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.

Q3061.

A sphere and a cylinder of equal radii have equal total surface areas. Find the ratio of cylinder height to radius.

समान त्रिज्या वाले एक गोले और एक बेलन के कुल पृष्ठीय क्षेत्रफल बराबर हैं। बेलन की ऊँचाई और त्रिज्या का अनुपात ज्ञात कीजिए।

A1 : 1
B7 : 4
C6 : 5
D2 : 5

Show answer

Concept: Sphere $TSA = 4 π r^{2}$ ; cylinder $TSA = 2 π r (r + h)$ .

$4 π r^{2} = 2 π r (r + h)$ .
$2 r = r + h \to h = r$ .
$h : r = 1 : 1$ .

Trick: Cancel $2 π r$ first: $2 r = r + h$ , done in one line.

Answer: (A) 1 : 1

अवधारणा: गोले का $TSA = 4 π r^{2}$ , बेलन का $TSA = 2 π r (r + h)$ ।

$4 π r^{2} = 2 π r (r + h)$
$2 r = r + h \to h = r$
$h : r = 1 : 1$

ट्रिक: पहले $2 π r$ काट दें – $2 r = r + h$ ।

उत्तर: (A) 1 : 1

SSC CGLQuantitative AptitudeSphere and cylinder – equal TSASSC CGL 2025 · 17 Sep Shift 3

Q3062.

A large cubical cake needs to be cut into 27 smaller, identical cubical pieces. What is the minimum number of straight cuts required?

एक बड़े घनाकार केक को 27 समान छोटे घनों में काटना है। न्यूनतम कितने सीधे कट आवश्यक हैं?

Show answer

Concept: To get n pieces along an edge you need $n - 1$ cuts in that direction.

$27 = 3 \times 3 \times 3$ → 3 pieces along each edge.
Cuts = $2 + 2 + 2 = 6$ .

Trick: For $n^{3}$ equal cubes, minimum cuts = $3 (n - 1)$ .

Answer: (B) 6

अवधारणा: किसी दिशा में n टुकड़ों के लिए $n - 1$ कट चाहिए।

$27 = 3 \times 3 \times 3$ → हर किनारे पर 3 टुकड़े
कट = $2 + 2 + 2 = 6$

ट्रिक: $n^{3}$ समान घनों के लिए न्यूनतम कट = $3 (n - 1)$ ।

उत्तर: (B) 6

SSC CGLQuantitative AptitudeCube cutting – minimum cutsSSC CGL 2025 · 17 Sep Shift 3

Q3063.

The radius of a cone is raised by 25%, while the height is reduced by 10%. What is the percentage change in the volume of the cone?

एक शंकु की त्रिज्या 25% बढ़ाई जाती है, जबकि ऊँचाई 10% घटाई जाती है। शंकु के आयतन में कितने प्रतिशत परिवर्तन होगा?

A27.4% decrease27.4% कमी
B40.6% increase40.6% वृद्धि
C40.8% increase40.8% वृद्धि
D51.2% increase51.2% वृद्धि

Show answer

Concept: $V \propto r^{2} h$ , so multiply the change factors.

Factor = $(1.25)^{2} \times 0.9 = 1.5625 \times 0.9 = 1.40625$ .
Change = $+ 40.625 % \approx 40.6 %$ increase.

Trick: Fractions: $(\frac{5}{4})^{2} \times \frac{9}{10} = \frac{225}{160} = \frac{45}{32} \to \frac{13}{32} \approx 40.6 %$ .

Answer: (B) 40.6% increase

अवधारणा: $V \propto r^{2} h$ , इसलिए परिवर्तन-गुणक गुणा करें।

गुणक = $(1.25)^{2} \times 0.9 = 1.40625$
परिवर्तन ≈ 40.6% वृद्धि

ट्रिक: भिन्न में: $(\frac{5}{4})^{2} \times \frac{9}{10} = \frac{45}{32} \to \frac{13}{32} \approx 40.6 %$ ।

उत्तर: (B) 40.6% वृद्धि

SSC CGLQuantitative AptitudeCone – percentage change in volumeSSC CGL 2025 · 17 Sep Shift 3

Q3064.

If sinA =

\frac{5}{13}

and A lies in the second quadrant, find the value of (sinA + cosA)².

