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Q861.

A box contains 25% yellow balls, 40% purple balls and the rest are green. If there are 70 green balls, what is the total number of balls?

एक डिब्बे में 25% पीली, 40% बैंगनी और शेष हरी गेंदें हैं। यदि 70 हरी गेंदें हैं, तो कुल गेंदें कितनी हैं?

A150
B205
C200
D180

Show answer

Concept: Remaining % = 100 − given %; then scale up.

Green = $100 - 25 - 40 = 35 %$ .
$35 % = 70 \to 1 % = 2 \to 100 % = 200$ .

Trick: 70 is 35% → total is just $2 \times 100 = 200$ .

Answer: (C) 200

अवधारणा: शेष % = 100 − दिए गए %; फिर अनुपात से बढ़ाएँ।

हल:

हरी = $100 - 25 - 40 = 35 %$ ।
$35 % = 70 \to 1 % = 2 \to 100 % = 200$ ।

ट्रिक: 70, कुल का 35% है → 1% = 2, कुल = 200।

उत्तर: (C) 200

SSC CGLQuantitative AptitudePercentage – finding the totalSSC CGL 2025 · 24 Sep Shift 1 →

Q862.

A student answered 80% of the questions in an examination. He answered 75% of the answered questions correctly, and he was also given grace marks for 25% of the questions he did not answer. If the total number of questions is 180, how many questions did he get correct (including grace)?

एक विद्यार्थी ने परीक्षा के 80% प्रश्न हल किए। हल किए गए प्रश्नों में से 75% सही थे, और जो प्रश्न उसने नहीं किए उनके 25% पर उसे कृपांक भी मिले। यदि कुल प्रश्न 180 हैं, तो उसके कुल कितने प्रश्न सही हुए (कृपांक सहित)?

A137
B117
C108
D145

Show answer

Concept: Work stage by stage on the right base.

Attempted = 80% of 180 = 144; correct = 75% of 144 = 108.
Not attempted = 36; grace = 25% of 36 = 9.
Total = $108 + 9 = 117$ .

Trick: Correct = $180 \times (0.8 \times 0.75 + 0.2 \times 0.25) = 180 \times 0.65 = 117$ .

Answer: (B) 117

अवधारणा: हर चरण सही आधार पर निकालें।

हल:

हल किए = 180 का 80% = 144; सही = 144 का 75% = 108।
छोड़े = 36; ग्रेस = 36 का 25% = 9।
कुल = $108 + 9 = 117$ ।

ट्रिक: सही = $180 \times (0.8 \times 0.75 + 0.2 \times 0.25) = 180 \times 0.65 = 117$ ।

उत्तर: (B) 117

SSC CGLQuantitative AptitudePercentage – attempted, correct and graceSSC CGL 2025 · 24 Sep Shift 1 →

Q863.

A trader gave a 25% discount on the marked price but still earned a 15% profit. What percentage was the marked price above the cost price?

एक व्यापारी ने अंकित मूल्य पर 25% छूट दी, फिर भी 15% लाभ कमाया। अंकित मूल्य क्रय मूल्य से कितने प्रतिशत अधिक था?

A45.5%
B60%
C43.34%
D53.34%

Show answer

Concept: $\frac{M P}{C P} = \frac{100 + profit %}{100 - discount %}$ .

$\frac{M P}{C P} = \frac{115}{75} = \frac{23}{15}$ .
Markup = $\frac{23 - 15}{15} \times 100 = \frac{800}{15} = 53.33 %$ (≈ 53.34%).

Trick: Let CP = 100: $S P = 115 = 75 %$ of $M P \to M P = 153.33$ → 53.33% above.

Answer: (D) 53.34%

अवधारणा: $\frac{अंकित मूल्य}{क्रय मूल्य} = \frac{100 + लाभ %}{100 - छूट %}$ ।

हल:

$\frac{M P}{C P} = \frac{115}{75} = \frac{23}{15}$ ।
वृद्धि = $\frac{23 - 15}{15} \times 100 = \frac{800}{15} = 53.33 %$ (≈ 53.34%)।

ट्रिक: CP = 100: $S P = 115$ = MP का 75% → $M P = 153.33$ → 53.33% अधिक।

उत्तर: (D) 53.34%

SSC CGLQuantitative AptitudeDiscount and profit – markup %SSC CGL 2025 · 24 Sep Shift 1 →

Q864.

