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# Most Important Reasoning Questions for SSC Stenographer Exam 2017

Here we are providing a Reasoning quiz in accordance with the syllabus of  SSC Stenographer Grade ‘C’ & ‘D’ Exam 2017 and IB (ACIO). This post comprises 15 Questions of Miscellaneous Questions which will really help you get more marks in the final exam.
Q1. In a certain code language, ROUTINE is written as VMRGFLI. How will CRUELTY be written in that code language?
(a) VOCVZRL
(b) VPCVZRL
(c) VPVCZRL
(d) None of these

Q2. In this question, jumbled letters of a meaningful word are given. You have to rearrange these letters and select from the given alternatives the word which is almost similar in meaning to the rearranged word.
H N A G S R I
(a) Impress
(b) Compose
(c) Decorate
(d) Impose

Q3. If we substitute 1 to 24 indicating hours on the dial of a clock day and night by the letters of the English alphabet in their order starting with ‘C’ which letter will represent 18 hours?
(a) W
(b) P
(c) T
(d) R

Q4.  In a certain code language, ‘481’ means ‘sky is blue’, ‘246’ means ‘sea is deep’ and ‘698’ means ‘sea looks blue’.  What number is the code for ‘deep’?
(a) 2
(b) 4
(c) 6
(d) 9

Q5. If it is possible to make a number which is perfect square of a two-digit odd number with the fourth, the sixth and the seventh digits of the number 187642539, which of the following is the digit in the unit’s place of that two-digit odd number?
(a) 1
(b) 7
(c) 9
(d) None of the above

Q6. In a row of boys, if A who is tenth from the left and B who is ninth from the right interchange their positions, A becomes fifteenth from the left. How many boys are there in the row?
(a) 23
(b) 27
(c) 28
(d) 31

Directions (7): In each question below are given two statements followed by two conclusions numbered I and II. You have to take the given two statements to be true even if they seem to be at variance from commonly known facts. Read the conclusion and then decide which of the given conclusions logically follows from the two given statements, disregarding commonly known facts.
Q7. Statements:
All poles are guns.
Some boats are not poles.
Conclusions:
I. All guns are boats.
II. Some boats are not guns.
(a) If only conclusion I follows
(b) If only conclusion II follows
(c) If either conclusion I or II follows
(d) If neither conclusion I nor II follows

Q8. Find the missing character in the below figure form the given alternatives?

(a) M

(b) P
(c) Q
(d) S

Q9.If ‘+’ means ‘÷’, ‘×’ means ‘+’, ‘-’ means ‘×’ and ‘÷’ means ‘-’ then which of the following will be the correct equation?
(a) 20 + 8 – 7 × 6 ÷ 4 = 25
(b) 20 × 5 + 12 ÷ 6 × 5 = 15
(c) 20 ÷ 5 + 6 × 12 – 4 = 67 1/6
(d) 50 – 4 + 8 × 18 ÷ 6 = 21

Q10. Unscramble the letters in the given words and find the odd one out:
(a) EIWNTR
(b) UMRSME
(c) PIGRSN
(d) LCUOD

Q11. A meaningful word starting with A is made from the first, the second the fourth, the fifth and the sixth letters of the word CONTRACT. Which of the following is the middle letter of that word?
(a) C
(b) O
(c) T
(d) R

Directions (12-13): In each of the following questions, a word has been given, followed by four other words, one of which cannot be formed by using the letters of the given word. Find that word.
Q12. ESTRANGEMENT
(a) TANGENT
(b) GERMAN
(c) TREATMENT
(d) ENTREAT

Q13. VENTURESOME
(a) RANDOM
(b) SERMON
(c) VENTURE
(d) TENURE

Q14. In this question, the given equation becomes correct due to the interchange of two signs. One of the four alternatives under it specifies the interchange of signs in the equation which when made will make the equation correct. Find the correct alternative.
2 × 3 + 6 – 12 ÷ 4 = 17
(a) × and +
(b) + and –
(c) + and ÷
(d) – and ÷

Q15. Five children take part in a tournament. Everyone has to play with one another. How many games must they play?
(a) 8
(b) 10
(c) 24
(d) 30

Solutions

S1. Ans.(d)
Sol. Each letter in the word is replaced by the letter which occupies the same position from the other end of the English alphabet, and the group of letters so obtained is then written in the reverse order to get the code. Thus, we have:
CRUELTY → XIFVOGB → BGOVFIX

S2. Ans.(c)
Sol. The word is GARNISH meaning ‘Decorate’.

S3. Ans.(c)
Sol. 18 hours = T
S4. Ans.(a)
Sol. ‘481’ – ‘sky is blue’,
‘246’ – ‘sea is deep’
‘698’ – ‘sea looks blue’
From 1st and 2nd statement, ‘is’-4
And from 2nd and 3rd statement, ‘sea’ – 6
So, ‘deep’ – 2

S5. Ans.(d)
Sol. The the fourth, the sixth and the seventh digits of the number 187642539 are 6,2 and 5 respectively. The perfect square of a two-digit odd number, formed using these digits, is 625. And, 625= 252
So, the required odd number is 25. Clearly, its units digits is 5

S6. Ans.(a)
Sol. Clearly, A’s new position is 15th form the left. But this is the same as B’s earlier position which is 9th from the right.
So number of boys in a row = (15 + 8) = 23

S7. Ans.(d)
Sol. From the venn diagram we infer that neither conclusion I nor II follows.

S8. Ans.(b)
Sol. Putting A = 1, B = 2, C = 3, D = 4, ……………, X = 24, Y = 25, Z = 26, we have:
F + P = 6 + 16 = 22; G + N = 7 + 14 = 21; J + E = 10 + 5 = 15.
Since K = 11, so value corresponding to missing letter = (27 – 11) = 16.
So, the missing letter is the 16th letter of the English alphabet, Which is P.

S9. Ans.(c)
Sol.
After, changing signs according to the question, the new equations will be:
(a) 24 ÷ 8 ×7 + 6 – 4 = 23
(b) 20 + 5 ÷ 12 – 6 + 5 =19 (5/12)
(c) 20 –5 ÷ 6 + 12 × 4 = 67 (1/6)
(d) 50 ×4 ÷ 8 + 18 – 6 = 37
From option (3),
20 – 5 ÷ 6 + 12 × 4
= 20-5/6+48=68-5/6=67 (1/6) (True)

S10. Ans.(d)
Sol. Rearranging the letters, the given words are:
(a) WINTER (b) SUMMER (c) SPRING (d) CLOUD
Clearly, all except CLOUD are names of seasons.

S11. Ans.(c)
Sol. The first, second, fourth, fifth and sixth letters of the word CONTRACT and C, O, T, R, A respectively. The word formed is ACTOR, in which the middle letter is T.

S12. Ans.(c)
Sol. TREATMENT cannot be formed by using the letters of ESTRANGEMENT.

S13. Ans.(a)
Sol. RANDOM cannot be formed by using the letters of VENTURESOME.

S14. Ans.(a)
Sol. On interchanging × and +, we get;
Given expression = 2 + 3 × 6 – 12 ÷ 4 = 2 + 3 × 6 – 3 = 2 + 18 – 3 = 17

S15. Ans.(b)
Sol. Number of games = (n(n-1))/2
= (5(5-1))/2 = 10

General Awareness section plays a crucial role in your overall performance in SSC Stenographer Grade ‘C’ & ‘D’ Exam 2017 & IB (ACIO) you can download Capsule by below link:-

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