Question bank

Q&A: practice questions with answers

Search questions on any topic. Filter by exam category, exam, subject and topic, try the question, then open the answer and explanation.

Free mock tests Download App

4,800 questions found.

Q3561.

A mobile shop owner buys 10 boxes of mobiles for the price of 8 boxes from a wholesaler. If he sells each box at a 2% discount on its marked price, what is his profit percentage?

एक मोबाइल दुकानदार थोक विक्रेता से 8 बक्सों की कीमत पर 10 बक्स खरीदता है। यदि वह प्रत्येक बक्स अंकित मूल्य पर 2% छूट देकर बेचता है, तो उसका लाभ प्रतिशत क्या है?

A10%
B24.36%
C22.5%
D23.86%

Show answer

Concept: Compare cost and sale value of the same 10 boxes.

Let each box's marked price = ₹100. Cost of 10 boxes = ₹800.
Sale of 10 boxes at 2% off = $10 \times 98 =$ ₹980.
Profit = 180 → $\frac{180}{800} \times 100 = 22.5 %$ .

Trick: Profit% = $\frac{10 \times 0.98 - 8}{8} \times 100 = \frac{1.8}{8} \times 100 = 22.5 %$ .

Answer: (C) 22.5%

अवधारणा: उन्हीं 10 डिब्बों का क्रय व विक्रय मूल्य तुलना करें।

अंकित मूल्य ₹100 मानें; 10 डिब्बों की लागत ₹800।
2% छूट पर बिक्री = $10 \times 98 =$ ₹980।
लाभ = 180 → 22.5%।

ट्रिक: $\frac{10 \times 0.98 - 8}{8} \times 100 = 22.5 %$ ।

उत्तर: (C) 22.5%

SSC CGLQuantitative AptitudeProfit – buy 10 for price of 8SSC CGL 2025 · 15 Sep Shift 1 →

Q3562.

A fruit seller offers mangoes to a retailer at a 30% discount on the marked price. However, they charge an additional 5% on the discounted price for shipping. The retailer sells the mangoes for ₹1620 more, earning a profit of 40%. What was the marked price of the mangoes?

एक फल विक्रेता खुदरा व्यापारी को अंकित मूल्य पर 30% छूट पर आम देता है, पर छूट वाली कीमत पर 5% अतिरिक्त शिपिंग लेता है। खुदरा व्यापारी आम ₹1620 अधिक में बेचकर 40% लाभ कमाता है। आमों का अंकित मूल्य क्या था?

A₹5,510.20
B₹9,254.62
C₹8,181.82
D₹20,654.67

Show answer

Concept: Retailer's cost = $MP \times 0.70 \times 1.05$ ; the extra ₹1620 is the 40% profit.

Cost = $0.735 \times MP$ .
40% of cost = 1620 → cost = ₹4050.
MP = $4050 \div 0.735 \approx$ ₹5,510.20.

Trick: Profit ₹1620 at 40% → cost = $1620 \times \frac{100}{40} = 4050$ ; then divide by 0.735.

Answer: (A) ₹5,510.20

अवधारणा: खुदरा लागत = $MP \times 0.70 \times 1.05$ ; ₹1620 अधिक = 40% लाभ।

लागत = $0.735 \times MP$ ।
लागत का 40% = 1620 → लागत = ₹4050।
MP = $4050 \div 0.735 \approx$ ₹5,510.20।

ट्रिक: $1620 \times \frac{100}{40} = 4050$ , फिर 0.735 से भाग।

उत्तर: (A) ₹5,510.20

SSC CGLQuantitative AptitudeDiscount, shipping and profitSSC CGL 2025 · 15 Sep Shift 1 →

Q3563.

A cistern contains 200 liters of pure syrup. 20 liters of syrup are drawn out and replaced by water. Then, 20 liters of the mixture are drawn out and replaced by water again. What is the ratio of water to syrup in the final mixture?

एक टंकी में 200 लीटर शुद्ध सिरप है। 20 लीटर सिरप निकालकर पानी भरा जाता है। फिर 20 लीटर मिश्रण निकालकर फिर पानी भरा जाता है। अंतिम मिश्रण में पानी और सिरप का अनुपात क्या है?

