# IBPS Clerk Prelims Reasoning Ability Questions 2019 (Day-16)

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Puzzles

Directions (1-5): Study the below information carefully and answer the given question:

There are 6 boxes arranged one above another in the form of a stack, the lowermost box is numbered as 1 and the box above it is numbered as 2 and so on. Each was in different colours viz. Blue, Pink, Green, Yellow, Red and Orange but not necessarily in the same order.

Blue box is at the even position but not at the top. Red box is not adjacent to the green or yellow box. Orange box is at the 5th position. There are two boxes between pink and green box where pink box is above the green one. Green box is not at the bottommost position.

1) At which position is pink box?

a) 4th

b) 3rd

c) 6th

d) 2nd

e) Cannot be determined.

2) What is the position of blue box?

a) 6th

b) 4th

c) 2nd

d) Either option (a) or (b)

e) None of the above

3) Which of the following two boxes are adjacent to each other?

a) Pink, green

b) Pink, yellow

c) Yellow, green

d) Blue, orange

e) Green, orange

4) Which one of the following colour box is above the yellow colour box?

a) Green

b) Blue

c) Red

d) Pink

e) None of the above

5) Which of these boxes is at odd numbered position?

a) Yellow

b) Green

c) Pink

d) Blue

e) Both option (a) and (b)

Numerical Series

Directions (6-10): Study the following arrangement carefully and answer the questions given below:

5 1 2 9 3 8 7 1 4 6 8 3 2 3 9 7 1 2 5 4 8 9 3 5 4 6 7 6

6) If all the odd numbers are dropped from the above sequence, what is the sum of the numbers present in the odd position of the sequence?

a) 28

b) 30

c) 24

d) 36

e) 32

7) How many such multiples of 2 are there in the above sequence which is immediately preceded by even number?

a) One

b) Two

c) Three

d) Four

e) Five

8) If all the odd numbers in the sequence are replaced by alphabets starting from A from the left, then which number is replaced by alphabet ‘G’?

a) 4

b) 9

c) 3

d) 6

e) 5

9) If every alternate number is dropped from the left end (1 is dropped), how many even numbers have even numbers as both of their neighbours?

a) None

b) One

c) Two

d) Three

e) Four

10) If all the even numbers are increased by 1 in the given sequence, then how many 3’s are there in the new sequence?

a) Three

b) Four

c) Five

d) Six

e) Seven

Directions (1-5):

Given that the orange box is at the 5th position so,

The blue is at even position but not at the top. So, it can either be at 4th or 2nd position.

CASE 1: For 4th position.

CASE 2: For 2nd Position.

There are two boxes between pink and green box where pink box is above the green one. So, in the CASE 1, pink box can be at eh position.

And in CASE 2 also at eh position.

(Since, the green box is not at the bottommost position so pink box cannot be at 4th position.)

Now in case 1, the 1st and 2nd box will be of yellow and red colour. but it is given that, Red box is not adjacent to the green or yellow box. So, this case is contradictory.

Moving ahead with CASE 2, Since, green is not adjacent to red, so red will be the 1st box and 4th box would be yellow:

Directions (6-10) :

5 1 2 9 3 8 7 1 4 6 8 3 2 3 9 7 1 2 5 4 8 9 3 5 4 6 7 6

If all the odd numbers are dropped from the above sequence, we get

2 8 4 6 8 2 2 4 8 4 6 6

Sum of the numbers in odd position of the sequence,

=2+4+8+2+8+6

= 30

5 1 2 9 3 8 7 1 4 6 8 3 2 3 9 7 1 2 5 4 8 9 3 5 4 6 7 6

Multiples of 2 in the above sequence which is immediately preceded by even number

46, 68, 48, 46

There are four such multiples of 2 are there in the above sequence which is immediately preceded by even number.

5 1 2 9 3 8 7 1 4 6 8 3 2 3 9 7 1 2 5 4 8 9 3 5 4 6 7 6

If all the odd numbers in the sequence are replaced by alphabets starting from A, we get

A B 2 C D 8 E F 4 6 8 G 2 H I J K 2 L 4 8 M N O 4 6 P 6

‘G’ is replacing ‘3’.

5 1 2 9 3 8 7 1 4 6 8 3 2 3 9 7 1 2 5 4 8 9 3 5 4 6 7 6

If every alternate number is dropped from the left end,

we get 5 2 3 7 4 8 2 9 1 5 8 3 4 7

Only one even number have even number on its neighbors.

5 1 2 9 3 8 7 1 4 6 8 3 2 3 9 7 1 2 5 4 8 9 3 5 4 6 7 6

If all the even numbers are increased by 1 in the given sequence, we get

5 1 3 9 3 9 7 1 5 7 9 3 3 3 9 7 1 3 5 5 9 9 3 5 5 7 7 7

There are seven 3’s in the new sequence.