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Q2261.

From a point A on the ground, the angle of elevation to the top of a building is 30°. From a point B, which is directly between A and the foot of the building, the angle of elevation is 45°. What is the ratio of the distance of A from the foot to the distance of B from the foot?

ज़मीन पर स्थित बिंदु A से किसी भवन के शीर्ष का उन्नयन कोण 30° है। A और भवन के आधार के बीच स्थित बिंदु B से उन्नयन कोण 45° है। भवन के आधार से A की दूरी और B की दूरी का अनुपात क्या है?

A $\sqrt{2}$ : 3
B $\sqrt{3}$ : 1
C $\sqrt{3}$ : $\sqrt{2}$
D2 : 3

Show answer

Concept: With the same height h, distance = $\frac{h}{\tan θ}$ .

From A: $\tan 30^{\circ} = \frac{h}{D A} \to D A = h \sqrt{3}$
From B: $\tan 45^{\circ} = \frac{h}{D B} \to D B = h$
$D A : D B = h \sqrt{3} : h = \sqrt{3} : 1$

Trick: Distance ratio = $\cot 30^{\circ} : \cot 45^{\circ} = \sqrt{3} : 1$ – no need to assume h.

Answer: (B) $\sqrt{3}$ : 1

अवधारणा: समान ऊँचाई h के लिए दूरी = $\frac{h}{\tan θ}$ ।

हल:

A से: $\tan 30^{\circ} = \frac{h}{D A} \to D A = h \sqrt{3}$
B से: $\tan 45^{\circ} = \frac{h}{D B} \to D B = h$
$D A : D B = h \sqrt{3} : h = \sqrt{3} : 1$

ट्रिक: दूरियों का अनुपात = $\cot 30^{\circ} : \cot 45^{\circ} = \sqrt{3} : 1$ – h मानने की ज़रूरत नहीं।

उत्तर: (B) $\sqrt{3}$ : 1

SSC CGLQuantitative AptitudeHeights & distances – ratio of distancesSSC CGL 2025 · 19 Sep Shift 2 →

Q2262.

If the curved surface area of a sphere is 154 cm², find its diameter.

यदि किसी गोले का वक्र पृष्ठीय क्षेत्रफल 154 cm² है, तो उसका व्यास ज्ञात कीजिए।

A9 cm
B8 cm
C7 cm
D5 cm

Show answer

Concept: Surface area of sphere = $4 π r^{2}$ .

$4 \times \frac{22}{7} \times r^{2} = 154$
$r^{2} = 154 \times \frac{7}{88} = 12.25$
$r = 3.5$ cm → diameter = 7 cm

Trick: $4 π r^{2} = π d^{2}$ , so $\frac{22}{7} d^{2} = 154 \to d^{2} = 49 \to d = 7$ directly.

Answer: (C) 7 cm

अवधारणा: गोले का पृष्ठीय क्षेत्रफल = $4 π r^{2}$ ।

हल:

$4 \times \frac{22}{7} \times r^{2} = 154$
$r^{2} = 154 \times \frac{7}{88} = 12.25$
$r = 3.5$ cm → व्यास = 7 cm

ट्रिक: $4 π r^{2} = π d^{2}$ , अतः $\frac{22}{7} d^{2} = 154 \to d^{2} = 49 \to d = 7$ सीधे।

उत्तर: (C) 7 cm

SSC CGLQuantitative AptitudeSphere – diameter from surface areaSSC CGL 2025 · 19 Sep Shift 2 →

Q2263.

A cone has height h and radius r. It is melted and recast into a smaller cone whose height is

\frac{h}{4}

and radius is

\frac{r}{2}

. What fraction of the original volume is unused?

एक शंकु की ऊँचाई h और त्रिज्या r है। उसे पिघलाकर ऐसा छोटा शंकु बनाया जाता है जिसकी ऊँचाई

\frac{h}{4}

और त्रिज्या

\frac{r}{2}

है। मूल आयतन का कितना भाग अप्रयुक्त रहता है?

A $\frac{17}{18}$
B $\frac{15}{16}$
C $\frac{12}{17}$
D $\frac{16}{19}$

Show answer

Concept: Cone volume ∝ $r^{2} h$ .

