numerical symbolic operations Model Questions & Answers, Practice Test for ssc cgl tier 1
ssc cgl tier 1 SYLLABUS WISE SUBJECTS MCQs
Analogies, Similarities and differences
Relationship concepts
Arithmetical reasoning, arithmetic number series
Verbal Coding and decoding
Statement conclusion
Semantic Analogy, Symbolic/ Number Analogy, Figural Analogy
Numerical Operations, symbolic Operations
Venn Diagrams
Classification of centre codes/ roll numbers
Visual memory, discrimination, observation
Directions:
In the following questions, the symbols ©, $, %, @ and # are used with the following meaning as illustrated below:
- 'P @ Q' means P is neither greater than nor equal to Q.
- 'P $ Q' means P is not smaller than 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.
- Give answer (b) if only Conclusion II is true.
- Give answer (c) if either Conclusion I or II is true.
- Give answer (d) if neither Conclusion I nor II is true.
- Give answer (e) if both Conclusions I and II are true.
Statements: V $ D, D © R, R % F
Conclusions:
I. R % V
II. V @ F
Answer: (c)
- P @ Q ⇒ P ≯ Q and P ≠ Q
hence P < Q - P $ Q ⇒ P ≮ Q
hence P > Q and P = Q ⇒ P ≥ Q - P # Q ⇒ P ≮ Q and P ≯ Q
hence P = Q - P © Q ⇒ P ≯ Q
hence P ≥ Q
or P ≤ Q - P % Q ⇒ P ≮ Q and P ≠ Q
hence P > Q
| @ ⇒ < | $ ⇒ ≤ |
| # ⇒ = | © ⇒ ≥ |
| % ⇒ > |
Statements :
V $ D ⇒ V ≥ D
D © R ⇒ D ≤ R
R % F ⇒ R > F
Hence, V ≥ D ≤ R > F
Conclusions :
I. R % V ⇒ R > V ( False)
II. V @ F ⇒ V < F ( False)
Statements: M © T, K % T, K @ N
Conclusions:
I. N % M
II. K % M
Answer: (e)
- P @ Q ⇒ P ≯ Q and P ≠ Q
hence P < Q - P $ Q ⇒ P ≮ Q
hence P > Q and P = Q ⇒ P ≥ Q - P # Q ⇒ P ≮ Q and P ≯ Q
hence P = Q - P © Q ⇒ P ≯ Q
hence P ≥ Q
or P ≤ Q - P % Q ⇒ P ≮ Q and P ≠ Q
hence P > Q
| @ ⇒ < | $ ⇒ ≤ |
| # ⇒ = | © ⇒ ≥ |
| % ⇒ > |
Statements :
M © T ⇒ M ≤ T
K % T K > T
K @ N K < N
Hence, M ≤ T < K < N
Conclusions :
I. N % M ⇒ N > M (True)
II. K % M ⇒ K > M (True)
Statements: B @ E, E # S, S $ Z
Conclusions:
I. Z @ E
II. E # Z
Answer: (b)
- P @ Q ⇒ P ≯ Q and P ≠ Q
hence P < Q - P $ Q ⇒ P ≮ Q
hence P > Q and P = Q ⇒ P ≥ Q - P # Q ⇒ P ≮ Q and P ≯ Q
hence P = Q - P © Q ⇒ P ≯ Q
hence P ≥ Q
or P ≤ Q - P % Q ⇒ P ≮ Q and P ≠ Q
hence P > Q
| @ ⇒ < | $ ⇒ ≤ |
| # ⇒ = | © ⇒ ≥ |
| % ⇒ > |
Statements :
B @ E ⇒ B < E
E # S ⇒ E = S
S $ Z ⇒ S ≥ Z
Hence, B < E = S ≥ Z
Conclusions :
I. Z @ E Z < E ( False)
II. E # Z E = Z ( False)
Z is either smaller then E or equal to E
Statements: H % M, N © M, N $ T
Conclusions:
I. H # T
II. H % T
Answer: (d)
- P @ Q ⇒ P ≯ Q and P ≠ Q
hence P < Q - P $ Q ⇒ P ≮ Q
hence P > Q and P = Q ⇒ P ≥ Q - P # Q ⇒ P ≮ Q and P ≯ Q
hence P = Q - P © Q ⇒ P ≯ Q
hence P ≥ Q
or P ≤ Q - P % Q ⇒ P ≮ Q and P ≠ Q
hence P > Q
| @ ⇒ < | $ ⇒ ≤ |
| # ⇒ = | © ⇒ ≥ |
| % ⇒ > |
Statements :
H % M ⇒ H > M
N © M ⇒ N ≤ M
N $ T ⇒ N ≥ T
Hence, H > M ≥ N ≥ T
Conclusions :
I. H # T ⇒ H = T (False)
II. H % T ⇒ H > T (True)
Statements: J # R, R % K, K @ D
Conclusions:
I. K @ J
II. D @ J
Answer: (a)
- P @ Q ⇒ P ≯ Q and P ≠ Q
hence P < Q - P $ Q ⇒ P ≮ Q
hence P > Q and P = Q ⇒ P ≥ Q - P # Q ⇒ P ≮ Q and P ≯ Q
hence P = Q - P © Q ⇒ P ≯ Q
hence P ≥ Q
or P ≤ Q - P % Q ⇒ P ≮ Q and P ≠ Q
hence P > Q
| @ ⇒ < | $ ⇒ ≤ |
| # ⇒ = | © ⇒ ≥ |
| % ⇒ > |
Statements :
J # R J = R
R % K R > K
K @ D K < D
Hence, J = R > K < D
Conclusions :
I. K @ J ⇒ K < J (True)
II. D @ J ⇒ D < J ( False)
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