A cricketer has played 10 innings with an average score of 50. In the next innings he scores 110. What will be the new average?

एक क्रिकेटर ने 10 पारियों में 50 के औसत से रन बनाए। अगली पारी में वह 110 रन बनाता है। नया औसत क्या होगा?

A50.45
B66.33
C55.45
D49.45

Show answer

Concept: $New average = old average + \frac{new score - old average}{new count}$ .

Excess over old average = $110 - 50 = 60$ .
Spread over 11 innings: $\frac{60}{11} = 5.45$ .
New average = $50 + 5.45 = 55.45$ .

Trick: This deviation method avoids big totals: $50 + \frac{60}{11}$ .

Answer: (C) 55.45

अवधारणा: $नया औसत = पुराना औसत + \frac{नया स्कोर - पुराना औसत}{नई संख्या}$ ।

हल:

पुराने औसत से अधिक = $110 - 50 = 60$ ।
11 पारियों में बाँटें: $\frac{60}{11} = 5.45$ ।
नया औसत = $50 + 5.45 = 55.45$ ।

ट्रिक: विचलन विधि से बड़े योग नहीं करने पड़ते: $50 + \frac{60}{11}$ ।

उत्तर: (C) 55.45

SSC CGLQuantitative AptitudeAverage – new innings addedSSC CGL 2025 · 24 Sep Shift 1 →

Q865.

In a quadrilateral PQRS, angles P, Q and R are in the ratio 2 : 3 : 4. If angle S = 90°, what is angle Q?

चतुर्भुज PQRS में कोण P, Q और R 2 : 3 : 4 के अनुपात में हैं। यदि कोण S = 90° है, तो कोण Q क्या है?

A90°
B72°
C45°
D120°

Show answer

Concept: Sum of angles of a quadrilateral $= 360^{\circ}$ .

$P + Q + R = 360 - 90 = 270^{\circ}$ .
Ratio parts = $2 + 3 + 4 = 9$ → 1 part $= 30^{\circ}$ .
$Q = 3 \times 30 = 90^{\circ}$ .

Trick: Q is the middle ratio term → Q = average of P, Q, R = $\frac{270}{3} = 90^{\circ}$ .

Answer: (A) 90°

अवधारणा: चतुर्भुज के कोणों का योग $= 360^{\circ}$ ।

हल:

$P + Q + R = 360 - 90 = 270^{\circ}$ ।
अनुपात भाग = 9 → 1 भाग $= 30^{\circ}$ ।
$Q = 3 \times 30 = 90^{\circ}$ ।

ट्रिक: Q मध्य पद है (2, 3, 4 समांतर) → Q = $\frac{270}{3} = 90^{\circ}$ ।

उत्तर: (A) 90°

SSC CGLQuantitative AptitudeQuadrilateral – angle sumSSC CGL 2025 · 24 Sep Shift 1 →

Q866.

A combo offer says, "Buy 3 shirts at ₹800 each, get a fourth shirt free." What is the effective price per shirt?

एक ऑफ़र कहता है, "₹800 प्रति की 3 कमीज़ें खरीदिए, चौथी मुफ़्त।" प्रति कमीज़ प्रभावी मूल्य क्या है?

A₹200
B₹600
C₹1,100
D₹400

Show answer

Concept: Effective price = amount paid ÷ items received.

Paid = $3 \times 800 =$ ₹2,400.
Items = 4 → $2400 \div 4 =$ ₹600.

Trick: Buy n get 1 free → price $\times \frac{n}{n + 1} = 800 \times \frac{3}{4} = 600$ .

Answer: (B) ₹600

अवधारणा: प्रभावी मूल्य = चुकाई राशि ÷ मिली वस्तुएँ।

हल:

चुकाया = $3 \times 800 =$ ₹2,400।
वस्तुएँ = 4 → $2400 \div 4 =$ ₹600।

ट्रिक: n खरीदें 1 मुफ़्त → मूल्य $\times \frac{n}{n + 1} = 800 \times \frac{3}{4} = 600$ ।

उत्तर: (B) ₹600

SSC CGLQuantitative AptitudeOffers – buy 3 get 1 freeSSC CGL 2025 · 24 Sep Shift 1 →

Q867.

