SSC CGL · 4,800 questions

SSC CGL 2025 PYQs

Real previous year questions, subject-wise. Try the question, then open the answer and the explanation in Hindi & English – free, no login.

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4,800 questions.

Q2961.

A vertical tree PQ has its base on the ground at Q. From a point A 20 m away from Q, the angle of elevation of P is x°. From another point B 60 m away from Q (and in the same line as A), the angle of elevation of P is y°. If the height of the tree is 20

\sqrt{3}

m, what is the value of x° + y°?

एक ऊर्ध्वाधर वृक्ष PQ का आधार Q पर है। Q से 20 मीटर दूर बिंदु A से P का उन्नयन कोण x° है। उसी रेखा में Q से 60 मीटर दूर बिंदु B से उन्नयन कोण y° है। यदि वृक्ष की ऊँचाई 20

\sqrt{3}

मीटर है, तो x° + y° का मान क्या है?

A60°
B75°
C90°
D120°

Show answer

Concept: tan(elevation) $= \frac{height}{horizontal distance}$ .

At A: $\tan x = \frac{20 \sqrt{3}}{20} = \sqrt{3} \to x = 60^{\circ}$
At B: $\tan y = \frac{20 \sqrt{3}}{60} = \frac{1}{\sqrt{3}} \to y = 30^{\circ}$
$x + y = 90^{\circ}$

Trick: If $h^{2} = d_{1} \times d_{2}$ ( $1200 = 20 \times 60$ ), the two angles are complementary → sum 90°.

Answer: (C) 90°

अवधारणा: tan(उन्नयन कोण) = ऊँचाई/क्षैतिज दूरी।

A से: $\tan x = \sqrt{3} \to x = 60^{\circ}$
B से: $\tan y = \frac{1}{\sqrt{3}} \to y = 30^{\circ}$
$x + y = 90^{\circ}$

ट्रिक: यदि $h^{2} = d_{1} \times d_{2}$ ( $1200 = 20 \times 60$ ), तो दोनों कोण पूरक → योग 90°।

उत्तर: (C) 90°

SSC CGLQuantitative AptitudeHeights & distances – sum of anglesSSC CGL 2025 · 18 Sep Shift 1

Q2962.

Two identical solid hemispheres, each with a radius of 27 cm, are melted and recast into a single sphere. What is the radius of the new sphere formed?

27 cm त्रिज्या वाले दो समान ठोस अर्धगोलों को पिघलाकर एक गोला बनाया जाता है। नए गोले की त्रिज्या क्या है?

A21 cm
B27 cm
C24 cm
D29 cm

Show answer

Concept: Volume stays the same when solids are melted and recast.

Two hemispheres: $2 \times \frac{2}{3} π r^{3} = \frac{4}{3} π r^{3}$
This equals a sphere of the same radius.
New radius = 27 cm

Trick: Two equal hemispheres always make a sphere of the same radius — no calculation needed.

Answer: (B) 27 cm

अवधारणा: पिघलाकर दोबारा ढालने पर आयतन समान रहता है।

दो अर्धगोले: $2 \times \frac{2}{3} π r^{3} = \frac{4}{3} π r^{3}$
यह उसी त्रिज्या के गोले का आयतन है।
नई त्रिज्या = 27 cm

ट्रिक: दो समान अर्धगोले = उसी त्रिज्या का गोला — गणना की जरूरत नहीं।

उत्तर: (B) 27 cm

SSC CGLQuantitative AptitudeMensuration – recasting solidsSSC CGL 2025 · 18 Sep Shift 1

Q2963.

A right circular cone is cut parallel to its base, forming a frustum. The height of the original cone is 20 cm, and the cut is made 5 cm from the top. What is the ratio of the volume of the frustum to the volume of the entire cone?

एक लंब वृत्तीय शंकु को आधार के समांतर काटकर छिन्नक बनाया जाता है। मूल शंकु की ऊँचाई 20 cm है और कटाव शीर्ष से 5 cm पर किया गया है। छिन्नक और पूरे शंकु के आयतनों का अनुपात क्या है?

A63 : 64
B62 : 63
C16 : 19
D14 : 15

Show answer

Concept: Similar cones: $volume ratio = (height ratio)^{3}$ .