A19 : 1
B19 : 14
C19 : 81
D18 : 19

Show answer

Concept: Pure liquid left = $V \times (1 - \frac{x}{V})^{n}$ .

Each draw keeps $\frac{9}{10}$ of the syrup present.
After two draws syrup = $200 \times \frac{81}{100} = 162$ litres; the remaining 38 litres of the cistern are water.
Cancelling by 2, the water-to-syrup ratio is 19 : 81.

Trick: Syrup fraction = $(\frac{9}{10})^{2} = \frac{81}{100}$ → water $\frac{19}{100}$ → 19 : 81 directly.

Answer: (C) 19 : 81

अवधारणा: बचा शुद्ध द्रव = $V \times (1 - \frac{x}{V})^{n}$ ।

सिरप = $200 \times (\frac{9}{10})^{2} = 162$ L।
पानी = 38 L।
पानी : सिरप = 19 : 81।

ट्रिक: सिरप का अंश $(\frac{9}{10})^{2} = \frac{81}{100}$ → पानी $\frac{19}{100}$ → 19 : 81।

उत्तर: (C) 19 : 81

SSC CGLQuantitative AptitudeReplacement of mixtureSSC CGL 2025 · 15 Sep Shift 1 →

Q3564.

A metal sphere having a radius of 20 centimeters is melted down and molded into 64 identical smaller solid spheres. What is the ratio of the surface area of the original sphere to the total surface area of all 64 smaller spheres?

20 cm त्रिज्या वाले एक धातु गोले को पिघलाकर 64 समान छोटे ठोस गोले बनाए जाते हैं। मूल गोले के पृष्ठीय क्षेत्रफल और सभी 64 छोटे गोलों के कुल पृष्ठीय क्षेत्रफल का अनुपात क्या है?

A1 : 2
B2 : 1
C1 : 4
D1 : 1

Show answer

Concept: Volume stays the same; surface area $\propto r^{2}$ .

$64 r^{3} = 20^{3} \to r = \frac{20}{4} = 5$ cm.
Big SA ∝ $20^{2} = 400$ ; small total ∝ $64 \times 25 = 1600$ .
Ratio = 400 : 1600 = 1 : 4.

Trick: n equal spheres from one: total SA becomes $\sqrt[3]{n}$ times $\to \sqrt[3]{64} = 4$ → 1 : 4.

Answer: (C) 1 : 4

अवधारणा: आयतन समान रहता है; पृष्ठीय क्षेत्रफल $\propto r^{2}$ ।

$64 r^{3} = 20^{3} \to r = 5$ सेमी।
बड़ा ∝ 400; 64 छोटे ∝ $64 \times 25 = 1600$ ।
अनुपात 1 : 4।

ट्रिक: n बराबर गोले बनें तो कुल पृष्ठ $\sqrt[3]{n}$ गुना $\to \sqrt[3]{64} = 4$ ।

उत्तर: (C) 1 : 4

SSC CGLQuantitative AptitudeSphere recast – surface areaSSC CGL 2025 · 15 Sep Shift 1 →

Q3565.

How many hemispheres with a radius of 6 cm can be produced by melting down a hemisphere with a radius of 24 cm?

24 cm त्रिज्या के अर्धगोले को पिघलाकर 6 cm त्रिज्या के कितने अर्धगोले बनाए जा सकते हैं?

Show answer

Concept: Number of pieces = big volume ÷ small volume = $(\frac{R}{r})^{3}$ .

$\frac{R}{r} = \frac{24}{6} = 4$ .
Number = $4^{3} = 64$ .

Trick: Same shape melted into same shape → just cube the radius ratio.

Answer: (B) 64

अवधारणा: टुकड़ों की संख्या = $(\frac{R}{r})^{3}$ ।

$\frac{R}{r} = \frac{24}{6} = 4$ ।
संख्या = $4^{3} = 64$ ।

ट्रिक: समान आकृति में ढालें तो त्रिज्या अनुपात का घन लें।

उत्तर: (B) 64

SSC CGLQuantitative AptitudeRecasting hemispheresSSC CGL 2025 · 15 Sep Shift 1 →

Q3566.

A cylinder and hemisphere have the same radius. Their combined height is 35 cm. If the cylinder and hemisphere have equal volumes, find the radius.