New cone: radius $\frac{r}{2}$ , height $\frac{h}{4}$
New volume = ${(\frac{1}{2})}^{2} \times \frac{1}{4} = \frac{1}{16}$ of original
Unused = $1 - \frac{1}{16} = \frac{15}{16}$

Trick: Scale factors multiply: ${(\frac{1}{2})}^{2} \times \frac{1}{4} = \frac{1}{16}$ used → $\frac{15}{16}$ left.

Answer: (B) $\frac{15}{16}$

अवधारणा: शंकु का आयतन ∝ $r^{2} h$ ।

हल:

नया शंकु: त्रिज्या $\frac{r}{2}$ , ऊँचाई $\frac{h}{4}$
नया आयतन = ${(\frac{1}{2})}^{2} \times \frac{1}{4}$ = मूल का $\frac{1}{16}$
अप्रयुक्त = $1 - \frac{1}{16} = \frac{15}{16}$

ट्रिक: गुणांक गुणा करें: ${(\frac{1}{2})}^{2} \times \frac{1}{4} = \frac{1}{16}$ प्रयुक्त → $\frac{15}{16}$ बचा।

उत्तर: (B) $\frac{15}{16}$

SSC CGLQuantitative AptitudeCone – recasting, unused volumeSSC CGL 2025 · 19 Sep Shift 2 →

Q2264.

If a regular hexagon and a square have the same side length, what will be the ratio of their areas?

यदि एक सम षट्भुज और एक वर्ग की भुजा समान हो, तो उनके क्षेत्रफलों का अनुपात क्या होगा?

A3 $\sqrt{3}$ : 2
B2 : 3 $\sqrt{3}$
C $\sqrt{3}$ : 1
D1 : 6

Show answer

Concept: A regular hexagon = 6 equilateral triangles: area = $6 \times \frac{\sqrt{3}}{4} a^{2} = \frac{3 \sqrt{3}}{2} a^{2}$ .

Hexagon area = $\frac{3 \sqrt{3}}{2} a^{2}$
Square area = $a^{2}$
Ratio = $\frac{3 \sqrt{3}}{2} : 1 = 3 \sqrt{3} : 2$

Trick: Hexagon area $\approx 2.6 a^{2}$ , more than the square's $a^{2}$ – so the ratio must exceed 1; only $3 \sqrt{3} : 2$ (≈ 2.6) and 3 : 1 qualify, and $2.6 \neq 3$ .

Answer: (A) 3 $\sqrt{3}$ : 2

अवधारणा: सम षट्भुज = 6 समबाहु त्रिभुज: क्षेत्रफल = $6 \times \frac{\sqrt{3}}{4} a^{2} = \frac{3 \sqrt{3}}{2} a^{2}$ ।

हल:

षट्भुज का क्षेत्रफल = $\frac{3 \sqrt{3}}{2} a^{2}$
वर्ग का क्षेत्रफल = $a^{2}$
अनुपात = $\frac{3 \sqrt{3}}{2} : 1 = 3 \sqrt{3} : 2$

ट्रिक: षट्भुज $\approx 2.6 a^{2}$ , वर्ग $a^{2}$ से बड़ा – अनुपात 1 से अधिक; $3 \sqrt{3} : 2$ ≈ 2.6 ही सही।

उत्तर: (A) 3 $\sqrt{3}$ : 2

SSC CGLQuantitative AptitudeAreas – regular hexagon vs squareSSC CGL 2025 · 19 Sep Shift 2 →

Q2265.

A prism has an equilateral triangular base with a side length of 6 cm. Its height starts at 1 cm and increases by 1 cm for each subsequent layer, forming 5 layers in total. What is the total volume of the prism? (Use

\sqrt{3}

= 1.732)

एक प्रिज़्म का आधार 6 cm भुजा वाला समबाहु त्रिभुज है। उसकी ऊँचाई 1 cm से शुरू होकर हर अगली परत में 1 cm बढ़ती है, कुल 5 परतें हैं। प्रिज़्म का कुल आयतन क्या है? (

\sqrt{3}

= 1.732 लीजिए)

A207.84 cm³
B233.82 cm³
C259.80 cm³
D303.10 cm³

Show answer

Concept: Volume of prism = base area × height; layers with the same base simply add their heights.