Two taps can fill a tub in 10 minutes and 15 minutes respectively. A pipe can empty it in 12 minutes. If all three are kept open simultaneously, when will the tub be full?

दो नल एक टब को क्रमशः 10 और 15 मिनट में भरते हैं। एक पाइप उसे 12 मिनट में खाली करता है। यदि तीनों एक साथ खुले रहें, तो टब कब भरेगा?

A3 min3 मिनट
B6 min6 मिनट
C9 min9 मिनट
D12 min12 मिनट

Show answer

Concept: Take capacity = LCM; inlets add, outlet subtracts.

Capacity = LCM(10, 15, 12) = 60 units.
Rates: +6, +4, −5 → net 5 units/min.
Time = $60 \div 5 = 12$ min.

Trick: $\frac{1}{10} + \frac{1}{15} - \frac{1}{12} = \frac{6 + 4 - 5}{60} = \frac{1}{12}$ → 12 min.

Answer: (D) 12 min

अवधारणा: क्षमता = LCM मानें; भरने वाले जोड़ें, खाली करने वाला घटाएँ।

हल:

क्षमता = LCM(10, 15, 12) = 60 इकाई।
दरें: +6, +4, −5 → शुद्ध 5 इकाई/मिनट।
समय = $60 \div 5 = 12$ मिनट।

ट्रिक: $\frac{1}{10} + \frac{1}{15} - \frac{1}{12} = \frac{5}{60} = \frac{1}{12}$ → 12 मिनट।

उत्तर: (D) 12 मिनट

SSC CGLQuantitative AptitudePipes and cisterns – with outletSSC CGL 2025 · 24 Sep Shift 1 →

Q868.

In a company, the average salary of 3 junior employees is ₹25,000 and that of 2 senior employees is ₹50,000. What is the combined average salary?

एक कंपनी में 3 कनिष्ठ कर्मचारियों का औसत वेतन ₹25,000 और 2 वरिष्ठ कर्मचारियों का ₹50,000 है। संयुक्त औसत वेतन क्या है?

A₹25,000
B₹48,000
C₹35,000
D₹35,500

Show answer

Concept: Combined average = total of all values ÷ total count.

Juniors: $3 \times 25, 000 = 75, 000$ ; seniors: $2 \times 50, 000 = 1, 00, 000$ .
Total = 1,75,000 for 5 people → ₹35,000.

Trick: Weighted average lies nearer to the bigger group: $25, 000 + (\frac{2}{5}) \times 25, 000 = 35, 000$ .

Answer: (C) ₹35,000

अवधारणा: संयुक्त औसत = सभी मानों का योग ÷ कुल संख्या।

हल:

जूनियर: $3 \times 25, 000 = 75, 000$ ; सीनियर: $2 \times 50, 000 = 1, 00, 000$ ।
योग = 1,75,000, 5 लोग → ₹35,000।

ट्रिक: भारित औसत बड़े समूह के पास रहता है: $25, 000 + (\frac{2}{5}) \times 25, 000 = 35, 000$ ।

उत्तर: (C) ₹35,000

SSC CGLQuantitative AptitudeAverage – combined groupsSSC CGL 2025 · 24 Sep Shift 1 →

Q869.

A container has an 80-litre mixture of milk and water in the ratio 3 : 1. 12 litres is removed and replaced with water. Then 12 litres is removed again and replaced with milk. What is the final percentage of water?

एक पात्र में 80 लीटर दूध-पानी का मिश्रण 3 : 1 के अनुपात में है। 12 लीटर निकालकर उतना पानी मिलाया जाता है। फिर 12 लीटर निकालकर उतना दूध मिलाया जाता है। अंत में पानी का प्रतिशत क्या है?

A21.25%
B30.81%
C25.25%
D20.25%

Show answer

Concept: Removing a part takes milk and water in the current ratio.

Start: milk 60 L, water 20 L. Remove 12 L ( $\frac{3}{20}$ of mix): milk −9, water −3; add 12 L water → milk 51, water 29.
Remove 12 L again: water $- 29 \times \frac{3}{20} = - 4.35 \to 24.65$ L; adding milk does not change water.
Water % = $\frac{24.65}{80} \times 100 = 30.81 %$ .