यदि sinA =

\frac{5}{13}

और A दूसरे चतुर्थांश में है, तो (sinA + cosA)² का मान ज्ञात कीजिए।

A0
B1
C $\frac{49}{169}$
D $\frac{39}{169}$

Show answer

Concept: In quadrant II, sin is positive and cos is negative.

$\sin A = \frac{5}{13}$ → triplet 5-12-13 → $\cos A = - \frac{12}{13}$ .
$\sin A + \cos A = \frac{5}{13} - \frac{12}{13} = - \frac{7}{13}$ .
Square: $\frac{49}{169}$ .

Trick: Add first, then square: $(- \frac{7}{13})^{2}$ – no need to expand $1 + 2 \sin A \cos A$ .

Answer: (C) $\frac{49}{169}$

अवधारणा: द्वितीय चतुर्थांश में sin धनात्मक, cos ऋणात्मक।

5-12-13 त्रिक → $\cos A = - \frac{12}{13}$
$\sin A + \cos A = - \frac{7}{13}$
वर्ग = $\frac{49}{169}$

ट्रिक: पहले जोड़ें, फिर वर्ग करें – विस्तार की ज़रूरत नहीं।

उत्तर: (C) $\frac{49}{169}$

SSC CGLQuantitative AptitudeTrigonometry – sign in second quadrantSSC CGL 2025 · 17 Sep Shift 3

Q3065.

If sinA =

\frac{p}{q}

, then what is the value of (1 + tan²A)?

यदि sinA =

\frac{p}{q}

है, तो (1 + tan²A) का मान क्या है?

A $\frac{q ²}{q ² - p ²}$
B $\frac{2 q ²}{q ² - p ²}$
C $\frac{q ²}{2 (q ² - p ²)}$
D $\frac{5 q ²}{q ² - p ²}$

Show answer

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

$\sin A = \frac{p}{q}$ → $\cos^{2} A = 1 - \frac{p^{2}}{q^{2}} = \frac{q^{2} - p^{2}}{q^{2}}$ .
$\sec^{2} A = \frac{q^{2}}{q^{2} - p^{2}}$ .

Trick: Use $\cos^{2} = 1 - \sin^{2}$ directly; no need to find the third side.

Answer: (A) $\frac{q ²}{q ² - p ²}$

अवधारणा: $1 + \tan^{2} A = \sec^{2} A = \frac{1}{\cos^{2} A}$

$\cos^{2} A = 1 - \frac{p^{2}}{q^{2}} = \frac{q^{2} - p^{2}}{q^{2}}$
$\sec^{2} A = \frac{q^{2}}{q^{2} - p^{2}}$

ट्रिक: $\cos^{2} = 1 - \sin^{2}$ सीधे लगाएँ, तीसरी भुजा की ज़रूरत नहीं।

उत्तर: (A) $\frac{q ²}{q ² - p ²}$

SSC CGLQuantitative AptitudeTrigonometric identity – sec²ASSC CGL 2025 · 17 Sep Shift 3

Q3066.

What is the slope of the line passing through points (3, 4) and (5, 8)?

बिंदुओं (3, 4) और (5, 8) से गुजरने वाली रेखा की प्रवणता क्या है?

Show answer

Concept: Slope $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$m = \frac{8 - 4}{5 - 3} = \frac{4}{2} = 2$ .

Trick: Rise over run: y goes up 4 while x goes up 2.

Answer: (A) 2

अवधारणा: ढाल $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

$m = \frac{8 - 4}{5 - 3} = \frac{4}{2} = 2$

ट्रिक: y में बढ़त ÷ x में बढ़त = $4 \div 2$ ।

उत्तर: (A) 2

SSC CGLQuantitative AptitudeSlope of a lineSSC CGL 2025 · 17 Sep Shift 3

Q3067.

A sector has a central angle of 180° and a radius of 24 cm. Another sector has a central angle of

\frac{π}{2}

radians and the same radius. What is the ratio of the area of the first sector to the area of the second sector?

एक त्रिज्यखंड का केंद्रीय कोण 180° और त्रिज्या 24 cm है। दूसरे त्रिज्यखंड का केंद्रीय कोण

\frac{π}{2}

रेडियन और त्रिज्या वही है। पहले और दूसरे त्रिज्यखंड के क्षेत्रफलों का अनुपात क्या है?