Base area = $\frac{\sqrt{3}}{4} \times 6^{2} = 9 \sqrt{3}$ cm²
Total height = $1 + 2 + 3 + 4 + 5 = 15$ cm
Volume = $9 \sqrt{3} \times 15 = 135 \times 1.732 = 233.82$ cm³

Trick: $135 \times 1.732$ : $135 \times 1.7 = 229.5$ , plus $135 \times 0.032 = 4.32$ → 233.82.

Answer: (B) 233.82 cm³

अवधारणा: प्रिज़्म का आयतन = आधार क्षेत्रफल × ऊँचाई; समान आधार वाली परतों की ऊँचाइयाँ जुड़ती हैं।

हल:

आधार क्षेत्रफल = $\frac{\sqrt{3}}{4} \times 6^{2} = 9 \sqrt{3}$ cm²
कुल ऊँचाई = $1 + 2 + 3 + 4 + 5 = 15$ cm
आयतन = $9 \sqrt{3} \times 15 = 135 \times 1.732 = 233.82$ cm³

ट्रिक: $135 \times 1.732$ : $135 \times 1.7 = 229.5$ + $135 \times 0.032 = 4.32$ → 233.82।

उत्तर: (B) 233.82 cm³

SSC CGLQuantitative AptitudePrism – layered volumeSSC CGL 2025 · 19 Sep Shift 2 →

Q2266.

What is the equation of the line through (3, −5) with slope −2?

प्रवणता −2 वाली, बिंदु (3, −5) से गुजरने वाली रेखा का समीकरण क्या है?

Ay = −2x − 5
By = −2x + 1
Cy = −2x − 2
Dy = −2x + 3

Show answer

Concept: Line through (x₁, y₁) with slope m: y − y₁ = m(x − x₁).

Put m = −2, point (3, −5): y − (−5) = −2(x − 3)
So y = −2x + 6 − 5
y = −2x + 1

Trick: All options have slope −2; just put x = 3: y must be −5 → −6 + 1 = −5 ✓ only for (B).

Answer: (B) y = −2x + 1

अवधारणा: बिंदु (x₁, y₁) और प्रवणता m वाली रेखा: y − y₁ = m(x − x₁)।

हल:

y − (−5) = −2(x − 3)
y + 5 = −2x + 6
y = −2x + 1

ट्रिक: सभी विकल्पों में प्रवणता −2 है; x = 3 रखें, y = −5 आना चाहिए → केवल (B): −6 + 1 = −5।

उत्तर: (B) y = −2x + 1

SSC CGLQuantitative AptitudeCoordinate geometry – point-slope formSSC CGL 2025 · 19 Sep Shift 2 →

Q2267.

If every exterior angle of a regular polygon is 72°, how many sides does the polygon have?

यदि किसी सम बहुभुज का प्रत्येक बाह्य कोण 72° है, तो उसकी कितनी भुजाएँ हैं?

Show answer

Concept: Sum of exterior angles of any polygon = 360°; n = 360°/exterior angle.

$n = \frac{360}{72}$
$n = 5$
It is a regular pentagon.

Yaad rakhein: Interior angle = $180^{\circ} - 72^{\circ} = 108^{\circ}$ , the well-known interior angle of a regular pentagon.

Answer: (B) 5

अवधारणा: किसी भी बहुभुज के बाह्य कोणों का योग = 360°; n = 360°/बाह्य कोण।

हल:

$n = \frac{360}{72}$
$n = 5$
यह सम पंचभुज है।

याद रखें: अंतः कोण = $180^{\circ} - 72^{\circ} = 108^{\circ}$ , जो सम पंचभुज का अंतः कोण है।

उत्तर: (B) 5

SSC CGLQuantitative AptitudePolygon – sides from exterior angleSSC CGL 2025 · 19 Sep Shift 2 →

Q2268.

