshwar and Farooq **d) Kishore and Farooq

2) Which group of students are sittting just next to the right of Gajendra?

a) Chetan, Hariom, Dayal, Eshwar b) Chetan, Hariom, Inayat, Baham

c) Chetan, Hariom, Inayat, Dayal **d) none of the above**

3) Who is sitting just in the middle?

**a) Inayat **b) Chetan c) Baham d) Jatin

4) Which of the five given statements is unnecessary?

a) (i) and (ii) b) (i) c) (iii) **d) All four are necesssary**

5) Which of the following is correct?

a) Eshwar and Anwar are the closest neighbourers of Jatin

**b) Gajendar, Inayat and Baham are sitting to the left of Chetan**

c) Hariom is in the middle of the line

d) Anwar, Kishore and Eshwar are to the right of Jatin

The Order will be –

**E K A J B I G C H D F**

**Ques. The sum of ages of 5 children born at intervals of 3 years each is 50 years. What is the age of the youngest child?**

Let x = the youngest child. Each of the other four children will then be x+3, x+6, x+9, x+12. We know that the sum of their ages is 50 so we can form an equation:

- x+(x+3)+(x+6)+(x+9)+(x+12) = 50
- therefore 5x+30 = 50
- therefore 5x=50-30
- therefore x=20/5 = 4

I wanted to go the church, I moved northwards and after covering to some distance turned to left and moved 5 km and reached the crossing. The road in front of me led to the Casino while the road to my left led to me Miranda college and the road to my right led to the church. In which direction the church is located with reference to starting point.?

**North**

North east

North west

South

The sum of the ages of 5 children born at the intervals of 3 years each is 50 years what is the age of the youngest child?

4 years

8 years

10 years

None of the above

- Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
- Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
- 5x = 20
- x = 4.
- Age of the youngest child = x = 4 years.

The question consists of two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. You must indicate whether –

a)Choice 1 if the question can be answered using one of the statements alone, while the other statement is not sufficient to answer the question.

- b) Choice 2 if the question can be answered using second of the statements alone while the other statement is not sufficient to answer the question.
- c) Choice 3 if both the statements together are needed to answer the question
- d) Choice 4 if both the statements independently or taken together are not sufficient to answer the question

e)Either of the statements taken individually are sufficient in answering the questions

Is X a prime number, given that X is a positive integer? (A) X^4 > 3000 (B) X^4 < 10,000

a

b

c

D

From statement A we know that X4 is greater than 3000. There are infinite values for X possible for which X4 is greater than 3000

From statement B we know that X4 is less than 10,000. This one is a lot better. There are only 9 integer values that satisfy this condition. But this still does not give us a unique answer.

Combining the two statements, we know that 2000 < X4 < 10000. The following values of X satisfy this condition -> 8 and 9. Both 8 and 9 are not prime. Hence, the question can be answered conclusively using the two statements. Hence answer choice (3)

How long will it take for two pipes A and B to fill an empty cistern if they worked alternately for an hour each?

(A) Working alone, Pipe A can fill the cistern in 40 hours (B) Pipe B is one third as efficient as Pipe A

a

b

c

D

From statement A, we know that Pipe A can fill the tank in 40 hours. However, this information is not sufficient as we do not have the data for Pipe B. Hence, statement A alone cannot answer the given question.

From statement B, we know that Pipe B is one third as efficient as pipe A. However, we do not know the rate at which Pipe A fills the tank. Hence, we will not be able to find the rate at which Pipe B fills the cistern. Therefore, statement B alone is not sufficient to answer the question.

Now, if we combine the two statements, we know that Pipe A take 40 hours to fill the cistern.

Pipe B takes 120 hours to fill the cistern.

If they worked alternately, then either Pipe A could have started the cycle or Pipe B could have started the cycle.

If Pipe A started the sequence of filling alternately, then at the end of two hours, the two pipes together would have filled th of the tank in an hour. Or the cistern will fill in 30 hours.

If Pipe B started the sequence, then at the end of 2 hours, the two pipes together would have filled th of the tank in an hour. Or the cistern will fill in 30 hours.

As the answer obtained irrespective of which pipe started the sequence is the same, the correct answer is (3) – i.e., both the statement are required to answer the questionThe set S of numbers has the following properties:

- I) If x is in S, then 1/x is in S. II) If both x and y are in S, then so is x + y. Is 3 in S? (A) 1/3 is in S. (B) 1 is in S.

a

b

c

D

Correct Answer – (2)

Solution:

Consider (1) alone. Since 1/3 is in S, we know from Property I that 1/(1/3) = 3 is in S. Hence, (1) is sufficient.

Consider (2) alone. Since 1 is in S, we know from Property II that 1 + 1 = 2 (Note, nothing in Property II prevents x and y from standing for the same number. In this case both stand for 1.) is in S. Applying Property II again shows that 1 + 2 = 3 is in S. Hence, (2) is also sufficient.

Is ‘0’ the smallest of five consecutive integers even?

(A)The product of the five integers is 0 (B) The arithmetic mean of the five integers is 0.

a

b

c

d

If the smallest of five consecutive integers is even, then the first, third and fifth integers will be even. From statement A, we know that one of the 5 numbers is 0. However, we will not be able to say which of the 5 numbers happen to be 0.

From statement B, we know the arithmetic mean of the 5 numbers is 0. The A.M of five consecutive integers is the third integer, which is 0. 0 is even. Hence, the smallest of the 5 consecutive integers is even. Hence statement B alone is sufficient and the answer is (1).

If CANDLE IS IGTXFY then FLAME is?

LRGPRS

LRGGYM

**LGMYGR**

LRGSKY

When Kumar saw Sujay, he recalled ” He is the son of the father of the mother of my daughter”?

Uncle

**Brother-in-Law**

Cousin

Cannot be determined.