Small top cone height : full cone height = 5 : 20 = 1 : 4
Volume ratio = $1^{3} : 4^{3} = 1 : 64$
Frustum = $64 - 1 = 63$ parts → 63 : 64

Trick: Cut at $\frac{1}{n}$ of height from top → frustum : cone = $(n^{3} - 1) : n^{3}$ .

Answer: (A) 63 : 64

अवधारणा: समरूप शंकुओं में $आयतन अनुपात = (ऊँचाई अनुपात)^{3}$ ।

ऊपर का छोटा शंकु : पूरा शंकु = 5 : 20 = 1 : 4
आयतन = 1 : 64
छिन्नक = 63 भाग → 63 : 64

ट्रिक: ऊपर से $\frac{1}{n}$ ऊँचाई पर कटे तो छिन्नक : शंकु = $(n^{3} - 1) : n^{3}$ ।

उत्तर: (A) 63 : 64

SSC CGLQuantitative AptitudeFrustum – volume ratioSSC CGL 2025 · 18 Sep Shift 1

Q2964.

Each side of a regular octagon is

\frac{3}{4}

of the side of a regular hexagon. If the perimeter of the hexagon is 24 cm, what will be the perimeter of the octagon?

एक सम अष्टभुज की प्रत्येक भुजा एक सम षट्भुज की भुजा का

\frac{3}{4}

है। यदि षट्भुज का परिमाप 24 cm है, तो अष्टभुज का परिमाप क्या होगा?

A18 cm
B20 cm
C24 cm
D16 cm

Show answer

Concept: Perimeter of a regular polygon = number of sides × side.

Hexagon side = $\frac{24}{6} = 4$ cm
Octagon side = $\frac{3}{4} \times 4 = 3$ cm
Octagon perimeter = $8 \times 3 = 24$ cm

Trick: Perimeter ratio = $(8 \times \frac{3}{4}) : 6 = 1 : 1$ → same 24 cm.

Answer: (C) 24 cm

अवधारणा: सम बहुभुज का परिमाप = भुजाओं की संख्या × भुजा।

षट्भुज की भुजा = $\frac{24}{6} = 4$ cm
अष्टभुज की भुजा = $\frac{3}{4} \times 4 = 3$ cm
परिमाप = $8 \times 3 = 24$ cm

ट्रिक: $8 \times \frac{3}{4} = 6$ → दोनों का परिमाप बराबर।

उत्तर: (C) 24 cm

SSC CGLQuantitative AptitudePerimeter – regular polygonsSSC CGL 2025 · 18 Sep Shift 1

Q2965.

A decorative column is made in the shape of a regular hexagonal prism. The side length of its base is 8 cm. The column is built from multiple sections, and their heights form an arithmetic progression: the first section is 8 cm tall, and each subsequent section is 3 cm taller than the previous one. If the column has 4 sections, what is its total volume?

एक सजावटी स्तंभ सम षट्भुजीय प्रिज्म के आकार का है। उसके आधार की भुजा 8 cm है। स्तंभ कई खंडों से बना है जिनकी ऊँचाइयाँ समांतर श्रेढ़ी में हैं: पहला खंड 8 cm ऊँचा है और हर अगला खंड पिछले से 3 cm ऊँचा है। यदि स्तंभ में 4 खंड हैं, तो उसका कुल आयतन क्या है?

A4200 $\sqrt{3}$ cm³
B4800 $\sqrt{3}$ cm³
C5500 $\sqrt{3}$ cm³
D4600 $\sqrt{3}$ cm³

Show answer

Concept: Prism volume = base area × total height; regular hexagon area = $\frac{3 \sqrt{3}}{2} a^{2}$ .

Heights 8, 11, 14, 17 → total = $\frac{4}{2} \times (8 + 17) = 50$ cm
Base area = $\frac{3 \sqrt{3}}{2} \times 64 = 96 \sqrt{3}$ cm²
Volume = $96 \sqrt{3} \times 50 = 4800 \sqrt{3}$ cm³

Trick: Hexagon = 6 equilateral triangles: $6 \times \frac{\sqrt{3}}{4} \times 64 = 96 \sqrt{3}$ .