एक बेलन और अर्धगोले की त्रिज्या समान है। उनकी संयुक्त ऊँचाई 35 cm है। यदि बेलन और अर्धगोले के आयतन बराबर हैं, तो त्रिज्या ज्ञात कीजिए।

A14 cm
B13 cm
C10 cm
D21 cm

Show answer

Concept: $π r^{2} h = (\frac{2}{3}) π r^{3}$ gives h in terms of r.

$h = \frac{2 r}{3}$ .
Total height = $h + r = \frac{2 r}{3} + r = \frac{5 r}{3}$ .
$\frac{5 r}{3} = 35 \to r = 21$ cm.

Trick: Options must make $35 \times \frac{3}{5}$ — only 21 works.

Answer: (D) 21 cm

अवधारणा: $π r^{2} h = (\frac{2}{3}) π r^{3}$ से h निकालें।

$h = \frac{2 r}{3}$ ।
कुल ऊँचाई = $\frac{2 r}{3} + r = \frac{5 r}{3}$ ।
$\frac{5 r}{3} = 35 \to r = 21$ सेमी।

ट्रिक: $35 \times \frac{3}{5}$ = 21 — सीधा उत्तर।

उत्तर: (D) 21 cm

SSC CGLQuantitative AptitudeCylinder and hemisphere – equal volumesSSC CGL 2025 · 15 Sep Shift 1 →

Q3567.

A pyramid has a base area of 120 cm² and height 18 cm. What is its volume?

एक पिरामिड का आधार क्षेत्रफल 120 cm² और ऊँचाई 18 cm है। इसका आयतन क्या है?

A720 cm³
B780 cm³
C540 cm³
D460 cm³

Show answer

Concept: $V = \frac{1}{3} \times$ base area $\times$ height.

$V = \frac{1}{3} \times 120 \times 18$ .
$= 120 \times 6 = 720$ cm³.

Trick: Divide the height by 3 first: $\frac{18}{3} = 6$ , then $120 \times 6$ .

Answer: (A) 720 cm³

अवधारणा: $V = \frac{1}{3} \times$ आधार क्षेत्रफल $\times$ ऊँचाई।

$V = \frac{1}{3} \times 120 \times 18$ ।
$= 120 \times 6 = 720$ सेमी³।

ट्रिक: पहले ऊँचाई को 3 से भाग दें: 6, फिर $120 \times 6$ ।

उत्तर: (A) 720 cm³

SSC CGLQuantitative AptitudeVolume of a pyramidSSC CGL 2025 · 15 Sep Shift 1 →

Q3568.

A circular disc of diameter 14 cm is inscribed inside an equilateral triangle. What is the approximate area of the remaining portion of the triangle?

14 cm व्यास की एक वृत्ताकार चकती एक समबाहु त्रिभुज के अंदर अंकित है। त्रिभुज के शेष भाग का अनुमानित क्षेत्रफल क्या है?

A155.6 cm²
B168.2 cm²
C250.3 cm²
D100.56 cm²

Show answer

Concept: Inradius $r = \frac{a}{2 \sqrt{3}}$ and area of triangle = $3 \sqrt{3} r^{2}$ .

$r = \frac{14}{2} = 7$ cm.
Triangle area = $3 \sqrt{3} \times 49 = 147 \sqrt{3} \approx 254.6$ cm².
Circle area = $\frac{22}{7} \times 49 = 154$ cm².
Remaining ≈ $254.6 - 154 = 100.6$ cm².

Trick: Equilateral triangle area from inradius = $3 \sqrt{3} r^{2}$ — skips finding the side.

Answer: (D) 100.56 cm²

अवधारणा: अंतःत्रिज्या $r = \frac{a}{2 \sqrt{3}}$ ; त्रिभुज क्षेत्रफल = $3 \sqrt{3} r^{2}$ ।

r = 7 सेमी।
त्रिभुज = $147 \sqrt{3} \approx 254.6$ सेमी²।
वृत्त = 154 सेमी²।
शेष ≈ 100.6 सेमी²।

ट्रिक: अंतःत्रिज्या से सीधे क्षेत्रफल = $3 \sqrt{3} r^{2}$ ।

उत्तर: (D) 100.56 cm²

SSC CGLQuantitative AptitudeIncircle of an equilateral triangleSSC CGL 2025 · 15 Sep Shift 1 →

Q3569.