Trick: Track only water: $20 \times \frac{17}{20} + 12 = 29$ ; then $29 \times \frac{17}{20} = 24.65$ .

Answer: (B) 30.81%

अवधारणा: कुछ भाग निकालने पर दूध और पानी वर्तमान अनुपात में निकलते हैं।

हल:

शुरू: दूध 60 L, पानी 20 L। 12 L (मिश्रण का $\frac{3}{20}$ ) निकालें: दूध −9, पानी −3; 12 L पानी डालें → दूध 51, पानी 29।
फिर 12 L निकालें: पानी $29 \times \frac{3}{20} = 4.35$ घटा → 24.65 L; दूध डालने से पानी नहीं बदलता।
पानी % = $\frac{24.65}{80} \times 100 = 30.81 %$ ।

ट्रिक: केवल पानी देखें: $20 \times \frac{17}{20} + 12 = 29$ ; फिर $29 \times \frac{17}{20} = 24.65$ ।

उत्तर: (B) 30.81%

SSC CGLQuantitative AptitudeMixture – removal and replacementSSC CGL 2025 · 24 Sep Shift 1 →

Q870.

If sinA =

\frac{35}{37}

and A is acute, find cotA.

यदि sinA =

\frac{35}{37}

है और A न्यून कोण है, तो cotA ज्ञात कीजिए।

A $\frac{5}{37}$
B $\frac{12}{35}$
C $\frac{5}{12}$
D $\frac{37}{35}$

Show answer

Concept: $\sin = \frac{perpendicular}{hypotenuse}$ , $\cot = \frac{base}{perpendicular}$ .

Perpendicular = 35, hypotenuse = 37.
Base = $\sqrt{37^{2} - 35^{2}} = \sqrt{72 \times 2} = \sqrt{144} = 12$ .
$\cot A = \frac{12}{35}$ .

Trick: Triplet 12-35-37 is standard; $a^{2} - b^{2} = (a + b) (a - b) = 72 \times 2$ .

Answer: (B) $\frac{12}{35}$

अवधारणा: $\sin = \frac{लंब}{कर्ण}$ , $\cot = \frac{आधार}{लंब}$ ।

हल:

लंब = 35, कर्ण = 37।
आधार = $\sqrt{37^{2} - 35^{2}} = \sqrt{72 \times 2} = \sqrt{144} = 12$ ।
$\cot A = \frac{12}{35}$ ।

ट्रिक: 12-35-37 मानक त्रिक है; $a^{2} - b^{2} = (a + b) (a - b) = 72 \times 2$ ।

उत्तर: (B) $\frac{12}{35}$

SSC CGLQuantitative AptitudeTrigonometry – cot from sinSSC CGL 2025 · 24 Sep Shift 1 →

Q871.

Convert

\frac{5 π}{6}

radians to degrees.

\frac{5 π}{6}

रेडियन को डिग्री में बदलिए।

A135°
B115°
C45°
D150°

Show answer

Concept: $π$ radians = $180^{\circ}$ .

$\frac{5 π}{6} \times \frac{180}{π} = 5 \times 30$ .
$= 150^{\circ}$ .

Trick: Replace $π$ by 180 directly: $5 \times \frac{180}{6} = 150^{\circ}$ .

Answer: (D) 150°

अवधारणा: $π$ रेडियन = $180^{\circ}$ ।

हल:

$\frac{5 π}{6} \times \frac{180}{π} = 5 \times 30$ ।
$= 150^{\circ}$ ।

ट्रिक: $π$ की जगह सीधे 180 रखें: $5 \times \frac{180}{6} = 150^{\circ}$ ।

उत्तर: (D) 150°

SSC CGLQuantitative AptitudeRadian to degreeSSC CGL 2025 · 24 Sep Shift 1 →

Q872.

If A + B = 90° and sinA =

\frac{3}{5}

, determine the value of sinA × sinB.

यदि A + B = 90° और sinA =

\frac{3}{5}

है, तो sinA × sinB का मान ज्ञात कीजिए।

A $\frac{16}{25}$
B $\frac{25}{12}$
C $\frac{12}{25}$
D $\frac{4}{5}$

Show answer

Concept: If $A + B = 90^{\circ}$ , then $\sin B = \cos A$ .