Answer: (B) 4800 $\sqrt{3}$ cm³

अवधारणा: प्रिज्म आयतन = आधार क्षेत्रफल × कुल ऊँचाई; षट्भुज क्षेत्रफल = $\frac{3 \sqrt{3}}{2} a^{2}$ ।

ऊँचाइयाँ 8, 11, 14, 17 → कुल 50 cm
आधार = $\frac{3 \sqrt{3}}{2} \times 64 = 96 \sqrt{3}$ cm²
आयतन = $96 \sqrt{3} \times 50 = 4800 \sqrt{3}$ cm³

ट्रिक: षट्भुज = 6 समबाहु त्रिभुज: $6 \times \frac{\sqrt{3}}{4} \times 64 = 96 \sqrt{3}$ ।

उत्तर: (B) 4800 $\sqrt{3}$ cm³

SSC CGLQuantitative AptitudePrism volume with AP heightsSSC CGL 2025 · 18 Sep Shift 1

Q2966.

Find the midpoint of the segment joining (1, 2) and (5, 8).

(1, 2) और (5, 8) को जोड़ने वाले रेखाखंड का मध्यबिंदु ज्ञात कीजिए।

A(3, 4)
B(5, 6)
C(3, 5)
D(6, 7)

Show answer

Concept: Midpoint = $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .

$x = \frac{1 + 5}{2} = 3$
$y = \frac{2 + 8}{2} = 5$
Midpoint = (3, 5)

Yaad rakhein: Midpoint is just the average of the coordinates.

Answer: (C) (3, 5)

अवधारणा: मध्य बिंदु = $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ ।

$x = \frac{1 + 5}{2} = 3$
$y = \frac{2 + 8}{2} = 5$
मध्य बिंदु = (3, 5)

याद रखें: मध्य बिंदु = निर्देशांकों का औसत।

उत्तर: (C) (3, 5)

SSC CGLQuantitative AptitudeCoordinate geometry – midpointSSC CGL 2025 · 18 Sep Shift 1

Q2967.

How many diagonals are there in a decagon?

एक दशभुज में कितने विकर्ण होते हैं?

Show answer

Concept: Diagonals of an n-sided polygon = $\frac{n (n - 3)}{2}$ .

Decagon: n = 10
Diagonals = $\frac{10 \times 7}{2} = 35$

Trick: Alternatively $^{10} C_{2} - 10 = 45 - 10 = 35$ (all joining lines minus sides).

Answer: (A) 35

अवधारणा: n भुजा वाले बहुभुज के विकर्ण = $\frac{n (n - 3)}{2}$ ।

दशभुज: n = 10
विकर्ण = $\frac{10 \times 7}{2} = 35$

ट्रिक: या $^{10} C_{2} - 10 = 45 - 10 = 35$ (कुल रेखाएँ − भुजाएँ)।

उत्तर: (A) 35

SSC CGLQuantitative AptitudePolygons – number of diagonalsSSC CGL 2025 · 18 Sep Shift 1

Q2968.

Triangle ABC has its centroid at point G(6, 7), and vertex A is located at (3, 5). If point D is the midpoint of side BC, what are the coordinates of D?

त्रिभुज ABC का केंद्रक G(6, 7) पर है और शीर्ष A(3, 5) पर है। यदि D भुजा BC का मध्यबिंदु है, तो D के निर्देशांक क्या हैं?

A(8.5, 9.5)
B(9.1, 10.5)
C(7.5, 8.0)
D(6.5, 9.5)

Show answer

Concept: Centroid G divides median AD as $A G : G D = 2 : 1$ .

$G = \frac{A + 2 D}{3} \to 2 D = 3 G - A$
x: $2 D_{x} = 18 - 3 = 15 \to D_{x} = 7.5$
y: $2 D_{y} = 21 - 5 = 16 \to D_{y} = 8 \to D = (7.5, 8.0)$

Trick: $D = G + \frac{G - A}{2}$ : $x = 6 + 1.5$ , $y = 7 + 1$ .