In triangle PQR, medians PS, QT, and RU intersect at the centroid G. What is the ratio of the area of triangle GPQ to the area of triangle PQR?

त्रिभुज PQR में माध्यिकाएँ PS, QT और RU केंद्रक G पर मिलती हैं। त्रिभुज GPQ और त्रिभुज PQR के क्षेत्रफलों का अनुपात क्या है?

A2 : 1
B1 : 3
C3 : 1
D3 : 4

Show answer

Concept: Joining the centroid to the three vertices splits a triangle into three equal areas.

$△ G P Q = △ G Q R = △ G R P$ .
Each = $\frac{1}{3}$ of $△ P Q R$ → ratio 1 : 3.

Trick: The 3 medians together make 6 small triangles of equal area; $△ G P Q$ contains 2 of them = $\frac{2}{6} = \frac{1}{3}$ .

Answer: (B) 1 : 3

अवधारणा: केन्द्रक को तीनों शीर्षों से मिलाने पर तीन बराबर क्षेत्रफल बनते हैं।

$△ G P Q = △ G Q R = △ G R P$ ।
प्रत्येक $= \frac{1}{3}$ → अनुपात 1 : 3।

ट्रिक: तीनों माध्यिकाएँ 6 बराबर त्रिभुज बनाती हैं; $△ G P Q$ में 2 हैं $= \frac{1}{3}$ ।

उत्तर: (B) 1 : 3

SSC CGLQuantitative AptitudeCentroid and areaSSC CGL 2025 · 15 Sep Shift 1 →

Q3570.

In △PQR, a line segment DE is parallel to QR, with D on PQ and E on PR. If the ratio of the area of △PDE to the area of the trapezoid DERQ is 4 : 21, what is the ratio of PD to DQ?

△PQR में रेखाखंड DE, QR के समांतर है (D, PQ पर और E, PR पर)। यदि △PDE और समलंब DERQ के क्षेत्रफलों का अनुपात 4 : 21 है, तो PD : DQ क्या है?

A2 : 3
B2 : 7
C2 : 5
D4 : 7

Show answer

Concept: For similar triangles, $area ratio = (side ratio)^{2}$ .

Small △ : whole △ = $4 : (4 + 21) = 4 : 25$ .
Side ratio = 2 : 5, so the cut-off side is 2 parts of 5.
Remaining part = 3 → PD : DQ = 2 : 3.

Trick: Area parts 4 and 25 are perfect squares — take roots (2, 5) and subtract: 2 : 3.

Answer: (A) 2 : 3

अवधारणा: समरूप त्रिभुजों में $क्षेत्रफल अनुपात = (भुजा अनुपात)^{2}$ ।

छोटा : पूरा = 4 : 25।
भुजा अनुपात = 2 : 5।
शेष भाग 3 → PD : DQ = 2 : 3।

ट्रिक: 4 और 25 के वर्गमूल 2, 5 लें; $2 : (5 - 2) = 2 : 3$ ।

उत्तर: (A) 2 : 3

SSC CGLQuantitative AptitudeSimilar triangles – area to side ratioSSC CGL 2025 · 15 Sep Shift 1 →

Q3571.

A right triangle has sides 6, 8, and 10. A smaller triangle is drawn inside it with its vertices on the sides of the larger triangle, such that it is similar to the larger triangle. If its perimeter is 12, what is its area?

एक समकोण त्रिभुज की भुजाएँ 6, 8 और 10 हैं। उसके अंदर एक छोटा त्रिभुज बनाया जाता है जो बड़े त्रिभुज के समरूप है। यदि उसका परिमाप 12 है, तो उसका क्षेत्रफल क्या है?

A3.5
B2.2
C3
D6

Show answer

Concept: Perimeter ratio = side ratio; area ratio = its square.

Big perimeter = 24, area = $\frac{1}{2} \times 6 \times 8 = 24$ .
Ratio of sides = $\frac{12}{24} = \frac{1}{2}$ .
Small area = $24 \times (\frac{1}{2})^{2} = 6$ .

Trick: Joining mid-points of sides gives exactly this half-size triangle with $\frac{1}{4}$ the area.