In a triangle PQR, the centroid is at G(3, 1). If the vertices P and Q are at (5, −2) and (−1, 3) respectively, what are the coordinates of vertex R?

त्रिभुज PQR का केंद्रक G(3, 1) है। यदि शीर्ष P और Q क्रमशः (5, −2) और (−1, 3) पर हैं, तो शीर्ष R के निर्देशांक क्या हैं?

A(5, 2)
B(2, 4)
C(1, 3)
D(2, 3)

Show answer

Concept: Centroid $G = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$ .

x: $\frac{5 - 1 + x}{3} = 3 \to 4 + x = 9 \to x = 5$
y: $\frac{- 2 + 3 + y}{3} = 1 \to 1 + y = 3 \to y = 2$
$R = (5, 2)$

Trick: Third vertex = $3 G - P - Q = (9 - 5 + 1, 3 + 2 - 3) = (5, 2)$ .

Answer: (A) (5, 2)

अवधारणा: केंद्रक $G = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$ ।

हल:

x: $\frac{5 - 1 + x}{3} = 3 \to 4 + x = 9 \to x = 5$
y: $\frac{- 2 + 3 + y}{3} = 1 \to 1 + y = 3 \to y = 2$
$R = (5, 2)$

ट्रिक: तीसरा शीर्ष = $3 G - P - Q = (9 - 5 + 1, 3 + 2 - 3) = (5, 2)$ ।

उत्तर: (A) (5, 2)

SSC CGLQuantitative AptitudeCentroid – finding the third vertexSSC CGL 2025 · 19 Sep Shift 2 →

Q2269.

Evaluate: (

\sqrt{13}

\sqrt{5}

)² − (

\sqrt{13}

−

\sqrt{5}

)²

मान ज्ञात कीजिए: (

\sqrt{13}

\sqrt{5}

)² − (

\sqrt{13}

−

\sqrt{5}

)²

A4 $\sqrt{65}$
B2 $\sqrt{65}$
C6 $\sqrt{65}$
D8 $\sqrt{65}$

Show answer

Concept: Square of a sum minus square of a difference equals 4 times the product.

Take the two terms as $\sqrt{13}$ and $\sqrt{5}$
Required value = $4 \times \sqrt{13} \times \sqrt{5} = 4 \sqrt{65}$

Trick: Never expand both squares – the cross term 2ab appears with opposite signs, leaving 4ab.

Answer: (A) 4 $\sqrt{65}$

अवधारणा: $(a + b)^{2} - (a - b)^{2} = 4 a b$ ।

हल:

यहाँ $a = \sqrt{13}$ और $b = \sqrt{5}$
$4 a b = 4 \times \sqrt{13} \times \sqrt{5} = 4 \sqrt{65}$

ट्रिक: दोनों वर्ग खोलने की ज़रूरत नहीं – 2ab विपरीत चिह्नों से आता है, केवल 4ab बचता है।

उत्तर: (A) 4 $\sqrt{65}$

SSC CGLQuantitative AptitudeIdentity (a + b)² − (a − b)² = 4abSSC CGL 2025 · 19 Sep Shift 2 →

Q2270.

Two circles with radii r₁ and r₂ have their centres separated by a distance d. If the length of a direct common tangent equals the distance between the centres, which of the following is true?

r₁ और r₂ त्रिज्या वाले दो वृत्तों के केंद्रों के बीच दूरी d है। यदि प्रत्यक्ष (सीधी) उभयनिष्ठ स्पर्श रेखा की लंबाई केंद्रों के बीच की दूरी के बराबर हो, तो निम्न में से कौन सत्य है?

Ar₁ × r₂ = 0
B(r₁ − r₂) = 0
C(r₁ + r₂) = 0
Dr₁ = 2r₂

Show answer

Concept: Length of the direct common tangent $L = \sqrt{d^{2} - (r_{1} - r_{2})^{2}}$ .

Given $L = d$ : $\sqrt{d^{2} - (r_{1} - r_{2})^{2}} = d$ .
Squaring: $d^{2} - (r_{1} - r_{2})^{2} = d^{2} \Rightarrow (r_{1} - r_{2})^{2} = 0$ .
So $r_{1} - r_{2} = 0$ , i.e. the circles have equal radii (the tangent is then parallel to $O_{1} O_{2}$ ).