A3 : 2
B2 : 1
C1 : 2
D1 : 3

Show answer

Concept: Same radius → sector area is proportional to central angle.

$\frac{π}{2} rad = 90^{\circ}$ .
Ratio = $180^{\circ} : 90^{\circ} = 2 : 1$ .

Trick: Radius (24 cm) is a distractor when radii are equal.

Answer: (B) 2 : 1

अवधारणा: त्रिज्या समान हो तो क्षेत्रफल केंद्रीय कोण के अनुपात में।

$\frac{π}{2}$ रेडियन $= 90^{\circ}$
अनुपात = $180 : 90 = 2 : 1$

ट्रिक: समान त्रिज्या में 24 cm का मान अनावश्यक है।

उत्तर: (B) 2 : 1

SSC CGLQuantitative AptitudeArea of sectors – angle ratioSSC CGL 2025 · 17 Sep Shift 3

Q3068.

What is the total measure of all interior angles in an 8-sided polygon?

आठ भुजाओं वाले बहुभुज के सभी अंतःकोणों का कुल माप क्या है?

A1540°
B1080°
C1220°
D900°

Show answer

Concept: Sum = $(n - 2) \times 180^{\circ}$ .

$n = 8 \to 6 \times 180^{\circ} = 1080^{\circ}$ .

Trick: Answer must be a multiple of 180° – 1080 is the only one fitting n = 8 (900 is for 7 sides).

Answer: (B) 1080°

अवधारणा: योग = $(n - 2) \times 180^{\circ}$

$n = 8 \to 6 \times 180^{\circ} = 1080^{\circ}$

ट्रिक: उत्तर 180 का गुणज होगा; 900° सात भुजाओं का है।

उत्तर: (B) 1080°

SSC CGLQuantitative AptitudeSum of interior angles of a polygonSSC CGL 2025 · 17 Sep Shift 3

Q3069.

What is the equation of the perpendicular bisector of the line segment connecting the points (4, 8) and (2, 6)?

बिंदुओं (4, 8) और (2, 6) को जोड़ने वाले रेखाखंड के लंब समद्विभाजक का समीकरण क्या है?

Ay = −x − 10
By = x − 3
Cy = −x + 10
Dy = x + 4

Show answer

Concept: It passes through the midpoint and has slope = $- \frac{1}{segment slope}$ .

Midpoint = $(\frac{4 + 2}{2}, \frac{8 + 6}{2}) = (3, 7)$ .
Segment slope = $\frac{6 - 8}{2 - 4} = 1$ → perpendicular slope = −1.
$y - 7 = - (x - 3) \to y = - x + 10$ .

Trick: Slope −1 leaves (A) and (C); plug (3, 7): only $y = - x + 10$ works.

Answer: (C) y = −x + 10

अवधारणा: यह मध्यबिंदु से गुज़रता है और $ढाल = - \frac{1}{रेखाखंड की ढाल}$ ।

मध्यबिंदु = (3, 7)
रेखाखंड की ढाल = 1 → लंब ढाल = −1
$y - 7 = - (x - 3) \to y = - x + 10$

ट्रिक: ढाल −1 से (A), (C) बचे; (3, 7) रखें – केवल (C) सही।

उत्तर: (C) y = −x + 10

SSC CGLQuantitative AptitudePerpendicular bisector of a segmentSSC CGL 2025 · 17 Sep Shift 3

Q3070.

The ratios of the perimeters of two similar triangles are expressed as 5 : 7. Together, their areas total 74 cm². What is the area of the larger triangle?

दो समरूप त्रिभुजों के परिमापों का अनुपात 5 : 7 है। उनके क्षेत्रफलों का योग 74 cm² है। बड़े त्रिभुज का क्षेत्रफल क्या है?

A39 cm²
B45 cm²
C49 cm²
D35 cm²

Show answer

Concept: Area ratio of similar triangles = $(ratio of perimeters/sides)^{2}$ .

Area ratio = 25 : 49 → total 74 parts.
74 parts = 74 cm² → 1 part = 1 cm².
Larger = 49 cm².

Trick: Total $74 = 25 + 49$ is a giveaway – the squares themselves are the areas.