Answer: (D) 6

अवधारणा: परिमाप अनुपात = भुजा अनुपात; क्षेत्रफल अनुपात = उसका वर्ग।

बड़ा परिमाप 24, क्षेत्रफल 24।
भुजा अनुपात $= \frac{1}{2}$ ।
छोटा क्षेत्रफल = $24 \times \frac{1}{4} = 6$ ।

ट्रिक: भुजाओं के मध्य-बिंदु मिलाने से यही त्रिभुज बनता है — क्षेत्रफल $\frac{1}{4}$ ।

उत्तर: (D) 6

SSC CGLQuantitative AptitudeSimilar triangles – perimeter and areaSSC CGL 2025 · 15 Sep Shift 1 →

Q3572.

Two circles have radii of 12 cm and 5 cm. If the length of a direct common tangent is 24 cm, what is the distance between their centers?

दो वृत्तों की त्रिज्याएँ 12 cm और 5 cm हैं। यदि उनकी अनुप्रस्थ उभयनिष्ठ स्पर्श रेखा की लंबाई 24 cm है, तो उनके केंद्रों के बीच की दूरी क्या है?

A16 cm
B22 cm
C19 cm
D25 cm

Show answer

Concept: Direct tangent $^{2} = d^{2} - (r_{1} - r_{2})^{2}$ .

$576 = d^{2} - 7^{2} = d^{2} - 49$ .
$d^{2} = 625 \to d = 25$ cm.

Trick: 7, 24, 25 is a Pythagorean triplet — answer at a glance.

Answer: (D) 25 cm

अवधारणा: प्रत्यक्ष स्पर्शरेखा $^{2} = d^{2} - (r_{1} - r_{2})^{2}$ ।

$576 = d^{2} - 49$ ।
$d^{2} = 625 \to d = 25$ सेमी।

ट्रिक: 7, 24, 25 पाइथागोरस त्रिक है — तुरंत उत्तर।

उत्तर: (D) 25 cm

SSC CGLQuantitative AptitudeDirect common tangentSSC CGL 2025 · 15 Sep Shift 1 →

Q3573.

The distance between the centers of two circles is D. The lengths of their direct and transverse common tangents are P and Q, respectively. If P² + Q² = 400 and the sum of the squares of their radii is 200, what is the value of D?

दो वृत्तों के केंद्रों के बीच की दूरी D है। उनकी अनुप्रस्थ और तिर्यक उभयनिष्ठ स्पर्श रेखाओं की लंबाइयाँ क्रमशः P और Q हैं। यदि P² + Q² = 400 और त्रिज्याओं के वर्गों का योग 200 है, तो D का मान क्या है?

A20
B20 $\sqrt{2}$
C5 $\sqrt{2}$
D30

Show answer

Concept: $P^{2} = D^{2} - (r_{1} - r_{2})^{2}, Q^{2} = D^{2} - (r_{1} + r_{2})^{2}$ .

Adding: $P^{2} + Q^{2} = 2 D^{2} - 2 (r_{1}^{2} + r_{2}^{2})$ .
$400 = 2 D^{2} - 400 \to D^{2} = 400$ .
D = 20.

Trick: Remember: $P^{2} + Q^{2} = 2 [D^{2} - (r_{1}^{2} + r_{2}^{2})] \to 200 = D^{2} - 200$ .

Answer: (A) 20

अवधारणा: $P^{2} = D^{2} - (r_{1} - r_{2})^{2}, Q^{2} = D^{2} - (r_{1} + r_{2})^{2}$ ।

जोड़ने पर: $P^{2} + Q^{2} = 2 D^{2} - 2 (r_{1}^{2} + r_{2}^{2})$ ।
$400 = 2 D^{2} - 400 \to D^{2} = 400$ ।
D = 20।

ट्रिक: सूत्र याद रखें: $\frac{P^{2} + Q^{2}}{2} = D^{2} - (r_{1}^{2} + r_{2}^{2})$ ।

उत्तर: (A) 20

SSC CGLQuantitative AptitudeDirect and transverse tangentsSSC CGL 2025 · 15 Sep Shift 1 →

Q3574.

Two parallel chords of a circle are 30 cm and 16 cm long. Both chords lie on the same side of the circle's center, and the distance between them is 7 cm. Find the diameter of the circle.