Trick: Direct tangent uses $(r_{1} - r_{2})$ , transverse uses $(r_{1} + r_{2})$ . Equal radii make the direct tangent equal to the distance between the centres.

Answer: (B) (r₁ − r₂) = 0

अवधारणा: प्रत्यक्ष उभयनिष्ठ स्पर्श रेखा की लंबाई $L = \sqrt{d^{2} - (r_{1} - r_{2})^{2}}$ ।

दिया है $L = d$ : $\sqrt{d^{2} - (r_{1} - r_{2})^{2}} = d$ ।
वर्ग करने पर: $d^{2} - (r_{1} - r_{2})^{2} = d^{2} \Rightarrow (r_{1} - r_{2})^{2} = 0$ ।
अतः $r_{1} - r_{2} = 0$ , अर्थात दोनों वृत्तों की त्रिज्याएँ बराबर हैं (तब स्पर्श रेखा $O_{1} O_{2}$ के समांतर होती है)।

याद रखें: प्रत्यक्ष स्पर्श रेखा में $(r_{1} - r_{2})$ , अनुप्रस्थ में $(r_{1} + r_{2})$ । बराबर त्रिज्याओं पर प्रत्यक्ष स्पर्श रेखा केंद्रों की दूरी के बराबर होती है।

उत्तर: (B) (r₁ − r₂) = 0

SSC CGLQuantitative AptitudeCircles – direct common tangentSSC CGL 2025 · 19 Sep Shift 2 →

Q2271.

A chord of a circle has a length of 2

\sqrt{17}

cm. The distance of the chord from the centre is 8 cm. What is the radius of the circle?

एक वृत्त की जीवा की लंबाई 2

\sqrt{17}

cm है। जीवा की केंद्र से दूरी 8 cm है। वृत्त की त्रिज्या क्या है?

A6.89 cm
B6 cm
C9.52 cm
D9 cm

Show answer

Concept: The perpendicular from the centre bisects the chord: $r^{2} = (half chord)^{2} + {distance}^{2}$ .

Chord = $2 \sqrt{17}$ → half chord $= \sqrt{17}$
$r^{2} = (\sqrt{17})^{2} + 8^{2} = 17 + 64 = 81$
$r = 9$ cm

Trick: The radius must be more than the distance 8 cm – only 9 cm and 9.52 cm qualify, and 81 is a perfect square.

Answer: (D) 9 cm

अवधारणा: केंद्र से लंब जीवा को समद्विभाजित करता है: $r^{2} = (आधी जीवा)^{2} + {दूरी}^{2}$ ।

हल:

जीवा = $2 \sqrt{17}$ → आधी जीवा $= \sqrt{17}$
$r^{2} = (\sqrt{17})^{2} + 8^{2} = 17 + 64 = 81$
$r = 9$ cm

ट्रिक: त्रिज्या दूरी 8 cm से बड़ी होगी – 9 और 9.52 बचते हैं; 81 पूर्ण वर्ग है, अतः 9।

उत्तर: (D) 9 cm

SSC CGLQuantitative AptitudeCircle – chord and radiusSSC CGL 2025 · 19 Sep Shift 2 →

Q2272.

If sinA + cosA = Y, then find the value of sin²A + cos²A + 2 sinA cosA.

यदि sinA + cosA = Y है, तो sin²A + cos²A + 2 sinA cosA का मान ज्ञात कीजिए।

AY
BY²
C1
DY + 1

Show answer

Concept: $a^{2} + b^{2} + 2 a b = (a + b)^{2}$ .

$\sin^{2} A + \cos^{2} A + 2 \sin A \cos A = (\sin A + \cos A)^{2}$
$= Y^{2}$

Trick: Don't replace $\sin^{2} A + \cos^{2} A$ by 1 here – that leads to $1 + 2 \sin A \cos A$ , which still equals $Y^{2}$ . Recognise the square pattern at once.

