SSC CGL 22 September 2025 · Shift 2 · General Intelligence & Reasoning — questions with answers · ShikshaSphere
Chapter 125 of 192
22 September 2025 · Shift 2
General Intelligence & Reasoning
25 questions ~25 min readFree
25 questions
A
1
In the sequence D, F, H, J, L…, which letter comes 6th?
A.M
B.L
C.N
D.J
Solution
Letter series (+2)
Each letter moves two places forward.
D(4), F(6), H(8), J(10), L(12).
The 6th term is 12 + 2 = 14.
The 14th letter is N — option (c).
2
Choose the address that is the same as the one given below.
Sample Address: HIG-21, Sector 5, Govindpura, Bhopal, Madhya Pradesh - 462022
Address matching
(a) moves the hyphen from HIG-21 to Sector-5.
(c) writes the sector in Roman numerals; (d) changes both the house number and the PIN.
Only (b) matches the sample exactly.
Hence option (b).
3
Find the smallest number which, when divided by 6, 7 and 12, leaves a remainder of 4 in each case.
A.88
B.173
C.85
D.874
Solution
LCM with remainder
The number must be 4 more than a common multiple of 6, 7 and 12.
LCM(6, 7, 12) = 84.
Required number = 84 + 4.
= 88 — option (a).
4
What mathematical symbols should replace A and B respectively to make the equation CORRECT?
11 − 60 ÷ 12 A 5 B 11 = 61
A.× and +
B.+ and ×
C.+ and −
D.× and −
Solution
Replace signs
Put + for A and × for B: 11 − 60 ÷ 12 + 5 × 11.
BODMAS: 60 ÷ 12 = 5 and 5 × 11 = 55.
11 − 5 + 55.
= 61 ✓ — option (b).
5
A statement is followed by two conclusions. Find which conclusion is true based on the given statement.Statement: A > B = C; D ≤ E < B; F ≥ C
Conclusions:
I. F > D
II. A > E
A.Only I follows
B.Only II follows
C.Both I and II follow
D.Neither I nor II follows
Solution
Coded inequality
F ≥ C = B > E ≥ D, so F is definitely greater than D — I is true.
A > B and B > E give A > E — II is true.
Both chains run in one direction throughout.
Both follow — option (c).
6
Which number is NOT a multiple of 3?
A.315
B.672
C.1014
D.725
Solution
Divisibility by 3
A number is divisible by 3 when its digit sum is divisible by 3.
315 → 9 ✓, 672 → 15 ✓, 1014 → 6 ✓.
725 → 7 + 2 + 5 = 14, which is not divisible by 3 ✗.
Hence option (d).
7
Which country does not belong to the group?
A.Germany
B.France
C.Italy
D.Japan
Solution
European countries
Germany, France and Italy are all European countries.
Japan lies in East Asia.
So it belongs to a different continent.
Hence option (d).
8
Find the odd one out.
A.Lata Mangeshkar
B.Asha Bhosle
C.A.R. Rahman
D.Sachin Tendulkar
Solution
Music vs cricket
Lata Mangeshkar, Asha Bhosle and A.R. Rahman all belong to the world of MUSIC.
Sachin Tendulkar is a cricketer.
So he stands apart from the rest.
Hence option (d).
9
A is the sister of B. B is the father of C. D is the mother of A. How is D related to C?
A.Aunt
B.Grandmother
C.Mother
D.Cousin
Solution
Blood relation
A and B are siblings, so D is the mother of B as well.
B is the father of C.
So D is one generation above C’s father.
D is C’s grandmother — option (b).
10
C is the brother of D. D is the son of E. How is C related to E?
A.Brother
B.Son
C.Father
D.Cousin
Solution
Blood relation
C and D are brothers, so they share the same parents.
D is the son of E.
Hence C is also a son of E.
Option (b).
11
Which group has letters with a position sum = 32?
A.P, Q, X
B.A, E, Y
C.B, E, Y
D.Q, N, O
Solution
Sum of place values
Write the place values: B = 2, E = 5, Y = 25.
2 + 5 + 25 = 32 ✓.
(a) gives 16+17+24 = 57, (b) gives 1+5+25 = 31 and (d) gives 17+14+15 = 46.
Hence B, E, Y — option (c).
12
Study the following information carefully and answer the question.
Six persons A, B, C, D, E and F were born in six different years — 1946, 1958, 1963, 1971, 1994 and 2006. (Age is calculated with respect to 2019.)
The sum of the ages of D and A is 86. D was born in an even-numbered year. D is older than A. B is older than D. E was born in an odd-numbered year. The sum of the ages of A and C is 81. F is younger than E.
Which of the following is the birth year of A?
A.1988
B.1994
C.1995
D.2004
Solution
Birth-year puzzle
Ages in 2019 are 73, 61, 56, 48, 25 and 13 for the six years listed.
D + A = 86 fits 61 + 25, and since D is older, D = 61 (born 1958, an even year) and A = 25.