किसी वृत्त की दो समांतर जीवाएँ 30 cm और 16 cm लंबी हैं। दोनों जीवाएँ केंद्र के एक ही ओर हैं और उनके बीच की दूरी 7 cm है। वृत्त का व्यास ज्ञात कीजिए।

A28 cm
B30 cm
C24 cm
D34 cm

Show answer

Concept: Perpendicular from centre bisects a chord: $r^{2} =$ (half chord) $^{2}$ + distance $^{2}$ .

Let 30 cm chord be x from centre, 16 cm chord at $x + 7$ .
Equal radius: $15^{2} + x^{2} = 8^{2} + (x + 7)^{2}$ , which simplifies to $14 x = 112$ , so x = 8.
Radius² = $15^{2} + 8^{2} = 289$ , radius 17 cm, so the diameter is 34 cm.

Trick: Triplets: (8, 15, 17) and (15, 8, 17) — distances 8 and 15 differ by 7 ✓.

Answer: (D) 34 cm

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

30 सेमी जीवा की दूरी x, 16 सेमी की $x + 7$ ।
$225 + x^{2} = 64 + (x + 7)^{2} \to x = 8$ ।
r = 17 → व्यास 34 सेमी।

ट्रिक: त्रिक (8, 15, 17): दूरी 8 और 15 का अंतर 7 ✓।

उत्तर: (D) 34 cm

SSC CGLQuantitative AptitudeParallel chords – same sideSSC CGL 2025 · 15 Sep Shift 1 →

Q3575.

What is the area of the segment formed by a chord in a circle of radius 14 cm, if the angle subtended at the center is 90°?

14 cm त्रिज्या वाले वृत्त में कोई जीवा केंद्र पर 90° का कोण अंतरित करती है। उससे बने वृत्तखंड का क्षेत्रफल क्या है?

A84π − 49
B49π − 98
C60π − 72
D36π − 72

Show answer

Concept: Segment = sector − triangle.

Sector = $(\frac{90}{360}) \times π \times 196 = 49 π$ .
Right triangle = $\frac{1}{2} \times 14 \times 14 = 98$ .
Segment = $49 π - 98$ .

Trick: For 90°: segment = $r^{2} (\frac{π}{4} - \frac{1}{2})$ .

Answer: (B) 49π − 98

अवधारणा: वृत्तखंड = त्रिज्यखंड − त्रिभुज।

त्रिज्यखंड = $\frac{1}{4} \times π \times 196 = 49 π$ ।
त्रिभुज = $\frac{1}{2} \times 14 \times 14 = 98$ ।
वृत्तखंड = $49 π - 98$ ।

ट्रिक: 90° के लिए: $r^{2} (\frac{π}{4} - \frac{1}{2})$ ।

उत्तर: (B) 49π − 98

SSC CGLQuantitative AptitudeArea of a minor segmentSSC CGL 2025 · 15 Sep Shift 1 →

Q3576.

Select the most appropriate option to substitute the underline/Bold segment in the given sentence.
She is junior than him in experience.

Ajunior from him
Bmore junior than him
Cjunior to him
Dyounger to him

Show answer

Concept: Words ending in '-ior' (junior, senior, superior, inferior, prior) are followed by 'to', not 'than'.

'junior than' is wrong.
Correct: 'She is junior to him in experience.'
'younger to' is also wrong — 'younger' takes 'than'.

Trick: '-ior' → 'to': senior to, superior to, prior to, junior to.

Answer: (C) junior to him

अवधारणा: '-ior' पर खत्म होने वाले शब्द (junior, senior, superior, inferior, prior) के बाद 'than' नहीं, 'to' आता है।

'junior than' गलत है।
सही: 'She is junior to him in experience.'
'younger' के साथ 'than' आता है, 'to' नहीं।

ट्रिक: '-ior' वाले शब्द → 'to'।

उत्तर: (C) junior to him

SSC CGLEnglish ComprehensionComparatives with 'to' – juniorSSC CGL 2025 · 15 Sep Shift 1 →

Q3577.

Select the correct indirect/direct form of the given sentence.
She said, "I am reading a novel right now."

AShe said that she is reading a novel right now.
BShe said that she was reading a novel at that time.
CShe said she had been reading a novel right now.
DShe said she reads a novel at that time.

Show answer

Concept: With a past reporting verb, present continuous becomes past continuous; 'now' becomes 'then/at that time'.

