Welcome to Online Quant Section in AffairsCloud.com. We are starting IBPS Clerk course 2015 and we are creating sample questions in Reasoning section, type of which will be asked in IBPS Clerk Main Exam.
Stratus – IBPS Clerk Course 2015
Directions (1-5): Study the following information to answer the given questions:
In a certain code, ‘always go with ideas’ is written as ‘ba ri dib gi’, ‘ideas and with approach’ is written as ‘ta gi ma ri’, ‘go approach and insights’ is written as ‘ma jo ba ta’, and ‘with and better mind’ is written as ‘ki ri to ta’.
- What is the code for ‘ideas’?
E) Cannot be determinedC) gi
With- ri, so ideas – gi
- What does ‘ta’ stand for?
E) mindD) and
‘ta’ and ‘and’ – both present in last three
- ’ta lo ba’ could be a code for which of the following?
A) approach and action
B) go and innovate
C) ideas and approach
D) go with mind
E) always better ideasB) go and innovate
From que 2, ta – and
ba – go from 1st and 3rd code
lo and innovate not present anywhere, so lo can be innovate
- What is code for ‘with’?
E) baB) ri
From 1st and last code, with – ri
- Which of the following may represent ‘insights always better’?
A) jo ki to
B) ki to ri
C) dib jo ri
D) to dib jo
E) dib to baD) to dib jo
Insights – jo, always – dib,
Better is ‘ki’ or ‘to’
So code should be ‘jo dib ki’ or ‘jo dib to’. ‘jo dib to’ is present in options
Directions (6-10): In all the questions that follow, relationship between different elements is shown in the statements. The statements are followed by two conclusions. Give answer accordingly.
(A) only 1st follows
(B) only 2nd follows
(C) either 1st or 2nd
(D) neither 1st nor 2nd
(E) both 1st and 2nd
- Statements: Z > B ≤ G = C, G ≤ P ≤ T, C ≤ M
I. T > M
II. T ≤ M(C) either 1st or 2nd
T ≥ G = C ≤ M, this gives both T and M less than equal to F
So either T will be > M or T will be ≤ M
- Statements: W ≥ S > K, S > F < C, P > W
I. W > F
II. P > C(A) only 1st follows
W ≥ S > F, so W > F
P > W ≥ S > F < C, so relation cant be determined between P and C
- Statements: N > V < B ≤ C, P ≤ M ≤ V, G < M
I. C > G
II. P < N(E) both 1st and 2nd
C > V ≥ M > G, so C > G
P ≤ V < N, so P < N
- Statements: H > M, M < N, N ≥ K
I. K < M II. H > N(D) neither 1st nor 2nd
K ≤ N > M, so no relationship between K and M
H > M < N, both H and N less than M so no relationship between H and N
- Statements: D ≥ E ≥ J, R < P, P > Q
I. Q > E
II. D > P(D) neither 1st nor 2nd
Q < P > R ≤ E, so relation cant be determined between Q and E
D ≥ R < P, so relation cant be determined between D and P