Answer: (C) (7.5, 8.0)

अवधारणा: केन्द्रक G माध्यिका AD को 2 : 1 में बाँटता है।

$G = \frac{A + 2 D}{3} \to 2 D = 3 G - A$
x: $2 D_{x} = 18 - 3 = 15 \to 7.5$
y: $2 D_{y} = 21 - 5 = 16 \to 8 \to D = (7.5, 8.0)$

ट्रिक: $D = G + \frac{G - A}{2}$ : $x = 6 + 1.5$ , $y = 7 + 1$ ।

उत्तर: (C) (7.5, 8.0)

SSC CGLQuantitative AptitudeCentroid – midpoint of opposite sideSSC CGL 2025 · 18 Sep Shift 1

Q2969.

Given

\sqrt{x + 5}

\sqrt{x - 4}

= 9, then find the value of x.

यदि

\sqrt{x + 5}

\sqrt{x - 4}

= 9 है, तो x का मान ज्ञात कीजिए।

Show answer

Concept: Pick x so both $(x + 5)$ and $(x - 4)$ become perfect squares.

The two squares differ by $(x + 5) - (x - 4) = 9$ .
Squares 25 and 16 differ by 9, and $5 + 4 = 9$ ✓.
$x + 5 = 25 \to x = 20$

Trick: Option test: $x = 20 \to \sqrt{25} + \sqrt{16} = 5 + 4 = 9$ ✓.

Answer: (B) 20

अवधारणा: x ऐसा लें कि $(x + 5)$ और $(x - 4)$ दोनों पूर्ण वर्ग बनें।

दोनों का अंतर 9 है।
25 और 16 का अंतर 9 तथा $5 + 4 = 9$ ✓।
$x + 5 = 25 \to x = 20$

ट्रिक: विकल्प जाँच: $x = 20 \to 5 + 4 = 9$ ✓।

उत्तर: (B) 20

SSC CGLQuantitative AptitudeSurd equation – value of xSSC CGL 2025 · 18 Sep Shift 1

Q2970.

The distance between the centres of two circles is 20 cm. The length of the internal common tangent joining both circles is 12 cm. If the radius of the larger circle is 10 cm, what will be the radius of the smaller circle?

दो वृत्तों के केंद्रों के बीच की दूरी 20 cm है। उनकी तिर्यक (आंतरिक) उभयनिष्ठ स्पर्श रेखा की लंबाई 12 cm है। यदि बड़े वृत्त की त्रिज्या 10 cm है, तो छोटे वृत्त की त्रिज्या क्या होगी?

A6 cm
B8 cm
C5 cm
D7 cm

Show answer

Concept: Transverse (internal) common tangent = $\sqrt{d^{2} - (r_{1} + r_{2})^{2}}$ .

$12^{2} = 20^{2} - (10 + r)^{2}$
$(10 + r)^{2} = 400 - 144 = 256$
$10 + r = 16 \to r = 6$ cm

Trick: 12, 16, 20 is a 3-4-5 triple (×4) → sum of radii = 16 directly.

Answer: (A) 6 cm

अवधारणा: अनुप्रस्थ उभयनिष्ठ स्पर्शरेखा = $\sqrt{d^{2} - (r_{1} + r_{2})^{2}}$ ।

$144 = 400 - (10 + r)^{2}$
$(10 + r)^{2} = 256$
$10 + r = 16 \to r = 6$ cm

ट्रिक: 12, 16, 20 = 3-4-5 त्रिक (×4) → त्रिज्याओं का योग 16।

उत्तर: (A) 6 cm

SSC CGLQuantitative AptitudeCircles – internal common tangentSSC CGL 2025 · 18 Sep Shift 1

Q2971.

In a cyclic quadrilateral PQRS, PR is the diameter of the circle. If ∠PRS = 55°, then what is the measure of ∠RPS?

चक्रीय चतुर्भुज PQRS में PR वृत्त का व्यास है। यदि ∠PRS = 55° है, तो ∠RPS की माप क्या है?

A40°
B35°
C45°
D90°

Show answer

Concept: An angle subtended by a diameter at the circumference is 90°.

PR is a diameter, so ∠PSR = 90°.
In △PRS: ∠RPS = 180° − 90° − 55° = 35°.