A + C = 81 gives C = 56 (born 1963); B older than D means B = 73 (1946); E in an odd year = 1971 and F, the youngest, = 2006.
So A was born in 1994 — option (b).
13
Study the following information carefully to answer the question.
'M & N' means 'M is the mother of N'.
'M $ N' means 'M is the brother of N'.
'M # N' means 'M is the sister of N'.
'M @ N' means 'M is the father of N'.
'M % N' means 'M is the daughter of N'.
'M * N' means 'M is the son of N'.
Which of the following means R is a brother of P?
A.P * Q * R
B.Q % P & R
C.R * Q & P
D.Q * R & P
Solution
Symbol relations
Take option (c): R * Q means R is the son of Q.
Q & P means Q is the mother of P.
So R and P are both children of Q, and R is male.
R is the brother of P — option (c).
14
Read the following statements carefully and identify the conclusion that follows.Statements:
All fruits are sweet.
Some sweet things are healthy.
No healthy thing is junk food.
Conclusions:
I. Some fruits are healthy.
II. Some sweet things are not junk food.
A.Only I follows
B.Only II follows
C.Both I and II follow
D.Neither I nor II follows
Solution
Syllogism
Some sweet things are healthy and no healthy thing is junk food — those sweet things are therefore not junk food, so II follows.
Fruits lie inside the sweet group, but nothing places them in the healthy part of it.
So I is only a possibility, not a certainty.
Only II — option (b).
15
In a certain code AQZHM is written as ZPYGL. How is GDZQS written?
A.FCYPR
B.FCYQR
C.GDYQR
D.GCZQS
Solution
Coding (−1)
Every letter moves one step BACK: A−1 = Z, Q−1 = P, Z−1 = Y, H−1 = G, M−1 = L.
Apply to GDZQS: G−1 = F, D−1 = C, Z−1 = Y.
Q−1 = P, S−1 = R.
Hence FCYPR — option (a).
16
What comes next in the series?
C, F, J, O, U, ?
A.B
B.A
C.C
D.D
Solution
Series — rising gaps
Positions: C(3), F(6), J(10), O(15), U(21) — the gaps are +3, +4, +5, +6.
So the next gap is +7: 21 + 7 = 28.
Counting cyclically, 28 − 26 = 2.
The 2nd letter is B — option (a).
17
From the given alternative words, select the word which can be formed using the letters of the given word.
CONSTRUCTION
A.JOINT
B.TRACK
C.COUNT
D.LIONS
Solution
Word from letters
CONSTRUCTION has C, O, N, S, T, R, U, C, T, I, O, N.
COUNT needs C, O, U, N, T — all available ✓.
JOINT needs J, TRACK needs A and K, and LIONS needs L — none are present.
Hence option (c).
18
Which three-letter combination is different from the rest?
A.ACD
B.BDF
C.EFG
D.EGK
Solution
Letter gaps
Look at the gap between the first two letters of each group.
ACD: A → C is +2; BDF: B → D is +2; EGK: E → G is +2.
EFG: E → F is only +1, so it breaks the pattern.
Hence option (c).
19
What comes next?
KN, JO, IP, HQ, ?
A.GR
B.RH
C.FG
D.GQ
Solution
Letter pairs (−1, +1)
1st letters: K, J, I, H — each −1 ⇒ next is G.
2nd letters: N, O, P, Q — each +1 ⇒ next is R.
So the pair is G and R.
Hence GR — option (a).
20
What should come at the place of the question mark?
20, 40, 38, 76, 74, ?
A.148
B.142
C.150
D.200
Solution
Series (×2, −2)
The rule alternates ×2 and −2.
20×2 = 40, 40−2 = 38, 38×2 = 76, 76−2 = 74 ✓.
Next comes a ×2 step: 74 × 2.
= 148 — option (a).
If the cost price of 25 articles is equal to the selling price of 20 articles, what is the profit percent?
A.25%
B.22%
C.21%
D.20%
Solution
CP of 25 = SP of 20
Take the cost of one article as ₹1, so CP of 25 = ₹25 = SP of 20 articles.
SP of one article = = ₹1.25.
Profit on one article = 1.25 − 1 = ₹0.25.
Profit% = × 100 = 25% — option (a).
24
If HBSFOO is coded as ICTHYP, how is ICTHYP coded?
A.JDUJIQ
B.ICUGPP
C.JDUJQL
D.JDUHIM
Solution
Coding pattern
HBSFOO → ICTHYP: H+1 = I, B+1 = C, S+1 = T, F+2 = H, O+10 = Y, O+1 = P.
So the shifts by position are +1, +1, +1, +2, +10, +1.
Apply the same to ICTHYP: I+1 = J, C+1 = D, T+1 = U, H+2 = J, Y+10 = I (cyclic), P+1 = Q.
Hence JDUJIQ — option (a).
25
In a class, Anil receives higher grades than Suresh but lower grades than Rajesh. Which of these statements is true if Suresh scored 70?
A.Anil's marks are 65, Rajesh's marks are 70.