$\sin A = \frac{3}{5} \to \cos A = \frac{4}{5}$ (3-4-5 triangle).
$\sin A \times \sin B = \sin A \times \cos A = \frac{3}{5} \times \frac{4}{5} = \frac{12}{25}$ .

Trick: The product $\sin A \cos A = \frac{\sin 2 A}{2}$ is always $\leq \frac{1}{2}$ , ruling out $\frac{25}{12}$ and $\frac{4}{5}$ at once.

Answer: (C) $\frac{12}{25}$

अवधारणा: यदि $A + B = 90^{\circ}$ तो $\sin B = \cos A$ ।

हल:

$\sin A = \frac{3}{5} \to \cos A = \frac{4}{5}$ (3-4-5 त्रिभुज)।
$\sin A \times \sin B = \sin A \times \cos A = \frac{3}{5} \times \frac{4}{5} = \frac{12}{25}$ ।

ट्रिक: $\sin A \cos A = \frac{\sin 2 A}{2}$ हमेशा $\leq \frac{1}{2}$ — $\frac{25}{12}$ और $\frac{4}{5}$ तुरंत बाहर।

उत्तर: (C) $\frac{12}{25}$

SSC CGLQuantitative AptitudeTrigonometry – complementary anglesSSC CGL 2025 · 24 Sep Shift 1 →

Q873.

In a right-angled triangle, the point of intersection of the altitudes is found at which of the following points?

समकोण त्रिभुज में शीर्षलंबों का प्रतिच्छेद बिंदु कहाँ होता है?

AAt the midpoint of the hypotenuseकर्ण के मध्यबिंदु पर
BAt the vertex where the right angle is formedउस शीर्ष पर जहाँ समकोण बनता है
CSomewhere on the hypotenuse, but not exactly in the middleकर्ण पर कहीं, पर ठीक बीच में नहीं
DAt the midpoint of any sideकिसी भी भुजा के मध्यबिंदु पर

Show answer

Concept: Orthocentre = meeting point of the three altitudes.

In a right triangle the two legs are perpendicular, so each leg is the altitude to the other.
They meet at the right-angle vertex — so the orthocentre lies there.
Midpoint of the hypotenuse is the circumcentre, not the orthocentre (trap A).

Yaad rakhein: Right triangle: orthocentre at 90° vertex, circumcentre at hypotenuse midpoint.

Answer: (B) At the vertex where the right angle is formed

अवधारणा: लंबकेंद्र = तीनों शीर्षलंबों का मिलन बिंदु।

हल:

समकोण त्रिभुज में दोनों भुजाएँ परस्पर लंब हैं, अतः हर भुजा दूसरी पर शीर्षलंब है।
ये समकोण वाले शीर्ष पर मिलती हैं — वहीं लंबकेंद्र।
कर्ण का मध्यबिंदु परिकेंद्र है, लंबकेंद्र नहीं (विकल्प A जाल)।

याद रखें: समकोण त्रिभुज: लंबकेंद्र 90° शीर्ष पर, परिकेंद्र कर्ण के मध्य में।

उत्तर: (B) उस शीर्ष पर जहाँ समकोण बनता है

SSC CGLQuantitative AptitudeTriangle centres – orthocentreSSC CGL 2025 · 24 Sep Shift 1 →

Q874.

A circle has a radius of 29 cm. A chord is drawn at a distance of 20 cm from the centre. What is the length of the chord?

एक वृत्त की त्रिज्या 29 cm है। केंद्र से 20 cm की दूरी पर एक जीवा खींची गई है। जीवा की लंबाई क्या है?

A34 cm
B42 cm
C28 cm
D24 cm

Show answer

Concept: Perpendicular from centre bisects the chord: $(half-chord)^{2} = r^{2} - d^{2}$ .

Half-chord = $\sqrt{29^{2} - 20^{2}} = \sqrt{841 - 400} = \sqrt{441} = 21$ .
Chord = $2 \times 21 = 42$ cm.

Trick: 20-21-29 is a Pythagorean triplet — half-chord is 21 instantly.