Trick: In a right triangle formed on a diameter, the two acute angles add to 90°: 90° − 55° = 35°.

Answer: (B) 35°

अवधारणा: व्यास द्वारा परिधि पर बना कोण 90° होता है।

PR व्यास है → ∠PSR = 90°।
△PRS में: ∠RPS = 180° − 90° − 55° = 35°।

ट्रिक: व्यास पर बने समकोण त्रिभुज में दोनों न्यून कोणों का योग 90°: 90° − 55° = 35°।

उत्तर: (B) 35°

SSC CGLQuantitative AptitudeCircles – angle in a semicircleSSC CGL 2025 · 18 Sep Shift 1

Q2972.

If sin⁴A − cos⁴A =

\frac{1}{2}

, find the value of cos²A.

यदि sin⁴A − cos⁴A =

\frac{1}{2}

है, तो cos²A का मान ज्ञात कीजिए।

A $\frac{1}{2}$
B $\frac{1}{4}$
C $\frac{3}{4}$
D $\frac{2}{3}$

Show answer

Concept: $a^{2} - b^{2} = (a - b) (a + b)$ and $\sin^{2} A + \cos^{2} A = 1$ .

$\sin^{4} A - \cos^{4} A = (\sin^{2} A - \cos^{2} A) (1) = \frac{1}{2}$
$(1 - \cos^{2} A) - \cos^{2} A = \frac{1}{2}$
$1 - 2 \cos^{2} A = \frac{1}{2} \to \cos^{2} A = \frac{1}{4}$

Trick: $\sin^{4} A - \cos^{4} A = - \cos 2 A = \frac{1}{2} \to \cos 2 A = - \frac{1}{2} \to A = 60^{\circ}, \cos^{2} 60^{\circ} = \frac{1}{4}$ .

Answer: (B) $\frac{1}{4}$

अवधारणा: $a^{2} - b^{2} = (a - b) (a + b)$ और $\sin^{2} A + \cos^{2} A = 1$ ।

$\sin^{4} A - \cos^{4} A = \sin^{2} A - \cos^{2} A = \frac{1}{2}$
$1 - 2 \cos^{2} A = \frac{1}{2}$
$\cos^{2} A = \frac{1}{4}$

ट्रिक: $A = 60^{\circ}$ लेकर जाँचें: $\sin^{4} - \cos^{4} = \frac{9}{16} - \frac{1}{16} = \frac{1}{2}$ ✓, $\cos^{2} 60^{\circ} = \frac{1}{4}$ ।

उत्तर: (B) $\frac{1}{4}$

SSC CGLQuantitative AptitudeTrigonometry – sin⁴A − cos⁴ASSC CGL 2025 · 18 Sep Shift 1

Q2973.

If cotA = x +

\frac{1}{x}

, find sec²A.

यदि cotA = x +

\frac{1}{x}

है, तो sec²A ज्ञात कीजिए।

A1 + (1 + $\frac{1}{x}$ )²
B2 + (x + $\frac{1}{x}$ )²
C1 + (x + $\frac{1}{x}$ )⁻²
D2 + (x + $\frac{1}{x}$ )⁻²

Show answer

Concept: $\sec^{2} A = 1 + \tan^{2} A$ and $\tan A = \frac{1}{\cot A}$ .

$\tan A = \frac{1}{x + \frac{1}{x}}$
$\tan^{2} A = (x + \frac{1}{x})^{- 2}$
$\sec^{2} A = 1 + (x + \frac{1}{x})^{- 2}$

Trick: $\sec^{2} A$ is always $> 1$ and here the added part must be the reciprocal square of cot A — only (C) has the "1 +" and the power −2.

Answer: (C) 1 + (x + $\frac{1}{x}$ )⁻²

अवधारणा: $\sec^{2} A = 1 + \tan^{2} A$ तथा $\tan A = \frac{1}{\cot A}$ ।

$\tan A = \frac{1}{x + \frac{1}{x}}$
$\tan^{2} A = (x + \frac{1}{x})^{- 2}$
$\sec^{2} A = 1 + (x + \frac{1}{x})^{- 2}$

ट्रिक: "1 +" और घात −2 दोनों केवल (C) में हैं।

उत्तर: (C) 1 + (x + $\frac{1}{x}$ )⁻²

SSC CGLQuantitative AptitudeTrigonometric identities – sec²ASSC CGL 2025 · 18 Sep Shift 1

Q2974.