Answer: (C) 49 cm²

अवधारणा: समरूप त्रिभुजों का क्षेत्रफल अनुपात = $(परिमाप अनुपात)^{2}$

अनुपात 25 : 49, कुल 74 भाग
74 भाग = 74 cm² → 1 भाग = 1 cm²
बड़ा = 49 cm²

ट्रिक: $25 + 49 = 74$ – वर्ग ही सीधे क्षेत्रफल हैं।

उत्तर: (C) 49 cm²

SSC CGLQuantitative AptitudeSimilar triangles – area ratioSSC CGL 2025 · 17 Sep Shift 3

Q3071.

18³ + 20³ − 29³ + 116 is equal to:

18³ + 20³ − 29³ + 116 किसके बराबर है?

A−10441
B−16789
C0
D1

Show answer

Concept: $18 + 20 - 29 \neq 0$ , so no identity shortcut – compute directly.

Positive part: $18^{3} + 20^{3} = 5832 + 8000 = 13832$ ; add 116 → 13948.
Negative part: $29^{3} = 24389$ .
Result = $- (24389 - 13948) = - 10441$ .

Trick: Unit digits: $2 + 0 - 9 + 6 = - 1$ → a negative answer ending in 1; −10441 fits, −16789 does not.

Answer: (A) −10441

अवधारणा: $18 + 20 - 29 \neq 0$ , इसलिए सर्वसमिका नहीं लगेगी – सीधे गणना करें।

धनात्मक भाग: $18^{3} + 20^{3} = 13832$ , इसमें 116 जोड़ें → 13948
ऋणात्मक भाग: $29^{3} = 24389$
परिणाम = $- (24389 - 13948) = - 10441$

ट्रिक: इकाई अंक: $2 + 0 - 9 + 6 = - 1$ → ऋणात्मक उत्तर जिसका अंतिम अंक 1 हो – −10441।

उत्तर: (A) −10441

SSC CGLQuantitative AptitudeEvaluating cubesSSC CGL 2025 · 17 Sep Shift 3

Q3072.

If tanA = x +

\frac{1}{x}

, find sec²A.

यदि tanA = x +

\frac{1}{x}

है, तो sec²A ज्ञात कीजिए।

Ax² + $\frac{1}{x ²}$ + 5
Bx² + $\frac{1}{x ²}$ + 2
Cx² + $\frac{1}{x ²}$ + 3
Dx² + $\frac{1}{x ²}$ + 4

Show answer

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

$\tan^{2} A = (x + \frac{1}{x})^{2} = x^{2} + \frac{1}{x^{2}} + 2$ .
$\sec^{2} A = x^{2} + \frac{1}{x^{2}} + 3$ .

Trick: $(x + \frac{1}{x})^{2}$ always gives +2; then add 1 → +3.

Answer: (C) x² + $\frac{1}{x ²}$ + 3

अवधारणा: $\sec^{2} A = 1 + \tan^{2} A$

$\tan^{2} A = x^{2} + \frac{1}{x^{2}} + 2$
$\sec^{2} A = x^{2} + \frac{1}{x^{2}} + 3$

ट्रिक: $(x + \frac{1}{x})^{2}$ में +2 आता है; 1 जोड़ें → +3।

उत्तर: (C) x² + $\frac{1}{x ²}$ + 3

SSC CGLQuantitative AptitudeTrigonometric identity – sec²A from tanASSC CGL 2025 · 17 Sep Shift 3

Q3073.

A tangent line PA is drawn from an external point P to a circle. If the radius is 7 cm and the distance from P to the centre is 25 cm, what is the length of the tangent PA?

बाहरी बिंदु P से एक वृत्त पर स्पर्श रेखा PA खींची गई है। यदि त्रिज्या 7 cm और P से केंद्र की दूरी 25 cm है, तो स्पर्श रेखा PA की लंबाई क्या है?

A19 cm
B26 cm
C24 cm
D28 cm

Show answer

Concept: Radius ⟂ tangent, so $P A^{2} = O P^{2} - r^{2}$ .

$P A^{2} = 25^{2} - 7^{2} = 625 - 49 = 576$ .
PA = 24 cm.

Trick: 7-24-25 is a Pythagorean triplet.

Answer: (C) 24 cm

अवधारणा: त्रिज्या स्पर्शरेखा पर लंब, अतः $P A^{2} = O P^{2} - r^{2}$

$P A^{2} = 625 - 49 = 576$
PA = 24 cm

ट्रिक: 7-24-25 पाइथागोरस त्रिक है।

उत्तर: (C) 24 cm

SSC CGLQuantitative AptitudeLength of tangent from external pointSSC CGL 2025 · 17 Sep Shift 3

Q3074.