'I am reading' → 'she was reading'.
'right now' → 'at that time'.
Only option (B) carries both changes.

Trick: Two quick checks: tense moved one step back + time word changed. Only (B) does both.

Answer: (B) She said that she was reading a novel at that time.

अवधारणा: Reporting verb भूतकाल में हो तो present continuous → past continuous, और 'now' → 'at that time'।

'I am reading' → 'she was reading'।
'right now' → 'at that time'।
दोनों बदलाव केवल (B) में हैं।

ट्रिक: दो जाँच: काल एक कदम पीछे + समय-सूचक बदला — केवल (B) में दोनों।

उत्तर: (B) She said that she was reading a novel at that time.

SSC CGLEnglish ComprehensionDirect to indirect – present continuousSSC CGL 2025 · 15 Sep Shift 1 →

Q3578.

Select the most appropriate synonym of the given word:
Carefree

ATense
BAnxious
CStressed
DNon-chalant

Show answer

Concept: Carefree = without worry; relaxed.

Nonchalant = calm, casually unconcerned → same sense.
Tense, anxious, stressed are opposites.

Trick: Three options share one meaning (worried) — the odd one out is the synonym.

Answer: (D) Non-chalant

अवधारणा: Carefree = बेफ़िक्र, चिंता-मुक्त।

Nonchalant = शांत, बेपरवाह → समान अर्थ।
Tense, anxious, stressed = चिंतित (विपरीत)।

ट्रिक: तीन विकल्प एक जैसे (चिंतित) हैं — अलग वाला ही उत्तर।

उत्तर: (D) Non-chalant

SSC CGLEnglish ComprehensionSynonym – CarefreeSSC CGL 2025 · 15 Sep Shift 1 →

Q3579.

Rearrange the following sentences to form a coherent paragraph:
1. This practice involves using genetic material to identify individuals and establish relationships, offering unprecedented accuracy.
2. Forensic science has been revolutionized by advancements in DNA profiling.
3. Its applications range from solving cold cases and identifying disaster victims to exonerating wrongly accused individuals.
4. Before DNA, forensic methods relied heavily on less precise evidence like fingerprints and eyewitness accounts.

A2, 4, 1, 3
B4, 2, 1, 3
C2, 1, 4, 3
D1, 2, 3, 4

Show answer

Concept: Start with the general topic, then background, explanation, and uses.

2 introduces the topic (DNA profiling changed forensics).
4 gives the earlier situation ('Before DNA…').
1 'This practice' explains the method; 3 'Its applications' lists uses.

Trick: Sentences starting with 'This/Its' cannot open a paragraph — eliminates (D).

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

अवधारणा: पहले विषय, फिर पृष्ठभूमि, व्याख्या और उपयोग।

2 विषय का परिचय देता है।
4 पहले की स्थिति बताता है ('Before DNA…')।
1 'This practice' विधि समझाता है; 3 'Its applications' उपयोग।

ट्रिक: 'This/Its' से शुरू होने वाला वाक्य पहला नहीं हो सकता।

उत्तर: (A) 2, 4, 1, 3

SSC CGLEnglish ComprehensionPara jumble – DNA profilingSSC CGL 2025 · 15 Sep Shift 1 →

Q3580.

Fill in the blanks with the antonym of the bracketed word.
He favoured a ____ (quixotic) approach to the problem.

ARomantic
BVisionary
CPragmatic
DIdealistic

Show answer

Concept: Quixotic = idealistic in an impractical way (from Don Quixote).

Opposite: pragmatic = practical, realistic.
Romantic, visionary, idealistic are near-synonyms of quixotic.

Trick: Three options mean 'dreamy' — the one that doesn't is the antonym.

Answer: (C) Pragmatic

अवधारणा: Quixotic = अव्यावहारिक रूप से आदर्शवादी (डॉन क्विक्ज़ोट से)।

विलोम: pragmatic = व्यावहारिक।
Romantic, visionary, idealistic समानार्थी हैं।

ट्रिक: तीन विकल्प 'सपनों वाले' अर्थ के — अलग वाला विलोम।

उत्तर: (C) Pragmatic

SSC CGLEnglish ComprehensionAntonym – QuixoticSSC CGL 2025 · 15 Sep Shift 1 →