Answer: (B) Y²

अवधारणा: $a^{2} + b^{2} + 2 a b = (a + b)^{2}$ ।

हल:

$\sin^{2} A + \cos^{2} A + 2 \sin A \cos A = (\sin A + \cos A)^{2}$
$= Y^{2}$

ट्रिक: पूर्ण वर्ग का पैटर्न तुरंत पहचानें – $(\sin A + \cos A)^{2} = Y^{2}$ ।

उत्तर: (B) Y²

SSC CGLQuantitative AptitudeTrigonometry – (sin A + cos A)²SSC CGL 2025 · 19 Sep Shift 2 →

Q2273.

If sinx =

\sqrt{3}

cosx, find sin⁴x + cos⁴x.

यदि sinx =

\sqrt{3}

cosx है, तो sin⁴x + cos⁴x ज्ञात कीजिए।

A $\frac{3}{8}$
B $\frac{5}{8}$
C $\frac{1}{8}$
D $\frac{1}{6}$

Show answer

Concept: $\sin x = \sqrt{3} \cos x \to \tan x = \sqrt{3} \to x = 60^{\circ}$ .

$\sin 60^{\circ} = \frac{\sqrt{3}}{2} \to \sin^{4} = \frac{9}{16}$
$\cos 60^{\circ} = \frac{1}{2} \to \cos^{4} = \frac{1}{16}$
Sum $= \frac{10}{16} = \frac{5}{8}$

Trick: $\sin^{4} + \cos^{4} = 1 - 2 \sin^{2} \cos^{2} = 1 - 2 \times \frac{3}{4} \cdot \frac{1}{4} = 1 - \frac{3}{8} = \frac{5}{8}$ .

Answer: (B) $\frac{5}{8}$

अवधारणा: $\sin x = \sqrt{3} \cos x \to \tan x = \sqrt{3} \to x = 60^{\circ}$ ।

हल:

$\sin 60^{\circ} = \frac{\sqrt{3}}{2} \to \sin^{4} = \frac{9}{16}$
$\cos 60^{\circ} = \frac{1}{2} \to \cos^{4} = \frac{1}{16}$
योग $= \frac{10}{16} = \frac{5}{8}$

ट्रिक: $\sin^{4} + \cos^{4} = 1 - 2 \sin^{2} \cos^{2} = 1 - 2 \times \frac{3}{4} \cdot \frac{1}{4} = \frac{5}{8}$ ।

उत्तर: (B) $\frac{5}{8}$

SSC CGLQuantitative AptitudeTrigonometry – sin⁴x + cos⁴xSSC CGL 2025 · 19 Sep Shift 2 →

Q2274.

If x + y + z = 0 and x = 0.5, y = 0.7, z = −1.2, then what is (x³ + y³ + z³) ÷ (3xyz)?

यदि x + y + z = 0 और x = 0.5, y = 0.7, z = −1.2 है, तो (x³ + y³ + z³) ÷ (3xyz) क्या है?

A−3
B1
C0
D−2

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Concept: If $x + y + z = 0$ , then $x^{3} + y^{3} + z^{3} = 3 x y z$ .

Check: $0.5 + 0.7 - 1.2 = 0$ ✓
So $x^{3} + y^{3} + z^{3} = 3 x y z$
Required value = $3 x y z \div 3 x y z = 1$

Trick: Whenever the sum is zero, $\frac{x^{3} + y^{3} + z^{3}}{3 x y z}$ is always 1 – no calculation needed.

Answer: (B) 1

अवधारणा: यदि $x + y + z = 0$ , तो $x^{3} + y^{3} + z^{3} = 3 x y z$ ।

हल:

जाँच: $0.5 + 0.7 - 1.2 = 0$ ✓
अतः $x^{3} + y^{3} + z^{3} = 3 x y z$
मान = $3 x y z \div 3 x y z = 1$

ट्रिक: योग शून्य हो तो $\frac{x^{3} + y^{3} + z^{3}}{3 x y z}$ सदैव 1 – गणना की ज़रूरत नहीं।

उत्तर: (B) 1

SSC CGLQuantitative AptitudeIdentity – x³ + y³ + z³ when x + y + z = 0SSC CGL 2025 · 19 Sep Shift 2 →

Q2275.