When two tangents, PA and PB, are drawn to a circle from a point P outside the circle, and the angle between them is 90°, what is the distance from P to the centre (O) of the circle, if the radius is 6 cm?

वृत्त के बाहर बिंदु P से दो स्पर्श रेखाएँ PA और PB खींची गई हैं और उनके बीच का कोण 90° है। यदि त्रिज्या 6 cm है, तो P से केंद्र O की दूरी क्या है?

A6 cm
B6 $\sqrt{2}$ cm
C12 cm
D6 $\sqrt{3}$ cm

Show answer

Concept: $O A ⟂ P A$ , $O B ⟂ P B$ and $∠ A P B = 90^{\circ}$ → OAPB is a square.

$∠ A O B = 360^{\circ} - 90^{\circ} - 90^{\circ} - 90^{\circ} = 90^{\circ}$ ; $O A = O B = 6$ and $P A = P B$ .
So OAPB is a square of side 6.
$O P = diagonal = 6 \sqrt{2}$ cm.

Trick: $∠ A P B = 90^{\circ} \Rightarrow O P = r \sqrt{2}$ (in general $O P = \frac{r}{\sin (\frac{1}{2} ∠ A P B)}$ ).

Answer: (B) 6 $\sqrt{2}$ cm

अवधारणा: $O A ⟂ P A$ , $O B ⟂ P B$ और $∠ A P B = 90^{\circ}$ → OAPB वर्ग है।

$∠ A O B = 90^{\circ}$ , सभी भुजाएँ 6
OAPB भुजा 6 का वर्ग
$O P =$ विकर्ण $= 6 \sqrt{2}$ cm

ट्रिक: $∠ A P B = 90^{\circ}$ हो तो $O P = r \sqrt{2}$ ।

उत्तर: (B) 6 $\sqrt{2}$ cm

SSC CGLQuantitative AptitudeTwo tangents at right angleSSC CGL 2025 · 17 Sep Shift 3

Q3075.

Multiply 0.48 by 0.7 and express the result as a decimal.

0.48 को 0.7 से गुणा कीजिए और परिणाम दशमलव में लिखिए।

A0.386
B0.336
C0.236
D0.254

Show answer

Concept: Multiply as whole numbers, then place decimal = total decimal digits.

$48 \times 7 = 336$ .
Decimal places: $2 + 1 = 3 \to 0.336$ .

Trick: Estimate: about $0.5 \times 0.7 = 0.35$ – slightly less → 0.336.

Answer: (B) 0.336

अवधारणा: पूर्णांक की तरह गुणा करें, फिर कुल दशमलव स्थान गिनें।

$48 \times 7 = 336$
दशमलव स्थान $2 + 1 = 3 \to 0.336$

ट्रिक: अनुमान: $0.5 \times 0.7 = 0.35$ से थोड़ा कम → 0.336।

उत्तर: (B) 0.336

SSC CGLQuantitative AptitudeDecimal multiplicationSSC CGL 2025 · 17 Sep Shift 3

Q3076.

Select the option that expresses the given sentence in the Active/Passive voice.
The students were made to rewrite the essay twice.

AThe teacher made the students rewrite the essay twice.
BThe students made the teacher rewrite the essay twice.
CThe students rewrote the essay twice.
DThe teacher was making the students rewrite the essay twice.

Show answer

Concept: Passive 'be made to + V1' becomes active 'make + object + V1' (no 'to').

Hidden doer = the teacher; tense = simple past → 'made'.
Active: The teacher made the students rewrite the essay twice.
(B) swaps roles, (C) loses the 'forcing' sense, (D) changes tense to past continuous.

Trick: 'To' appears only in the passive of make/let/see; it disappears in the active.

Answer: (A) The teacher made the students rewrite the essay twice.

अवधारणा: Passive 'be made to + V1' का active रूप 'make + object + V1' होता है (to हटता है)।

कर्ता = the teacher, काल = simple past → 'made'
सही: The teacher made the students rewrite the essay twice.
(B) में भूमिका उलटी, (C) में 'मजबूर करना' का भाव गायब, (D) में काल बदला।

ट्रिक: make/let/see के passive में 'to' आता है, active में हट जाता है।

उत्तर: (A) The teacher made the students rewrite the essay twice.

