Reasoning Quiz – Inequalities – IBPS Clerk Prelims 2018 – 1

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Hello and welcome to exampundit. Here is a set of Reasoning Quiz for IBPS Clerk Prelims 2018 on Inequalities.

The following reasoning quiz has 11 questions from Inequalities with 10 minutes.

[WpProQuiz 218]

Reasoning Quiz – Inequalities – IBPS Clerk Prelims 2018 – 1

Directions (Qs.1-5): In the following questions, the symbols @, #, $, * and % are used as illustrated below:

‘P @ Q’ means ‘P is not smaller than Q’.

‘P # Q’ means ‘P is neither greater than nor equal to Q’.

‘P $ Q’ means ‘P is neither smaller than nor greater than Q’.

‘P * Q’ means ‘P is not greater than Q’.

‘P % Q’ means ‘P is neither smaller than nor equal to Q’.

Now, in each of the following questions assuming the given statements to be true, find which of the two conclusions I and II given below them is/are definitely true? Give answer

(a) if only Conclusion I is true.

(b) if only Conclusion II is true.

(d) if neither Conclusion I nor II is true.

(e) if both Conclusions I and II are true.

Statements:

M $ K, D * K, R # K

Conclusions:

(I) D $ M

(II) M % D

Show Solution

(c) M = K .…. (i);
D ≤ K ….. (ii);
R < K ..… (iii)
From (i) and (ii), we get
M = K ≥ D –> M ≥ D
Hence, either M > D (conclusion II) or M = D(conclusion I) is true

Statements:

F * M, M % R, E @ F

Conclusions:

(I) M % E

(II) R @ E

Show Solution

(d) F ≤ M… (i); M > R… (ii); E ≥ F… (iii)
From (i) and (iii), no specific relation can be obtained between M and E. Similarly, no specific relation can be obtained between R and E.

Statements:

H $ K, T # H, W * T

Conclusions:

(I) K % W

(II) T # K

Show Solution

(e) H = K… (i); T < H…(ii),
W≤T …(iii)
From (i), (ii) and (iii), we get
K = H > T ≥ W –> K > W (conclusion I) and
T < K (conclusion II).

Statements:

N % A, A # L, F $ N

Conclusions:

(I) L % F

(II) F % A

Show Solution

(b) N > A… (i), A < L… (ii), F = N…(iii)
From (i) and (iii), we get
F = N > A–> F > A (conclusion II). But no specific relation can be obtained between L and F. Hence, conclusion I is not necessarily true.

Statements:

B * D, D $ M, F % M

Conclusions:

(I) B # M

(II) F % B

Show Solution

(b) B < D…(i); D = M…(ii);
F > M …(iii)
From (i), (ii) and (iii), we get
F > M = D > B –> B < M and F > B (conclusion II).
Since, B < M, therefore, conclusion I is not necessarily true.

Directions (Qs. 6-11): In the following questions the symbols +, ×, ?, @ and $ are used with the following meanings:

P + Q means P is neither smaller nor greater than Q.

P × Q’means P is neither equal to nor smaller than Q.

P ? Q means P is neither greater than nor equal to Q.

P @ Q means P is either greater than or equal to Q.

P $ Q means P is not equal to Q.

Now, in each of the following questions assuming the given statements to be true, find which of the two conclusions I and II given below them is/are definitely true. Give answer

(a) if only conclusion I is true;

(b) if only conclusion II is true;

(d) if neither I nor II is true; and

(e) if both I and II are true.

Statement:

P $ Q, Q × R, P + R

Conclusions:

(I) Q × P

(II) P ? Q

Show Solution

(e) P ≠ Q …(i), Q > R … (ii), P = R …(iii)
From (ii) and (iii), we get Q > R = P –> Q > P. Hence,
Both I and II are true.

Statement:

A + B, B $ C, C ? A

Conclusions:

(I) C $ A

(II) B + C

Show Solution

(a) A = B … (i), B≤ C … (ii), C < A … (iii) From (iii), conclusion I is true. II contradicts statement (ii), hence, it is not true.

Statement:

Y @ Z, Z × Q, Q $ P

Conclusions:

(I) Y ? Q

(II) Y ? P

Show Solution

(d) Y≥ Z …(i), Z > Q … (ii), Q ÷ P …(iii)
From (i) and (ii), we get Y > Z > Q –> Y > Q … (A)
Hence, I is not true. From (iii), two possible relationships between P and Q are;
Case I: When P > Q
Now, using (A), we get Y > Q < P –> no conclusion.
Case II: When Q > P
using (A), we get Y > Q > P –> Y > P. Hence, II is not true.

Statement:

E × F, F @ L, L+ N

Conclusions:

(I) N + F

(II) E × L

Show Solution

(b) E > F …. (i), F > L … (ii), L = N …(iii)
From (ii) and (iii), we get F≥ L= N –> F≥ N or N′ F.
Hence, I may be true but not necessarily so.
From (i) and (ii), we get E > F > L –> E > L
Hence, II is true.

Statement:

H @ J. J ? K, K × M

Conclusions:

(I) H @ M

(II) M × J

Show Solution

(d) H ≥ J … (i), J < K … (ii), K > M … (iii)
From (ii) and (iii), we get J < K > M fi no relationship
between J and M can be established. Hence, II can’t
be established. Again, combining all we can’t conclude
Thee relationship between H and M. Hence, I is not true.

Statement:

M @ T, T + V, V ? E

Conclusions:

(I) V + M

(II) V ? M

Show Solution

(c) M ≥ T … (i), T = V …. (ii), V < E …(iii)
From (i) and (ii), we get
M ≥ T = Vfi M ≥ V –> ei–>er V = M or V < M is true.

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This post was last modified on June 2, 2019 12:01 pm