If p = 0.7 and q = 0.05, find the value of

\frac{(3 p) ³ + (3 q) ³}{p ³ + q ³}

यदि p = 0.7 और q = 0.05 है, तो

\frac{(3 p) ³ + (3 q) ³}{p ³ + q ³}

का मान ज्ञात कीजिए।

Show answer

Concept: $(k p)^{3} = k^{3} p^{3}$ , so a common factor can be taken out.

$(3 p)^{3} + (3 q)^{3} = 27 p^{3} + 27 q^{3} = 27 (p^{3} + q^{3})$
Divide by $(p^{3} + q^{3})$
= 27 (values of p, q are not needed)

Trick: Scaling every term by 3 scales a cubic expression by $3^{3} = 27$ .

Answer: (B) 27

अवधारणा: $(k p)^{3} = k^{3} p^{3}$ , अतः सार्व गुणनखंड बाहर निकालें।

हल:

$(3 p)^{3} + (3 q)^{3} = 27 p^{3} + 27 q^{3} = 27 (p^{3} + q^{3})$
$(p^{3} + q^{3})$ से भाग दें
= 27 (p, q के मान की आवश्यकता नहीं)

ट्रिक: हर पद को 3 गुना करने पर घन व्यंजक $3^{3} = 27$ गुना हो जाता है।

उत्तर: (B) 27

SSC CGLQuantitative AptitudeAlgebra – common factor in cubesSSC CGL 2025 · 19 Sep Shift 2 →

Q2276.

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 committee, comprising (1)/ experts from various fields, (2)/ have submitted its report to the governing body. (3)/ No error (4)

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

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Concept: A collective noun acting as one unit takes a singular verb; a phrase between commas does not change the subject.

Subject = 'The committee' (one body); 'comprising experts…' is only a descriptive phrase.
The pronoun 'its' also shows the singular sense.
Correct: '…has submitted its report…' → error in part (3).

Trick: Cover the words between the commas and read 'The committee have submitted its…' – the mismatch with 'its' is obvious.

Answer: (C) (3)

अवधारणा: एक इकाई के रूप में प्रयुक्त समूहवाचक संज्ञा के साथ एकवचन क्रिया आती है; अल्पविरामों के बीच का वाक्यांश subject नहीं बदलता।

हल:

Subject = 'The committee' (एक समूह); 'comprising experts…' केवल वर्णन है।
सर्वनाम 'its' भी एकवचन बताता है।
सही: '…has submitted its report…' → भाग (3) में त्रुटि।

ट्रिक: कॉमा के बीच के शब्द हटाकर पढ़ें: 'The committee have … its' – 'its' के साथ 'have' बेमेल है।

उत्तर: (C) (3)

SSC CGLEnglish ComprehensionError spotting – collective noun agreementSSC CGL 2025 · 19 Sep Shift 2 →

Q2277.

Select the sentence containing the homonym of the highlighted word:
The medieval cleric carefully transcribed the codex.

AThe copyist set the codex next to an early printed book.
BThe archivist scanned a valuable codex originating in Florence.
CThe software engineer refined the codex to improve processing speed.
DThe codex was covered with vellum made from goatskin.

Show answer

Concept: A homonym has the same spelling but a different meaning.

In the given sentence, 'codex' = an old manuscript book (handwritten, bound pages).
Options (A), (B), (D) – copyist, archivist, vellum cover – all use this same 'old book' sense.
Option (C): a software engineer refining a 'codex' = a code/program system – a different meaning → homonym.

Trick: Look for the option whose context (software, speed) is from a completely different field.

Answer: (C) The software engineer refined the codex to improve processing speed.

अवधारणा: समनाम (homonym) शब्द की वर्तनी समान, पर अर्थ भिन्न होता है।

हल:

दिए वाक्य में 'codex' = प्राचीन हस्तलिखित पुस्तक।
(A), (B), (D) – copyist, archivist, vellum – सभी में यही 'पुरानी पुस्तक' अर्थ।
(C) में software engineer का 'codex' = कोड/प्रोग्राम प्रणाली – भिन्न अर्थ → homonym।

ट्रिक: जिस विकल्प का संदर्भ (software, speed) बिल्कुल अलग क्षेत्र का हो, वही उत्तर।

उत्तर: (C) The software engineer refined the codex to improve processing speed.