What is the value of (0.11³ + 0.08³) ÷ (0.22³ + 0.16³)?

(0.11³ + 0.08³) ÷ (0.22³ + 0.16³) का मान क्या है?

A0.125
B0.5
C0.25
D1

Show answer

Concept: If every term doubles, the sum of cubes becomes $2^{3} = 8$ times.

$0.22 = 2 \times 0.11, 0.16 = 2 \times 0.08$
Denominator = $8 ({0.11}^{3} + {0.08}^{3})$
Value = $\frac{1}{8} = 0.125$

Trick: Ratio of bases is 1 : 2 → answer = $(\frac{1}{2})^{3} = \frac{1}{8}$ .

Answer: (A) 0.125

अवधारणा: हर पद दोगुना हो तो घनों का योग $2^{3} = 8$ गुना हो जाता है।

$0.22 = 2 \times 0.11, 0.16 = 2 \times 0.08$
हर = $8 ({0.11}^{3} + {0.08}^{3})$
मान = $\frac{1}{8} = 0.125$

ट्रिक: आधारों का अनुपात 1 : 2 → उत्तर $(\frac{1}{2})^{3} = \frac{1}{8}$ ।

उत्तर: (A) 0.125

SSC CGLQuantitative AptitudeSimplification – common factor of cubesSSC CGL 2025 · 18 Sep Shift 1

Q2975.

Simplify: {(0.6)³ + (0.06)³ + (0.006)³} ÷ {(0.2)³ + (0.02)³ + (0.002)³}

सरल कीजिए: {(0.6)³ + (0.06)³ + (0.006)³} ÷ {(0.2)³ + (0.02)³ + (0.002)³}

Show answer

Concept: If each numerator term is k times the matching denominator term, the ratio of cube sums is $k^{3}$ .

$\frac{0.6}{0.2} = \frac{0.06}{0.02} = \frac{0.006}{0.002} = 3$
Numerator = $3^{3} \times denominator$
Value = 27

Trick: Spot a constant term-by-term ratio and cube it — no cubing of decimals needed.

Answer: (C) 27

अवधारणा: अंश का हर पद हर के संगत पद का k गुना हो, तो घन-योगों का अनुपात $k^{3}$ ।

$\frac{0.6}{0.2} = \frac{0.06}{0.02} = \frac{0.006}{0.002} = 3$
अंश = $3^{3} \times$ हर
मान = 27

ट्रिक: पदों का स्थिर अनुपात देखें और उसका घन लें।

उत्तर: (C) 27

SSC CGLQuantitative AptitudeSimplification – ratio of sums of cubesSSC CGL 2025 · 18 Sep Shift 1

Q2976.

Select the most appropriate option to fill in the blank.
The historian often alluded ____ ancient myths in his lectures, enriching the narrative.

Aat
Bbut
Cto
Dtowards

Show answer

Concept: Some verbs take a fixed preposition; "allude" (to refer indirectly) always takes "to".

Correct: "The historian often alluded to ancient myths…"
"at", "towards" and "but" never pair with allude.

Yaad rakhein: Same family: refer to, allude to, object to, adhere to.

Answer: (C) to

अवधारणा: कुछ क्रियाओं के साथ निश्चित preposition आता है; "allude" (अप्रत्यक्ष संकेत करना) के साथ हमेशा "to"।

सही: "The historian often alluded to ancient myths…"
"at", "towards", "but" allude के साथ नहीं आते।

याद रखें: इसी तरह: refer to, object to, adhere to।

उत्तर: (C) to

SSC CGLEnglish ComprehensionPrepositions – allude toSSC CGL 2025 · 18 Sep Shift 1

Q2977.

The following sentence has been divided into parts. One of them contains an error. Select the part that contains the error from the given options.
After years of hard work (1)/ he finally decided to call off from (2)/ his struggling business and start anew (3)/ in a different field. (4)

A(1)
B(2)
C(3)
D(4)

Show answer

Concept: "Call off" is a complete phrasal verb (= cancel/abandon) and takes its object directly.