Answer: (B) 42 cm

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

हल:

आधी जीवा = $\sqrt{29^{2} - 20^{2}} = \sqrt{441} = 21$ ।
जीवा = $2 \times 21 = 42$ cm।

ट्रिक: 20-21-29 पाइथागोरस त्रिक है — आधी जीवा सीधे 21।

उत्तर: (B) 42 cm

SSC CGLQuantitative AptitudeCircle – chord lengthSSC CGL 2025 · 24 Sep Shift 1 →

Q875.

A tangent is drawn from a point 20 cm from the centre of a circle of radius 12 cm. Find the length of the tangent.

12 cm त्रिज्या वाले वृत्त के केंद्र से 20 cm दूर स्थित बिंदु से स्पर्श रेखा खींची गई है। स्पर्श रेखा की लंबाई ज्ञात कीजिए।

A20 cm
B17 cm
C11 cm
D16 cm

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Concept: Radius $⟂$ tangent: ${tangent}^{2} = d^{2} - r^{2}$ .

Tangent = $\sqrt{20^{2} - 12^{2}} = \sqrt{400 - 144} = \sqrt{256}$ .
$= 16$ cm.

Trick: 12-16-20 = 4 × (3-4-5) triplet.

Answer: (D) 16 cm

अवधारणा: त्रिज्या $⟂$ स्पर्शरेखा: स्पर्शरेखा² $= d^{2} - r^{2}$ ।

हल:

स्पर्शरेखा = $\sqrt{20^{2} - 12^{2}} = \sqrt{256}$ ।
$= 16$ cm।

ट्रिक: 12-16-20 = 4 × (3-4-5) त्रिक।

उत्तर: (D) 16 cm

SSC CGLQuantitative AptitudeCircle – tangent lengthSSC CGL 2025 · 24 Sep Shift 1 →

Q876.

Select the option that expresses the given sentence in the Active/Passive voice.
The children will finish their homework by dinner time.

ATheir homework will be finished by the children by dinner time.
BTheir homework is finished by the children by dinner time.
CTheir homework would be finished by the children by dinner time.
DTheir homework has been finished by the children by dinner time.

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Concept: Simple future passive = will be + V3.

Active: 'will finish' (simple future); object 'their homework' becomes subject.
Passive: Their homework will be finished by the children by dinner time.
'is finished' (present), 'would be' (conditional), 'has been' (present perfect) change the tense.

Trick: Tense never changes in voice change — keep 'will', add 'be' + V3.

Answer: (A) Their homework will be finished by the children by dinner time.

अवधारणा: Simple future का passive = will be + V3।

हल:

Active में 'will finish' (simple future) है; object 'their homework' अब subject बनेगा।
Passive: Their homework will be finished by the children by dinner time.
'is finished' (present), 'would be', 'has been' — ये tense बदल देते हैं।

ट्रिक: Voice बदलने पर tense नहीं बदलता — 'will' रखें, 'be' + V3 जोड़ें।

उत्तर: (A) Their homework will be finished by the children by dinner time.

SSC CGLEnglish ComprehensionVoice – simple future to passiveSSC CGL 2025 · 24 Sep Shift 1 →

Q877.

Select the correct spelling to fill in the blank.
The scientist's interpretation was based on a ____ understanding of nature.

Asupurficial
Bsupperficial
Csuperficial
Dsupperficiall

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Concept: Superficial = shallow, only on the surface.

Correct form: s-u-p-e-r-f-i-c-i-a-l (super + ficial, single 'p', single 'l').
'supurficial' wrong vowel; 'supperficial/ supperficiall' double letters.

Trick: Think 'SUPER + face' — on the surface (super = above, ficies = face).

Answer: (C) superficial

अवधारणा: Superficial = सतही, केवल ऊपर-ऊपर का।

हल:

सही रूप: s-u-p-e-r-f-i-c-i-a-l (super + ficial, एक 'p', एक 'l')।
'supurficial' में गलत स्वर; 'supperficial/ supperficiall' में दोहरे अक्षर।

ट्रिक: 'SUPER + face' सोचें — सतह के ऊपर (super = ऊपर, ficies = चेहरा)।

उत्तर: (C) superficial

SSC CGLEnglish ComprehensionSpelling – superficialSSC CGL 2025 · 24 Sep Shift 1 →

Q878.