SSC CGLEnglish ComprehensionVoice – causative 'make' (passive to active)SSC CGL 2025 · 17 Sep Shift 3

Q3077.

Select the most appropriate option to substitute the underline/Bold segment in the given sentence.
She made a decision after thinking for a long time.

Adid a decision
Btook a decision
Chad a decision
Dbrought a decision

Show answer

Concept: 'Decision' pairs with take or make; 'do/have/bring a decision' are not idiomatic.

Among the options only 'took a decision' is a natural collocation.
Sentence: She took a decision after thinking for a long time.

Trick: Learn noun–verb pairs: take a decision, make a choice, pay attention, do homework.

Answer: (B) took a decision

अवधारणा: 'Decision' के साथ take या make आता है; do/have/bring नहीं।

विकल्पों में केवल 'took a decision' सही collocation है।
वाक्य: She took a decision after thinking for a long time.

ट्रिक: संज्ञा–क्रिया जोड़ियाँ याद करें: take a decision, pay attention, do homework।

उत्तर: (B) took a decision

SSC CGLEnglish ComprehensionCollocation – take a decisionSSC CGL 2025 · 17 Sep Shift 3

Q3078.

Select the most appropriate meaning of the highlighted idiom.
He cries stinking fish about his achievements.

AUndervaluing one's own efforts
BMaking fun of others
CFeeling proud without showing it
DPraising oneself loudly

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Concept: To 'cry stinking fish' = to run down or belittle one's own work or goods.

Picture a fish-seller shouting that his own fish stinks – he damages himself.
So he belittles/undervalues his own achievements.

Trick: Opposite idiom: 'blow one's own trumpet' (option D's meaning) – don't confuse them.

Answer: (A) Undervaluing one's own efforts

अवधारणा: 'Cry stinking fish' = अपनी ही चीज़/काम को बुरा बताना, खुद को कम आँकना।

मछली बेचने वाला खुद कहे कि उसकी मछली सड़ी है – अपना ही नुकसान।
यानी वह अपनी उपलब्धियों को कम आँकता है।

ट्रिक: उल्टा मुहावरा: 'blow one's own trumpet' (अपनी तारीफ़) – (D) वाला अर्थ।

उत्तर: (A) Undervaluing one's own efforts

SSC CGLEnglish ComprehensionIdiom – cry stinking fishSSC CGL 2025 · 17 Sep Shift 3

Q3079.

Choose the correct spelling.

ACreede
BCreed
CCread
DCreid

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Concept: Correct spelling: C-R-E-E-D (double e, ends in d).

Meaning: a set of religious or guiding beliefs.
Creede adds an extra e; Cread and Creid change the vowel.

Trick: Think of 'need', 'seed', 'greed' – creed rhymes and is spelt the same way.

Answer: (B) Creed

अवधारणा: सही वर्तनी: C-R-E-E-D।

अर्थ: धार्मिक/मूल विश्वासों का समूह (मत, पंथ)।
Creede में अतिरिक्त e, Cread व Creid में स्वर गलत।

ट्रिक: need, seed, greed जैसा ही – creed।

उत्तर: (B) Creed

SSC CGLEnglish ComprehensionSpelling – CreedSSC CGL 2025 · 17 Sep Shift 3

Q3080.

Choose the correctly spelt word.

ADeleterius
BDelitterous
CDeliterous
DDeleterious

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Concept: Correct form: DE-LE-TE-RI-OUS (three e's, then -rious).

Meaning: harmful, damaging (e.g., deleterious effects of smoking).
Other options drop the 'o', change 'e' to 'i', or double the 't'.

Trick: Split it as 'dele-terious' – like 'delete' + 'rious'.

Answer: (D) Deleterious

अवधारणा: सही वर्तनी: DE-LE-TE-RI-OUS।

अर्थ: हानिकारक।
बाकी विकल्पों में o गायब, e की जगह i या दोहरा t है।

ट्रिक: 'delete' + 'rious' की तरह तोड़कर याद रखें।

उत्तर: (D) Deleterious

SSC CGLEnglish ComprehensionSpelling – DeleteriousSSC CGL 2025 · 17 Sep Shift 3