B.Anil's marks are 75, Rajesh's marks are 80.
C.Anil's marks are 60, Rajesh's marks are 70.
D.Anil's marks are 85, Rajesh's marks are 75.
Solution
Ordering by marks
The order must be Rajesh > Anil > Suresh(70).
So Anil must be above 70 and Rajesh above Anil.
75 and 80 satisfy both conditions ✓.
Hence option (b).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Each letter moves two places forward.
D(4), F(6), H(8), J(10), L(12).
The 6th term is 12 + 2 = 14.
The 14th letter is N — option (c).
The number must be 4 more than a common multiple of 6, 7 and 12.
LCM(6, 7, 12) = 84.
Required number = 84 + 4.
= 88 — option (a).
Put + for A and × for B: 11 − 60 ÷ 12 + 5 × 11.
BODMAS: 60 ÷ 12 = 5 and 5 × 11 = 55.
11 − 5 + 55.
= 61 ✓ — option (b).
A number is divisible by 3 when its digit sum is divisible by 3.
315 → 9 ✓, 672 → 15 ✓, 1014 → 6 ✓.
725 → 7 + 2 + 5 = 14, which is not divisible by 3 ✗.
Hence option (d).
A and B are siblings, so D is the mother of B as well.
B is the father of C.
So D is one generation above C’s father.
D is C’s grandmother — option (b).
C and D are brothers, so they share the same parents.
D is the son of E.
Hence C is also a son of E.
Option (b).
Write the place values: B = 2, E = 5, Y = 25.
2 + 5 + 25 = 32 ✓.
(a) gives 16+17+24 = 57, (b) gives 1+5+25 = 31 and (d) gives 17+14+15 = 46.
Hence B, E, Y — option (c).
Ages in 2019 are 73, 61, 56, 48, 25 and 13 for the six years listed.
D + A = 86 fits 61 + 25, and since D is older, D = 61 (born 1958, an even year) and A = 25.
A + C = 81 gives C = 56 (born 1963); B older than D means B = 73 (1946); E in an odd year = 1971 and F, the youngest, = 2006.
So A was born in 1994 — option (b).
Take option (c): R * Q means R is the son of Q.
Q & P means Q is the mother of P.
So R and P are both children of Q, and R is male.
R is the brother of P — option (c).
Some sweet things are healthy and no healthy thing is junk food — those sweet things are therefore not junk food, so II follows.
Fruits lie inside the sweet group, but nothing places them in the healthy part of it.
So I is only a possibility, not a certainty.
Only II — option (b).
Every letter moves one step BACK: A−1 = Z, Q−1 = P, Z−1 = Y, H−1 = G, M−1 = L.
Apply to GDZQS: G−1 = F, D−1 = C, Z−1 = Y.
Q−1 = P, S−1 = R.
Hence FCYPR — option (a).
Positions: C(3), F(6), J(10), O(15), U(21) — the gaps are +3, +4, +5, +6.
So the next gap is +7: 21 + 7 = 28.
Counting cyclically, 28 − 26 = 2.
The 2nd letter is B — option (a).
CONSTRUCTION has C, O, N, S, T, R, U, C, T, I, O, N.
COUNT needs C, O, U, N, T — all available ✓.
JOINT needs J, TRACK needs A and K, and LIONS needs L — none are present.
Hence option (c).
Look at the gap between the first two letters of each group.
ACD: A → C is +2; BDF: B → D is +2; EGK: E → G is +2.
EFG: E → F is only +1, so it breaks the pattern.
Hence option (c).
1st letters: K, J, I, H — each −1 ⇒ next is G.
2nd letters: N, O, P, Q — each +1 ⇒ next is R.
So the pair is G and R.
Hence GR — option (a).
The rule alternates ×2 and −2.
20×2 = 40, 40−2 = 38, 38×2 = 76, 76−2 = 74 ✓.
Next comes a ×2 step: 74 × 2.
= 148 — option (a).
Rule: a & a = a² × 3.
Check: 3² × 3 = 27 ✓; 4² × 3 = 48 ✓.
5 & 5 = 5² × 3 = 25 × 3.
= 75 — option (d).
Each term = previous × 2 − 1.
3×2−1 = 5, 5×2−1 = 9, 9×2−1 = 17, 17×2−1 = 33, 33×2−1 = 65 ✓.
Next: 65 × 2 − 1 = 130 − 1.
= 129 — option (b).
HBSFOO → ICTHYP: H+1 = I, B+1 = C, S+1 = T, F+2 = H, O+10 = Y, O+1 = P.
So the shifts by position are +1, +1, +1, +2, +10, +1.
Apply the same to ICTHYP: I+1 = J, C+1 = D, T+1 = U, H+2 = J, Y+10 = I (cyclic), P+1 = Q.
Hence JDUJIQ — option (a).
The order must be Rajesh > Anil > Suresh(70).
So Anil must be above 70 and Rajesh above Anil.
75 and 80 satisfy both conditions ✓.
Hence option (b).