Sentences of a paragraph are given below in jumbled order. Arrange the sentences in the correct order to form a meaningful and coherent paragraph.
1. As a result, the demand for renewable energy sources has grown significantly in recent years.
2. The excessive use of fossil fuels has led to environmental degradation and global warming.
3. Renewable sources like solar and wind power are now being adopted worldwide as cleaner alternatives.
4. Governments and organisations are investing heavily in sustainable energy technologies.

A2, 1, 4, 3
B1, 2, 3, 4
C2, 1, 3, 4
D4, 3, 2, 1

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Concept: Start with the cause; linkers like 'As a result' must follow it.

2 states the problem: fossil fuels → global warming.
1 ('As a result') shows rising demand for renewables; 3 names those sources (solar, wind).
4 ends with heavy investment in such technology → 2-1-3-4.

Trick: A sentence beginning with 'As a result' can never be first — removes (B).

Answer: (C) 2, 1, 3, 4

अवधारणा: कारण से शुरू करें; 'As a result' जैसे linker उसके बाद ही आते हैं।

हल:

2 समस्या बताता है: fossil fuels → global warming।
1 ('As a result') renewable की बढ़ती माँग; 3 उन स्रोतों (solar, wind) के नाम।
4 निवेश के साथ समाप्त → 2-1-3-4।

ट्रिक: 'As a result' से शुरू होने वाला वाक्य पहला नहीं हो सकता — (B) बाहर।

उत्तर: (C) 2, 1, 3, 4

SSC CGLEnglish ComprehensionPara jumble – cause and effectSSC CGL 2025 · 24 Sep Shift 1 →

Q879.

Sentences of a paragraph are given below in jumbled order. Arrange the sentences in the correct order to form a meaningful and coherent paragraph.
1. This enables machines to learn from data without explicit programming.
2. Machine learning, a subset of Artificial Intelligence, is rapidly gaining prominence.
3. Algorithms are trained on vast datasets to recognize patterns and make predictions.
4. Consequently, it finds applications in diverse fields, from healthcare to finance.

A2-1-3-4
B2-3-1-4
C3-1-2-4
D1-2-3-4

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Concept: Introduce the topic, explain the process, then its outcome.

2 introduces machine learning.
3 explains training on datasets; 1 ('This enables…') refers back to that training.
4 ('Consequently') gives applications → 2-3-1-4.

Trick: 'This' in 1 needs a noun/process before it — that is sentence 3, so 3 → 1 is a fixed pair.

Answer: (B) 2-3-1-4

अवधारणा: पहले विषय, फिर प्रक्रिया, फिर परिणाम।

हल:

2 machine learning का परिचय देता है।
3 datasets पर training बताता है; 1 ('This enables…') उसी की ओर संकेत करता है।
4 ('Consequently') उपयोग बताता है → 2-3-1-4।

ट्रिक: 1 का 'This' अपने से पहले प्रक्रिया माँगता है — वह 3 है, अतः 3 → 1 जोड़ी पक्की।

उत्तर: (B) 2-3-1-4

SSC CGLEnglish ComprehensionPara jumble – definition, process, resultSSC CGL 2025 · 24 Sep Shift 1 →

Q880.

Select the most appropriate antonym of the given word: ADVANCED

AMature
BIncipient
CDeveloped
DProgressed

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Concept: Advanced = far developed, ahead in progress.

Incipient = in an early stage, just starting → opposite.
Mature, Developed, Progressed all mean nearly the same as 'advanced'.

Trick: When three options are synonyms of the word, the odd fourth is the antonym.

Answer: (B) Incipient

अवधारणा: Advanced = बहुत विकसित, प्रगति में आगे।

हल:

Incipient = प्रारंभिक अवस्था में, अभी शुरू होता हुआ → विलोम।
Mature, Developed, Progressed — तीनों 'advanced' के समानार्थी हैं।

ट्रिक: जब तीन विकल्प समानार्थी हों, बचा हुआ चौथा विलोम होता है।

उत्तर: (B) Incipient

SSC CGLEnglish ComprehensionAntonym – AdvancedSSC CGL 2025 · 24 Sep Shift 1 →