Error in part (2): "call off from" — the extra "from" is wrong.
Remove "from": the object (his struggling business) follows the phrasal verb directly.

Trick: Phrasal verbs already contain their particle; do not add a second preposition after them.

Answer: (B) (2)

अवधारणा: "Call off" (रद्द करना/छोड़ देना) पूर्ण phrasal verb है; इसके बाद object सीधे आता है।

भाग (2) में त्रुटि: "call off from" — "from" अनावश्यक।
सही: "…decided to call off his struggling business…"

ट्रिक: Phrasal verb के बाद एक और preposition न जोड़ें।

उत्तर: (B) (2)

SSC CGLEnglish ComprehensionError spotting – phrasal verb "call off"SSC CGL 2025 · 18 Sep Shift 1

Q2978.

The following sentence has been divided into parts. One of them contains an error. Select the part that contains the error from the given options.
The museum, which located near the river, (1)/ attracts thousands of visitors (2)/ every year for its rare collection (3)/ of ancient artefacts. (4)

A(1)
B(2)
C(3)
D(4)

Show answer

Concept: A relative clause beginning with "which" must have its own finite verb.

Part (1): "which located near the river" lacks a verb.
"Located" here is passive, so add "is": "The museum, which is located near the river,…"

Trick: Either keep "which is located" or drop "which" entirely ("The museum, located near the river,…").

Answer: (A) (1)

अवधारणा: "which" से शुरू होने वाले clause में अपनी मुख्य क्रिया होनी चाहिए।

भाग (1): "which located near the river" में क्रिया नहीं है।
सही: "The museum, which is located near the river,…"

ट्रिक: या तो "which is located" लिखें या "which" हटा दें।

उत्तर: (A) (1)

SSC CGLEnglish ComprehensionError spotting – relative clause needs a verbSSC CGL 2025 · 18 Sep Shift 1

Q2979.

Select the most appropriate synonym of the given word: Convincing

AWeak
BDubious
CCogent
DVague

Show answer

Concept: "Convincing" = persuasive, able to make people believe.

"Cogent" = clear, logical and persuasive → same meaning.
Weak, Dubious (doubtful), Vague (unclear) are nearly opposite.

Yaad rakhein: Cogent argument = convincing argument — a common exam collocation.

Answer: (C) Cogent

अवधारणा: "Convincing" = विश्वास दिलाने वाला, प्रभावशाली।

"Cogent" = तर्कसंगत और प्रभावशाली → समानार्थी।
Weak, Dubious (संदिग्ध), Vague (अस्पष्ट) — उल्टे अर्थ।

याद रखें: "Cogent argument" = ठोस, विश्वसनीय तर्क — परीक्षा में बार-बार आता है।

उत्तर: (C) Cogent

SSC CGLEnglish ComprehensionSynonym – ConvincingSSC CGL 2025 · 18 Sep Shift 1

Q2980.

Choose the correct meaning of the idiom:
The young athlete won his laurels at the Olympic games.

AAchieved recognition or fame
BLost all possessions
CChanged his view points
DSaved a lot of money

Show answer

Concept: Laurels (a wreath of bay leaves) were the ancient prize for victory, so "laurels" stands for honour.

"Won his laurels" = gained recognition or fame.
The Olympic context supports fame from victory.

Yaad rakhein: Rest on one's laurels = stop trying after early success; look to one's laurels = guard one's position.

Answer: (A) Achieved recognition or fame

अवधारणा: Laurel (तेजपत्ते का मुकुट) प्राचीन काल में विजय का पुरस्कार था, अतः laurels = सम्मान।

"Won his laurels" = ख्याति या सम्मान पाया।
ओलंपिक का संदर्भ भी जीत से मिली ख्याति दर्शाता है।

याद रखें: Rest on one's laurels = पुरानी सफलता पर संतुष्ट होकर प्रयास छोड़ देना।

उत्तर: (A) Achieved recognition or fame

SSC CGLEnglish ComprehensionIdiom – win one's laurelsSSC CGL 2025 · 18 Sep Shift 1