SSC CGLEnglish ComprehensionHomonym – codexSSC CGL 2025 · 19 Sep Shift 2 →

Q2278.

Select the most appropriate option to substitute the underline/Bold segment in the given sentence.
The committee debated about the new policy for hours without a resolution.

Adebated on
Bdebated over
Cdebated
Ddebated regarding

Show answer

Concept: 'Debate' (verb) is transitive: it takes its object directly.

We debate something, not debate about/on/over it (in formal usage).
Correct: 'The committee debated the new policy for hours without a resolution.'

Yaad rakhein: Similar verbs that take no preposition: discuss, describe, emphasise, order, resemble, request ('discuss about' is wrong).

Answer: (C) debated

अवधारणा: 'Debate' (क्रिया) सकर्मक है: इसके बाद object सीधे आता है।

हल:

औपचारिक अंग्रेज़ी में 'debate something' – 'debate about/on/over' नहीं।
सही: 'The committee debated the new policy for hours without a resolution.'

याद रखें: discuss, describe, emphasise, order, resemble, request के बाद भी पूर्वसर्ग नहीं लगता ('discuss about' गलत)।

उत्तर: (C) debated

SSC CGLEnglish ComprehensionSentence improvement – 'debate' without prepositionSSC CGL 2025 · 19 Sep Shift 2 →

Q2279.

Select the most appropriate option to substitute the underlined/Bold segment in the given sentence.
She has been waiting for you since two hours.

Afrom two hours
Bfor two hours
Csince the two hours
Dfrom the last two hours

Show answer

Concept: Perfect continuous tense: 'for' + period of time, 'since' + point of time.

'Two hours' is a duration, not a starting point.
So 'since two hours' is wrong; use 'for two hours'.
'From' is not used with the perfect continuous for duration.

Trick: Ask 'since when?' → answer is a point (since 5 p.m., since Monday); 'for how long?' → a period (for two hours).

Answer: (B) for two hours

अवधारणा: Perfect continuous tense में: अवधि के साथ 'for', समय-बिंदु के साथ 'since'।

हल:

'Two hours' एक अवधि है, आरंभ बिंदु नहीं।
अतः 'since two hours' गलत; 'for two hours' सही।
अवधि बताने के लिए यहाँ 'from' प्रयुक्त नहीं होता।

ट्रिक: 'कब से?' → बिंदु (since 5 p.m.); 'कितनी देर से?' → अवधि (for two hours)।

उत्तर: (B) for two hours

SSC CGLEnglish ComprehensionSentence improvement – since vs forSSC CGL 2025 · 19 Sep Shift 2 →

Q2280.

Choose the one-word substitute for the below sentence.
Wild animals are kept in a special enclosed place for exhibition.

AAviary
BSanctuary
CMenagerie
DPark

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Concept: Menagerie = a collection of wild animals kept in captivity for exhibition.

Aviary – a large enclosure for birds only.
Sanctuary – protected area where animals live freely.
Park – public open ground. So 'menagerie' fits.

Yaad rakhein: Apiary = bees, aquarium = fish, aviary = birds, menagerie = wild animals on display.

Answer: (C) Menagerie

अवधारणा: Menagerie = प्रदर्शन हेतु बाड़ों में रखे गए जंगली जानवरों का संग्रह।

हल:

Aviary – केवल पक्षियों का बाड़ा।
Sanctuary – संरक्षित क्षेत्र जहाँ पशु स्वतंत्र रहते हैं।
Park – सार्वजनिक खुला स्थान। अतः 'menagerie' सही।

याद रखें: Apiary = मधुमक्खी, aquarium = मछली, aviary = पक्षी, menagerie = प्रदर्शन वाले जंगली पशु।

उत्तर: (C) Menagerie

SSC CGLEnglish ComprehensionOne-word substitution – menagerieSSC CGL 2025 · 19 Sep Shift 2 →